i
Applications of Group Theory
to the Physics of Solids
M. S. Dresselhaus
8.510J
6.734J
SPRING 2002
ii
Contents
1 Basic Mathematical Background 1
1.1 Definition of a Group . . . . . . . . . . . . . . . . . . . . 1
1.2 Simple Example of a Group . . . . . . . . . . . . . . . . 2
1.3 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Rearrangement Theorem . . . . . . . . . . . . . . . . . . 6
1.5 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.6 Conjugation and Class . . . . . . . . . . . . . . . . . . . 8
1.6.1 Self-Conjugate Subgroups . . . . . . . . . . . . . 9
1.7 Factor Groups . . . . . . . . . . . . . . . . . . . . . . . . 10
1.8 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 11
2 Representation Theory 15
2.1 Important Definitions . . . . . . . . . . . . . . . . . . . . 15
2.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.3 Irreducible Representations . . . . . . . . . . . . . . . . 18
2.4 The Unitarity of Representations . . . . . . . . . . . . . 20
2.5 Schur’s Lemma (Part I) . . . . . . . . . . . . . . . . . . 23
2.6 Schur’s Lemma (Part 2) . . . . . . . . . . . . . . . . . . 24
2.7 Wonderful Orthogonality Theorem . . . . . . . . . . . . 27
2.8 Representations and Vector Spaces . . . . . . . . . . . . 30
2.9 Suggested Problems . . . . . . . . . . . . . . . . . . . . . 31
3 Character of a Representation 33
3.1 Definition of Character . . . . . . . . . . . . . . . . . . . 33
3.2 Characters and Class . . . . . . . . . . . . . . . . . . . . 34
3.3 Wonderful Orthogonality Theorem for Character . . . . . 36
3.4 Reducible Representations . . . . . . . . . . . . . . . . . 38
iii
iv CONTENTS
3.5 The Number of Irreducible Representations . . . . . . . . 40
3.6 Second Orthogonality Relation for Characters . . . . . . 41
3.7 Regular Representation . . . . . . . . . . . . . . . . . . . 43
3.8 Setting up Character Tables . . . . . . . . . . . . . . . . 47
3.9 Symmetry Notation . . . . . . . . . . . . . . . . . . . . . 51
3.10 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 73
4 Basis Functions 75
4.1 Symmetry Operations and Basis Functions . . . . . . . . 75
4.2 Basis Functions for Irreducible Representations . . . . . 77
4.3 Projection Operators ˆP
(Γn)
kl . . . . . . . . . . . . . . . . . 82
4.4 Derivation of ˆP
(Γn)
k . . . . . . . . . . . . . . . . . . . . . 83
4.5 Projection Operations on an Arbitrary Function . . . . . 84
4.6 Linear Combinations for 3 Equivalent Atoms . . . . . . . 86
4.7 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 92
5 Group Theory and Quantum Mechanics 95
5.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
5.2 The Group of Schr¨odinger’s Equation . . . . . . . . . . . 96
5.3 The Application of Group Theory . . . . . . . . . . . . . 98
5.4 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 100
6 Application to Crystal Field Splitting 103
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 103
6.2 Comments on the Form of Crystal Fields . . . . . . . . . 106
6.3 Characters for the Full Rotation Group . . . . . . . . . . 109
6.4 Example of a Cubic Crystal Field Environment . . . . . 113
6.5 Comments on Basis Functions . . . . . . . . . . . . . . . 119
6.6 Characters for Other Symmetry Operators . . . . . . . . 124
6.7 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 126
7 Application to Selection Rules 129
7.1 Summary of Important Results for Basis Functions . . . 131
7.2 Direct Product of Two Groups . . . . . . . . . . . . . . . 133
7.3 Direct Product of Two Irreducible Representations . . . 134
7.4 Characters for the Direct Product of Groups and Representations
. . . . . . . . . . . . . . . . . . . . . . . . . . 135
CONTENTS v
7.5 The Selection Rule Concept in Group Theoretical Terms 138
7.6 Selection Rules for Electric Dipole Transitions . . . . . . 140
7.7 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 144
8 Electronic States of Molecules 147
8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 147
8.2 General Concept of Equivalence . . . . . . . . . . . . . . 151
8.3 Directed Valence Bonding . . . . . . . . . . . . . . . . . 153
8.4 Diatomic Molecules . . . . . . . . . . . . . . . . . . . . . 154
8.4.1 Homonuclear Diatomic Molecules in General . . . 154
8.4.2 The Hydrogen Molecule H2 . . . . . . . . . . . . 156
8.4.3 The Helium Molecule He2 . . . . . . . . . . . . . 157
8.4.4 Heterogeneous Diatomic Molecules . . . . . . . . 157
8.5 Electronic Orbitals for Multi-atomic Molecules . . . . . . 162
8.5.1 The NH3 Molecule . . . . . . . . . . . . . . . . . 162
8.5.2 The CH4 Molecule . . . . . . . . . . . . . . . . . 163
8.5.3 The Hypothetical SH6 Molecule . . . . . . . . . . 171
8.5.4 The SF6 Molecule . . . . . . . . . . . . . . . . . . 175
8.5.5 The B12H12 Molecule . . . . . . . . . . . . . . . . 178
8.6 Bond Strengths . . . . . . . . . . . . . . . . . . . . . . . 181
8.7 σ- and π-bonds . . . . . . . . . . . . . . . . . . . . . . . 184
8.8 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 192
9 Molecular Vibrations 195
9.1 Molecular Vibrations – Background . . . . . . . . . . . . 195
9.2 Application of Group Theory to Molecular Vibrations . . 197
9.3 Molecular Vibrations in H2O . . . . . . . . . . . . . . . . 200
9.4 Overtones and Combination Modes . . . . . . . . . . . . 203
9.5 Infrared Activity . . . . . . . . . . . . . . . . . . . . . . 203
9.6 Vibrations for Linear Molecules . . . . . . . . . . . . . . 206
9.6.1 The CO Molecule . . . . . . . . . . . . . . . . . . 206
9.6.2 The O2 Molecule . . . . . . . . . . . . . . . . . . 208
9.6.3 The CO2 Molecule . . . . . . . . . . . . . . . . . 209
9.6.4 The C2H2 Molecule . . . . . . . . . . . . . . . . . 210
9.7 Molecular Vibrations in Other Molecules . . . . . . . . . 212
9.7.1 Vibrations of the NH3 Molecule . . . . . . . . . . 212
9.7.2 Vibrations of the CH4 Molecule . . . . . . . . . . 214
vi CONTENTS
9.7.3 Vibrations of the B12H12 Molecule . . . . . . . . . 215
9.8 Raman Effect . . . . . . . . . . . . . . . . . . . . . . . . 218
9.8.1 The Raman Effect for H2 . . . . . . . . . . . . . . 221
9.8.2 The Raman Effect for H2O . . . . . . . . . . . . . 221
9.8.3 The Raman Effect for NH3 . . . . . . . . . . . . . 221
9.8.4 The Raman Effect for CH4 . . . . . . . . . . . . . 222
9.8.5 The Raman Effect for CO2 and C2H2 . . . . . . . 222
9.8.6 The Raman Effect for Planar XH3 . . . . . . . . . 223
9.8.7 The Raman Effect for B12H12 . . . . . . . . . . . 223
9.9 Rotational Energy Levels . . . . . . . . . . . . . . . . . . 224
9.10 Vibrational-Rotational Interaction . . . . . . . . . . . . . 227
9.11 Wigner–Eckart Theorem and Selection Rules . . . . . . . 230
9.12 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 232
10 Permutation Groups 235
10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 236
10.2 Classes of Permutation Groups . . . . . . . . . . . . . . 239
10.3 Number of Irreducible Representations . . . . . . . . . . 242
10.4 Basis Functions of Permutation Groups . . . . . . . . . . 243
10.5 Pauli Principle in Atomic Spectra . . . . . . . . . . . . . 245
10.5.1 Two-Electron States . . . . . . . . . . . . . . . . 246
10.5.2 Three-Electron States . . . . . . . . . . . . . . . 250
10.5.3 Four-Electron States . . . . . . . . . . . . . . . . 255
10.5.4 Five-Electron States . . . . . . . . . . . . . . . . 258
10.6 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . 260
10.7 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 262
11 Transformation of Tensors 267
11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 267
11.2 Independent Components of Tensors . . . . . . . . . . . 270
11.3 Tensors under Permutations . . . . . . . . . . . . . . . . 271
11.4 Independent Components of Tensors . . . . . . . . . . . 276
11.5 Tensors Arising in Non-Linear Optics . . . . . . . . . . . 277
11.5.1 Cubic Symmetry – Oh . . . . . . . . . . . . . . . 277
11.5.2 Tetrahedral Symmetry – Td . . . . . . . . . . . . 280
11.5.3 Hexagonal Symmetry . . . . . . . . . . . . . . . . 281
11.5.4 Hexagonal Symmetry . . . . . . . . . . . . . . . . 282
CONTENTS vii
11.6 Elastic Modulus Tensor . . . . . . . . . . . . . . . . . . . 283
11.6.1 Full Rotational Symmetry: 3D Isotropy . . . . . . 284
11.6.2 Icosahedral Symmetry . . . . . . . . . . . . . . . 288
11.6.3 Cubic Symmetry . . . . . . . . . . . . . . . . . . 289
11.6.4 Full Axial Symmetry . . . . . . . . . . . . . . . . 291
11.6.5 Hexagonal Symmetry . . . . . . . . . . . . . . . . 293
11.6.6 Other Symmetry Groups . . . . . . . . . . . . . . 295
11.7 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 295
12 Space Groups 299
12.1 Simple Space Group Operations . . . . . . . . . . . . . . 299
12.2 Space Groups and Point Groups . . . . . . . . . . . . . . 306
12.3 Compound Space Group Operations . . . . . . . . . . . . 308
12.4 Incompatibility of Five-Fold Symmetry . . . . . . . . . . 311
12.5 Two Dimensional Space Groups . . . . . . . . . . . . . . 315
12.5.1 Five Two-dimensional Bravais Lattices . . . . . . 315
12.5.2 Notation . . . . . . . . . . . . . . . . . . . . . . . 315
12.5.3 Listing of the Space Groups . . . . . . . . . . . . 316
12.5.4 2D Oblique Space Groups . . . . . . . . . . . . . 318
12.5.5 2D Rectangular Space Groups . . . . . . . . . . . 318
12.5.6 2D Square Space Group . . . . . . . . . . . . . . 327
12.5.7 2D Hexagonal Space Groups . . . . . . . . . . . . 335
12.6 Three Dimensional Space Groups . . . . . . . . . . . . . 335
12.6.1 Examples of Non-Symmorphic 3D Space Groups . 336
12.7 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 342
13 Group of the Wave Vector and Bloch’s Theorem 345
13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 345
13.2 Bloch’s Theorem . . . . . . . . . . . . . . . . . . . . . . 346
13.3 Group of the Wave Vector . . . . . . . . . . . . . . . . . 349
13.3.1 Reciprocal Lattice . . . . . . . . . . . . . . . . . 352
13.4 Simple Cubic Lattice . . . . . . . . . . . . . . . . . . . . 353
13.5 High Symmetry Points and Axes . . . . . . . . . . . . . 359
13.6 Group Operations on Bloch Functions . . . . . . . . . . 365
13.7 Compatibility Relations . . . . . . . . . . . . . . . . . . 368
13.7.1 Irreducible Representations . . . . . . . . . . . . 371
13.8 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 372
viii CONTENTS
14 Applications to Lattice Vibrations 375
14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 375
14.2 Lattice Modes Relative to Molecular Vibrations . . . . . 379
14.3 Zone Center Phonon Modes . . . . . . . . . . . . . . . . 381
14.3.1 In the NaCl Structure . . . . . . . . . . . . . . . 381
14.3.2 In the Perovskite Structure . . . . . . . . . . . . 383
14.3.3 Phonons in the Diamond Lattice . . . . . . . . . 387
14.3.4 Phonons in the Zincblende Structure . . . . . . . 391
14.4 Lattice Modes Away From k = 0 . . . . . . . . . . . . . 392
14.4.1 Phonons in NaCl at the X point k = π
a
(100) . . . 393
14.4.2 Phonons in BaTi3 at the X point . . . . . . . . . 394
14.5 Phonons in Te and Quartz . . . . . . . . . . . . . . . . . 399
14.5.1 Phonons in Tellurium . . . . . . . . . . . . . . . . 400
14.5.2 Phonons in α-Quartz . . . . . . . . . . . . . . . . 407
14.5.3 Effect of Uniaxial Stress on Phonons . . . . . . . 414
14.6 Lattice Modes in High Tc Related Materials . . . . . . . 417
14.6.1 The K2NiF4 Structure . . . . . . . . . . . . . . . 417
14.6.2 Phonons in the YBa2Cu3O6 Structure . . . . . . . 419
14.6.3 In The YBa2Cu3O7 Structure . . . . . . . . . . . 421
14.7 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 424
15 Use of Standard Reference Texts 425
15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 425
15.2 Determination of the Crystal Structure . . . . . . . . . . 426
15.3 Determination of the Space Group . . . . . . . . . . . . 429
15.4 Character Tables for Groups of the Wave Vector . . . . . 434
15.5 Phonons in Graphite . . . . . . . . . . . . . . . . . . . . 437
15.5.1 Phonons in Ordinary Hexagonal Graphite . . . . 439
15.5.2 Phonons in Puckered Graphite . . . . . . . . . . . 441
15.6 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 444
16 Energy Levels in Cubic Crystals 447
16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 447
16.2 Plane Wave Solutions at k = 0 . . . . . . . . . . . . . . . 450
16.3 Symmetrized Plane Waves at ∆ . . . . . . . . . . . . . . 456
16.4 Plane Wave Solutions at the X Point . . . . . . . . . . . 459
16.5 Effect of Glide Planes and Screw Axes . . . . . . . . . . 465
CONTENTS ix
16.6 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 474
17 Energy Band Models Based on Symmetry 477
17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 477
17.2 k · p Perturbation Theory . . . . . . . . . . . . . . . . . 478
17.3 k · p perturbation theory in sc lattice . . . . . . . . . . . 480
17.4 Two Band Model in Perturbation Theory . . . . . . . . . 483
17.5 Degenerate k · p Perturbation Theory . . . . . . . . . . . 489
17.6 Non-Degenerate k · p Perturbation Theory . . . . . . . . 497
17.7 Optical Matrix Elements . . . . . . . . . . . . . . . . . . 498
17.8 Fourier Expansion of Energy Bands . . . . . . . . . . . . 499
17.8.1 Contributions at d = 0: . . . . . . . . . . . . . . . 508
17.8.2 Contributions at d = 1: . . . . . . . . . . . . . . . 508
17.8.3 Contributions at d = 2: . . . . . . . . . . . . . . . 509
17.8.4 Contributions at d = 3: . . . . . . . . . . . . . . . 509
17.8.5 Other Degenerate Levels . . . . . . . . . . . . . . 510
17.8.6 Summary . . . . . . . . . . . . . . . . . . . . . . 512
17.9 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 513
18 Application to Valley-Orbit Interactions 515
18.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 515
18.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . 516
18.3 Impurity States for Multivalley Semiconductors . . . . . 518
18.4 The Valley-Orbit Interaction . . . . . . . . . . . . . . . . 520
18.5 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 527
19 Spin Orbit Interaction in Solids 529
19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 529
19.2 Crystal Double Groups . . . . . . . . . . . . . . . . . . . 534
19.3 Double Group Properties . . . . . . . . . . . . . . . . . . 537
19.4 Crystal Field Including Spin-Orbit . . . . . . . . . . . . 544
19.5 Use of the Koster et al. Reference . . . . . . . . . . . . . 550
19.6 Plane Wave Functions for Double Groups . . . . . . . . . 554
19.7 Use of Reference Books . . . . . . . . . . . . . . . . . . . 561
19.8 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 571
x CONTENTS
20 Application to Energy Bands with Spin 573
20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 573
20.2 k · p Perturbation with Spin-Orbit . . . . . . . . . . . . . 574
20.3 Basis Functions for Double Group Representations . . . . 578
20.4 Basis Functions for j = 3/2 and 1/2 States . . . . . . . . 580
20.5 Basis Functions for Other Γ+
8 States . . . . . . . . . . . . 583
20.6 E(k) Including Spin-Orbit Interaction . . . . . . . . . . . 585
20.7 E(k) for Degenerate Bands . . . . . . . . . . . . . . . . . 587
20.8 Effective g–Factor . . . . . . . . . . . . . . . . . . . . . . 591
20.9 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 603
21 Time Reversal Symmetry 605
21.1 The Time Reversal Operator . . . . . . . . . . . . . . . . 605
21.2 Time Reversal Operator . . . . . . . . . . . . . . . . . . 606
21.3 Effect of ˆT on E(k) . . . . . . . . . . . . . . . . . . . . . 610
21.4 Including the Spin-Orbit Interaction . . . . . . . . . . . 615
21.5 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 622
22 Magnetic Groups 623
22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 623
22.2 Types of Elements . . . . . . . . . . . . . . . . . . . . . 624
22.3 Types of Magnetic Point Groups . . . . . . . . . . . . . 624
22.4 58 Magnetic Point Groups . . . . . . . . . . . . . . . . . 627
22.5 Examples of Magnetic Structures . . . . . . . . . . . . . 631
22.5.1 Orthorhombic Ferromagnetic Unit Cell . . . . . . 631
22.5.2 Antiferromagnets with the Rutile Structure . . . 633
22.5.3 The Magnetic States of EuSe . . . . . . . . . . . 635
22.6 Selected Problems . . . . . . . . . . . . . . . . . . . . . . 637
23 Fullerenes and Carbon Nanotubes 639
23.1 Icosahedral Symmetry Operations . . . . . . . . . . . . . 640
23.2 Symmetry of Vibrational Modes . . . . . . . . . . . . . . 647
23.3 Symmetry for Electronic States . . . . . . . . . . . . . . 652
23.4 Going from Higher to Lower Symmetry . . . . . . . . . . 659
23.4.1 Symmetry Considerations for C70 . . . . . . . . . 659
23.4.2 Symmetry for Higher Mass Fullerenes . . . . . . . 670
23.5 Symmetry for Isotopic Effects . . . . . . . . . . . . . . . 671
CONTENTS 1
23.6 Symmetry Properties of Carbon Nanotubes . . . . . . . . 675
23.6.1 Relation between Nanotubes and Fullerenes . . . 675
23.6.2 Specification of Lattice Vectors in Real Space . . 681
23.6.3 Symmetry for Symmorphic Carbon Nanotubes . . 685
23.6.4 Symmetry for Nonsymmorphic Nanotubes . . . . 688
23.6.5 Reciprocal Lattice Vectors . . . . . . . . . . . . . 692
23.7 Suggested Problems . . . . . . . . . . . . . . . . . . . . . 694
Chapter 1
Basic Mathematical
Background – Introduction
In this chapter we consider mainly mathematical definitions and concepts
that are basic to group theory and the classification of symmetry
properties.
1.1 Definition of a Group
A collection of elements A, B, C, . . . form a group when the following
four conditions are satisfied:
1. The product of any two elements of the group is itself an element
of the group. For example, relations of the type AB = C are
valid for all members of the group.
2. The associative law is valid – i.e., (AB)C = A(BC).
3. There exists a unit element E (also called the identity element)
such that the product of E with any group element leaves that
element unchanged AE = EA = A.
4. For every element there exists an inverse, A−1
A = AA−1
= E.
It is not necessary that elements of the group commute. In general, the
elements will not commute AB = BA. But if all elements of a group
commute, the group is then called an Abelian group.
1
2 CHAPTER 1. BASIC MATHEMATICAL BACKGROUND
1 3
2
0
Figure 1.1: The symmetry operations
on an equilateral triangle,
are the rotations by ±2π/3 about
the origin 0 and the rotations by π
about the axes 01, 02, and 03.
1.2 Simple Example of a Group
As a simple example of a group, consider the permutation group for
three elements, P(3). Below are listed the 3!=6 possible permutations
that can be carried out; the top row denotes the initial arrangement of
the three numbers and the bottom row denotes the final arrangement.
E =
1 2 3
1 2 3
A =
1 2 3
2 1 3
B =
1 2 3
1 3 2
C =
1 2 3
3 2 1
D =
1 2 3
3 1 2
F =
1 2 3
2 3 1
(1.1)
This group is identical with the symmetry operations on a equilateral
triangle shown in Fig. 1.1. What then are the symmetry operations
of an equilateral triangle?
We can also think of the elements in Eq. 1.1 in terms of the 3 points
of the triangle in the initial state and the bottom line as the effect of
the six distinct symmetry operations that can be performed on these
three points. We can call each symmetry operation an element of the
group.
1.2. SIMPLE EXAMPLE OF A GROUP 3
Table 1.1: Multiplication†
table for permutation group of 3 elements;
P(3)
E A B C D F
E E A B C D F
A A E D F B C
B B F E D C A
C C D F E A B
D D C A B F E
F F B C A E D
†
AD = B defines use of multiplication table.
It is convenient to classify the products of group elements. We write
these products using a multiplication table. In Table 1.1 a multiplication
table is written out for the symmetry operations on an equilateral
triangle or equivalently for the permutation group of 3 elements. It can
easily be shown that the symmetry operations given in Eq. 1.1 satisfy
the four conditions in §1.1 and therefore form a group. Each symmetry
element of the permutation group P(3) has a one-to-one correspondence
to the symmetry operations of an equilateral triangle and we
therefore say that these two groups are isomorphic to each other. We
furthermore can use identical group theoretical procedures in dealing
with physical problems associated with either of these groups, even
though the two groups arise from totally different physical situations.
It is this generality that makes group theory so useful as a general way
to classify symmetry operations arising in physical problems.
We illustrate the use of the notation in Table 1.1 by verifying the
associative law (AB)C = A(BC) for a few elements:
(AB)C = DC = B
A(BC) = AD = B
(1.2)
Often, when we deal with symmetry operations in a crystal, the geometrical
visualization of repeated operations becomes difficult. Group
theory is designed to help with this problem. Suppose that the symmetry
operations in practical problems are elements of a group; this is
4 CHAPTER 1. BASIC MATHEMATICAL BACKGROUND
generally the case. Then if we can associate each element with a matrix
that obeys the same multiplication table as the elements themselves,
that is, if the elements obey AB = D, then the matrices representing
the elements must obey
M(A) M(B) = M(D). (1.3)
If this relation is satisfied, then we can carry out all geometrical operations
analytically in terms of arithmetic operations on matrices, which
are usually easier to perform. The one-to-one identification of a generalized
symmetry operation with a matrix is the basic idea of a representation
and why group theory plays such an important role in the
solution of practical problems.
A set of matrices that satisfy the multiplication table (Table 1.1)
for the group P(3) are:
E =
1 0
0 1
A =
1 0
0 −1
B =
−1
2
√
3
2√
3
2
1
2
C =
−1
2
−
√
3
2
−
√
3
2
1
2
D =
−1
2
√
3
2
−
√
3
2
−1
2
F =
−1
2
−
√
3
2√
3
2
−1
2
(1.4)
We note that the matrix corresponding to the identity operation is
always a unit matrix. The matrices in Eq. 1.4 constitute a matrix representation
of the group that is isomorphic to P(3) and to the symmetry
operations on an equilateral triangle.
1.3 Basic Definitions
Definition: The order of a group ≡ the number of elements in the
group. We will be mainly concerned with finite groups. As an
example, P(3) is of order 6.
Definition: A subgroup ≡ a collection of elements within a group
that by themselves form a group.
1.3. BASIC DEFINITIONS 5
Examples of subgroups in P(3):
E (E, A) (E, D, F)
(E, B)
(E, C)
Theorem: If in a finite group, an element X is multiplied by itself
enough times (n), the identity Xn
= E is eventually recovered.
Proof: If the group is finite, and any arbitrary element is multiplied
by itself repeatedly, the product will eventually give rise to a
repetition. For example, for P(3) which has six elements, seven
multiplications must give a repetition. Let Y represent such a
repetition:
Y = Xp
= Xq
where p > q. (1.5)
Then let p = q + n so that
Xp
= Xn
Xq
= Xq
= EXq
(1.6)
from which it follows that
Xn
= E. (1.7)
Definition: The order of an element ≡ the smallest value of n in
the relation Xn
= E.
We illustrate the order of an element using P(3) where:
• E is of order 1
• A, B, C are of order 2
• D, F are of order 3
Definition: The period of an element X ≡ collection of elements
E, X, X2
, . . . , Xn−1
where n is the order of the element. The
period forms an Abelian subgroup.
Some examples of periods based on the group P(3) are:
E, A
E, B
E, C
E, D, F = E, D, D2
(1.8)
6 CHAPTER 1. BASIC MATHEMATICAL BACKGROUND
1.4 Rearrangement Theorem
The rearrangement theorem is fundamental and basic to many theorems
to be proven subsequently.
Rearrangement Theorem: If E, A1, A2, . . . , Ah are the elements of a
group, and if Ak is an arbitrary group element, then the assembly
of elements
AkE, AkA1, . . . , AkAh (1.9)
contains each element of the group once and only once.
Proof: 1. We show first that every element is contained.
Let X be an arbitrary element. If the elements form a
group there will be an element Ar = A−1
k X. Then AkAr =
AkA−1
k X = X. Thus we can always find X after multiplication
of the appropriate group elements.
2. We now show that X occurs only once. Suppose that X
appears twice in the assembly AkE, AkA1, . . . , AkAh, say
X = AkAr = AkAs. Then by multiplying on the left by
A−1
k we get Ar = As which implies that two elements in the
original group are identical, contrary to the original listing
of the group elements.
Because of the rearrangement theorem, every row and column of a
multiplication table contains each element once and only once.
1.5 Cosets
Definition: If B is a subgroup of the group G, and X is an element of
G, then the assembly EX, B1X, B2X, . . . , BgX is the right coset
of B, where B consists of E, B1, B2, . . . , Bg.
A coset need not be a subgroup. A coset will be a subgroup if X is an
element of B.
Theorem: Two right cosets of given subgroup either contain exactly
the same elements, or else have no elements in common.
1.5. COSETS 7
Proof: Clearly two right cosets either contain no elements in common
or at least one element in common. We show that if there is one
element in common, all elements are in common.
Let BX and BY be two right cosets. If BkX = B Y = one
element that the two cosets have in common, then
B−1
Bk = Y X−1
(1.10)
and Y X−1
is in B, since the product on the right hand side of
Eq. 1.10 is in B. And also contained in B is EY X−1
, B1Y X−1
,
B2Y X−1
,. . ., BgY X−1
. Furthermore, according to the rearrangement
theorem, these elements are, in fact, identical with B except
for possible order of appearance. Therefore the elements of BY
are identical to the elements of BY X−1
X which are also identical
to the elements of BX so that all elements are in common.
We now give some examples of cosets using the group P(3).
Let B = E, A be a subgroup. Then the right cosets of B are
(E, A)E → E, A (E, A)C → C, F
(E, A)A → A, E (E, A)D → D, B
(E, A)B → B, D (E, A)F → F, C
(1.11)
so that there are three distinct right cosets of (E, A), namely
(E, A) which is a subgroup
(B, D) which is not a subgroup
(C, F) which is not a subgroup.
Similarly there are three left cosets of (E, A):
(E, A)
(C, D)
(B, F)
(1.12)
To multiply two cosets, we multiply constituent elements of each coset
in proper order. Such multiplication either yields a coset or joins two
cosets.
8 CHAPTER 1. BASIC MATHEMATICAL BACKGROUND
Theorem: The order of a subgroup is a divisor of the order of the
group.
Proof: If an assembly of all the distinct cosets of a subgroup is formed
(n of them), then n multiplied by the number of elements in a
coset, C, is exactly the number of elements in the group. Each
element must be included since cosets have no elements in com-
mon.
For example, for the group P(3), the subgroup (E, A) is of order
2, the subgroup (E, D, F) is of order 3 and both 2 and 3 are
divisors of 6, which is the order of P(3).
1.6 Conjugation and Class
Definition: An element B conjugate to A is by definition B ≡ XAX−1
,
where X is an arbitrary element of the group.
For example,
A = X−1
BX = Y BY −1
where BX = XA and AY = Y B.
The elements of an Abelian group are all self-conjugate.
Theorem: If B is conjugate to A and C is conjugate to B, then C is
conjugate to A.
Proof: By definition of conjugation, we can write
B=XAX−1
C=Y BY −1
.
Thus, upon substitution we obtain
C=Y XAX−1
Y −1
= Y XA(Y X)−1
.
Definition: A class is the totality of elements which can be obtained
from a given group element by conjugation.
1.6. CONJUGATION AND CLASS 9
For example in P(3), there are three classes:
(i) E; (ii) A, B, C; (iii) D, F.
Consistent with this class designation is
ABA−1
= AF = C (1.13)
DBD−1
= DA = C (1.14)
Note that each class corresponds to a physically distinct kind of
symmetry operation such as rotation of π about equivalent twofold
axes, or rotation of 2π/3 about equivalent three-fold axes.
The identity symmetry element is always in a class by itself. An
Abelian group has as many classes as elements. The identity
element is the only class forming a group, since none of the other
classes contain the identity.
Theorem: All elements of the same class have the same order.
Proof: The order of an element n is defined by An
= E. An arbitrary
conjugate of A is B = XAX−1
. Then Bn
= (XAX−1
)(XAX−1
) . . .
n times gives XAn
X−1
= XEX−1
= E.
1.6.1 Self-Conjugate Subgroups
Definition: A subgroup B is self-conjugate if XBX−1
is identical
with B for all possible choices of X in the group.
For example (E, D, F) forms a self-conjugate subgroup of P(3), but
(E, A) does not. The subgroups of an Abelian group are self-conjugate
subgroups. We will denote self-conjugate subgroups by N. To form a
self-conjugate subgroup, it is necessary to include entire classes in this
subgroup.
Definition: A group with no self-conjugate subgroups ≡ a simple
group.
Theorem: The right and left cosets of a self-conjugate subgroup N
are the same.
10 CHAPTER 1. BASIC MATHEMATICAL BACKGROUND
Proof: If Nk is an arbitrary element of the group, then the left coset
is found by elements XNk = XNkX−1
X = NjX, where the right
coset is formed by the elements NjX where Nj = XNkX−1
.
For example in the group P(3), one of the right cosets is (E, D, F)A =
(A, C, B) and one of the left cosets is A(E, D, F)=(A, B, C) and
both cosets are identical except for the listing of the elements.
Theorem: The multiplication of the elements of two right cosets of a
self-conjugate subgroup gives another right coset.
Proof: Let NX and NY be two right cosets. Then multiplication of
two right cosets gives
(NX)(NY ) ⇒ NkXN Y = Nk(XN )Y
= Nk(NmX)Y = (NkNm)(XY ) ⇒ N(XY )
(1.15)
and N(XY ) denotes a right coset.
The elements in one right coset of P(3) are (E, D, F)A = (A, C, B)
while (E, D, F)D = (D, F, E) is another right coset. The product
(A, C, B)(D, F, E) is (A, B, C) which is a right coset. Also the product
of the two right cosets (A, B, C)(A, B, C) is (D, F, E) which is a right
coset.
1.7 Factor Groups
Definition: The factor group of a self-conjugate subgroup is the collection
of cosets of the self-conjugate subgroup, each coset being
considered an element of the factor group. The factor group satisfies
the four rules of §1.1 and is therefore a group.
1. multiplication – (NX)(NY ) = NXY
2. associative law – holds because it holds for the elements.
3. identity – EN where E is the coset that contains the identity
element
4. inverse – (XN)(X−1
N) = (NX)(X−1
N) = N 2
= EN
1.8. SELECTED PROBLEMS 11
Definition: The index of a subgroup ≡ total number of cosets =
(order of group)/(order of subgroup).
The order of the factor group is the index of the self-conjugate subgroup.
In §1.6 we saw that (E, D, F) forms a self-conjugate subgroup, N.
The only other coset of this subgroup N is (A, B, C), so that the order
of this factor group = 2. Let (A, B, C) = A and (E, D, F) = E be the
two elements of the factor group. Then the multiplication table for this
factor group is
E A
E E A
A A E
which is also the multiplication table for the group for the permutation
of 2 objects P(2). E is the identity element of this factor group. E
and A are their own inverses. From this illustration you can see how
the four group properties (see §1.1) apply to the factor group. The
multiplication table is easily found by taking an element in each coset,
carrying out the multiplication of the elements and finding the coset of
the resulting element.
1.8 Selected Problems
1. (a) Show that the trace of an arbitrary square matrix X is invariant
under a similarity transformation UXU−1
.
(b) Given a set of matrices that represent the group G, denoted
by D(R) (for all R in G), show that the matrices obtainable
by a similarity transformation UD(R)U−1
also are a
representation of G.
2. (a) Show that the operations of P(3) in Eq. 1.1 of the class notes
form a group, referring to the rules in §1.1.
(b) Multiply the two left cosets of subgroup (E, A): (B, F) and
(C, D), referring to §1.5 of the class notes. Is the result
another coset?
12 CHAPTER 1. BASIC MATHEMATICAL BACKGROUND
(c) Prove that in order to form a normal subgroup it is necessary
to include entire classes in this subgroup.
(d) Demonstrate that the normal subgroup of P(3) includes entire
classes.
3. (a) What are the symmetry operations for the molecule AB4,
where the B atoms lie at the corners of a square and the A
atom is at the center and is not coplanar with the B atoms.
(b) Find the multiplication table.
(c) List the subgroups. Which subgroups are self-conjugate?
(d) List the classes.
(e) Find the multiplication table for the factor group for the
self-conjugate subgroup(s) of (c).
4. The group defined by the permutations of 4 objects, P(4), is
isomorphic with the group of symmetry operations of a regular
tetrahedron (Td). The symmetry operations of this group are sufficiently
complex so that the power of group theoretical methods
can be appreciated. For notational convenience, the elements of
this group are listed below.
e = (1234) g = (3124) m = (1423) s = (4213)
a = (1243) h = (3142) n = (1432) t = (4231)
b = (2134) i = (2314) o = (4123) u = (3412)
c = (2143) j = (2341) p = (4132) v = (3421)
d = (1324) k = (3214) q = (2413) w = (4312)
f = (1342) l = (3241) r = (2431) y = (4321)
Here we have used a shorthand notation to denote the elements:
for example j = (2341) denotes
1 2 3 4
2 3 4 1
that is, the permutation which takes objects in the order 1234
and leaves them in the order 2341.
1.8. SELECTED PROBLEMS 13
(a) What is the product vw? wv?
(b) List the subgroups of this group which correspond to the
symmetry operations on an equilateral triangle.
(c) List the right and left cosets of the subgroup (e, a, k, l, s, t).
(d) List all the symmetry classes for P(4), and relate them to
symmetry operations on a regular tetrahedron.
(e) Find the factor group and multiplication table formed from
the self-conjugate sub-group (e, c, u, y). Is this factor group
isomorphic to P(3)?
14 CHAPTER 1. BASIC MATHEMATICAL BACKGROUND
Chapter 2
Representation Theory and
Basic Theorems
In this chapter we introduce the concept of a representation of an abstract
group and prove a number of important theorems relating to irreducible
representations, including the “Wonderful Orthogonality The-
orem”.
2.1 Important Definitions
Definition: Two groups are isomorphic or homomorphic if there
exists a correspondence between their elements such that
A → ˆA
B → ˆB
AB → ˆA ˆB
where the plain letters denote elements in one group and the
letters with carets denote elements in the other group. If the two
groups have the same order (same number of elements), then they
are isomorphic.
For example, the permutation group of three numbers P(3) is isomorphic
to the symmetry group of the equilateral triangle and homomorphic
to its factor group, as shown in Table 2.1. Thus, the
15
16 CHAPTER 2. REPRESENTATION THEORY
Table 2.1: Table of homomorphic mapping.
Permutation group element Factor group
E, D, F → E
A, B, C → A
homomorphic representations in Table 2.1 are unfaithful. Isomorphic
representations are faithful, because they maintain the one-to-one cor-
respondence.
Definition: A representation of an abstract group is a substitution
group (matrix group with square matrices) such that the substitution
group is homomorphic (or isomorphic) to the abstract group.
We assign a matrix D(A) to each element A of the abstract group
such that D(AB) = D(A)D(B).
The matrices of Eq. 1.4 are an isomorphic representation of the
permutation group P(3). In considering the representation
E
D
F



→ (1)
A
B
C



→ (−1)
the one-dimensional matrices (1) and (–1) are a homomorphic representation
of P(3) and an isomorphic representation of the factor group
E, A (see §1.7). The homomorphic one-dimensional representation (1)
is a representation for any group, though an unfaithful one.
In quantum mechanics, the matrix representation of a group is important
for several reasons. First of all, we will find that the eigenfunctions
for a quantum mechanical problem will transform under a symmetry
operation according to some matrix representation of a group.
Secondly, quantum mechanical operators are usually written in terms
of a matrix representation, and thus it is convenient to write symmetry
operations using the same kind of matrix representation. Finally, matrix
algebra is often easier to manipulate than geometrical symmetry
operations.
2.2. MATRICES 17
2.2 Matrices
Definitions: Hermitian matrices are defined by: ˜A = A∗
, ˜A∗
= A,
or A†
= A (where the symbol ∗ denotes complex conjugation, ∼
denotes transposition, and † denotes taking the adjoint)
A =




a11 a12 · · ·
a21 a22 · · ·
...
...



 (2.1)
˜A =




a11 a21 · · ·
a12 a22 · · ·
...
...



 (2.2)
Unitary matrices are defined by: ˜A∗ = A†
= A−1
Orthonormal matrices are defined by: ˜A = A−1
Definition: The dimensionality of a representation is equal to the dimensionality
of each of its matricies, which is in turn equal to the
number of rows or columns of the matrix.
These representations are not unique. For example, by performing
a similarity (or equivalence, or canonical) transformation UD(A)U−1
we generate a new set of matrices which provides an equally good representation.
We can also generate another representation by taking one
or more representations and combining them according to
D(A) O
O D (A)
(2.3)
where O = (m×n) matrix of zeros, not necessarily a square zero matrix.
The matrices D(A) and D (A) can be either two distinct representations
or they can be identical representations.
To overcome the difficulty of non-uniqueness of a representation
with regard to a similarity transformation, we often just deal with the
traces of the matrices which are invariant under similarity transformations.
The trace of a matrix is defined as the sum of the diagonal
matrix elements.
18 CHAPTER 2. REPRESENTATION THEORY
2.3 Irreducible Representations
To overcome the difficulty of the ambiguity of representations in general,
we introduce the concept of irreducible representations. Consider
the representation made up of two distinct or identical representations
for every element in the group
D(A) O
O D (A)
.
This is a reducible representation because the matrix corresponding to
each and every element of the group is in the same block form. We
could now carry out a similarity transformation which would mix up
all the elements so that the matrices are no longer in block form. But
still the representation is reducible. Hence the definition:
Definition: If by one and the same equivalence transformation, all
the matrices in the representation of a group can be made to
acquire the same block form, then the representation is said to
be reducible; otherwise it is irreducible. Thus, an irreducible
representation cannot be expressed in terms of representations of
lower dimensionality.
We will now consider three irreducible representations for the permutation
group P(3):
E A B
Γ1 : (1) (1) (1)
Γ1 : (1) (−1) (−1)
Γ2 :
1 0
0 1
1 0
0 −1
−1
2
√
3
2√
3
2
1
2
C D F
Γ1 : (1) (1) (1)
Γ1 : (−1) (1) (1)
Γ2 :
−1
2
−
√
3
2
−
√
3
2
1
2
−1
2
√
3
2
−
√
3
2
−1
2
−1
2
−
√
3
2√
3
2
−1
2
(2.4)
2.3. IRREDUCIBLE REPRESENTATIONS 19
A reducible representation containing these three irreducible representations
is:
E A B
ΓR :





1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1










1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 −1











1 0 0 0
0 −1 0 0
0 0 −1
2
√
3
2
0 0
√
3
2
1
2






etc.
(2.5)
where ΓR is of the form



Γ1 0 O
0 Γ1 O
O O Γ2


 . (2.6)
It is customary to list the irreducible representations contained in a
reducible representation ΓR as
ΓR = Γ1 + Γ1 + Γ2. (2.7)
In working out problems of physical interest, each irreducible representation
describes the transformation properties of a set of eigenfunctions
and corresponds to a distinct energy eigenvalue. Assume ΓR
is a reducible representation for some group G but an irreducible representation
for some other group G . If ΓR contains the irreducible
representations Γ1 + Γ1 + Γ2 as illustrated above for the group P(3),
this indicates that some interaction is breaking up a four-fold degenerate
level in group G into three energy levels in group G: two nondegenerate
ones and a doubly degenerate one. Group theory doesn’t
tell us what these energies are, nor their ordering. Group theory only
specifies the symmetries and degeneracies of the energy levels. In general,
the higher the symmetry, the higher the degeneracy. Thus when
a perturbation is applied to lower the symmetry, the degeneracy of the
energy levels tends to be reduced. Group theory provides a systematic
method for determining how the degeneracy is lowered.
Representation theory is useful for the treatment of physical problems
because of certain orthogonality theorems which we will now discuss.
To prove the orthogonality theorems we need first to prove some
other theorems (including the unitarity of representations in §2.4 and
the two Schur lemmas in §2.5 and §2.6.)
20 CHAPTER 2. REPRESENTATION THEORY
2.4 The Unitarity of Representations
This theorem which shows that in most physical cases, the elements
of a group can be represented by unitary matrices. This theorem is
then used to prove lemmas leading to the proof of the “Wonderful
Orthogonality Theorem”.
Theorem: Every representation with matrices having non-vanishing
determinants can be brought into unitary form by an equivalence
transformation.
Proof: By unitary form we mean that the matrix elements obey the
relation (A−1
)ij = A†
ij = A∗
ji where A is an arbitrary matrix
of the representation. The proof is carried out be finding the
corresponding unitary matrices if the Aij matrices are not already
unitary matrices.
Let A1, A2, · · · , Ah denote matrices of the representation. We
start by forming the matrix sum
H =
h
x=1
AxA†
x (2.8)
where the sum is over all the elements in the group and where the
adjoint of a matrix is the transposed complex conjugate matrix
(A†
x)ij = (Ax)∗
ji. The matrix H is Hermitian because
H†
=
x
(AxA†
x)†
=
x
AxA†
x. (2.9)
Any Hermitian matrix can be diagonalized by a suitable unitary
transformation. Let U be a unitary matrix made up of the orthonormal
eigenvectors which diagonalize H to give the diagonal
matrix d:
d = U−1
HU =
x
U−1
AxA†
xU =
x
U−1
AxUU−1
A†
xU =
x
ˆAx
ˆA†
x
(2.10)
where we define ˆAx = U−1
AxU for all x. The diagonal matrix d is
a special kind of matrix and contains only real, positive diagonal
2.4. THE UNITARITY OF REPRESENTATIONS 21
elements since
dkk = x j( ˆAx)kj( ˆA†
x)jk
= x j( ˆAx)kj( ˆAx)∗
kj
= x j |( ˆAx)kj|2
.
(2.11)
One can form out of the diagonal matrix d two matrices (d1/2
and
d−1/2
) such that
d1/2
≡




√
d11 O√
d22
O
...



 (2.12)
and
d−1/2
≡




1√
d11
O
1√
d22
O
...



 (2.13)
where d1/2
and d−1/2
are real, diagonal matrices. We note that
the generation of d−1/2
from d1/2
requires that none of the dkk
vanish. These matrices clearly obey the relations
(d1/2
)†
= d1/2
(2.14)
(d−1/2
)†
= d−1/2
(2.15)
(d1/2
)(d1/2
) = d (2.16)
so that
d1/2
d−1/2
= d−1/2
d1/2
= 1 = unit matrix. (2.17)
We can also from Eq. 2.10 write
d = d1/2
d1/2
=
x
ˆAx
ˆA†
x. (2.18)
We now define a new set of matrices
ˆˆAx ≡ d−1/2 ˆAxd1/2
(2.19)
and
ˆA†
x = (U−1
AxU)†
= U−1
A†
xU (2.20)
22 CHAPTER 2. REPRESENTATION THEORY
ˆˆA
†
x = (d−1/2 ˆAxd1/2
)†
= d1/2 ˆ
A†
xd−1/2
. (2.21)
We now show that the matrices
ˆˆAx are unitary:
ˆˆAx
ˆˆA
†
x = (d−1/2 ˆAxd1/2
)(d1/2 ˆA†
xd−1/2
)
= d−1/2 ˆAxd ˆA†
xd−1/2
= d−1/2
y
ˆAx
ˆAy
ˆA†
y
ˆA†
xd−1/2
= d−1/2
y
( ˆAx
ˆAy)( ˆAx
ˆAy)†
d−1/2
= d−1/2
z
ˆAz
ˆA†
z d−1/2
(2.22)
by the rearrangement theorem. But from the relation
d =
z
ˆAz
ˆA†
z (2.23)
it follows that
ˆˆAx
ˆˆA
†
x = 1 so that
ˆˆAx is unitary.
Therefore we have demonstrated how we can always construct a
unitary representation by the transformation:
ˆˆAx = d−1/2
U−1
AxUd1/2
(2.24)
where
H =
h
x=1
AxA†
x (2.25)
d =
h
x=1
ˆAx
ˆA†
x (2.26)
U is the unitary matrix that diagonalizes the Hermitian matrix
H and ˆAx = U−1
AxU.
Note: On the other hand, not all symmetry operations can be represented
by a unitary matrix; an example of an operation which cannot be
represented by a unitary matrix is the time inversion operator. Time inversion
symmetry is represented by an anti-unitary matrix rather than
an unitary matrix. It is thus not possible to represent all symmetry
operations by a unitary matrix. Time inversion symmetry is discussed
later in the book.
2.5. SCHUR’S LEMMA (PART I) 23
2.5 Schur’s Lemma (Part I)
Schur’s lemmas on irreducible representations are proved in order to
prove the “Wonderful Orthogonality Theorem” in §2.7.
Lemma: A matrix which commutes with all matrices of an irreducible
representation is a constant matrix, i.e., a constant times the unit
matrix. Therefore, if a non-constant commuting matrix exists,
the representation is reducible; if none exists, the representation
is irreducible.
Proof: Let M be a matrix which commutes with all the matrices of
the representation A1, A2, . . . , Ah
MAx = AxM. (2.27)
Take the adjoint of both sides of Eq. 2.27 to obtain
A†
xM†
= M†
A†
x. (2.28)
Since Ax can in all generality be taken to be unitary (see §2.4), multiply
on the right and left of Eqs. 2.28 by Ax to yield
M†
Ax = AxM†
(2.29)
so that if M commutes with Ax so does M†
, and so do the Hermitian
matrices H1 and H2 defined by
H1=M + M†
H2=i(M − M†
),
(2.30)
HjAx = AxHj where j = 1, 2. (2.31)
We will now show that a commuting Hermitian matrix is a constant
matrix from which it follows that M = H1 − iH2 is also a constant
matrix.
Since Hj (j = 1, 2) is a Hermitian matrix, it can be diagonalized.
Let U be the matrix that diagonalizes Hj (for example H1) to give the
diagonal matrix d
d = U−1
HjU. (2.32)
24 CHAPTER 2. REPRESENTATION THEORY
We now perform the unitary transformation on the matrices Ax of
the representation ˆAx = U−1
AxU. From the commutation relations
Eqs. 2.27, 2.28 and 2.31, a unitary transformation on all matrices
HiAx = AxHi yields
(U−1
HjU)
d
(U−1
AxU)
ˆAx
= (U−1
AxU)
ˆAx
(U−1
HjU)
d
. (2.33)
So now we have a diagonal matrix d which commutes with all the
matrices of the representation. We now show that this diagonal matrix
d is a constant matrix, if the ˆAx matrices (and thus also the Ax matrices)
form an irreducible representation. Thus, starting with Eq. 2.33
d ˆAx = ˆAxd (2.34)
we take the ij element of both sides of Eq. 2.34
dii( ˆAx)ij = ( ˆAx)ijdjj (2.35)
so that
( ˆAx)ij(dii − djj) = 0 (2.36)
for all the matrices Ax.
If dii = djj, so that the matrix d is not a constant diagonal matrix,
then ( ˆAx)ij must be 0 for all the ˆAx. This means that the similarity
transformation U−1
AxU has brought all the matrices of the representation
into the same block form, showing that the representation Ax is
reducible. But we have assumed the Ax to be irreducible – therefore
dii = djj and Schur’s lemma part 1 is proved.
2.6 Schur’s Lemma (Part 2)
Lemma: If the matrix representations D(1)
(A1), D(1)
(A2), . . . , D(1)
(Ah)
and D(2)
(A1), D(2)
(A2), . . . , D(2)
(Ah) are two irreducible representations
of a given group of dimensionality 1 and 2, respectively,
then, if there is a matrix of 1 columns and 2 rows M such that
MD(1)
(Ax) = D(2)
(Ax)M (2.37)
2.6. SCHUR’S LEMMA (PART 2) 25
for all Ax, then M must be the null matrix (M = O) if 1 = 2. If
1 = 2, then either M = O or the representations D(1)
(Ax) and
D(2)
(Ax) differ from each other by an equivalence or similarity
transformation.
Proof: Since the matrices which form the representation can always
be transformed into unitary form, we can in all generality assume
that the matrices of both representations D(1)
(Ax) and D(2)
(Ax)
have already been brought into unitary form.
Assume 1 ≤ 2, and take the adjoint of Eq. 2.37
[D(1)
(Ax)]†
M†
= M†
[D(2)
(Ax)]†
. (2.38)
The unitary property of the representation implies [D(Ax)]†
= [D(Ax)]−1
=
D(A−1
x ), since the matrices form a substitution group for the elements
Ax of the group. Therefore we can write Eq. 2.38 as
D(1)
(A−1
x )M†
= M†
D(2)
(A−1
x ). (2.39)
Then multiplying Eq. 2.39 on the left by M yields:
MD(1)
(A−1
x )M†
= MM†
D(2)
(A−1
x ) = D(2)
(A−1
x )MM†
(2.40)
which follows from applying Eq. 2.37 to the element A−1
x which is also
an element of the group:
MD(1)
(A−1
x ) = D(2)
(A−1
x )M. (2.41)
We have now shown that if MD(1)
(Ax) = D(2)
(Ax)M then MM†
commutes with all the matrices of representation (2) and M†
M commutes
with all matrices of representation (1). But if MM†
commutes
with all matrices of a representation, then by Schur’s lemma (part 1),
MM†
is a constant matrix of dimensionality ( 2 × 2):
MM†
= c ˆ1, (2.42)
where ˆ1 is the unit matrix.
26 CHAPTER 2. REPRESENTATION THEORY
First we consider the case 1 = 2. Then M is a square matrix, with
an inverse:
M−1
=
M†
c
, c = 0. (2.43)
Then if M−1
= O, multiplying Eq. 2.37 by M−1
on the left yields:
D(1)
(Ax) = M−1
D(2)
(Ax)M (2.44)
and the two representations differ by an equivalence transformation.
However, if c = 0 then we cannot write Eq. 2.43, but instead we
have to consider MM†
= 0:
k
MikM†
kj = 0 =
k
MikM∗
jk (2.45)
for all ij elements. In particular, for i = j we can write
k
MikM∗
ik =
k
|Mik|2
= 0 (2.46)
Therefore each element Mik = 0 and M is a null matrix. This completes
proof of the case 1 = 2 and M = O.
Finally we prove that for 1 = 2, then M = O. Suppose that
1 = 2, then we can arbitrarily take 1 < 2. Then M has 1 columns
and 2 rows. We can make a square ( 2 × 2) matrix out of M by adding
( 2 − 1) columns of zeros
1 columns
2 rows









0 0 0
0 0 0
M 0 0 0
...
...
...
0 0 0









= N = square ( 2 × 2) matrix.
(2.47)
The adjoint of Eq. 2.47 is then written as











M†
0 0 0 · · · 0
0 0 0 · · · 0
...
...
...
...
0 0 0 · · · 0











= N†
(2.48)
2.7. WONDERFUL ORTHOGONALITY THEOREM 27
so that
NN†
= MM†
= c ˆ1 dimension ( 2 × 2). (2.49)
k NikN†
ki = k NikN∗
ik = c
ik NikN∗
ik = c 2.
But if we sum over i we see by direct computation k,i NikN∗
ik = 0, so
that c = 0. But this implies that every element Nik = 0 and therefore
also Mik = 0, so that M is a null matrix, completing the proof of
Schur’s lemma (part 2).
2.7 Wonderful Orthogonality Theorem
The orthogonality theorem which we now prove is so central to the
application of group theory to quantum mechanical problems that it
was named the “Wonderful Orthogonality Theorem” by Van Vleck,
and is widely known by this name.
Theorem: The orthonormality relation
R
D( 2)
µν (R)D
( 1)
ν µ (R−1
) =
h
1
δΓ1,Γ2 δµ,µ δν,ν (2.50)
is obeyed for all the inequivalent, irreducible representations of a
group, where the summation is over all h group elements A1, A2, . . . , Ah
and i (i = 1, 2) is the dimensionality of representation Γi. If the
representations are unitary, the orthonormality relation becomes
R
D( 2)
µν (R) D
( 1)
µ ν (R)
∗
=
h
1
δΓ1,Γ2 δµ,µ δν,ν . (2.51)
Proof: Consider the 2 × 1 matrix
M =
R
D( 2)
(R)XD( 1)
(R−1
) (2.52)
where X is an arbitrary matrix with 2 rows and 1 columns
so that M is a rectangular matrix of dimensionality ( 2 × 1).
28 CHAPTER 2. REPRESENTATION THEORY
Multiply M by D( 2)
(S) for some element S in the group:
D( 2)
(S)M
2× 1
=
R
D( 2)
(S)D( 2)
(R) X D( 1)
(R−1
) (2.53)
Then carrying out the multiplication of two elements in a group
D( 2)
(S)M
2× 1
=
R
D( 2)
(SR) X D( 1)
(R−1
S−1
)D( 1)
(S) (2.54)
where we have used the group properties of the representations
Γ1 and Γ2. By the rearrangement theorem, the above equation
can be rewritten
D( 2)
(S)M =
R
D( 2)
(R) X D( 1)
(R−1
)
M
D( 1)
(S) = M D( 1)
(S).
(2.55)
Now apply Schur’s lemma part 2 for the various cases.
Case 1 1 = 2 or if 1 = 2, and the representations are not
equivalent.
Since D( 2)
(S)M = MD( 1)
(S), then by Schur’s lemma part 2, M
must be a null matrix. From the definition of M we have
0 = Mµµ =
R γ,λ
D( 2)
µγ (R)XγλD
( 1)
λµ (R−1
). (2.56)
But X is an arbitrary matrix. By choosing X to have an entry 1
in the νν position and 0 everywhere else, we write:
X =









0 0 0 0 0 0 · · ·
0 0 0 1 0 0 · · ·
0 0 0 0 0 0 · · ·
0 0 0 0 0 0 · · ·
...
...
...
...
...
...









. Xγλ = δγνδλν (2.57)
It then follows by substituting Eq. 2.57 into Eq. 2.56 that
0 =
R
D( 2)
µν (R)D
( 1)
ν µ (R−1
). (2.58)
2.7. WONDERFUL ORTHOGONALITY THEOREM 29
Case 2 1 = 2 and the representations Γ1 and Γ2 are equivalent
If the representations Γ1 and Γ2 are equivalent, then 1 = 2 and
Schur’s lemma part 1 tells us that M = cˆ1. The definition for M
in Eq. 2.52 gives
Mµν = cδµµ =
R γ,λ
D( 2)
µγ (R)XγλD
( 2)
λµ (R−1
). (2.59)
Choose X in Eq. 2.57 as above to have a non-zero entry at νν
and 0 everywhere else. Then Xγλ = cνν δγνδλν so that
cνν δµµ =
R
D( 2)
µν (R) D
( 2)
ν µ (R−1
) (2.60)
where cνν = c/cνν . To evaluate cνν choose µ = µ in Eq. 2.60
and sum on µ:
cνν
µ
δµµ
2
=
R µ
D( 2)
µν (R) D
( 2)
ν µ (R−1
) =
R
D
( 2)
ν ν (R−1
R)
(2.61)
since D( 2)
(R) is a representation of the group and follows the
multiplication table for the group. Therefore we can write
cνν 2 =
R
D
( 2)
ν ν (R−1
R) =
R
D
( 2)
ν ν (E) = D
( 2)
ν ν (E)
R
1. (2.62)
But D
( 2)
ν ν (E) is a unit ( 2 × 2) matrix and the ν ν matrix element
is δν ν. The sum of unity over all the group elements is h.
Therefore we obtain
cνν =
h
2
δνν . (2.63)
Substituting Eq. 2.63 into Eq. 2.60 gives:
h
2
δµµ δνν =
R
D( 2)
µν (R) D
( 2)
ν µ (R−1
). (2.64)
We can write the results of case 1 and case 2 in compact form
R
D( j)
µν (R) D
( j )
ν µ (R−1
) =
h
j
δΓj,Γj
δµµ δνν . (2.65)
30 CHAPTER 2. REPRESENTATION THEORY
For a unitary representation Eq. 2.65 can also be written as:
R
D( j)
µν (R) D
( j )∗
µ ν (R) =
h
j
δΓj,Γj
δµµ δνν . (2.66)
This completes the proof of the wonderful orthogonality theorem.
2.8 Representations and Vector Spaces
Let us spend a moment and consider what the representations in
Eq. 2.66 mean as an orthonormality relation in a vector space of
dimensionality h. Here h is the order of the group which equals
the number of group elements. In this space, the representations
D
( j)
µν (R) can be considered as elements in this h-dimensional
space:
V ( j)
µ,ν = D( j)
µν (A1), D( j)
µν (A2), . . . , D( j)
µν (Ah) . (2.67)
The three indices Γ j
, µ, ν label a particular vector. All distinct
vectors in this space are orthogonal. Thus two representations are
orthogonal if any one of their three indices is different. But in an
h-dimensional vector space, the maximum number of orthogonal
vectors is h. We now ask how many vectors V
( j)
µ,ν can we make?
For each representation, we have j choices for µ and ν so that
the total number of vectors we can have is j
2
j where we are
now summing over representations. This argument yields the
important result
j
2
j ≤ h. (2.68)
We will see later that it is the equality that holds in Eq. 2.68.
The result in Eq. 2.68 is extremely helpful in finding the totality
of irreducible (non-equivalent) representations. In our example of
P(3) we have h = 6. Therefore j
2
j = 6. The representations we
found in §2.3 were two one-dimensional and one two-dimensional
representation. Therefore j
2
j = 12
+ 12
+ 22
= 1 + 1 + 4 = 6.
This tells us that no matter how hard we try, we will not find any
more irreducible representations for P(3) – we have them all.
2.9. SUGGESTED PROBLEMS 31
2.9 Suggested Problems
1. Show that every symmetry operator for every group can be represented
by the (1 × 1) unit matrix. Is it also true that every
symmetry operator for every group can be represented by the (2
× 2) unit matrix? If so, does such a representation satisfy the
Wonderful Orthogonality Theorem? Why?
32 CHAPTER 2. REPRESENTATION THEORY
Chapter 3
Character of a
Representation
We have already discussed the arbitrariness of a representation with regard
to similarity or equivalence transformations. Namely, if D( j)
(R)
is a representation of a group, so is U−1
D( j)
(R)U. To get around this
arbitrariness we introduce the use of the trace (or character) of a matrix
representation which remains invariant under a similarity transformation.
In this chapter we define the character of a representation, derive
the most important theorems for the character, summarize the conventional
notations used to denote symmetry operations and groups and
list some of the most important character tables for the point groups.
3.1 Definition of Character
Definition: The character of the matrix representation χ j
(R) for
a symmetry operation R in a representation D(R) is the trace
(or the sum over diagonal matrix elements) of the matrix of the
representation:
χ( j)
(R) = trace D( j)
(R) =
j
µ=1
D( j)
(R)µµ (3.1)
where j is the dimensionality of the representation Γj and j is a
representation index. From the definition, it follows that repre-
33
34 CHAPTER 3. CHARACTER OF A REPRESENTATION
sentation Γj will have h characters, one for each element in the
group. Since the trace of a matrix is invariant under a similarity
transformation, the character is invariant under such a transfor-
mation.
3.2 Characters and Class
We relate concepts of class (see §1.6) and character by the following
theorem.
Theorem: The character for each element in a class is the same.
Proof: Let A and B be elements in the same class. By the definition
of class this means that A and B are related by conjugation (see
§1.6)
A = Y −1
BY (3.2)
where Y is an element of the group. Each element can always be
represented by a unitary matrix D (see §2.4), so that
D(A) = D(Y −1
) D(B) D(Y ) = D−1
(Y ) D(B) D(Y ). (3.3)
And since a similarity transformation leaves the trace invariant,
we have the desired result for characters in the same class: χ(A) =
χ(B), which completes the proof.
The property that all elements in a class have the same character is
responsible for what Van Vleck called “the great beauty of character”.
If two elements of a group are in the same class, this means
that they correspond to similar symmetry operations – e.g., the class
of two-fold axes of rotation of the equilateral triangle, or the class of
three-fold rotations for the equilateral triangle.
Sometimes a given group will have more than one kind of two-fold
symmetry axis. To test whether these two kinds of axes are indeed
symmetrically inequivalent, we check whether or not not they have the
same characters.
We summarize the information on the characters of the representations
of a group in the celebrated character table. In a character
3.2. CHARACTERS AND CLASS 35
Table 3.1: Character table for the permutation group P(3): Group
“D3”.
Class → C1 3C2 2C3
irreducible χ(E) χ(A, B, C) χ(D, F)
representation Γ1 1 1 1
↓ Γ1 1 –1 1
Γ2 2 0 –1
Table 3.2: Classes for the permutation group P(3): Group “D3”.
D3 P(3)
Class 1 E (identity) 1C1 (identity class) (1)(2)(3)
Class 2 A, B, C (3 elements) 3C2 (rotation of π about 2-fold axis) (1)(23)
Class 3 D, F (2 elements) 2C3 (rotation of 120◦
about 3-fold axis) (123)
table we list the representations in column form (for example, the left
hand column of the character table) and the class as rows (top row
labels the class). For example, the character table for the permutation
group P(3) (see §1.2) is shown in Table 3.1. (Sometimes you will see
character tables with the columns and rows interchanged relative to
this display.) We will later see that the name for this point group is D3
(Schoenflies notation). In Table 3.1 the notation NkCk is used in the
character table to label each class Ck, and Nk is the number of elements
in Ck. If a representation is irreducible, then we say that its character
is primitive. In a character table we limit ourselves to the primitive
characters. The classes for group D3 and P(3) are listed in Table 3.2.
Now that we have introduced character and character tables, let us
see how to use the character tables. To appreciate the power of the
character tables we present a few fundamental theorems for character.
36 CHAPTER 3. CHARACTER OF A REPRESENTATION
3.3 Wonderful Orthogonality Theorem for
Character
The “Wonderful Orthogonality Theorem” for character follows directly
from the wonderful orthogonality theorem (see §2.7). There is also a
second orthogonality theorem for character which is discussed below
(see §3.6).
Theorem: The primitive characters of an irreducible representation
obey the orthogonality relation
R
χ(Γ1)
(R−1
) χ(Γ2)
(R) = hδΓ1,Γ2 (3.4)
or
R
χ( j)
(R)∗
χ( j )
(R) = hδΓj,Γj
(3.5)
where Γj denotes irreducible representation j with dimensionality
j. This theorem says that unless the representations are identical
or equivalent, the characters are orthogonal in h-dimensional
space, where h is the order of the group.
Proof: The proof of the wonderful orthogonality theorem for character
follows from the Wonderful Orthogonality Theorem (see §2.7)
itself. Consider the wonderful orthogonality theorem (Eq. 2.51)
R
D( j)
µν (R)D
( j )
ν µ (R−1
) =
h
j
δΓj,Γj
δµ,µ δν,ν . (3.6)
Take the diagonal elements of Eq. 3.6:
R
D( j)
µµ (R)D
( j )
µ µ (R−1
) =
h
j
δΓj,Γj
δµµ δµ µ. (3.7)
Now sum Eq. 3.7 over µ and µ to calculate the traces or characters
R µ
D( j)
µµ (R)
µ
D
( j )
µ µ (R−1
) =
h
j
δΓj,Γj
µµ
δµµ δµ µ (3.8)
3.3. WONDERFUL ORTHOGONALITY THEOREM FOR CHARACTER37
where we note that
µµ
δµµ δµ µ =
µ
δµµ = j (3.9)
so that
R
χ( j)
(R)χ( j )
(R−1
) = hδΓj,Γj
, (3.10)
completing the proof. Equation 3.10 implies that the primitive
characters of an irreducible representation form a set of orthogonal
vectors in group-element space. Since any arbitrary representation
is equivalent to some unitary representation (and the
character is preserved under a unitary transformation), Eq. 3.10
can also be written as
R
χ( j)
(R) χ( j )
(R)
∗
= hδΓj,Γj
. (3.11)
Since the character is the same for each element in the class, the
summation in Eq. 3.11 can be written over classes k to obtain
k
Nkχ( j)
(Ck) χ( j )
(Ck)
∗
= hδΓj,Γj
(3.12)
where Nk denotes the number of elements in class k.
The importance of the results in Eqs. 3.10, 3.11, and 3.12 cannot be
over-emphasized:
1. Character tells us if a representation is irreducible or not. If a
representation is reducible then the characters are not primitive
and will generally not obey this orthogonality relation (and other
orthogonality relations that we will soon discuss).
2. Character tells us whether or not we have found all the irreducible
representations. For example, the permutation group P(3) could
not contain a three-dimensional irreducible representation, since
by Eq. 2.68
j
2
j ≤ h, (3.13)
38 CHAPTER 3. CHARACTER OF A REPRESENTATION
and if P(3) contained one 3D irreducible representation then:
12
+ 32
> 6 (3.14)
contrary to Eq. 3.13.
We will now demonstrate the use of the Wonderful Orthogonality
Theorem for Character for the permutation group P(3). Let Γ j
= Γ1
and Γ j
= Γ1 . Then use of Eq. 3.12 yields
k Nkχ( j)
(Ck) χ( j )
(Ck)
∗
= (1)(1)(1)
class of E
+ (3)(1)(−1)
class of A,B,C
+ (2)(1)(1)
class of D,F
= 1 − 3 + 2 = 0.
(3.15)
It can likewise be verified that the Wonderful Orthogonality Theorem
works for all possible combinations of Γ j
and Γ j
in the group P(3).
Character allows us to check the uniqueness of an irreducible representation,
using the following theorem.
Theorem: A necessary and sufficient condition that two irreducible
representations be equivalent is that the characters be the same.
Proof: Necessary condition: If they are equivalent, then the characters
are the same – we have demonstrated this already since
the trace of a matrix is invariant under an equivalence transfor-
mation.
Sufficient condition: If the characters are the same, the vectors
for each of the irreducible representations in h-dimensional space
can’t be orthogonal, so the representations must be equivalent.
3.4 Reducible Representations
We now prove a theorem that forms the basis for setting up the characters
of a reducible representation in terms of the primitive characters
for the irreducible representations. This theoretical background will
also be used in constructing irreducible representations and character
tables, and is essential to most of the practical applications of group
theory to solid state physics.
3.4. REDUCIBLE REPRESENTATIONS 39
Theorem: The reduction of any reducible representation into its irreducible
constituents is unique.
Thus, if χ(Ck) is the character for some class in a reducible representation,
then this theorem claims that we can write the character
for the reducible representation χ(Ck) as a linear combination
of characters for the irreducible representations of the group
χ(Γi)
(Ck)
χ(Ck) =
Γi
aiχ(Γi)
(Ck) (3.16)
where the ai coefficients are non-negative integers which denote
the number of times the irreducible representation Γi is contained
in the reducible representation. Furthermore we show here that
the ai coefficients are unique.
Proof: In proving that the ai coefficients are unique, we explicitly
determine the ai’s which constitute the characters for a reducible
representation.
Consider the sum over classes k:
k
Nk χ(Γj)
(Ck)
∗
χ(Ck) = Sj. (3.17)
Since χ(Ck) is reducible, we write the linear combination for χ(Ck)
in Eq. 3.17 using Eq. 3.16:
Sj= k Nk χ(Γj)
(Ck)
∗
Γi
aiχ(Γi)
(Ck)
= Γi
ai k Nk χ(Γj)
(Ck)
∗
χ(Γi)
(Ck) .
(3.18)
We now apply the Wonderful Orthogonality Theorem for Characters
Eq. 3.12 to get
Γi
aihδΓi,Γj
= ajh =
k
Nk χ(Γj)
(Ck)
∗
χ(Ck) = Sj (3.19)
yielding the decomposition relation
aj =
1
h k
Nk χ(Γj)
(Ck)
∗
χ(Ck) =
Sj
h
(3.20)
40 CHAPTER 3. CHARACTER OF A REPRESENTATION
and completing the proof of the theorem. Thus the coefficients ai
in Eq. 3.16 are uniquely determined. In other words, the number
of times the various irreducible representations are contained in a
given reducible representation can be obtained directly from the
character table for the group.
This sort of decomposition of the character for a reducible representation
is important for the following type of physical problem.
Consider a cubic crystal. A cubic crystal has many symmetry
operations and therefore many classes and many irreducible
representations. Now suppose that we squeeze this crystal and
lower its symmetry. Let us further suppose that the energy levels
for the cubic crystal are degenerate for certain points in the Brillouin
zone. This squeezing would most likely lift some of the level
degeneracies. To find out how the degeneracy is lifted, we take
the representation for the cubic group that corresponds to the
unperturbed energy and treat this representation as a reducible
representation in the group of lower symmetry. Then the decomposition
formulae (Eqs. 3.16 and 3.20) tell us immediately the degeneracy
and symmetry types of the split levels in the perturbed
or stressed crystal.
3.5 The Number of Irreducible Represen-
tations
We now come to another extremely useful theorem.
Theorem: The number of irreducible representations is equal to the
number of classes.
Proof: The Wonderful Orthogonality Theorem for character is
k
k =1
Nk χ(Γi)
(Ck )
∗
χ(Γj)
(Ck ) = h δΓi,Γj
(3.21)
or
k
k =1
Nk
h
χ(Γi)
(Ck )
∗ Nk
h
χ(Γj)
(Ck ) = δΓi,Γj
. (3.22)
3.6. SECOND ORTHOGONALITY RELATION FOR CHARACTERS41
Each term Nk
h
χ(Γi)
(Ck ) in Eq. 3.22 gives the k th component
of a k-dimensional vector. There can be only k such vectors in
a k-dimensional space, since the (k + 1)st vector would be linearly
dependent on the other k vectors. If there were less than
k such vectors, then the number of independent vectors would
not be large enough to span the k-dimensional space. To express
a reducible representation in terms of its irreducible components
requires that the vector space be spanned by irreducible representations.
Therefore the number of irreducible representations
must be k, the number of classes.
For our example of the permutation group of three objects, we
have three classes and therefore only three irreducible representations.
We have already found these irreducible representations
and we now know that any additional representations that we
might find are either equivalent to these representations or they
are reducible. Knowing the number of distinct irreducible representations
is very important in setting up character tables.
3.6 Second Orthogonality Relation for Char-
acters
We now prove a second orthogonality theorem for characters which
sums over the irreducible representations and is extremely valuable for
constructing character tables.
Theorem: The summation over all irreducible representations
Γj
χ( j)
(Ck) χ( j)
(Ck )
∗
Nk = hδk,k (3.23)
yields a second orthogonality relation for the characters. Thus,
the Wonderful Orthogonality Theorem for Character yields an
orthogonality relation between rows in the character table while
the second orthogonality theorem gives a similar relation between
the columns of the character table.
42 CHAPTER 3. CHARACTER OF A REPRESENTATION
Proof: Construct the matrix
Q =






χ(1)
(C1) χ(1)
(C2) · · ·
χ(2)
(C1) χ(2)
(C2) · · ·
χ(3)
(C1) χ(3)
(C2)
...
...






(3.24)
where the irreducible representations label the rows and the classes
label the columns. Q is a square matrix since by Eq. 3.22 the
number of classes (designating the column index) is equal to the
number of irreducible representations (designating the row index).
We now also construct the square matrix
Q =
1
h






N1χ(1)
(C1)∗
N1χ(2)
(C1)∗
· · ·
N2χ(1)
(C2)∗
N2χ(2)
(C2)∗
· · ·
N3χ(1)
(C3)∗
N3χ(2)
(C3)∗
· · ·
...
...






(3.25)
where the classes label the rows, and the irreducible representations
label the columns. The ij matrix element of the product
QQ summing over classes is then
(QQ )ij =
k
Nk
h
χ(Γi)
(Ck) χ(Γj)
(Ck)
∗
= δΓi,Γj
(3.26)
using the Wonderful Orthogonality Theorem for Character (Eq. 3.12).
Therefore QQ = ˆ1 or Q = Q−1
and Q Q = ˆ1 since QQ−1
=
Q−1
Q = ˆ1 where ˆ1 is the unit matrix. Now we will write Q Q
in terms of components, but now summing over the irreducible
representations
(Q Q)kk = δkk =
Γi
Nk
h
χ(Γi)
(Ck) χ(Γi)
(Ck )
∗
(3.27)
so that
Γi
χ(Γi)
(Ck) χ(Γi)
(Ck )
∗
=
h
Nk
δk,k (3.28)
which completes the proof of the second orthogonality theorem.
3.7. REGULAR REPRESENTATION 43
Table 3.3: Multiplication table for the group P(3) used to generate the
regular representation.
E A B C D F
E = E−1
E A B C D F
A = A−1
A E D F B C
B = B−1
B F E D C A
C = C−1
C D F E A B
F = D−1
F B C A E D
D = F−1
D C A B F E
3.7 Regular Representation
The regular representation provides a recipe for finding all the irreducible
representations of a group. It is not always the fastest method
for finding the irreducible representations, but it will always work.
The regular representation is found directly from the multiplication
table by rearranging the rows and columns so that the identity
element is always along the main diagonal. When this is done, the
group elements label the columns and their inverses label the rows. We
will illustrate this with the permutation group of three objects P(3) for
which the multiplication table is given in Table 1.1. Application of the
rearrangement theorem to give the identity element along the main diagonal
gives Table 3.3. Then the matrix representation for an element
X in the regular representation is obtained by putting 1 wherever X
appears in the multiplication Table 3.3 and 0 everywhere else. Thus we
obtain
Dreg
(E) =










1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1










(3.29)
which is always the unit matrix of dimension (h × h). For one of
44 CHAPTER 3. CHARACTER OF A REPRESENTATION
the other elements in the regular representation we obtain
Dreg
(A) =










0 1 0 0 0 0
1 0 0 0 0 0
0 0 0 0 0 1
0 0 0 0 1 0
0 0 0 1 0 0
0 0 1 0 0 0










(3.30)
and so on. By construction, only Dreg
(E) has a non-zero trace!
We now show that the regular representation is indeed a representation.
This means that the regular representation obeys the multiplication
table (either Table 1.1 or 3.3). Let us for example show
Dreg
(BC) = Dreg
(B)Dreg
(C). (3.31)
It is customary to denote the matrix elements of the regular representation
directly from the definition Dreg
(X)A−1
k
,Ai
where A−1
k labels the
rows and Ai labels the columns using the notation
Dreg
(X)A−1
k
,Ai
=



1 if A−1
k Ai = X
0 otherwise.
(3.32)
Using this notation, we have to show that:
Dreg
(BC)A−1
k
,Ai
=
Aj
Dreg
(B)A−1
k
,Aj
Dreg
(C)A−1
j ,Ai
. (3.33)
Now look at the rearranged multiplication table given in Table 3.3. By
construction, we have for each of the matrices:
Dreg
(B)A−1
k
,Aj
=



1 if A−1
k Aj = B
0 otherwise
(3.34)
Dreg
(C)A−1
j ,Ai
=



1 if A−1
j Ai = C
0 otherwise
(3.35)
3.7. REGULAR REPRESENTATION 45
Therefore in the sum of Eq. 3.33 over Aj, namely Aj
Dreg
(B)A−1
k
,Aj
Dreg
(C)A−1
j ,Ai
,
we have only non-zero entries when
BC = (A−1
k Aj)(A−1
j
1
Ai) = A−1
k Ai. (3.36)
But this coincides with the definition of Dreg
(BC):
Dreg
(BC)A−1
k
,Ai
=



1 if A−1
k Ai = BC
0 otherwise
(3.37)
Therefore Dreg
is, in fact, a representation of the group A1, . . . Ah, completing
the proof.
The following theorem allows us to find all the irreducible representations
from the regular representation.
Theorem: The regular representation contains each irreducible representation
a number of times equal to the dimensionality of the
representation. (For the group P(3), this theorem says that Dreg
contains D(Γ1)
once, D(Γ1 )
once and D(Γ2)
twice so that the regular
representation of P(3) would be of dimensionality 6.)
Proof: Since Dreg
is a reducible representation, we can write for the
characters (see Eq. 3.16)
χreg
(Ck) =
Γi
aiχ(Γi)
(Ck) (3.38)
where Γi
is the sum over the irreducible representations and the
ai coefficients have been shown to be unique (Eq. 3.20) and given
by
ai =
1
h k
Nk χ(Γi)
(Ck)
∗
χreg
(Ck). (3.39)
We note that NE = 1 for the identity element, which is in a class
by itself. But by construction χreg
(Ck) = 0 unless Ck = E in
which case χreg
(E) = h. Therefore ai = χΓi
(E) = i, where χΓi
is the trace of an i dimensional unit matrix, thereby completing
the proof.
46 CHAPTER 3. CHARACTER OF A REPRESENTATION
The theorem (Eq. 3.38) that we have just proven tells us that the
regular representation contains each irreducible representation of
the group at least once. To obtain these irreducible representations
explicitly, we have to carry out a similarity transformation
which brings the matrices of the regular representation into block
diagonal form. It turns out to be very messy to extract the matrices
of the regular representation – in fact, it is so tedious to
do this operation that it doesn’t even make an instructive homework
problem. It is much easier to write down the matrices which
generate the symmetry operations of the group directly.
Consider for example the permutation group of three objects P(3)
which is isomorphic to the symmetry operations of a regular triangle.
The matrices for D and F generate rotations by ±2π/3
about the z axis, which is ⊥ to the plane of the triangle. The
A matrix represents a rotation by ±π about the x axis while the
B and C matrices represent rotations by ±π about axes in the
x − y plane which are ±120◦
away from the x axis. In setting up
a representation, it is advantageous to write down those matrices
which can be easily written down – such as E, A, D, F. The
remaining matrices such as B and C can then be found through
the multiplication table.
We will now make use of the regular representation to prove a useful
theorem for setting up character tables. This is the most useful
application of the regular representation for our purposes.
Theorem: The order of a group h and the dimensionality j of its
irreducible representations Γj are related by
j
2
j = h. (3.40)
We had previously found (Eq. 2.68) that j
2
j ≤ h. The regular
representation allows us to prove that it is the equality that
applies.
Proof: By construction, the regular representation is of dimensionality
h which is the number of elements in the group and in the multiplication
table. But each irreducible representation of the group
3.8. SETTING UP CHARACTER TABLES 47
is contained j times in the regular representation (see Eq. 3.38)
so that
χreg
(E) = h =
Γj
aj
j
χΓj
(E)
j
=
Γj
j
2
(3.41)
where the first j is the number of times each irreducible representation
is contained in the regular representation and the second
j is the dimension of the irreducible representation Γj.
We thus obtain the result:
j
2
j = h. (3.42)
where j is the sum over irreducible representations. For example
for P(3), we have 1 = 1, 1 = 1, 2 = 2 so that 2
j = 6 = h.
3.8 Setting up Character Tables
For many applications it is sufficient to know just the character table
without the actual matrix representations for a particular group. So
far, we have only set up the character table by taking traces of the
irreducible representations – i.e., from the definition of χ. For the
most simple cases, the character table can be constructed using the
results of the theorems we have just proved – without knowing the
representations themselves. In practice, the character tables that are
needed to solve a given problem are found either in books or in journal
articles. The examples in this section are thus designed to show the
reader how character tables are constructed, should this be necessary.
Our goal is further to give some practice in using the theorems proven
in Chapter 3.
A summary of useful rules for the construction of character tables
is given below.
1. The number of irreducible representations is equal to the number
of classes. (§3.5) The number of classes is found most conveniently
from the classification of the symmetry operations of the
group. Another way to find the classes is to compute all possible
48 CHAPTER 3. CHARACTER OF A REPRESENTATION
conjugates for all group elements using the group multiplication
table.
2. The dimensionalities of the irreducible representations are found
from i
2
i = h (see Eq. 3.42). For most cases, this relation
uniquely determines the dimensionalities of the irreducible representations.
For example, the permutation group of three objects
P(3) has three classes and therefore three irreducible representations.
One of these must be 1-dimensional (i.e., the matrix for
each element of the group is unity). So this gives 12
+?2
+?2
= 6.
This equation only has one solution, namely 12
+ 12
+ 22
= 6. No
other solution works!
3. There is always a whole row of 1’s in the character table for the
identity representation.
4. The first column of the character table is always the trace for
the unit matrix representing the identity element or class. This
character is always 1, the dimensional of the ( i × i) unit matrix.
Therefore, the first column of the character table is also filled in.
5. For all representations other than the identity representation Γ1,
the following relation is satisfied:
k
Nkχ(Γi)
(Ck) = 0. (3.43)
where k denotes the sum on classes. Equation 3.43 follows from
the wonderful orthogonality theorem for character and taking the
identity representation Γ1 as one of the irreducible representa-
tions.
If there are only a few classes in the group, Eq. 3.43 often uniquely
determines the characters for several of the irreducible representations;
particularly for the 1-dimensional representations.
6. The Wonderful Orthogonality Theorem for Character works on
rows of the character table:
k
χ(Γi)
(Ck)
∗
χ(Γj)
(Ck)Nk = hδΓi,Γj
(3.44)
3.8. SETTING UP CHARACTER TABLES 49
This theorem can be used both for orthogonality (different rows)
or for normalization (same rows) of the characters in an irreducible
representation.
7. The second orthogonality theorem works for columns of the character
table:
Γi
χ(Γi)
(Ck)
∗
χ(Γi)
(Ck ) =
h
Nk
δk,k . (3.45)
This relation can be used both for orthogonality (different columns)
or normalization (same columns), as the wonderful orthogonality
theorem for character.
8. From the second orthogonality theorem for character, and from
the character for the identity class
χ(Γi)
(E) = i (3.46)
we see that the characters for all the other classes obey the rela-
tion
Γi
χ(Γi)
(Ck) i = 0 (3.47)
where Γi
denotes the sum on irreducible representations and i
is the dimensionality of representation Γi. Equation 3.47 follows
from the wonderful orthogonality theorem for character, and it
uses as one of the irreducible representations, any but the identity
representation (Γi = Γ1).
With all this machinery it is often possible to complete the character
tables for simple groups without an explicit determination of the
matrices for a representation.
Let us illustrate the use of the rules for setting up character tables
with the permutation group of three objects, P(3).
We fill in the first row and first column of the character table immediately
from rules #3 and #4 in the above list.
C1 3C2 2C3
Γ1 1 1 1
Γ1 1
Γ2 2
C1 3C2 2C3
Γ1 1 1 1
Γ1 1 –1 1
Γ2 2
50 CHAPTER 3. CHARACTER OF A REPRESENTATION
In order to satisfy #5, we know that χ(Γ1 )
(C2) = −1 and χ(Γ1 )
(C3) = 1,
which we add to the character table.
Now apply the second orthogonality theorem using columns 1 and 2
and then again with columns 1 and 3, and this completes the character
table, thereby obtaining:
C1 3C2 2C3
Γ1 1 1 1
Γ1 1 –1 1
Γ2 2 0 –1
Let us give another example of a character table which illustrates
another principle – not all entries in a character table need to be real.
Such a situation can occur in the case of cyclic groups. Consider a
group with three symmetry operations:
• E – identity
• C3 – rotation by 2π
3
• C2
3 – rotation by 4π
3
See Table 3.4 for the multiplication Table for this group. All three
operations in this cyclic group are in separate classes as can be easily
seen by conjugation of the elements. Hence there are three classes and
three irreducible representations to write down. The character table
we start with is obtained by following Rules #3 and #4.
E C3 C2
3
Γ1 1 1 1
Γ2 1 a b
Γ3 1 c d
Orthogonality of Γ2 to Γ1 yields the algebraic relation: 1 + a + b = 0.
Since C2
3 = C3 and C2
3 C3 = E, it follows that b = a2
and ab = a3
=
1, so that a = e
2πi
3 . From this information we can readily complete the
character table
3.9. SYMMETRY NOTATION 51
Table 3.4: Multiplication table for the cyclic group of 3 rotations by
2π/3 about a common axis.
E C3 C2
3
E E C3 C2
3
C3 C3 C2
3 E
C2
3 C2
3 E C2
3
E C3 C2
3
Γ1 1 1 1
Γ2 1 ω ω2
Γ3 1 ω2
ω
where ω = exp[2πi
3
]. In a physical problem with time inversion, the
energy levels corresponding to Γ2 and Γ3 are degenerate.
This idea of the cyclic group can be applied to a 4-element group:
E, C2, C4, C3
4 – to a 5-element group: E, C5, C2
5 , C3
5 , C4
5 – and to a
6-element group: E, C6, C3, C2, C2
3 , C5
6 , etc. For the case of Bloch’s
theorem we have an N-element group with characters that comprise
the Nth roots of unity ω = exp[2πi
N
].
All these cyclic groups are Abelian so that each element is in a class
by itself. The representations for these groups correspond to the multiplication
tables, which therefore contain the appropriate collections of
roots of unity.
3.9 Symmetry Notation
We make use of the following point group notation for the symmetry
operations in the character tables printed in books and journals:
• E = Identity
• Cn = rotation through 2π/n . For example C2 is a rotation of
180◦
. Likewise C3 is a rotation of 120◦
, while C2
6 represents a
rotation of 60◦
followed by another rotation of 60◦
about the
52 CHAPTER 3. CHARACTER OF A REPRESENTATION
 ¢¡
£
¤
¥
Figure 3.1: Schematic illustration
of a dihedral symmetry axis. The
reflection plane containing the diagonal
of the square and the fourfold
axes is called a dihedral plane.
For this geometry σd(x, y, z) =
(−y, −x, z).
same axis so that C2
6 = C3. In a Bravais lattice it can be shown
that n in Cn can only assume values of n=1, 2, 3, 4, and 6.
The observation of a diffraction pattern with five-fold symmetry
in 1984 was therefore completely unexpected, and launched the
field of quasicrystals.
• σ = reflection in a plane.
• σh = reflection in a “horizontal” plane. The reflection plane here
is perpendicular to the axis of highest rotational symmetry.
• σv = reflection in a “vertical” plane. The reflection plane here
contains the axis of highest symmetry.
• σd = reflection in a diagonal plane. The reflection plane here is a
vertical plane which bisects the angle between the two fold axes ⊥
to the principal symmetry axis. An example of a diagonal plane
is shown in Fig. 3.1. σd is also called a dihedral plane.
• i = inversion which takes



x → −x
y → −y
z → −z
• Sn = improper rotation through 2π/n, which consists of a rotation
by 2π/n followed by a reflection in a horizontal plane.
3.9. SYMMETRY NOTATION 53
• iCn = compound rotation-inversion, which consists of a rotation
followed by an inversion.
In addition to these point group symmetry operations, there are several
space group symmetry operations, such as translations, glide planes,
screw axes, etc., which are discussed in Chapter 12. The notation used
in the list above for the symmetry operations is called the Schoenflies
notation. Based on this symmetry notation is a point group notation.
There are 32 common point groups and the character tables for
these 32 point groups are given in any standard group theory text. For
convenience we also list the character tables for the 32 point groups in
the notes at the end of Chapter 3 (see Tables 3.8 – 3.34 on pp. 63-70).
For example, groups C1, C2, . . . , C6 only have n-fold rotations about a
simple symmetry axis Cn (see Tables 3.8 – 3.13 on pp. 63-64). Groups
Cnv have, in addition to the n-fold axes, vertical reflection planes σv (see
Tables 3.14 – 3.18 on pp. 64-66). Groups Cnh have, in addition to the
n-fold axes (see Tables 3.19 and 3.21 pp. 66-67), horizontal reflection
planes σh and include each operation Cn together with the compound
operations Cn followed by σh. The groups S2, S4 and S6 have mostly
(see Tables 3.22 – 3.23) compound operations. The groups denoted
by Dn are dihedral groups and have (see Tables 3.24 – 3.28 pp. 68-69)
non-equivalent symmetry axes in perpendicular planes. The group of
the operations of a square is D4 and has in addition to the principal
four-fold axes, two sets of non-equivalent two-fold axes. When nonequivalent
axes are combined (see Tables 3.29 – 3.31 on p. 69) with
mirror planes we get groups like D2h, D3h, etc. There are 5 cubic groups
T, O, Td, Th and Oh. These groups have no principal axis but instead
have four three-fold axes (see Tables 3.32 – 3.34).
There is also a second notation for symmetry operations and groups
– namely the Hermann–Mauguin or international notation, referring
to the International Tables for X-Ray Crystallography, a standard
structural and symmetry reference book. The international notation is
what is usually found in crystallography textbooks and various materials
science journals, and for that reason it is also necessary to become
familiar with this notation. The general correspondence between the
two notations is shown in Table 3.5 for rotations and mirror planes.
The Hermann–Mauguin notation ¯n means iCn which is equivalent to a
54 CHAPTER 3. CHARACTER OF A REPRESENTATION
Table 3.5: Comparison between Schoenflies and Hermann-Mauguin no-
tation.
Schoenflies Hermann-Mauguin
rotation Cn n
rotation-inversion iCn ¯n
mirror plane σ m
horizontal reflection
plane ⊥ to n − fold axes
σh n/m
n − fold axes in
vertical reflection plane
σv nm
two non − equivalent
vertical reflection planes
σv nmm
rotation followed by or preceded by an inversion. A string of numbers
like 422 (see Table 3.26 on p. 68) means that there is a four-fold major
symmetry axis (C4 axis), and perpendicular to this axis are 2 inequivalent
two-fold axes C2 and C2 , such as occur in the group of the square
(D4). If there are several inequivalent horizontal mirror planes like
2
m
, 2
m
, 2
m
,
an abbreviated notation mmm is sometimes used [see notation for the
group D2h below (Table 3.29 on p. 69)]. The notation 4mm (see Table
3.16 on p. 65) denotes a four-fold axis and two sets of vertical mirror
planes, one set through the axes C4 and denoted by 2σv and the other
set through the bisectors of the 2σv planes and denoted by the dihedral
vertical mirror planes 2σd.
Table 3.6 is useful in relating the two kinds of notations for rotations
and improper rotations.
Some useful relations on the commutativity of symmetry operations
are:
1. Inversion commutes with all point symmetry operations.
2. All rotations about the same axis commute.
3. All rotations about an arbitrary rotation axis commute with reflections
across a plane perpendicular to this rotation axis.
3.9. SYMMETRY NOTATION 55
Table 3.6: Comparison of notation for proper and improper rotations
in the Schoenflies and International systems.
Proper Rotations Improper Rotations
International Schoenflies International Schoenflies
1 C1
¯1 S2
2 C2
¯2 ≡m σ
3 C3
¯3 S−1
6
32 C−1
3
¯32 S6
4 C4
¯4 S−1
4
43 C−1
4
¯43 S4
6 C6
¯6 S−1
3
65 C−1
6
¯65 S3
4. Two two-fold rotations about perpendicular axes commute.
5. Two reflections in perpendicular planes will commute.
6. Any two of the symmetry elements σh, S2, Cn (n = even) implies
the third.
If we have a major symmetry axis Cn(n ≥ 2) and there are either
two-fold axes C2 or vertical mirror planes σv, then there will generally
be more than one C2 or σv.
The classification of the 32 point symmetry groups shown in Table
3.7 is often useful in making practical applications of character
tables in textbooks and journal articles to specific materials. In Table
3.7 the first symbol in the Hermann-Mauguin notation denotes the
principal axis or plane. The second symbol denotes an axis (or plane)
perpendicular to this axis, except for the cubic groups where the second
symbol refers to a 111 axis. The third symbol denotes an axis or
plane that is ⊥ to the first axis and at an angle of π/n with respect to
the second axis.
It is also convenient to picture group symmetries with stereograms
(e.g., Tinkham p.55), which are here reproduced in Fig. 3.2. The stereogram
is a mapping of a general point on a sphere onto a plane going
through the center of the sphere. If the point on the sphere is above
56 CHAPTER 3. CHARACTER OF A REPRESENTATION
Table 3.7: The 32 point groups and their symbols. In the Hermann–
Mauguin notation, the symmetry axes parallel to and the symmetry
planes perpendicular to each of the “principal” directions in the crystal
are named in order. When there is both an axis parallel to and a plane
normal to a given direction, these are indicated as a fraction; thus
6/m means a six-fold rotation axis standing perpendicular to a plane
of symmetry, while ¯4 denotes a four-fold rotary inversion axis. In some
classifications, the rhombohedral (trigonal) groups are listed with the
hexagonal groups.
The 32 Point Groups and Their Symbols
System Schoenflies Hermann-Mauguin symbol Examples
symbol Full Abbreviated
Triclinic C1 1 1
Ci, (S2) ¯1 ¯1 Al2SiO5
Monoclinic C2v, (C1h), (S1) m m KNO2
C2 2 2
C2h 2/m 2/m
Orthorhombic C2v 2mm mm
D2, (V ) 222 222
D2h, (Vh) 2/m 2/m 2/m mmm I, Ga
Tetragonal S4 ¯4 ¯4
C4 4 4
C4h 4/m 4/m CaWO4
D2d, (Vd) ¯42m ¯42m
C4v 4mm 4mm
D4 422 42
D4h 4/m 2/m 2/m 4/mmm TiO2, In, β − Sn
Rhombohedral C3 3 3 AsI3
C3i, (S6) ¯3 ¯3 FeTiO3
C3v 3m 3m
D3 32 32 Se
D3d
¯32/m ¯3m Bi, As, Sb, Al2O3
Hexagonal C3h, (S3) ¯6 ¯6
C6 6 6
C6h 6/m 6/m
D3h
¯62m ¯62m
C6v 6mm 6mm ZnO, NiAs
D6 622 62 CeF3
D6h 6/m 2/m 2/m 6/mmm Mg, Zn, graphite
Cubic T 23 23 NaClO3
Th 2/m¯3 m3 FeS2
Td
¯43m ¯43m ZnS
O 432 43 β-Mn
Oh 4/m ¯3 2/m m3m NaCl, diamond, Cu
3.9. SYMMETRY NOTATION 57
Figure 3.2: Stereographic projections of simple point groups. For the
case of icosahedral symmetry projections for C5, C5v, D5h, and D5d are
useful.
58 CHAPTER 3. CHARACTER OF A REPRESENTATION
the plane it is indicated as a +, if below as a ◦. In general, the polar
axis of the stereogram coincides with the principal axis of symmetry.
The 5 stereograms on the first row pertaining to groups with a
single axis of rotation show the effect of 2-, 3-, 4-, and 6-fold rotation
axes on a point +. These groups are cyclic groups with only n-fold
axes. Note the symmetry of the central point for each group. On the
second row we have added vertical mirror planes which are indicated
by the solid lines. Since the “vertical” and “horizontal” planes are not
distinguishable for C1, the addition of a mirror plane to C1 is given in
the third row, showing the groups which result from the first row upon
addition of horizontal planes. The symbols ⊕ indicate the coincidence
of the projection of points above and below the plane, characteristic of
horizontal mirror planes.
If instead of proper rotations as in the first row, we have improper
rotations, then the groups on row 4 are generated. Since S1 is identical
with C1h, it is not shown separately; this also applies to S3 ⇒ C3h. It
is of interest to note that S2 and S6 have inversion symmetry but S4
does not.
The addition of two-fold axes ⊥ to the principal symmetry axis for
the groups in the first row yields the stereograms of the fifth row where
the two-fold axes appear as dashed lines. Here we see that the higher
the symmetry of the principal symmetry axis, the greater the number
of two-fold axes.
The addition of two-fold axes to the groups on the 4th row yields
the stereograms of the 6th row, where D2d comes from S4, and D3d from
S6. The addition of two-fold axes to S2 results in C2h. The stereograms
on the last row are obtained by adding two-fold axes ⊥ to Cn to the
stereograms on the 3rd row. The effect of adding a two-fold axis to C1h
is to produce C2v.
The remaining 5 point symmetry groups not shown in Fig. 3.2
have higher symmetry and have no single principal axis. The resulting
stereograms are very complicated and for this reason are not given
in Fig. 3.2. We give some of the symmetry elements for these groups
below.
The group T (or 23 using the International notation) has 12 symmetry
elements which include:
3.9. SYMMETRY NOTATION 59
 
¡
¢
£
¤
¥
¦
§
Figure 3.3: Schematic diagram
for the symmetry operations
of the group Td.
1 identity
3 two-fold axes (x, y, z)
4 three-fold axes (body diagonals–positive rotation)
4 three-fold axes (body diagonals–negative rotations)
12 symmetry elements
The point group Th (denoted by m3 in the abbreviated International
notation or by 2/m3 in the full International notation) contains all
the symmetry operations of T and inversion as well, and is written as
Th ≡ T ⊗ i, indicating the direct product of the group T and the group
Ci having 2 symmetry elements E, i. This is equivalent to adding a
horizontal plane of symmetry, hence the notation 2/m; the symbol 3
means a three-fold axis (see Table 3.6). Thus Th has 24 symmetry
elements.
The point group Td (¯43m) contains the symmetry operations of
the regular tetrahedron (see Fig. 3.3), which correspond to the point
symmetry for diamond and the zincblende (III–V and II–VI) structures.
We list below the 24 symmetry operations of Td.
Symmetry Operations of Td
• Identity
• 8 C3 about body diagonals corresponding to rotations of ±2π
3
• 3 C2 about x, y, z directions
60 CHAPTER 3. CHARACTER OF A REPRESENTATION
 
¡ ¢¤£ ¥
¢§¦ ¨
© ¢¤§© ¢ ¦ ¨



¥
¢§
Figure 3.4: Schematic for
the symmetry operations of
the group O.
• 6 S4 about x, y, z corresponding to rotations of ±π
2
• 6 σd planes that are diagonal reflection planes
The cubic groups are O (432) and Oh (m3m) and are shown schematically
in Fig. 3.4. The operations for group O are shown in Fig. 3.4
and are E, 8C3, 3C2 = 3C2
4 , 6C2 and 6C4. To get Oh we combine these
24 operations with inversion to give 48 operations in all. We note that
the second symbol in the Hermann-Mauguin (International) notation
for all 5 cubic groups is for the 111 axes rather than for an axis ⊥ to
the principal symmetry axis.
At this point we are ready to understand the notation used in the
character tables of the texts (e.g., Tinkham) and journal articles. As
we examine the character tables we observe that not all the entries are
real. However, in the 11 point groups where there are representations
with complex characters, there is always another representation that is
the complex conjugate of the first.
In addition to the 32 point groups, the character tables contain
listings for C∞v and D∞h which have full rotational symmetry around
a single axis, and therefore have an ∞ number of symmetry operations
and classes. These two groups are sometimes called the semi-infinite
groups because they have an infinite number of operations about the
major symmetry axis. An example of the C∞v group is the CO molecule
shown in Fig. 3.5. Here the symmetry operations are E, 2Cφ and σv.
The notation Cφ denotes an axis of full rotational symmetry and σv
denotes the corresponding infinite array of vertical planes. The group
D∞h has in addition the inversion operation which is compounded with
each of the operations in C∞v, and this is written as D∞h = C∞v ⊗ i.
3.9. SYMMETRY NOTATION 61
O
C
CO
O
 ¢¡¤£
¥ ¡§¦
Figure 3.5: Schematic diagram of the CO molecule with symmetry C∞v
and symmetry operations E, 2Cφ, σv, and the linear CO2 molecule in
which the inversion operation is also present to give the group D∞h.
¨
¨©
©
¨ © ¨
©
© ©
¨
¨

 
 
 
Figure 3.6: Schematic diagram of an X3O3 molecule.
An example of a molecule with D∞h symmetry in the CO2 molecule
(see Fig. 3.5).
To make use of group theory for describing physical properties of
molecules, we classify the symmetry operations of a molecule in terms
of a point group. We illustrate this problem by considering a molecule
X3O3 shown in the schematic diagram of Fig. 3.6.
62 CHAPTER 3. CHARACTER OF A REPRESENTATION
The symmetry operations for the molecule in Fig. 3.6 are:
E
C3
σh (only for planar molecule)
σv
σd (only for planar molecule)
i (X goes into O)
S6 (X goes into O)
iC3 (X goes into O)
Thus, if the X and O atoms are distinct, the appropriate group is C3v.
If X and O are the same, the group is D3d. If the A3B3 molecule is
planar but X and O are distinct the appropriate group is D3h and if X
and O are the same the group is D6h.
In Tables 3.8 – 3.34, we give the character tables for the 32 point
groups, following Tinkham’s tables. These are followed by Tables 3.35
and 3.36 for the semi-infinite groups C∞v and D∞h. Tables 3.37–3.40
are for groups with five-fold symmetry axes not readily found in group
theory books, but have recently become important because of the discovery
of quasi-crystals and C60 and related molecules. Note that the
tables for five-fold symmetry are: C5 (Table 3.12); C5v (Table 3.17);
C5h ≡ C5 ⊗ σh; D5 (Table 3.27); D5d (Table 3.37); D5h (Table 3.38);
I (Table 3.39); and Ih (Table 3.40). Recurrent in these tables is the
“golden mean”, τ = (1 +
√
5)/2 where τ − 1 = 2 cos(2π/5) = 2 cos 72◦
.
At this point there are many features of the character tables which
have not yet been explained in the book. Future chapters that will
address these features are on the topics of basis functions (Chapter 4),
direct products (Chapter 4), linear molecules (Chapter 4) and icosahedral
symmetry (Chapter 4). Furthermore, as we use the various character
tables in physical applications, the notation will become more clear
and more familiar to the reader.
3.9. SYMMETRY NOTATION 63
Table 3.8: Character Table for Group C1
C1 (1) E
A 1
Table 3.9: Character Table for Group C2
C2 (2) E C2
x2
, y2
, z2
, xy Rz, z A 1 1
xz, yz
(x, y)
(Rx, Ry)
B 1 −1
Table 3.10: Character Table for Group C3
C3(3) E C3 C2
3
x2
+ y2
, z2
Rz, z A 1 1 1
(xz, yz)
(x2
− y2
, xy)
(x, y)
(Rx, Ry)
E
1
1
ω
ω2
ω2
ω
where ω = e2πi/3
Table 3.11: Character Table for Group C4
C4 (4) E C2 C4 C3
4
x2
+ y2
, z2
Rz, z A 1 1 1 1
x2
− y2
, xy B 1 1 −1 −1
(xz, yz)
(x, y)
(Rx, Ry)
E
1
1
−1
−1
i
−i
−i
i
64 CHAPTER 3. CHARACTER OF A REPRESENTATION
Table 3.12: Character Table for Group C5
C5 (5) E C5 C2
5 C3
5 C4
5
x2
+ y2
, z2
Rz, z A 1 1 1 1 1
(xz, yz)
(x, y)
(Rx, Ry)
E
1
1
ω
ω4
ω2
ω3
ω3
ω2
ω4
ω
(x2
− y2
, xy) E
1
1
ω2
ω3
ω4
ω
ω
ω4
ω3
ω2
where ω = e2πi/5
Table 3.13: Character Table for Group C6
C6 (6) E C6 C3 C2 C2
3 C5
6
x2
+ y2
, z2
Rz, z A 1 1 1 1 1 1
B 1 −1 1 −1 1 −1
(xz, yz)
(x, y)
(Rx, Ry)
E
1
1
ω
ω5
ω2
ω4
ω3
ω3
ω4
ω2
ω5
ω
(x2
− y2
, xy) E
1
1
ω2
ω4
ω4
ω2
1
1
ω2
ω4
ω4
ω2
where ω = e2πi/6
Table 3.14: Character Table for Group C2v
C2v (2mm) E C2 σv σv
x2
, y2
, z2
z A1 1 1 1 1
xy Rz A2 1 1 −1 −1
xz Ry, x B1 1 −1 1 −1
yz Rx, y B2 1 −1 −1 1
3.9. SYMMETRY NOTATION 65
Table 3.15: Character Table for Group C3v
C3v (3m) E 2C3 3σv
x2
+ y2
, z2
z A1 1 1 1
Rz A2 1 1 –1
(x2
− y2
, xy)
(xz, yz)
(x, y)
(Rx, Ry)
E 2 −1 0
Table 3.16: Character Table for Group C4v
C4v (4mm) E C2 2C4 2σv 2σd
x2
+ y2
, z2
z A1 1 1 1 1 1
Rz A2 1 1 1 −1 −1
x2
− y2
B1 1 1 −1 1 −1
xy B2 1 1 −1 −1 1
(x2
− y2
, xy)
(xz, yz)
(x, y)
(Rx, Ry)
E 2 −2 0 0 0
Table 3.17: Character Table for Group C5v
C5v (5m) E 2C5 2C2
5 5σv
x2 + y2, z2, z3, z(x2 + y2) z A1 1 1 1 1
Rz A2 1 1 1 −1
z(x, y), z2(x, y), (x2 + y2)(x, y)
(x, y)
(Rx, Ry)
E1 2 2 cos α 2 cos 2α 0
(x2 − y2, xy), z(x2 − y2, xy), [x(x2 − 3y2), y(3x2 − y2)] E2 2 2 cos 2α 2 cos 4α 0
where α = 2π
5
= 72◦
.
Note that τ = (1 +
√
5)/2 so that τ = −2 cos 2α = −2 cos 4π/5 and
τ − 1 = 2 cos α = 2 cos 2π/5.
66 CHAPTER 3. CHARACTER OF A REPRESENTATION
Table 3.18: Character Table for Group C6v
C6v (6mm) E C2 2C3 2C6 3σd 3σv
x2
+ y2
, z2
z A1 1 1 1 1 1 1
Rz A2 1 1 1 1 −1 −1
B1 1 −1 1 −1 −1 1
B2 1 −1 1 −1 1 –1
(xz, yz)
(x, y)
(Rx, Ry)
E1 2 −2 −1 1 0 0
(x2
− y2
, xy) E2 2 2 −1 −1 0 0
Table 3.19: Character Table for Group C1h
C1h(m) E σh
x2
, y2
, z2
, xy Rz, x, y A 1 1
xz, yz Rx, Ry, z A 1 −1
Table 3.20: Character Table for Group C2h
C2h (2/m) E C2 σh i
x2
, y2
, z2
, xy Rz Ag 1 1 1 1
z Au 1 1 −1 −1
xz, yz Rx, Ry Bg 1 −1 −1 1
x, y Bu 1 −1 1 −1
3.9. SYMMETRY NOTATION 67
Table 3.21: Character Table for Group C3h
C3h = C3 ⊗ σh (¯6) E C3 C2
3 σh S3 (σhC2
3 )
x2
+ y2
, z2
Rz A 1 1 1 1 1 1
z A 1 1 1 −1 −1 −1
(x2
− y2
, xy) (x, y) E
1
1
ω
ω2
ω2
ω
1
1
ω
ω2
ω2
ω
(xz, yz) (Rx, Ry) E
1
1
ω
ω2
ω2
ω
−1
−1
−ω
−ω2
−ω2
−ω
where ω = e2
πi/3
C4h = C4 ⊗ i (4/m)
C5h = C5 ⊗ σh
¯(10)
C6h = C6 ⊗ i (6/m)
Table 3.22: Character Table for Group S2
S2 (¯1) E i
x2
, y2
, z2
, xy, xz, yz Rx, Ry, Rz Ag 1 1
x, y, z Au 1 −1
Table 3.23: Character table for Group S4
S4 (¯4) E C2 S4 S3
4
x2
+ y2
, z2
Rz A 1 1 1 1
z B 1 1 −1 −1
(xz, yz)
(x2
− y2
, xy)
(x, y)
(Rx, Ry)
E
1
1
−1
−1
i
−i
−i
i
S6 = C3 ⊗ i (¯3)
68 CHAPTER 3. CHARACTER OF A REPRESENTATION
Table 3.24: Character Table for Group D2
D2 (222) E Cz
2 Cy
2 Cx
2
x2
, y2
, z2
A1 1 1 1 1
xy Rz, z B1 1 1 −1 −1
xz Ry, y B2 1 −1 1 −1
yz Rx, x B3 1 −1 −1 1
Table 3.25: Character Table for Group D3
D3 (32) E 2C3 3C2
x2
+ y2
, z2
A1 1 1 1
Rz, z A2 1 1 −1
(xz, yz)
(x2
− y2
, xy)
(x, y)
(Rx, Ry)
E 2 −1 0
Table 3.26: Character Table for Group D4
D4 (422) E C2 = C2
4 2C4 2C2 2C2
x2
+ y2
, z2
A1 1 1 1 1 1
Rz, z A2 1 1 1 −1 −1
x2
− y2
B1 1 1 −1 1 −1
xy B2 1 1 −1 −1 1
(xz, yz)
(x, y)
(Rx, Ry)
E 2 −2 0 0 0
Table 3.27: Character Table for Group D5
D5 (52) E 2C5 2C2
5 5C2
x2
+ y2
, z2
A1 1 1 1 1
Rz, z A2 1 1 1 −1
(xz, yz)
(x, y)
(Rx, Ry)
E1 2 2cos α 2cos 2α 0
(x2
− y2
, xy) E2 2 2cos 2α 2cos 4α 0
where α = 2π/5 = 72◦
.
Note that τ = (1 +
√
5)/2 so that τ = −2 cos 2α = −2 cos 4π/5 and
τ − 1 = 2 cos α = 2 cos 2π/5.
3.9. SYMMETRY NOTATION 69
Table 3.28: Character Table for Group D6
D6 (622) E C2 2C3 2C6 3C2 3C2
x2
+ y2
, z2
A1 1 1 1 1 1 1
Rz, z A2 1 1 1 1 −1 −1
B1 1 −1 1 −1 1 −1
B2 1 −1 1 −1 −1 1
(xz, yz)
(x, y)
(Rx, Ry)
E1 2 −2 −1 1 0 0
(x2
− y2
, xy) E2 2 2 −1 −1 0 0
Table 3.29: Character Table for Group D2d
D2d (¯42m) E C2 2S4 2C2 2σd
x2
+ y2
, z2
A1 1 1 1 1 1
Rz A2 1 1 1 −1 −1
x2
− y2
B1 1 1 −1 1 –1
xy z B2 1 1 −1 −1 1
(xz, yz)
(x, y)
(Rx, Ry)
E 2 −2 0 0 0
D2h = D2 ⊗ i (mmm)
D5d = D5 ⊗ i (¯5m) see Table 3.37
Table 3.30: Character Table for Group D3d
D3d = D3 ⊗ i (¯3m) E 2C3 3C2 i 2iC3 3iC2
x2
+ y2
, z2
A1g 1 1 1 1 1 1
Rz A2g 1 1 −1 1 1 −1
(xz, yz),(x2
− y2
, xy) (Rx, Ry) Eg 2 −1 0 2 −1 0
A1u 1 1 1 −1 −1 −1
z A2u 1 1 −1 −1 −1 1
(x, y) Eu 2 −1 0 −2 1 0
Table 3.31: Character Table for Group D3h
D3h = D3 ⊗ σh (¯6m2) E σh 2C3 2S3 3C2 3σv
x2
+ y2
, z2
A1 1 1 1 1 1 1
Rz A2 1 1 1 1 −1 −1
A1 1 −1 1 −1 1 −1
z A2 1 −1 1 −1 −1 1
(x2
− y2
, xy) (x, y) E 2 2 −1 −1 0 0
(xz, yz) (Rx, Ry) E 2 −2 −1 1 0 0
70 CHAPTER 3. CHARACTER OF A REPRESENTATION
D4h = D4 ⊗ i (4/mmm)
D5h = D5 ⊗ σh (10m2) see Table 3.38
D6h = D6 ⊗ i (6/mmm)
Table 3.32: Character Table for Group T
T (23) E 3C2 4C3 4C3
A 1 1 1 1
E
1
1
1
1
ω
ω2
ω2
ω
(Rx, Ry, Rz)
(x, y, z)
T 3 −1 0 0
where ω = exp(2πi/3)
Th = T ⊗ i (m3)
Table 3.33: Character Table for Group O
O (432) E 8C3 3C2 = 3C2
4 6C2 6C4
(x2
+ y2
+ z2
) A1 1 1 1 1 1
A2 1 1 1 −1 −1
(x2
− y2
, 3z2
− r2
) E 2 −1 2 0 0
(Rx, Ry, Rz)
(x, y, z)
T1 3 0 −1 −1 1
(xy, yz, zx) T2 3 0 −1 1 −1
Oh = O ⊗ i (m3m)
Table 3.34: Character Table for Group Td
Td (¯43m) E 8C3 3C2 6σd 6S4
A1 1 1 1 1 1
A2 1 1 1 −1 −1
E 2 −1 2 0 0
(Rx, Ry, Rz) T1 3 0 −1 −1 1
(x, y, z) T2 3 0 −1 1 −1
3.9. SYMMETRY NOTATION 71
Table 3.35: Character Table for Group C∞v
C∞v (∞m) E 2Cφ σv
(x2
+ y2
, z2
) z A1(Σ+
) 1 1 1
Rz A2(Σ−
) 1 1 −1
(xz, yz)
(x, y)
(Rx, Ry)
E1(Π) 2 2 cos φ 0
(x2
− y2
, xy) E2(∆) 2 2 cos 2φ 0
...
...
...
...
Table 3.36: Character Table for Group D∞h
D∞h (∞/mm) E 2Cφ C2 i 2iCφ iC2
x2
+ y2
, z2
A1g(Σ+
g ) 1 1 1 1 1 1
A1u(Σ−
u ) 1 1 1 −1 −1 −1
Rz A2g(Σ−
g ) 1 1 −1 1 1 −1
z A2u(Σ+
u ) 1 1 −1 −1 −1 1
(xz, yz) (Rx, Ry) E1g(Πg) 2 2 cos φ 0 2 2 cos φ 0
(x, y) E1u(Πu) 2 2 cos φ 0 −2 −2 cos φ 0
(x2
− y2
, xy) E2g(∆g) 2 2 cos 2φ 0 2 2 cos 2φ 0
E2u(∆u) 2 2 cos 2φ 0 −2 −2 cos 2φ 0
...
...
...
...
...
...
...
Table 3.37: Character table for D5d.
D5d E 2C5 2C2
5 5C2 i 2S−1
10 2S10 5σd (h = 20)
A1g +1 +1 +1 +1 +1 +1 +1 +1 (x2 + y2), z2
A2g +1 +1 +1 −1 +1 +1 +1 −1 Rz
E1g +2 τ − 1 −τ 0 +2 τ − 1 −τ 0 z(x + iy, x − iy)
E2g +2 −τ τ − 1 0 +2 −τ τ − 1 0 [(x + iy)2, (x − iy)2]
A1u +1 +1 +1 +1 −1 −1 −1 −1
A2u +1 +1 +1 −1 −1 −1 −1 +1 z
E1u +2 τ − 1 −τ 0 −2 1−τ +τ 0 (x + iy, x − iy)
E2u +2 −τ τ − 1 0 −2 +τ 1−τ 0
Note: D5d = D5 ⊗ i, iC5 = S−1
10 and iC2
5 = S10. Also iC2 = σd.
where α = 2π/5 = 72◦
.
Note also that τ = (1 +
√
5)/2 so that τ = −2 cos 2α = −2 cos 4π/5
and τ − 1 = 2 cos α = 2 cos 2π/5.
72 CHAPTER 3. CHARACTER OF A REPRESENTATION
Table 3.38: Character table for D5h.
D5h (10m2) E 2C5 2C2
5 5C2 σh 2S5 2S3
5 5σv (h = 20)
A1 +1 +1 +1 +1 +1 +1 +1 +1 x2 + y2, z2
A2 +1 +1 +1 −1 +1 +1 +1 −1 Rz
E1 +2 τ − 1 −τ 0 +2 τ − 1 −τ 0 (x, y), (xz2, yz2), [x(x2 + y2), y(x2 + y2)]
E2 +2 −τ τ − 1 0 +2 −τ τ − 1 0 (x2 − y2, xy), [y(3x2 − y2), x(x2 − 3y2)]
A1 +1 +1 +1 +1 −1 −1 −1 −1
A2 +1 +1 +1 −1 −1 −1 −1 +1 z, z3, z(x2 + y2)
E1 +2 τ − 1 −τ 0 −2 1−τ +τ 0 (Rx, Ry), (xz, yz)
E2 +2 −τ τ − 1 0 −2 +τ 1−τ 0 [xyz, z(x2 − y2)]
Note that τ = (1 +
√
5)/2 so that τ = −2 cos 2α = −2 cos 4π/5 and
τ − 1 = 2 cos α = 2 cos 2π/5.
Table 3.39: Character table for I.
I (532) E 12C5 12C2
5 20C3 15C2 (h = 60)
A +1 +1 +1 +1 +1 x2
+ y2
+ z2
F1 +3 +τ 1−τ 0 −1 (x, y, z); (Rx, Ry, Rz)
F2 +3 1−τ +τ 0 −1
G +4 −1 −1 +1 0
H +5 0 0 −1 +1



2z2
− x2
− y2
x2
− y2
xy
xz
yz
Note that τ = (1 +
√
5)/2 so that τ = −2 cos 2α = −2 cos 4π/5 and
τ − 1 = 2 cos α = 2 cos 2π/5.
3.10. SELECTED PROBLEMS 73
Table 3.40: Character table for Ih.
Ih E 12C5 12C2
5 20C3 15C2 i 12S3
10 12S10 20S6 15σ (h = 120)
Ag +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 x2 + y2 + z2
F1g +3 +τ 1−τ 0 −1 +3 τ 1 − τ 0 −1 Rx, Ry, Rz
F2g +3 1−τ +τ 0 −1 +3 1 − τ τ 0 −1
Gg +4 −1 −1 +1 0 +4 −1 −1 +1 0
Hg +5 0 0 −1 +1 +5 0 0 −1 +1



2z2 − x2 − y2
x2 − y2
xy
xz
yz
Au +1 +1 +1 +1 +1 −1 −1 −1 −1 −1
F1u +3 +τ 1−τ 0 −1 −3 −τ τ − 1 0 +1 (x, y, z)
F2u +3 1−τ +τ 0 −1 −3 τ − 1 −τ 0 +1
Gu +4 −1 −1 +1 0 –4 +1 +1 −1 0
Hu +5 0 0 −1 +1 –5 0 0 +1 −1
where τ = (1 +
√
5)/2.
Note: C5 and C−1
5 are in different classes, labeled 12C5 and 12C2
5 in
the character table. Then iC5 = S−1
10 and iC−1
5 = S10 are in the classes
labeled 12S3
10 and 12S10, respectively. Also iC2 = σv.
3.10 Selected Problems
1. (a) Explain the symmetry operations pertaining to each class of
the point group D3h. You may find the stereograms on p. 42
useful.
(b) Prove that the following irreducible representations are or-
thonormal:
• E1 and E2 in the group D5 (see Table 3.27).
• F2g and Gu in the group Ih (see Table 3.39).
(c) Given the group T (see Table 3.31), verify that the equality
j
2
j = h
is satisfied. What is the meaning of the two sets of characters
given for the two-dimensional irreducible representation E?
Are they orthogonal to each other or are they part of the
same irreducible representation?
74 CHAPTER 3. CHARACTER OF A REPRESENTATION
(d) Which symmetry operation results from multiplying the operations
σv and σd in group C4v? Can you obtain this information
from the character table? If so, how?
2. Make stereographic sketches for groups C5, C5v, D5h, and D5d,
such as are given in Fig. 3.2.
3. Consider the point group D6
(a) Construct the character table for D6h = D6 ⊗ i (see Table
3.28).
(b) How many two-dimensional irreducible representations are
there in D6? and in D6h?
(c) Consider the groups D3h, C3, and C3v as subgroups of D6h:
for which group (or groups) are the two two-dimensional
representations of D6 no longer irreducible? If the representations
are reducible, into which irreducible representations
of the lower symmetry group do they reduce?
4. (a) What are the symmetry operations of a regular hexagon?
(b) Find the classes. Why are not all the 2-fold axes in the same
class?
(c) Find the self-conjugate subgroups, if any.
(d) Identify the appropriate character table.
(e) For some representative cases (two cases are sufficient), check
the validity of the “Wonderful Orthogonality and Second
Orthogonality Theorems” on character, using the character
table in (d).
5. Suppose that you have the following set of characters: χ(E) =
4, χ(σh) = 2, χ(C3) = 1, χ(S3) = −1, χ(C2) = 0, χ(σv) = 0.
(a) Do these characters correspond to a representation of the
point group D3h? Is it irreducible?
(b) If the representation is reducible, find the irreducible representations
contained therein.
(c) Give an example of a molecule with D3h symmetry.
Chapter 4
Basis Functions
In the previous Chapters we have discussed symmetry elements, their
matrix representations and the properties of the characters of these
representations. In this discussion we saw that the matrix representations
are not unique though their characters are unique. Because
of the uniqueness of the characters of each irreducible representation,
the characters for each group are tabulated in character tables. Also
associated with each irreducible representation are “basis functions”
which can be used to generate the matrices that represent the symmetry
elements of a particular irreducible representation. Because of the
importance of basis functions, it is customary to list the most important
basis functions in the character tables.
4.1 Symmetry Operations and Basis Func-
tions
Suppose that we have a group G with symmetry elements R and symmetry
operators ˆPR. We denote the representations by Γnj where n (or
Γn) labels the representation and j labels the component or partner of
the representation — e.g., if we have a two-dimensional representation
then j = 1, 2. We can then define a set of basis vectors denoted by
|Γnj . These basis vectors relate the symmetry operator ˆPR with its
75
76 CHAPTER 4. BASIS FUNCTIONS
matrix representation denoted by DΓn
(R) through the relation
ˆPR|Γnα =
j
DΓn
(R)jα|Γnj . (4.1)
Each vector |Γnj of an irreducible representation Γn is called a partner
and all partners collectively generate the matrix representation DΓn
(R).
The basis vectors can be abstract vectors; a very important type of basis
vector is a basis function which we define here as a basis vector
expressed explicitly in coordinate space. Wavefunctions in quantum
mechanics which are basis functions for symmetry operators are a special
but important example of such basis functions.
In quantum mechanics, each energy eigenvalue of Schr¨odinger’s equation
is labeled according to its symmetry classification, which is specified
according to an irreducible representation of a symmetry group. If
the dimensionality of the representation is j > 1, the energy eigenvalue
will correspond to a j-fold degenerate state, with j linearly independent
wave-functions. The effect of the symmetry operator ˆPR on one of
these wavefunctions (e.g., the αth
wavefunction) will generally be the
formation of a linear combination of the j wavefunctions, as is seen in
Eq. 4.1.
Like the matrix representations and the characters, the basis vectors
also satisfy orthogonality relations
Γnj|Γn j = δn,n δj,j , (4.2)
as is shown in Tinkham p. 41-2 and will be proved in §7.1 in connection
with selection rules. In quantum (wave) mechanics, this orthogonality
relation would be written in terms of the orthogonality for the wave-
functions
ψ∗
n,j(r)ψn ,j (r)d3
r = δn,n δj,j (4.3)
where the wave functions ψn,j and ψn ,j correspond to different energy
eigenvalues (n, n ) and to different components (j, j ) of a particular
degenerate state, and the integration is performed in 3D space. The
orthogonality relation (Eq. 4.3) allows us to generate matrices for an
irreducible representation from a complete set of basis vectors, as is
demonstrated in §4.2.
4.2. BASIS FUNCTIONS FOR IRREDUCIBLE REPRESENTATIONS77
4.2 Use of Basis Functions to Generate Irreducible
Representations
In this section we demonstrate how basis functions can be used to
generate the matrices for an irreducible representation.
Multiplying Eq. 4.1 on the left by the basis vector Γn j | (corresponding
in wave mechanics to ψ∗
n ,j (r)), we obtain using the orthogonality
relation for basis functions (Eq. 4.2):
Γn j | ˆPR|Γnα =
j
D(Γn)
(R)jα Γn j |Γnj = D(Γn )
(R)j αδnn (4.4)
from which we see that the matrix elements of a symmetry operator
are diagonal in each irreducible representation, which correspond to
symmetry quantum numbers. From Eq. 4.4 we obtain a relation between
each matrix element of D(Γn)
(R)jα and the effect of the symmetry
operation on the basis functions:
D(Γn)
(R)jα = Γnj| ˆPR|Γnα . (4.5)
Thus by taking matrix elements of a symmetry operator ˆPR between all
possible partners of an irreducible representation as shown by Eq. 4.5
the matrix representation DΓn
(R)jα can be generated. In practice, this
turns out to be the easiest way to obtain these matrix representations
for the symmetry elements.
As an example of how basis vectors or basis functions can generate
the matrices for an irreducible representation, consider a planar
molecule with three-fold symmetry such that the symmetry operations
are isomorphic to those of an equilateral triangle and also isomorphic to
P(3). Thus there are 6 symmetry operations and 6 operators ˆPR. (See
§3.2. The proper point group to describe the symmetry operations of a
regular planar triangle is D3h.) Group theory tells us that the energy
levels can never be more than two-fold degenerate. Thus no three-fold
or six-fold levels can occur because the largest dimensionality of an
irreducible representation of P(3) is 2. For the 1-dimensional representation
Γ1, the operator ˆPR leaves every basis vector invariant. Thus any
constant such as 1 forms a suitable basis function. For many practical
problems we like to express our basis functions in terms of functions
78 CHAPTER 4. BASIS FUNCTIONS
x
y
1
2 3
Figure 4.1: Symmetry operations
of an equilateral triangle.
The notation of this diagram
defines the symmetry
operations in Table 4.1.
Table 4.1: Symmetry operations of the group of the equilateral triangle
on basis functions.
ˆPR/f(x, y, z) x y z x2 y2 z2
E = E x y z x2 y2 z2
C3 = F 1
2
(−x +
√
3y) 1
2
(−y −
√
3x) z 1
4
(x2 + 3y2 − 2
√
3xy) 1
4
(y2 + 3x2 + 2
√
3xy) z2
C−1
3 = D 1
2
(−x −
√
3y) 1
2
(−y +
√
3x) z 1
4
(x2 + 3y2 + 2
√
3xy) 1
4
(y2 + 3x2 − 2
√
3xy) z2
C2(1) = A −x y −z x2 y2 z2
C2(2) = B 1
2
(x −
√
3y) 1
2
(−y −
√
3x) −z 1
4
(x2 + 3y2 − 2
√
3xy) 1
4
(y2 + 3x2 + 2
√
3xy) z2
C2(3) = C 1
2
(x +
√
3y) 1
2
(−y +
√
3x) −z 1
4
(x2 + 3y2 + 2
√
3xy) 1
4
(y2 + 3x2 − 2
√
3xy) z2
of the coordinates (x, y, z). Some explanation is needed here about
the meaning of (x, y, z) as a basis function. To satisfy the orthogonality
requirement, the basis functions are vectors with unit length and
the matrices which represent the symmetry operations are unitary matrices.
Thus (x, y, z) would generally correspond to the x, y, and z
components of a vector of unit length. The transformation properties
of the x, y, and z components of an arbitrary vector under the symmetry
operations of the group are the same as those for the unit vectors
x, y, and z.
In this connection it is convenient to write out a basis function
table such as Table 4.1. On the top row we list the functions to be
investigated; in the first column we list all the symmetry operations of
the group (see Fig. 4.1 for notation). If we denote the entries in the
table by f (x, y, z), then Table 4.1 can be summarized as:
ˆPRf(x, y, z) = f (x, y, z) (4.6)
4.2. BASIS FUNCTIONS FOR IRREDUCIBLE REPRESENTATIONS79
where the symmetry operations ˆPR label the rows. From Table 4.1 we
can then write down the matrix representations for each irreducible
representation. In the trivial case of the identity representation, the
(1 × 1) matrix 1 satisfies ˆPR1 = 1 for all ˆPR so that this homomorphic
representation always applies, i.e., |Γ1 = 1.
To find the basis functions for the Γ1 representation (i.e., the representation
of the factor group for P(3)), we note that (E, D, F) leaves
z invariant while (A, B, C) takes z into −z, so that z forms a suitable
basis function for Γ1 , which we write as |Γ1 = z. Then application of
Eq. 4.5 yields the matrices for the irreducible representation Γ1
z|(E, D, F)|z = 1 z|(A, B, C)|z = −1. (4.7)
Thus the characters (1) and (−1) for the (1 × 1) irreducible representations
are obtained for Γ1 . We note that in the case of (1 × 1)
representations, the characters and the representations are
identical.
To find the 2-dimensional representation Γ2 we note that all the
group operations take (x, y) into (x , y ). Table 4.1 can be used to find
the matrix representation for Γ2 by taking as basis functions |Γ2, 1 =
|x and |Γ2, 2 = |y . We now illustrate the use of Table 4.1 to generate
the matrix D(Γ2)
(C3 = F) where F is a counter clockwise rotation of
2π/3 about the z axis:
F|x = 1
2
(−x +
√
3y) yields first column of matrix representation
F|y = −1
2
(y +
√
3x) yields second column of matrix representation
so that
F
x
y
= (x y)DΓ2
(F) gives the matrix D(Γ2)
(F) using Eq. 4.1:
D(Γ2)
(C3 = F) =
−1
2
−
√
3
2√
3
2
−1
2
(4.8)
To clarify how we obtain all the matrices for the irreducible representations
with Γ2 symmetry, we repeat the operations leading to Eq. 4.8
for each of the symmetry operations ˆPR. We thus obtain for the other
5 symmetry operations of the group ˆPR using the same basis functions
80 CHAPTER 4. BASIS FUNCTIONS
(x, y) and the notation of Fig. 4.1:
D(Γ2)
(E) =
1 0
0 1
(4.9)
D(Γ2)
(C2(2) = B) =
1
2
−
√
3
2
−
√
3
2
−1
2
(4.10)
D(Γ2)
(C−1
3 = D) =
−1
2
√
3
2
−
√
3
2
−1
2
(4.11)
D(Γ2)
(C2(1) = A) =
−1 0
0 1
(4.12)
D(Γ2)
(C2(3) = C) =
1
2
√
3
2√
3
2
−1
2
(4.13)
As mentioned before, x and y are both basis functions for representation
Γ2 and are called the partners of this irreducible representation. The
number of partners is equal to the dimensionality of the representation.
In Table 4.1 we have included entries for ˆPRx2
, ˆPRy2
, ˆPRz2
and
these entries are obtained as illustrated below by the operation F = C3:
Fx2
= (−
x
2
+
√
3
2
y)2
= (
x2
4
−
√
3
2
xy +
3
4
y2
) (4.14)
Fy2
= (−
y
2
−
√
3
2
x)2
= (
y2
4
+
√
3
2
xy +
3
4
x2
) (4.15)
F(x2
+ y2
) = x2
+ y2
(4.16)
F(xy) = (−
x
2
+
√
3
2
y)(−
y
2
−
√
3
2
x)
=
1
4
(−2xy +
√
3[x2
− y2
]) (4.17)
F(x2
− y2
) = −
1
4
(2[x2
− y2
] + 4
√
3xy) (4.18)
F(xz) = (−
x
2
+
√
3
2
y)z (4.19)
F(yz) = (−
y
2
−
√
3
2
x)z (4.20)
4.2. BASIS FUNCTIONS FOR IRREDUCIBLE REPRESENTATIONS81
Using Eq. 4.1 we see that ˆPR(x2
+ y2
) = (x2
+ y2
) for all ˆPR so that
(x2
+y2
) is a basis function for Γ1 or as we often say transforms according
to the irreducible representation Γ1. Correspondingly z(x2
+ y2
)
transforms as Γ1 and z2
transforms as Γ1. These transformation properties
will be used extensively for many applications of group theory. It
is found that many important basis functions are given directly in the
published character tables. Like the matrix representations, the basis
functions are not unique. However corresponding to a given set of basis
functions, the matrix representation which is generated by these basis
functions will be unique.
As before, the characters for a given representation are found by
taking the sum of the diagonal elements of each matrix in a given
representation:
χ(Γn)
(R) ≡ tr D(Γn)
(R) =
j
D(Γn)
(R)jj =
j
Γnj| ˆPR|Γnj . (4.21)
Since the trace is invariant under a similarity transformation, the character
is independent of the particular choice of basis functions or matrix
representations.
If instead of a basis function (which generates irreducible representations)
we use an arbitrary function f, then a reducible representation
will result, in general. We can express an arbitrary function as a linear
combination of the basis functions. For example, any linear function
of x, y, z such as f(x, y, z) can be expressed in terms of linear combinations
of basis vectors x, y, z and likewise any quadratic function is
expressed in terms of basis functions which transform as irreducible
representations of the group. For example for the group P(3) (see Table
4.1), quadratic forms which serve as basis functions are (x2
+ y2
)
and z2
which both transform as Γ1; z transforms as Γ1 ; (xz, yz) and
(xy, x2
− y2
) both transform as Γ2.
If we now inspect the character table D3(32) found in Table 3.25 we
find that these basis functions are listed. The basis functions labeled
Rα represent the angular momentum component around axis α. For
the two dimensional irreducible representations both partners of the
basis functions are listed, for example (xz, xy) and (x2
− y2
, xy), etc.
The reason why (x, y, z) and (Rx, Ry, Rz) often transform as different
irreducible representations (not the case for the group D3(32)) is that
82 CHAPTER 4. BASIS FUNCTIONS
x, y, z transforms as a radial vector (such as coordinate, momentum)
while Rx, Ry, Rz transforms as an axial vector (such as angular momentum
r × p).
D3(32) E 2C3 3C2
x2
+ y2
, z2
A1 1 1 1
Rz, z A2 1 1 −1
(xz, yz)
(x2
− y2
, xy)
(x, y)
(Rx, Ry)
E 2 −1 0
(The full symmetry of an equilateral triangle is D3h = D3⊗σh. Since the
triangle is a 2D object, the horizontal mirror plane is not an important
symmetry operation and we can simplify the algebra by using the group
D3.)
4.3 Projection Operators ˆP
(Γn)
kl
The previous discussion of basis vectors assumed that we already knew
how to write down the basis vectors. In many cases, representative
basis functions are tabulated in the character tables. However, suppose
that we have to find basis functions for the following cases:
1. an irreducible representation for which no basis functions are
listed in the character table; or
2. an arbitrary function.
In such cases the basis functions can often be found using projection
operators ˆPk , not to be confused with the symmetry operators
ˆPR. We define the projection operator ˆP
(Γn)
k as transforming one basis
vector |Γn into another basis vector |Γnk of the same irreducible
representation Γn:
ˆP
(Γn)
k |Γn ≡ |Γnk . (4.22)
The utility of projection operators is mainly to project out basis functions
for a given partner of a given irreducible representation from an
arbitrary function.
The discussion of this topic focuses on the following issues:
4.4. DERIVATION OF ˆP
(ΓN )
K 83
1. The relation of the projection operator to symmetry operators of
the group and to the matrix representation of these symmetry
operators for an irreducible representation.
2. The effect of projection operators on an arbitrary function.
As an example, we illustrate in §4.6 how to find basis functions from an
arbitrary function for the case of the group of the equilateral triangle
(see §4.2).
4.4 Derivation of an Explicit Expression
for ˆP
(Γn)
k
In this section we find an explicit expression for the projection operators
ˆP
(Γn)
kl by considering the relation of the projection operator to
symmetry operators of the group. We will find that the coefficients of
this expression give the matrix representations of each of the symmetry
elements.
Let the projection operator ˆP
(Γn)
k be written as a linear combination
of the symmetry operators ˆPR:
ˆP
(Γn)
k =
R
Ak (R) ˆPR (4.23)
where the Ak (R) are arbitrary expansion coefficients to be determined.
Substitution of Eq. 4.23 into Eq. 4.22 yields
ˆP
(Γn)
k |Γn ≡ |Γnk =
R
Ak (R) ˆPR|Γn . (4.24)
Multiply Eq. 4.24 on the left by Γnk| to yield
Γnk|Γnk = 1 =
R
Ak (R) Γnk| ˆPR|Γn
D(Γn)(R)k
. (4.25)
But the Wonderful Orthogonality Theorem (Eq. 2.50) specifies that
R
D(Γn)
(R)∗
k D(Γn)
(R)k =
h
n
(4.26)
84 CHAPTER 4. BASIS FUNCTIONS
where h is the number of symmetry operators in the group and n is
the dimensionality of the irreducible representation Γn, so that we can
identify Ak (R) with the matrix element of the representation for the
symmetry element R:
Ak (R) =
n
h
D(Γn)
(R)∗
k . (4.27)
Thus the projection operator is explicitly given in terms of the symmetry
operators of the group by the relation:
ˆP
(Γn)
k =
n
h R
D(Γn)
(R)∗
k
ˆPR. (4.28)
From the explicit form for ˆP
(Γn)
k in Eq. 4.28 and from Eq. 4.22 we see
how to find the partners of an irreducible representation Γn from any
single known basis vector, provided that the matrix representation for
all the symmetry operators D(Γn)
(R) is known.
As a special case, the projection operator ˆP
(Γn)
kk transforms |Γnk
into itself and can be used to check that |Γnk is indeed a basis function.
We note that the relation of ˆP
(Γn)
kk to the symmetry operators ˆPR
involves only the diagonal elements of the matrix representations
(though not the trace):
ˆP
(Γn)
kk =
n
h R
D(Γn)
(R)∗
kk
ˆPR (4.29)
where
ˆP
(Γn)
kk |Γnk ≡ |Γnk . (4.30)
4.5 The Effect of Projection Operations
on an Arbitrary Function
The projection operators ˆP
(Γn)
kk defined in Eq. 4.30 are of special importance
because they can project the kth
partner of irreducible representation
Γn from an arbitrary function. Any arbitrary function F can
4.5. PROJECTION OPERATIONS ON AN ARBITRARY FUNCTION85
be written as a linear combination of a complete set of basis functions
f
(Γn )
j :
F =
Γn j
f
(Γn )
j |Γn j . (4.31)
We can then write from Eq. 4.29:
ˆP
(Γn)
kk F =
n
h R
D(Γn)
(R)∗
kk
ˆPRF (4.32)
and substitution of Eq. 4.31 into 4.32 then yields
ˆP
(Γn)
kk F =
n
h R Γn j
f
(Γn )
j D(Γn)
(R)∗
kk
ˆPR|Γn j (4.33)
But substitution of Eq. 4.1 into 4.33 and use of the Wonderful Orthogonality
Theorem (Eq. 2.50):
R
D(Γn )
(R)jj D(Γn)
(R)∗
kk =
h
n
δΓnΓn
δjkδj k (4.34)
yields
ˆP
(Γn)
kk F = f
(Γn)
k |Γnk (4.35)
where
ˆP
(Γn)
kk =
n
h R
D(Γn)
(R)∗
kk
ˆPR. (4.36)
We note that the projection operator does not yield normalized
basis functions. One strategy to find basis functions is to start with an
arbitrary function F.
1. We then use ˆP
(Γn)
kk to project out one basis function |Γnk .
2. We can then use the projection operator ˆP
(Γn)
k to project out all
other partners |Γn orthogonal to |Γnk in irreducible representation
Γn. Or alternatively we can use ˆP
(Γn)
to project out each
of the partners of the representation, whichever method works
most easily in a given case.
86 CHAPTER 4. BASIS FUNCTIONS
If we do not know the explicit representations D
(Γn)
k (R)∗
, but only
know the characters, then we can still project out basis functions which
transform according to the irreducible representations (using the argument
given below in §4.5), though we cannot in this case project out
specific partners but only linear combinations of the partners of these
irreducible representations.
If we only know the characters of an irreducible representation Γn,
we define the projection operator for this irreducible representation as
ˆP(Γn)
:
ˆP(Γn)
≡
k
ˆP
(Γn)
kk =
n
h R k
D(Γn)
(R)∗
kk
ˆPR (4.37)
so that
ˆP(Γn)
=
n
h R
χ(Γn)
(R)∗ ˆPR (4.38)
and using Eq. 4.35 we then obtain
ˆP(Γn)
F =
k
ˆP
(Γn)
kk F =
k
f
(Γn)
k |Γnk (4.39)
which projects out a function transforming as Γn but not a specific
partner of Γn.
In dealing with physical problems it is useful to use physical insight
in guessing at an appropriate “arbitrary function” to initiate this
process for finding the basis functions and matrix representations for
specific problems.
4.6 Linear Combinations of Atomic Orbitals
for 3 Equivalent Atoms at the
Corners of an Equilateral Triangle
As an example of finding basis functions for an arbitrary function,
we will consider forming linear combinations of atomic orbitals which
transform as irreducible representations of the symmetry group.
In many of the applications that we will be making of group theory
to solid state physics, we will have equivalent atoms at different sites.
We use the symmetry operations of the group to show which irreducible
4.6. LINEAR COMBINATIONS FOR 3 EQUIVALENT ATOMS 87
 ¢¡¤£ ¥§¦
¥§¨
¥§©
¥¥
 £
 ¢£
Figure 4.2: Equilateral triangle
and arbitrary functions
defining the notation used in
§4.6.
representations result when the equivalent atoms transform into each
other under the symmetry operations of the group. The discussion
of projection operators of an arbitrary function applies to this very
important case.
As an example of this application, suppose that we have 3 equivalent
atoms at the 3 corners of an equilateral triangle (see Figure 4.2)
and that each atom is in the same spherically symmetric ground state
described by a wave function ψ0(ri), where the subscript i is a site index
which can apply to any of the 3 sites. As a short-hand notation
for ψ0(ra), ψ0(rb), ψ0(rc) we will here use a, b, c so that the arbitrary
function is written as
F =



a
b
c


 . (4.40)
We now want to combine these atomic orbitals to make a molecular
orbital that transforms according to the irreducible representations of
the group. To do this we use the results on the projection operator to
find out which irreducible representations are contained in the function
F. According to the above discussion, we can project out a basis function
for representation Γn by considering the action of ˆP
(Γn)
kk on one of
the atomic orbitals, as for example orbital a:
ˆP
(Γn)
kk a =
n
h R
D(Γn)
(R)∗
kk
ˆPRa = f
(Γn)
k |Γnk (4.41)
88 CHAPTER 4. BASIS FUNCTIONS
in which we have used the definition for ˆP
(Γn)
kk given by Eq. 4.35 and the
expression for ˆP
(Γn)
kk given by Eq. 4.36. If the representation Γn is onedimensional,
then we can obtain D(Γn)
(R) directly from the character
table, and Eq. 4.41 then becomes
ˆP(Γn)
a =
n
h R
χ(Γn)
(R)∗ ˆPRa = f(Γn)
|Γn (4.42)
For the appropriate symmetry operators for this problem we refer to
§1.2 where we have defined: E ≡ identity; (A, B, C) ≡ π rotations
about two-fold axes in the plane of triangle; (D, F) ≡ 2π/3 rotations
about the three-fold axis ⊥ to the plane of the triangle. These symmetry
operations are also indicated in Fig. 4.2.
For the identity representation Γ1 the characters and matrix representations
are all unity so that
ˆP(Γ1)
a =
1
6
( ˆPEa + ˆPAa + ˆPBa + ˆPCa + ˆPDa + ˆPF a)
=
1
6
(a + b + a + c + b + c)
=
1
3
(a + b + c), (4.43)
a result which is intuitively obvious. Each atom site must contribute
equally to the perfectly symmetrical molecular representation Γ1. This
example illustrates how starting with an arbitrary function a (or ψ(ra))
we have found a linear combination that transforms as Γ1. Likewise,
we obtain the same result by selecting b or c as the arbitrary function
ˆP(Γ1)
b = ˆP(Γ1)
c =
1
3
(a + b + c). (4.44)
We now apply a similar analysis for representation Γ1 to illustrate
another important point. In this case the matrix representations and
characters are +1 for (E, D, F), and −1 for (A, B, C). Thus
ˆP(Γ1 )
a =
1
6
( ˆPEa − ˆPAa − ˆPBa − ˆPCa + ˆPDa + ˆPF a)
=
1
6
(a − b − a − c + c + b) = 0 (4.45)
4.6. LINEAR COMBINATIONS FOR 3 EQUIVALENT ATOMS 89
which states that no molecular orbital with Γ1 symmetry can be made
by taking a linear combination of the a, b, c orbitals. This is verified by
considering
ˆP(Γ1 )
b = ˆP(Γ1 )
c = 0. (4.46)
The same approach also yields the 2-dimensional irreducible representations.
To carry out the evaluations, we use the (11) and (22)
elements of the matrix representation given by Eq. 1.4 in §1.2 for Γ2 and
the symmetry operations. We will see below that only the irreducible
representations of Γ1 + Γ2 are contained in the linear combination of
atomic orbitals for a, b, c. This makes sense since we have 3 atomic
orbitals which split into a non-degenerate and a 2-dimensional representation
in trigonal symmetry through the symmetry operations ˆPR
on the equivalent site functions a, b, c:
ˆPR



a
b
c


 . (4.47)
Equations 4.36 and 4.39 then yield:
ˆP
(Γ2)
11 a =
2
6
1 ˆPEa + 1 ˆPAa −
1
2
ˆPBa −
1
2
ˆPCa −
1
2
ˆPDa −
1
2
ˆPF a
=
1
3
a + b −
1
2
a −
1
2
c −
1
2
b −
1
2
c
=
1
3
1
2
a +
1
2
b − c = |Γ21 (4.48)
and
ˆP
(Γ2)
22 a =
2
6
1 ˆPEa − 1 ˆPAa +
1
2
ˆPBa +
1
2
ˆPCa −
1
2
ˆPDa −
1
2
ˆPF a
=
1
3
a − b +
1
2
a +
1
2
c −
1
2
b −
1
2
c
=
1
3
3
2
a −
3
2
b =
1
2
[a − b] = |Γ22 . (4.49)
The orthogonality of the basis functions |Γ21 and |Γ22 can be checked
by inspection. The representations corresponding to these basis functions
are found from the main definition (Eq. 4.1). For example, the
90 CHAPTER 4. BASIS FUNCTIONS
operation of ˆPB on the basis functions found in Eqs. 4.48 and 4.49 also
provide basis functions for ˆPB as shown explicitly in Eqs. 4.50 and 4.51
ˆPB(ψ1 ψ2) = (ψ1 ψ2)
−1
2
3
2
1
2
1
2
(4.50)
ˆPB
1
6
a +
1
6
b −
1
3
c =
1
6
a +
1
6
c −
1
3
b = −
1
2
1
6
a +
1
6
b −
1
3
c +
1
2
1
2
a −
1
2
b
ˆPB
1
2
a −
1
2
b =
1
2
a −
1
2
c =
3
2
1
6
a +
1
6
b −
1
3
c +
1
2
1
2
a −
1
2
b (4.51)
These basis functions are applied equally well to all the other group
elements: C, D, F, E. In terms of the basis functions |Γ21 and |Γ22
given in Eqs. 4.48 and 4.49 we obtain by a similar procedure, as was
used to obtain Eq. 4.51, the following matrix representation:
D(Γ2)
(E) =
1 0
0 1
(4.52)
D(Γ2)
(A) =
1 0
0 −1
(4.53)
D(Γ2)
(B) =
−1
2
3
2
1
2
1
2
(4.54)
D(Γ2)
(C) =
−1
2
−3
2
−1
2
1
2
(4.55)
D(Γ2)
(D) =
−1
2
−3
2
1
2
−1
2
(4.56)
D(Γ2)
(F) =
−1
2
3
2
−1
2
−1
2
. (4.57)
It is readily checked that these matrices obey the multiplication table
given in Table 1.1 on p. 3.
The basis functions given by Eqs. 4.48 and 4.49 and the corresponding
representation given by Eqs. 4.50 and 4.52–4.57 are not unique. Two
sets of equally good basis functions are:
1
6
b +
1
6
c −
1
3
a and
1
2
[b − c] (4.58)
4.6. LINEAR COMBINATIONS FOR 3 EQUIVALENT ATOMS 91
or
1
6
c +
1
6
a −
1
3
b and
1
2
[c − a] . (4.59)
All of these basis functions lack aesthetic symmetry and do not give
rise to unitary matrix representations.
To obtain a more symmetrical set of basis functions for this problem,
we start with an arbitrary function that is like one of the symmetry
operations of the group (e.g., a three-fold rotation ˆPD)
|Γ2α = a + ωb + ω2
c (4.60)
where ω = e2πi/3
and we note that ˆPD|Γ2α = ω2
|Γ2α
Thus |Γ2α is already a basis function. Clearly the partner of |Γ2α
is |Γ2α ∗
since ˆPD|Γ2α ∗
= ˆPD(a+ω2
b+ωc) = ω(a+ω2
b+ωc) = ω|Γ2β
where we have used the notation (α, β) to denote another set of partners
of the Γ2 representation:
|Γ2α = a + ωb + ω2
c |Γ2β = a + ω2
b + ωc. (4.61)
The two partners in Eq. 4.61 are complex conjugates of each other.
Corresponding to these basis functions, the matrix representation for
each of the group elements is simple and symmetrical
E =
1 0
0 1
A =
0 ω2
ω 0
B =
0 1
1 0
(4.62)
C =
0 ω
ω2
0
D =
ω2
0
0 ω
F =
ω 0
0 ω2 . (4.63)
By inspection the representation given by Eq. 4.61 is unitary while
that corresponding to the basis functions |Γ21 and |Γ22 is not. By
inspection we further see that the basis functions |Γ21 and |Γ22 are
linear combinations of the basis functions |Γ2α and |Γ2β :
|Γ21 =
1
6
(a + b − 2c) = −
1
6
[ω(a + ωb + ω2
c) + ω2
(a + ω2
b + ωc)]
= −
1
6
ω|Γ2α + ω2
|Γ2β , (4.64)
|Γ22 =
1
2
(a − b) = −
i
2
√
3
[ω(a + ωb + ω2
c) − ω2
(a + ω2
b + ωc)]
= −
i
2
√
3
ω|Γ2α − ω2
|Γ2β (4.65)
92 CHAPTER 4. BASIS FUNCTIONS
or
|Γ2α = −3ω2
|Γ21 + i
√
3ω2
|Γ22 (4.66)
|Γ2β = −3ω|Γ21 − i
√
3ω|Γ22 = |Γ2α ∗
(4.67)
In general, the more symmetric the choice of basis functions, the easier
the use of the representation.
Icosahedral symmetry will be covered in Chapter 23.
4.7 Selected Problems
1. (a) What are the matrix representations for (xy, x2
− y2
) and
(Rx, Ry) in the point group D3?
(b) Using the results in (a), find the unitary transformation
which transforms the matrices for the representation corresponding
to the basis functions (xy, x2
− y2
) into the representation
corresponding to the basis functions (x, y).
(c) Using projection operators, check that xy forms a proper
basis function of the two-dimensional irreducible representation
Γ2 in point group D3. Using the matrix representation
found in (a) and projection operators, find the partner of xy.
(d) Using the basis functions in the character table for D3h, write
a set of (2 × 2) matrices for the two 2-dimensional representations
E and E .
2. (a) Explain the Hermann–Manguin notation Td(¯43m).
(b) What are the irreducible representations and partners of the
following basis functions in Td symmetry? (i) ωx2
+ω2
y2
+z2
,
where ω = exp(2πi/3); (ii) xyz; and (iii) x2
yz.
(c) Using the results of (b) and the basis functions in the character
table for the point group Td, give one set of basis functions
for each irreducible representation of Td.
(d) Using the basis function ωx2
+ ω2
y2
+ z2
and its partner (or
partners), find the matrix for an S4 rotation about the x axis
in this irreducible representation.
4.7. SELECTED PROBLEMS 93
3. Consider the cubic group Oh. Find the basis functions for all the
irreducible representations of the point group Oh.
4. Consider the hypothetical molecule CH4 where the four H atoms
are at the corners of a square (±a, 0, 0) and (0, ±a, 0) while the
C atom is at (0, 0, z), where z a. What are the symmetry ele-
ments?
(a) Identify the appropriate character table.
(b) Using the basis functions in the character table, write down
a set of (2 × 2) matrices which provide a representation for
the two-dimensional irreducible representation of this group.
(c) Find the 4 linear combinations of the four H orbitals (assume
identical s-functions at each H site) that transform as
the irreducible representations of the group. What are their
symmetry types?
(d) What are the basis functions that generate the irreducible
representations.
(e) Check that xz forms a proper basis function for the two
dimensional representation of this point group and find its
partner.
(f) What are the irreducible representations and partners of the
following basis functions in the point group (the hydrogens
lie in the xy plane): (i) xyz, (ii) x2
y, (iii) x2
z, (iv) x + iy.
(g) What additional symmetry operations result in the limit
that all H atoms are coplanar with atom C? What is now the
appropriate group and character table? (The stereograms in
Table 3.2 of the class notes may be useful.)
5. Consider a molecule AB6 where the A atom lies in the central
plane and three B atoms indicated by “o” lie in a plane at a
94 CHAPTER 4. BASIS FUNCTIONS
distance c above the central plane and the B atoms indicated by
“×” lie in a plane below the central plane at a distance −c. When
projected onto the central plane, all B atoms occupy the corners
of a hexagon.
A
B
B
B
B
B
B
(a) Find the symmetry elements and classes.
(b) Construct the character table. To which point group (Chapter
3 does this molecule correspond? How many irreducible
representations are there? How many are one-dimensional
and how many are of higher dimensionality?
(c) Using the basis functions in the character table for this point
group, find a set of matrices for each irreducible representation
of the group.
(d) Find the linear combinations of the six s-orbitals of the B
atoms that transform as the irreducible representations of
the group.
(e) What additional symmetry operations result in the limit
that all B atoms are coplanar with A? What is now the appropriate
group and character table for this more symmetric
molecule?
(f) Indicate which stereogram in Fig. 3.2 is appropriate for the
case where the B atoms are not coplanar with A, and the
case where they are coplanar.
Chapter 5
Group Theory and Quantum
Mechanics
In this brief chapter we consider the connection between group theory
and quantum mechanics. For many practical problems we calculate the
solution of Schr¨odinger’s equation for a Hamiltonian displaying various
symmetries, which implies that the system is invariant under the action
of both the Hamiltonian H and the symmetry operation ˆPR.
5.1 Overview
We define the “Group of Schr¨odinger’s Equation” as the group of all
ˆPR such that
H, ˆPR = 0. (5.1)
If H and ˆPR commute, and if ˆPR is a Hermitian operator, then H and
ˆPR can be simultaneously diagonalized. In this Chapter we show that:
1. The elements ˆPR satisfy the group property.
2. If ψn is an eigenfunction corresponding to the energy eigenvalue
En then ˆPRψn and ˆPR
ˆPSψn, etc. are also eigenfunctions corresponding
to the same eigenvalue.
3. If En is a k-fold degenerate level, then the matrix representation
(irreducible) of ˆPR is given by
95
96CHAPTER 5. GROUP THEORY AND QUANTUM MECHANICS
ˆPRψnα =
k
j=1
ψnj D(n)
(R)jα. (5.2)
4. The operation of ˆPR on a general vector consisting of a complete
set of eigenfunctions yields a matrix representation of R in block
diagonal form.
5.2 The Group of Schr¨odinger’s Equation
We have now learned enough to start making applications of group theory
to physical problems. In such problems we typically have a system
described by a Hamiltonian which may be very complicated. Symmetry
often allows us to make certain simplifications, without knowing
the detailed Hamiltonian. To make a connection between group
theory and quantum mechanics, we consider the group of symmetry
operators ˆPR which leave the Hamiltonian invariant. These operators
ˆPR are symmetry operations of the system and the ˆPR commute with
the Hamiltonian (see Eq. 5.1). The operators ˆPR are said to form
the group of Schr¨odinger’s equation.
We now show that these operators form a group. The identity element
clearly exists (leaving the system unchanged). Each symmetry
operator ˆPR has an inverse ˆP−1
R to undo the operation ˆPR and from
physical considerations the element ˆP−1
R is also in the group. The product
of 2 operators of the group is still an operator of the group, since
we can consider these separately as acting on the Hamiltonian. The associative
law clearly holds. Thus the requirements for forming a group
are satisfied.
Whether the operators ˆPR be rotations, reflections, translations or
permutations, these symmetry operations do not alter the Hamiltonian
or its eigenvalues. If Hψn = Enψn is a solution to Schr¨odinger’s equation
and H and ˆPR commute, then
ˆPRHψn = ˆPREnψn = H( ˆPRψn) = En( ˆPRψn) (5.3)
Thus ˆPRψn is as good an eigenfunction of H as ψn itself. Furthermore,
both ψn and ˆPRψn correspond to the same eigenvalue En. Thus,
5.2. THE GROUP OF SCHR ¨ODINGER’S EQUATION 97
starting with a particular eigenfunction, we can generate all other eigenfunctions
of the same degenerate set (same energy) by applying all the
symmetry operations that commute with the Hamiltonian (or leave it
invariant). Similarly, if we consider the product of two symmetry operators
we again generate an eigenfunction of the Hamiltonian H
ˆPR
ˆPSH=H ˆPR
ˆPS
ˆPR
ˆPSHψn= ˆPR
ˆPSEnψn = En( ˆPR
ˆPSψn) = H( ˆPR
ˆPSψn)
(5.4)
in which ˆPR
ˆPSψn is an eigenfunction of H. We also note that the
action of ˆPR on an arbitrary vector consisting of eigenfunctions, yields
a × matrix representation of ˆPR that is in block diagonal form,
with each block having the dimensions of the corresponding irreducible
representation.
Suppose En is a k-fold degenerate level of the group of Schr¨odinger’s
equation. Then any linear combination of the eigenfunctions ψn1,ψn2, . . . , ψnk
is also a solution of Schr¨odinger’s equation. We can write the operation
ˆPRψnα on one of these eigenfunctions as
ˆPRψnα =
j
ψnjD(n)
(R)jα (5.5)
where D(n)
(R)jα is an irreducible matrix which defines the linear combination,
n labels the energy index, α labels the degeneracy index.
Equation 5.5 is identical with the more general equation for a basis
function (Eq. 4.1) where the states |Γnα and |Γnj are written symbolically
rather than explicitly as they are in Eq. 5.5.
We show here that the matrices D(n)
(R) form an n dimensional
irreducible representation of the group of Schr¨odinger’s equation where
n denotes the degeneracy of the energy eigenvalue En. Let R and S
be two symmetry operations which commute with the Hamiltonian and
let RS be their product. Then from Eq. 5.5 we can write
ˆPRSψnα= ˆPR
ˆPSψnα = ˆPR j ψnjD(n)
(S)jα
= jk ψnkD(n)
(R)kjD(n)
(S)jα = k ψnk D(n)
(R)D(n)
(S)
kα
(5.6)
98CHAPTER 5. GROUP THEORY AND QUANTUM MECHANICS
after carrying out the indicated matrix multiplication. But by definition,
the product operator RS can be written as:
ˆPRSψnα =
k
ψnkD(n)
(RS)kα (5.7)
so that
D(n)
(RS) = D(n)
(R)D(n)
(S) (5.8)
and the matrices D(n)
(R) form a representation for the group. We label
quantum mechanical states typically by a state vector (basis vector)
|α, Γn, j where Γn labels the irreducible representation, j the component
or partner of the irreducible representation, and α labels the other
quantum numbers that do not involve the symmetry of the ˆPR opera-
tors.
The dimension of the irreducible representation is equal to the degeneracy
of the eigenvalue En. The representation D(n)
(R) generated
by ˆPRψnα is an irreducible representation if all the ψnk correspond to a
single eigenvalue En. For otherwise it would be possible to form linear
combinations of the type
ψn1, ψn2, . . . , ψns
subset 1
ψn,s+1, . . . , ψnk
subset 2
(5.9)
whereby the linear combinations within the subsets would transform
amongst themselves. But if this happened, then the eigenvalues for
the 2 subsets would be different, except for the rare case of accidental
degeneracy. Thus, the transformation matrices for the symmetry operations
form an irreducible representation for the group of Schr¨odinger’s
equation.
5.3 The Application of Group Theory
It is convenient at this point to classify the ways that group theory
is used to solve quantum mechanical problems. Group theory is used
both to obtain exact results and in applications of perturbation theory.
In the category of exact results, we have as examples:
5.3. THE APPLICATION OF GROUP THEORY 99
high symmetry low symmetry
degenerate state
Figure 5.1: The effect of lowering
the symmetry often results
in a lowering of the degeneracy
of degenerate energy
states.
1. The irreducible representations of the symmetry group of Schr¨odinger’s
equation label the states and specify their degeneracies
(e.g., an atom in a crystal field).
2. Group theory is useful in following the changes in the degeneracies
of the energy levels as the symmetry is lowered. This case can
be thought of in terms of a Hamiltonian
H = H0 + H (5.10)
where H0 has high symmetry corresponding to the group G, and
H is a perturbation having lower symmetry and corresponding to
a group G of lower order (fewer symmetry elements). Normally
group G is a subgroup of Group G. Here we find first which symmetry
operations of G are contained in G ; the irreducible representations
of G label the states of the lower symmetry situation
exactly. In going to lower symmetry we want to know what happens
to the degeneracy of the various states in the initial higher
symmetry situation (see Fig. 5.1). We say that in general the irreducible
representation of the higher symmetry group forms
reducible representations for the lower symmetry group.
The degeneracy of states may either be lowered as the symmetry
is lowered or the degeneracy may be unchanged. Group theory
tells us exactly what happens to these degeneracies. We are also
interested in finding the basis functions for the lower symmetry
group G . For those states where the degeneracy is unchanged, the
basis functions are generally unchanged. When the degeneracy is
reduced, then by proper choice of the form of the partners, the basis
functions for the degenerate state will also be basis functions
100CHAPTER 5. GROUP THEORY AND QUANTUM MECHANICS
for the states in the lower symmetry situation; e.g., if (x, y, z)
are basis functions for a 3-dimensional representation in the cubic
group, then lowering the symmetry to tetragonal with z as
the main symmetry direction will give a 2-dimensional representation
with basis functions (x, y) and a one-dimensional representation
with basis function z. However, if the symmetry is lowered
to tetragonal along a z direction (different from z), then linear
combinations of (x, y, z) must be taken to obtain a vector along
z and two others that are mutually orthogonal. The lowering of
degeneracy is a very general topic and will enter the discussion of
many applications of group theory.
3. Group theory is helpful in finding the correct linear combination
of wavefunctions that is needed to diagonalize the Hamiltonian.
This procedure involves the concept of equivalence which
applies to situations where equivalent atoms sit at symmetrically
equivalent sites.
We elaborate on these concepts in the following chapters.
5.4 Selected Problems
1. Consider the hypothetical XH12 molecule which has Ih icosahedral
symmetry, and the X atom is at the center. The lines connecting
the X and H atoms are 5-fold axes.
(a) Suppose that we stretch the XH12 molecule along one of the
5-fold axes. What are the resulting symmetry elements of
the stretched molecule?
5.4. SELECTED PROBLEMS 101
(b) What is the appropriate point group?
(c) Consider the Gu and Hg irreducible representations of group
Ih as a reducible representation of the lower symmetry group.
Find the symmetries of the lower symmetry group that were
contained in a 4-fold energy level that transforms as Gu and
in a 5-fold level that transforms as Hg in the Ih group. Assuming
the basis functions given in the character table for
the Ih point group, give the corresponding basis functions for
each of the levels in the multiplets for the stretched molecule.
(d) Suppose that the symmetry of the XH12 molecule is described
in terms of hydrogen atoms placed at the center of
each pentagon of a regular dodecahedron. A regular dodecahedron
has 12 regular pentagonal faces, 20 vertices and 30
edges. What are the symmetry classes for the regular dodecahedron.
Suppose that the XH12 molecule is stretched along
one of its 5-fold axes as in (a). What are the symmetry elements
of the stretched XH12 molecule when viewed from the
point of view of a distortion from dodecahedral symmetry?
102CHAPTER 5. GROUP THEORY AND QUANTUM MECHANICS
Chapter 6
Application of Group Theory
to Crystal Field Splitting
This is the first of several chapters aimed at presenting some general
applications of group theory while further developing theoretical concepts
and amplifying on those given in the first 5 chapters. The first
application of group theory is made to crystal field splittings because
of the relative simplicity of this application and because it provides a
good example of going from higher to lower symmetry, a procedure used
very frequently in applications of group theory to solid state physics.
In this chapter we also consider irreducible representations of the full
rotation group.
6.1 Introduction
The study of crystal field theory is relevant for physics and engineering
applications in situations where it is desirable to exploit the sharp,
discrete energy levels that are characteristic of atomic systems together
with the larger atomic densities that are typical of solids. As an example,
consider the variety of powerful lasers whose operation is based
on the population inversion of impurity levels of rare earth ions in a
transparent host crystal. The energy levels of an electron moving in the
field of an ion embedded in such a solid are approximately the same as
for an electron moving in the field of a free ion. The interaction of the
103
104CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
impurity ion with the host crystal is small enough so that the electrons
can be localized and identified in a tight binding sense with an impurity
ion. Thus the interaction between the ion and the host crystal can
be treated in perturbation theory. Group theory plays a major role in
finding the degeneracy and the symmetry types of the electronic levels
in the crystalline field.
Recently the topic of crystal field splittings has attracted considerable
attention with the very important technical breakthrough of
erbium-doped silica-based optical glass fiber amplifiers for use in optical
communications systems. With the arrival of a wave packet of photons
having a frequency within the linewidth of an optical transition, optical
amplification occurs when there are more ions in the upper state than
in the lower state. The ground state of Er3+
is a 4
I15/2 level (s = 3/2,
l = 6, j = 15/2) and in the silica host the only radiative transition is the
4
I13/2 →4
I15/2 transition. In the actual optical fibers, this transition is
a broad line centered at 1.55 µm, and is homogeneously broadened. The
optical amplifier can be pumped by InGaAs or InGaAsP laser diodes
to create the population inversion. Some characteristics of the erbiumdoped
fiber amplifier include: low insertion loss (< 0.5 dB), high gain
(30–45 dB) with no polarization sensitivity, high saturation power outputs
(> 10 dBm), slow gain dynamics (100 µs – 1 ms) giving negligible
cross talk at frequencies greater than 100 kHz, and quantum-limited
noise figures. This technological breakthrough provides added motivation
for understanding the splitting of the energy levels of an impurity
ion in a crystal field.
In this chapter the point group symmetry of an impurity ion in a
crystal is presented. The crystal potential Vxtal determines the point
group symmetry. Following the discussion on the form of the crystal
potential, some properties of the full rotation group are given, most
importantly the characters χ( )
(α) for rotations through an angle α
and χ( )
(i) for inversions. The expression for χ( )
(α) is given by
χ( )
(α) =
sin( + 1/2)α
sin(α/2)
for rotations. (6.1)
In general, inversion is not equivalent to a C2 operation. For the inversion
operation i, we have
iY m(θ, φ) = Y m(π − θ, π + φ) = (−1) Y m(θ, φ) (6.2)
6.1. INTRODUCTION 105
and therefore
χ( )
(i) =
m=
m=−
(−1) = (−1) (2 + 1), (6.3)
where denotes the angular momentum quantum number.
Irreducible representations of the full rotation group are generally
found to be reducible representations of a point group of lower symmetry
which is a subgroup of the higher symmetry group. If the representation
is reducible, then crystal field splittings of the energy levels
occur. If, however, the representation is irreducible, then no crystal
field splittings occur. Examples of each type of representation are pre-
sented.
We focus explicitly on giving examples of going from higher to lower
symmetry. In so doing, we consider the:
1. Splitting of the energy levels.
2. Symmetry types of the split levels.
3. Choice of basis functions to bring the Hamiltonian H into block
diagonal form. Spherical symmetry results in spherical harmonics
Y m(θ, φ) for basis functions. Proper linear combinations of the
spherical harmonics Y m(θ, φ) are taken to make appropriate basis
functions for the point group of lower symmetry.
In crystal field theory we write down the Hamiltonian for the impurity
ion in a crystalline solid as
H =
i



p2
i
2m
−
Ze2
riµ
+
j
e2
rij
+
j
ξij i · sj + γiµji · Iµ



+ Vxtal (6.4)
where the first term is the kinetic energy of the electrons associated with
the ion and the second term represents the Coulomb attraction of the
electrons of the ion to their nucleus and the third term represents the
mutual Coulomb repulsion of the electrons associated with the impurity
ion, and the sum on j denotes a sum on pairs of electrons. These three
quantities are denoted by H0 the electronic Hamiltonian of the free
atom without spin-orbit interaction. H0 is the dominant term in the
106CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
total Hamiltonian H. The remaining terms are treated in perturbation
theory in some order. Here ξij i · sj is the spin-orbit interaction of
electrons on the impurity ion and γiµji · Iµ is the hyperfine interaction
of electrons on the ion. The perturbing crystal potential Vxtal of the
host ions acts on the impurity ion and lowers its spherical symmetry.
Because of the various perturbation terms appearing in Eq. 6.4, it
is important to distinguish the two limiting cases of weak and strong
crystal fields.
1. Weak field case – In this case, the perturbing crystal field Vxtal is
considered to be small compared with the spin-orbit interaction.
In this limit, we find the energy levels of the free impurity ion with
spin-orbit interaction and at this point we consider the crystal
field as an additional perturbation. These approximations are
appropriate to rare earth ions in ionic host crystals. We will deal
with the group theoretical aspects of this case in §19.4, after we
have learned how to deal with the spin on the electron in the
context of group theory.
2. Strong field case – In this case, the perturbing crystal field Vxtal
is strong compared with the spin-orbit interaction. We now consider
Vxtal as the major perturbation on the energy levels of H0.
Examples of the strong field case are transition metal ions (Fe,
Ni, Co, Cr, etc.) in a host crystal. It is this limit that we will
consider first, and is the focus of this Chapter.
6.2 Comments on the Form of Crystal Fields
To construct the crystal field, we consider the electrostatic interaction
of the neighboring host ions at the impurity site. To illustrate how
this is done, consider the highly symmetric case of an impurity ion in a
cubic environment provided by ions at x = ±a, y = ±a, z = ±a. The
contribution from an ion at x = −a at the field point (x, y, z) is (see
Fig. 6.1):
Vx=−a =
e
a (1 + x/a)2 + (y/a)2 + (z/a)2
=
e
a
√
1 + ε
(6.5)
6.2. COMMENTS ON THE FORM OF CRYSTAL FIELDS 107
  ¡
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Figure 6.1: Coordinate
system used for
expansion of the impurity
ion potential.
where e is the charge on the electron and ε is a small dimensionless
quantity. Then using the binomial expansion:
(1 + ε)− 1
2 = 1 −
1
2
ε +
3
8
ε2
−
5
16
ε3
+
35
128
ε4
+ ... (6.6)
we obtain the contribution to the potential for charges e at x = a and
x = −a:
Vx=−a + Vx=a = 2e
a
[1 − 1
2
(r2
/a2
) + 3
2
(x2
/a2
) + 3
8
(r4
/a4
)
− 15
4
(x2
/a2
)(r2
/a2
) + 35
8
(x4
/a4
) + ...].
(6.7)
For a cubic field with charges e at x = ±a, y = ±a, z = ±a we get for
Vtotal = Vxtal:
Vtotal =
2e
a
[3 +
35
8a4
(x4
+ y4
+ z4
) −
21
8
(r4
/a4
) + ...] (6.8)
108CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
so that the perturbation that will lift the degeneracy of the free atom
is of the form
Vcubic =
35e
4a5
(x4
+ y4
+ z4
) −
3
5
r4
. (6.9)
From these expressions it also follows that for a rhombic field where
the charges are at x = ±a, y = ±b, z = ±c (where a = b = c)
Vtotal =
2e
a
+
2e
b
+
2e
c
+ex2 2
a3
−
1
b3
−
1
c3
+ey2 2
b3
−
1
a3
−
1
c3
+ez2 2
c3
−
1
a3
−
1
b3
(6.10)
so that the perturbation that will lift the degeneracy of the free atom
is of the form
Vrhombic = Ax2
+ By2
− (A + B)z2
. (6.11)
We note that Vcubic contains no terms of order x2
whereas Vrhombic does.
It is sometimes useful to express the crystal field potential in terms
of spherical harmonics since the unperturbed states for the free impurity
ion are expressed in that way. Here we make use of the fact that the
crystal field potential is generated by a collection of point sources and
in the intervening space we are “outside” the field sources so that the
potential must satisfy the Laplace equation 2
V = 0. Solutions to
Laplace’s equation are of the form r Y m(θ, φ). Let us then recall the
form of the spherical harmonics Y m(θ, φ) which are the basis functions
for the full rotation group:
Y m(θ, φ) =
2 + 1
4π
( − |m|)!
( + |m|)!
1
2
Pm
(cos θ)eimφ
(6.12)
where the associated Legendre polynomial is written as
Pm
(x) = (1 − x2
)
1
2
|m| d|m|
dx|m|
P (x) (6.13)
and the Legendre polynomial P (x) is generated by
1/
√
1 − 2sx + s2 =
∞
=0
P (x)s . (6.14)
6.3. CHARACTERS FOR THE FULL ROTATION GROUP 109
From these definitions it is clear that for a cubic field, the only
spherical harmonics that will enter Vcubic are Y4,0, Y4,4 and Y4,−4 since
(z/4)4
involves only Y4,0 while
[(x/4)4
+ (y/4)4
]
involves only Y4,4 and Y4,−4. Following this example, we conclude that
the crystal field potential Vxtal can be written in terms of spherical
harmonics, the basis functions normally used to describe the potential
of the free ion which has full spherical symmetry.
More generally, we can write any function (e.g., any arbitrary Vxtal)
in terms of a complete set of basis functions, such as the spherical harmonics.
One important role of group theory in the solution of quantum
mechanical problems is to determine the degeneracy of the eigenvalues
and which choice of basis functions yields the eigenvalues most directly.
This information is available without the explicit diagonalization of the
Hamiltonian. In the case of the crystal field problem, we determine Vxtal
for a specific crystal symmetry using the appropriate basis functions for
the relevant point group.
We note that the crystal potential Vxtal lowers the full rotational
symmetry of the free atom to cause level splittings relative to those of
the free atom. We show here how group theory tells us the degeneracy
of the resulting levels and the appropriate basis functions to use
in diagonalizing the crystal field Hamiltonian. The first step in this
process is to consider the irreducible representation of the higher symmetry
group (the full rotation group) as a reducible representation of
the lower symmetry group (the crystal field group). We now consider
in §6.3 some of the fundamental properties of the full rotation group.
These results are liberally used in later Chapters.
6.3 Characters for the Full Rotation Group
The free atom has full rotational symmetry and the number of symmetry
operations which commute with the Hamiltonian is infinite. That is,
all Cφ rotations about any axis are symmetry operations of the full rotation
group. We are not going to discuss infinite or continuous groups
in any detail, but we will adopt results that we have used in quantum
110CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
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§  "!$#
Figure 6.2: Polar coordinate system
defining the angles θ and φ.
mechanics without rigorous proofs. It can be shown that the spherical
harmonics (angular momentum eigenfunctions) can be written in the
form:
Y ,m(θ, φ) = CPm
(θ) eimφ
(6.15)
where C is a normalization constant and P m
(θ) is an associated Legendre
polynomial given explicitly in Eq. 6.12. The coordinate system
used to define the polar and azimuthal angles is shown in Fig. 6.2.
The Y ,m(θ, φ) spherical harmonics generate odd-dimensional representations
of the rotation group and these representations are irreducible
representations. For = 0, we have a 1-dimensional representation;
= 1(m = 1, 0, −1) gives a 3-dimensional irreducible representation;
= 2(m = 2, 1, 0, 1, −2) gives a 5-dimensional representation, etc. So
for each value of the angular momentum, the spherical harmonics provide
us with a representation of the proper 2 + 1 dimensionality.
These irreducible representations are found from the so-called addition
theorem for spherical harmonics which tells us that if we
change the polar axis (i.e., the axis of quantization), then the “old”
spherical harmonics Y ,m(θ, φ) and the “new” Y ,m (θ , φ ) are related
by a linear transformation of basis functions when = :
ˆPRY ,m(θ , φ ) =
m
Y ,m (θ, φ)D( )
(R)m m (6.16)
6.3. CHARACTERS FOR THE FULL ROTATION GROUP 111
where ˆPR denotes a rotation operator that changes the polar axis, and
the matrix D( )
(R)m m provides an -dimensional matrix representation
of element R in the full rotation group. Let us assume that the reader
has previously seen this expansion for spherical harmonics which is a
major point in the development of the irreducible representations of the
rotation group.
In any system with full rotational symmetry, the choice of the z
axis is arbitrary. We thus choose the z axis as the axis about which the
operator ˆPα makes the rotation α. Because of the form of the spherical
harmonics Y ,m(θ, φ) (see Eq. 6.15) and the choice of the z axis, the
action of ˆPα on the Y m(θ, φ) basis functions only affects the φ dependence
of the spherical harmonic (not the θ dependence). The effect of
this rotation on the function Y ,m(θ, φ) is equivalent to a rotation of the
axes in the opposite sense by the angle −α
ˆPαY ,m(θ, φ) = Y ,m(θ, φ − α) = e−imα
Y ,m(θ, φ) (6.17)
in which the second equality results from the explicit form of Y ,m(θ, φ).
But Eq. 6.17 gives the linear transformation of Y ,m(θ, φ) resulting from
the action by the operator ˆPα. Thus by comparing Eqs. 6.16 and 6.17,
we see that the matrix D( )
(α)m m is diagonal in m so that we can write
D( )
(α)m m = e−imα
δm m where − ≤ m ≤ , yielding
D( )
(α) =






e−i α
O
e−i( −1)α
...
O ei α






(6.18)
where O represents all the zero entries in the off-diagonal positions.
The character of the rotations Cα is given by the geometric series
χ( )
(α) = trace D( )
(α) = e−i α
+ ... + ei α
= e−i α
1 + eiα
+ ... + e2i α
= e−i α
2
k=0
(eikα
)
= e−i α ei(2 +1)α
− 1
eiα − 1
112CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
=
ei( + 1
2
)α
− e−i( + 1
2
)α
eiα/2 − e−iα/2
=
sin( + 1
2
)α
sin(1
2
)α
. (6.19)
Thus we show that the character for rotations α about the z axis is
χ( )
(α) =
sin( + 1
2
)α
sin α/2
(6.20)
which is identical with Eq. 6.1. To obtain the character χ( )
(i) for the
inversion operator, we use the result (−1) (2 + 1) given in Eq. 6.3.
We now give a general result for an improper rotation defined by
Sn = Cn ⊗ σh (6.21)
Sn can be a product of C(α) ⊗ i, as for example, S6 = C3 ⊗ i, or
S3 = C6 ⊗ i, ... etc. where ⊗ denotes the direct product of the two
symmetry operations appearing at the left and right of the symbol ⊗.
If we now apply Eqs. 6.1 and 6.2, we obtain
χ( )
(Sn) = χ( )
(C(α) ⊗ i) = (−1)
sin( + 1
2
)α
sin α/2
. (6.22)
In the case of mirror planes, such as σh, σd, or σv we can make use of
relations such as
σh = C2 ⊗ i (6.23)
to obtain the character for mirror planes in the full rotation group.
But we are free to choose the z axis in any arbitrary way in the
full rotation group. Therefore, the formula for the character given by
Eq. 6.20 is applicable to any arbitrary rotation of α about any axis
whatsoever. Furthermore, it can be shown that there are no equivalent
irreducible representations of odd order for the full rotation group and
thus the character given by Eq. 6.20 is unique.
Now we are going to place our free ion into a crystal field which
does not have full rotational symmetry operations, but rather has the
symmetry operations of a crystal which include rotations about finite
angles, inversions and a finite number of reflections. The full rotation
group contains all these symmetry operations. Therefore, the
representation D( )
(α) given above is a representation of the crystal
6.4. EXAMPLE OF A CUBIC CRYSTAL FIELD ENVIRONMENT113
point group if α is a symmetry operation in that point group. But
D( )
(α) is, in general, a reducible representation of the lower symmetry
group. Therefore the (2 + 1) fold degeneracy of each level will in
general be partially lifted. We can find out how the degeneracy of
each level is lifted by asking what are the irreducible representations
contained in D( )
(α) in terms of the group of lower symmetry for the
crystal. The actual calculation of the crystal field splittings depends
on setting up a suitable Hamiltonian and solving it, usually in some
approximation scheme. But the energy level degeneracy does not
depend on the detailed Hamiltonian, but only on its symmetry. Thus,
the decomposition of the level degeneracies in a crystal field is exact
and is a consequence of the symmetry of the crystal field.
6.4 Example of a Cubic Crystal Field Environment
for a Paramagnetic Transition
Metal Ion
Imagine that we place our paramagnetic ion (e.g., an iron impurity)
in a cubic host crystal. Assume further that this impurity goes into a
substitutional lattice site, and is surrounded by a regular octahedron
of negative ions - symmetry type O (see Fig. 6.3). The character table
for O is shown in Table 6.1 (see also Table 3.33). From all possible
rotations on a sphere only 24 symmetry operations remain in the group
O.
Reviewing the notation in Table 6.1, the Γ notations for the irreducible
representations are the usual ones used in solid state physics
and are due to Bouckaert, Smoluchowski and Wigner, Phys. Rev. 50,
58 (1936). The second column in Table 6.1 follows the notation usually
found in molecular physics and chemistry applications, which also make
lots of use of symmetry and group theory. The key to the notation is
that A denotes 1 dimensional representations, E denotes 2 dimensional
representations, and T denotes 3 dimensional representations. Papers
on lattice dynamics of solids often use the A, E, T symmetry notation
to make contact with the molecular analog. The subscripts in Table 6.1
refer to the conventional indexing of the representations. The pertinent
114CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
Table 6.1: Character table for O
O E 8C3 3C2 = 3C2
4 6C2 6C4
Γ1 A1 1 1 1 1 1
Γ2 A2 1 1 1 −1 −1
Γ12 E 2 −1 2 0 0
Γ15 T1 3 0 −1 −1 1
Γ25 T2 3 0 −1 1 −1
Γ =0 A1 1 1 1 1 1
Γ =1 T1 3 0 −1 1 −1
Γ =2 E + T2 5 –1 1 1 –1
Γ =3 A1 + T1 + T2 7 –1 –1 1 1
Γ =4 A1 + E + T1 + T2 9 0 1 1 1
Γ =5 E + 2T1 + T2 11 –1 –1 –1 1
symmetry operations can be found from Fig. 6.3, and the classes associated
with these symmetry operations label the various columns where
the characters in Table 6.1 appear. The various types of rotational
symmetry operations are listed as:
• The 8C3 rotations are about the axes through face centroids of
the octahedron.
• The 6C4 rotations are about the corners of the octahedron.
• The 3C2 rotations are also about the corners of the octahedron,
with C2 = C2
4 .
• The 6C2 rotations are two-fold rotations about a (110) axis passing
through the midpoint of the edges (along the 110 directions
of the cube).
To be specific, suppose that we have a magnetic impurity atom
with angular momentum = 2. We first find the characters for all
the symmetry operations which occur in the O group for an irreducible
representation of the full rotation group. The representation of the full
6.4. EXAMPLE OF A CUBIC CRYSTAL FIELD ENVIRONMENT115
x
y
z
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Figure 6.3: A regular octahedron
inscribed in a cube, illustrating
the symmetry operations
of group O.
Table 6.2: Classes and characters for the group O.
Class α χ(2)
(α)
8C3 2π/3 sin(5/2)·(2π/3)
sin((2π)/(2·3))
= (−
√
3/2)/(
√
3/2) = −1
6C4 2π/4 sin(5/2)·(π/2)
sin(π/4)
= (−1/
√
2)/(1/
√
2) = −1
3C2 and 6C2 2π/2 sin(5/2)π
sin(π/2)
= 1
rotation group will be a representation of group O, but in general this
representation will be reducible.
Since the character for a general rotation α in the full rotation group
is found using Eq. (6.20), the identity class (or α = 0) yields get the
characters
χ( )
(0) =
+ 1
2
1/2
= 2 + 1. (6.24)
For our case l = 2, and by applying Eq. 6.24 to the symmetry operations
in each class we obtain the results summarized in Table 6.2.
To compare with the character table for group O (Table 6.1), we list
the characters found in Table 6.2 for the Γ
(2)
rot of the full rotation group
116CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
(l = 2) according to the classes listed in the character table for group
O (see Tables 6.1 and 3.33):
E 8C3 3C2 6C2 6C4
Γ
(2)
rot 5 −1 1 1 −1
We note that Γ
(2)
rot is a reducible representation of group O because
group O has no irreducible representations with dimensions n > 3.
To find the irreducible representations contained in Γ
(2)
rot we use the
decomposition formula for reducible representations Eq. 3.20:
aj =
1
h k
Nkχ(Γj)
(Ck)∗
χreducible
(Ck) (6.25)
where we have used the expression
χreducible
(Ck) =
j
ajχ(Γj)
(Ck) (6.26)
in which χ(Γj)
is an irreducible representation and the characters for the
reducible representation Γ
(2)
rot are written as χreducible
(Ck) ≡ χΓ
(2)
rot (Ck).
We now ask how many times is A1 contained in Γ
(2)
rot? Using Eq. 6.25
we obtain:
aA1 =
1
24
[5 − 8 + 3 + 6 − 6] = 0 (6.27)
which shows that the irreducible representation A1 is not contained in
Γ
(2)
rot. We then apply Eq. 6.25 to the other irreducible representations
of group O:
A2: aA2 = 1
24
[5 – 8 + 3 – 6 + 6] = 0
E: aE = 1
24
[10 + 8 + 6 + 0 – 0] = 1
T1: aT1 = 1
24
[15 + 0 – 3 – 6 – 6] = 0
T2: aT2 = 1
24
[15 + 0 – 3 + 6 + 6] = 1
so that finally we write
Γ
(2)
rot = E + T2
which means that the reducible representation Γ
(2)
rot breaks into the irreducible
representations E and T2 in cubic symmetry. In other words,
6.4. EXAMPLE OF A CUBIC CRYSTAL FIELD ENVIRONMENT117
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6 #
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W Q¢RTSUS B4VD R
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Figure 6.4: The splitting of the
d–levels (five-fold) in an octahedral
crystal field.
an atomic d–level in a cubic Oh crystal field splits into an Eg and a
T2g level, where the g denotes evenness under inversion. Group theory
doesn’t provide any information about the ordering of the levels
(see Fig. 6.4). For general utility, we have included in Table 6.1 the
characters for the angular momentum states = 0, 1, 2, 3, 4, 5 for the
full rotation group expressed as reducible representations of the group
O. The splittings of these angular momentum states in cubic group O
symmetry are also included in Table 6.1.
We can now carry the passage from higher to lower symmetry by going
one step further. Suppose that the presence of the impurity strains
the crystal. Let us further imagine (for the sake of argument) that the
new local symmetry of the impurity site is D4 (see Table 3.26), which
is a proper subgroup of the full rotation group. Then the levels E and
T2 given above may be split further in D4 (tetragonal) symmetry (for
example by stretching the molecule along the 4-fold axis). We now
apply the same technique to investigate this tetragonal field splitting.
We start again by writing the character table for the group D4 which
is of order 8. We then consider the representations E and T2 of the
group O as reducible representations of group D4 and write down the
characters for the E, C4, C2
4 , C2 and C2 operations from the character
table for O given above, noting that the C2 in the group D4 refers to
three of the (110) axes 6C2 of the cubic group O:
118CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
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X`Ybadcfehg §'ip¢§©
W
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t ¥ t ©dQuD t  Yvg¢e 
 Q¡w£¦¥¨§©§xQQ§
y€ IQP@RVR 9dTA P
e  ehYvg uD¥ tdg §
‚8ƒ
t ¥ t ©dQuD t  Yvg¢e 
t ¥ t ©dQuD t  Yvg¢e 6„
ƒ
„
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Figure 6.5: d-level splitting in octahedral and D4 crystal fields
Character Table for D4 E C2 = C2
4 2C4 2C2 2C2
Γ1 A1 1 1 1 1 1
Γ1 A2 1 1 1 −1 −1
Γ2 B1 1 1 −1 1 −1
Γ2 B2 1 1 −1 −1 1
Γ3 E 2 −2 0 0 0
reducible representations from O group
E 2 2 0 2 0 ≡ A1 + B1
T2 3 −1 −1 −1 1 ≡ E + B2
Using the decomposition theorem, Eq. 3.20, we find that E splits into
the irreducible representations A1 + B1 in the group D4 while T2 splits
into the irreducible representations E + B2 in the group D4, as summarized
in Fig. 6.5.
We note that the C2 operations in D4 is a π rotation about the z
axis and the 2C2 are π rotations about the x and y axes. The 2C2 are
π rotations about (110) axes. To check the decomposition of the = 2
level in D4 symmetry, we add up the characters for A1 + B1 + B2 + E
for group D4 and we get
E C2 2C4 2C2 2C2
Γ
(2)
rot 5 1 −1 1 1 A1 + B1 + B2 + E
which are the characters for the spherical harmonics considered as a
reducible representation of group D4, so that this result checks.
6.5. COMMENTS ON BASIS FUNCTIONS 119
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uƒ‚„x …‡† ˆ‰x
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Figure 6.6: d-level splitting in various crystal fields
Suppose now that instead of applying a strain along a (001) direction,
we apply a strain along a (110) direction. This will give a two-fold
axis along a (110) direction and another two-fold axis at right angles in
the (¯110) direction. The other right angle direction (001) now also becomes
a two-fold axis to give the group D2 (see Table 6.3 and Fig. 6.7)
which has only 1-dimensional representations. Therefore, application
of a stress along a (110) direction will lift all the degeneracies of the
levels in the octahedral field, while the tetragonal distortion will not.
Figure 6.6 shows the splitting of the = 2 level in going from full rotational
symmetry to various lower symmetries, including D∞h, Td, Oh,
and D2h, showing in agreement with the above discussion, the lifting of
all the degeneracy of the = 2 level in D2h symmetry. The symmetry
axis and stereographic projections for the group D2 (222) are shown in
Fig. 6.7.
6.5 Comments on Basis Functions
Although group theory tells us how the impurity ion energy levels are
split by the crystal field, it doesn’t tell us the ordering of these levels.
Often a simple physical argument can be given to decide which levels
ought to lie lower. Consider the case of a d–electron in a cubic field,
where the host ions are at x = ±a, y = ±a, z = ±a. Assume that the
impurity ion enters the lattice substitutionally, so that it is replacing
one of the cations. Then the nearest neighbor host ions are all anions.
The charge distributions for the d–states are shown in Fig. 6.8. Re-
120CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
Table 6.3: Character table for the group D2 (222).
D2(222) E Cz
2
(a)
Cy
2
(b)
Cx
2
(c)
x2
, y2
, z2
A1 1 1 1 1
xy Rz, z B1 1 1 −1 −1
xz Ry, y B2 1 −1 1 −1
yz Rz, x B3 1 −1 −1 1
χOh
E 2 2 0 0 → A1 + B1
χOh
T2
3 −1 1 1 → A1 + B2 + B3
Γ
(2)
rot 5 1 1 1 → 2A1 + B1 + B2 + B3
(a)
C2(001) corresponds to Cz
2
(b)
C2(110) corresponds to Cy
2
(c)
C2(110) corresponds to Cx
2
 ¢¡
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Figure 6.7: Symmetry axes and
stereograph for D2 symmetry.
6.5. COMMENTS ON BASIS FUNCTIONS 121
 ¢¡¤£¦¥¨§©£ £¦¥¨£


¡




¡


  ¡


¥ ¥
¥¥
¥
¥
 


¥¥
© ¢ 
¢! ¢"¦
Figure 6.8: The angular parts of d–wave functions in cubic crystals
are shown as labeled by the basis functions for the partners of the
irreducible representations.
(a) xy/r2
⇒ (Γ+
25, T2g)
(b) yz/r2
⇒ (Γ+
25, T2g)
(c) (x2
− y2
)/r2
⇒ (Γ+
12, Eg)
(d) (3z2
− r2
)/r2
⇒ (Γ+
12, Eg).
ferring to the basis functions for O which are listed in Table 3.33, we
see that for the irreducible representation Γ+
12 we have basis functions
(x2
− y2
, 3z2
− r2
) and for Γ+
25 we have basis functions (xy, yz, zx).
For the basis functions which transform as the Γ+
25 representation, the
charge distributions do not point to the host ions and hence the crystal
field interaction is relatively weak. For the d–functions which transform
as Γ+
12, the interaction will be stronger since the charge distributions
now do point to the host ion sites. If, however, the interaction is repulsive,
then the E level will lie higher than the T2 level. A more quantitative
way to determine the ordering of the levels is to solve the eigenvalue
problem explicitly. In carrying out this solution it is convenient to use
basis functions that transform as the irreducible representations of the
crystal field group.
122CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
 ¡
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23 "!$#&%'#)(0$(1#
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¢¡¡¡¡£
¤§



¤
4 "!$#&%5#)(0$(1#
6
¦879
@BA3CEDGF
Figure 6.9: Schematic energy level splitting for the five-fold = 2 level
into two-fold and three-fold levels in an environment where bonding is
along the {110} directions. The Eg and T2g levels are labeled by the
basis functions (partners) for each if the irreducible representations of
Oh.
We now look at the basis functions which provide irreducible representations
for these cases of lower symmetry. In going from the full
rotation group to the cubic group Oh, we obtain the irreducible representations
Eg and T2g shown in Fig. 6.9 in terms of the basis functions
for these irreducible representations. We note that these basis functions
bring the crystal field potential into block form, but need not completely
diagonalize the Hamiltonian. There are various forms of the
crystal field potential that have Oh symmetry (e.g., octahedral sites,
cubic sites, etc.), and in each case the appropriate set of basis functions
that transform as irreducible representations of the group will bring the
secular equation into block form.
Upon lowering the symmetry further to D4 symmetry, the T2g and
Eg levels split further according to T2g → E + B2 and Eg → A1 + B1
(see Fig. 6.5). The appropriate basis functions for these levels can be
identified with the help of the character table for group D4 in Table 3.26:
E
yz
zx
, B2 xy , B1 x2
− y2
, A1 {z2
. (6.28)
As a further example we can consider the case of going from cubic
Oh symmetry to D2 symmetry by applying a stress along the (1 1 0)
direction. Here the appropriate coordinate system is chosen with the
z–axis taken along the (0 0 1) direction, and the (y, x)-axes taken along
the (1 1 0) and (¯1 1 0) directions, respectively, as shown in Fig. 6.7.
6.5. COMMENTS ON BASIS FUNCTIONS 123
 
¡£¢
¤¦¥
§¨¥
§ ¥
¤©
¤ ¢
Figure 6.10: Schematic diagram of level splitting of the d-levels in going
from cubic Oh symmetry to D2 symmetry.
The character table for the group D2 is given in Table 3.24 and in
Table 6.3. To help with the assignment of the two-fold axes, we use
the stereogram for D2 from Fig. 3.2, which is shown in more detail in
Fig. 6.7.
We can see immediately from Fig. 6.10 how the irreducible representations
Eg and T2g of the cubic group Oh become reducible representations
for the group D2 with the following basis functions referred
to the axes shown in Fig. 6.7:
A1 x2
− y2
A1 3z2
− r2
B1 xy
B2 xz
B3 yz
Because all these basis functions ultimately relate to the = 2 level
Y2,m(θ, φ) with full rotational symmetry, we have the opportunity to
choose the coordinate system for D2 group symmetry in any way that
is convenient.
In §6.4 and §6.5 we consider the spherical harmonics for = 2
as reducible representations of the point groups Oh, O, D4, and D2.
In this connection, Table 6.4 gives the decomposition of the various
spherical harmonics for angular momentum ≤ 15 into irreducible
124CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
representations of the cubic group Oh. We note that the lowest angular
momentum state to contain the A1g irreducible representation of
Oh is = 4, consistent with Eq. (6.9). The corresponding table for
icosahedral symmetry is Table 6.5, where it is seen that the angular
momentum state for = 6 is the lowest value to contain the A1g
irreducible representation of the Ih group.
6.6 Characters for Other Symmetry Operators
in the Rotation Group
In dealing with the symmetry operations of the full rotation group, the
inversion operation frequently occurs. This operation also occurs in
the lower symmetry point groups either as a separate operation i or
in conjunction with other compound operations (e.g., S6 = i ⊗ C−1
3 ).
A compound operation (like an improper rotation or a mirror plane)
can be represented as a product of a proper rotation followed by inversion.
The character for the inversion operation is +1 for even angular
momentum states ( = even in Y ,m(θ, φ)) and −1 for odd angular momentum
states (see Eq. 6.3). This idea of compound operations will
become clearer after we have discussed in Chapter 7 the direct product
groups and direct product representations. A listing of the decompositions
of the spherical harmonics for various values into irreducible
representations of the icosahedral is given below (for ≤ 6) and in
Table 6.5 for higher values of ≤ 10.
Γ =0 −→ (Ag)Ih
Γ =1 −→ (F1u)Ih
Γ =2 −→ (Hg)Ih
Γ =3 −→ (F2u)Ih
+ (Gu)Ih
Γ =4 −→ (Gg)Ih
+ (Hg)Ih
Γ =5 −→ (F1u)Ih
+ (Gu)Ih
Γ =6 −→ (A1g)Ih
+ (Gg)Ih
(6.29)
6.6. CHARACTERS FOR OTHER SYMMETRY OPERATORS 125
Table 6.4: Splitting of angular momentum in cubic symmetry Oh.
A1g A2g Eg T1g T2g A1u A2u Eu T1u T2u
0 1
1 1
2 1 1
3 1 1 1
4 1 1 1 1
5 1 2 1
6 1 1 1 1 2
7 1 1 2 2
8 1 2 2 2
9 1 1 1 3 2
10 1 1 2 2 3
11 1 2 3 3
12 2 1 2 3 3
13 1 1 2 4 3
14 1 1 3 3 4
15 1 2 2 4 4
Table 6.5: Splitting of angular momentum in icosahedral symmetry Ih.
Ag T1g T2g Gg Hg Au T1u T2u Gu Hu
0 1
1 1
2 1
3 1 1
4 1 1
5 1 1 1
6 1 1 1 1
7 1 1 1 1
8 1 1 2
9 1 1 2 1
10 1 1 1 1 2
126CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
6.7 Selected Problems
1. Consider the hydrogen atom, described by the Schr¨odinger equa-
tion
HΨn m = −
¯h2
2m
2
r −
L2
r2
+ V (r) Ψn m = En Ψn m
(a) Does H commute with any arbitrary rotation about the origin?
Explain your answer.
(b) If the electron is in a d-orbital ( = 2) described by the
eigenfunction
Ψn2m(r, θ, φ) = Rn(r)Y2,m(θ, φ)
where Y2,m(θ, φ) is a spherical harmonic for = 2. What is
the effect on Ψn2m(r, θ, φ) of rotating the system by θ around
the origin? Is the new wave function still an eigenfunction
of the Hamiltonian with the same eigenvalue? Explain.
2. Suppose that an iron (Fe) impurity is introduced into a twodimensional
honeycomb lattice of an insulating host material.
A honeycomb lattice is a hexagonal lattice with atoms at the
hexagon corners but not at the center. Suppose that the Fe impurity
is placed first in a substitutional location and second in an
interstitial location at the center of the hexagon.
(a) What is the difference in crystal potential (include only nearest
neighbors) between the substitutional and interstitial lo-
cations?
6.7. SELECTED PROBLEMS 127
(b) For the interstitial case, express your result in part (a) in
terms of spherical harmonics for the lowest order terms with
angular dependencies.
(c) What is the proper point group symmetry and character
table in each case?
(d) Give the crystal field splitting of the 5-fold d-levels of the Fe
impurity in the crystal fields in part (a).
(e) Identify the basis functions associated with each of the levels
in part (d).
(f) Since the bonding orbitals lie lower in energy than the antibonding
orbitals, indicate how the ordering of the levels
might indicate whether the Fe impurity is located substitutionally
or interstitially in the honeycomb lattice.
3. Suppose that an iron (Fe) impurity is introduced into a substitutional
site in the high Tc superconductor YCu2Ba3O7. Assume
that the material is in the tetragonal phase (a = b = c).
(a) What is the difference in crystal potential (include only nearest
neighbors) between substitutional and interstitial sites?
(b) What is the proper point group symmetry and character
table for a single molecule of YCu2Ba3O7?
(c) Give the crystal field splitting of the 5-fold d-levels of the
Fe impurity in the crystal field in (a). Consider all three
cases regarding the relative size of the crystal field and the
spin-orbit interaction.
(d) Identify the basis functions associated with each of the levels
of the Fe impurity in (a).
(e) Suppose that you measure the reflectivity of this material
with polarized light and determine the energy levels. How
could you distinguish if the Fe impurity is in a Cu(1) site, a
Cu(2) site or an Y site, making use of the crystal symmetry?
4. Show (by finding the characters of the rotation group) that the
d-level for a transition metal impurity in a metal cluster with Ih
point symmetry is not split by the icosahedral crystal field.
128CHAPTER 6. APPLICATION TO CRYSTAL FIELD SPLITTING
Chapter 7
Application of Group Theory
to Selection Rules and Direct
Products
Our second general application of group theory to physical problems
will be to selection rules. In considering selection rules we always
involve some interaction H matrix that couples two states ψα and
ψβ. Group theory is often invoked to decide whether or not these
states are indeed coupled and this is done by testing whether or not
the matrix element (ψα, H ψβ) vanishes by symmetry. The simplest
case to consider is the one where the perturbation H does not destroy
the symmetry operations of the group of the Schr¨odinger equation.
Since these matrix elements transform as scalars (numbers),
(ψα, H ψβ) must exhibit the full group symmetry, and must transform
as the fully symmetric representation Γ1. These matrix elements of
the interaction Hamiltonian must be invariant under all the symmetry
operations of the group of Schr¨odinger’s equation. Thus, if (ψα, H ψβ)
does not transform as a number, it vanishes. To exploit these symmetry
properties, we thus choose ψ∗
α and ψβ to be symmetry eigenfunctions
for the unperturbed Hamiltonian - i.e., basis functions for irreducible
representations of the group of Schr¨odinger’s equation. We then determine
how H ψβ transforms - i.e., according to which irreducible representations
of the group. (This involves the direct product of two
representations and the theory behind the direct product of two repre-
129
130 CHAPTER 7. APPLICATION TO SELECTION RULES
sentations will be given in this chapter.) If H ψβ is orthogonal to ψα,
then the matrix element vanishes by symmetry; otherwise the matrix
element need not vanish, and the transition may occur.
In considering various selection rules that arise in physical problems,
we often have to consider matrix elements of a perturbation Hamiltonian
which lowers the symmetry of the unperturbed problem, as for
example Hem describing the electromagnetic field
Hem = −
e
mc
p · A. (7.1)
Such a perturbation Hamiltonian is generally not invariant under the
symmetry operations of the group of Schr¨odinger’s equation which is
determined by the unperturbed Hamiltonian H0. Therefore we must
consider the transformation properties of H ψβ where ψβ is an eigenfunction
that is chosen to transform as one partner ψ
(Γi)
j of an irreducible
representation Γi of the unperturbed Hamiltonian H0. In general,
the action of H on ψ
(Γi)
j will mix in all other partners of the representation
Γi since any arbitrary function can be expanded in terms
of a complete set of functions such as the ψ
(Γi)
j . In group theory, the
transformation properties of H ψ
(Γi)
j are handled through what is called
the direct product. Even though H need not transform as the totally
symmetric representation (e.g., Hem transforms as a vector x, y, z ), the
matrix element (ψi(Γi), H ψj(Γi)) may not vanish, since it may contain
a term that transforms as a scalar (number).
The discussion of the selection rules in this chapter is organized
around the following topics:
1. Summary of the important symmetry rules for basis functions
2. Theory of the Direct Product of Groups and Representations
3. The Selection Rule concept in Group Theoretical Terms
4. Example of Selection Rules for electric dipole transitions in a
system with Oh symmetry.
7.1. SUMMARY OF IMPORTANT RESULTS FOR BASIS FUNCTIONS131
7.1 Summary of Important Results for Basis
Functions
The basis functions ψ(i)
α for a given irreducible representation i are
defined by (see Eq. 4.1 of notes):
ˆPRψ(i)
α =
i
j=1
ψ
(i)
j D(i)
(R)jα (7.2)
where ˆPR is the symmetry operator, ψ(i)
α denotes the basis functions
for an li-dimensional irreducible representation (i) and D(i)
(R)jα is the
matrix representation for symmetry element R in irreducible representation
(i).
To exploit the symmetry properties of a given problem, we want to
find eigenfunctions which form basis functions for the irreducible representations
of the group of Schr¨odinger’s equation. We can find such
eigenfunctions using the symmetry operator and projection operator
techniques discussed in §4.1 and §4.3. In this chapter, we will then
assume that the eigenfunctions have been chosen to transform as irreducible
representations of the group of Schr¨odinger’s equation for H0.
The application of group theory to selection rules then depends on the
following orthogonality theorem. This orthogonality theorem can be
considered as the selection rule for the identity operator.
Theorem: Two basis functions which belong either to different irreducible
representations or to different columns (rows) of the same
representation are orthogonal.
Proof: Let φ(i)
α and ψ
(i )
α be two basis functions belonging respectively
to irreducible representations (i) and (i ) and corresponding to
columns α and α of their respective representations. By defini-
tion:
ˆPRφ(i)
α =
i
j=1
φ
(i)
j D(i)
(R)jα
ˆPRψ
(i )
α =
i
j =1
ψ
(i )
j D(i )
(R)j α . (7.3)
132 CHAPTER 7. APPLICATION TO SELECTION RULES
Because the scalar product (or the matrix element of unity taken
between the two states) is independent of coordinate system, we
can write the scalar product:
(φ(i)
α , ψ
(i )
α ) = ( ˆPR)φ(i)
α , ˆPRψ
(i )
α )
=
j,j
D(i)
(R)∗
jαD(i )
(R)j α (φ
(i)
j , ψ
(i )
j )
=
1
h j,j R
D(i)
(R)∗
jαD(i )
(R)j α (φ
(i)
j , ψ
(i )
j )(7.4)
since the left hand side of Eq. 7.4 is independent of R, and h is
the order of the group. Now apply the Wonderful Orthogonality
Theorem
1
h R
D(i)
(R)∗
jαD(i )
(R)j α =
1
i
δii δjj δαα (7.5)
to Eq. 7.4, which yields:
(φ(i)
α , ψ
(i )
α ) =
1
i
δi,i δα,α
i
j=1
(φ
(i)
j , ψ
(i)
j ). (7.6)
Thus according to Eq. 7.6, if the basis functions φ(i)
α and ψ
(i )
α
correspond to two different irreducible representations i = i they
are orthogonal. If they correspond to the same representation
(i = i ), they are still orthogonal if they correspond to different
columns (or rows) – i.e., if they correspond to different partners.
We further note that the right hand side of Eq. 7.6 is independent
of α so that the scalar product is the same for all components
α, thereby completing the proof.
In general, selection rules deal with the matrix elements of an operator
different from the identity operator. Clearly if the operator is invariant
under all of the symmetry operations of the group of Schr¨odinger’s
equation then it transforms like the identity operator. For example, if
H0ψ
(i )
α = E
(i )
α ψ
(i )
α (7.7)
7.2. DIRECT PRODUCT OF TWO GROUPS 133
then E
(i )
α is a number which is independent of any coordinate system.
If ψ
(i )
α and φ(i)
α are both eigenfunctions of the Hamiltonian H0 and are
also basis functions for irreducible representations (i ) and (i), then
the matrix element (φ(i)
α , H0ψ
(i )
α ) vanishes unless i = i and α = α ,
which is a result familiar to us from quantum mechanics. In the more
general case when we have a perturbation H , the perturbation need not
have the full symmetry of H0. In general H ψ transforms differently
from ψ.
7.2 Direct Product of Two Groups
We now define the direct product of two groups. Let GA =
E,A2, . . ., Aha and GB = E, B2, . . . , Bhb
be two groups such that all
operators AR commute with all operators BS. Then the direct product
group is
GA ⊗ GB = E, A2, . . . , Aha , B2, A2B2, . . . , Aha B2, . . . , Aha Bhb
(7.8)
and has (ha × hb) elements. It is easily shown that if GA and GB are
groups, then the direct product group GA ⊗GB is a group. Examples of
direct product groups that are frequently encountered involve products
of groups with the group of inversions (group Ci with two elements E, i)
and reflections (group Cσ with two elements E, σ). For example, we can
make a direct product group D3d from the group D3 by compounding
all the operations of D3 with (E, i) where i is the inversion operation
(see Table 3.30). An example of the group D3d is a triangle with finite
thickness. We write the direct product group when compounding the
initial group with the inversion operation
D3d = D3 ⊗ i (7.9)
or with the mirror reflection in a horizontal plane (see Table 3.31):
D3h = D3 ⊗ σ. (7.10)
The full cubic group Oh is a direct product group of O ⊗ i.
134 CHAPTER 7. APPLICATION TO SELECTION RULES
7.3 Direct Product of Two Irreducible Rep-
resentations
In addition to direct product groups we have the direct product of
two representations which is defined in terms of the direct product
of two matrices. From algebra, we have the definition of the direct
product of two matrices A ⊗ B = C, whereby every element of A is
multiplied by every element of B. Thus, the direct product matrix C
has a double set of indices
AijBk = Cik,j (7.11)
Thus, if A is a (2 × 2) matrix and B is a (3 × 3) matrix, then C is a
(6 × 6) matrix.
Theorem: The direct product of the representations of the groups A
and B forms a representation of the direct product group.
Proof: To prove this theorem we need to show that
Da⊗b
(AkB ) Da⊗b
(Ak B ) = Da⊗b
(AiBj) (7.12)
where
Ai = AkAk Bj = B B . (7.13)
Since the elements of group A commute with those of group B
by the definition of the direct product group, the multiplication
property of elements in the direct product group is
AkB Ak B = AkAk B B = AiBj (7.14)
where AkB is a typical element of the direct product group. We
must now show that the representations reproduce this multiplication
property. By definition:
Da⊗b
(AkB ) Da⊗b
(Ak B ) = D(a)
(Ak)⊗D(b)
(B ) D(a)
(Ak )⊗D(b)
(B ) .
(7.15)
7.4. CHARACTERS FOR THE DIRECT PRODUCT OF GROUPS AND REPRESENTATIONS135
To proceed with the proof, we write Eq. (7.15) in terms of com-
ponents:
Da⊗b
(AkB )Da⊗b
(Ak B )
ip,jq
= sr D(a)
(Ak) ⊗ D(b)
(B )
ip,sr
× D(a)
(Ak ) ⊗ D(b)
(B )
sr,jq
= s D
(a)
is (Ak)D
(a)
sj (Ak ) × r D(b)
pr (B )D(b)
rq (B )
= D
(a)
ij (Ai)D(b)
pq (Bj) = D(a⊗b)
(AiBj)
ip,jq
(7.16)
This completes the proof.
It can be further shown that the direct product of two irreducible
representations of groups GA and GB yields an irreducible representation
of the direct product group so that all irreducible representations
of the direct product group can be generated from the irreducible representations
of the original groups before they are joined. We can also
take direct products between 2 representations of the same group. Essentially
the same proof as given in this section shows that the direct
product of two representations of the same group is also a representation
of that group, though in general, it is a reducible representation.
The proof proceeds by showing
[D( 1⊗ 2)
(A)D( 1⊗ 2)
(B)]ip,jq = D( 1⊗ 2)
(AB)ip,jq (7.17)
where 1 and 2 denote irreducible representations with the corresponding
dimensionalities. The direct product representation D( 1⊗ 2)
(R) will
in general be reducible even though 1 and 2 are irreducible.
7.4 Characters for the Direct Product of
Groups and Representations
In this section we find the characters for direct product groups and
representations.
136 CHAPTER 7. APPLICATION TO SELECTION RULES
Theorem: The simplest imaginable formulae are assumed by the characters
in direct product groups or in taking the direct product of
two representations.
1. If the direct product occurs between two groups, then the
characters for the irreducible representations in the direct
product group are obtained by multiplication of the characters
of the irreducible representations of the original groups
according to:
χ(a⊗b)
(AkB ) = χ(a)
(Ak) χ(b)
(B ) (7.18)
2. If the direct product is taken between two representations
of the same group, then the character for the direct product
representation is written as
χ( 1⊗ 2)
(R) = χ( 1)
(R) χ( 2)
(R). (7.19)
Proof: Consider the diagonal matrix element of an element in the direct
product group. From the definition of the direct product of
two groups, we write
D(a⊗b)
(AkB )ip,jq = D
(a)
ij (Ak)D(b)
pq (B ). (7.20)
Taking the diagonal matrix elements of Eq. 7.20 and summing
over these matrix elements, we obtain
ip
D(a⊗b)
(AkB )ip,ip =
i
D
(a)
ii (Ak)
p
D(b)
pp (B ) (7.21)
which can be written in terms of the traces:
χ(a⊗b)
(AkB ) = χ(a)
(Ak)χ(b)
(B ). (7.22)
This completes the proof of the theorem for the direct product of two
groups.
The result of Eq. 7.22 holds equally well for classes (i.e., R → C),
and thus can be used to find the character tables for direct product
groups as is explained below.
7.4. CHARACTERS FOR THE DIRECT PRODUCT OF GROUPS AND REPRESENTATIONS137
Exactly the same proof as given above can be applied to find for
the direct product of two representations of the same group
χ( 1⊗ 2)
(R) = χ( 1)
(R)χ( 2)
(R). (7.23)
The direct product representation is irreducible only if χ( 1⊗ 2)
is identical
to one of the irreducible representations of the group 1 ⊗ 2. In
general, if we take the direct product between 2 irreducible representations
of a group, then the resulting direct product representation will
be reducible. If it is reducible, the direct product can then be written
as a linear combination of the irreducible representations of the group:
χ(λ)
(R)χ(µ)
(R) =
ν
aλµνχ(ν)
(R) (7.24)
where from Eq. 3.20 we can write the coefficients aλµν as:
aλµν =
1
h Cα
NCα χ(λ
)(Cα)χ(µ)
(Cα) χ(ν)
(Cα)∗
(7.25)
where Cα denotes classes and NCα denotes the number of elements in
class Cα. In applications of group theory to selection rules, constant
use is made of Eqs. 7.24 and 7.25.
Finally we use the result of Eq. 7.22 to show how the character
tables for the original groups GA and GB are used to form the character
table for the direct product group. First we form the elements
and classes of the direct product group and then we use the character
tables of GA and GB to form the character table for GA ⊗GB. In many
important cases one of the groups (e.g., GB) has only two elements
(such as the group Ci with elements E, i) and two irreducible representations
Γ1 with characters (1,1) and Γ1 with characters (1, −1). We
illustrate such a case below for the direct product group C4h = C4 ⊗ i.
In the character table for group C4h shown below we use the notation g
to denote representations that are even (German, gerade) under inversion,
and u to denote representations that are odd (German, ungerade)
under inversion.
138 CHAPTER 7. APPLICATION TO SELECTION RULES
C4h ≡ C4 ⊗ i (4/m)
E C2 C4 C3
4 i iC2 iC4 iC3
4
Ag 1 1 1 1 1 1 1 1
Bg 1 1 −1 −1 1 1 −1 −1 even under
Eg
1
1
−1
−1
i
−i
−i
i
1
1
−1
−1
i
−i
−i
i
inversion (g)
Au 1 1 1 1 −1 −1 −1 −1
Bu 1 1 −1 −1 −1 −1 1 1 odd under
Eu
1
1
−1
−1
i
−i
−i
i
−1
−1
1
1
−i
i
i
−i
inversion (u)
We note that the upper left hand quadrant contains the character table
for the group C4. The 4 classes obtained by multiplication of the
classes of C4 by i are listed on top of the upper right columns. The
characters in the upper right hand and lower left hand quadrants are
the same as in the upper left hand quadrant, while the characters in
the lower right hand quadrant are all multiplied by (−1) to produce
the odd (ungerade) irreducible representations.
7.5 The Selection Rule Concept in Group
Theoretical Terms
Having considered the background for taking direct products, we are
now ready to consider the selection rules for the matrix element
(ψ
(i )
α , H φ(i)
α ). (7.26)
This matrix element can be computed by integrating the indicated
scalar product over all space. Group theory then tells us that when
any or all the symmetry operations of the group are applied, this matrix
element must transform as a constant. Conversely, if the
matrix element is not invariant under the symmetry operations which
form the group of Schr¨odinger’s equation, then the matrix element must
vanish. We will now express the same physical concepts in terms of the
direct product formalism.
7.5. THE SELECTION RULE CONCEPT IN GROUP THEORETICAL TERMS139
Let the wave functions φ(i)
α and ψ
(i )
α transform, respectively, as partners
α and α of irreducible representations Γi and Γi , and let H transform
as representation Γj. Then if the direct product Γj ⊗Γi is orthogonal
to Γi the matrix element vanishes, or equivalently if Γi ⊗ Γj ⊗ Γi
does not contain the fully symmetrical representation Γ1, the matrix
element vanishes. In particular, if H transforms as Γ1 (i.e., the perturbation
does not lower the symmetry of the system), then φ(i)
α and ψ
(i )
α
must correspond to the same irreducible representation and the same
partners of that representation because of the orthogonality theorem
for basis functions.
To illustrate the meaning of these statements for a more general
case, we will apply these selection rule concepts to the case of electric
dipole transitions in §7.6 below. First we express the perturbation H
(in this case due to the electromagnetic field) in terms of the irreducible
representations that H contains in the group of Schr¨odinger’s equation:
H =
j,β
f
(j)
β H
(j)
β (7.27)
where j denotes the irreducible representations Γj of the Hamiltonian
H and β denotes the partners of Γj. Then H φ(i)
α transforms as the
direct product representation formed by taking direct products H
(j)
β ⊗
φ(i)
α in accordance with Eq. 7.27. The matrix element (ψ
(i )
α , H φ(i)
α )
vanishes if and only if ψ
(i )
α is orthogonal to all the basis functions that
occur in the decomposition of H φ(i)
α into irreducible representations.
An equivalent expression of the same concept is obtained by considering
the triple direct product ψ
(i )
α ⊗ H
(j)
β ⊗ φ(i)
α . In order for the matrix
element in Eq. 7.26 to be non-zero, this triple direct product must
contain a term that transforms as a scalar or a constant number – i.e.,
according to the irreducible representation Γ1.
140 CHAPTER 7. APPLICATION TO SELECTION RULES
7.6 Example of Selection Rules for Electric
Dipole Transitions in a System
with Oh Symmetry
The electromagnetic interaction giving rise to electric dipole transitions
is
Hem = −
e
2mc
p · A (7.28)
in which p is the momentum of the electron and A is the vector potential
of an external electromagnetic field. The momentum operator
is part of the physical “system” under consideration while the vector
A acts like the “bath” or “reservoir” in a thermodynamic sense. Thus
p acts like an operator with respect to Schr¨odinger’s equation but A
does not. Therefore, in terms of group theory, Hem for the electromagnetic
interaction transforms like a vector in the context of the group of
Schr¨odinger’s equation for the unperturbed system H0ψ = Eψ. If we
have unpolarized radiation, we must then consider all three components
of the vector p (i.e., px, py, pz). In cubic symmetry, all 3 components of
the vector transform as the same irreducible representation. If instead
we had a system which exhibits tetragonal symmetry, then px and py
would transform as one of the two-dimensional irreducible representations
and pz would transform as one of the one-dimensional irreducible
representations.
To find the particular irreducible representations that are involved
in cubic symmetry we consult the character table for Oh = O ⊗ i.
In the cubic group Oh the vector (x, y, z) transforms according to the
irreducible representation T1u and so does (px, py, pz), because both are
radial vectors and both are odd under inversion. We note that the
character table for Oh (Table 3.33) gives the irreducible representation
for vectors, and the same is true for most of the other character tables
in Chapter 3. To obtain the character table for the direct product
group Oh = O ⊗ i we note that each symmetry operation in O is also
compounded with E and i to yield 48 symmetry operations and 10
classes.
For the Oh group there will then be 10 irreducible representations,
5 of which are even and 5 are odd. For the even irreducible represen-
7.6. SELECTION RULES FOR ELECTRIC DIPOLE TRANSITIONS141
Table 7.1: Character Table for O (432)
O(432) E 8C3 3C2 = 3C2
4 6C2 6C4
A1 1 1 1 1 1
A2 1 1 1 −1 −1
(x2
− y2
, 3z2
− r2
) E 2 −1 2 0 0
(Rx, Ry, Rz)
(x, y, z)
T1 3 0 −1 −1 1
(xy, yz, zx) T2 3 0 −1 1 −1
Oh = O ⊗ i
tations, the same characters are obtained for class C and class iC. For
the odd representations the characters for C and iC have opposite signs.
Even representations are denoted by the subscript g and odd representations
by the subscript u. The radial vector p transforms as an odd
irreducible representation since p → –p under inversion.
To find selection rules we must also specify the initial and final
states. For example, if the system is initially in a state with symmetry
T2g then the direct product Hem ⊗ ψT2g contains the irreducible
representations found by taking the direct product χT1u ⊗ χT2g . The
characters for χT1u ⊗ χT2g are given below:
E 8C3 3C2 6C2 6C4 i 8iC3 3iC2 6iC2 6iC4
9 0 1 −1 −1 –9 0 −1 1 1
We consider χT1u ⊗ χT2g as a reducible representation of the group Oh.
Then using the decomposition formula Eq. 7.25 yields:
T1u ⊗ T2g = A2u + Eu + T1u + T2u. (7.29)
Thus we obtain the selection rules that electric dipole transitions to
a state T2g can only be made from states with A2u, Eu, T1u, and T2u
symmetry. Furthermore, since Hem is an odd function, electric dipole
transitions will couple only states with opposite parity. The same arguments
as given above can be used to find selection rules between any
142 CHAPTER 7. APPLICATION TO SELECTION RULES
initial and final states for the case of cubic symmetry. For example,
from Table 7.1, we can write the following direct products as:
Eg ⊗ T1u = T1u + T2u
T1u ⊗ T1u = A1g + Eg + T1g + T2g
.
Suppose that we now consider the situation where we lower the
symmetry from Oh to D4h. Referring to the character table for D4 in
Table 3.26 and below, we can form the direct product group D4h by
taking the direct product D4h = D4 ⊗ i.
D4 (422) E C2 = C2
4 2C4 2C2 2C2
x2
+ y2
, z2
A1 1 1 1 1 1
Rz, z A2 1 1 1 −1 −1
x2
− y2
B1 1 1 −1 1 −1
xy B2 1 1 −1 −1 1
(xz, yz)
(x, y)
(Rx, Ry)
E 2 −2 0 0 0
We note here the important result that the vector in D4h = D4 ⊗ i
symmetry does not transform as a single irreducible representation but
rather as the irreducible representations:
z → A2u
(x, y) → Eu
so that T1u in Oh symmetry goes into: A2u + Eu in D4h symmetry.
Furthermore a state with symmetry T2g in the Oh group goes into
states with Eg + B2g symmetries in D4h (see discussion in §6.4). Thus
for the case of the D4h group, electric dipole transitions will only couple
an A1g state to states with Eu and A2u symmetries. For a state with Eg
symmetry according to group D4h the direct product with the vector
yields
Eg ⊗ (A2u + Eu) = Eu + (A1u + A2u + B1u + B2u), (7.30)
so that for the D4h group, electric dipole transitions from an Eg state
can be made to any odd parity state. This analysis points out that as
7.6. SELECTION RULES FOR ELECTRIC DIPOLE TRANSITIONS143
we reduce the amount of symmetry, there are fewer selection rules and
more transitions become allowed.
Polarization effects also are significant when considering selection
rules. For example, if the electromagnetic radiation is polarized along
the z direction in the case of the D4h group, then the electromagnetic
interaction involves only pz which transforms according to A2u. With
the pz polarization, the following states are coupled by electric dipole
radiation (i.e., by matrix elements of pz):
initial state final state
A1g A2u
A2g A1u
B1g B2u
B2g B1u
Eg Eu
A1u A2g
A2u A1g
B1u B2g
B2u B1g
Eu Eg
If, on the other hand, the radiation is polarized in the x direction, then
the basis function is a single partner x of the Eu representation. Then if
the initial state has A1g symmetry, the electric dipole transition will be
to a state which transforms as the x partner of the Eu representation. If
the initial state has A2u symmetry (transforms as z), then the general
selection rule gives A2u ⊗ Eu = Eg while polarization considerations
indicate that the transition couples the A2u level with the xz partner of
the Eg representation. If the initial state has Eu symmetry, the general
selection rule gives
(Eu ⊗ Eu) = A1g + A2g + B1g + B2g. (7.31)
The polarization x couples the partner Ex
u to Ax2+y2
1g and Bx2−y2
1g while
the partner Ey
u couples to Axy−yx
2g and Bxy
2g . Thus polarization effects
further restrict the states that are coupled in electric dipole transitions.
If the polarization direction is not along one of the (x, y, z) directions,
144 CHAPTER 7. APPLICATION TO SELECTION RULES
Hem will transform as a linear combination of the irreducible representations
A2u + Eu even though the incident radiation is polarized.
Selection rules can be applied to a variety of perturbations H other
than the electric dipole interactions, such as uniaxial stress, hydrostatic
pressure and the magnetic dipole interaction. In these cases, the special
symmetry of H in the group of Schr¨odinger’s equation must be
considered.
7.7 Selected Problems
1. (a) The (2 × 2) matrices B and C form the direct product A =
B ⊗ C, where
B =
b11 b12
b21 b22
and C =
c11 c12
c21 c22
to give a 4 × 4 matrix labeled A.
(b) Show that if GA with elements E, A2, . . . , Aha and GB with
elements E, B2, . . . , Bhb
are groups, then the direct product
group GA ⊗ GB is also a group. Use the notation BijCkl =
(B⊗C)ik,jl to label the rows and columns of the direct product
matrix.
(c) In going from higher to lower symmetry, if the inversion operation
is preserved, show that even representations remain
even and the odd representations remain odd.
2. (a) Consider electric dipole transitions in full cubic Oh symmetry
for transitions between an initial state with A1g symmetry
(s-state) and a final state with T1u symmetry (p-state).
[Note that one of these electric dipole matrix elements is
proportional to a term (1|px|x), where |1) denotes the sstate
and |x) denotes the x partner of the p-state.] Of the 9
possible matrix elements that can be formed, how many are
non-vanishing? Of those that are non-vanishing, how many
are equivalent?
7.7. SELECTED PROBLEMS 145
(b) If the initial state has Eg symmetry (rather than A1g symmetry),
repeat part (a). You will find it convenient to use as basis
functions for the Eg level the two partners x2
+ωy2
+ω2
z2
and x2
+ ω2
y2
+ ωz2
where ω = exp(2πi/3).
(c) Repeat part (a) for the case of electric dipole transitions from
an s-state to a p-state in tetragonal D4h symmetry. Consider
the light polarized first along the z direction and then in the
x − y plane. Note that as the symmetry is lowered, the
selection rules become less stringent.
146 CHAPTER 7. APPLICATION TO SELECTION RULES
Chapter 8
Electronic States of
Molecules and Directed
Valence
In this chapter we consider the electronic states of molecules, the formation
of molecular bonds and the simplifications that are introduced
through the use of group theory. We organize our discussion in this
chapter in terms of a general discussion of molecular energy levels; the
general concept of equivalence; the concept of directed valence bonding;
the application of the directed valence bond concept to various
molecules, including H2, CO, NH3, CH4, “SH6”, SF6, and B12H12; bond
strengths in directed valence bonds; and finally σ and π bonding.
8.1 Introduction
The energy levels of molecules are basically more complicated than
those for atoms because:
1. there are several centers of positive charge which serve to attract
a given electron,
2. these center are themselves in relative motion.
Since the nuclei are very massive relative to the electrons, we can
utilize the Born-Oppenheimer approximation which separates out the
147
148 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
electronic motion from the nuclear or ionic motion. In this approximation,
the electrons move in a potential generated by the equilibrium
positions of the nuclei. As for the nuclei, they can be involved in some
form of relative motion, giving rise to molecular vibrations in the isolated
molecules or to lattice modes in the solid. Also possible in the
case of isolated molecules are molecular rotations or translations. Since
the translational motion corresponds to plane wave solutions and the
eigenvalues form a continuous spectrum, these solutions do not give rise
to molecular spectra and need not be considered further. We are thus
left with 3 kinds of molecular motion:
1. electronic-most energetic
2. vibrational-less energetic
3. rotational-least energetic
If these motions are in fact independent and can be decoupled (this
is not always the case), then we can write for the wave functions and
the energies as:
ψtotal = ψelectronic × ψvibrational × ψrotational (8.1)
and
Etotal = Eelectronic + Evibrational + Erotational. (8.2)
In this chapter we consider the electronic energy levels of some typical
molecules, and in Chapter 9 we consider the vibrational and rotational
levels of molecules
The effective one-electron potential V (r) for an electron in a molecule
must be invariant under all symmetry operations which leave the molecule
invariant. If we did not exploit the symmetry explicitly through group
theory, we would then solve the Schr¨odinger equation to find the energy
eigenvalues and the corresponding eigenfunctions of the molecule
taking into account all the valence electrons for all the atoms in the
molecule. This would require solution of a large secular equation of the
form:
| ψi|H|ψj − Eδij| = 0. (8.3)
8.1. INTRODUCTION 149
The utilization of symmetry (as for example using group theoretical
methods) allows us to choose our basis functions wisely, so that many
of the matrix elements in the secular equation vanish through symmetry
arguments and the secular equation breaks up into block diagonal
form. Thus by using symmetry, we have to solve much smaller secular
equations for only those states which transform according to the
same irreducible representations, because it is only those states of like
symmetry that are coupled in the secular equation.
Group theory is used in yet another way for solving the electronic
problem. Many molecules contain more than one equivalent atom.
Symmetry is used to simplify the secular equation by forming linear
combinations of atomic orbitals that transform according to the irreducible
representations of the group of Schr¨odinger’s equation. Using
such linear combinations of atomic orbitals, the secular equation can
more readily be brought into block diagonal form. In this chapter we
show how to form linear combinations of atomic orbitals that transform
as irreducible representations of the appropriate symmetry group,
and we will show how the equivalence concept is used in forming these
linear combinations.
In the free atom, the electronic orbitals display the symmetry of a
(1/r) potential, and therefore the free-atom orbitals are eigenfunctions
which transform according to irreducible representations of the full rotation
group. In a molecule or in a solid, the electrons tend to spend
more time between the ion cores in the bonding state and the increased
probability of finding the electron between two nuclei (see Fig. 8.1) is
called a chemical bond. These bonds display the known symmetry of
the molecule (or the solid). For this reason, the wavefunctions for the
electrons in the molecule (or the solid) transform as irreducible representations
of the appropriate symmetry group, which in general will be
of lower symmetry than the full rotation group. From elementary considerations,
we know that molecular bonds arise from the exchange
interaction whose magnitude depends on the extent of the overlap
of the charge clouds between neighboring atoms. Because these orbitals
concentrate the charge along preferred directions, the bonding is
called directed valence bonding, and exhibits the symmetry of the
molecule or of the solid. We use the directed valence bonding concepts
to identify the kind of symmetries needed to make the desired orbitals.
150 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
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Figure 8.1: Electronic wave functions for a diatomic molecule. The
formation of bonding and antibonding states is indicated. To find the
energy splitting between the bonding and antibonding states (indicated
schematically), the solution of Schr¨odinger’s equation is necessary.
8.2. GENERAL CONCEPT OF EQUIVALENCE 151
Symmetry enters the electronic problem of molecules in yet another
way, namely through the Pauli principle and the effect of permutation
of the electrons on the electron wavefunctions. This topic is discussed
in Chapter 10 for many-electron states.
8.2 General Concept of Equivalence
Equivalent bonding orbitals are required to transform into
one another under all the symmetry operations of the point
group with no more change than a possible change of phase.
The transformation which takes one equivalent function into another
generates a representation for the point group called the equivalence
transformation. The equivalence representation will in general be
reducible. We denote the representation that generates the transformation
between equivalent atom sites by Γatom sites
and its characters
by χatom sites
. (To save space, we sometimes use the abbreviation a.s. ≡
atomic sites.) In this section we present the equivalence concept, show
how to find the irreducible representations contained in the equivalence
representation (i.e., χatom sites
) and then give a few examples.
The matrices D(atom sites)
(R)ji for the equivalence representation Γatom sites
are found from the general definition
ˆPRψi =
j
ψjD(atom sites)
(R)ji (8.4)
or written in matrix form
ˆPR(ψ1, ψ2, . . . ψn) = (ψ1, ψ2, . . . ψn)(D(atom sites)
(R)). (8.5)
Explicitly, the D(atom sites)
(R)ij matrices are found by entering unity into
the i, j position in the matrix if ˆP(R) takes site i into an equivalent
site j and zero otherwise. From this argument we readily see that the
characters for the equivalence representation can be found by counting
the number of points which are left unaffected by the symmetry operation,
because it is only those points that will give a contribution to
the matrix on diagonal positions and contribute to the character for
that representation. To obtain the characters for χatom sites
we take a
152 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
representative member of each class and consider the number of points
that are left unchanged under action of the representative symmetry
operator.
The representation Γatom sites
is in general reducible. The pertinent
symmetry types for the problem are then found by decomposing
Γatom sites
into its irreducible representations. To illustrate this concept,
consider the example of 3 identical atoms at the corners of an equilateral
triangle as for example the 3 hydrogen atoms in the NH3 molecule. The
symmetry group is C3v, and the character table is given in Table 8.2.
Referring to Fig. 4.2, where the 3 equivalent sites are labeled by (a, b, c)
we obtain D(atom sites)
(R) for some typical symmetry operators:
D(a.s.)
(E) =



1 0 0
0 1 0
0 0 1


 (8.6)
D(a.s.)
(C3) =



0 1 0
0 0 1
1 0 0


 (8.7)
D(a.s.)
(σv) =



1 0 0
0 0 1
0 1 0


 (8.8)
in which the rows and columns correspond to the sequence of atoms
(a, b, c). From these matrices we can compute the characters for each of
the classes for the Γatom sites
representation in group C3v(3m). The character
χatom sites
(R) is always the number of sites that are left unchanged
by the operation ˆPR so that χatom sites
(E) = 3, χatom sites
(C3) = 0 and
χatom sites
(σv) = 1. These results are summarized below and
E 2C3 3σv
Γatom sites
3 0 1 ⇒ Γ1 + Γ2 = A1 + E
from Table 8.2 we see immediately that Γatom sites
= Γ1 + Γ2, in agreement
with the explicit orbitals found in §4.6. The orbitals on the nitrogen
atom are then chosen to bond to the atomic orbitals of the 3
hydrogen atoms as discussed in §8.5.1.
8.3. DIRECTED VALENCE BONDING 153
Table 8.1: Character Table for the Group C1h
C1h(m) E σh
x2
, y2
, z2
, xy Rz, x, y A (Γ1) 1 1
xz, yz Rx, Ry, z A (Γ1) 1 −1
χatom sites
2 0 ⇒ Γ1 + Γ1 ≡ A + A
8.3 Directed Valence Bonding
For diatomic molecules we know immediately, without recourse to group
theory, how to make a bond out of the atomic orbitals. We need simply
to take the symmetrical combination (ψa + ψb) to pile up charge in the
directed valence bond (see Fig. 8.1).
For the case of the homopolar diatomic molecule, we thus form
a bonding state (ψa + ψb) and an antibonding state of higher energy
(ψa − ψb) which is generally unoccupied. Suppose that this diatomic
molecule only has 2 symmetry operations, the identity E and the mirror
plane reflections m. These are the 2 symmetry elements of the group
C1h. (In §8.4 we will consider the semi-infinite groups D∞h and C∞v
which give the full symmetry of typical homogeneous and heterogeneous
diatomic molecules.) Taking ψa as an arbitrary function, and noting
that ˆPmψa = ψb, the projection operator for 1-dimensional irreducible
representations (see Eq. 4.38) can be written as:
ˆP(Γn)
=
ln
h R
χ(Γn)
(R)∗ ˆPR. (8.9)
The basic formula (Eq. 8.9) for finding linear combinations of atomic
orbitals when acting on the wave function ψa yields:
ˆP(Γ1)
ψa=1
2
[(1) ˆPEψa + (1) ˆPmψa] = 1
2
[ψa + ψb] bonding
ˆP(Γ1)
ψa=1
2
[(1) ˆPEψa + (−1) ˆPmψa] = 1
2
[ψa − ψb] antibonding
(8.10)
for the bonding and anti-bonding states, so that the bonding orbitals
will have Γ1 symmetry and the antibonding orbitals Γ1 symmetry. Since
154 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
there are only two initial wave functions ψa and ψb, the combinations
in Eq. 8.10 are all the independent linear combinations that can be
formed.
Our discussion of the use of projection operators (see §4.5 and §4.6)
illustrated how linear combinations of atomic orbitals could be found
such that the resulting orbitals transform according to irreducible representations
of the point group. This process is simplified by using the
directed valence representation ΓD.V. which introduces two kinds
of simplifications:
1. ΓD.V. gives all the irreducible representations for the molecular
orbitals before the molecular orbitals are found explicitly. This
saves time because the projection operator ˆP(Γn)
need not then be
applied to irrelevant representations, but only to those irreducible
representations contained in ΓD.V..
2. If we are only interested in finding the number of distinct eigenvalues
and their degeneracies, this follows directly from the characters
χD.V. of the representation ΓD.V.. To obtain this kind of
information it is not necessary to solve Schr¨odinger’s equation or
even to find the linear combination of molecular orbitals as in
§4.3.
8.4 Diatomic Molecules
8.4.1 Homonuclear Diatomic Molecules in General
The simplest molecules are the homonuclear diatomic molecules. For
homonuclear molecules (such as H2) the appropriate symmetry group
is D∞h and the character table for D∞h is shown below. We now summarize
the main points about this character table.
8.4. DIATOMIC MOLECULES 155
D∞h (∞/mm) E 2Cφ C2 i 2iCφ iC2
x2
+ y2
, z2
A1g(Σ+
g ) 1 1 1 1 1 1
A1u(Σ−
u ) 1 1 1 −1 −1 −1
Rz A2g(Σ−
g ) 1 1 −1 1 1 −1
z A2u(Σ+
u ) 1 1 −1 −1 −1 1
(xz, yz) (Rx, Ry) E1g(Πg) 2 2 cos φ 0 2 2 cos φ 0
(x, y) E1u(Πu) 2 2 cos φ 0 −2 −2 cos φ 0
(x2
− y2
, xy) E2g(∆g) 2 2 cos 2φ 0 2 2 cos 2φ 0
E2u(∆u) 2 2 cos 2φ 0 −2 −2 cos 2φ 0
...
...
...
...
...
...
...
Cφ denotes an arbitrary rotation about the linear molecular axis (zaxis)
and C2 is a two-fold axis ⊥ to Cφ. In the group D∞h, each of the
operations E, Cφ and C2 is also combined with inversion. We further
note that σv is a plane through the molecular axis, so that σv = iC2 .
The subscripts g and u refer to the evenness and oddness of functions
under the inversion operation while the superscripts + and – refer to the
evenness and oddness of functions under reflection in a mirror plane.
The characters for σv in the D∞h group are found most conveniently by
considering the effect of the operation σv on the basis functions which
correspond to a given irreducible representation —e.g., σv changes (x, y)
into (−x, y) yielding a transformation matrix
D(σv) =
−1 0
0 1
(8.11)
and the corresponding character for σv is χ(σv) = 0 for the E1u irreducible
representation.
For a homogeneous diatomic molecule (such as H2) we have the following
characters for the equivalence transformation:
E 2Cφ C2 = iσv i 2iCφ iC2 = σv
χatom sites
2 2 0 0 0 2
⇒ A1g + A2u
⇒ Σ+
g + Σ+
u
When forming LCAO from s functions on the two equivalent atomic
156 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
sites (see §8.3), the normalized bonding orbital ψS = (ψa + ψb)/
√
2
has Σ+
g or A1g symmetry and the normalized antibonding orbital ψA =
(ψa − ψb)/
√
2 has Σ+
u or A2u symmetry. Using the equivalence concept
in §8.3, we can construct a linear combination of atomic orbitals which
transform as irreducible representations of the group of Schr¨odinger’s
equation. Thus ψS and ψA form such basis functions and the Hamiltonian
for the homogeneous diatomic molecule will not couple states ψS
and ψA to each other. This follows from the argument that the product
(HψS) transforms as A1g since H transforms as A1g and so does
ψS; also ψA transforms as A2u. The selection rules thus tell us that
the matrix element (ψA|H|ψS) must vanish. Thus to bring the secular
equation into block diagonal form, we have to make a unitary transformation
on the atomic basis functions (ψa, ψb) to bring them into the
form (ψS, ψA):
ψS
ψA
= U
unitary matrix
ψa
ψb
=
1√
2
1√
2
1√
2
− 1√
2
ψa
ψb
. (8.12)
Applying the unitary transformation UHU†
to the original matrix (written
in terms of the original ψa and ψb) will bring the secular matrix into
block diagonal form. Bringing the secular equation into block diagonal
form greatly simplifies the solution of the secular equation – in this
simple case from a coupled (2 × 2) secular equation to two decoupled
(1 × 1) secular equations.
8.4.2 The Hydrogen Molecule H2
In this case we can put each electron in a (σg1s) orbital and construct
bonding and antibonding orbitals. For H2, the bonding orbital σg is
occupied with electrons having opposite spin states and the antibonding
σu orbital is unoccupied. The (σg1s) state is symmetric under
both inversion i and reflection σv. Hence the symmetry for each of
the separated atoms is Σ+
g so that the symmetry for the molecule is
Σ+
g ⊗ Σ+
g = Σ+
g . We write this state as 1
Σ+
g where the superscript 1
denotes a singlet (s = 0) with a total spin degeneracy of (2s + 1) = 1.
By making spatial bonding orbitals that are symmetric under exchange
8.4. DIATOMIC MOLECULES 157
of the electrons, the spin state must be antisymmetric:
1
√
2
[α(1)β(2) − α(2)β(1)] . (8.13)
8.4.3 The Helium Molecule He2
Suppose we could make a bound diatomic molecule out of 2 helium
atoms and containing 4 electrons. This molecule would have, on the
separated atom model, a configuration (σg1s)2
(σu1s)2
and would be in
a 1
Σ+
g state, all spins being antiparallel in pairs. Here σu denotes the
antibonding orbital for the 1s states. In this case, both the bonding
and antibonding states need to be occupied to make a He2 molecule
and the resulting symmetry is Σ+
g ⊗ Σ+
g ⊗ Σ+
u ⊗ Σ+
u = Σ+
g . Since the
He2 molecule is not formed under ordinary circumstances we know that
the antibonding state lies sufficiently high in energy so that it is not
energetically favorable to form the He2 molecule. On the other hand,
H−
2 involves occupation of an antibonding state and does indeed form
a bound state. Group Theory gives us the symmetry designation for
each molecule, but does not give definitive information as to whether
or not a bound state is formed.
8.4.4 Heterogeneous Diatomic Molecules
We illustrate the case of a linear heterogeneous diatomic molecule with
CO. Since the electronic wave functions on each site are not equivalent
(see Fig. 8.2), there is no inversion symmetry. The appropriate
symmetry group for CO is C∞v which has the following character table:
C∞v (∞m) E 2Cφ σv
(x2
+ y2
, z2
) z A1(Σ+
) 1 1 1
Rz A2(Σ−
) 1 1 −1
(xz, yz)
(x, y)
(Rx, Ry)
E1(Π) 2 2 cos φ 0
(x2
− y2
, xy) E2(∆) 2 2 cos 2φ 0
...
...
...
...
158 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
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' "§ £  () 0¡§ £¦
132 154
132"6¨154
132¦7814
9@BADCE
9 @GFAIH P¢C¤FRQSH FST
U V8W
WYX
V8W
`
9 P¢C¤FRQSH FST
U V&W
WaX
V8W
`
b V
`
Figure 8.2: The wave functions for a heteropolar diatomic molecule and
their formation of bonding and antibonding states. If 2V3 is the energy
separation between the anion and cation for large interatomic distance,
the splitting resulting from an interaction energy 2V2 is shown.
8.4. DIATOMIC MOLECULES 159
The symmetry operations of C∞v have already been covered when discussing
the symmetry operations of D∞h (see §8.4.1). Using the equivalence
operation on the carbon and oxygen atoms in CO, we have the
result χatom sites
= 2A1 (see also χatom sites
for H2 with D∞h symmetry
in §8.4.2). Now the C atom has the electronic configuration 2s2
2p2
while O has the configuration 2s2
2p4
. We will then make bonding and
antibonding molecular orbitals from 2s, 2pz and 2px,y atomic orbitals.
From the basis functions given in the character table for C∞v we see
that the irreducible representations for these atomic orbitals (for group
C∞v) are:
2s → A1
2pz → A1
2px,y → E1
To find the direct products using the character table for C∞v we note
that
cos2
φ =
1
2
(1 + cos 2φ)
which allows us to evaluate the direct product π ⊗ π or E1 ⊗ E1 to
obtain
π ⊗ π = 1
Σ
+
+ 3
Σ
−
+ 1
∆
+
or
E1 ⊗ E1 = A1 + A2 + E2.
The secular equation implied by the interactions in Fig. 8.2 is
V3 − E V2
V2 −V3 − E
= 0 (8.14)
which gives −(V 2
3 − E2
) − V 2
2 = 0 or E2
= V 2
2 + V 2
3 so that
E = ± V 2
2 + V 2
3 (8.15)
as shown in Fig. 8.2.
Referring to Fig. 8.3 the number of electrons which form bonds
in CO are 4+6=10. We note from Fig. 8.3 that the occupied levels
160 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
 ¢¡¤£¦¥¨§ ©§ ©

   
 


 
 
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)0 ) 
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)0
)0
1$2 1$2
 43 1527681@9A6 ¡B3 15276 1@9C
189AD189 D
189AE7FG189 E7FG
HPIRQS¡
T¦¡¢QU¡
Figure 8.3: Bonding and antibonding molecular levels for the CO
molecule.
include the 2s A1 bonding and antibonding orbitals and the 2p A1 and
E1 bonding orbitals. The 2p A1 and E1 antibonding orbitals will remain
unoccupied. Since the pz orbitals are directed along the molecular axis,
the bonding-antibonding interaction (and level splitting) will be largest
for the pz orbitals, as shown in Fig. 8.3.
The symmetry of the s-function orbitals for a diatomic molecule are
found directly from the transformation properties of χatom sites
. However,
for p-function orbitals we must take the direct product of χatom sites
⊗
χvector
since the p-functions transform as vectors. For the case of the
heterogeneous CO molecule with C∞v symmetry χatom sites
= 2A1 =
2Σ+
and χvector
= A1 + E1 = Σ+
+ Π. With regard to the pz orbital,
both the bonding and antibonding orbitals (see Fig. 8.3) have A1 or
Σ+
symmetry. For the bonding pz orbital, there is a maximum of the
charge accumulation between the C and O atoms as shown in Fig. 8.3.
For the (px, py) orbitals, the bonding and antibonding levels both have
E1 or Π symmetry (see Character Table 3.35 for notation).
The symmetry types of each of the molecular orbitals determines
the form of the secular equation. The block structure of the secular
equation then assumes the form shown in Fig. 8.4, in which the coupling
terms appear in the blocks indicated.
8.4. DIATOMIC MOLECULES 161
 
 
 
 
 
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¡£¢¥¤¦§ ¨ ¤§ ¨1 ¡£¢¥¤¦§©¨ ¤§¨ 1 ##
¡2¢¥¤¦§ ¨ ¤§ ¨1  #
Figure 8.4: Schematic diagram of the matrix Hamiltonian for molecular
orbitals for the CO molecule.
162 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
Table 8.2: Character Table for Group C3v (3m)
C3v(3m) E 2C3 3σv
x2
+ y2
, z2
z A1 (Γ1) 1 1 1
Rz A2 (Γ1) 1 1 –1
(x2
− y2
, xy)
(xz, yz)
(x, y)
(Rx, Ry)
E (Γ2) 2 –1 0
8.5 Electronic Orbitals for Multi-atomic
Molecules
In this section we consider the electronic levels for various multi-atomic
molecules, each selected for particular pedagogic purposes.
8.5.1 The NH3 Molecule
We have already seen in §4.6 how to construct LCAO’s for the three
equivalent atoms at the corners of an equilateral triangle (e.g., the hydrogen
atoms in NH3). In this case we use group C3v (see Fig. 8.5) and
obtain the irreducible representations A1 + E for the directed valence
representation.
To bond to the H atoms, the N atom must make orbitals directed to
the 3 hydrogens. We refer to this as the directed valence bonds of the
nitrogen atoms. The directed valence bonds ΓD.V. for the nitrogen must
therefore exhibit the symmetry of Γatom sites
for the hydrogens. Thus
ΓD.V. for the nitrogen atom is written as ΓD.V. = Γ1 + Γ2 or A1 + E.
We now explore the orbitals that can be made at the nitrogen site.
Nitrogen has the electronic configuration 1s2
2s2
2p3
. The 1s and 2s
electrons will lie low in energy, and bonding orbitals to the hydrogens
will be made with the three p electrons. Since p electrons have angular
momentum l = 1, they transform like the vector (x, y, z) and the character
table for C3v shows that the px and py functions will transform as
E(Γ2) and the pz as A1(Γ1) (see Fig. 8.5). The states with like symmetries
will interact to form bonding and antibonding orbitals, as shown
8.5. ELECTRONIC ORBITALS FOR MULTI-ATOMIC MOLECULES163
 
¡
¢¤£
¢¤£
¢¤£
¡
¡
Figure 8.5: Schematic diagram of
the symmetry operations for an
NH3 molecule (group C3v) where
the three hydrogen atoms are at
the corners of an equilateral triangle
and the N atom is along the normal
through the midpoint of this
triangle but not coplanar with the
hydrogens.
schematically in Fig. 8.6. States with unlike symmetries do not
interact. Thus the Nitrogen has three p electrons for bonding and the
H3 likewise has 3 electrons for bonding. The A1 bonding states will
hold 2 electrons and the E bonding state will hold 4 electrons. These
bonds then can accommodate all 6 electrons. All the antibonding states
will be unoccupied.
8.5.2 The CH4 Molecule
In this example we consider generally how carbon atoms will form tetrahedral
bonds. One example of such tetrahedral bonds for carbon is in
the diamond structure. This problem is identical to the formation of
tetrahedral valence bonds in the CH4 molecule. The methane molecule
forms a regular tetrahedron (see Fig. 3.3), where the carbon atom is at
the center of the tetrahedron, and the four H atoms are at the tetrahedral
vertices; this structure has Td point symmetry (see Table 3.34).
What are the features of the electronic configuration that will produce
this bond geometry? The ground state of the carbon atom is
1s2
2s2
2p2
. Can tetrahedral bonds be formed in this ground state, or
does the C atom have to go into an excited state?
We can see by inspection that a bond can be formed by superim-
164 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
(a)
(b)
Figure 8.6: (a)Schematic energy level diagram for the XH3 molecule
(N=X for the ammonia molecule). Energy states are shown for the
atomic orbitals (AO) of nitrogen and the molecular orbitals (MO) for
the cluster of three hydrogen atoms. Finally on the right are shown
the molecular orbitals for the NH3 molecule. The spin ↑ and spin ↓
electrons in the nitrogen 2s state are paired. The three electrons in the
p state form bonds to the three hydrogen atoms. The higher levels are
all antibonding states. HOMO denotes the “highest occupied molecular
orbital” and LUMO denotes the “lowest unoccupied molecular orbital”.
(b) Same plot from Herzberg’s book.
8.5. ELECTRONIC ORBITALS FOR MULTI-ATOMIC MOLECULES165
posing (or making a linear combination of) an s and a p wave function
shown schematically in Fig. 8.7. In superimposing s and p functions we
can make linear combinations of s, px, py and pz functions so that the
bonds will lie along each of the (111) directions. The matrices, which
transform the directed valence orbitals ψi (i = 1, . . . 4) associated with
the carbon atom into one another, form a 4-dimensional directed valence
representation of group Td. The matrices for the directed valence
representation from the 4 hydrogen atoms to the central carbon atom
are found by considering the permutations of points a, b, c, d in Fig. 3.3.
The 24 symmetry operations of Td are described in §3.9 and in
Fig. 3.3. If we now consider each of the symmetry operations the group
Td acting on the points a, b, c, d (see Fig. 3.3) we obtain the equivalence
representation for the hydrogen orbital Γatom sites
or equivalently
the directed valence representation from the hydrogens to the carbon
atom at the center of the tetrahedron. Some typical matrices for the
symmetry operations of Td in the equivalence representation Γatom sites
are:
Da.s.
(E) =





1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1





(8.16)
Da.s.
(C3) =





1 0 0 0
0 0 1 0
0 0 0 1
0 1 0 0





(8.17)
where the rows and columns relate to the array (a b c d) of Fig. 3.3.
The results for the characters of the equivalence representation formed
from transforming the atom sites χa.s.
are summarized below just under
the character table for Td and are related to irreducible representations
of Td:
166 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
Figure 8.7: Schematic picture of the
constant charge contours of (a) a
2px orbital and (b) a directed orbital
arising from the superposition
of [(1/2)ψs + (
√
3/2)ψpx ].
8.5. ELECTRONIC ORBITALS FOR MULTI-ATOMIC MOLECULES167
Td(43m) E 8C3 3C2 6σd 6S4
A1 1 1 1 1 1
A2 1 1 1 −1 −1
E 2 −1 2 0 0
(Rx, Ry, Rz) T1 3 0 −1 −1 1
(x, y, z) T2 3 0 −1 1 −1
χa.s.
4 1 0 2 0 ⇒ A1 + T2
The characters χa.s.
are for the atom sites for the 4 hydrogen atoms H4
and χD.V. = χa.s.
gives the characters for the directed valence representation
for C at the center of the regular tetrahedron.
From the point of view of the 4 hydrogens, χa.s.
gives the symmetries
for the linear combination of atomic orbitals (LCAO). The appropriate
LCAO’s in normalized form are found using the procedure given in §4.6
yielding for the A1 representation:
ψ(A1) =
1
2
(ψa + ψb + ψc + ψd) (8.18)
and for the three degenerate partners of the T2 representation:
ψ1(T2)=1
2
(ψa + ψb − ψc − ψd)
ψ2(T2)=1
2
(−ψa + ψb + ψc − ψd)
ψ3(T2)=1
2
(ψa − ψb + ψc − ψd).
(8.19)
How did we get this result?
The linear combination that transforms as A1 is clearly the sum
of the atomic orbitals. The T2 orbitals must be orthogonal and are
obtained from Eq. 8.9 using the characters for the T2 irreducible repre-
sentation:
ˆP(Γn)
a =
3
24
3a − (b + c + d) + (a + c + a + d + a + b)
− (d + b + b + c + d + c)
=
3
24
6a − 2b − 2c − 2d . (8.20)
In reduced form Eq. 8.20 is written as
ˆP(Γn)
a =
1
4
[3a − b − c − d] (8.21)
168 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
Taking cyclic permutations, we write:
ˆP(Γn)
b =
1
4
[3b − c − d − a] (8.22)
ˆP(Γn)
c =
1
4
[3c − d − a − b] (8.23)
ˆP(Γn)
d =
1
4
[3d − a − b − c]. (8.24)
To obtain LCAO’s that are more symmetric and orthogonal, consider:
ˆP(Γn)
a + ˆP(Γn)
c = 1
2
[a − b + c − d]
ˆP(Γn)
a + ˆP(Γn)
b = 1
2
[a + b − c − d]
− ˆP(Γn)
a − ˆP(Γn)
d = 1
2
[−a + b + c − d]
(8.25)
which are (except for a normalization factor) the results given in Eq. 8.19.
Since the symmetries for the directed valence orbitals from the central
carbon atom are the same as those from the four hydrogen orbitals,
interactions between orbitals with like symmetry will occur so
that bonding and antibonding orbitals will be produced. These symmetries
in the directed valence orbitals can be related conveniently to
angular momentum states for carbon. This is done for the general case
by considering the characters for rotations and inversions (see Eqs. 6.1
and 6.3):
χ(α) =
sin( + 1
2
)
sin(α/2)
for pure rotations
χ(iα) = (−1)
sin( + 1
2
)α
sin(α/2)
for improper rotations.
We thus obtain the characters for the angular momentum states in the
Td group and list them in Table 8.3, where we have made use of the
fact that
σd = iC2
S4 = iC4.
The results in Table 8.3 could equally well have been obtained by looking
at the character table for group Td (see Table 3.34) and making the
following identifications:
8.5. ELECTRONIC ORBITALS FOR MULTI-ATOMIC MOLECULES169
Table 8.3: Characters and symmetries for the angular momentum states
in Td symmetry.
E 8C3 3C2 6σd 6S4
χ = 0 1 1 1 1 1 A1 A1 → s state
χ = 1 3 0 −1 1 −1 T2 T2 → p state
χ = 2 5 −1 1 1 −1 E + T2
basis functions
= 0 s-state 1
= 1 p-state (x, y, z)
= 2 d-state (xy, yz, zx
T2
, x2
− y2
, 3z2
− r2
)
E
and associating the various basis functions of the angular momentum
states with the appropriate irreducible representations for the Td group.
If we now apply this discussion to the CH4 molecule we see that the
directed valence orbitals for the carbon are made from one 2s (A1) state
and three 2p (T2) states (see Fig. 8.8) and involve 8 valence electrons
for the molecule. A set of 4 mutually orthogonal functions for the linear
combination of hydrogen orbitals is given in Eq. 8.19. Using the same
form as these orbitals, we obtain the corresponding directed valence
orbitals emanating from the carbon atom:
Ψ(1, 1, 1) = 1
2
(ψs + ψpx + ψpy + ψpz )
Ψ(1, −1, −1) = 1
2
(ψs + ψpx − ψpy − ψpz )
Ψ(−1, 1, −1) = 1
2
(ψs − ψpx + ψpy − ψpz )
Ψ(−1, −1, 1) = 1
2
(ψs − ψpx − ψpy + ψpz ) (8.26)
Equations 8.26 also represent normalized functions for tetrahedral bonding
orbitals in common semiconductors. Bonding states are made between
the A1 carbon orbital and the A1 orbital of the four hydrogens
170 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
(a)
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Figure 8.8: (a)Schematic diagram for the energy levels in the XH4
molecule (C=X) as they are formed from C and H4 orbitals. In this
diagram the A1 state is labeled a1 and the T2 state in labeled f2 to follow
the Herzberg notation, and both bonding and antibonding states are
shown. The molecular orbitals labeled a1 and f2 can accommodate the
8 valence electrons of CH4. (b) Same plot from Herzberg’s book.
8.5. ELECTRONIC ORBITALS FOR MULTI-ATOMIC MOLECULES171
and between the corresponding T2 carbon and hydrogen orbitals (see
Fig. 8.8). Although the carbon electrons must be promoted to the excited
sp3
configuration to satisfy the bonding orbitals in the molecule,
the bonding energy due to the CH4 bonds more than compensates for
the electronic excitation. The directed valence can only be considered
as approximate since the electronic orbitals in the molecular states are
strongly hybridized, rather than being atomic-like, as is assumed in
forming LCAO’s.
8.5.3 The Hypothetical SH6 Molecule
Consider a hypothetical molecule SH6 where the six identical H atoms
are arranged on a regular hexagon (e.g., the benzene ring has this basic
symmetry) and the sulfur is at the center. For the hydrogens, we have 6
distinct atomic orbitals. To simplify the secular equation we use group
theory to make appropriate linear combinations of atomic orbitals:










ψa
ψb
ψc
ψd
ψe
ψf










(8.27)
so that the transformed linear combinations are proper basis functions
for irreducible representations of the point symmetry group D6h which
applies to this problem. We see that the largest dimension for an irreducible
representation in D6h is n = 2. We show below that the use
of symmetry will result in a secular equation with block diagonal form,
having blocks with dimensions no greater than (2 × 2).
The characters for χatom sites
(R) for the six H orbitals in D6h symmetry
are found by considering how many atom sites go into each other under
the various symmetry operations of the group. The results are given
at the bottom of the character table below for D6 where D6h = D6 ⊗ i.
172 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
 
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Figure 8.9: Geometry of the hypothetical
SH6 molecule with 6 hydrogens
at the corners of a hexagon and
the sulfur atom at the center (D6h
symmetry).
D6 E C2 2C3 2C6 3C2 3C2
x2
+ y2
, z2
Γ1(A1) 1 1 1 1 1 1
z Γ2(A2) 1 1 1 1 −1 −1
Γ3(B1) 1 −1 1 −1 1 −1
Γ4(B2) 1 −1 1 −1 −1 1
(x2
− y2
, xy) Γ5(E2) 2 2 −1 −1 0 0
(xz, yz), (x, y) Γ6(E1) 2 −2 −1 1 0 0
χatom sites
6 0 0 0 2 0 ⇒ Γ1 + Γ3 + Γ5 + Γ6
We now set up the appropriate linear combinations. This can be
done by projection operators or by inspection (see §4.3), utilizing the
correspondence of this problem with the nth
roots of unity, in this
case, the 6th
roots of unity. We will denote the 6th
roots of unity
by 1, Ω, ω, −1, ω2
, Ω5
where ω = e2πi/3
and Ω = e2πi/6
. For simplicity
we will denote the atomic orbitals at a site α by ψα and use the abbreviated
notation α. In terms of the site notation (a, b, c, d, e, f), the 6
orthogonal linear combinations formed by taking the 6th
roots of unity
are:
ψ1 a + b + c + d + e + f transforms as Γ1
ψ2 a + Ωb + ωc − d + ω2
e + Ω5
f
ψ3 a + ωb + ω2
c + d + ωe + ω2
f
8.5. ELECTRONIC ORBITALS FOR MULTI-ATOMIC MOLECULES173
ψ4 a − b + c − d + e − f transforms as Γ3
ψ5 a + ω2
b + ωc + d + ω2
e + ωf
ψ6 a + Ω5
b + ω2
c − d + ωe + Ωf
where Ω = e2πi/6
and ω = e2πi/3
.
To obtain the symmetries of the functions ψ1, . . . , ψ6 we examine
Rψi where R is a symmetry operation in group D6. Clearly ψ2 and ψ6
are partners since ψ∗
2 = ψ6, and similarly ψ3 and ψ5 are partners since
ψ∗
3 = ψ5, so these provide good candidates for representing the Γ5 and
Γ6 irreducible representations. By inspection ψ1 is invariant under all
the symmetry operations of the group and thus ψ1 transforms as Γ1. As
for ψ4, application of C6(ψ4) = −ψ4, and C3ψ4 = ψ4 etc., verifies that
ψ4 transforms as Γ3. Inspection of the character table shows differences
between Γ5 and Γ6 under the operations in classes C2 and 2C6. It is
clear that the basis formed by ψ2 and ψ6 transforms under C6 as:
C6(ψ2, ψ6) = (ψ2, ψ6)
Ω5
0
0 Ω
(8.28)
since a → b, b → c, c → d etc. Thus the trace of the matrix is
Ω + Ω5
= e2πi/6
+ e−2πi/6
= 2 cos
2π
6
= 1 (8.29)
which is the proper character for Γ6. As a check, we see that C2(ψ2, ψ6)
results in a trace = Ω3
+ Ω15
= Ω3
+ Ω3
= 2 cos π = −2, and this also
checks.
Similarly we see that the transformation matrix for
C6(ψ3, ψ5) = (ψ3, ψ5)DΓ5
(C6)
again sends a → b, b → c, c → d etc. and yields a trace of ω + ω2
= −1
while C2(ψ3, ψ5) yields a trace of ω3
+ω6
= 2. The unitary transformation
U which takes the original basis a, b, c, d, e, f, g into a basis that
exhibits D6 symmetry
U










a
b
c
d
e
f










=










ψ1
ψ4
ψ2
ψ6
ψ3
ψ5










(8.30)
174 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
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Figure 8.10: Schematic of the secular equation for 6 hydrogen orbitals
at the corners of a regular hexagon. Outside of the block structure, all
entries are zeros. The Γ1 and Γ3 are 1-dimensional representations and
the Γ5 and Γ6 are 2-dimensional representations.
brings the one-electron molecular secular matrix into the block diagonal
form shown in Fig. 8.10, and zeros in all the off-diagonal positions
coupling these blocks.
Just as we used some intuition to write down the appropriate basis
functions, we can use physical arguments to guess at the ordering of the
energy levels. The fully symmetric state yields a maximum charge density
between the atom sites and therefore results in maximum bonding.
On the other hand, the totally antisymmetric state yields a minimum
bonding and therefore should be the highest energy state. The doubly
degenerate levels have an intermediate amount of wave function
overlap.
Since the total energy of the states in the 6-atom system is conserved,
the center of gravity of the energy of the LCAO’s is at the
center of the atomic orbitals (see Fig. 8.11). The 6 symmetric orbitals
that we make can be populated by 12 electrons. But we only have 6
electrons at our disposal and these go into the lowest energy states. For
this reason, the molecule produces a lower energy state than the free
atoms. One can then make directed valence orbitals from the S at the
8.5. ELECTRONIC ORBITALS FOR MULTI-ATOMIC MOLECULES175
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Figure 8.11: Energies of the LCAO’s formed by six hydrogen atoms at
the corners of a hexagon. Also shown is a schematic summary of the
wave functions for the various orbitals.
center of the hexagon to the 6 hydrogens.
An isolated S atom is in a 1s2
2s2
2p6
3s2
3p4
configuration which is
compatible with the Γ1+Γ2+Γ6 irreducible representations of group D6.
For the directed valence orbitals from the sulfur atom to the 6 hydrogen
atoms in a plane, higher angular momentum states are needed to obtain
orbitals with Γ3 and Γ5 symmetry. To obtain Γ5 symmetry a d function
( = 2) is needed while Γ3 symmetry requires an f function ( = 3)
which first occurs in a 4f state, since the = 2 angular momentum
state in full rotational symmetry reduces to the Γ1 +Γ5 +Γ6 irreducible
representations in D6 symmetry, and = 3 reduces to Γ2 + Γ3 + Γ4 +
Γ5 +Γ6 in D6 symmetry. Therefore the planar geometry is not favorable
for forming the SH6 molecule on the basis of directed valence bonding
arguments.
8.5.4 The SF6 Molecule
We next give an example of SF6 with a molecular configuration that involves
octahedral bonding (see Fig. 8.12). The octahedral configuration
is very common in solid state physics.
If we now use the symmetry operations of Oh we get the characters
for the equivalence representation χatom sites
for the six atoms which sit
176 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
F
S
F
FF
F
F
6
2
5
1
3
4
y
x
z
Figure 8.12: Schematic diagram of
the SF6 molecule which exhibits octahedral
bonding.
at the corners of the octahedron (see Fig. 8.12):
E 8C3 3C2 6C2 6C4 i 8iC3 3iC2 6iC2 6iC4
χa.s. 6 0 2 0 2 0 0 4 2 0 ⇒ A1g + Eg + T1u
The decomposition of the reducible representation χatom sites
for the six
equivalent fluorine atoms gives
χa.s.
= A1g + Eg + T1u (8.31)
If we (hypothetically) put s-functions on each of the six fluorine sites,
then we can write χs(F6) = χa.s.
. However, if we put p-functions on
each fluorine site then χp(F6) = χa.s.
⊗ χT1u . This general concept of
taking the direct product of the transformation of the sites with the
symmetry of the orbital on each site is frequently used in applications
of the equivalence principle.
8.5. ELECTRONIC ORBITALS FOR MULTI-ATOMIC MOLECULES177
Table 8.4: Characters for angular momentum states and their irreducible
representations in Oh symmetry.
E 8C3 3C2 6C2 6C4 i 8iC3 3iC2 6iC2 6iC4
= 0 1 1 1 1 1 1 1 1 1 1 ⇒ A1g
= 1 3 0 −1 −1 1 –3 0 1 1 −1 ⇒ T1u
= 2 5 −1 1 1 −1 5 −1 1 1 −1 ⇒ Eg + T2g
= 3 7 1 −1 −1 −1 –7 −1 1 1 1 ⇒ A2u + T1u + T2u
= 4 9 0 1 1 1 –9 0 –1 –1 –1 ⇒ A1g + Eg + T1g + T2g
O(432) E 8C3 3C2 = 3C2
4 6C2 6C4
A1 1 1 1 1 1
A2 1 1 1 −1 −1
(x2
− y2
, 3z2
− r2
) E 2 −1 2 0 0
(Rx, Ry, Rz)
(x, y, z)
T1 3 0 −1 −1 1
(xy, yz, zx) T2 3 0 −1 1 −1
Oh ≡ O(m3m) ⊗ i
We will now look at the orbitals for electrons on the sulfur site.
Bonding orbitals are found by setting the directed valence representation
equal to the symmetries found from the equivalence transformation.
For simplicity let us assume that χs(F6) = χa.s.
= χD.V.. We then
need to identify the irreducible representations contained in χD.V. with
angular momentum states. The characters for the angular momentum
states in Oh symmetry are then found from:
χ(α) =
sin( + 1
2
)α
sin(α/2)
(8.32)
and using the character table for Oh. The results are tabulated in
Table 8.4. To produce χD.V. = A1g + Eg + T1u as in Eq. 8.31 we can
use an s state = 0 for the A1g symmetry, a p state ( = 1) for the
T1u symmetry, and a d state ( = 2) for the Eg symmetry in Eq. 8.31.
Thus sp3
d2
orbitals are required for the directed valence of the sulfur
ion, which ordinarily has an atomic ground state configuration 3s2
3p4
.
178 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
Thus to make the necessary bonding, we must promote the S atom to
an excited state, namely to a 3d state. This type of excitation is called
configuration mixing. Still this configuration is more stable than that
for a planar SH6 molecule.
8.5.5 The B12H12 Molecule
Since the point groups for icosahedral symmetry have become of interest
recently, and these groups are not discussed in many of the standard
group theory text books, we give an example here of this symmetry
group for pedagogic reasons.
Five-fold rotation axes are known to occur in molecules (e.g., IF7
with symmetry D5h; Fe(C2H5)2 with symmetry D5d; and (B12H12) with
icosahedral symmetry Ih). In 1985 interest in the icosahedral group was
kindled by the observation in a new type of matter called quasi-crystals
which require two sets of basis vectors rather than the single set that is
used to describe crystals following one of the 230 possible space groups.
The discovery of how to prepare gram quantities of C60 (in 1990), an
insulating form of carbon with icosahedral symmetry, has made fivefold
symmetry an important current topic in condensed matter physics
and materials science. These C60 species are found to be very stable
in the gas phase and form a “soccerball” structure (see Fig. 8.13) with
a diameter of ∼7 ˚A. This arrangement of the carbon atoms at the 60
corners of a truncated icosahedron leaves no dangling bonds. In some
cases the “soccerball” is filled with a single transition metal or rare
earth metal atom; such carbon cages for metal atoms have now been
reported for C60La, but are also found for other metal atoms (Ca, Ba,
and Sr).
Five-fold symmetry also occurs in clusters and in liquids. For example,
DNA is known to exhibit five-fold coordination. However, it was
commonly believed that there would be no five-fold axes in condensed
matter physics because it is not possible to fill a Bravais lattice with
structures based on five-fold point-group symmetry. (We will prove this
theorem in a later chapter, when we discuss space groups for periodic
lattices.)
The molecule B12H12 is a relatively simple molecule with icosahedral
symmetry. The molecule is shown schematically in Fig. 8.14 in terms of
8.5. ELECTRONIC ORBITALS FOR MULTI-ATOMIC MOLECULES179
Figure 8.13: Truncated icosahedron
showing pentagonal and hexagonal
faces (the “soccerball”). This is the
proposed structure for C60, a sixty
atom stable carbon molecule. A
carbon atom is located at each vertex,
which occurs at the intersection
of one pentagonal face with two
hexagonal faces.
Figure 8.14: The B12H12 molecule has Ih icosahedral symmetry (see
text). Shown in the figure is a regular icosahedron before the truncation
that forms the figure in Fig. 8.13. The big bullets denote the 12 boron
atoms and the small bullets denote the 12 hydrogens.
180 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
the regular icosahedron, where each B atom is at one of the 12 corners
of the icosahedron. The B12H12 molecule could also be described in
terms of the regular dodecahedron where each B atom (or H atom) sits
over the center of a pentagonal face. The hydrogens are all sticking
out from the borons on either side of each of the six C5 axes (see
Fig. 8.14). The symmetry operations of the group Ih include: six C5
axes, ten C3 axes, fifteen C2 axes, inversion, six S10 axes (coincident
with the C5 axes) and ten S6 axes (coincident with the C3 axes) and
fifteen mirror planes. The symmetry operations are summarized in the
character table for Ih shown in Table 3.40. The symmetry operations
of the regular dodecahedron and the regular icosahedron are the same,
and each of these figures is said to be the dual of the other. The
golden mean τ = (1 +
√
5)/2 = 1.618 appears in the characters for
the 3–dimensional representations of group Ih and arises from the fivefold
rotations which involve the angle of 2π/5 = 72◦
. We note here
that τ − 1 = 2 cos(2π/5). The number of symmetry elements in the
group Ih is 120. We note that the icosahedral group contains some
representations of dimensionality 4 and 5. The number of elements of
the icosahedral group I is 60 which is equal to the sum of the squares
of the dimensionality of the representations h = i n2
i = 12
+ 32
+ 32
+ 42
+ 52
= 60.
From Fig. 8.14, we can compute the characters Γatom sites
for the 12
hydrogens (or the 12 borons) corresponding to each of the classes for
this reducible representation in group Ih, and the results are given in
Table 8.5. From the characters in Table 8.5 and the character table for
Ih (Table 3.40), we use Eq. 3.20 to obtain the irreducible representation
for χa.s.
for the 12 hydrogen orbitals in the B12H12 molecule:
Γatom sites
= Ag + Hg + F1u + F2u.
In the B12H12 molecule the local symmetry at the B site is C5v.
Also included in Table 8.5 are χa.s.
for 20 and 30 atoms, respectively,
at the 20 vertices and the centers of the 30 edges of the dodecahedron.
With regard to the regular icosahedron, C20 and C30 would respectively
have atoms at the centers of each of the 20 triangular faces or at the
centers of the 30 edges (see Fig. 8.14). Also χa.s.
is given in Table 8.5
for 60 atoms at the 60 vertices of the regular truncated icosahedron to
form the C60 molecule.
8.6. BOND STRENGTHS 181
Table 8.5: Characters for the equivalence transformation Γatom sites
of
various LCAO’s in icosahedral symmetry.
Ih E 12C5 12C2
5 20C3 15C2 i 12S10 12S3
10 20S6 15σ
H12 12 2 2 0 0 0 0 0 0 4 ⇒
Ag + Hg
+F1u + F2u
C20 20 0 0 2 0 0 0 0 0 4 ⇒
Ag + Gg + Hg
+F1u + F2u + Gu
C30 30 0 0 0 2 0 0 0 0 4 ⇒
Ag + Gg + 2Hg
+F1u + F2u + Gu + Hu
C60 60 0 0 0 0 0 0 0 0 4 ⇒
Ag + F1g + F2g + 2Gg + 3Hg
+2F1u + 2F2u + 2Gu + 2Hu
In the C20 molecule the local symmetry at the C site in C3v. The
directed valence representation for this symmetry has χd.v. = A + E
which suggests that the sp2
atomic configuration would give the necessary
directed orbitals. The irreducible representations contained in
Γatom sites
for the C20 molecule are tabulated in Table 8.5.
8.6 Bond Strengths
In this section we show that the amplitudes of the wave functions are
in fact maxima along the bond directions, consistent with the concept
of a directed valence bond. Let us then consider the angular parts of
the wave functions and demonstrate that the directed valence bond is
a maximum in the bond direction. The (1, 1, 1) directed valence bond
for CH4 is written as the first equation in Eq. 8.26. We express each
of the terms of this equation in terms of spherical harmonics, using the
coordinate system of Fig. 8.15. For angular momentum l = 0 and l = 1
the spherical harmonics yield
ψs = 1 ψpy =
√
3 sin θ sin φ
ψpx =
√
3 sin θ cos φ ψpz =
√
3 cos θ.
(8.33)
We can thus write the angular dependence of the directed valence bond
182 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
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Figure 8.15: Polar coordinate system
defining the angles θ and φ.
along (111) as:
Ψ(1, 1, 1)|(θ,φ) =
f(r)
2
1 +
√
3 sin θ(cos φ + sin φ) +
√
3 cos θ . (8.34)
By differentiation with respect to φ, we can find the maximum:
∂Ψ(1, 1, 1)
∂φ
= 0 = − sin φ + cos φ f(r)
√
3 sin θ, (8.35)
so that at the maximum:
tan φ = 1 or φ =
π
4
(8.36)
from which it follows that
sin φ = cos φ =
1
√
2
. (8.37)
Then differentiating Ψ(1, 1, 1) with respect to θ to get the maximum
with respect to θ gives
∂Ψ(1, 1, 1)
∂θ
= 0 = f(r)
√
3 cos θ(cos φ + sin φ) − sin θ (8.38)
8.6. BOND STRENGTHS 183
which when evaluated at φ = π
4
yields
tan θ =
2
√
2
=
√
2 and θ = tan−1
√
2. (8.39)
But these are the values of (θ, φ) which denote the (1, 1, 1) direction,
along which
ψpx =
√
3 sin θ cos φ =
√
3(
2
3
)(
1
√
2
) = 1 = ψpy (8.40)
ψpz =
√
3 cos θ =
√
3(
1
√
3
) = 1. (8.41)
Thus the maximum value for Ψ(1, 1, 1) is along a (1,1,1) direction:
Ψ(1, 1, 1)|(1,1,1) =
f(r)
2
(1 + 1 + 1 + 1) = 2f(r) (8.42)
and 2f(r) is the maximum value that Ψ(1, 1, 1) can have.
If we consider the value of Ψ(1, 1, 1) along a different {1,1,1} direction,
we will get a smaller amplitude. For example, along a (¯1, ¯1, ¯1)
direction:
Ψ(1, 1, 1)|(¯1,¯1,¯1) =
f(r)
2
[1 +
√
3(
2
3
)(−
1
√
2
−
1
√
2
) +
√
3(
−1
√
3
)] = −f(r)
(8.43)
and along a (¯1, ¯1, 1) direction:
Ψ(1, 1, 1)|(¯1,¯1,1) =
f(r)
2
[1+
√
3(
2
3
)(−
1
√
2
−
1
√
2
)+
√
3(
1
√
3
)] = 0 (8.44)
and along a (¯1, 1, 1) direction:
Ψ(¯1, 1, 1)|(¯1,1,1) =
f(r)
2
[1 +
√
3(
2
3
)(−
1
√
2
+
1
√
2
) +
√
3(
1
√
3
)] = f(r).
(8.45)
This analysis shows that Ψ(1, 1, 1) has a large lobe along (1,1,1). Since
the derivatives of Ψ(1, 1, 1) along the (¯1, ¯1, ¯1), (¯1, ¯1, 1) and (¯1, 1, 1) all
vanish, Ψ(1, 1, 1) has 3 smaller lobes, making tetrahedral angles with
184 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
one another along (¯1, ¯1, ¯1), (¯1, 1, 1), (1, ¯1, 1) and (1, 1, ¯1). In the directions
where the other three functions of Eq. 8.30 have their largest lobes,
Ψ(1, 1, 1) has a node (i.e., vanishes)– namely along (¯1, ¯1, 1), (¯1, 1, ¯1) and
(1, ¯1, ¯1). Furthermore, the wave function for this tetrahedrally directed
valence bond is larger than for other types of bonding schemes for CH4.
For example, we will soon investigate the trigonal bonding of carbon.
In this connection we will see that tetrahedral bonding is stronger than
trigonal bonding. Also we will see that for trigonal bonding, the planar
(px, py) bonding is stronger than the pz bonding. Hence, diamond
(tetrahedral bonds) has strong binding in 3 dimensions, while
graphite (trigonal bonds) only has strong binding in the layer
planes.
8.7 σ- and π-bonds
We now discuss the difference between σ- and π-bonds which are defined
in the diagram in Fig. 8.16. The situation which we have considered
until now is bonding by s-functions or by p-functions in the direction
of the bond and this is denoted by σ-bonding, as shown in Fig. 8.16.
We can also obtain some degree of bonding by directing our p-functions
⊥ to the bond direction, as also shown in Fig. 8.16, and this is called
π-bonding. We note that there are two equivalent mutually perpendicular
directions that are involved in π-bonding. From considerations
of overlapping wavefunctions, we would expect π-bonding to be much
weaker than σ-bonding.
Just as group theory tells us which linear combinations of atomic orbitals
(LCAO) are needed to form σ-bonds, group theory also provides
corresponding information about the linear combination of atomic orbitals
that form π-bonds. We will describe the procedure for finding
both σ-bonds and π-bonds in this section.
Let us first review the situation for the σ-bonds. To find a σ-bond,
we consider the atomic wave function at each equivalent site to be degenerate
with the corresponding wave functions on the other sites and
we find the transformation matrices that transform equivalent sites into
one another according to the symmetry operations of the group. To find
out if an entry in this matrix is 1 or 0 we ask the question whether or not
8.7. σ- AND π-BONDS 185
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325
6
327
!
Figure 8.16: Schematic diagram
of σ-bonding by s-functions
and by longitudinally oriented
p-functions. π-bonding with
transverse p-functions is also
shown.
186 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
a site goes into itself under a particular symmetry operation. If it goes
into itself we produce a 1 on the diagonal, otherwise a 0. Therefore
by asking how many sites go into themselves, we obtain the character
for each symmetry operation. This is the procedure we have used
throughout the chapter to find χatom sites
which denotes the equivalence
transformation.
To find the characters for a π-bond, we have to consider how many
vectors normal to the bond direction remain invariant under the symmetry
operations of the group. The simplest way to obtain the characters
for the σ-bonds and π-bonds is to consider the transformation as
the product of 2 operations: the transformation of one equivalent site
into another, followed by the transformation of the vector on a site.
The character for such a transformation is most easily found from the
theory of the direct product of two representations:
χ(R)atom sites
⊗ χ(R)general vector = χ(R)atom sites
⊗ χ(R)vector⊥ to σ−bonds
+ χ(R)atom sites
⊗ χ(R)vector to σ−bonds
(8.46)
But
χ(R)D.V. σ−bonds ≡ χ(R)atom sites
⊗ χ(R)(vector to σ−bonds).
Thus:
χ(R)D.V. π−bonds = χ(R)atom sites
⊗ χ(R)general vector − χ(R)D.V. σ−bonds
(8.47)
and we thus obtain the desired result:
χ(R)D.V.π−bonds = χ(R)atom sites
⊗ χ(R)vector ⊥ to σ−bonds. (8.48)
As an example of σ-bonds and π-bonds let us consider the problem
of trigonal bonding of a hypothetical C4 cluster where one carbon
atom is at the center of an equilateral triangle and the other 3 carbon
atoms are at the corners of the triangle, as shown in Fig. 8.17. The
pertinent character table is D3h which is given below. For this group
σh denotes an x-y reflection plane and σv denotes a reflection plane
containing the three-fold axis and one of the two-fold axes.
8.7. σ- AND π-BONDS 187
 
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Figure 8.17: Schematic diagram of
a carbon atom forming bonds to 3
other carbon atoms at the corners
of an equilateral triangle.
D3h(6m2) ≡ D3 ⊗ σh E σh 2C3 2S3 3C2 3σv
x2
+ y2
, z2
A1 1 1 1 1 1 1
Rz A2 1 1 1 1 −1 −1
A1 1 −1 1 −1 1 −1
z A2 1 −1 1 −1 −1 1
(x2
− y2
, xy) (x, y) E 2 2 −1 −1 0 0
(xz, yz) (Rx, Ry) E 2 –2 −1 1 0 0
Consider the linear combination of atomic orbitals made out of the 3
carbon atoms at the corners of the equilateral triangle. From the equivalence
transformation for these 3 carbons, we obtain χatom sites
:
E σh 2C3 2S3 3C2 3σv
χatom sites
3 3 0 0 1 1 ⇒ A1 + E
Clearly if each of the orbitals at the corners of the equilateral triangle
were s-functions, then the appropriate linear combination of atomic
orbitals would transform as A1 + E
A1 : ψ1 + ψ2 + ψ3 (8.49)
E :
ψ1 + ωψ2 + ω2
ψ3
ψ1 + ω2
ψ2 + ωψ3
(8.50)
188 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
where
ω = exp(
2πi
3
). (8.51)
In transforming wavefunctions corresponding to higher angular momentum
states, we must include the transformation of a tensor (vector)
on each of the equivalent sites. This is done formally by considering the
direct product of χatom sites
with χtensor, where χtensor gives the transformation
properties of the orbital: a scalar for s-functions, a vector for
p-functions, a tensor for d-functions, etc.
We now illustrate the construction of LCAO’s from s and p-functions,
noting that from the character table for the group D3h, s-functions
transform as A1, pz functions as A2 and (px, py) functions as E . We
thus obtain for the transformation properties of the three s-functions
at the corners of an equilateral triangle:
χatom sites
⊗ χs = (A1 + E ) ⊗ A1 = A1 + E (8.52)
for the pz functions which transform as A2 we have for the direct prod-
uct:
χatom sites
⊗ χpz = (A1 + E ) ⊗ A2 = A2 + E . (8.53)
Finally for the px,y functions which transform as E we obtain
χatom sites
⊗ χpx,py = (A1 + E ) ⊗ E = A1 + A2 + 2E . (8.54)
We will see below that the A1 + E symmetries correspond to σ-bonds
and the remaining (A2 + E ) + (A2 + E ) correspond to π-bonds, as
shown in Fig. 8.18.
For the carbon atom at the center of the equilateral triangle (see
Fig. 8.17) we make directed valence orbitals to the carbon atoms at
sites (1), (2), and (3) from states with A1 and E symmetry, which in
accordance with the character table for D3h, transform as the ψs and
ψpx , ψpy wave functions. The directed orbitals from the central carbon
atom are thus:
ψ1 = αψs + βψpx
ψ2 = αψs + β −
1
2
ψpx +
√
3
2
ψpy
ψ3 = αψs + β −
1
2
ψpx −
√
3
2
ψpy . (8.55)
8.7. σ- AND π-BONDS 189
 
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Figure 8.18: Schematic diagram
for the σ-bonds and the
in-plane π-bonds for carbon
atoms at the corners of a triangle
to a carbon atom at the
center of the triangle.
The orthonormality condition on the three waves functions in Eqs. 8.55,
gives
α2
+ β2
= 1 β2
= 2α2
(8.56)
or
α =
1
√
3
β =
2
3
(8.57)
so that
ψ1 =
1
3
ψs +
2
3
ψpx
ψ2 =
1
3
ψs −
1
6
ψpx +
1
2
ψpy (8.58)
ψ3 =
1
3
ψs −
1
6
ψpx −
1
2
ψpy .
Using the basis functions in the character table for D3h and the classification
of angular momentum states in Table 8.6, we can make σbonding
orbitals with the following orbitals for the central carbon atom,
neglecting for the moment the energetic constraints on the problem:
190 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
Table 8.6: Characters for the angular momentum states and their irreducible
representations for the group D3h.
E σh = iC2 2C3 2S3 = 2iC6 3C2 3σv = iC2
= 0 1 1 1 1 1 1 A1
= 1 3 1 0 –2 −1 1 A2 + E
= 2 5 1 −1 1 1 1 A1 + E + E
= 3 7 1 1 1 −1 1 A1 + A2 + A2 + E + E
2s2p2
s + (px, py)
2s3d2
s + (dxy, dx2−y2 )
3d2p2
d3z2−r2 + (px, py)
3d3
d3z2−r2 + (dxy, dx2−y2 ) .
It is clear from the list in Table 8.6 that the lowest energy σ-bond is
made with the 2s2p2
configuration. The carbon atom has 4 valence
electrons, 3 of which make the in-plane trigonal σ-bonds. The 4th
electron is free to bond in the z direction. This bonding involves π-
bonds.
To obtain π-bonds from the central carbon atom to the atoms at
the corners of the triangle, we look at the character table to see how
the vector (x, y, z) transforms. From the character table, we have the
result
χvector = E + A2. (8.59)
We then take the direct product:
χa.s.
⊗ χvector =
χatom sites
(A1 + E ) ⊗ (E + A2)
χvector
= (A1 ⊗ E ) + (A1 ⊗ A2) + (E ⊗ E ) + (E ⊗ A2)
= (E ) + (A2) + (E + A1 + A2) + (E )
= (A1 + E ) + (E + A2 + A2 + E ). (8.60)
Since the irreducible representations for the σ-bonds are A1 and E ,
we have the desired result that the irreducible representations for the
8.7. σ- AND π-BONDS 191
π-bonds are:
E + A2 + A2 + E .
We can now go one step further by considering the polarization of the
π-bonds by considering the irreducible representations that are even
and odd under the horizontal mirror plane σh:
χD.V. π−bonds =
even under σh
A2 + E + A2 + E
odd under σh
. (8.61)
The above polarization analysis identifies the bonds given in Eqs. 8.52–
8.54.
To find the representations contained in the directed valence for the
π-bonds we have to go to rather high angular momentum states: = 2
for an E state and = 3 for an A2 state. Such high angular momentum
states correspond to much higher energy. Therefore π-bonding will be
much weaker than σ-bonding. The irreducible representations A2 + E
correspond to π-bonding in the z direction while the A2 + E representations
correspond to π-bonding in the plane of the triangle, but ⊥ to
the σ-bonding directions. We further note that the symmetries A2 +E
correspond to pz and dxz, dyz orbitals for angular momentum 1 and 2,
respectively. On the other hand, the symmetries A2 + E require = 3
states, and therefore correspond to higher energies than the A2 + E
orbitals. A diagram showing the orbitals for the σ-bonds and π-bonds
for the various carbon atoms is given in Fig. 8.18.
192 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
8.8 Selected Problems
1. Consider a hypothetical SF6 molecule with octahedral symmetry
(p. 171 of class notes).
F
S
F
FF
F
F
6
2
5
1
3
4
y
x
z
(a) Using χatom sites
, construct the linear combination of atomic
orbitals for the six fluorine atoms which transform according
to the 3 irreducible representations contained in χatom sites
,
assuming for the moment s functions on the six fluorine sites.
(b) What are the symmetries for the six LCAO’s in (a) if we
assume that we have p-functions on each of the fluorine sites?
(c) What are the irreducible representations corresponding to
σ-bonds and π-bonds for the central sulfur atom to the 6
fluorine atoms? Sketch the orientation of these bonding or-
bitals.
(d) What are the angular momentum states required to bond
the sulfur to the fluorine.
2. C2H4 (ethylene) is a planar molecule which has the configuration
shown on the diagram below:
8.8. SELECTED PROBLEMS 193
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(a) Identify the appropriate point group for C2H4.
(b) Find χatom sites
for the 2 carbon atoms and for the 4 hydrogen
atoms.
(c) Give the block diagonal structure for the secular equation
for the electronic energy levels of ethylene.
(d) How do the carbon atoms satisfy their bonding requirements?
Which angular momentum states are needed to form
bonding orbitals from each carbon atom?
3. Suppose that we have a hypothetical CdH12 molecule where the
Cd (4s2
4p6
4d10
5s2
) is located at the center of a regular truncated
icosahedron and 12 hydrogens are at the centers of the pentagonal
faces.
(a) Find a linear combination of atomic orbitals for the 12 hydrogen
atoms that transform as irreducible representations
of group Ih.
(b) To make chemical bonds to the hydrogen atoms, what are
the symmetries and angular momentum states that are needed
to form the directed valence orbitals from the Cd atom to
the 12 hydrogen atoms? Would you expect the hypothetical
CdH12 molecule to be stable and why?
4. (a) Consider the hypothetical molecule XH12 where the 12 hydrogen
atoms are at the vertices of a regular icosahedron
and the atom X is at the center of the icosahedron. What
are the symmetries and degeneracies of the 12 linear combinations
of atomic orbitals (LCAO’s) associated with the 12
equivalent hydrogen atoms?
194 CHAPTER 8. ELECTRONIC STATES OF MOLECULES
(b) Give the linear combination of atomic orbitals (LCAO’s) for
the hydrogen atoms.
(c) What are the angular momentum states involved with each
of the directed valence orbitals from the central atom X?
Chapter 9
Molecular Vibrations,
Infrared, and Raman
Activity
In this chapter we review molecular vibrations and present the use
of group theory to identify the symmetry and degeneracy of the normal
modes. Selection rules for infrared and Raman activity are also
discussed and are illustrated for a variety of molecules selected for pedagogic
purposes.
9.1 Molecular Vibrations – Background
A molecule having its ions at their equilibrium sites is in an energy
minimum. If the ions are displaced from their equilibrium positions, a
restoring force will be exerted which will tend to bring the ions back
to equilibrium. If the displacement is small, the restoring forces will be
harmonic. The vibrational motion of the molecule will thus be harmonic
in the limit of small displacements.
Suppose that a molecule contains N ions and suppose further that
the potential function is known from solution of the electronic problem
(see Chapter 8). The potential function for the electronic problem is expressed
in terms of the 3N coordinates for the N ions, V (R1, . . . , RN ).
We are particularly interested in V (R1, . . . , RN ) about its equilibrium
195
196 CHAPTER 9. MOLECULAR VIBRATIONS
value at R◦
1, . . . , R◦
N . In solving the molecular vibration problem, we
expand V about its equilibrium value, utilizing the fact that a minimum
in energy implies the vanishing of the first derivative of the potential.
We can conveniently take our zero of energy at the potential minimum
and obtain a Hamiltonian for molecular vibrations in terms of displacements
from equilibrium:
H =
k
1
2
mk
˙ξ2
k
kinetic energy
+
k,
1
2
∂2
V
∂ξk∂ξ
ξkξ
potential energy
(9.1)
where mk denotes the mass of the kth
ion, ξk denotes its displacement
coordinate, and the potential energy depends of the second derivative
of V (R1, . . . , RN ). The Hamiltonian in Eq. 9.1 gives rise to a (3N ×3N)
secular equation. The roots of this secular equation are the eigenfrequencies
ω2
k and the eigenvector is the normal mode linear combination
of the displacements.
The usual procedure for finding the normal modes involves one
transformation:
qk =
√
mk ξk (9.2)
to eliminate the mass term in the kinetic energy, and a second transformation
to express qk in terms of the normal mode coordinates QK:
qk =
K
akKQK (9.3)
where akK denotes the amplitude of each normal mode QK that is
contained in qk.
Thus, by proper choice of the akK amplitudes we can use Eqs. 9.2
and 9.3 to reduce the potential energy to a sum of squares of the form
ω2
KQ2
K/2. These transformations yield for the potential function in
Eq. 9.1:
V =
1
2
k,
K, L
∂2
V
∂qk∂q
akKa LQKQL =
1
2 K
ω2
KQ2
K (9.4)
9.2. APPLICATION OF GROUP THEORY TO MOLECULAR VIBRATIONS197
Table 9.1: Correspondence between the electronic problem and the
molecular vibration problem.
Quantity Electronic Molecular Vibration
Matrix element Hk
∂2V
∂qk∂q
Eigenvalue En ω2
K
Eigenfunction†
ψn(r) akK
†
For the molecular vibration problem, the eigenfunction is the normal
mode amplitude while for the electronic problem it is the wavefunction.
——where
the coefficients akK are chosen to form a unitary matrix satisfying
Eq. 9.4. Thus we obtain the relations a†
Kk = a−1
Kk = akK if the
matrix elements of akK are real. The akK coefficients are thus chosen
to solve the eigenvalue problem defined in Eq. 9.4. To achieve the
diagonalization implied by Eq. 9.4 we must solve the secular equation
k,
a−1
Kk
∂2
V
∂qk∂q
a L = ω2
KδKL. (9.5)
Solution of the secular equation (Eq. 9.5) yields the eigenvalues or normal
mode frequencies ω2
K and the eigenfunctions or normal mode amplitudes
akK for K = 1, . . . , 3N. From the form of the secular equation we
can immediately see the correspondence between the electronic problem
and the molecular vibration problem shown in Table 9.1.
9.2 Application of Group Theory to Molecular
Vibrations
In an actual solution to a molecular vibration problem, group theory
helps us to classify the normal modes and to find out which modes are
coupled when electromagnetic radiation is incident on the molecule,
either through electric dipole transitions (infrared activity) or in light
scattering (Raman effect). We discuss all of these issues in this chapter.
198 CHAPTER 9. MOLECULAR VIBRATIONS
We make use of the symmetry of the molecule by noting that the
molecule remains invariant under a symmetry operation of the group of
Schr¨odinger’s equation. Therefore application of a symmetry operation
ˆPR to an eigenfunction of a normal mode fq just produces a linear
combination of other normal modes of the same frequency ωq. That is,
fq forms a basis for a representation for the symmetry operators ˆPR
ˆPRf(i)
q =
q
f
(i)
q D(i)
(R)q q (9.6)
where D(i)
(R)q q is the matrix representation for symmetry operator R
and i is the label for the irreducible representation which labels both
the matrix and the basis function (normal mode coordinate in this
case). Since different irreducible representations do not couple to each
other, group theory helps to bring the normal mode matrix into block
diagonal form, with each eigenvalue and its corresponding normal mode
labeled by its appropriate irreducible representation. This is in concept
similar to the solution of the electronic eigenvalue problem discussed in
Chapter 8, except that every atom (or ion) in the molecule has 3 degrees
of freedom, so that the vibration itself must transform as a vector. Thus
the molecular vibration problem is analogous to the electronic problem
for p-functions, since the p-functions also transform as a vector.
Therefore to find the normal modes for the vibration problem, we
carry out the following steps:
1. Identify the symmetry operations and appropriate symmetry group
G to describe the molecule in its equilibrium position.
2. Find χequivalence = χatom sites
, the characters for the equivalence
transformation, which represents the number of atoms that are
invariant under the symmetry operations of the group. But since
χatom sites
is in general a reducible representation in group G, we
must decompose χatom sites
into its irreducible representations.
3. We now use the concept that a molecular vibration involves the
transformation properties of a vector. In group theoretical terms,
this means that the molecular vibrations are found by taking the
direct product of χatom sites
with the irreducible representations
9.2. APPLICATION OF GROUP THEORY TO MOLECULAR VIBRATIONS199
for a polar vector. The characters for the molecular vibrations
are thus found according to the relation
χmolecular vibrations = (χatom sites
⊗ χvector) − χtranslations − χrotations
(9.7)
where the simple translations of the center of mass or rotations
of the molecule about the center of mass do not contribute to the
degrees of freedom for the molecular vibrations themselves. The
characters found from Eq. 9.7 in general correspond to a reducible
representation of Group G. We therefore express χmolecular vibrations
in terms of the irreducible representations of group G to obtain
the normal modes. Each eigenmode is labeled by one of these
irreducible representations, and the degeneracy of each eigenfrequency
is the dimensionality of the corresponding irreducible representation.
The characters for χtranslation are found by identifying
the irreducible representations of the group G corresponding to
the basis functions (x, y, z) for the radial vector. The characters
for χrotation are found by identifying the irreducible representations
corresponding to the basis functions (Rx, Ry, Rz) for the
axial vector (e.g., angular momentum). Every standard character
table normally lists the irreducible representations for the 6
basis functions for (x, y, z) and (Rx, Ry, Rz).
4. From the characters for the irreducible representations for the
molecular vibrations, we find the normal modes. The normal
modes for a molecule are constrained to contain no translations
or rotations and to be orthogonal to each other.
5. We use the techniques for selection rules (see Chapter 7) to find
out whether or not each of the vibrational modes is infrared active
(can be excited by electromagnetic radiation, see §9.5) or Ramanactive
(see §9.8).
To illustrate the procedure for finding molecular vibrations, we will
consider the molecular vibrations of an isolated H2O water molecule.
200 CHAPTER 9. MOLECULAR VIBRATIONS
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132
Figure 9.1: Normal modes
for the H2O molecule with
3 vibrational degrees of freedom.
(a) The breathing
mode with symmetry A1,
which changes only bond
lengths. (b) The symmetric
stretch mode of H2O with
A1 symmetry, which changes
bond angles. (c) The antisymmetric
stretch mode with
B1 symmetry.
9.3 Molecular Vibrations in H2O
The four symmetry operations for the H2O molecule (see Fig. 9.1) include
E the identity operation, a 180◦
rotation C2 around the z-axis,
a reflection plane σv in the plane of molecule and a σv reflection perpendicular
to the plane of the molecule. The σv plane is a vertical
reflection plane rather than a σh plane since the xz plane contains the
highest symmetry axis C2 rather than being ⊥ to the highest symmetry
axis and similarly for the σv plane. The reflection plane σv which goes
through C2 is ⊥ to the plane of the molecule. In labelling the axes, the
plane of the H2O molecule is denoted by xz, with the x axis parallel to
a line going through the two hydrogens, and the perpendicular y axis
goes through the oxygen atom. The appropriate point group for the
H2O molecule is the group C2v and the character table is given below.
9.3. MOLECULAR VIBRATIONS IN H2O 201
C2v (2mm) E C2 σv σv
x2
, y2
, z2
z A1 1 1 1 1
xy Rz A2 1 1 −1 −1
xz Ry, x B1 1 −1 1 −1
yz Rx, y B2 1 −1 −1 1
Next we find χatom sites
. For H2O we have to consider the transformation
of three atoms under the symmetry operations of the group.
In writing down χatom sites
we recall that for each site that is invariant
under a symmetry operation, a contribution of +1 is made to the
character of that operation; otherwise the contribution is zero. Thus,
we obtain χatom sites
(H2O) for all 3 atoms in the H2O molecule as given
below:
E C2 σv σv
χatom sites
(H2O) 3 1 3 1 ⇒ 2A1 + B1
χatom sites
(2H) 2 0 2 0 ⇒ A1 + B1
Also given in the above listing is χatom sites
(2H) for the two hydrogens
only. In both cases, the irreducible representations of group C2v
contained in χatom sites
are listed. We note that for the 2 hydrogens
considered by themselves, χatom sites
(2H) = A1 + B1, so that for the
electronic problem, the appropriate LCAOs for the two hydrogen atoms
are the bonding orbital (ψH1 + ψH2 ) which transforms as A1 and the
antibonding orbital (ψH1 − ψH2 ) which transforms as B1.
From the character table we see that the vector transforms as
χvector = A1 + B1 + B2
where z, x, y, respectively, transform as A1, B1 and B2. Likewise the
irreducible representations for the rotations are A2+B1+B2 corresponding
to Rz, Ry, and Rx, respectively. We then calculate the irreducible
representations contained in the molecular vibrations:
χmolecular vibrations = χatom sites
⊗ χvector − χtranslations − χrotations
= (2A1 + B1) ⊗ (A1 + B1 + B2) − (A1 + B1 + B2)
202 CHAPTER 9. MOLECULAR VIBRATIONS
−(A2 + B1 + B2)
= [3A1 + 3B1 + 2B2 + A2] − (A1 + B1 + B2)
−(A2 + B1 + B2)
χmolecular vibrations = 2A1 + B1
The 3 modes in χmolecular vibrations are all 1-dimensional irreducible
representations and therefore have non-degenerate or distinct vibrational
frequencies. The 2 normal modes with A1 symmetry must leave
the symmetry undisturbed and this can be accomplished by the stretching
of bonds and bond angles. These modes are the breathing and
symmetric stretch modes (see Fig. 9.1).
We must now find the normal modes corresponding to each eigenfrequency.
All molecules have a “breathing” mode which leaves the
symmetry unchanged. To get the eigenvectors for the breathing mode
of the H2O molecule, assume that one of the hydrogen atoms is displaced
in some way (as shown below). With A1 symmetry this implies
(under operation C2) that the other H atom must be correspondingly
displaced (see Fig. 9.1a). To prevent translations and rotations of the
molecule, O must be displaced as shown. (The actual vibration amplitude
for each atom is constrained to avoid translation and rotation of
the molecule).
The same arguments can be applied to obtain the A1 symmetric
stretch mode shown in Fig. 9.1b. The H atom motion is taken so that
the two A1 modes are orthogonal. Since the breathing mode and symmetric
stretch mode have the same symmetry they can mix (or couple
to each other) and for this reason the directions of the H atom motion
for each of the modes in Fig. 9.1a and Fig. 9.1b are not uniquely
specified.
To obtain the normal mode for B1 symmetry, we observe that the
character for the C2 operation is −1, so that the two hydrogen atoms
must move in opposite directions relative to the OH bond. Likewise,
the motion of the O atom must be odd under C2. These arguments
determine the normal B1 mode shown in Fig. 9.1c.
As mentioned above, all molecules have a breathing mode which
transforms as A1 and preserves the molecular symmetry. As a practical
matter in checking whether or not the calculated normal modes are
proper normal modes, it is useful to verify that the normal mode motion
9.4. OVERTONES AND COMBINATION MODES 203
preserves the center of mass (conservation of linear momentum), that
no torques are applied (angular momentum must be conserved), and
that all normal modes are orthogonal to each other.
To identify which normal modes are infrared-active, we must consider
the selection rules for the electromagnetic interaction. This is
described in §9.5 in general terms, and is then illustrated for several
types of molecules for illustrative purposes. In §9.8 the Raman selection
rules are discussed and several illustrations are given.
9.4 Overtones and Combination Modes
The non-linear elastic constants in the equations of motion for the
molecular vibrations give rise to overtones and combination modes
which can be observed by either infrared or Raman spectroscopy.
The mode frequencies for the overtone modes (or harmonics) are at
∼ 2ωΓi
and the symmetries are given by the direct product Γi⊗Γi where
Γi corresponds to the symmetry of the fundamental mode at frequency
ωΓi
. The combination modes are at frequencies ∼ (ωΓi
+ ωΓj
) and have
symmetries given by Γi ⊗ Γj. Some of these modes for the methane
molecule CH4 are given in Table 9.2 (see §9.8.4).
9.5 Infrared Activity
If electromagnetic radiation is incident on a molecule in its ground
state, then the radiation will excite those vibrational modes which give
rise to a dipole moment. In the ground state, the molecule has A1
symmetry in accordance with the invariance of the Hamiltonian for the
electronic ground state under the symmetry operations of the group of
Schr¨odinger’s equation. We can use group theory to decide whether or
not an electromagnetic transition will occur. The perturbation Hamiltonian
for the interaction of the molecule with the electromagnetic (infrared)
interaction is
H infrared = −E · u = −
e
mc
p · A (9.8)
204 CHAPTER 9. MOLECULAR VIBRATIONS
Table 9.2: Observed vibrational frequencies for the methane molecule.
Assignment Symmetry mode Frequency (cm−1
)
ν1(A1) A1 fundamental 2914.2
ν2(E) E fundamental 1526
ν3(T2) T2 fundamental 3020.3
ν4(T2) T2 fundamental 1306.2
2ν2 A1 + E overtone†
3067.0
2ν3 (A1 + E) + T2 overtone†
6006
3ν3 (A1 + T1) + 2T2 overtone‡
9047
2ν4 (A1 + E) + T2 overtone†
2600
ν4 − ν3 T2 combination 1720
ν2 + ν4 (T1) + T2 combination 2823
†
For overtones, only the symmetric combinations are Raman allowed.
‡
For 3ν3 the symmetric combinations correspond to the angular momentum
states L = 1 which transforms as T2 and L = 3 which transforms
as A1 + T1 + T2.
—
9.5. INFRARED ACTIVITY 205
where E is the incident oscillating electric field, u is the induced dipole
moment arising from atomic displacements, and p is the electronic momentum
of the molecule. In this interaction, u and p transform like
vectors. To find out whether incident electromagnetic radiation will
excite a particular vibrational mode, we must examine the selection
rule for the process. This means that we must see whether or not the
matrix element for the excitation (ψf |u|ψi) vanishes, where ψf denotes
the normal mode which we are trying to excite, u is the vector giving
the transformation properties of H infrared, and ψi denotes the initial
state of the molecule, which for most cases is the ground state. The
ground state has no vibrations and is represented by the totally symmetric
state A1.
We use the usual methods for finding out whether or not a matrix
element vanishes. That is, we ask whether the direct product Γvector⊗Γi
contains the representation Γf ; if (Γvector⊗Γi) does not contain Γf , then
the matrix element ≡ 0. Since molecular vibrations are typically excited
at infrared frequencies we say that a molecule is infrared active
if any molecular vibrations can be excited by the absorption of electromagnetic
radiation. The particular modes that are excited are called
infrared-active modes. Correspondingly, the modes that cannot be
optically excited are called infrared inactive.
As applied to the H2O molecule (see §9.3) we have the following
identification of terms in the electromagnetic matrix element: ψ1 has
A1 symmetry for the unexcited molecule, the vector u transforms as
u → A1 + B1 + B2
corresponding to the transformation properties of z, x, y, respectively.
Thus, we obtain
χH infrared
⊗ χψi
= (A1 + B1 + B2) ⊗ (A1) = A1 + B1 + B2 (9.9)
Therefore the two A1 modes and the B1 mode of water are all
infrared-active. Each of the three vibrations corresponds to an oscillating
dipole moment. As far as polarization selection rules are concerned,
we can excite either of the two A1 modes with an optical electric field
in the z direction, the two-fold axis of the molecule. To excite the B1
206 CHAPTER 9. MOLECULAR VIBRATIONS
mode, the optical electric field must be along the x direction, the direction
of a line connecting the two hydrogen atoms. An electric field
in the y direction (perpendicular to the plane of the molecule) does not
excite any vibrational modes. Since all vibrational modes of the water
molecule can be excited by an arbitrarily directed E field, all the vibrational
modes of the water molecule are infrared-active. It is not always
the case that all vibrational modes of a molecule are infrared-active. It
can also happen that for some molecules only a few of the modes are
infrared-active. This situation occurs in molecules having a great deal
of symmetry.
To observe infrared activity in the second-order infrared spectra,
we require that the combination of two vibrational modes be infraredactive.
From a group theoretical standpoint, the symmetry of the combination
mode arising from constituent modes of symmetries Γi and Γj
is given by the direct product Γi ⊗Γj. Since groups containing inversion
symmetry have only odd parity infrared-active modes, such symmetry
groups have no overtones in the second-order infrared spectrum.
9.6 Vibrations for Linear Molecules
The procedure for dealing with the molecular vibrations of linear molecules
such as CO (Fig. 9.2) is slightly different from what has been described
above. We now present a method for handling the linear molecules and
give some examples.
For a linear molecule, the irreducible representations for the rotations
just involves the rotations Rx and Ry, assuming the molecular
axis to be along ˆz. Thus for the linear molecule, only two degrees of
freedom are removed by χrotations. We now illustrate this point with
the CO molecule, followed by other linear molecules, each selected to
demonstrate additional issues.
9.6.1 The CO Molecule
The appropriate symmetry group for CO is C∞v (see §8.4.4). The
symmetry operations 2Cφ denote rotations about the ˆz axis in clockwise
and counter-clockwise senses by an arbitrary angle φ. Thus Cφ is a class
9.6. VIBRATIONS FOR LINEAR MOLECULES 207
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Figure 9.2: CO molecule having only an A1 breathing mode.
with an ∞ number of members. The symmetry plane σv is a vertical
plane through the molecular axis at an angle φ with respect to an
arbitrary direction denoted by φ = 0. Since the 2Cφ and σv classes
are of infinite order, the number of irreducible representations is also
infinite.
The first step in finding χmolecular vibrations for a linear molecule is to
compute χatom sites
. For the CO molecule χatom sites
becomes:
E 2Cφ σv
χatom sites
2 2 2 ⇒ 2A1
from which we find the irreducible representations for the molecular
vibrations of CO remembering that χrotations only contains rotations in
the xy plane:
χmolecular vibrations = χatom sites
⊗ χvector − χtranslations − χrotations
= (2A1) ⊗ (A1 + E1) − (A1 + E1) − E1 = A1.
The A1 mode is the breathing mode for the CO molecule (see Fig. 9.2).
208 CHAPTER 9. MOLECULAR VIBRATIONS
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Figure 9.3: The O2 molecule having only an A1g breathing mode.
9.6.2 The O2 Molecule
If we now consider the O2 molecule (see Fig. 9.3), we have a homonuclear
molecule following the symmetry group D∞h (see character Table
3.36). In this case
E 2Cφ 2C2 i 2iCφ iC2
χatom sites
2 2 0 0 0 2 ⇒ A1g + A2u
Thus the irreducible representations for the molecular vibrations of O2
become:
χmolecular vibrations = χatom sites
⊗ χvector − χtranslations − χrotations
χmolecular vibrations = (A1g + A2u) ⊗ (A2u + E1u) − (A2u + E1u) − E1g = A1g
Because of the inversion symmetry of the O2 molecule, all the normal
modes have either even (gerade) or odd (ungerade) symmetries.
Thus, for O2 the breathing mode (see Fig. 9.3) has A1g symmetry
and is infrared-inactive. From simple physical considerations the
breathing mode for O2 has no oscillating dipole moment and hence
should not couple to an electromagnetic field through an electric dipole
9.6. VIBRATIONS FOR LINEAR MOLECULES 209
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Figure 9.4: The
CO2 molecule having 3 vibrational
normal modes: (a) the
breathing mode of A1g symmetry,
(b) the antisymmetric
stretch mode of A2u symmetry,
and (c) the doubly degenerate
E1u mode where the
mode displacements for the
two partners are orthogonal.
interaction, in agreement with our group theoretical result. In contrast,
for the case of the CO molecule, there is an oscillating dipole moment
and thus the CO molecule would be expected to be infrared-active, also
in agreement with the group theoretical result.
9.6.3 The CO2 Molecule
The CO2 molecule is chosen for discussion to show the various types
of modes that can be expected for linear molecules involving 3 or more
atoms. In §9.6.4 we consider another molecule (C2H2) described by the
same symmetry group D∞h but having slightly more complexity.
For the case of CO2 (see Fig. 9.4), we again have a linear molecule
with D∞h symmetry and now χatom sites
corresponds to a three-dimensional
representation:
210 CHAPTER 9. MOLECULAR VIBRATIONS
E 2Cφ C2 i 2iCφ iC2
χatom sites
3 3 1 1 1 3 ⇒ 2A1g + A2u
so that
χmolecular vibrations = χatom sites
⊗ χvector − χtranslations − χrotations
= (2A1g + A2u) ⊗ (A2u + E1u) − (A2u + E1u) − E1g
= A1g + A2u + E1u
In this case the A2u and E1u modes are infrared-active while the
symmetric A1g mode is infrared-inactive. The normal modes for CO2
are easily found with the help of the character table, and are shown
in Fig. 9.4. The A1g mode is the breathing mode, the A2u mode is the
antisymmetric stretch mode and the E1u mode is a doubly degenerate
bending mode where the displacements of the carbon and the two
oxygens are normal to the molecular axis for each partner of the E1u
bending mode.
9.6.4 The C2H2 Molecule
For the case of the linear C2H2 molecule also following group D∞h
symmetry, we obtain:
χatom sites
= 2A1g + 2A2u (9.10)
using the result for O2 obtained in §9.6.2. Thus χmolecular vibrations (χm.v.)
for the C2H2 molecule becomes:
χm.v. = (2A1g + 2A2u) ⊗ (A2u + E1u) − (A2u + E1u) − E1g
= 2A1g + A2u + E1u + E1g
The five normal modes for the molecular vibrations of C2H2 are shown
in Fig. 9.5, again illustrating the breathing, antisymmetric stretch and
bending modes corresponding to 5 different vibrational frequencies.
These concepts can of course be generalized to give normal modes for
more complex linear molecules.
9.6. VIBRATIONS FOR LINEAR MOLECULES 211
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Figure 9.5: Schematic diagram of the normal modes of the linear C2H2
molecule: (a) two breathing modes of A1g symmetry, (b) an antisymmetric
stretch mode of A2u symmetry, and (c) and (d) two doublydegenerate
bending modes of E1g and E1u symmetries.
212 CHAPTER 9. MOLECULAR VIBRATIONS
9.7 Molecular Vibrations in Other Molecules
In this section we illustrate how symmetry is used in molecular vibration
problems for several cases of pedagogic interest.
9.7.1 Vibrations of the NH3 Molecule
To illustrate some features of degenerate normal modes, let us consider
the NH3 molecule (see Fig. 9.6). The hydrogen atoms in NH3 are at the
corners of an equilateral triangle and the nitrogen atom is either above
or below the center of the triangle. If the molecule were planar it would
have D3h symmetry, but because the N atom is not coplanar with the
hydrogens, the appropriate symmetry group is C3v (see Table 3.15). To
obtain the molecular vibrations we find χatom sites
= χa.s. first for all
four atoms and then for the three hydrogen atoms separately:
E 2C3 3σv
χtotal
a.s. 4 1 2 ⇒ 2A1 + E
χH
a.s. 3 0 1 ⇒ A1 + E
χmolecular vibrations = χatom sites
⊗ χvector − χtranslations − χrotations
= (2A1 + E) ⊗ (A1 + E) − (A1 + E) − (A2 + E)
= 2A1 + 2E
The two modes of NH3 with A1 symmetry are breathing modes both
in-plane and in the z-direction as shown in Fig. 9.6.
• One mode of the NH3 molecule with A1 symmetry is the breathing
mode where the nitrogen atom is at rest and the equilateral
triangle expands and contracts.
• For the A1 out-of-plane breathing mode, the H atoms move in the
+z direction while the N atom moves in the −z direction such
that no translation results.
9.7. MOLECULAR VIBRATIONS IN OTHER MOLECULES 213
 
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Figure 9.6: Normal modes for
the NH3 molecule: (a) the
in-plane breathing mode, (b)
the z-axis breathing mode,
and (c) one partner of the
in-plane mode of E symmetry;
the second partner
(complex conjugate of the
first) is not shown. Also
the other doubly-degenerate
E mode for z-axis motion is
not shown.
214 CHAPTER 9. MOLECULAR VIBRATIONS
• One of the E modes is a doubly-degenerate in-plane mode. One
eigenvector is made from the linear combination of hydrogen atom
motions, Ha +ωHb +ω2
Hc where the motion of each H atom bears
a phase relation of ω = e2πi/3
relative to the next H atom. The
second eigenvector is Ha +ω2
Hb +ωHc which is orthogonal to the
first. The nitrogen atom moves in the xy plane in such a way as
to prevent translation of the center of mass and rotation of the
molecule.
• For the second doubly degenerate E mode, the hydrogen atoms
move along the z-direction with a phase difference between adjacent
hydrogen atoms of ω = e2πi/3
for one partner and ω2
= e4πi/3
for the other partner. The nitrogen atom again moves in the x−y
plane to prevent translations or rotations of the molecule.
The molecular vibrations for the NH3 molecule illustrate the concept
of phase relations between the motions of various atoms in executing
a normal mode. Though it should be emphasized that in the case
of degenerate modes, the normal mode (basis function) picture is not
unique, and linear combinations of modes of the same symmetry are
possible.
Since the normal modes for the NH3 molecules have A1 and E symmetries
and since χvector = A1 + E, all the vibrational modes for NH3
are infrared-active with one of the two A1 modes excited by polarization
E ˆz, the other being excited by polarization E⊥ˆz and likewise
for the two E modes.
9.7.2 Vibrations of the CH4 Molecule
The CH4 molecule is chosen for discussion to show that not all modes
are infrared-active and to give more practice with Td symmetry because
of the importance of this symmetry to semiconductor physics.
In the case of the CH4 molecule χatom sites
for all the hydrogens and
the carbon atom (see Fig. 9.7) contains the irreducible representations
of the point group Td: (2A1 + T2) (see §8.2). In Td symmetry, the
vector transforms as T2 while the angular momentum transforms as
T1. We thus get for the symmetry types in the molecular vibrations
9.7. MOLECULAR VIBRATIONS IN OTHER MOLECULES 215
χmolecular vibrations = χm.v. (see Fig. 9.7):
χm.v. = [(2A1 + T2) ⊗ (T2)] − T2
translations
−
rotations
T1
= 2T2 + (T1 + T2 + E + A1) − T2 − T1
= A1 + E + 2T2
For many molecules of interest, the normal modes are given in
Herzberg’s book on IR and Raman Spectra, though you may find the
diagrams difficult to understand. We give in Fig. 9.7 the normal modes
found in Herzberg, as well as our own version.
9.7.3 Vibrations of the B12H12 Molecule
In this subsection we consider the molecular vibrations for a molecule
with many atoms and this example also provides more practice with
the use of the icosahedral group Ih.
Referring to Table 3.40 and §8.5.5, where the symmetry of the electronic
states of the B12H12 molecule is discussed, we find that the equivalence
transformation for the twelve B atoms or the 12 H atoms is the
same and each gives:
χatom sites
= Ag + Hg + F1u + F2u (9.11)
so that together the B12H12 molecule gives 2(Ag + Hg + F1u + F2u)
for χatom sites
. From the character table for icosahedral symmetry (Table
3.40), we see that the vector in Ih symmetry transforms as F1u. We
can now find the vibrational modes for B12H12 for χmolecular vibrations =
χm.v. as
χm.v.(B12H12) = χatom sites
⊗ χvector − χtranslations − χrotations
= 2(Ag + Hg + F1u + F2u) ⊗ F1u − F1u − F1g
Taking the indicated direct products we obtain
χm.v.(B12H12) = 2(Ag ⊗ F1u) + 2(Hg ⊗ F1u) + 2(F2u ⊗ F1u)
216 CHAPTER 9. MOLECULAR VIBRATIONS
 
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Figure 9.7: Normal vibrations of a tetrahedral XY4 molecule such as
methane taken from Herzberg’s book. The three two-fold axes (dotdash
lines) are chosen as the x, y, and z axes.
9.7. MOLECULAR VIBRATIONS IN OTHER MOLECULES 217
(a) (b)
Figure 9.8: Displacement for the (a) Ag (B and H vibrating in phase)
and (b) Ag (B and H vibrating out of phase) normal modes for the
B12H12 molecule.
+2(F1u ⊗ F1u) − F1u − F1g
= 2F1u + 2(F1u + F2u + Gu + Hu) + 2(Ag + Hg + F1g)
+2(Gg + Hg) − F1u − F1g
= 2Ag + F1g + 2Gg + 4Hg
+3F1u + 2F2u + 2Gu + 2Hu (9.12)
representing 66 degrees of freedom. The symmetries of these 66 modes
are tabulated in Table 9.3 where we also have listed the symmetries of
the normal modes for the vibrations of C20, C30, and C60 clusters. The
symmetries in χatom sites
for these clusters are tabulated in Table 8.5. For
the B12H12 molecule, we have 3 infrared-active F1u modes (also called
T1u in some references), each of which is three-fold degenerate. Therefore
polarization selection rules will be important for the IR spectra of
B12H12. The Ag normal mode (see Fig. 9.8) is a breathing mode where
all atoms are moving in phase and this mode is not infrared active.
Equation 9.12 shows that for the 66 modes, we have 18 distinct
frequencies, only three of which will be seen on the infrared spectra. In
§9.8 we discuss the symmetries of the Raman-active modes.
218 CHAPTER 9. MOLECULAR VIBRATIONS
Table 9.3: Vibrational modes for clusters in Ih symmetry
Molecule Ag F1g F2g Gg Hg Au F1u F2u Gu Hu
B12 1 1 2 1 1 1 1
B12H12 2 1 2 4 3 2 2 2
C20 1 1 2 3 1 2 2 2
C30 1 1 2 3 4 2 3 3 3
C60 2 3 4 6 8 1 4 5 6 7
—
9.8 Raman Effect
In the Raman effect we look at the light scattered from a system placed
in an electromagnetic field. The induced dipole moment is
u =
↔
α ·E0 cos ωt (9.13)
where
↔
α is the polarizability tensor, a second rank symmetric tensor.
Because the incident light can excite molecular vibrations, the polarization
tensor will have frequency dependent contributions at the molecular
vibration frequencies ωv
↔
α=
↔
α0 +
↔
∆α cos ωvt (9.14)
so that
u =
↔
α0 +
↔
∆α cos ωvt · E0 cos ωt
=
↔
α0 ·E0 cos ωt +
↔
∆α
2
[cos(ω − ωv)t + cos(ω + ωv)t] · E0
where the first term is the Rayleigh component at incident frequency
ω, the second term is the Stokes component at frequency (ω − ωv), and
the third term is the anti-Stokes component at frequency (ω + ωv). In
observing the first-order Raman effect, the scattered light is examined
for the presence of Stokes components at frequencies (ω − ωv) and of
9.8. RAMAN EFFECT 219
anti-Stokes components at frequencies (ω+ωv). Not all normal modes of
the molecule will yield scattered light at (ω±ωv), although if the Stokes
component is excited, symmetry requires the anti-Stokes component to
be present also.
To find the selection rules for the Raman effect we observe that the
polarizability
↔
α in Eq. 9.13 is a second rank symmetric tensor and
has the same transformation properties as a general quadratic form–
e.g., x2
, y2
, z2
, xy, yz, zx. We note here that the symmetric off-diagonal
components correspond to combinations (xy + yx)/2 and the corresponding
terms for yz and zx. The antisymmetric terms correspond to
(xy − yx)/2 and its partners, which transform as the angular momentum.
The basis functions for the angular momentum are given in the
character table by Rx, Ry, and Rz.
In this section we will find the Raman activity for the molecules
considered previously in this chapter. To find whether or not a vibrational
mode is Raman active, we ask whether or not the matrix element
for the Raman perturbation Hamiltonian vanishes. The Raman perturbation
Hamiltonian is of the −u · E form and using Eq. 9.13, H Raman
is written as:
H Raman = −
↔
∆α
2
EE cos(ω ± ωv)t (9.15)
where the transformation properties of HRaman are those of a second
rank symmetric tensor, since the tensor (EE) is external to the molecular
system and it is only the polarizability tensor that pertains to the
molecule. To find out whether a given normal mode is Raman active
we consider the matrix element:
(ψf |H Raman|ψi) (9.16)
where ψf is the final state corresponding to a normal mode we are trying
to excite, H Raman is the Raman perturbation Hamiltonian which has
the transformation properties of a symmetric 2nd rank tensor, and ψi
is the initial state generally taken as the ground state which has the
full symmetry of the group of Schr¨odinger’s equation. A vibrational
mode is Raman active if the direct product (Γi ⊗Γ2nd rank symmetric tensor)
contains the irreducible representation for the final state Γf . This is
the basic selection rule for Raman activity.
220 CHAPTER 9. MOLECULAR VIBRATIONS
In quantum mechanics, the Raman process is a second order process
involving an intermediate state. This means that the Raman process
involves the coupling by a vector interaction V of an initial state i to
an intermediate state m and then from this intermediate state m to a
final state f : (i|V |m)(m|V |f). In terms of the spectroscopy of molecular
systems with inversion symmetry, the Raman effect is especially
important because it is a complementary technique to infrared
spectroscopy. Since the infrared excitation is a first-order process
and the dipole operator in HRaman transforms as a vector, selection
rules for a vector interaction couple states with opposite parity. On the
other hand, the Raman process being a second-order process is characterized
by an interaction Hamiltonian H Raman which is even under
inversion and therefore couples an initial and final state of similar parity.
Thus infrared spectroscopy probes molecular vibrations with odd
parity, while Raman spectroscopy probes modes with even parity.
The use of polarized light plays a major role in the assignment of
experimentally observed Raman lines to specific Raman-active modes.
In Raman experiments with polarized light, it is customary to use the
notation: ki(Ei Es)ks to denote the incident propagation direction ki,
the incident and scattered polarization directions (EiEs) and the scattered
propagation direction ks. From Eq. 9.15 we see that the Raman
interaction Hamiltonian H Raman depends on both Ei and Es and on the
change in the polarizability tensor
↔
∆α, where Ei and Es are, respectively,
the incident and the scattered electric fields. It is customary to
designate the scattered light as having diagonal Raman components
(Ei Es), or off-diagonal Raman components (Ei⊥Es).
In the second-order Raman spectra, a combination mode or overtone
will be observable if Γi ⊗ Γj for modes i and j contains irreducible
representations that are themselves Raman-active. Some silent modes
that cannot be found in the first-order spectrum can thus be observed
in the second-order spectrum.
In the following subsections we discuss the Raman effect for the
specific molecules previously discussed in this chapter, and in so doing,
we will also include comments on the polarization selection rules.
9.8. RAMAN EFFECT 221
9.8.1 The Raman Effect for H2
The Raman effect is simplest for the homo-polar diatomic molecule H2.
The symmetry group in this case is D∞h. The only vibrational mode
in this case has A1g symmetry. Since x2
+ y2
and z2
all transform as
A1g, this mode will be Raman-active and the Raman tensor will have
only diagonal components (Ei Es), thereby giving the polarization
selection rules.
9.8.2 The Raman Effect for H2O
This molecule has C2v point group symmetry. We take the initial state
Γi to be the unexcited state which has A1 symmetry, since the unexcited
state always has the full symmetry of the molecule, which in the case
of H2O is the group C2v (see §9.3). From the character table for C2v
in §9.3, we can write down the irreducible representations contained in
the 2nd rank symmetric tensor (denoted here by ΓRaman):
ΓRaman = (3A1 + A2 + B1 + B2) (9.17)
where 3A1 arises from the x2
, y2
, z2
components, A2 from the xy component,
B1 from the xz component, and B2 from the yz component.
Since (Γi ⊗ ΓRaman) contains all the symmetry types of group C2v, all
the normal modes for the H2O molecule are Raman-active. Specifically
the symmetries of the molecular vibrations for H2O are 2A1 + B1 and
from Eq. 9.17 all of these modes are Raman-active.
The A1 modes are observed when the incident (i) and scattered
(s) light have polarization directions (Ei Es), while the B1 mode is
observed for Ei ˆx and Es ˆz or for Ei ˆz and Es ˆx. We thus say
that the Raman tensor for the A1 modes is diagonal, while for the B1
mode it is off-diagonal.
9.8.3 The Raman Effect for NH3
For the case of the NH3 molecule which has C3v symmetry (see §9.7.1),
the Raman-active modes have the symmetries A1 for x2
+ y2
, z2
and E
for (x2
− y2
, xy) and (xz, yz) so that all the normal modes for the NH3
molecule (2A1 + 2E) are Raman-active. Polarization selection rules
222 CHAPTER 9. MOLECULAR VIBRATIONS
imply that the A1 modes are diagonal (i.e., scattering occurs when
Ei Es) while the E modes are off-diagonal (i.e., scattering occurs
when Ei ⊥ Es).
9.8.4 The Raman Effect for CH4
The appropriate symmetry group in this case is Td (see §8.5.2 for the
character table and notation). From §9.7.2, the molecular vibrations
have A1 + E + 2T2 symmetries. Here again the unexcited state Γi has
A1 symmetry and the irreducible representations contained within the
Raman second rank symmetric tensor include:
ΓRaman = T2 + E + A1 (9.18)
where (xy, yz, zx) transforms as T2 and the basis functions x2
−y2
, and
3z2
− r2
transform as E while r2
transforms as A1. Thus we obtain
Γi ⊗ ΓRaman = A1 + E + T2 (9.19)
which contains all the symmetries of the molecular vibrations. Thus
all the normal modes are Raman-active for CH4. The modes A1 and E
have diagonal Raman tensor components while the T2 modes are fully
off-diagonal. It is not always the case that all the normal modes are
Raman-active. We illustrate this below. Examples of harmonics and
combination modes that can be observed in the second-order Raman
and infrared spectra are given in Table 9.2. In this table the frequencies
of the four lines in the Raman spectra are given. We note that the T2
modes are observed in the first-order infrared spectrum for CH4. Some
of the direct products of importance in interpreting the second-order
spectra are
E ⊗ E = A1 + A2 + E
and
T2 ⊗ T2 = A1 + E + T1 + T2.
9.8.5 The Raman Effect for CO2 and C2H2
From §9.6.3 and §9.6.4 we see that χmolecular vibrations for the two linear
molecules CO2 and C2H2 contains the irreducible representations for
9.8. RAMAN EFFECT 223
the group D∞h:
CO2 : A1g + A2u + E1u
C2H2 : 2A1g + A2u + E1u + E1g
(9.20)
From the character table for D∞h, the Raman tensor transforms as
A1g(x2
+ y2
, z2
) + E1g(xz, yz) + E2g(x2
− y2
, xy) (9.21)
where the subscripts on the coefficients give the pertinent symmetry
types associated with the indicated basis functions. Thus for CO2 the
only Raman-active mode is A1g, while for C2H2, the two A1g modes
and the E1g mode are all Raman-active. Both of these molecules have
inversion symmetry and therefore the modes A2u + E1u are odd and
are not Raman-active. These modes however are infrared-active. All
Raman-active modes have even parity for systems with inversion sym-
metry.
9.8.6 The Raman Effect for Planar XH3
Consider a planar version of the ammonia molecule NH3, which we
denote for illustrative purposes by XH3. The symmetry group in this
case is D3h. From §9.8.3, the irreducible representations contained in
χa.s.
are: A1 + E and those in χvector are A2 + E . Then we can write
for the irreducible representations for the normal modes:
χmolecular vibrations = χa.s.
⊗ χvector − χtrans − χrot
= (A1 + E ) ⊗ (A2 + E ) − (A2 + E ) − (A1 + E )
= A1 + A2 + 2E .
Of these modes, the E mode is IR-active and the A1 and E modes
are Raman-active. The A2 mode is neither IR nor Raman-active, and
is called a silent mode.
9.8.7 The Raman Effect for B12H12
The vibrational levels were discussed in §9.7.3. The normal modes for
the 66 degrees of freedom were found to be
2Ag + F1g + 2Gg + 4Hg + 3F1u + 2F2u + 2Gu + 2Hu.
224 CHAPTER 9. MOLECULAR VIBRATIONS
From the character table for group Ih we see that ΓRaman contains the
irreducible representations Ag + Hg so that only six of the 18 eigenfrequencies
of B12H12 are Raman active. The Ag mode is seen in the
polarization Ei Es, whereas the Hg mode is seen for all polarizations.
Since the three F1u mode frequencies correspond to infrared active vibrations,
nine of the vibrational frequencies for B12H12 are silent in the
first-order Raman and infrared spectra.
9.9 Rotational Energy Levels
In practice all molecules have rotational levels (labeled by quantum
number J) very close in energy to the vibrational states. Since it is
the total symmetry of the rotational-vibrational state that enters the
selection rule, one expects to observe many Raman lines associated
with each molecular vibration. We discuss the rotational energy levels
in this section and the coupling between the rotational and vibrational
levels in §9.10.
In the approximation that we can discuss the rotational motion
as distinct from the translational motion, we consider the rotational
motion of molecules to be much slower than the vibrational motion,
and of course very much slower than the electronic motion. Typical
rotational energies are of the order of ∼ 1 meV and occur at far-infrared
frequencies.
For some problems, it is adequate to consider the molecule as a
rigid rotator, neglecting the effect of the molecular vibrations. The
Hamiltonian is then written as:
H =
J2
x
2Ix
+
J2
y
2Iy
+
J2
z
2Iz
(9.22)
where Ix, Iy, Iz are the principal moments of inertia and Jx, Jy, Jz are
the angular momentum components. The coordinates x, y, z are chosen
so that the z axis is along the figure axis or the main symmetry axis of
rotation of the molecule. If we have a diatomic molecule, one principal
moment of inertia vanishes while the other two become equal. In this
9.9. ROTATIONAL ENERGY LEVELS 225
case the Hamiltonian is simply
H =
J2
2I
(9.23)
and has eigenvalues
Ej = ¯h2
j(j + 1)/2I.
Unlike the vibrational energy levels which are all equally spaced with a
level separation ¯hωv, the rotational energy levels are unequally spaced:
Ej+1 − Ej = C[(j + 1)(j + 2) − j(j + 1)] = 2C(j + 1). (9.24)
and the level spacing depends on the quantum number j (see Fig. 9.9).
If the molecule contains a permanent electric dipole moment, then it is
possible to excite the molecule into higher rotational energy states by
electric dipole transitions. For light polarized along the principal axis
of rotation, the selection rule for electric dipole transitions is ∆j = 0
while for light polarized in the plane ⊥ to this axis, the selection rule
is ∆j = ±1 (Wigner–Eckart theorem, which is discussed in §9.11).
Thus, the first rotational transition will have energy 2C, the second
4C, the third 6C, etc. This pattern is indicated in Fig. 9.9, and is in
contrast with the vibrational levels which have a constant separation
energy of ¯hωv. It is clear that diatomic molecules like H2 have a center
of inversion and hence no permanent dipole moment. Thus,
molecules of this type do not exhibit any rotational infrared spectra.
On the other hand, molecules like CO can exhibit rotational infrared
spectra.
Also of interest are molecules which are more complex than the
linear molecules. One class of molecules that are of interest and have
sufficient symmetry to make the use of group theory helpful is the
symmetric top molecule. Here we have 2 principal moments of
inertia that are equal, Iy = Ix, and a third non-vanishing moment
of inertia Iz that is unequal to the others. When we now quantize
the angular momentum, we must not only quantize the total angular
momentum J2
, but we must also quantize the component of angular
momentum about the figure axis Jz. The quantum numbers that are
used are K for Jz along the figure axis
(jK|Jz|jK) = ¯hK (9.25)
226 CHAPTER 9. MOLECULAR VIBRATIONS
 
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Figure 9.9: (a) Rotational levels of a diatomic molecule. (b) Energy
separation between sequential rotational levels. (c) The rotational absorption
spectrum for gaseous HCl.
where the integer K can assume values from −j up to j. The total
angular momentum obeys the relation
jK|J2
|jK = ¯h2
j(j + 1) (9.26)
so that by writing the Hamiltonian for the symmetric top molecules as
H =
J2
x + J2
y
2Ix
+
J2
z
2Iz
=
J2
− J2
z
2Ix
+
J2
z
2Iz
(9.27)
we obtain the energy eigenvalues
E(j, K) =
¯h2
j(j + 1)
2Ix
+
¯h2
K2
2
1
Iz
−
1
Ix
. (9.28)
The selection rules for electric dipole transitions in symmetric top
molecules occurring between purely rotational states are ∆K = 0, along
with ∆j = 0, ±1, as for the case of diatomic molecules.
From a group theoretical standpoint the various angular momentum
states will correspond to different irreducible representations and
9.10. VIBRATIONAL-ROTATIONAL INTERACTION 227
the selection rules follow from determination of the vanishing or nonvanishing
of the matrix element coupling the initial state (no rotation,
J = 0) to the final rotational state.
9.10 Vibrational-Rotational Interaction
Since the nuclei of a molecule are actually in vibrational motion, there is
consequently an interaction between the vibrational and rotational motions.
Let us illustrate this coupling in terms of the diatomic molecule,
where we write for the Hamiltonian
H =
p2
2µ
+
J2
2µR2
+ a2ξ2
+ a3ξ3
(9.29)
in which the first term is the kinetic energy (and µ is the reduced mass
of the molecule). The second term denotes the rotational energy of the
molecule, while a2ξ2
is the harmonic restoring force for the vibrational
energy, and a3ξ3
is an anharmonic restoring term arising in the vibrational
problem. The distance between the nuclei is now modified by
the vibrational displacements from equilibrium
R − Req
Req
= ξ where R = Req(1 + ξ). (9.30)
We therefore write
1
R2
=
1
R2
eq(1 + ξ)2
=
1
R2
eq
1 − 2ξ + 3ξ2
+ . . . (9.31)
so that we can express the Hamiltonian in terms of an unperturbed
term H0 and a perturbation term H :
H = H0 + H (9.32)
where
H0=p2
2µ
+ BeqJ2
+ a2ξ2
(9.33)
and
Beq =
1
2µR2
eq
. (9.34)
228 CHAPTER 9. MOLECULAR VIBRATIONS
The first term in Eq. 9.33 denotes the kinetic energy and the second
term defines the rotational energy when the molecule is in its equilibrium
configuration, while the third term denotes the vibrational potential
energy for the harmonic restoring forces. Thus H0 gives the
energies for the vibrational and rotational motion in the limit where
the vibrational and rotational motions are decoupled. For the H0 limit
the selection rules are the same as if the vibrations and rotations occurred
independently. The perturbation Hamiltonian then becomes
H = a3ξ3
− 2BeqξJ2
+ 3Beqξ2
J2
(9.35)
where the first term is an anharmonic term that gives rise to overtones
and combination modes in the vibrational spectrum. The second and
third terms in Eq. 9.35 are associated with coupling between rotational
and vibrational levels and give corrections to the rotational levels. The
term in ξJ2
makes a contribution in 2nd-order perturbation theory,
while the term in ξ2
J2
makes a contribution in 1st-order perturbation
theory which is proportional to
n +
1
2
¯hωvj(j + 1).
Thus, the application of perturbation theory results in energy levels for
the vibrational-rotational problem:
En,j = ¯hωv n +
1
2
pure vibrational
+ A1j(j + 1)
pure rotational
+ A2 ¯hωv n +
1
2
j(j + 1) + . . .
interaction terms
(9.36)
in which A1 and A2 are constants. For the diatomic molecule A1 =
(¯h/2I) in accordance with Eq. 9.23. From a group theoretical point of
view, the interaction terms modify the selection rules and new features
in the IR and Raman specta can be seen. In general the symmetry
of an interacting vibrational and rotational level is given by the direct
product Γvib ⊗ Γrot.
In making rotational transitions on absorption between different vibrational
levels, we not only can have ∆j = 1 (the R branch) but
we also can have ∆j = −1 (the P branch). This is illustrated in
the vibrational-rotational spectrum shown in Figure 9.10 for the HCl
9.10. VIBRATIONAL-ROTATIONAL INTERACTION 229
 
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Figure 9.10: P and R branches of
the rotational structure of the HCl
vibrational-rotational band near
2885 cm−1
.
molecule. We note here that the spectral lines in the R branch (upshifted
in frequency) are not symmetrically spaced with respect to the
down-shifted P branch. The Q branch (∆j = 0) occurs very close
to the central frequency ν0, and would in fact be coincident with ν0 if
the moment of inertia would be independent of the vibrational state.
Study of the Q branch requires high resolution laser spectroscopy.
If there were no vibrational-rotational interaction, the spacing of all
spectral lines (shown in the top portion of Figure 9.10) would be the
same for all vibrational levels n. For the case of diatomic molecules
and for the polarization where E is along the molecular axis, then
the selection rules ∆n = +1 and ∆j = 0 determine the vibrationalrotational
spectrum, while for E ⊥ to the main symmetry axis of the
molecule, the selection rules are ∆n = 0 and ∆j = +1.
Rotational Raman Spectra are also observed. Here the transitions
with ∆j = 2 are excited for the pure rotational transitions,
∆n = 0 (see Figs. 9.9 and 9.10). This series is called the S branch.
230 CHAPTER 9. MOLECULAR VIBRATIONS
When vibrational-rotational Raman spectra are excited, transitions
with ∆j = 0 and ∆j = −2 are also possible and these are called
the O branches. Because of the anharmonic terms in the Hamiltonian,
there are vibrational-rotational spectra which occur between
vibrational states separated by ∆n = 2, 3, . . . , etc. These anharmonic
transitions would be expected to have lower intensity.
The above discussion focused on the vibrational degrees of freedom.
There are in addition the electronic levels which generally are separated
by much greater energies than are the vibrational and rotational
levels. There is however some interaction also between the vibrational
and rotational states and the electronic levels. Interactions between
the electronic and rotational levels give rise to “Λ-doubling”
of the rotational levels, and coupling between the electronic and
vibrational levels gives rise to vibronic levels.
9.11 Wigner–Eckart Theorem and Selection
Rules
For proof of the Wigner–Eckart theorem see Tinkham p. 131-2. This
theorem deals with the matrix elements of a tensor T ω
µ where ω is the
rank of the tensor and µ is a component index, to be discussed further
below. The theorem is discussed for angular momentum states which
correspond (through the group of Schr¨odinger’s equation) to the full
rotation group.
The full rotation group has only odd–dimensional representations:
1-dimensional = 0 s-states
3-dimensional = 1 p-states
5-dimensional = 2 d-states
Thus, a scalar ( = 0) corresponds to a tensor with ω = 0 and µ = 0.
A vector corresponds to a tensor with ω = 1, = 1 and µ = ±1, 0, the
three m values for = 1. A second rank tensor can be considered as
the direct product
Γ =1
⊗ Γ =1
= Γ =0
+ Γ =1
+ Γ =2
(9.37)
9.11. WIGNER–ECKART THEOREM AND SELECTION RULES231
having dimensions 3 × 3 = 1 + 3 + 5 = 9. Thus the second rank
tensor will have a part which transforms as ω = 0 and µ = 0, another
part which transforms as ω = 1, µ = ±1, 0 and a third part which
transforms as ω = 2, µ = ±2, 1, 0. The part that transforms as ω = 0
and ω = 2 constitute the symmetric components and the parts that
transforms as ω = 1 corresponds to the anti-symmetric components of
a second rank tensor.
The Wigner–Eckart Theorem gives the selection rules for a tensor
operator Tω
µ between states having full rotational symmetry
N j m Tω
µ Njm = Ajωj
mµ δm ,m+µ (N j ||Tω
||Nj) (9.38)
where j lies in the range
|j − ω| ≤ j ≤ (j + ω). (9.39)
In Eq. 9.38, N and N are principal quantum numbers, j and j are
quantum numbers for the total angular momentum, while m and m
are quantum numbers for the z component of the angular momentum.
The coefficients Ajωj
mµ are called Wigner coefficients and are tabulated
in group theory texts. The reduced matrix element (N j ||Tω
||Nj) in
Eq. 9.38 is independent of µ, m and m and can therefore be found for
the simplest case µ = m = m = 0.
Thus the Wigner–Eckart Theorem gives selection rules on both j
and m. Rewriting the restrictions implied by Eqs. 9.38 and 9.39 yields
the selection rules
|∆j| = |j − j | ≤ ω
|∆m| = |m − m| = µ ≤ ω. (9.40)
We now write down special cases of Eq. 9.40. For electric dipole transitions
(ω = 1) for which the dipole is induced, we have the selection
rules:
∆j = 0, ±1
∆m = 0 for E ˆz
∆m = ±1 for E ⊥ ˆz (9.41)
232 CHAPTER 9. MOLECULAR VIBRATIONS
where E ˆz refers to linear polarization along the quantization axis
and E ⊥ ˆz refers to circular polarization about the quantization axis.
For Raman transitions (where HRaman transforms as a second rank symmetric
tensor) we have either ω = 0 or ω = 2 and the corresponding
selection rules:
ω = 0 : ∆j = 0 ∆m = 0
ω = 2 : ∆j = 0, ±1, ±2 ∆m = 0, ±1, ±2 (9.42)
In specific geometries, not all of these transitions are possible.
In applying the Wigner–Eckart theorem to the rotational selection
rules for a linear diatomic molecule, we know that the dipole
moment must be along the molecular z axis, so that only µ = 0 applies.
In this case the Wigner–Eckart Theorem gives the selection rules
∆j = 0, ±1; ∆m = 0 for I.R. activity
∆j = 0, ±2; ∆m = 0 for Raman activity (9.43)
Since µ = 0 is the only applicable component for the symmetric
top molecule, the same selection rules (Eq. 9.43) as given for the linear
diatomic molecule also apply to the symmetric top molecule. The case
∆j = ±1 does not occur for Raman scattering because the Raman
tensor only contains the symmetric components of the second rank
tensor, whereas ∆j = ±1 corresponds to the antisymmetric components
of the second rank tensor.
9.12 Selected Problems
1. C2H4 (ethylene) is a planar molecule which has the configuration
shown on the diagram below:
9.12. SELECTED PROBLEMS 233
 ¢¡¤£¦¥ £
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(a) Using the point group and χatom sites
found in Problem 2 of
§8.8, find the symmetries of the allowed molecular vibra-
tions.
(b) Sketch the normal mode displacements for each of the allowed
molecular vibrations in (a).
(c) Which modes are infrared-active? Which are Raman-active?
What are the polarization selection rules?
2. Both CO2 and N2O are linear molecules, but have different equilibrium
arrangements:
§©¨
 ¨
(a) What are the appropriate point groups for CO2 and N2O?
(b) What symmetries are involved for the bonding and antibonding
states for these molecules?
(c) What are the differences in the symmetries of the normal
modes for these two molecules?
(d) Show schematically the atomic displacements for the normal
modes of each molecule.
(e) What are the expected differences in their IR spectra? Raman
spectra?
234 CHAPTER 9. MOLECULAR VIBRATIONS
3. Study the proof of the Wigner-Eckart theorem (e.g., Tinkham,
p. 131-2).
4. (a) We will now find the molecular vibrations for the hypothetical
molecule XH12 (see Problem 4 of §8.8) where the 12
hydrogen atoms are at the vertices of a regular icosahedron
and the atom X is at the center of the icosahedron. Find
χatom sites
for XH12 for the icosahedral group Ih.
(b) What are the symmetries for the normal modes? Which are
infrared-active? Raman active?
(c) What are the polarization selection rules for infrared? for
Raman?
Chapter 10
Permutation Groups and
Many-Electron States
In this chapter we discuss the properties of permutation groups, which
are known as the “Symmetric Group” in the mathematics literature.
Although permutation groups are quite generally applicable to manybody
systems, they are used in this chapter to classify the symmetry
in many-electron states. This discussion applies to the symmetries of
both the spin and orbital states. In Chapter 11 we apply the results of
Chapter 10 for the permutation groups to help classify the symmetry
properties for tensors.
The main application of the permutation group in this chapter is to
isolated atoms with full rotational symmetry. We give explicit results
for two, three, four and five electron systems. Whereas two electron systems
can be handled without group theory, the power of group theory
is evident for three, four, five, and even larger electron systems. With a
5-electron system, we can treat all multi-electron states arising from s,
p, and d electrons, since 5 electrons fill half of a d level, and a more than
half-filled level is equivalent to the presence of hole states. To deal with
all multi-electron states that could be made with f electrons we would
need to also consider the permutation groups for 6 and 7 objects. In the
solid state, multi-electron states occur predominantly in the context of
crystal fields, as for example the substitution of a transition metal ion
(having d electrons) on a crystal site with cubic symmetry. The crystal
field lowers the full rotational symmetry of the free ion giving rise to
235
236 CHAPTER 10. PERMUTATION GROUPS
crystal field splittings. In this case the effect of the crystal field must
be considered once the symmetry of the electronic configuration of the
free ion has been determined using the permutation groups discussed
in this chapter.
10.1 Introduction
In the physics of a many-electron atom or molecule we are interested
in solutions to a Hamiltonian of the form
H(r1, ..., rn) =
n
i=1
p2
i
2m
+ V (ri) +
1
2 i=j
e2
rij
(10.1)
where V (ri) is a one-electron potential and the Coulomb electronelectron
interaction term is explicitly included. The one-electron potential
determines the rotational and translational symmetry of the
Hamiltonian.
In addition to symmetry operations in space, the Hamiltonian in
Eq. 10.1 is invariant under interchanges of electrons, i.e., permutation
operations P of the type
P =
1 2 . . . n
a1 a2 . . . an
, (10.2)
where the operation P replaces 1 by a1, 2 by a2, etc. We have already
seen that these permutation operations form a group, i.e., there exists
the inverse operation
P−1
=
a1 a2 ... an
1 2 ... n
, (10.3)
and the identity element is given by
E =
1 2 ... n
1 2 ... n
(10.4)
which leaves the n electrons unchanged. Multiplication involves sequential
permutation operations of the type given by Eq. 10.2. The
10.1. INTRODUCTION 237
number of symmetry operations in a permutation group of n objects is
n! which gives the order of the permutation group to be n!.
The solutions of the many-electron Hamiltonian (Eq. 10.1) are denoted
by ΨΓi
(r1, ..., rn). Since all electrons are indistinguishable, the
permutation P commutes with the Hamiltonian, and we therefore can
classify the wave functions of the group of the Schr¨odinger equation
according to an irreducible representation Γi of the permutation or the
symmetric group. Some permutations give rise to symmetric states,
others to antisymmetric states, and the remainder are neither. In some
cases, all possible states are either symmetric or antisymmetric, and
there are no states that are neither fully symmetric nor fully antisym-
metric.
For the permutation group of n objects amongst the various possible
irreducible representations there are two special 1-dimensional irreducible
representations: one that is symmetric and one that is antisymmetric
under the interchange of two particles. The basis function for
the symmetric representation Γs
1 of an orbital state is just the product
wave function
ΨΓs
1
(r1, r2, . . . , rn) =
1
√
n! permutations
ψ1(r1)ψ2(r2) . . . ψn(rn). (10.5)
The total wave function for a many-electron system is the product of
the orbital and spin wave functions. The basis function for the antisymmetric
representation Γa
1 is conveniently written in terms of the
Slater determinant:
ΨΓa
1
(x1, x2 . . . , xn) =
1
√
n!
ψ1(r1, σ1) ψ1(r2, σ2) . . . ψ1(rn, σn)
ψ2(r1, σ1) ψ2(r2, σ2) . . . ψ2(rn, σn)
...
...
...
...
ψn(r1, σ1) ψn(r2, σ2) . . . ψn(rn, σn)
(10.6)
where xi denotes a generalized coordinate, consisting of ri, the spatial
coordinate and σi, the spin coordinate. When written in this form,
the many-body wave function automatically satisfies the Pauli Principle
since the repetition of either a row or a column results in a zero
determinant thereby guaranteeing that every electron is in a different
state.
238 CHAPTER 10. PERMUTATION GROUPS
The higher dimensional irreducible representations of the permutation
group are also important in determining many-electron states
which satisfy the Pauli principle. For example, in the L · S coupling
scheme, one must take combinations of n spins to get a total S. These
must be combined with the orbital angular momentum combinations to
get a total L. Both the spin states and the orbital states will transform
as some irreducible representation of the permutation group. When
combined to make a total J, only those combinations with transform
as the antisymmetric representation Γa
1 are allowed by the Pauli principle.
We will illustrate these concepts with several examples in this
chapter including the 3-electron p3
state and the 4-electron p4
state.
In this chapter we will use the permutation groups to yield information
about the symmetry and the degeneracy of the states for a
many-electron system. We emphasize that in contrast to the case of
rotational invariance, the ground state of Eq. 10.1 does not transform
as the totally symmetric representation of the permutation group Γs
1.
But rather for electrons (or half integral spin (Fermions) particles), the
ground state and all allowed excited states transform as the antisymmetric
one-dimensional irreducible representation Γa
1 since any physical
permutation H will not distinguish between like particles. The perturbation
itself transforms as the totally symmetric irreducible representation
of the permutation group. Only integral spin particles (Bosons)
have ground states that transform as the totally symmetric irreducible
representation Γs
1.
Mathematicians are presently studying a new aspect of permutations
called braids. A schematic picture for braids is shown in Fig. 10.1
where both the permutation and the ordered sequence of the permutation
is part of the definition of the group element. At present, the mathematicians
cannot express the irreducible representations for braids and
the physics implications of braids are also unexplored at this time.
In this chapter we first discuss the classes of the permutation groups
(§10.2), their irreducible representations (§10.3), and their basis functions
(§10.4). Applications of the permutation groups are then made
(§10.5) to classify 2-electron, 3-electron, 4-electron and 5-electron states.
10.2. CLASSES OF PERMUTATION GROUPS 239
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¢¤£¦¥
¥§£ ¢
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¥!£"¢
#%$'&'(' 
Figure 10.1: The symmetry operations
on a braid of two strands.
The solid curve (line) is over the
dotted line (curve).
10.2 Classes of Permutation Groups
Of particular interest to the symmetry properties of permutation groups
are cyclic permutations. If we have q objects, an example of a cyclic
permutation of q objects is:
1 2 3 ... (q − 1) q
2 3 4 ... q 1
≡ (23 . . . q1),
where (123 . . . q) denotes the identity element. It is clear that the cyclic
permutations of q identical objects are all related to one another by an
equivalence transformation
(123...q) = (234...q1) = (34...q12) = etc. (10.7)
since all of these group elements imply that 1 → 2, 2 → 3, 3 → 4, etc.,
and therefore represent the same physics.
Any permutation can be decomposed into cycles. For example the
permutation
Pi =
1 2 3 4 5 6 7
4 3 2 5 7 6 1
≡ (1457)(23)(6) (10.8)
240 CHAPTER 10. PERMUTATION GROUPS
can be decomposed into 3 cycles as indicated in Eq. 10.8. The decomposition
of a permutation into cycles is unique, since different arrangements
of cycles correspond to different permutations.
Let us assume that a permutation of q objects is decomposed into
cycles as follows: there are λ1 cycles of length 1, λ2 cycles of length 2,
..., λn cycles of length n:
q = λ1 + 2λ2 + ... + nλn. (10.9)
It is easily seen that there are
q!
1λ1 λ1! 2λ2 λ2!...nλn λn!
(10.10)
permutations that have the same cycle structure.
As an example consider the cycle (abc)(d) of the permutation group
P(4). The class (abc)(d) which in the isomorphic point group Td for
the symmetry operations of a regular tetrahedron, corresponds to the
rotation about a 3-fold axis. The number of symmetry operations in
this class according to Eq. 10.10 is
4!
(11)(1!)(31)(1!)
= 8.
Another example is finding the number of symmetry operations in the
class (ab)(cd) of the point group P(4), corresponding to the two-fold
axes aroung x, y, z. From Eq. 10.10 the number of elements in this class
is
4!
(22)(2!)
= 3.
Theorem: Permutations with the same cycle structure belong to the
same class.
Proof: Consider two permutations P and P with the same cycle structure
given by
P = (a1a2 . . . aλ1 )(b1b2 . . . bλ2 ) . . . (d1d2 . . . dλr )
P = (a1a2 . . . aλ1
)(b1b2 . . . bλ2
) . . . (d1d2 . . . dλr
). (10.11)
10.2. CLASSES OF PERMUTATION GROUPS 241
Here P takes a1 → a2, etc., b1 → b2, etc., d1 → d2, etc. while
P does the corresponding permutation for the primed quantities.
Now we introduce the permutation operation T which takes the
primed quantities into the unprimed quantities (e.g., ai → ai)
T =
a1 . . . aλ1
b1 . . . bλ2
. . . d1 . . . dλr
a1 . . . aλ1 b1 . . . bλ2 . . . d1 . . . dλr
(10.12)
and T−1
takes ai → ai. Thus T−1
PT does the following sequence:
ai → ai, ai → ai+1 and finally ai+1 → ai+1. But this is equivalent
to ai → ai+1 which is precisely the permutation P . Therefore
T−1
PT = P
and P is related to P by conjugation, thus completing the proof
of the theorem. The number of elements in each class is found
from Eq. 10.10.
From the above theorem it follows that the number of different
classes (and hence the number of irreducible representations) of the
permutation group of q objects is the number of different cycle structures
that can be formed. Thus, the number of classes is just the
number of ways in which the number q can be written as the sum of
positive integers. For example, q = 4 objects can be constituted into 5
different cycle structures as enumerated below:
q = 4 4 = 4 (1, 2, 3, 4)
4 = 3 + 1 (1, 2, 3)(4)
4 = 2 + 1 + 1 (1, 2)(3)(4)
4 = 2 + 2 (1, 2)(3, 4)
4 = 1 + 1 + 1 + 1 (1)(2)(3)(4)
(10.13)
giving rise to 5 classes. As an example of the notation, 4 = 3 + 1
denotes a cycle structure (123)(4). In the same way, q = 5 objects can
242 CHAPTER 10. PERMUTATION GROUPS
be constituted in 7 different cycle structures giving rise to 7 classes:
q = 5 5 = 5 (1, 2, 3, 4, 5)
5 = 4 + 1 (1, 2, 3, 4)(5)
5 = 3 + 2 (1, 2, 3)(4, 5)
5 = 3 + 1 + 1 (1, 2, 3)(4)(5)
5 = 2 + 1 + 1 + 1 (1, 2)(3)(4)(5)
5 = 2 + 2 + 1 (1, 2)(3, 4)(5)
5 = 1 + 1 + 1 + 1 + 1 (1)(2)(3)(4)(5)
(10.14)
Correspondingly q = 6 gives rise to 11 classes, q = 7 gives rise to 15
classes, q = 8 gives rise to 22 classes, etc.
10.3 The Number of Irreducible Representations
of Permutation Groups
Since permutation groups are finite groups, we can appeal to our experience
regarding finite groups and use the theorem (Eq. 3.40)
h =
i
2
i (10.15)
where i is the dimensionality of the representation i, and h is the order
of the group. For a permutation group of q objects, the order of the
group is h = q!. Since the number of classes is equal to the number
of irreducible representations, we can construct Table 10.1 where P(q)
labels the permutation group of q objects. From Table 10.1 we note that
P(3) is isomorphic with group C3v or alternatively group D3. Similarly
P(4) is isomorphic with the tetrahedral group Td. Although the groups
P(5) and Ih both have 120 symmetry operations, P(5) is not isomorphic
to the icosahedral group Ih since the two groups have different numbers
of classes. The number of classes of P(5) is 7 while the number of classes
of Ih is 10. The dimensions i of the 7 classes in the group P(5) are
listed in Table 10.1, and include two irreducible representations with
i = 1, two with i = 4, two with i = 5, and one with i = 6. The
10 irreducible representations of Ih have the following dimensionalities:
2[1 + 3 + 3 + 4 + 5]. Making use of the isomorphism of P(3) and P(4)
mentioned above, matrix representations for the symmetry operations
of these groups are easily written down.
10.4. BASIS FUNCTIONS OF PERMUTATION GROUPS 243
Table 10.1: The number of classes and a listing of the dimensionalities
of the irreducible representations.
Group Classes Number of group elements i
2
i
P(1) 1 1! = 12 = 1
P(2) 2 2! = 12 + 12 = 2
P(3) 3 3! = 12 + 12 + 22 = 6
P(4) 5 4! = 12 + 12 + 22 + 32 + 32 = 24
P(5) 7 5! = 12 + 12 + 42 + 42 + 52 + 52 + 62 = 120
P(6) 11 6! = 12 + 12 + 52 + 52 + 52 + 52 + 92 + 92 + 102 + 102 + 162 = 720
P(7) 15 7! = 12 + 12 + 62 + 62 + 142 + 142 + 142 + 142 + 152 + 152 + 212 + 212+
+352 + 352 + 202 = 5040
P(8) 22 8! = 12 + 12 + 72 + 72 + 142 + 142 + 202 + 202 + 212 + 212 + 282 + 282+
+352 + 352 + 562 + 562 + 642 + 642 + 702 + 702 + 422 + 902 = 40320
.
.
.
10.4 Basis Functions of Permutation Groups
The one-electron Hamiltonian
H0(r1) =
p2
1
2m
+ V (r1) (10.16)
has one-electron solutions ψ0(r1), ψ1(r1), etc. Thus the solutions of the
many-electron problem can be expanded in terms of products of the
one-electron wave functions for the Hamiltonian in Eq. 10.16. Below,
we write down the ground state many-electron wave function formed by
putting all electrons in the ground state, and the lowest excited states
formed by putting 1 electron in an excited state.
Ground State: (Boson gas)
The many-particle ground state wave function Ψ0 is found by
putting all the particles into the one-particle ground state:
Ψ0 = ψ0(r1)ψ0(r2) . . . ψ0(rn) → Γs
1 (10.17)
and from a group theoretical point of view this orbital state transforms
at the totally symmetric representation Γs
1.
Single Excitation: (e.g., “phonons” or “magnons”)
To form the first excited state, consider the functions gi found by
244 CHAPTER 10. PERMUTATION GROUPS
placing the ith particle in the first excited state ψ1(ri):
ψ1(r1)ψ0(r2) . . . ψ0(rn) = g1
ψ0(r1)ψ1(r2) . . . ψ0(rn) = g2
...
ψ0(r1)ψ0(r2) . . . ψ1(rn) = gn (10.18)
The basis functions given by Eqs. 10.18 transform as an n-dimensional
reducible representation. Decomposition of this reducible representation
yields:
Γn(reducible) = Γs
1 + Γn−1,
where Γs
1 refers to the totally symmetric representation. Denoting the
many-body wave function for the excited state by Ψ to distinguish it
from the ground state function in Eq. 10.17, the basis functions for the
totally symmetric combination with Γs
1 symmetry is:
ΨΓs
1
=
1
√
n
n
i=1
gi → Γs
1 (10.19)
and the other representation depends on the ensemble of phase factors
forming all possible nth
roots of unity:
ΨΓn−1
=



1√
n
n
i=1 ω(i−1)
gi
1√
n
n
i=1 ω2(i−1)
gi
...
1√
n
n
i=1 ωn(i−1)
gi



→ Γn−1 (10.20)
where ω are phase factors given by ω = e
2πi
n .
For the special case n = 2, where ω = −1, we obtain
ΨΓ1
=
1
√
2
[ψ1(r1)ψ0(r2) − ψ0(r1)ψ1(r2)].
For the case n = 3, where ω = e2πi/3
, we obtain
ΨΓ2
=
1
√
3
(ψ1(r1)ψ0(r2)ψ0(r3)+ωψ0(r1)ψ1(r2)ψ0(r3)+ω2
ψ0(r1)ψ0(r2)ψ1(r3))
10.5. PAULI PRINCIPLE IN ATOMIC SPECTRA 245
and its partner
ΨΓ2
=
1
√
3
(ψ1(r1)ψ0(r2)ψ0(r3)+ω2
ψ0(r1)ψ1(r2)ψ0(r3)+ωψ0(r1)ψ0(r2)ψ1(r3))
for the two dimensional irreducible representation. The (n − 1) cyclic
permutations (1)(2 3 . . . n), (1)(n23 . . . (n − 1)), . . . all commute with
each other. Hence the eigenfunctions can be chosen so that these matrices
are diagonal - the (n − 1) eigenvalues being
e
2πi
n
( n−2
2
)
, . . . , e
−2πi
n
( n−2
2
)
Regular Representation:
If all n functions in a Slater determinant are distinct, then the
Slater determinant does not vanish. The Slater determinant is the
unique basis function for the antisymmetric representation Γa
1. For
the case where all n functions are distinct, the n! functions form a
regular representation of the permutation group and the character for
the identity element for the regular representation is the order of the
group and according to Eq. 3.42 we have
χregular
=
q
j
j χΓj
= h = n! (10.21)
where j is the dimension of the irreducible representation Γj and each
representation occurs a number of times which is equal to the dimension
of the representation, and h is the order of the group = n!. If two of
the n functions are identical then the irreducible representation that is
odd under the exchange of the two identical particles (namely Γa
1) does
not occur.
10.5 Pauli Principle in Atomic Spectra
We will in the following subsections of §10.5 apply the results in §10.4
to specify the symmetry of many-body wavefunctions formed by two
electrons, three electrons, etc.
246 CHAPTER 10. PERMUTATION GROUPS
10.5.1 Two-Electron States
For the case of two electrons, the use of group theory is not especially
needed for selecting the proper linear combinations of wave functions
and the same results can be found just from consideration of even and
odd states, since there are only two classes and two irreducible representations
for P(2). We discuss this case here largely for review and
pedagogic reasons.
The Slater determinant for the two-electron problem can be written
as:
Ψ(x1, x2) =
1
√
2
ψ1(r1, σ1) ψ1(r2, σ2)
ψ2(r1, σ1) ψ2(r2, σ2)
(10.22)
where Ψ(x1, x2) denotes the many-electron wave function for the manyelectron
problem and ψ2(r1, σ1) denotes a one-electron wave function
that has an orbital part and a spin part. We use the vector xi to denote
both the orbital and spin variables (ri, σi).
The lowest energy state for the two-electron problem is achieved
by putting both electrons in 1s states, taking the symmetric (s) linear
combination of spatial orbitals and taking the spins antiparallel.
Multiplying out the Slater determinant in this case yields:
Ψ(1, 2) =
1
√
2
ψ1s
s (1)ψ1s
s (2)[α(1)β(2) − α(2)β(1)] (10.23)
where the spin up state is denoted by α and the spin down state by
β, and Ψ(1, 2) denotes the two-electron ground state. The function
[α(1)β(2) − α(2)β(1)] denotes the antisymmetric spin function for the
two electrons.
Let us now consider the transformation properties of these two electrons
more generally, including their excited states. The possible spin
states for two electrons are S = 0, 1. The phase factor for the twoelectron
problem is ω = e2πi/2
= −1 so that the linear combinations
simply involve ±1. For the two-electron problem we can form a symmetric
and an antisymmetric combination of α and β as given in Table
10.2. For the antisymmetric combination (S = 0) as in Eq. 10.23,
we can have only MS = 0 and the corresponding linear combination
of spin states is given in Table 10.2. For the symmetric combination
(S = 1), we can have 3 linear combinations. Only the MS = 1
10.5. PAULI PRINCIPLE IN ATOMIC SPECTRA 247
Table 10.2: Transformation properties of two-electron states under permutations.
The symmetries of the irreducible representations of the
permutation group P(2) label the various spin and orbital angular momentum
states. To obtain the states allowed by the Pauli Principle the
direct product of the symmetries between the orbital and spin states
must contain Γa
1.
Configuration State Irreducible Allowed
Representations States
(α1β2 − β1α2)/
√
2 S = 0 Γa
1
(α1α2 + α2α1)/
√
2, . . . S = 1 Γs
1
s2
L = 0 Γs
1
1
S
1s2s L = 0 Γs
1 + Γa
1
1
S, 3
S
sp L = 1 Γs
1 + Γa
1
1
P, 3
P
p2
L = 0 Γs
1
1
S
p2
L = 1 Γa
1
3
P
p2
L = 2 Γs
1
1
D
d2
L = 0 Γs
1
1
S
d2
L = 1 Γa
1
3
P
d2
L = 2 Γs
1
1
D
d2
L = 3 Γa
1
3
F
d2
L = 4 Γs
1
1
G
f2
L = 0 Γs
1
1
S
f2
L = 1 Γa
1
3
P
f2
L = 2 Γs
1
1
D
f2
L = 3 Γa
1
3
F
f2
L = 4 Γs
1
1
G
f2
L = 5 Γa
1
3
H
f2
L = 6 Γs
1
1
I
248 CHAPTER 10. PERMUTATION GROUPS
combination (α1α2 + α2α1)/
√
2 is listed explicitly in Table 10.2. The
MS = 0 combination (α1β2 + β1α2)/
√
2 and the MS = −1 combination
(β1β2 + β2β1)/
√
2 do not appear in the table.
We also make entries in Table 10.2 for the symmetries of the orbital
angular momentum states. If the two electrons are in the same symmetric
orbital s state (L = 0), then the spin functions must transform as an
antisymmetric linear combination Γa
1 in Table 10.2 and corresponding
to the spectroscopic notation 1
S as in Eq. 10.23. However, if the two s
electrons have different principal quantum numbers, then we can make
both a symmetric and an antisymmetric combination of orbital states,
as is illustrated here for the two electrons occupying 1s and 2s states,
where the symmetric and antisymmetric combinations are
(ψ1s(r1)ψ2s(r2) + ψ1s(r2)ψ2s(r1))/
√
2
which transforms at Γs
1 and
(ψ1s(r1)ψ2s(r2) − ψ1s(r2)ψ2s(r1))/
√
2
which transforms at Γa
1. Because of the Pauli principle, the orbital Γs
1
combination goes with the Γa
1 spin state leading to an 1
S level, while
the Γa
1 orbital state goes with the Γs
1 spin state leading to an 3
S level
(see Table 10.2).
We now consider the next category of entries in Table 10.2. If one
electron is in an s state and the second is in a p state (configuration
labeled sp), the total L value must be L = 1. We however have two
choices for the orbital states: a symmetric Γs
1 state or an antisymmetric
Γa
1 state. The symmetric combination of orbital wave functions (Γs
1)
must then correspond to the S = 0 antisymmetric spin state (Γa
1), resulting
in the 1
P level, whereas the antisymmetric orbital combination
(transforming as Γa
1) goes with the symmetric triplet Γs
1 spin state and
yields the 3
P level (see Table 10.2).
Placing two electrons in p states with the same principal quantum
number (configuration p2
in Table 10.2) allows for a total angular momentum
of L = 0 (which must have Γs
1 symmetry), of L = 1 (with Γa
1
symmetry) and of L = 2 (again with Γs
1 symmetry). Each electron can
be in one of the 3 states (p+
, p0
, p−
), corresponding to ml = 1, 0, −1,
10.5. PAULI PRINCIPLE IN ATOMIC SPECTRA 249
respectively for each one-electron state. Combining the p+
p+
product
yields an ML = 2 state which belongs exclusively to the L = 2
multiplet, whereas combining the p+
p0
states symmetrically yields the
ML = 1 state of the L = 2 multiplet. (We use the notation p+
p0
to denote ψp+ (r1)ψp0 (r2).) However combining p+
p0
antisymmetrically
yields the ML = 1 state of the L = 1 multiplet. The formation of the
two-electron states for the various L and ML values occurring for the p2
configuration is given below. Since the orbital functions for the L = 1
state transform as Γa
1 the spin functions transform as Γs
1 and the L = 1
multiplet is a triplet spin state 3
P. The L = 0 and L = 2 states both
transform as Γs
1 and thus the allowed spin states must be the singlet
spin state S = 0 (see Table 10.2).
The wave functions for the p2
configuration sketched above can be
found in many standard Quantum mechanics text books and are:
L = 2 symmetry (Γs
1) going with Γa
1 for the spins to yield a 1
D state
Ψ(L = 2, ML = 2)=(p+
p+
)
Ψ(L = 2, ML = 1)=(p0
p+
+ p+
p0
)/
√
2
Ψ(L = 2, ML = 0)=[(p0
p0
) + (p+
p−
+ p−
p+
)/
√
2]/
√
2
Ψ(L = 2, ML = −1)=(p0
p−
+ p−
p0
)/
√
2
Ψ(L = 2, ML = −2)=(p−
p−
)
(10.24)
L = 1 symmetry (Γa
1) going with a symmetric spin state (Γs
1) to yield a
3
P state.
Ψ(L = 1, ML = 1)=(p0
p+
− p+
p0
)/
√
2
Ψ(L = 1, ML = 0)=(p+
p−
− p−
p+
)/
√
2
Ψ(L = 1, ML = −1)=(p0
p−
− p−
p0
)/
√
2
(10.25)
L = 0 symmetry (Γs
1) going with an antisymmetric spin state (Γa
1) to
yield a 1
S state.
Ψ(L = 0, ML = 0) = [(p0
p0
) − (p+
p−
+ p−
p+
)/
√
2]/
√
2 (10.26)
Following this explanation for the p2
configuration, the reader can now
fill in the corresponding explanations for the states formed from twoelectron
states derived from the pd, d2
or f2
configurations listed in
Table 10.2.
250 CHAPTER 10. PERMUTATION GROUPS
Table 10.3: Extended character table for permutation group P(3).
χ(E) χ(A,B,C) χ(D,F)
P(3) (13
) 3(2, 1) 2(3)
Γs
1 1 1 1
Γa
1 1 –1 1
Γ2 2 0 –1
χperm.(ψ1ψ1ψ1) 1 1 1 ⇒ Γs
1
χperm.(ψ1ψ1ψ2) 3 1 0 ⇒ Γs
1 + Γ2
χperm.(ψ1ψ2ψ3) 6 0 0 ⇒ Γs
1 + Γa
1 + 2Γ2
10.5.2 Three-Electron States
For the case of three electrons, the use of group theory becomes more
important. In this case we have the permutation group of three objects
P(3) which has 6 elements, 3 classes and 3 irreducible representations
(see Table 10.3). In the extended character table above, we label the
class (13
) to denote the cyclic structure (1)(2)(3) and class (2, 1) to denote
the cyclic structure (12)(3) and class (3) to denote the cyclic structure
(123). The correspondence between the elements E, A, B, C, D, F
and these three classes is immediate and is given in the table explicitly.
Also given below the character table proper are all the possible symmetries
of the permutations for three-electron wave functions. Because
of these additional listings, we call this an extended character table.
The first possibility for the 3-electron state is that all the one-electron
states are the same (ψ1ψ1ψ1). This function is invariant under any of
the 6 permutations of the group, so that all characters are 1 and the
function (ψ1ψ1ψ1) transforms as Γs
1. In the second possible case, one of
the electrons is in a different state (ψ1ψ1ψ2), and since there are 3 possible
combinations that can be formed with the ψ2 one-electron wave
function, we have three distinct functions that can be obtained from
permutation of the electrons. Hence (ψ1ψ1ψ2) transforms as a threedimensional
reducible representation of the permutation group P(3)
with 3 partners for this state. The identity leaves the three partners
invariant so we get a character 3. Each of the permutation operations
10.5. PAULI PRINCIPLE IN ATOMIC SPECTRA 251
[3(2, 1)] leaves one of the partners invariant, so we get a character of 1,
while the cyclic permutations change all partners yielding a character
of 0. The reduction of this reducible representation to its irreducible
components yields Γs
1 + Γ2 as indicated on the table. Finally, we consider
the case when all 3 electrons are in different states (ψ1ψ2ψ3).
This gives rise to 6 partners, and it is only the identity operation which
leaves the partners (ψ1ψ2ψ3) invariant. This reducible representation
(like the regular representation can be expressed in terms of its irreducible
constituents using the relation h = i
2
i ) contains Γs
1+Γa
1 +2Γ2,
which can be directly verified by adding the characters. In the following
chapter, there will be a great deal more discussion of the equivalence
principle that is used here to form the reducible representations given
in Table 10.3 which is the extended character table for P(3).
Let us now look at the spin states that can be made from 3 electrons.
Referring to §10.4 we can make a symmetric state
(α1α2α3)
with symmetry Γs
1 that corresponds to the S = 3/2 and MS = 3/2 spin
state. To obtain the linear combination of spin states for the three other
MS values (MS = 1/2, −1/2, −3/2), we must apply lowering operators
to the MS = 3/2 state (α1α2α3). With regard to the S = 1/2 state,
Eq. 10.18 tells us that this state is a 2-dimensional representation with
partners:
ΨΓ2
=



(g1 + ωg2 + ω2
g3)
(g1 + ω2
g2 + ωg3)
(10.27)
where ω = exp(2πi/3) and where the functions gi are assembled by
sequentially selecting the spin down state β at each of the sites 1, 2 or
3. This explains the first two entries in Table 10.4.
Now let us examine the spatial states. Putting all three electrons in
the same s state would yield a state with L = 0, ML = 0 and having Γs
1
symmetry. Taking the direct product between Γs
1 for the orbital L = 0
state and either of the spin states Γs
1 ⊗ (Γs
1 + Γ2) does not yield a state
with Γa
1 symmetry, and therefore the s3
configuration is not allowed
because of the Pauli principle. This is a group theoretical statement
of the fact that a particular s level can only accommodate one spin up
252 CHAPTER 10. PERMUTATION GROUPS
Table 10.4: Transformation properties of three-electron states under
permutations. The symmetries of the irreducible representations of
the permutation group P(3) label the various spin and orbital angular
momentum states. To obtain the states allowed by the Pauli Principle
the direct product of the symmetries between the orbital and spin states
must contain Γa
1.
Configuration State Irreducible Representation Allowed State
(↑↑↓) S = 1/2 Γ2
(↑↑↑) S = 3/2 Γs
1
s3 L = 0 Γs
1 –
1s22s L = 0 Γs
1 + Γ2
2S
s2p L = 1 Γs
1 + Γ2
2P
sp2 L = 0 Γs
1 + Γ2
2S
sp2 L = 1 Γa
1 + Γ2
2P, 4P
sp2 L = 2 Γs
1 + Γ2
2D
(2p)2(3p) L = 0 Γa
1 + Γ2
2S, 4S
(2p)2(3p) L = 1 2Γs
1 + Γa
1 + 3Γ2
2P, 2P, 2P, 4P
(2p)2(3p) L = 2 Γs
1 + Γa
1 + 2Γ2
2D, 2D, 4D
(2p)2(3p) L = 3 Γs
1 + Γ2
2F
p3 L = 0 Γa
1
4S
p3 L = 1 Γs
1 + Γ2
2P
p3 L = 2 Γ2
2D
p3 L = 3 Γs
1 –
d3 L = 0 Γs
1 –
d3 L = 1 Γa
1 + Γ2
2P, 4P
d3 L = 2 Γs
1 + 2Γ2
2D, 2D
d3 L = 3 Γs
1 + Γa
1 + Γ2
2F, 4F
d3 L = 4 Γs
1 + Γ2
2G
d3 L = 5 Γ2
2H
d3 L = 6 Γs
1 –
f3 L = 0 Γa
1
4S
f3 L = 1 Γs
1 + Γ2
2P
f3 L = 2 Γa
1 + 2Γ2
2D, 2D, 4D
f3 L = 3 2Γs
1 + Γa
1 + 2Γ2
2F, 2F, 4F
f3 L = 4 Γs
1 + Γa
1 + 2Γ2
2G, 2G, 4G
f3 L = 5 Γs
1 + 2Γ2
2H, 2H
f3 L = 6 Γs
1 + Γa
1 + Γ2
2I, 4I
f3 L = 7 Γs
1 + Γ2
2J
f3 L = 8 Γ2
2K
f3 L = 9 Γs
1 –
10.5. PAULI PRINCIPLE IN ATOMIC SPECTRA 253
and one spin down electron. If now one of the electrons is promoted to
a 2s state, then we can make an Γs
1 state and a Γ2 state in accordance
with §10.4 and with the character table for P(3) in Table 10.3, taking
g1 = ψ2s(r1)ψ1s(r2)ψ1s(r3), etc. and forming states such as given in
Eqs. 10.19 and 10.20. The direct product
Γ2 ⊗ Γ2 = Γs
1 + Γa
1 + Γ2
then ensures that a state with Γa
1 symmetry can be assembled to satisfy
the Pauli principle. Since the spin state with Γ2 symmetry corresponds
to a Pauli-allowed component S = 1/2, the allowed 1s2
2s state will be
a doublet 2
S state as shown in Table 10.4. Similar arguments apply to
the formation of s2
p states with L = 1.
For the sp2
configuration the orbital angular momentum can be
L = 0, L = 1 and L = 2. This corresponds to (2 × 6 × 6 = 72) possible
states in the multiplet. We show below using the Pauli principle
and group theory arguments that the number of allowed states is 30.
The spatial states for the sp2
configuration with L = 2 are formed from
products of the type sp+
p+
for the ML = 2 state (see Eqs. 10.24–10.26).
Once again from the character table (Table 10.3) for P(3), the symmetries
which are contained in the three-electron wave function sp+
p+
(denoting ψs(r1)ψp+ (r2)ψp+ (r3)) are Γs
1 and Γ2 just as was obtained for
the 1s2
2s configuration. The only possible allowed state for L = 2 has
S = 1/2 which results in the 2
D state listed in the table. The ML = 1
states are linear combinations of the sp+
p0
functions which have the
symmetries Γs
1 + Γa
1 + 2Γ2, since this case corresponds to (ψ1ψ2ψ3) in
the character table. Of these symmetry types the Γs
1 + Γ2 states are
associated with the ML = 1 state of the L = 2 multiplet, since the irreducible
representation is specified by the quantum number L and the
ML only specify the partners of that irreducible representation. After
this subtraction has been performed the symmetry types Γa
1 + Γ2 for
the L = 1, ML = 1 level are obtained.
Referring to Table 10.4, the symmetry for the L = 0 state of the sp2
configuration could arise from a sp0
p0
state which is of the (ψ1ψ1ψ2)
form and therefore transforms according to Γs
1 + Γ2 symmetry [see the
character table (Table 10.3) for P(3)]. These orbital states go with the
spin states Γa
1.
254 CHAPTER 10. PERMUTATION GROUPS
For the L = 1 state, the orbital Γa
1 irreducible representation goes
with the Γs
1 spin 3/2 state to give rise to a quartet 4
P state while the
Γ2 orbital state can only go with the Γ2 spin state to give a Γa
1 state
when taking the direct product of the symmetries of the orbital and
spin states (Γ2 ⊗ Γ2).
The case of the p3
configuration is an instructive example where
we can see how group theory can be used to simplify the analysis of
the symmetries of multi-electron states. As the number of electrons
increases, the use of group theory becomes essential to keep track of
the symmetries that are possible by the addition of angular momentum
and the symmetries that are allowed by the Pauli principle. For the
p3
configuration, we can have a total of 6 × 6 × 6 = 216 states. We
will show below that if all electrons have the same principal quantum
number only 20 of these states are allowed by the Pauli principle and
we will here classify their symmetry types.
For the p3
configuration we can have L = 3, 2, 1 and 0 total
orbital angular momentum states. In the discussion that follows we
will assume that all electrons have the same principal quantum number
(e.g., 2p3
). For the L = 3 state to be allowed, we must be able to put
all 3 electrons into a (p+
p+
p+
) state to make the ML = 3 state. From
the extended character table (Table 10.3) for P(3), we see that L = 3
must transform as Γs
1. Since the direct product of the orbital and spin
states Γs
1 ⊗ (Γs
1 + Γ2) does not contain Γa
1 this state is not allowed. The
L = 2 multiplet is constructed from an ML = 2 state having p+
p+
p0
combinations which [from the character table (Table 10.3) for P(3) on
p. 250] transform as Γs
1 + Γ2. Since ML = 2 also contributes to the
L = 3 state with symmetry Γs
1, we must subtract Γs
1 from Γs
1 + Γ2 to
get the symmetry Γ2 for the L = 2 state. If we take a direct product
of the orbital and spin states for this case, we obtain
Γ2 ⊗ (Γs
1 + Γ2) = Γs
1 + Γa
1 + 2Γ2,
but it is only the direct product Γ2 ⊗ Γ2 which contributes a state with
Γa
1 symmetry that is allowed by the Pauli principle. Thus only the
2
D state is symmetry-allowed as indicated in Table 10.4. To get the
symmetry of the L = 1 state, consider the combinations p+
p0
p0
and
p+
p+
p−
which contribute to the ML = 1 state. In this case the ML = 1
10.5. PAULI PRINCIPLE IN ATOMIC SPECTRA 255
Table 10.5: Character Table for Group P(4)
P(4) (14) 8(3, 1) 3(22) 6(2, 12) 6(4)
Γs
1 1 1 1 1 1
Γa
1 1 1 1 −1 −1
Γ2 2 −1 2 0 0
Γ3 3 0 −1 1 −1
Γ3 3 0 −1 −1 1
χperm.(ψ1ψ1ψ1ψ1) 1 1 1 1 1 ⇒ Γs
1
χperm.(ψ1ψ1ψ1ψ2) 4 1 0 2 0 ⇒ Γs
1 + Γ3
χperm.(ψ1ψ1ψ2ψ2) 6 0 2 2 0 ⇒ Γs
1 + Γ2 + Γ3
χperm.(ψ1ψ1ψ2ψ3) 12 0 0 2 0 ⇒ Γs
1 + Γ2 + 2Γ3 + Γ3
χperm.(ψ1ψ2ψ3ψ4) 24 0 0 0 0 ⇒ Γs
1 + Γa
1 + 2Γ2 + 3Γ3 + 3Γ3
state contains irreducible representations 2(Γs
1 +Γ2). Since ML = 1 also
appears for L = 2 and L = 3, we need to subtract (Γs
1 + Γ2) to obtain
(Γs
1 +Γ2) for the symmetries of the orbital L = 1 state (see Table 10.4).
For the ML = 0 levels we have the combinations p0
p0
p0
and p+
p−
p0
,
the first transforming as Γs
1 and the second as Γs
1 + Γa
1 + 2Γ2 to give a
total of 2Γs
1 +Γa
1 +2Γ2. However ML = 0 is also present in the L = 3, 2
and 1 multiplets, so we must subtract the irreducible representations
(Γs
1) + (Γ2) + (Γs
1 +Γ2) to obtain Γa
1 for the L = 0 state. For an orbital
angular momentum with symmetry Γa
1, it is only the S = 3/2 Γs
1 spin
state that is allowed by the Pauli principle (see Table 10.4).
The same procedure can be used to obtain all the other entries in
Table 10.4, as well as the many 3-electron states not listed. As the angular
momentum increases (e.g., for the case of d3
or f3
configurations),
group theoretical concepts become increasingly important.
10.5.3 Four-Electron States
In consideration of the 4-electron problem we must consider the permutation
group P(4). The character table for the group P(4) is given in
Table 10.5. Also shown in Table 10.5 are the transformation properties
for the various products of functions. These transformation properties
256 CHAPTER 10. PERMUTATION GROUPS
are obtained in the same way as for the case of the group P(3) discussed
in §10.5.2. The various 4-electron states of a free ion or atom that are
consistent with the Pauli principle are formed with the help of these
tables.
We first consider the possible spin states for the 4-electron configuration.
The transformation of the spin states under the operations
of the permutation group are shown in Table 10.6. The 4 spins
can be arranged to give a total spin of S = 2, S = 1 and S = 0.
The fully symmetric (α1α2α3α4) state, which appears in Table 10.5 as
χperm.(ψ1ψ1ψ1ψ1), has S = 2 and clearly transforms as Γs
1. The S = 1
state is formed from a combination (α1α2α3β4) with MS = 1 and is of
the form χperm.(ψ1ψ1ψ1ψ2), which from the extended character table in
Table 10.5 transforms as Γs
1 + Γ3. But MS = 1 also contributes to the
S = 2 state which transforms as Γs
1. Thus by subtraction, S = 1 transforms
as Γ3. Likewise, the S = 0 state is formed from a configuration
(α1α2β3β4) with MS = 0 which from the extended character Table 10.5
is of the form χperm.(ψ1ψ1ψ2ψ2) and transforms as Γs
1 + Γ2 + Γ3. Upon
subtraction of the symmetry types for the S = 1 and S = 2 states
(Γ3 + Γs
1), we obtain the symmetry Γ2 for the S = 0 state, as shown
in Table 10.6. This completes the discussion for the spin entries to
Table 10.5.
With regard to the symmetries of the s4
, 1s3
2s and 1s2
2s2
orbital
states, these follow from the discussion in §10.5.2. Some similarity is
also found for the sp3
states in Table 10.6.
We illustrate the 4-electron problem with the p4
electron configuration,
assuming the same principal quantum number for all 4 electrons
as for example in a (2p4
) state. Here we can have L = 4, 3, 2, 1 and 0
(see Table 10.6). Starting with the L = 4 multiplet, the ML = 4 state
p+
p+
p+
p+
would have Γs
1 symmetry. This state is forbidden by the
Pauli principle since the direct product of the orbital and spin states
Γs
1 ⊗ (Γs
1 + Γ2 + Γ3) does not contain Γa
1 symmetry. To find the symmetry
for the L = 3 multiplet, we consider the ML = 3 states which arise
from a p+
p+
p+
p0
configuration and from Table 10.5 [giving the character
table for P(4)], we see that (ψ1ψ1ψ1ψ2) contains the irreducible
representations Γs
1 + Γ3. Thus subtracting Γs
1 for the L = 4 state gives
the symmetry Γ3 for the L = 3 multiplet. The direct product of the
10.5. PAULI PRINCIPLE IN ATOMIC SPECTRA 257
Table 10.6: Transformation properties of four-electron states under permutations.
The symmetries of the irreducible representations of the
permutation group P(4) label the various spin and orbital angular momentum
states. To obtain the states allowed by the Pauli Principle the
direct product of the symmetries between the orbital and spin states
must contain Γa
1.
Configuration State Irreducible Representation Allowed State
(↑↑↓↓) S = 0 Γ2
(↑↑↑↓) S = 1 Γ3
(↑↑↑↑) S = 2 Γs
1
s4
L = 0 Γs
1 –
1s3
2s L = 0 Γs
1 + Γ3 –
1s2
2s2
L = 0 Γs
1 + Γ2 + Γ3
1
S
sp3
L = 0 Γa
1 + Γ3
3
S, 5
S
sp3
L = 1 Γs
1 + Γ2 + 2Γ3 + Γ3
1
P , 3
P
sp3
L = 2 Γ2 + Γ3 + Γ3
1
D, 3
D
sp3
L = 3 Γs
1 + Γ3 –
(2p)3
(3p) L = 0 Γs
1 + Γ2 + 2Γ3 + Γ3
1
S, 3
S
(2p)3
(3p) L = 1 Γs
1 + Γa
1 + 2Γ2 + 3Γ3 + 3Γ3
1
P , 1
P , 3
P , 3
P , 3
P , 5
P
(2p)3
(3p) L = 2 2Γs
1 + 2Γ2 + 4Γ3 + 2Γ3
1
D, 1
D, 3
D, 3
D
(2p)3
(3p) L = 3 Γs
1 + Γ2 + 2Γ3 + Γ3
1
F , 3
F
(2p)3
(3p) L = 4 Γs
1 + Γ3 –
p4
L = 0 Γs
1 + Γ2
1
S
p4
L = 1 Γ3 + Γ3
3
P
p4
L = 2 Γs
1 + Γ2 + Γ3
1
D
p4
L = 3 Γ3 –
p4
L = 4 Γs
1 –
d4
L = 0 Γs
1 + 2Γ2
1
S, 1
S
d4
L = 1 2Γ3 + 2Γ3
3
P , 3
P
d4
L = 2 2Γs
1 + Γa
1 + 2Γ2 + 2Γ3 + Γ3
1
D, 1
D, 3
D, 5
D
d4
L = 3 Γ2 + 3Γ3 + 2Γ3
1
F , 3
F , 3
F
d4
L = 4 2Γs
1 + 2Γ2 + 2Γ3 + Γ3
1
G, 1
G, 3
G
d4
L = 5 Γs
1 + 2Γ3 + Γ3
3
H
d4
L = 6 Γs
1 + Γ2 + Γ3
1
I
d4
L = 7 Γ3 –
d4
L = 8 Γs
1 –
f4
L = 0 2Γs
1 + Γa
1 + 3Γ3
5
S
f4
L = 1 2Γ2 + 3Γ3 + 3Γ3
1
P , 1
P , 3
P , 3
P , 3
P
f4
L = 2 2Γs
1 + Γa
1 + 4Γ2 + 3Γ3 + 2Γ3
1
D, 1
D, 1
D, 1
D, 3
D, 3
D, 5
D
f4
L = 3 Γs
1 + Γa
1 + Γ2 + 5Γ3 + 4Γ3
1
F , 3
F , 3
F , 3
F , 3
F , 5
F
f4
L = 4 3Γs
1 + Γa
1 + 4Γ2 + 4Γ3 + 3Γ3
1
G, 1
G, 1
G, 1
G, 3
G, 3
G, 3
G, 5
G
f4
L = 5 Γs
1 + 2Γ2 + 5Γ3 + 4Γ3
1
H, 1
H, 3
H, 3
H, 3
H
f4
L = 6 3Γs
1 + Γa
1 + 3Γ2 + 4Γ3 + 2Γ3
1
I, 1
I, 1
I, 3
I, 3
I, 5
I
f4
L = 7 Γs
1 + Γ2 + 4Γ3 + 2Γ3
1
J, 3
J, 3
J
f4
L = 8 2Γs
1 + 2Γ2 + 2Γ3 + Γ3
1
K, 1
K, 3
K
f4
L = 9 Γs
1 + 2Γ3 + Γ3
3
L
f4
L = 10 Γs
1 + Γ2 + Γ3
1
M
f4
L = 11 Γ3 –
.
.
.
.
.
.
.
.
.
f4
L = 12 Γs
1 –
258 CHAPTER 10. PERMUTATION GROUPS
orbital and spin states
Γ3 ⊗ (Γs
1 + Γ2 + Γ3) = Γs
1 + Γ2 + 3Γ3 + 2Γ3
again does not contain Γa
1 and therefore is not allowed by the Pauli
principle. However the L = 2 state is allowed and gives rise to a 1
D level
since ML = 2 arises from p+
p+
p0
p0
or p+
p+
p+
p−
which respectively
correspond to the symmetries
(Γs
1 + Γ2 + Γ3) + (Γs
1 + Γ3).
Thus subtracting the contributions of ML = 2 to the L = 3 and L = 4
states gives (Γa
1 + Γ2 + Γ3). Now taking the direct product between the
orbital and spin states
(Γs
1 + Γ2 + Γ3) ⊗ (Γs
1 + Γ2 + Γ3) = 3Γs
1 + Γa
1 + 4Γ2 + 5Γ3 + 3Γ3
does contain the Γa
1 symmetry arising from the direct product of Γ2 ⊗Γ2
and corresponding to the S = 0 spin state which is a singlet state.
Likewise the symmetries of the 3
P and 1
S states for L = 1 and L = 0,
respectively, can be found, and the results are given in Table 10.6. Since
a p4
electron configuration is equivalent to a p2
hole configuration the
allowed states for p4
should be the same as for p2
. This can be verified
by comparing p2
in Table 10.2 with p4
in Table 10.6.
It is left to the reader to verify the other entries in Table 10.6 and
to explore the symmetries of other 4-electron combinations not listed.
In finding these entries it should be noted that
Γ2 ⊗ Γ2 = Γs
1 + Γa
1 + Γ2
and
Γ3 ⊗ Γ3 = Γa
1 + Γ2 + Γ3 + Γ3
so that the spatial functions with Γa
1, Γ2 and Γ3 all can give rise to
states allowed by the Pauli principle.
10.5.4 Five-Electron States
The character table for the permutation group of 5 particles is shown in
Table 10.7. Also listed in Table 10.7 are the characters for all possible
10.5. PAULI PRINCIPLE IN ATOMIC SPECTRA 259
Table 10.7: Character table for the permutation group of five particles
P(5) (the symmetric group of order 5)
P(5) or S5 (15
) 10(2, 13
) 15(22
, 1) 20(3, 12
) 20(3, 2) 30(4, 1) 24(5)
Γs
1 1 1 1 1 1 1 1
Γa
1 1 –1 1 1 –1 –1 1
Γ4 4 2 0 1 –1 0 –1
Γ4 4 –2 0 1 1 0 –1
Γ5 5 1 1 –1 1 –1 0
Γ5 5 –1 1 –1 –1 1 0
Γ6 6 0 –2 0 0 0 1
χperm.(ψ1ψ1ψ1ψ1ψ1) 1 1 1 1 1 1 1
χperm.(ψ1ψ1ψ1ψ1ψ2) 5 3 1 2 0 1 0
χperm.(ψ1ψ1ψ1ψ2ψ2) 10 4 2 1 1 0 0
χperm.(ψ1ψ1ψ1ψ2ψ3) 20 6 0 2 0 0 0
χperm.(ψ1ψ1ψ2ψ2ψ3) 30 6 2 0 0 0 0
χperm.(ψ1ψ1ψ2ψ3ψ4) 60 6 0 0 0 0 0
χperm.(ψ1ψ2ψ3ψ4ψ5) 120 0 0 0 0 0 0
260 CHAPTER 10. PERMUTATION GROUPS
distinct products of 5 functions considered within the equivalence representation.
The irreducible representations of P(5) contained in the
decomposition of the reducible equivalence representation are listed be-
low:
S5 Irreducible representations
χperm.(ψ1ψ1ψ1ψ1ψ1) ⇒ Γs
1
χperm.(ψ1ψ1ψ1ψ1ψ2) ⇒ Γs
1 + Γ4
χperm.(ψ1ψ1ψ1ψ2ψ2) ⇒ Γs
1 + Γ4 + Γ5
χperm.(ψ1ψ1ψ1ψ2ψ3) ⇒ Γs
1 + 2Γ4 + Γ5 + Γ6
χperm.(ψ1ψ1ψ2ψ2ψ3) ⇒ Γs
1 + 2Γ4 + 2Γ5 + Γ5 + Γ6
χperm.(ψ1ψ1ψ2ψ3ψ4) ⇒ Γs
1 + 3Γ4 + Γ4 + 3Γ5 + 2Γ5 + 3Γ6
χperm.(ψ1ψ2ψ3ψ4ψ5) ⇒ Γs
1 + Γa
1 + 4Γ4 + 4Γ4 + 5Γ5 + 5Γ5 + 6Γ6
With the help of these tables, the entries in Table 10.8 can be obtained
for the spin and orbital symmetries of a number of 5-electron states that
are listed in this table. The possible spin states are S = 1/2 which occurs
5 times, the S = 3/2 which occurs 4 times and the S = 5/2 which
occurs once. In making the antisymmetric combinations it should be
noted that
Γ4 ⊗ Γ4 = Γa
1 + Γ4 + Γ5 + Γ6
and
Γ5 ⊗ Γ5 = Γa
1 + Γ4 + Γ4 + Γ5 + Γ5 + Γ6
so that the spatial functions with Γa
1, Γ4 and Γ5 all give rise to states
that are allowed by the Pauli Principle. Five-electron states occur in a
half-filled d level. Such half-filled d levels are important in describing
the magnetic ions in magnetic semiconductors formed by the substitution
of Mn2+
for Cd in CdTe or CdSe.
10.6 Discussion
The Pauli-allowed states for n electrons in a more than half filled p shell
and for 6 − n holes are the same. For example, referring to Table 10.8,
the only Pauli-allowed state for p5
is an L = 1, 2
P state. But this
corresponds to a single hole in a p-shell, which has the same allowed
angular momentum states as a single p electron (S = 1/2) in a p-shell.
10.6. DISCUSSION 261
Table 10.8: Transformation properties of five-electron states under permutations.
The symmetries of the irreducible representations of the
permutation group P(5) label the various spin and orbital angular momentum
states. To obtain the states allowed by the Pauli Principle the
direct product of the symmetries between the orbital and spin states
must contain Γa
1.
Configuration State Irreducible Representation Allowed State
(↑↑↑↓↓) S = 1/2 Γ5
(↑↑↑↑↓) S = 3/2 Γ4
(↑↑↑↑↑) S = 5/2 Γs
1
s5
L = 0 Γs
1 –
1s4
2s L = 0 Γs
1 + Γ4 –
1s2
2s2
3s L = 0 Γs
1 + 2Γ4 + 2Γ5 + Γ5 + Γ6
2
S
p5
L = 0 Γ6 –
p5
L = 1 Γs
1 + Γ4 + Γ5 + Γ5
2
P
p5
L = 2 Γ4 + Γ5 + Γ6 –
p5
L = 3 Γs
1 + Γ4 + Γ5 –
p5
L = 4 Γ4 –
p5
L = 5 Γs
1 –
d5
L = 0 Γa
1 + Γ4 + Γ5 + Γ6
2
S, 6
S
d5
L = 1 Γs
1 + 2Γ4 + Γ4 + 3Γ5 + Γ5 + 2Γ6
2
P, 4
P
d5
L = 2 2Γs
1 + 3Γ4 + Γ4 + 4Γ5 + 3Γ5 + 2Γ6
2
D, 2
D, 2
D, 4
D
d5
L = 3 Γs
1 + 4Γ4 + Γ4 + 3Γ5 + 2Γ5 + 4Γ6
2
F, 2
F, 4
F
d5
L = 4 2Γs
1 + 4Γ4 + Γ4 + 4Γ5 + 2Γ5 + 2Γ6
2
G, 2
G, 4
G
d5
L = 5 Γs
1 + 3Γ4 + 3Γ5 + Γ5 + 3Γ6
2
H
d5
L = 6 2Γs
1 + 3Γ4 + 2Γ5 + Γ5 + Γ6
2
I
d5
L = 7 Γs
1 + 2Γ4 + Γ5 + Γ6 –
d5
L = 8 Γs
1 + Γ4 + Γ5 –
d5
L = 9 Γ4 –
d5
L = 10 Γs
1 –
262 CHAPTER 10. PERMUTATION GROUPS
We can denote both of these states by p1
corresponding to the level
designation 2
P. Using the same arguments p2
and p4
have the same
allowed states. Similarly, the states for the d6
electron configuration
are identical to those for the d4
hole configuration which are worked
out in the Table 10.6, etc. In this sense, the tables that are provided
in this chapter are sufficient to handle all atomic s, p and d levels. To
treat the f levels completely we would need to construct tables for the
permutation groups P(6) and P(7).
In solids and molecules where point group symmetry rather than full
rotational symmetry applies, the application of permutation groups to
the many-electron states is identical. Thus the 3d levels of a transition
metal ion in a crystal field of cubic symmetry are split into a Eg and a
T2g level and the allowed d2
levels would be either a 1
Eg or a 1
T2g, 3
T2g
level. In general, crystal field splittings are applied to the many-electron
states whose symmetries are given in Tables 10.2, 10.4, 10.6 and 10.8.
The d states in icosahedral symmetry do not experience any crystal
field splitting and all the arguments of this chapter apply directly.
10.7 Selected Problems
1. Use the following character table for the permutation group P(5).
P (5) or S5 (15
) 10(2, 13
) 15(22
, 1) 20(3, 12
) 20(3, 2) 30(4, 1) 24(5)
Γs
1 1 1 1 1 1 1 1
Γa
1 1 –1 1 1 –1 –1 1
Γ4 4 2 0 1 –1 0 –1
Γ4 4 –2 0 1 1 0 –1
Γ5 5 1 1 –1 1 –1 0
Γ5 5 –1 1 –1 –1 1 0
Γ6 6 0 –2 0 0 0 1
χperm.(ψ1ψ1ψ1ψ1ψ1) 1 1 1 1 1 1 1
χperm.(ψ1ψ1ψ1ψ1ψ2) 5 3 1 2 0 1 0
χperm.(ψ1ψ1ψ1ψ2ψ2) 10 4 2 1 1 0 0
χperm.(ψ1ψ1ψ1ψ2ψ3) 20 6 0 2 0 0 0
χperm.(ψ1ψ1ψ2ψ2ψ3) 30 6 2 0 0 0 0
χperm.(ψ1ψ1ψ2ψ3ψ4) 60 6 0 0 0 0 0
χperm.(ψ1ψ2ψ3ψ4ψ5) 120 0 0 0 0 0 0
(a) Show that there are 10 symmetry elements in the class (2, 13
)
and 20 symmetry elements in class (3, 2) for this permutation
group.
(b) What are the characters for the equivalence transformation
for a state where 3 of the 5 electrons are in one state (e.g.,
10.7. SELECTED PROBLEMS 263
a d-state) and 2 electrons are in another state (e.g., a pstate)?
Explain how you obtained your result. What irreducible
representations are contained in this equivalence
transformation?
2. (a) Consider the addition of Mn2+
as a substitutional magnetic
impurity for CdTe. Since Mn2+
has five 3d electrons, use
the permutation group P(5) to find the Pauli-allowed states
for the Mn2+
ion in CdTe. Of these Pauli-allowed d5
states,
which is the ground state based on Hund’s rule?
(b) Using the electric dipole selection rule for optical transitions,
find the allowed transitions from the ground state in (a) to
Pauli-allowed states in the 3d4
4p configuration.
3. Use the following character table for the permutation group P(6).
P (6) 1 15 45 15 40 120 40 90 90 144 120
(16
) (2, 14
) (22
, 12
) (23
) (3, 13
) (3, 2, 1) (32
) (4, 12
) (4, 2) (5, 1) (6)
Γs
1 1 1 1 1 1 1 1 1 1 1 1
Γa
1 1 –1 1 –1 1 –1 1 –1 1 1 –1
Γ5 5 3 1 –1 2 0 –1 1 –1 0 –1
Γ5 5 –3 1 1 2 0 –1 –1 –1 0 1
Γ5 5 1 1 –3 –1 1 2 –1 –1 0 0
Γ5 5 –1 1 3 –1 –1 2 1 –1 0 0
Γ9 9 3 1 3 0 0 0 –1 1 –1 0
Γ9 9 –3 1 –3 0 0 0 1 1 –1 0
Γ10 10 2 –2 –2 1 –1 1 0 0 0 1
Γ10 10 –2 –2 2 1 1 1 0 0 0 –1
Γ16 16 0 0 0 –2 0 –2 0 0 1 0
(a) Show that there are 45 symmetry elements in the class (22
, 12
)
and 40 symmetry elements in class (3, 13
).
(b) Show that the irreducible representations Γ5 and Γ9 as given
in the character table are orthogonal. (This is a check that
the character table is correct.) Which of the four 5-dimensional
irreducible representations correspond to the basis functions
ΨΓn−1
in Eq. 10.19?
(c) What are the symmetries for the spin angular momentum
states S = 3, 2, 1, 0? To solve this problem, you will have
to find the equivalence transformations corresponding to selected
permutations of spin configurations that are needed to
construct the various spin angular momentum states. (See
264 CHAPTER 10. PERMUTATION GROUPS
Tables 10.5 and 10.6 for the permutation group P(5) to provide
guidance for solving this problem for P(6).)
(d) According to Hund’s rule, what are the S, L and J values
for placing 6 electrons in a d6
electronic configuration. To
which irreducible representations of P(6) do the spin and
spatial parts of this Hund’s rule ground state correspond?
4. Review the proof of the Wigner-Eckart theorem (e.g., Tinkham,
p. 131-2). No written work is expected.
5. Both CO2 and N2O are linear molecules, but have different equilibrium
arrangements:
 ¢¡¤£
¥ £¦¡
(a) What are the appropriate point groups for CO2 and N2O?
(b) What are the differences in the symmetries of the normal
mode vibrations for these two molecules?
(c) Show schematically the atomic displacements for the normal
mode vibrations of each molecule.
(d) What are the expected differences in their IR vibrational
spectra? Raman vibrational spectra?
(e) What are the expected differences in the rotational spectra
of these two molecules?
(f) Which of these rotational modes can be excited by infrared
or Raman spectroscopy?
6. (a) We will now find the molecular vibrations for the hypothetical
molecule XH12 (see Problem 4 of §3) where the 12 hydrogen
atoms are at the vertices of a regular icosahedron
10.7. SELECTED PROBLEMS 265
and the atom X is at the center of the icosahedron. Find
χatom sites
for XH12 for the icosahedral group Ih.
(b) What are the symmetries for the normal modes? Which are
infrared-active? Raman active?
(c) What are the polarization selection rules for the infraredactive
modes? for the Raman-active modes?
7. Use the following character table for the permutation group P(5).
P(5) or S5 (15
) 10(2, 13
) 15(22
, 1) 20(3, 12
) 20(3, 2) 30(4, 1) 24(5)
Γs
1 1 1 1 1 1 1 1
Γa
1 1 –1 1 1 –1 –1 1
Γ4 4 2 0 1 –1 0 –1
Γ4 4 –2 0 1 1 0 –1
Γ5 5 1 1 –1 1 –1 0
Γ5 5 –1 1 –1 –1 1 0
Γ6 6 0 –2 0 0 0 1
χperm.(ψ1ψ1ψ1ψ1ψ1) 1 1 1 1 1 1 1
χperm.(ψ1ψ1ψ1ψ1ψ2) 5 3 1 2 0 1 0
χperm.(ψ1ψ1ψ1ψ2ψ2) 10 4 2 1 1 0 0
χperm.(ψ1ψ1ψ1ψ2ψ3) 20 6 0 2 0 0 0
χperm.(ψ1ψ1ψ2ψ2ψ3) 30 6 2 0 0 0 0
χperm.(ψ1ψ1ψ2ψ3ψ4) 60 6 0 0 0 0 0
χperm.(ψ1ψ2ψ3ψ4ψ5) 120 0 0 0 0 0 0
(a) Multiply element
Pi =
1 2 3 4 5
3 2 1 4 5
by element
Pj =
1 2 3 4 5
4 2 5 1 3
to form PiPj and PjPi. Are your results consistent with the
character table?
(b) Show that there are 10 symmetry elements in the class (2, 13
)
and 20 symmetry elements in class (3, 2) for this permutation
group. Give an example of a symmetry element in each class.
266 CHAPTER 10. PERMUTATION GROUPS
(c) Verify the results given in Table 10.6 for the Pauli-allowed
states for p5
by considering unfilled states (holes) in the p-
shell.
(d) Referring to Table 10.6, what are the irreducible representations
for the spin configuration (↑↑↓↓↓)?
(e) Using Hund’s rule, what are the total spin and orbital angular
momenta values for the d5
configuration? Explain how
to find the irreducible representations for this configuration
from Table 10.6. You will make use of some of the results in
(e) when you work parts (h) and (i).
(f) What are the characters for the equivalence transformation
for a state where 3 of the 5 electrons are in a p-state and 2
electrons are in a d-state? Explain how you obtained your
result. What irreducible representations are contained in
this equivalence transformation?
(g) What are the Pauli allowed states (as would be given in Table
10.6) with the largest L value for the p3
d2
configuration?
(h) Consider the addition of Mn2+
as a substitutional magnetic
impurity for CdTe. Since Mn2+
has five 3d electrons, use
the permutation group P(5) to find the Pauli-allowed states
for the Mn2+
ion in CdTe. Of these Pauli-allowed d5
states,
which is the ground state based on Hund’s rule (see part e)?
(i) Using the electric dipole selection rule for optical transitions,
find the allowed transitions from the ground state for Mn2+
in CdTe in (a) to Pauli-allowed states in the 3d4
4p configu-
ration.
Chapter 11
Transformation of Tensors
In theories and experiments involving physical systems with high symmetry,
one frequently encounters the question of how many independent
terms are required by symmetry to specify a tensor of given rank for
each symmetry group. These questions have simple group theoretical
answers. This chapter deals with the transformation properties of tensors,
with particular attention given to those tensors of rank 2 and
higher that arise in non-linear optics and in elasticity theory. In this
analysis we consider the symmetry implied by the permutation group
which gives the number of independent components in the case of no
point group symmetry. We then consider the additional symmetry that
is introduced by the presence of symmetry elements such as rotations,
reflections and inversions. We explicitly discuss full rotational symmetry
and several point group symmetries.
Reference:
“Tensors and group theory for physical properties of crystals”, W.A.
Wooster, Clarendon Press, Oxford.
11.1 Introduction
We start by listing a few commonly occurring examples of tensors of
rank 2, 3, and 4 that occur in solid state physics. Second rank symmetric
tensors occur in the constitutive equations of Electromagnetic
Theory, as for example in the linear equations relating the current den-
267
268 CHAPTER 11. TRANSFORMATION OF TENSORS
sity to the electric field intensity
J(1)
=
↔
σ
(1)
· E (11.1)
where the conductivity
↔
σ
(1)
is a symmetric second rank tensor. A
similar situation holds for the relation between the polarization and
the electric field
P(1)
=
↔
α
(1)
· E (11.2)
where the polarizability
↔
α
(1)
is a symmetric second rank tensor, and
where
↔
α
(1)
≡
↔
χ
(1)
E is often called the electrical susceptibility. A similar
situation also holds for the relation between the magnetization and the
magnetic field
M(1)
=
↔
χ
(1)
H ·H (11.3)
where the magnetic susceptibility
↔
χ
(1)
H is a symmetric second rank tensor.
These relations all involve second rank tensors:
↔
σ
(1)
,
↔
α
(1)
and
↔
χ
(1)
H .
Each second rank tensor Tij has 9 components (6 symmetric and 3 antisymmetric
combinations under the interchange of the indices i and
j). Thus, a symmetric second rank tensor, such as the polarizability
tensor or the Raman tensor, has only 6 independent components. In
this chapter we are concerned with the symmetry properties of these
tensors under permutations and point group symmetry operations.
As an example of higher rank tensors, consider non-linear optical
phenomena, where the polarization in Eq. 11.2 is further expanded to
higher order terms in E as
P = P(1)
+
↔
α
(2)
· EE +
↔
α
(3)
· EEE + . . . (11.4)
where we can consider the polarizability tensor to be field dependent
↔
α =
↔
α
(1)
+
↔
α
(2)
· E +
↔
α
(3)
· EE + . . . (11.5)
More will be said about the symmetry of the various
↔
α
(i)
tensors under
permutations and point group operations in §11.2. Similar expansions
can be made for Eqs. 11.1 and 11.3.
11.1. INTRODUCTION 269
Let us now consider the number of tensor components. As stated
above
↔
α
(1)
has 32
= 9 coefficients (6 for the symmetric components,
αij = αji). There are 33
= 27 coefficients (10 symmetric) in
↔
α
(2)
,
34
= 81 coefficients (only 15 symmetric) in
↔
α
(3)
, and 35
= 243 coefficients
(21 symmetric) in
↔
α
(4)
, etc. How many tensor components are independent?
Which components are related to one another? How many
independent experiments must be carried out to completely characterize
these tensors? These are important questions that occur in many
fields of physics and materials science. We address these questions in
this chapter.
As another example, consider the piezoelectric tensor which is a
3rd
rank tensor relating polarization per unit volume P to the strain
tensor,
↔
e, where P is given by
P =
↔
d
(2)
·
↔
e, (11.6)
which can be rewritten to show the rank of each tensor explicitly
Pk =
i,j
dkij
ui
xj
(11.7)
in which u is the change in the length of xi. We note that there are 27
components in the tensor
↔
d
(2)
without taking into account symmetry
under permutation operations. A frequently used 4th
rank tensor is the
elastic constant tensor
↔
C
(3)
defined by
↔
T=
↔
C
(3)
·
↔
e (11.8)
where the stress tensor
↔
T and strain tensor
↔
e (i.e., the gradient of
the displacement) are related through the fourth rank elastic constant
tensor
↔
C
(3)
(or Cijkl) which neglecting permutation symmetry would
have 81 components. More will be said about the elastic constant
tensor below (see §11.6).
These tensors and many more are discussed in “Physical Properties
of Crystals” by J.F. Nye. The discussion of tensors which we give
270 CHAPTER 11. TRANSFORMATION OF TENSORS
here is group theoretical, whereas Nye’s book gives tables of the tensors
which summarize many of the results which we can deduce from
our group theoretical analysis. In §11.2, we discuss the reduction in
the number of independent components of tensors obtained from point
group symmetry (rotations, reflections and inversion) while in §11.3 we
discuss the corresponding reduction arising from symmetries associated
with the permutation of tensor indices. The number of independent coefficients
for the case of complete isotropy (full rotational symmetry) is
considered in §11.4, while lower point group symmetries are treated in
§11.5. The independent coefficients of the elastic modulus tensor Cijkl
are discussed in §11.6.
11.2 Independent Components of Tensors
– Point Symmetry Groups
In this section we discuss a very general group theoretical result for
tensor components arising from point group symmetry operations such
as rotations, reflections and inversions. These symmetry operations
greatly reduce the number of independent coefficients that need to be
introduced for the various tensors in crystals having various point group
symmetry. In §11.3 we consider the reduction in the number of independent
coefficients through symmetry under permutation group oper-
ations.
Let us consider a relation between a tensor of arbitrary rank Jij...
and another tensor Fi Fj . . . also of arbitrary rank and arbitrary form
Jij... =
i j ...
{tij...,i j ...}Fi Fj . . . (11.9)
The number of independent non-zero tensor components tij...,i j ... allowed
by point group symmetry in Eq. 11.9 is determined by finding
the irreducible representations contained in both {ΓJij...
} = αiΓi and
{ΓFi Fj ...} = βjΓj. The only non-vanishing couplings between JΓi
and {F, F . . .}Γj
are between partners transforming according to the
same irreducible representation because only these lead to matrix elements
that are invariant under the symmetry operations of the group.
11.3. TENSORS UNDER PERMUTATIONS 271
We therefore transform Eq. 11.9 to make use of the symmetrized form
{J}Γi
= tΓ+
1
{F, F . . .}Γi
(11.10)
where the JΓi
and {F, F . . .}Γi
correspond to the same partners of the
same irreducible representation and tΓ+
1
transforms as a scalar which
has Γ+
1 symmetry. Thus the number of independent matrix elements
in the tensor tij...i j ... is the number of times the scalar representation
Γ+
1 occurs in the decomposition of
{ΓJ } ⊗ {ΓF,F...} =
i
αiΓi ⊗
j
βjΓj =
k
γkΓk. (11.11)
In most cases of interest, permutational symmetry requirements on the
products {F, F, . . .} further limit the number of independent matrix
elements of a tensor matrix, as discussed below (§11.3).
11.3 Independent Components of Tensors
under Permutation Group Symme-
try
In this section we consider the effect of permutation symmetry on reducing
the number of independent components of tensors. For example,
second-rank symmetric tensors occur frequently in solid state physics.
In this case, the permutation symmetry αij = αji restricts the offdiagonal
matrix elements to follow this additional relation, thereby reducing
the number of allowed off-diagonal elements from 6 to 3 since the
symmetric combinations (αij + αji)/2 are allowed and the (αij − αji)/2
vanish by symmetry. Furthermore, the three elements (αij − αji)/2
constitute the 3 components of an antisymmetric 2nd
rank tensor, also
called an axial vector; the angular momentum (listed in character tables
as Ri) is an example of an antisymmetric second-rank tensor. To deal
with the symmetry of a second-rank tensor under permutation operations,
group theory is not needed. However, as the rank of the tensor
increases, group theory becomes increasingly helpful in this symmetry
classification.
272 CHAPTER 11. TRANSFORMATION OF TENSORS
Table 11.1: Transformation properties of various tensors under permutations.
The irreducible representations associated with the designated
permutation group, configuration and state are listed.
Tensor Configuration State Irreducible Representation Group
SS L = 0 Γs
1 P(2)
SD L = 2 Γs
1 + Γa
1 P(2)
DD L = 0 Γs
1 P(2)
C(ij)(kl) DD L = 1 Γa
1 P(2)
DD L = 2 Γs
1 P(2)
DD L = 3 Γa
1 P(2)
DD L = 4 Γs
1 P(2)
pS L = 1 Γs
1 + Γa
1 P(2)
di(jk) pD L = 1 Γs
1 + Γa
1 P(2)
pD L = 2 Γs
1 + Γa
1 P(2)
pD L = 3 Γs
1 + Γa
1 P(2)
p2 L = 0 Γs
1 P(2)
α(1) p2 L = 1 Γa
1 P(2)
p2 L = 2 Γs
1 P(2)
p3 L = 0 Γa
1 P(3)
p3 L = 1 Γs
1 + Γ2 P(3)
α(2) p3 L = 2 Γ2 P(3)
p3 L = 3 Γs
1 P(3)
p4 L = 0 Γs
1 + Γ2 P(4)
p4 L = 1 Γ3 + Γ3 P(4)
α(3) p4 L = 2 Γs
1 + Γ2 + Γ3 P(4)
p4 L = 3 Γ3 P(4)
p4 L = 4 Γs
1 P(4)
p5 L = 0 Γ6 P(5)
p5 L = 1 Γs
1 + Γ4 + Γ5 + Γ5 P(5)
α(4) p5 L = 2 Γ4 + Γ5 + Γ6 P(5)
p5 L = 3 Γs
1 + Γ4 + Γ5 P(5)
p5 L = 4 Γ4 P(5)
p5 L = 5 Γs
1 P(5)
p6 L = 0 Γs
1 + Γ5 + Γ9 P(6)
p6 L = 1 Γ5 + Γ5 + Γ10 + Γ16 P(6)
p6 L = 2 Γs
1 + Γ5 + 2Γ9 + Γ16 P(6)
α(5) p6 L = 3 Γ5 + Γ5 + Γ9 + Γ10 P(6)
p6 L = 4 Γs
1 + Γ5 + Γ9 P(6)
p6 L = 5 Γ5 P(6)
p6 L = 6 Γs
1 P(6)
11.3. TENSORS UNDER PERMUTATIONS 273
For illustrative purposes, we now consider the case of the second
rank tensor from the point of view of permutation group symmetry. Referring
to Table 11.1 (which is constructed from tables in Chapter 10),
we see that a second rank symmetric tensor is represented by pp, which
we can consider as the generic prototype of a second rank symmetric
tensor. From the discussion of Chapter 10, we found that p2
could have
angular momentum states L = 0, 1, 2 with the indicated permutation
group symmetries, labeled “irreducible representations” in Table 11.1,
and yielding a total number of states equal to 1 + 3 + 5 = 9. From
the table, it is seen that the symmetric states arise from the L = 0
and L =2 entries, corresponding to 1+5=6 states. Thus we obtain 6
independent coefficients for a symmetric second rank tensor based on
permutation symmetry alone. The number of independent coefficients
for the 2nd
rank antisymmetric tensor (transforming Γa
1) is correspondingly
equal to 3, and the antisymmetric contribution arises from the
L =1 state.
A third rank symmetric tensor (such as
↔
α
(2)
) is more interesting
from a group theoretical standpoint. Here we need to consider permutations
of the type p3
, so that p3
can be considered as the appropriate
basis function of the permutation group P(3) for the permutation symmetry
of
↔
α
(2)
. Referring to Eq. 11.4, we note that the EE fields are
clearly symmetric under interchange of E ↔ E; but since Eq. 11.5 defines
the general non-linear polarizability tensor
↔
α, then all terms in the
expansion of
↔
α must be symmetric under interchange of P and E. From
Table 11.1, we see that p3
consists of L = 0, 1, 2, 3 angular momentum
states. The entries for the p3
configuration in Table 11.1 come from Table
10.4 on p. 252 which contains a variety of configurations that can
be constructed from 3 electrons (or more generally 3 interchangeable
vectors). The total number of states in the p3
configuration is found
by considering the degeneracy of each angular momentum state which
for the L = 0, 1, 2, 3 multiplets is
(1)(1) + 3(1 + 2) + 5(2) + 7(1) = 27
which includes all 33
combinations. Of this total, the number of symmetric
combinations that go with Γs
1 is only 3(1)+7(1)=10. Similarly
Table 11.1 shows that there is only one antisymmetric combination. Of
274 CHAPTER 11. TRANSFORMATION OF TENSORS
Table 11.2: Number of independent components for various tensors for
the listed group symmetries
# of independent coefficients
Group Repr.(a) angular momentum values(b) C(ij)(kl) dk(ij) α(1) α(2) α(3) α(4)
R∞
(c) Γl=0 l = 0 2 0 1 0 1 0
Ih A1g l = 0, 6, 10, . . . 2 0 1 0 1 0
Oh A1g l = 0, 4, 6, 8, 10, . . . 3 0 1 0 2 0
Td A1 l = 0, 3, 4, 6, 7, 8, 9, . . . 3 1 1 1 2 1
D∞h A1g l = 0, 2, 4, 6, . . . 5 1 2 0 3 0
C∞v A1 l = 0, 1, 2, 3, 4, 5, . . . 5 4 2 2 3 3
D6h A1g l = 0, 2, 4, 6, . . . 5 1 2 0 3 0
C1 A1 l = 0, 1, 2, 3, 4, 5, . . .(d) 21 18 6 10 15 21
(a)
The notation for the totally symmetric irreducible representation for
each group is given.
(b)
The angular momentum states that contain the A1 (or A1g) irreducible
representation for the various symmetry groups (see Table 11.1).
(c)
The full rotational symmetry group is denoted by R∞.
(d)
For this lowest point group symmetry, the A1 representation occurs
2l + 1 times. For the other groups in this table, there is only one occurrence
of A1 for each listed l value. However, for higher l’s, multiple
occurrences of A1 may arise (e.g., in Oh symmetry, the l = 12 state has
two A1g modes).
—
interest (and perhaps not appreciated by many workers in the field) is
the large number of combinations that are neither symmetric nor antisymmetric:
3(2)+5(2)=16. Thus, Table 11.1 shows that on the basis
of permutation symmetry, there are only 10 independent coefficients
for
↔
α
(2)
, assuming no additional point group symmetry. This result is
summarized in Table 11.2.
As the next example, consider
↔
α
(3)
which is a fourth rank tensor that
couples P and EEE symmetrically. The generic tensor for this case is
p4
in Table 11.1 (taken from Table 10.6 on p. 257 for 4 electrons) with
81 coefficients neglecting permutational and point group symmetries,
obtained as follows:
(1)(1 + 2) + (3)(3 + 3) + 5(1 + 2 + 3) + 7(3) + 9(1) = 81.
Of these, 1+5+9=15 are symmetric (transforms as Γs
1) and this entry
11.3. TENSORS UNDER PERMUTATIONS 275
is included in Table 11.2. There are no antisymmetric combinations
(i.e., there is no Γa
1 for p4
in P(4)).
Another commonly occurring tensor in solid state physics is the
elastic modulus tensor Cijkl = C(ij)(kl) which is the direct product of two
symmetric tensors, each having 6 independent components, and thus
leading to 6×6=36 components for the product. But C(ij)(kl) is further
symmetric under interchange of ij ↔ kl, reducing the 30 off-diagonal
components of the 6×6 matrix into 15 symmetric and 15 antisymmetric
combinations, in addition to the 6 diagonal symmetric components,
as is explained in standard solid state physics texts. From a group
theoretical standpoint, the (ij) and (kl) are each treated as p2
units
which form total angular momentum states of L = 0 (labeled S in
Table 11.1) and L =2 (labeled D). Under the permutation group P(2),
we can make one SS combination (L = 0) and one symmetric and one
antisymmetric SD combination (L =2), and DD combinations with
L = 0, 1, 2, 3, 4. Adding up the total number of combinations that
can be made from C(ij)(kl) we get
(1)(1) + 5(1 + 1) + 1(1) + 3(1) + 5(1) + 7(1) + 9(1) = 36,
in agreement with the simple argument given above. Of these 21 are
symmetric (i.e., go with Γs
1) while 15 are antisymmetric (i.e., go with
Γa
1), and the number 21 appears in Table 11.2. If we had instead used
p4
in Table 11.1 as the basis function for the permutation of the elastic
tensor Cijkl, we would have neglected the symmetric interchange of the
stress and strain tensors (ij) ↔ (kl).
The final tensor that we will consider is the piezoelectric tensor
di(jk) formed as the symmetric direct product of a vector and a symmetric
second rank tensor (3×6=18 components). The symmetries are
calculated following the pS and pD combinations, using the concepts
discussed for the transformation properties of the
↔
α
(1)
and C(ij)(kl) tensors.
This discussion yields 18 independent coefficients for di(jk) under
permutation symmetry.
In summary, each second rank symmetric tensor is composed of
irreducible representations L = 0 and L = 2 of the full rotation group,
the third rank symmetric tensor from L = 1 and L = 3, the fourth rank
symmetric tensor from L = 0, L = 2 and L = 4, the elastic tensor from
276 CHAPTER 11. TRANSFORMATION OF TENSORS
L = 0, 2L = 2 and L = 4, and the piezoelectric tensor from 2L = 1,
L = 2 and L = 3.
We use these results to find the number of independent coefficients
for each symmetry group.
11.4 Independent Components of Tensors
under Full Rotational Symmetry
The highest point group symmetry is the full isotropy provided by the
full rotation group R∞. The number of the independent components
can also be found from Table 11.2. In §11.2, we showed that the number
of independent coefficients in a tensor formed by the direct product of
two tensors is the number of times this direct product contains Γs
1 in
the fully symmetric irreducible representation L = 0.
Referring to Table 11.1, we find that for the second rank tensor
(ΓL=0 + ΓL=2), ΓL=0 is contained once, so that α11 = α22 = α33 and
α12 = α23 = α31 = 0 and the result for the number of independent
components is given in Table 11.2.
Likewise Table 11.1 shows that there are no independent coefficients
for
↔
α
(2)
in full rotational symmetry and this tensor vanishes by symmetry
(as well as all tensors of odd rank of this type). With regard to
the 4th
rank tensor,
↔
α
(3)
, Table 11.1 shows only one independent coefficient.
In contrast, the C(ij)(kl) 4th
rank tensor contains 2 independent
coefficients in full rotational symmetry and the components of di(jk) all
vanish by symmetry.
This completes the discussion for the form of the various tensors
in Table 11.2 under full rotational symmetry. Also listed in the table
are the number of independent coefficients for several point group symmetries,
including Ih, Oh, Td, D∞h, C∞v, D6h, and C1. These results
can be derived by considering these groups are subgroups of the full
rotational group, and going from higher to lower symmetry. Some illustrative
examples of the various point group symmetries are given in
the following sections.
11.5. TENSORS ARISING IN NON-LINEAR OPTICS 277
Table 11.3: Character Table and Bases for the Cubic Group Oh
Repr. Basis E 3C2
4 6C4 6C2 8C3 i 3iC2
4 6iC4 6iC2 8iC3
Γ+
1 1 1 1 1 1 1 1 1 1 1 1
Γ+
2
x4(y2 − z2)+
y4(z2 − x2)+
z4(x2 − y2)
1 1 -1 -1 1 1 1 -1 -1 1
Γ+
12
x2 − y2
2z2 − x2 − y2 2 2 0 0 -1 2 2 0 0 -1
Γ+
15
xy(x2 − y2)
yz(y2 − z2)
zx(z2 − x2)
3 -1 1 -1 0 3 -1 1 -1 0
Γ+
25 xy, yz, zx 3 -1 -1 1 0 3 -1 -1 1 0
Γ−
1
xyz[x4(y2 − z2)+
y4(z2 − x2)+
z4(x2 − y2)]
1 1 1 1 1 -1 -1 -1 -1 -1
Γ−
2 xyz 1 1 -1 -1 1 -1 -1 1 1 -1
Γ−
12
xyz(x2 − y2)
xyz(2z2 − x2 − y2)
2 2 0 0 -1 -2 -2 0 0 1
Γ−
15 x, y, z 3 -1 1 -1 0 -3 1 -1 1 0
Γ−
25
z(x2 − y2)
x(y2 − z2)
y(z2 − x2)
3 -1 -1 1 0 -3 1 1 -1 0
11.5 Tensors Arising in Non-Linear Op-
tics
In this section we consider tensors arising in non-linear optics, including
symmetric 2nd
rank, 3rd
rank and 4th
rank tensors.
11.5.1 Cubic Symmetry – Oh
The character table for group Oh is shown in Table 3.33 on p. 70.
Table 11.3 gives the same character table for the cubic group, but using
solid state physics notation and the table furthermore contains more
basis functions. We first consider the transformation properties of the
linear response tensor
↔
α
(1)
and the non-linear polarizability tensors
↔
α
(2)
and
↔
α
(3)
.
Consider for example the second rank tensor
↔
α
(1)
defined by
P =
↔
α
(1)
· E. (11.12)
278 CHAPTER 11. TRANSFORMATION OF TENSORS
Both P and E transform as Γ−
15 which gives
ΓP ⊗ ΓE = Γ−
15 ⊗ Γ−
15 = Γ+
1 + Γ+
12 + Γ+
15 + Γ+
25. (11.13)
But since the Γ+
15 is antisymmetric under interchange of i ↔ j in any
of the symmetric second rank tensors, we write
Γ
(s)
↔
e
= Γ+
1 + Γ+
12 + Γ+
25, Γ
(a)
↔
e
= Γ+
15 (11.14)
showing the 6 partners for the second rank symmetric tensor, and the
3 partners for the second rank antisymmetric tensor. These results
can also be obtained starting from the full rotation group, considering
the decomposition of the L = 0 and L = 2 states for the symmetric
partners and the L = 1 states for the antisymmetric partners.
Since Γ+
1 is contained only once in the direct product Γ−
15 ⊗ Γ−
15 in
cubic Oh symmetry, there is only one independent tensor component
for
↔
α
(1)
and we can write
↔
α
(1)
= α0
↔
1 where
↔
1 is the unit tensor and α0
is a constant. As a consequence of this general result, the conductivity
in cubic symmetry (Oh or Td) is independent of the direction of the
fields relative to the crystal axes and only one experiment is required
to measure the polarizability or the conductivity on an unoriented cubic
crystal.
In non-linear optics the lowest order non-linear term is
↔
α
(2)
· EE in
Eq. 11.4 where
↔
α
(2)
is a third rank tensor. Since (EE) is symmetric
under interchange, then (EE) transforms as
Γ
(s)
EE
= Γ+
1 + Γ+
12 + Γ+
25, (11.15)
where we have thrown out the Γ+
15 term because it is antisymmetric
under interchange of EiEj −→ EjEi. Thus, we obtain the irreducible
representation contained in the direct product:
ΓP ⊗ Γ
(s)
EE
= Γ−
15 ⊗ {Γ+
1 + Γ+
12 + Γ+
25}
= (Γ−
2 + 2Γ−
15 + Γ−
25)(s)
+ (Γ−
12 + Γ−
15 + Γ−
25)
(11.16)
11.5. TENSORS ARISING IN NON-LINEAR OPTICS 279
Table 11.4: Decomposition of angular momentum in terms of irreducible
representation for the cubic group Oh.
Γ+
1 Γ+
2 Γ+
12 Γ+
15 Γ+
25 Γ−
1 Γ−
2 Γ−
12 Γ−
15 Γ−
25
0 1
1 1
2 1 1
3 1 1 1
4 1 1 1 1
5 1 2 1
6 1 1 1 1 2
7 1 1 2 2
8 1 2 2 2
9 1 1 1 3 2
10 1 1 2 2 3
11 1 2 3 3
12 2 1 2 3 3
13 1 1 2 4 3
14 1 1 3 3 4
15 1 2 2 4 4
—
yielding 18 partners, 10 of which are symmetric, in agreement with the
general result in Table 11.1. Of particular significance is the fact that
none of the 10 symmetric irreducible representations have Γ+
1 symmetry.
Thus there are no non-vanishing tensor components for a third
rank tensor (such as
↔
α
(2)
) in Oh symmetry, a result which could also be
obtained by going from full rotational symmetry to Oh symmetry. The
10 symmetric partners are found from Table 11.1 and includes angular
momentum states L = 1 (corresponding to Γ−
15) and L = 3 (corresponding
to Γ−
2 + Γ−
15 + Γ−
25) (see Table 11.4). We will now use the symmetric
partners of the third rank tensor to discuss the 4th
rank tensors.
The next order term in Eq. 11.4 for the non-linear response to a
strong optical beam (e.g., multiple photon generation) is the fourth
rank tensor
↔
α
(3)
defined by
P(3)
=
↔
α
(3)
· EEE. (11.17)
280 CHAPTER 11. TRANSFORMATION OF TENSORS
If we consider the product EEE to arise from the symmetric combination
for a 3rd
rank tensor (see Eq. 11.16), then
Γ
(s)
EEE
= Γ−
2 + 2Γ−
15 + Γ−
25 (11.18)
in cubic Oh symmetry, and
ΓP ⊗ Γ
(s)
EEE
=Γ−
15 ⊗ {Γ−
2 + 2Γ−
15 + Γ−
25}
=2Γ+
1 + Γ+
2 + 3Γ+
12 + 3Γ+
15 + 4Γ+
25.
(11.19)
Referring to Table 11.1 we see that the symmetric partners for p4
correspond
to L = 0 (giving Γ+
1 ), L = 2 (giving Γ+
12 + Γ+
25) and L = 4
(giving Γ+
1 + Γ+
12 + Γ+
15 + Γ+
25) yielding the 15 symmetric partners
(2Γ+
1 + 2Γ+
12 + Γ+
15 + 2Γ+
25)(s)
.
Since Γ+
1 is contained twice among the 15 symmetric partners in cubic
Oh symmetry, the symmetric 4th
rank tensor
↔
α
(3)
has 2 independent
coefficients.
11.5.2 Tetrahedral Symmetry – Td
The character table for Td symmetry is given in Table 3.34 on p. 70 and
the correspondence between irreducible representations in the Oh and
Td groups is given in Table 11.5. The group Td has half the number
of symmetry operations as the group Oh. In particular, the inversion
operation is no longer a symmetry element. The major change in the
symmetry of the tensor components of Td crystals relative to crystals
with Oh symmetry involves the 3rd
rank tensors
↔
α
(2)
. Explicitly, we
see from Table 11.5 that since Γ−
2 (Oh) → Γ1(Td), Eq. 11.16 shows that
there exists one nonvanishing tensor component in Td symmetry for
a 3rd
rank tensor
↔
α
(2)
. This means that zinc-blende structures such
as (GaAs and InSb) can have non-vanishing non-linear optical terms
in
↔
α
(2)
because in Td symmetry, the symmetric partners of the direct
product transform as:
(ΓP ⊗ Γ
(s)
EE
)(s)
= Γ1 + 2Γ25 + Γ15 (11.20)
and the Γ1 representation is contained once (see Table 11.2).
11.5. TENSORS ARISING IN NON-LINEAR OPTICS 281
Table 11.5: The compatibility relations given the correspondence between
groups Oh and Td
Oh Td
Γ+
1 , Γ−
2 Γ1
Γ−
1 , Γ+
2 Γ2
Γ±
12 Γ12
Γ−
25, Γ+
15 Γ15
Γ−
15, Γ+
25 Γ25
—
11.5.3 Hexagonal Symmetry – D6h
The character table for D6h (hexagonal symmetry) is shown in Table
3.28 on p. 69. In this subsection we will use the notation found in
this character table. Vector forces in hexagonal symmetry decompose
into 2 irreducible representations
Γvector = A2u + E1u. (11.21)
Thus the 2nd
rank conductivity tensor requires consideration of
ΓP ⊗ ΓE = (A2u + E1u) ⊗ (A2u + E1u)
= 2A1g + A2g + 2E1g + E2g
= (2A1g + E1g + E2g)(s)
+ (A2g + E1g)(a)
.
(11.22)
Equation 11.22 indicates that there are two independent components
for a symmetric second rank tensor such as the conductivity tensor.
Hence, one must measure both in-plane and out-of-plane conductivity
components to determine the conductivity tensor.
The symmetric tensor components (6 partners) of Eq. 11.22 are
Γ
(s)
EE
= 2A1g + E1g + E2g (11.23)
and the antisymmetric components (3 partners) are (A2g +E1g). Hence
282 CHAPTER 11. TRANSFORMATION OF TENSORS
for the symmetric 3rd
rank tensor we can write
ΓP ⊗ Γ
(s)
EE
= (A2u + E1u) ⊗ (2A1g + E1g + E2g)
= (A1u + A2u + B1u + B2u + 2E1u + E2u)(s)
+ (2A2u + 4E1u + E2u)
(11.24)
and there are thus no non-vanishing 3rd
rank tensor components in
hexagonal D6h symmetry because of parity considerations.
For the 4th
rank tensor we have
ΓP ⊗ Γ
(s)
EEE
= (A2u + E1u) ⊗ (A1u + A2u + B1u + B2u + 2E1u + 2E2u)
= (3A1g + B1g + B2g + 2E1g + 3E2g)(s)
+ (3A2g + 2B1g + 2B2g + 4E1g + 3E2g)
(11.25)
and there are three independent tensor components. This result could
also be obtained by going from full rotational symmetry (L = 0, L = 2,
and L = 4), yielding the identical result
[A1g + (A1g + E1g + E2g) + (A1g + B1g + B2g + E1g + 2E2g)].
The results for D6h and D∞h (see Table 11.2) show great similarity
between the behavior of all the tensors that are enumerated in this
table.
11.5.4 Hexagonal Symmetry – D3h
The character table for D3h is shown in Table 3.31 on p. 69, and again,
it is clear that D3h has no inversion symmetry. Comparison of the character
tables for D6h and D3h shows that Γ±
i (D6h) → Γi(D3h). Thus the
only difference between the tensor properties in D6h and D3d symmetries
involves odd rank tensors. Referring to Eq. 11.24 we see that for
D3h there are two non-vanishing 3rd
rank tensor components and once
again piezoelectric phenomena are symmetry allowed.
11.6. ELASTIC MODULUS TENSOR 283
11.6 Elastic Modulus Tensor
The elastic modulus tensor represents a special case of a fourth rank
tensor (see Eq. 11.8). The elastic energy is written as
W =
1
2
Cijkleijekl (11.26)
where W transforms as a scalar, the eij strain tensors transform as
second rank symmetric tensors, and the Cijkl matrices transform as a
fourth rank tensor formed by the direct product of two symmetric second
rank tensors. The symmetry of Cijkl with regard to permutations
was considered in §11.3. With regard to point group symmetry, we have
the result following Eq. 11.11 that the maximum number of independent
components of the Cijkl tensor is the number of times the totally
symmetric representation A1g is contained in the direct product of the
symmetric part of Γeij
⊗Γekl
. In this section we provide a review of the
conventions used to describe the Cijkl tensor and then give results for a
few crystal symmetries.
To make a connection between the elastic constants found in Nye’s
book and in conventional solid state physics books, we introduce a
contracted notation for the stress tensor and the strain tensor:
σ1 = σ11
σ2 = σ22
σ3 = σ33
σ4 = (σ23 + σ32)/2
σ5 = (σ13 + σ31)/2
σ6 = (σ12 + σ21)/2
ε1 = e11
ε2 = e22
ε3 = e33
ε4 = (e23 + e32)
ε5 = (e13 + e31)
ε6 = (e12 + e21)
. (11.27)
Since both the stress and strain tensors are symmetric, then Cijkl can
have no more than 36 components. We further note from Eq. 11.26 that
the Cijkl are symmetric under the interchange of ij ↔ kl, thereby reducing
the number of independent components to 21 for a crystal with
no symmetry operations beyond translational symmetry of the lattice.
Crystals with non-trivial symmetry operations such as rotations, reflections
and inversions will have fewer than 21 independent coefficients.
Using the notation of Eq. 11.27 for the stress and strain tensors, the
284 CHAPTER 11. TRANSFORMATION OF TENSORS
stress-strain relations can be written as










σ1
σ2
σ3
σ4
σ5
σ6










=










C11 C12 C13 C14 C15 C16
C22 C23 C24 C25 C26
C33 C34 C35 C36
C44 C45 C46
C55 C56
C66




















ε1
ε2
ε3
ε4
ε5
ε6










(11.28)
where the contracted Cij matrix is symmetric, with the 21 independent
coefficients containing 15 off-diagonal components and 6 diagonal
components. In the most compact form, we write
σi = Cijεj i, j = 1, . . . 6 (11.29)
where the Cij components are normally used in the description of the
mechanical properties of solids. The introduction of additional symmetry
operations reduces the number of independent components from
the maximum of 21 for a monoclinic crystal group C1 with no symmetry
to a much smaller number (e.g., 2 for the full rotational group
R∞). We consider here the case of full rotational symmetry, icosahedral
symmetry, cubic symmetry, full axial symmetry, and hexagonal
symmetry.
Fiber reinforced composites represent an interesting application of
these symmetry forms. If the fibers are oriented in three dimensional
space in the six directions prescribed by icosahedral symmetry, then
isotropy of the elastic modulus tensor will be obtained. In the corresponding
two dimensional situation, if the fibers are oriented at 60◦
intervals, then isotropy is obtained in the plane. It is standard practice
in the field of fiber composites to use fiber composite sheets stacked
at 60◦
angular intervals to obtain “quasi–planar isotropy”. Recent research
on quasi–crystals has emphasized the connection of the icosahedral
symmetry to describe the elastic properties of quasi–crystals.
11.6.1 Full Rotational Symmetry: 3D Isotropy
The highest overall symmetry for an elastic medium is the full rotation
group which corresponds to “jellium”. For the case of full rotational
symmetry, a second rank tensor
↔
T transforms according to the
11.6. ELASTIC MODULUS TENSOR 285
representation Γ↔
T
where Γ↔
T
can be written as a symmetric and an
antisymmetric part
Γ↔
T
= Γ
(s)
↔
T
+ Γ
(a)
↔
T
. (11.30)
In Eq. 11.30 the symmetric components for full rotational symmetry
transform as the irreducible representations
Γ
(s)
↔
T
= Γl=0 + Γl=2 (11.31)
and the antisymmetric components transform as
Γ
(a)
↔
T
= Γl=1, (11.32)
in which the irreducible representations of the full rotation group are
denoted by their total angular momentum values l. Since the stress
tensor · F ≡
↔
X and the strain tensor
↔
e are symmetric second rank
tensors, both Xα and eij transform according to (Γl=0 +Γl=2) in full rotational
symmetry, where Xα denotes a force in the x direction applied
to a plane whose normal is in the α direction.
The fourth rank symmetric Cijkl tensor of Eq. 11.26 transforms according
to the symmetric part of the direct product of two second rank
symmetric tensors Γ
(s)
↔
e
⊗ Γ
(s)
↔
e
yielding
(Γl=0+Γl=2)⊗(Γl=0+Γl=2) = (2Γl=0+2Γl=2+Γl=4)(s)
+(Γl=1+Γl=2+Γl=3)(a)
,
(11.33)
in which the direct product has been broken up into the 21 partners
that transform as symmetric irreducible representations (s) and the
15 partners for the antisymmetric irreducible representations (a). In
general eij is specified by 6 constants and the Cijkl tensor by 21 constants
because Cijkl is symmetrical under the interchange of ij ↔ kl. Since
all the symmetry groups of interest are subgroups of the full rotation
group, the procedure of going from higher to lower symmetry can be
used to determine the irreducible representations for less symmetric
groups that correspond to the stress and strain tensors and to the elastic
tensor Cijkl.
In the case of full rotational symmetry, Eq. 11.33 shows that the
totally symmetric representation (Γl=0) is contained only twice in the
286 CHAPTER 11. TRANSFORMATION OF TENSORS
direct product of the irreducible representations for two second rank
symmetric tensors, indicating that only two independent non–vanishing
constants are needed to describe the 21 constants of the Cijkl tensor in
full rotational symmetry, a result that is well known in elasticity theory
for isotropic media and discussed above (see §11.4). As stated in §11.2
and in §11.4, the number of times the totally symmetric representation
(e.g., Γl=0 for the full rotational group) is contained in the irreducible
representations of a general matrix of arbitrary rank determines the
number of independent non–vanishing constants needed to specify that
matrix.
We denote the two independent non-vanishing constants needed to
specify the Cijkl tensor by C0 for Γl=0 and by C2 for Γl=2 symmetry. We
then use these two constants to relate symmetrized stresses and strains
labeled by the irreducible representations Γl=0 and Γl=2 in the full rotation
group. The symmetrized stress–strain equations are first written
in full rotational symmetry, using basis functions for the partners of
the pertinent irreducible representations (one for l = 0 and five for the
l = 2 partners):
(Xx + Yy + Zz)=C0(exx + eyy + ezz) for l = 0, m = 0
(Xx − Yy + iYx + iXy)=C2(exx − eyy + iexy + ieyx) for l = 2, m = 2
(Zx + Xz + iYz + iZy)=C2(ezx + exz + ieyz + iezy) for l = 2, m = 1
(Zz − 1
2
(Xx + Yy))=C2(ezz − 1
2
(exx + eyy)) for l = 2, m = 0
(Zx + Xz − iYz − iZy)=C2(ezx + exz − ieyz − iezy) for l = 2, m = −1
(Xx − Yy − iYx − iXy)=C2(exx − eyy − iexy − ieyx) for l = 2, m = −2
(11.34)
From the first, second, fourth and sixth relations in Eqs. 11.34 we solve
for Xx in terms of the strains, yielding
Xx =
C0
3
+
2C2
3
exx+
C0
3
−
C2
3
eyy + ezz . (11.35)
Likewise five additional relations are then written down for the other 5
stress components in Eqs. 11.34.
Yy =
C0
3
+
2C2
3
eyy+
C0
3
−
C2
3
ezz + exx , (11.36)
Zz =
C0
3
+
2C2
3
ezz+
C0
3
−
C2
3
exx + eyy , (11.37)
11.6. ELASTIC MODULUS TENSOR 287
Zy + Yz = C2 ezy + eyz , (11.38)
Yx + Xy = C2 eyx + exy , (11.39)
Zx + Xz = C2 ezx + exz . (11.40)
In the notation that is commonly used in elastic theory, we write
the stress–strain relations as
σi =
j=1,6
Cijεj, (11.41)
where the 6 components of the symmetric stress and strain tensors are
written in accordance with Eq. 11.27 as
σ1=Xx
σ2=Yy
σ3=Zz
σ4=1
2
(Yz + Zy)
σ5=1
2
(Zx + Xz)
σ6=1
2
(Xy + Yx)
and
ε1=exx
ε2=eyy
ε3=ezz
ε4=(eyz + ezy)
ε5=(ezx + exz)
ε6=(exy + eyx)
(11.42)
and Cij is the 6 × 6 elastic modulus matrix. In this notation the 21
partners that transform as (2Γl=0+2Γl=2+Γl=4) in Eq. 11.33 correspond
to the symmetric coefficients of Cij. From the six relations for the six
stress components (given explicitly by Eqs. 11.35 through 11.40), the
relations between the C0 and C2 and the Cij coefficients follow:
C11=1
3
(C0 + 2C2) = C22 = C33
C12=1
3
(C0 − C2) = C13 = C23
C44=1
2
C2 = C55 = C66
Cij=Cji
(11.43)
from which we construct the Cij matrix for a 3D isotropic medium and
288 CHAPTER 11. TRANSFORMATION OF TENSORS
involves only two independent constants C11 and C12
Cij =










C11 C12 C12 0 0 0
C11 C12 0 0 0
C11 0 0 0
1
2
(C11 − C12) 0 0
1
2
(C11 − C12) 0
1
2
(C11 − C12)










.
(11.44)
11.6.2 Icosahedral Symmetry
Any subgroup of the full rotation group for which the 5–fold Γl=2 level
degeneracy is not lifted will leave the form of the Cij matrix invariant.
The icosahedral group with inversion symmetry Ih, which is a subgroup
of the full rotation group, and the icosahedral group without inversion
I, which is a subgroup of both the full rotation group and the group
Ih, are two examples of groups which preserve the 5–fold degenerate
level of the full rotation group and hence retain the form of the Cij
matrix given by Eq. 11.44. This result follows from at least two related
arguments. The first argument relates to the compatibility relations
between the full rotation group and the Ih group for which the basis
functions follow the compatibility relations
Γl=0 −→ (Ag)Ih
Γl=2 −→ (Hg)Ih
.
(11.45)
We thus show that for the icosahedral group the irreducible representations
for a symmetric second rank tensor are
Γ
(s)
↔
e
= (Ag)Ih
+ (Hg)Ih
. (11.46)
From Eq. 11.46 we see that with respect to second rank tensors no
lifting of degeneracy occurs in going from full rotational symmetry to
Ih symmetry from which it follows that the number of non–vanishing
independent constants in the Cij matrix remains at 2 for Ih (and I)
symmetry.
The same conclusion follows from the fact that the basis functions
for Γl=0 and Γl=2 for the full rotation group can also be used as basis
11.6. ELASTIC MODULUS TENSOR 289
functions for the Ag and Hg irreducible representations of Ih. Therefore
the same stress–strain relations are obtained in Ih symmetry as are
given in Eq. 11.34. It therefore follows that the form of the Cij matrix
will also be the same for group Ih and full rotational symmetry, thereby
completing the proof.
Clearly, the direct product Γ
(s)
↔
e
⊗ Γ
(s)
↔
e
given by Eq. 11.33 is not invariant
as the symmetry is reduced from full rotational symmetry to Ih
symmetry since the 9-fold representation Γl=4 in Eq. 11.33 splits into
the irreducible representations (Gg + Hg) in going to the lower symmetry
group Ih. But this is not of importance to the linear stress-strain
equations which are invariant to this lowering of symmetry. It might
be worth mentioning here that when non-linear effects are taken into
account and perturbations from Eq. 11.26 are needed to specify the
non-linear stress–strain relations, different mechanical behavior would
be expected to occur in Ih symmetry in comparison to the full rotation
group. In such a case, the compatibility relations between the irreducible
representations of the full rotation group and the Ih group can
be used to relate the terms in the non-linear elastic matrix for the two
symmetries. This generalization is similar in concept to that discussed
for the case of non-linear optics in §11.5.
11.6.3 Cubic Symmetry
It should be noted that all symmetry groups forming Bravais lattices
in solid state physics have too few symmetry operations to preserve the
5-fold degeneracy of the l = 2 level of the full rotation group. For example,
the Bravais lattice with the highest symmetry is the cubic group
Oh. The l = 2 irreducible representation in full rotational symmetry
corresponds to a reducible representation of group Oh which splits into
a 3-fold and a 2-fold level (the T2g and Eg levels), so that in this case
we will see below, 3 elastic constants are needed to specify the 6 × 6
matrix for Cij.
Since eij (where i, j = x, y, z) is a symmetric second rank tensor,
the irreducible representations for eij, in cubic symmetry are found as
Γ
(s)
↔
e
= Γ+
1 + Γ+
12 + Γ+
25. (11.47)
290 CHAPTER 11. TRANSFORMATION OF TENSORS
From the direct product we obtain
Γ
(s)
↔
e
⊗Γ
(s)
↔
e
= (Γ+
1 +Γ+
12+Γ+
25)⊗(Γ+
1 +Γ+
12+Γ+
25) = 3Γ+
1 +Γ+
2 +4Γ+
12+3Γ+
15+5Γ+
25,
(11.48)
which has 21 symmetric partners (3Γ+
1 + 3Γ+
12 + Γ+
15 + 3Γ+
25) and 15
antisymmetric partners (Γ+
2 +Γ+
12 +2Γ+
15 +2Γ+
25) and three independent
Cij coefficients. These results could also be obtained by going from
higher (full rotational R∞) symmetry to lower (Oh) symmetry using
the cubic field splittings of the angular momenta shown in Table 11.4.
Forming basis functions for the irreducible representations of the
stress and strain tensors in cubic Oh symmetry, we can then write the
symmetrized elastic constant equations as
(Xx + Yy + Zz)=CΓ+
1
(exx + eyy + ezz) for Γ+
1
(Xx + ωYy + ω2
Zz)=CΓ+
12
(exx + ωeyy + ω2
ezz) for Γ+
12
(Xx + ω2
Yy + ωZz)=CΓ+
12
(exx + ω2
eyy + ωezz) for Γ+∗
12
(Yz + Zy)=CΓ+
25
(eyz) for Γ+
25x
(Zx + Xz)=CΓ+
25
(exz) for Γ+
25y
(Xy + Yx)=CΓ+
25
(exy) for Γ+
25z
(11.49)
From the first three relations in Eq. 11.49 we obtain
Xx =
C+
Γ1
+ 2C+
Γ12
3
exx+
C+
Γ1
− C+
Γ12
3
eyy + ezz (11.50)
Yy =
C+
Γ1
+ 2C+
Γ12
3
eyy+
C+
Γ1
− C+
Γ12
3
ezz + exx (11.51)
Zz =
C+
Γ1
+ 2C+
Γ12
3
ezz+
C+
Γ1
− C+
Γ12
3
exx + eyy (11.52)
From Eqs. 11.49–11.52, we obtain the connections between the 3 symmetrybased
elastic constants C+
Γ1
, C+
Γ12
and C+
Γ25
and the C11, C12 and C44 in
Nye’s book (and other solid state physics books)
C11=(C+
Γ1
+ 2C+
Γ12
)/3
C12=(CΓ1 − C+
Γ12
)/3
C44=C+
Γ25
/2
(11.53)
11.6. ELASTIC MODULUS TENSOR 291
yielding an elastic tensor for cubic symmetry in the form
Cij =










C11 C12 C12 0 0 0
C11 C12 0 0 0
C11 0 0 0
C44 0 0
C44 0
C44










(11.54)
11.6.4 Full Axial Symmetry
One simple method for finding the irreducible representations for full
axial symmetry is to make use of the compatibility relations between
the full rotation group and the group D∞h (see character Table 3.36 on
p. 71):
Γl=0 −→ A1g
Γl=1 −→ A2u + E1u
Γl=2 −→ A1g + E1g + E2g
Γl=3 −→ A2u + E1u + E2u + E3u
Γl=4 −→ A1g + E1g + E2g + E3g + E4g.
(11.55)
Thus the symmetric 2nd
rank tensor eij transforms according to the
sum Γl=0 + Γl=2
Γ
(s)
↔
e
= A1g + (A1g + E1g + E2g) = 2A1g + E1g + E2g. (11.56)
From the symmetric terms in Eq. 11.33 and Eq. 11.55, we find that the
Cijkl tensor transforms according to 2Γl=0 + 2Γl=2 + Γl=4:
ΓCijkl
=(2A1g) + (2A1g + 2E1g + 2E2g) + (A1g + E1g + E2g + E3g + E4g)
=5A1g + 3E1g + 3E2g + E3g + E4g.
(11.57)
The same result as in Eq. 11.57 can be obtained by taking the direct
product of (A1g + E1g + E2g) ⊗ (A1g + E1g + E2g) and retaining only
the symmetric terms. From Eq. 11.57, we see that there are only 5
independent elastic constants.
The splitting of the five–dimensional representation l = 2 into three
irreducible representations (see Eq. 11.55) in D∞h symmetry increases
292 CHAPTER 11. TRANSFORMATION OF TENSORS
the number of independent coefficients by two and an additional independent
coefficient is needed to describe the off–diagonal coupling
between the two diagonal blocks with A1g symmetry.
To find the form of the elasticity matrix Cij we note that the basis
functions going with the irreducible representations of the second rank
symmetric stress and strain tensors are:
(Xx + Yy + Zz) (exx + eyy + ezz) l = 0, m = 0 A1g
(Xx − Yy + iYx + iXy) (exx − eyy + iexy + ieyx) l = 2, m = 2 E2g
(Xx − Yy − iYx − iXy) (exx − eyy − iexy − ieyx) l = 2, m = −2 E2g
(Zx + Xz + iYz + iZy) (ezx + exz + ieyz + iezy) l = 2, m = 1 E1g
(Zx + Xz − iYz − iZy) (ezx + exz − ieyz − iezy) l = 2, m = −1 E1g
(Zz − 1
2
(Xx + Yy)) (ezz − 1
2
(exx + eyy)) l = 2, m = 0 A1g.
(11.58)
From the basis functions in Eq. 11.58 we write the stress-strain relations
coupling basis functions of similar symmetry:
Xx + Yy + Zz=CA1g,1(exx + eyy + ezz) + CA1g,3[ezz − 1
2
(exx + eyy)]
Zz − 1
2
(Xx + Yy)=CA1g,2[ezz − 1
2
(exx + eyy)] + CA1g,4[exx + eyy + ezz]
Xx − Yy=CE2g (exx − eyy)
Xy + Yx=CE2g (exy + eyx)
Yz + Zy=CE1g (eyz + ezy)
Zx + Xz=CE1g (ezx + exz).
(11.59)
We then solve Eqs. 11.59 for Xx, Yy and Zz and require Cij = Cji. In
the case of D∞h, the requirement that C31 = C13 = C32 = C23 yields
the additional constraint CA1g,3 = 2CA1g,4 which is needed to obtain the
5 independent symmetry coefficients as required by Eq. 11.57: CA1g,1,
CA1g,2, CA1g,3, CE1g and CE2g . The relations between these symmetry
11.6. ELASTIC MODULUS TENSOR 293
coefficients and the Cij coefficients are:
C11=C22 = 1
2
[2
3
CA1g,1 + 1
3
CA1g,2 − 2
3
CA1g,3 + CE2g ]
C12=C21 = 1
2
[2
3
CA1g,1 + 1
3
CA1g,2 − 2
3
CA1g,3 − CE2g ]
C13=C23 = 1
3
[CA1g,1 − CA1g,2 + 1
2
CA1g,3]
C33=1
3
[CA1g,1 + 2CA1g,2 + 2CA1g,3]
C44=C55 = 1
2
CE1g
C66=1
2
CE2g = 1
2
(C22 − C21) = 1
2
(C11 − C12).
(11.60)
Combining the non-vanishing Cij coefficients then yields the matrix for
full axial D∞h symmetry:
Cij =










C11 C12 C13 0 0 0
C11 C13 0 0 0
C33 0 0 0
C44 0 0
C44 0
1
2
(C11 − C12)










. (11.61)
We see that the Cij matrix in this case has the same form as for the
hexagonal D6h symmetry group described in the next subsection.
11.6.5 Hexagonal Symmetry
Since eij is a symmetric second rank tensor, the irreducible representations
for eij, in hexagonal D6h symmetry are found to be:
Γ↔
e
= (A2u + E1u) ⊗ (A2u + E1u) = 2A1g + A2g + 2E1g + E2g (11.62)
The antisymmetric terms are A2g and E1g, so that the symmetric combination
is
Γ
(s)
↔
e
= 2A1g + E1g + E2g (11.63)
Using Eq. 11.33 and the irreducible representations contained in the
angular momentum states l = 0, l = 2, and l = 4 in D6h symmetry, we
294 CHAPTER 11. TRANSFORMATION OF TENSORS
get
Γl=0→A1g
Γl=1→A2u + E1u
Γl=2→A1g + E1g + E2g
Γl=3→A2u + B1u + B2u + E1u + E2u
Γl=4→A1g + B1g + B2g + E1g + 2E2g
(11.64)
which gives the symmetric irreducible representations for a symmetric
4th
rank tensor
ΓC(ij)(kl)
=2A1g + 2(A1g + E1g + E2g) + (A1g + B1g + B2g + E1g + 2E2g)
=5A1g + B1g + B2g + 3E1g + 4E2g
(11.65)
yielding 5 independent Cij coefficients. Since the basis functions for all
the quadratic forms for D∞h and D6h symmetry are the same, Eq. 11.58
also provides the basis functions for the stress and strain tensor components
in D6h symmetry. Likewise, Eq. 11.59 also applies for the stress
and strain relations in D6h symmetry, giving rise to Eq. 11.60 for the
relation between the Cij coefficients and the 5 independent symmetry
coefficients CA1g,1, CA1g,2, CA1g,3, CE1g and CE2g . Thus the form of the
elastic tensor in D6h symmetry is given by Eq. 11.61. We note the
correspondence here between D6h and D∞h with regard to the elastic
properties of solids, although D6h is not a subgroup of D∞h.
It must be emphasized that although the form of Cijkl is the same
for both D∞h and D6h symmetries, tensors of rank higher than 2 (such
as the non-linear elastic coefficients) will in general be expected to be
different in D∞h and D6h symmetries.
If we now consider the elastic tensor for the group D4h, we immediately
see that there is only one 2-dimensional irreducible representation
in group D4h so that the irreducible representations contained in the
second rank tensor Γ↔
e
are not the same as for D∞h. Thus the Cij
tensor is expected to have more independent coefficients for the case of
D4h symmetry.
11.7. SELECTED PROBLEMS 295
11.6.6 Other Symmetry Groups
The form of the elastic tensor for other lower symmetry groups is also
of importance in considering the mechanical properties of solids. We
only give results in these cases, leaving the derivation of the results to
the reader. The derivations can either be found by direct calculation
as for the case of the cubic group considered above, or by going from
higher to lower symmetry, as was illustrated for the case of the group
Ih derived from the group with full rotational symmetry.
For D2h group symmetry which is the case of symmetry with respect
to three mutually orthogonal planes (called orthotropy in the engineering
mechanics literature), there remain nine independent components
of Cij. The Cij tensor in this case assumes the form
Cij =










C11 C12 C13 0 0 0
C22 C23 0 0 0
C33 0 0 0
C44 0 0
C55 0
C66










(11.66)
The lowest non-trivial symmetry group for consideration of the elastic
tensor is group C2h with a single symmetry plane. In this case Cij
has 13 independent components and assumes the form
Cij =










C11 C12 C13 0 0 C16
C22 C23 0 0 C26
C33 0 0 C36
C44 C45 0
C55 0
C66










(11.67)
11.7 Selected Problems
1. Suppose that stress is applied to fcc aluminium in the (100) direction,
and suppose that the effect of the resulting strain is to
lower the symmetry of aluminum from cubic Oh symmetry to
tetragonal D4h symmetry. The situation outlined here arises in
296 CHAPTER 11. TRANSFORMATION OF TENSORS
the fabrication of superlattices using the molecular beam epitaxy
technique.
(a) How many independent elastic constants are there in the
stressed Al?
(b) What is the new symmetrized form of the stress-strain relations
(see Eq. 11.34 of the notes)?
(c) What is the form of the Cijkl tensor (see Eq. 11.44 of the
notes)?
2. (a) Now assume that the material in Problem 1 is non-linear
elastic material and the stress-strain relation is of the form
σij = C
(1)
ijklεkl + C
(2)
ijklmnεklεmn + . . .
Consider the symmetry of the non-linear tensor coefficient
C
(2)
ijklmn explicitly. How many independent constants are
there in C
(2)
ijklmn assuming that the point group symmetry
is C1 (i.e., no rotational symmetry elements other than the
identity operation)?
(b) How many independent constants are there when taking into
account permutation symmetry only?
(c) How many independent constants are there when taking
into account both permutation and crystal (Oh) symmetry?
(Note: To do this problem, you may have to make a new
entry to Table 11.1.)
3. Suppose that we prepare a quantum well using as the constituents
GaAs and GaAs1−xPx. The lattice mismatch introduces lattice
strain and lowers the symmetry. Denote by ˆz the direction normal
to the layer. Find the number of independent coefficients in the
polarizability tensor, including
↔
α
(1)
,
↔
α
(2)
, and
↔
α
(3)
, for:
(a) ˆz (100)
(b) ˆz (111)
(c) ˆz (110)
11.7. SELECTED PROBLEMS 297
Using these results, how can infrared and Raman spectroscopy
be used to distinguish between the crystalline orientation of the
quantum well?
4. Consider the third rank tensor di(jk)
(a) Show from Table 11.1 that there are exactly 18 independent
coefficients.
(b) How many independent coefficients are there for group Td?
298 CHAPTER 11. TRANSFORMATION OF TENSORS
Chapter 12
Space Groups
In this chapter we introduce the concept of translational symmetry
and the space groups used to classify the translational symmetry. In
addition to the point group and translation operations, we consider
the compound operations of glide planes and screw axes and the space
groups associated with these compound symmetry operations. The
properties of the two-dimensional space groups are discussed in detail.
12.1 Simple Space Group Operations
According to the one-electron Hamiltonian for the electronic energy
band structure for solids, we write Schr¨odinger’s equation as
−
¯h2
2m
2
+ V (r) ψ(r) = Eψ(r) (12.1)
where V (r) is a periodic potential which exhibits translational symmetry
in addition to point group symmetry operations. The symmetry
group of the one-electron Hamiltonian (Eq. 12.1) and of the periodic
potential is the space group of the crystal lattice, which consists of
both point group symmetry operations and translational symmetry
operations. All of these symmetry operations leave the Hamiltonian
invariant and consequently the operators representing these symmetry
operations will commute with the Hamiltonian, and provide quantum
numbers for labeling the energy eigenvalues and eigenfunctions.
299
300 CHAPTER 12. SPACE GROUPS
The point group and translation symmetry operations which carry
the crystal into itself form a group called the space group. A common
notation for space group operators is
{Rα|τ} (12.2)
where Rα denotes point group operations such as rotations, reflections,
improper rotations, inversions and τ denotes translation operations.
Pure rotations and pure translations are special cases of space group
operations:
{ε|0} = identity
{α|0} = pure rotations or more generally point group operations
{ε|τ} = pure translations by vector τ
We can relate the operator {α|τ} for the space group to a coordinate
transformation
r =
↔
α ·r + τ (12.3)
where
↔
α denotes the transformation matrix for rotation and τ denotes
a translational transformation. Multiplication of two space group operators
proceeds from this identification:
{β|τ }{α|τ} =
↔
β ·
↔
α ·r + τ + τ
=
↔
β ·
↔
α ·r +
↔
β ·τ + τ
= {βα|βτ + τ }
to yield the result for the multiplication of two space group operators
{β|τ } {α|τ} = {βα|βτ + τ } (12.4)
where {α|τ} is the first space group operator and {β|τ } is the second.
Likewise:
{α|τ} {β|τ } =
↔
α ·
↔
β ·r+
↔
α ·τ + τ (12.5)
so that commutation of these two space group operators requires that
↔
α ·
↔
β=
↔
β ·
↔
α and
↔
β ·τ + τ =
↔
α ·τ + τ (12.6)
12.1. SIMPLE SPACE GROUP OPERATIONS 301
which is not generally valid. Thus we conclude that although simple
translations commute with each other, general space group operations
do not commute.
To get the inverse of {α|τ} we make use of the multiplication relation
above to obtain
{α|τ}−1
= {α−1
| − α−1
τ} (12.7)
since
{α|τ}{α|τ}−1
= {αα−1
|α(−α−1
τ) + τ} = {ε|0}. (12.8)
Having specified the identity operation {ε|0}, the rules for multiplication
and the rules for specifying the inverse operation, and noting that
the associative law applies, we see that the elements {α|τ} form a
space group.
To write the space group operations in matrix form, a basis
1
r
is introduced where 1 is a number and r is a column vector consisting
for example of 


x
y
z


 .
The matrix representation for the space group operator is then
{α|τ} =
1 0
τ
↔
α
(12.9)
where 1 is a number, 0 denotes a row of three zeros, τ is a column
vector, and
↔
α is a (3 × 3) rotation matrix. The transformation on the
coordinate system then is written as
1 0
τ
↔
α
1
r
=
1
τ+
↔
α ·r
=
1
r
. (12.10)
To check that the matrix of Eq. 12.9 is a representation for the space
group operator {α|τ}, let us examine the multiplication and inverse
302 CHAPTER 12. SPACE GROUPS
transformations. Multiplication of two matrices yields
1 0
τ
↔
β
1 0
τ
↔
α
=
1 0
τ +
↔
β ·τ
↔
β ·
↔
α
(12.11)
which checks out since
{β|τ }{α|τ} = {βα|βτ + τ }. (12.12)
The inverse operation also checks out since:
1 0
−
↔
α
−1
·τ
↔
α
−1
1 0
τ
↔
α
=
1 0
0 1
. (12.13)
If all the operations of the space group are simply point group operations
on to which we add translation operations, we have a simple or
symmorphic space group. There are 73 symmorphic space groups in
all (see Table 12.1). We will illustrate the idea of symmorphic space
groups using an example based on the D2d point group (see Table 3.29).
Suppose that we have a molecule with atoms arranged in a D2d point
group configuration as shown in Fig. 12.2b. We see that the D2d point
group has classes E, C2 about the z axis, 2S4 also about the z axis,
2σd passing through the z axis and each of the dumbbell axes, and 2C2
in (110) directions in the median plane, as shown in Fig. 12.2a, which
is the top view of the molecule shown in Fig. 12.2b.
We could put such X4 molecules into a solid in many ways and still
retain the point group symmetry of the molecule. To illustrate how
different space groups can be produced with a single molecular configuration,
we will put the X4 molecule with D2d symmetry into two
different symmorphic space groups, as shown in Fig. 12.3. Another interesting
illustration of symmetry-lowering by adding a molecular configuration
on a lattice site can be found for the case of the icosahedral
C60 molecule in a FCC crystal lattice.
We note that with either of the placements of the molecule in
Fig. 12.3, all the point group operations of the molecule are
also operations of the space lattice. Conversely, if the symmetry
axes of the molecule do not coincide with the symmetry axes of the lattice
in which they are embedded, the combined space group symmetry
is lowered.
12.1. SIMPLE SPACE GROUP OPERATIONS 303
Table 12.1: The 73 symmorphic space groups. Here P, I, F, and B,
respectively, denote primitive, body centered, face centered and base
centered Bravais lattices (see Fig. 12.1).
Crystal system Bravais lattice Space group
Triclinic P P1, P¯1
Monoclinic P P2, Pm, P2/m
B or A B2, Bm, B2/m (1st setting)
Orthorhombic P P222, Pmm2, Pmmm
C, A, or B C222, Cmm2, Amm2∗
, Cmmm
I I222, Imm2, Immm
F F222, Fmm2, Fmmm
Tetragonal P P4, P¯4, P4/m, P422, P4mm,
P42m, P¯4m2∗
, P4/mmm
I I4, I¯4, I4/m, I422, I4mm
I¯42m, I¯4m2∗
, I4/mmm
Cubic P P23, Pm3, P432, P¯43m, Pm3m
I I23, Im3, I432, I¯43m, Im3m
F F23, Fm3, F432, F¯43m, Fm3m
Trigonal P P3, P¯3 P312, P321∗
, P3m1,
P1, P¯1
(Rhombohedral) R R3, R¯3 R32, R3m, R¯3m,
Hexagonal P P6, P¯6, P6/m, P622, P6mm,
P¯6m2, P¯6m2∗
, P6/mmm
∗
The asterisks mark the seven extra space groups that are generated
when the orientation of the point group operations is taken into account
with respect to the Bravais cell.
304 CHAPTER 12. SPACE GROUPS
Figure 12.1: The fourteen Bravais space lattices illustrated by a unit
cell of each: (1) triclinic, simple; (2) monoclinic, simple; (3) monoclinic,
base centered; (4) orthorhombic, simple; (5) orthorhombic, base
centered; (6) orthorhombic, body centered; (7) orthorhombic, face centered;
(8) hexagonal; (9) rhombohedral; (10) tetragonal, simple; (11)
tetragonal, body centered; (12) cubic, simple; (13) cubic, body centered;
and (14) cubic, face centered.
12.1. SIMPLE SPACE GROUP OPERATIONS 305
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Figure 12.2: (a) Top view of a
molecule X4 with D2d symmetry.
The symmetry axes are indicated.
(b) Schematic diagram of an X4
molecule with point group D2d
(42m) symmetry.
Figure 12.3: Tetragonal Bravais lattice with two possible orientations
of the molecule with D2d symmetry resulting in two different
3-dimensional space groups. The maximum symmetry that the tetragonal
Bravais lattice can support is D4h = D4 ⊗ i (4/mmm).
306 CHAPTER 12. SPACE GROUPS
In an elementary course in solid state physics we learn that there
are 14 three-dimensional Bravais lattices. Particular point group
operations are appropriate to specific Bravais lattices, but the connection
is homomorphic rather then isomorphic. For example, the point
group operations T, Td, Th, O and Oh leave each of the simple cubic,
face centered cubic and body centered cubic Bravais lattices invariant.
Even though a given Bravais lattice is capable of supporting a high
symmetry point group (e.g., the FCC structure), if we have a lower
symmetry structure at each of the lattice sites (e.g., the structure in
Fig. 12.2b), then the point symmetry is lowered to correspond to that
structure—in this case the point group symmetry will be D2d. On the
other hand, the highest point group symmetry that is possible in a crystal
lattice is that which is consistent with the Bravais lattice, so that
the group Oh will be the appropriate point group for an FCC structure
with spherical balls at each lattice site.
12.2 Space Groups and Point Groups
The space group G consists of all operations {Rα|τ} which leave a given
lattice invariant. The point group g is obtained from G by placing
τ = 0 for all Rα elements in G. If, with a suitable choice of origin in
the direct lattice, we find that all the elements of the point group g are
also elements of the space group G, then the space group G is called a
symmorphic group.
All the elements of the space group G that are of the form {ε|τ} constitute
the translation group T. Symmetry elements of the group T are
defined by the translation vectors Rn = Σniai, which leave the lattice
invariant. The translation group is a self-conjugate or invariant
or normal subgroup of G since
{Rα|τ}{ε|t}{Rα|τ}−1
= {Rα|τ}{ε|t}{R−1
α | − R−1
α τ}
= {Rα|τ}{R−1
α | − R−1
α τ + t}
= {ε| − RαR−1
α τ + Rαt + τ}
= {ε|Rαt}. (12.14)
But Rαt is just another translation vector in group T and therefore the
operation {ε|Rαt} is a symmetry operation of the translation group.
12.2. SPACE GROUPS AND POINT GROUPS 307
Although the translation group T is an invariant subgroup of G, we
cannot say that the space group G is a direct product of a translation
group with a point group, since the elements {ε|τ} and {Rα|0} do not
commute:
{ε|τ}{Rα|0} = {Rα|τ}
{Rα|0}{ε|τ} = {Rα|Rατ}. (12.15)
However, since the translation group is an invariant subgroup of G,
it is of interest to study the cosets of the factor group which it defines.
The right coset of the translation group considered as a subgroup of G
is then
Cα = [{ε|τ }{Rα|0}] = [{Rα|τ }] (12.16)
where the bracket in Eq. 12.16 denotes all the terms in the coset that
can be formed using all possible values of τ . We note that Cα is also
a left coset of the translation group because T is a self-conjugate (or
normal) subgroup of G. We see that the cosets Cα form a factor
group by considering the multiplication rule for the cosets:
CαCβ = [{Rα|τ1}{Rβ|τ2}] = [{RαRβ|Rατ2 + τ1}] = [{Rγ|τ3}] = Cγ
(12.17)
where RαRβ = Rγ defines the group property in the point group and
τ3 = Rατ2 +τ1 is a translation of the lattice. Since τ1 and τ2 range over
all possible translation vectors, the vector τ3 also spans all possible
translations.
These considerations show that the operations Rα apply to the
translation vectors in accordance with the definition of the space group
operations, and that the symmetry operations of the factor group G/T
for symmorphic space groups are isomorphic with the point group g.
Thus irreducible representations of the factor group G/T are also irreducible
representations of g and are likewise irreducible representations
of G. It can be shown that all irreducible representations of G can be
compounded from irreducible representations of g and T, even though
G is not a direct product group of g and T (see Koster’s group theory
article in the Seitz–Turnbull Solid State Physics series, vol. 5).
308 CHAPTER 12. SPACE GROUPS
 ¢¡¤£¦¥¨§ © § ¡  ¢£!#"%$&%'(¡0)#© 1
2 2 2
243 253
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Figure 12.4: (a) The glide plane operation: (b) Right and left-hand
screw axes. The right and left hand screw axes belong to closely related
but different space groups.
12.3 Compound Space Group Operations
Non-symmorphic space groups are also found and there are 157 of
these (230–73=157). In the non-symmorphic space groups we find instead
of simple translation operations, compound translation operations
such as glide planes and screw axes.
A glide plane consists of a translation parallel to a given plane
followed by a reflection in that plane (see Fig. 12.4a). There are in fact
three different types of glide planes that are identified: the axial glide
along a symmetry axis (a, b, or c), the diagonal glide or n-glide in
two or three directions (e.g., (a + b)/2 or (a + b + c)/2) and finally the
diamond glide corresponding to (a + b)/4 or (a + b + c)/4).
A screw axis is a translation along an axis about which a rotation
is simultaneously occurring. In Fig. 12.4b we show a 3-fold screw axis,
where a is the lattice constant. The tellurium and selenium structures
have 3-fold screw axes similar to those shown in Fig. 12.4b. A summary
of the various possible screw axes and the crystallographic notation for
each is shown in Fig. 12.5. A familiar example of a non-symmorphic
space group is the diamond structure shown in Fig. 12.6, where we
note that there are two atoms per unit cell (the dark and light atoms).
12.3. COMPOUND SPACE GROUP OPERATIONS 309
Figure 12.5: A summary of all possible screw axes.
310 CHAPTER 12. SPACE GROUPS
Figure 12.6: (a) Diamond structure
Fd3m (O7
h, #227) showing
a unit cell with 2 distinct atom
site locations. For the zincblende
structure (see Fig. 14.7) the atoms
on the two sites are distinct. (b)
The screw axis in the diamond
structure.
The symmetry operations of Td represent all the point group operations
that take light atoms into light atoms and dark atoms into dark atoms.
In addition, each of the operations of Td can be compounded with a
translation along (111)/4 which takes a light atom into a dark atom
and vice versa. Because of these additional symmetry operations which
are not point group operations of Td, the diamond structure is not a
Bravais lattice and is non-symmorphic. The screw axis pertinent to
the diamond structure is shown in Fig. 12.6b. Further examples of nonsymmorphic
operations in three-dimensional space groups are given in
§12.6.1 after discussion of the simpler situation occurring in 2D space
groups, in §12.5.
To summarize, there are 14 Bravais lattices, 32 point groups,
230 space groups of which 73 are symmorphic and 157 are
non-symmorphic. To demonstrate symmorphic and non-symmorphic
space groups, we enumerate in §12.5 the 17 two-dimensional space
groups which are important for surface science and for quasi-two dimensional
systems. In §12.4 we show that a Bravais lattice can only
support 1-fold, 2-fold, 3-fold, 4-fold and 6-fold rotations; a Bravais lattice
is not consistent with a 5-fold (or n-fold for n > 6) point group
symmetry.
12.4. INCOMPATIBILITY OF FIVE-FOLD SYMMETRY 311
 ¢¡¤£
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Figure 12.7: Translation–rotation symmetry for a 4–fold axis (a), and
a 3–fold axis (b).
12.4 Incompatibility of Five-Fold Symmetry
and Bravais Lattices
We show here that the requirements of translational symmetry limit the
possible rotation angles of a Bravais lattice and in particular restrict
the possible rotation axes to 1-fold, 2-fold, 3-fold, 4-fold and 6-fold.
Five-fold axes do not occur. When rotational symmetry does occur in
crystals, then severe restrictions on the rotation angle are imposed by
the simultaneous occurrence of the repetition of the unit cells through
rotations and translations.
To understand this restriction, consider the interplay of a 4–fold
axis and a translation, shown in Fig. 12.7a. The 4–fold axis A requires
the 4–fold repetition of the translation about itself. The translated
axis A also requires a set of translations at 90◦
with respect to each
other. Each of these translations, in turn, requires the repetition of a
4–fold axis at the translation distance τ from another 4–fold axis. Now
consider the two axes B and B . These are derived by translation from
the same axis A. Therefore B and B are translation–equivalent, so
that BB is a translation. In this example BB is exactly the same
translation as the original τ = AA . If the same reasoning is applied
to the interplay of a 3–fold axis and a translation, see Fig. 12.7b, one
finds that BB = 2AA = 2τ. In these two examples, the interplay
of a rotation and a translation produces new translations which are
312 CHAPTER 12. SPACE GROUPS
   ¢¡
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Figure 12.8: Translation–rotation symmetry for a 5–fold axis. BB is
not an integral multiple of τ, and is not a lattice vector.
consistent with the original translation.
But this is not true for all rotation angles, α. For example, it is not
true for a 5–fold axis, as shown in Fig. 12.8. The 5–fold axis at A is
required by translation τ to be repeated at A . Furthermore, the rotational
symmetry of A requires the translation AA = τ to be repeated
at an angular interval α = 2π/5 = 72◦
at AB, while the rotational
symmetry of A requires the translation AA = τ to be repeated at
α = 72◦
for A B . Since BA, AA , and A B are all translations, BB
is a translation having the same direction as the original translation
AA . Yet the lengths of BB and AA are irrational with respect to one
another. Thus BB is a new translation τ in the same direction as τ
but inconsistent with it. This inconsistency can also be expressed by
stating that BB violates the initial hypothesis that τ is the shortest
translation in that direction.
In general, the only acceptable values of α are those that cause BB
in Fig. 12.9 to be an integer multiple of the original translation, τ, in
order to be consistent with this translation. Thus the translation b
involves the condition that
b = mτ, (12.18)
12.4. INCOMPATIBILITY OF FIVE-FOLD SYMMETRY 313
Figure 12.9: Geometric construction
showing the translations τ and
b and the angle of rotation α (see
text).    ¢¡
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where m is an integer. But b is a simple function of α and τ, as shown
in Fig. 12.9 so that
b = τ − 2τ cos α. (12.19)
This can be compared with Eq. 12.18, giving
mτ = τ − 2τ cos α (12.20)
so that
m = 1 − 2 cos α. (12.21)
Consequently
2 cos α = 1 − m = M (12.22)
where M is also an integer. The permitted values for M, and the
corresponding values of α, m, and b, are listed in Table 12.2. The
permissible values of b are illustrated in Fig. 12.10.
314 CHAPTER 12. SPACE GROUPS
Table 12.2: Solutions of M = 2 cos α for permissible periods of crystallographic
axes.
M cos α α n = 2π/α b = τ − 2τ cos α
−3 −1.5 – – –
−2 –1 π 2 3τ
–1 –0.5 2π/3 3 2τ
0 0 π/2 4 τ
1 0.5 π/3 6 0
2 1 0 ∞ −τ
3 1.5 – – –
   
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Figure 12.10: Magnitudes of b in Eq. 12.18 for various crystallographic
values for n (see text).
12.5. TWO DIMENSIONAL SPACE GROUPS 315
Table 12.3: Five two-dimensional Bravais lattices.
oblique parallelogram a = b φ = 90◦
2
square square a = b φ = 90◦
4mm
hexagonal 60◦
rhombus a = b φ = 120◦
6mm
primitive rectangular rectangle a = b φ = 90◦
2mm
centered rectangular rectangle a = b φ = 90◦
2mm
12.5 Two Dimensional Space Groups
12.5.1 Five Two-dimensional Bravais Lattices
There are five distinct Bravais lattices in two-dimensions as shown in
Table 12.3. If we consider a, b to be the two primitive translation vectors
and φ to be the angle between a and b, then the five lattice types are
summarized in Table 12.3, and the two-dimensional space groups are
given in Table 12.4.
12.5.2 Notation
The notation used for the two-dimensional space groups is illustrated by
the example p4gm. The initial symbol (“p” in this example) indicates
that the unit cell is either a primitive (p) unit cell or a centered (c)
unit cell. The next symbol “4” indicates rotational symmetry about
an axis perpendicular to the plane of the two-dimensional crystal. The
possible n-fold rotations for a space group are 1, 2, 3, 4, and 6, as
shown in §12.4. The symbols used to denote such axes are shown in
Fig. 12.11. The last two symbols, when present, indicate additional
symmetries for the two inequivalent in-plane axes, where “g” denotes
a glide plane through the primary axis, “m” denotes a mirror plane
through the primary axis, and “1” indicates that there is no additional
symmetry.
We now discuss the notation for 3D groups as enumerated in Table
12.1. In the case of rectangular lattices, the inequivalent axes are
both parallel to the sides of the conventional rectangular unit cell. In
316 CHAPTER 12. SPACE GROUPS
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Figure 12.11: Space group symbols used at lattice points for 2-fold,
3-fold, 4-fold and 6-fold rotations (x = n for an n-fold rotation).
the case of square lattices, the first set of axes is parallel to the sides
and the second set is along the diagonals. In the case of hexagonal
lattices, one axis is 30◦
away from a translation vector. Other symbols
that are used for 3D space groups (see Table 12.1) include A or B for
monoclinic groups, and C, A or B, I, F for orthorhombic groups. Here
I refers to body centered and F to face centered groups, with F also
occurring for tetragonal and cubic groups. The symbol R is used for
rhombohedral groups.
12.5.3 Listing of the Space Groups
If we add two-dimensional objects, e.g., a set of atoms, to each cell of a
Bravais lattice, we can change the symmetry of the lattice. If the object,
sometimes called a motif, lowers the symmetry to that of another group,
then the resulting symmetry space group is identified with the lower
symmetry space group. Altogether there are 17 two-dimensional space
groups (see Table 12.4). We give below the symmetries of each of these
space groups, classified in terms of the 5 Bravais lattices given in §12.5.1.
Listings from the “International Tables for X-Ray Crystallography” are
given in Figs. 12.12–12.26. A listing of the 73 symmorphic space groups
is given in Table 12.1.
12.5. TWO DIMENSIONAL SPACE GROUPS 317
Table 12.4: Summary of the 17 two-dimensional space groups.
Point Lattice Type International(a)
Notation Type Notation
Group Table Number Full Short
1 Oblique 1 p1 symmorphic p1
2 2 p211 symmorphic p2
m Rectangular 3 p1m1 symmorphic pm
(p or c) 4 p1g1 non-symmorphic pg
5 c1m1 symmorphic cm
2mm 6 p2mm symmorphic pmm
7 p2mg non-symmorphic pmg
8 p2gg non-symmorphic pgg
9 c2mm symmorphic cmm
4 Square p 10 p4 symmorphic p4
4mm 11 p4mm symmorphic p4m
12 p4gm non-symmorphic p4g
3 Hexagonal 13 p3 symmorphic p3
3m 14 p3m1 symmorphic p3m1
15 p31m symmorphic p31m
6 16 p6 symmorphic p6
6mm 17 p6mm symmorphic p6m
(a)
International Tables for X-Ray Crystallography, published by the
International Union of Crystallography, Kynoch Press, Birmingham,
England (1952). See also G. Burns and A. M. Glazer, “Space Groups
for Solid State Scientists”, Academic Press, Inc.
—
318 CHAPTER 12. SPACE GROUPS
12.5.4 2D Oblique Space Groups
There are 2 oblique space groups as shown in Fig. 12.12. The lowest
symmetry two-dimensional space group (#1) only has translational
symmetry (p1) and no point group operations. The diagram for p1
shows only one general point (x, y) with translations by lattice vectors
(1,0), (0,1), and (1,1). Open circles on the left hand diagram in
Fig. 12.12 are used to denote the 3 open circles obtained from the first
open circle by these 3 translations.
However, by placing a motif with two-fold rotational symmetry normal
to the plane, the p211 space group (#2) is obtained, as shown in the
symmetry diagram from the International Tables for X-Ray Crystallography.
The two-fold axis through the center of the rhombus (indicated
by a football-shaped symbol on the right of Fig. 12.12) takes a general
point (x, y) into (−x, −y), shown as point symmetry type e in the table.
Points obtained by rotations are indicated by open circles in Fig. 12.12.
For the special points (1/2, 1/2), (1/2, 0), (0, 1/2), (0, 0), the two-fold
rotation takes the point into itself or into an equivalent point separated
by a lattice vector. The site symmetry for these points is listed in the
table as having a two-fold axis. A general point under the action of the
2-fold axis and translation by (1,0), (0,1), and (1,1) yields the 8 open
points in the figure.
12.5.5 2D Rectangular Space Groups
Primitive Lattices
Of the 7 rectangular 2D space groups, 5 are primitive and 2 are centered
(see Table. 12.4). We consider these together as is done in the
International Tables for X-Ray Crystallography. Of the 5 primitive
rectangular space groups only two are symmorphic, and three are nonsymmorphic.
In general, the full rectangular point symmetry is 2mm
(C2v). The point group 2mm has elements E, C2z, σx, σy: the identity;
a two-fold axis C2z perpendicular to the plane; and mirror planes parallel
to the x and y axes through C2z. The corresponding space group
listed as space group #6 is p2mm (see Fig. 12.15). When introducing
a lower symmetry motif, the resulting group must be a subgroup
of the original group. The lower symmetry rectangular space group
12.5. TWO DIMENSIONAL SPACE GROUPS 319
Figure 12.12: The two oblique two-dimensional space groups p1 and p2
(p211).
320 CHAPTER 12. SPACE GROUPS
Figure 12.13: The two-dimensional space groups pm and pg, #3 and
#4, respectively.
12.5. TWO DIMENSIONAL SPACE GROUPS 321
Figure 12.14: The two-dimensional space group cm, #5.
322 CHAPTER 12. SPACE GROUPS
Figure 12.15: The rectangular two-dimensional space group pmm, #6.
12.5. TWO DIMENSIONAL SPACE GROUPS 323
Figure 12.16: The rectangular two-dimensional space group pmg, #7.
324 CHAPTER 12. SPACE GROUPS
Figure 12.17: The two-dimensional rectangular space group pgg, #8.
12.5. TWO DIMENSIONAL SPACE GROUPS 325
Figure 12.18: The two-dimensional rectangular space group cmm, #9.
326 CHAPTER 12. SPACE GROUPS
p1m1 has point group operations (E, σx) and is listed as space group
#3 (see Fig. 12.13). We note that (E, σy) is equivalent to (E, σx) by an
interchange of axes and each corresponds to point group m (C1h).
Under a mirror plane operation (see Fig. 12.13) the symbols and
are used; the mirror plane is represented on the right by a solid
horizontal line. The use of a comma inside the circle provides a sense
of orientation that is preserved under translations. The three kinds
of site symmetries (the general point c and the points a and b on the
mirror planes) are also listed in the table for space group #3.
So far we have dealt with space groups where the point group operations
are separable from the translation group operations. Such groups
are symmorphic space groups.
In the case of the rectangular primitive lattice, mirror operations can
be replaced by glide reflections. The glide planes are denoted by dashed
lines (see diagram for space group #4 in Fig. 12.13). No distinct screw
operations are possible in two-dimensions. A glide reflection symmetry
operation is a compound operation consisting of a reflection combined
with a fractional translation, not a primitive unit cell translation. The
resulting space group is non-symmorphic.
Replacing m by g in p1m1 (space group #3) gives p1g1 (space
group #4) where the translation τ1/2 is compounded with the reflection
operation; this translation can be followed by comparing the symbols
for space groups #3 and #4.
For the case of space group #6, replacing one of the mirror planes
by a glide plane gives the non-symmorphic group p2mg (#7) as shown
in Fig. 12.13. When both mirror planes of space group #6 are replaced
by glide planes, we get p2gg (#8) which has the fractional translation
1
2
τ1 + 1
2
τ2, as shown in Fig. 12.17. The compound mirror plane translation
operations can be denoted by [σx|1
2
τ1 + 1
2
τ2], [σy|1
2
τ1 + 1
2
τ2].
Centered Rectangular Lattices
The centered rectangular lattice with the full centered rectangular symmetry
(see Fig. 12.18) is the symmorphic space group c2mm (#9). The
lower symmetry subgroup, related to p1m1 is c1m1 (#5) (shown in
Fig. 12.14). We note that the centering is equivalent to introducing a
τ1/2 glide plane as indicated in Fig. 12.13 for space group c1m1 (#5).
There are no non-symmorphic centered rectangular lattices. As a more
12.5. TWO DIMENSIONAL SPACE GROUPS 327
Figure 12.19: The square two-dimensional space groups p4 (#10).
interesting example of a centered rectangular space group, let us look
at space group #9 which is denoted by c2mm. This space group has
2 equivalent positions (0,0) and (1/2, 1/2). The symmetry operations
include a two-fold axis along the z-direction and two sets of intersecting
mirror planes. The table shows that c2mm can be realized through 6
different kinds of site symmetries.
12.5.6 2D Square Space Group
There are three 2D square space groups. The square lattice space with
the full 4mm point group symmetry is p4mm (space group #11), which
is shown in Fig. 12.20. The point group symmetry elements are E, C+
4z,
C−
4z, C2z, σy, σx, σda, σdb corresponding to C4v. The only distinct
328 CHAPTER 12. SPACE GROUPS
Figure 12.20: The square two-dimensional space group p4m (#11).
12.5. TWO DIMENSIONAL SPACE GROUPS 329
Figure 12.21: The square two-dimensional space group p4g (#12).
330 CHAPTER 12. SPACE GROUPS
Figure 12.22: Hexagonal two-dimensional space group
12.5. TWO DIMENSIONAL SPACE GROUPS 331
Figure 12.23: Hexagonal two-dimensional space group p3m1 (#14).
332 CHAPTER 12. SPACE GROUPS
Figure 12.24: Hexagonal two-dimensional space group p31m (#15).
12.5. TWO DIMENSIONAL SPACE GROUPS 333
Figure 12.25: Hexagonal two-dimensional space group p6 (#16).
334 CHAPTER 12. SPACE GROUPS
Figure 12.26: Hexagonal two-dimensional space group p6m (#17).
12.6. THREE DIMENSIONAL SPACE GROUPS 335
subgroup of C4v is C4 which has symmetry elements E, C+
4z, C−
4z, C2z.
In this case, the space group is p4 (#10 in International Tables for
X-Ray Crystallography). The 4-fold axis is clearly seen on the left
hand diagram in Fig. 12.19. The points in #11 are obtained by
adding mirror planes to #10. In the diagram on the right we see lattice
locations with 4-fold and with 2-fold axes, a feature found in all three
2D square lattices (see Figs. 12.19, 12.20, and 12.21).
By combining the translation 1
2
τ1 + 1
2
τ2, where 1
2
τ1 and 1
2
τ2 are translation
vectors, with the mirror planes σx, σy, σda, σdb we obtain the glide
reflections [σx|1
2
τ1 + 1
2
τ2], [σy|1
2
τ1 + 1
2
τ2], [σda|1
2
τ1 + 1
2
τ2], [σdb|1
2
τ1 + 1
2
τ2].
These glide reflections are used to form the non-symmorphic square
lattice p4gm (#12). We note there are mirror planes along the square
diagonals and also mirror planes through the x and y axes. Space group
#12 is obtained from space group #11 by translation of the comma
points by 1
2
τ1 + 1
2
τ2.
12.5.7 2D Hexagonal Space Groups
There are five 2D hexagonal space groups, and all are symmorphic.
The hexagonal space group with the full 6mm point group symmetry
is p6mm (#17). The point group symmetry elements are E, C+
6 , C−
6 ,
C+
3 , C−
3 , C2, σd1, σd2, σd3, σv1, σv2, σv3. The diagram for p6mm (#17) is
shown in Fig. 12.26.
The four subgroups of C6v are C6, C3v, C3d, C3, giving rise, respectively,
to space groups p6 (#16), p3m1 (#14), p31m (#15), and p3
(#13), as summarized in Table 12.5. The symmetry diagrams for the
five 2D hexagonal space groups are shown in Figs. 12.22, 12.23, 12.24,
12.25, and 12.26.
12.6 Three Dimensional Space Groups
The fourteen Bravais space lattices (see Fig. 12.1) illustrated by a unit
cell of each: (1) triclinic, simple; (2) monoclinic, simple; (3) monoclinic,
base centered; (4) orthorhombic, simple; (5) orthorhombic, base
centered; (6) orthorhombic, body centered; (7) orthorhombic, face centered;
(8) hexagonal; (9) rhombohedral; (10) tetragonal, simple; (11)
336 CHAPTER 12. SPACE GROUPS
Table 12.5: Summary of the symmetry operations of two-dimensional
space groups with 3-fold symmetry.
Space Group Point Group Elements
p3 E, C+
3 , C−
3
p3m1 E, C+
3 , C−
3 , σv1, σv2, σv3
p31m E, C+
3 , C−
3 , σd1, σd2, σd3
p6 E, C+
6 , C−
6 , C+
3 , C−
3 , C2
tetragonal, body centered; (12) cubic, simple; (13) cubic, body centered;
and (14) cubic, face centered.
When combined with the various point group symmetry operations
there result 230 different 3D space groups. Of these 73 are symmorphic
(see Table 12.1) and 157 are non-symmorphic.
12.6.1 Examples of Non-Symmorphic 3D Space Groups
Some examples of space groups with screw axes are given in Fig. 12.27
for space groups P41 (C2
4 ) #75, P42 (C3
4 ) #77 and P43 (C4
4 ) #78.
Each has point group C4 symmetry, but a different 4-fold screw axis
(41, 42, 43) is present in each case. The atom locations are given in the
left hand diagrams and the symmetry operations include screw axes in
the right hand diagrams.
Screw axes may also occur normal to the c-axis, as is shown in
Fig. 12.28 for space group P421m (D3
2d) #113. Diamond glide planes
along 110 directions also occur for this space group. The D2d operations
result in the occurrence of sites (x, y, z), (−y, x, −z), (−x, −y, z)
and (y, −x, −z).
Another rather complicated space group diagram is shown in Fig. 12.29
for the space group R3c (C6
3v) #161. The R symbol indicates a rhombohedral
Bravais lattice. Two unit cells are indicated on the left hand
figure, a trigonal unit cell and a rhombohedral unit cell. For this space
group we see 6 point group operations (x, y, z), (−y, x − y, z), (y −
x, −x, z), (−y, −x, 1
2
+z), (x, x−y, 1
2
+z), (y −x, y, 1
2
+z) compounded
12.6. THREE DIMENSIONAL SPACE GROUPS 337
Figure 12.27: Examples of space groups with screw axes. The three
examples are (a) P41 (C2
4 ) #76, (b) P42 (C3
4 ) #77 and (c) P43 (C4
4 )
#78.
338 CHAPTER 12. SPACE GROUPS
Figure 12.28: Example of a space group with a screw axis in the plane
of the figure: P421m (D3
2d) (#113).
with 3 centerings (0, 0, 0), (1/3, 2/3, 2/3) and (2/3, 1/3, 1/3) to give 18
positions. c-axis glide planes pass through the 3-fold axes in Fig. 12.29.
The reader is referred to texts such as Burns and Glazer who give a
detailed treatment of space group symmetries.
12.6. THREE DIMENSIONAL SPACE GROUPS 339
Figure 12.29: Example of a space group with many atom sites R3c
(C6
3v) (#161).
340 CHAPTER 12. SPACE GROUPS
Figure 12.30: Example of cubic lattices. Here (a), (b), (c) pertain to
space group #225; (d) pertains to #221 and (e) to #229; whild (f) and
(g) are for #227; and (h) is for #223.
12.6. THREE DIMENSIONAL SPACE GROUPS 341
Figure 12.31: Example of 3 cubic lattices with the space group O1
h
(Pm3m) (see Table ??). (a) Simple cubic, (b) body centered cubic,
and (c) perovskite structure.
342 CHAPTER 12. SPACE GROUPS
12.7 Selected Problems
1. (a) For the crystal structure shown below on the left, identify
the space group and list the symmetry elements.
(b) Identify the high symmetry points (and axes), and list the
group of the wave vector at these high symmetry points (and
axes).
(c) Using the space group identified in (a), explain the diagrams
for this space group as shown in the International Crystallography
Tables.
(d) Using the tables in (c), find the atom sites and site symmetries
for the structure shown in (a).
2. (a) List the real space symmetry operations of the non-symmorphic
two-dimensional square space group p4gm (#12).
(b) Explain the diagrams and the point symmetry entries for
space group #12 (p4gm) in Fig. 12.20 which was taken from
the International Crystallography Tables.
3. Show that in the diamond structure, the product of two symmetry
operations involving translations τ yields a symmetry element
with no translations
{α|τ}{β|τ} = {γ|0}
12.7. SELECTED PROBLEMS 343
where τ = (1, 1, 1)a/4. What is the physical significance of this
result?
4. Consider the crystal structure in the diagram for Nb3Sn, a prototype
superconductor with the A–15 (or β–W) structure used for
high field superconducting magnet applications.
(a) List the symmetry elements of the space group.
(b) What is the space group designation? (Use the notation of
the International Crystallography Tables.)
344 CHAPTER 12. SPACE GROUPS
Chapter 13
Group of the Wave Vector
and Bloch’s Theorem
One of the most important applications of group theory to solid state
physics relates to the symmetries and degeneracies of the dispersion
relations, especially at high symmetry points in the Brillouin zone. The
classification of these symmetry properties involves the group of the
wave vector, which is the subject of this chapter. The group of the
wave vector is important because it is the way in which both the point
group symmetry and the translational symmetry of the crystal
lattice are incorporated into the formalism that describes elementary
excitations in a solid.
13.1 Introduction
If we have symmetry operations {Rα|τ} that leave the periodic potential
V (r) invariant,
{Rα|τ}V (r) = V (r) (13.1)
this has important implications on the form of the wave function ψ(r)
which satisfies Bloch’s theorem (see §13.2). In particular if we consider
only translation operations {ε|τ}, we have an Abelian group because
all translations
ˆP{ε|τ}ψ(r) = ψ(r + τ) (13.2)
345
346CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
commute with each other.
Definition:
Since the translation operation τ can be written in terms of translations
over unit vectors ai
τ =
3
i=1
niai,
we can think of the translation operators in each of the ai directions as
the commuting operators:
{ε|τ} = {ε|τ1}{ε|τ2}{ε|τ3} (13.3)
where τi = niai. The commutativity of the {ε|τi} operations gives
three commuting cyclic subgroups. It is convenient to use periodic
boundary conditions for these cyclic subgroups. Since these subgroups
are Abelian, all irreducible representations are 1-dimensional and have
matrix representations or characters which are appropriate roots of
unity—e.g., eik1n1a1
. Then the wave vector k1 = 2πm1/L1 describes
the irreducible representation where m1 is an integer, and L1 is the
length of the crystal in direction a1. In this context, the wave vector
serves as a quantum number for the translation operator. With this
background, we can prove Bloch’s theorem.
13.2 Bloch’s Theorem
Theorem:
If an eigenfunction ψk transforms under the translation group according
to the irreducible representation labeled by k, then ψk(r)
obeys the relation
{ε|τ}ψk(r) = ψk(r + τ) = eik·τ
ψk(r) (13.4)
and ψk(r) can be written in the form
ψk(r) = eik·r
uk(r) (13.5)
where uk(r + τ) = uk(r) has the full translational symmetry of
the crystal.
13.2. BLOCH’S THEOREM 347
Proof:
Since the translation group is Abelian, all the elements of the
group commute and all the irreducible representations are 1-dimensional.
The requirement of the periodic boundary condition can be
written as
{ε|τ1 + NL1} = {ε|τ1} (13.6)
where N is an integer and L1 is the length of the crystal along
basis vector a1. This results in the one-dimensional matrix representation
for translation operator τi = niai of
Dk1
(n1a1) = eik1n1a1
= eik1τ1
(13.7)
since
ˆPRψk(r) = ψk(r)Dk
(R) (13.8)
where k1 = 2πm1/L1 corresponds to the mth
1 irreducible representation
and m1 = 1, 2, . . . , L1/a1. For each m1, there is a unique
k1, so that each irreducible representation is labeled by either m1
or k1, as indicated above.
We now extend these arguments to three dimensions. For a general
translation
τ =
3
i=1
niai (13.9)
the matrix representation or character for the (m1m2m3)th
irreducible
representation is
Dk1
(n1a1)Dk2
(n2a2)Dk3
(n3a3) = eik1n1a1
eik2n2a2
eik3n3a3
= eik·τ
(13.10)
since
{ε|τ} = {ε|τ1}{ε|τ2}{ε|τ3}. (13.11)
Thus our basic formula PRψj = α ψαD(R)αj yields
{ε|τ}ψ(r) = ψ(r)eik·τ
= eik·τ
ψ(r) = ψ(r + τ) (13.12)
since the representations are all 1-dimensional. This result is
Bloch’s theorem where we often write τ = Rn in terms of the
lattice vector Rn. This derivation shows that the phase factor
eik·τ
is the eigenvalue of the translation operator {ε|τ}.
348CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Because of Bloch’s theorem, the wave function ψ(r) can be written
in the form
ψk(r) = eik·r
uk(r) (13.13)
where uk(r) exhibits the full translational symmetry of the crystal. This
result follows from:
ψk(r + Rn) = eik·(r+Rn)
uk(r + Rn) = eik·Rn
eik·r
uk(r) (13.14)
where the first equality in Eq. 13.14 is obtained simply by substitution
in Eq. 13.13 and the second equality follows from Bloch’s theorem. In
these terms, Bloch’s theorem is simply a statement of the translational
symmetry of a crystal.
The wave vector k has a special significance as the quantum number
of translation and provides a label for the irreducible representations
of the translation group. If the crystal has a length Li on a side
so that n0 different lattice translations can be made for each direction
ai, then the number of k vectors must be limited to
kx, ky, kz = 0, ±
2π
n0a
, ±
4π
n0a
, . . . , ±
π
a
(13.15)
in order to insure that the number of irreducible representations is equal
to the number of classes. Since the group for translations is Abelian,
every group element is in a class by itself, so that the number of
irreducible representations must equal the number of possible
translations. Since the number of translation operators is very
large (∼ 1023
), the quantum numbers for translations are discrete, but
very closely spaced, and form a quasi-continuum of points in reciprocal
space.
We note that all of these k-vectors are contained within the 1st
Brillouin zone. Thus, if we consider a vector in the extended zone
k + K where K is a reciprocal lattice vector, the appropriate phase
factor in Bloch’s theorem is
ei(k+K)·Rn
= eik·Rn
(13.16)
since K · Rn = 2πN where N is an integer.
13.3. GROUP OF THE WAVE VECTOR 349
13.3 Symmetry of k Vectors and the Group
of the Wave Vector
It is convenient to deal with the symmetries of the reciprocal lattice
because the quantum number of translation k is measured in reciprocal
space. From the definition of the reciprocal lattice vector Kj we have
Rn · Kj = 2πNnj = 2πN1 (13.17)
where Nnj is an integer depending on n, j.
If α is a symmetry operator of the point group of the crystal, then
αRn leaves the crystal invariant. If Rn is a translation operator, then
αRn is also a translation operator (lattice vector). Likewise αKj is a
translation operator in reciprocal space. Since αRn is a lattice vector
(αRn) · Kj = 2πN2 (13.18)
where N2 is an integer, not necessarily the same integer as N1 in
Eq. 13.17. Since α−1
is also a symmetry operation of the group, we
have
(α−1
Rn) · Kj = 2πN3 (13.19)
and again N3 is not necessarily the same integer as N1 or N2. Furthermore,
any scalar product (being a constant) must be invariant under
any point symmetry operation. Thus if we perform the same symmetry
operation on each member of the scalar product in Eq. 13.19, then the
scalar product remains invariant
α(α−1
Rn) · (αKj) = 2πN3 = Rn · (αKj). (13.20)
Equations 13.18–13.20 tell us that if Kj is a reciprocal lattice vector and
if Rn and α−1
Rn are lattice vectors, then αKj is also a reciprocal lattice
vector. Furthermore, the effect of α on a direct lattice vector
Rn is equivalent to the operation α−1
on the corresponding
reciprocal lattice vector Kj.
Let us now consider the action of the point group operations on a
general vector k in reciprocal space, not necessarily a reciprocal lattice
vector. The set of wave vectors k which are obtained by carrying
350CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
out all the point group operations on k is called the star of k. If k
is a general point in the Brillouin zone, there will be only one
symmetry element, namely the identity, which takes k into itself. On
the other hand, if the k-vector under consideration lies on a symmetry
axis or is at a high symmetry point in the Brillouin zone, then perhaps
several of the point group operations will transform k into itself or into
an equivalent k-vector k + Ki. The set of point group operations
which transform k into itself or into an equivalent k-vector
form the group of the wave vector. Clearly, all the symmetry
operations of the point group take the point k = 0 into itself so that
the point group itself forms the group of the wave vector at k = 0.
The group of the wave vector for non-zone center k-vectors (k = 0) is
a subgroup of the point group for k = 0.
An informative example for the formation of the group of the wave
vector for various k-vectors is provided by the 2-dimensional square
lattice. Here the point group is D4 and the symmetry operations are
E, C2 = 2C2
4 , 2C4, 2C2, 2C2 (diagonals). The various k-vectors
in the star of k are indicated in the diagrams in Fig. 13.1 for the 2dimensional
square lattice. The group elements for the group of the
wave vector in each case are indicated within the parenthesis. The
top three diagrams are for k-vectors to interior points within the first
Brillouin zone and the lower set of three diagrams are for k-vectors to
the Brillouin zone boundary. Thus the star of k shown in Fig. 13.1 is
formed by consideration of αk for all α in the point group. The group
of the wave vector is formed by those α for which αk = k + Kj, where
Kj is a reciprocal lattice vector (including Kj = 0).
We will now consider the effect of the symmetry operations on k
with regard to the eigenfunctions of Schr¨odinger’s equation. We already
know from Bloch’s theorem that the action of any pure translation
operator T(Rn) yields a wave function eik·Rn
ψk(r). There will
be as many wave functions of this form as there are translation vectors,
each corresponding to the energy E(k). These Bloch functions
provide basis functions for irreducible representations for the group of
the wave vector. If k is a general point in the Brillouin zone, then the
star of k contains wave vectors which are all equivalent to k from a
physical standpoint. The space group for a general wave vector k will
13.3. GROUP OF THE WAVE VECTOR 351
 
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Figure 13.1: Illustration of the star of k for various wave vectors in a
simple 2D square Brillouin zone. The top 3 diagrams are for k-vectors to
an interior point in the Brillouin zone, while the bottom 3 diagrams are
for wave vectors extending to the Brillouin zone boundary. In each case
the elements for the group of the wave vector are given in parentheses.
352CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Table 13.1: Summary of the real and reciprocal lattice vectors for the
two-dimensional Bravais lattices.
Translation Vectors Reciprocal Lattice Vectors
Type τ1 τ2 g1 g2
oblique, p (a, 0) b(cosθ, sinθ) 2π
a
(1, − cot θ) 2π
b
(0, csc θ)
rectangular, p (a, 0) (0, b) 2π
a
(1, 0) 2π
b
(0, 1)
rectangular, c (a
2
, b
2
) (−a
2
, b
2
) 2π(1
a
, 1
b
) 2π(−1
a
, 1
b
)
square, p (a, 0) (0, a) 2π
a
(1, 0) 2π
a
(0, 1)
hexagonal, p (0,−a) a(
√
3
2
, 1
2
) 2π
a
( 1√
3
, −1) 2π
a
( 2√
3
, 0)
however contain only the symmetry elements {ε|Rn}, since in this case
all the k-vectors are distinct. For a wave vector with higher symmetry
where the operations βk = k + Kj transform k into an equivalent wave
vector, the space group of the wave vector contains {β|Rn} and the
energy at equivalent k points must be equal. If the point group of the
wave vector contains irreducible representations that have more than
one dimension, then degeneracy in the energy bands will occur. Thus
bands tend to “stick together” along high symmetry axes and at high
symmetry points.
We will now illustrate the group of the wave vector for the 3dimensional
simple cubic lattice Pm3m (O1
h) #221, the BCC lattice
Im3m (O9
h) #229 and the FCC lattice Fm3m (O5
h) #225.
13.3.1 Reciprocal Lattice
Reciprocal Lattice Vectors
If τ1 and τ2 are the primitive translation vectors, then the reciprocal
lattice vectors g1 and g2 are determined by the relation
gi · τj = 2πδij(i, j = 1, 2) (13.21)
and the wave vector k = k1g1+k2g2. Table 13.1 contains the translation
vectors and the reciprocal lattice vectors for the five two-dimensional
Bravais lattices. Vectors τ1 and τ2 are expressed in terms of unit vectors
13.4. SIMPLE CUBIC LATTICE 353
 
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Figure 13.2: Brillouin zone
for a simple cubic lattice
showing the high symmetry
points and axes.

 
 
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.
Figure 13.3: Symmetry operations
for the group O.
along the orthogonal x and y directions. Vectors g1 and g2 are expressed
in terms of unit vectors along the orthogonal kx and ky directions.
13.4 Simple Cubic Lattice
In Fig. 13.3 we see the Brillouin zone for the simple cubic lattice belonging
to the space group Pm3m (O1
h) #221. The high symmetry
points and axes are labeled using the standard notation. The symmetry
operations of the point group are the symmetry operations of the
Oh group indicated in Fig. 13.3 compounded with full inversion symmetry,
Oh = O ⊗ i. The point group corresponding to the group of the
354CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Table 13.2: Character table for the cubic group Oh.
Repr. Basis Functions E 3C2
4 6C4 6C2 8C3 i 3iC2
4 6iC4 6iC2 8iC3
Γ1 1 1 1 1 1 1 1 1 1 1 1
Γ2
x4(y2 − z2)+
y4(z2 − x2)+
z4(x2 − y2)
1 1 -1 −1 1 1 1 −1 −1 1
Γ12
x2 − y2
2z2 − x2 − y2 2 2 0 0 −1 2 2 0 0 −1
Γ15 x, y, z 3 −1 1 −1 0 –3 1 −1 1 0
Γ25 z(x2 − y2), etc. 3 −1 −1 1 0 –3 1 1 −1 0
Γ1
xyz[x4(y2 − z2)+
y4(z2 − x2)+
z4(x2 − y2)]
1 1 1 1 1 −1 −1 −1 −1 −1
Γ2 xyz 1 1 −1 −1 1 −1 −1 1 1 −1
Γ12 xyz(x2 − y2), etc. 2 2 0 0 −1 −2 −2 0 0 1
Γ15 xy(x2 − y2), etc. 3 −1 1 −1 0 3 −1 1 −1 0
Γ25 xy, yz, zx 3 −1 −1 1 0 3 −1 −1 1 0
wave vector at k = 0 is the group Oh itself. The character table for
Oh along with the basis functions for all the irreducible representations
is given in Table 13.2. For the simple cubic structure this character
table applies to the group of the wave vector for the Γ point and the R
point, both of which have the full symmetry operations of the Oh point
group (see the Brillouin zone in Fig. 13.2). We notice that to obtain
basis functions for all the irreducible representations of the group Oh
we need to include up to 6th order polynomials. The notation used in
Table 13.2 is that traditionally used in the solid state physics literature.
In this notation, Γ15 and Γ25 are odd while Γ15 and Γ25 are even. To get
around this apparent non-uniformity of notation, we often use Γ±
i (e.g.,
Γ±
15) to emphasize the parity of a wavefunction for the cubic groups.
The group of the wave vector at a point along the ∆ axis (for example)
has fewer symmetry operations than the group of the wave vector
k = 0. The symmetry operations for a point along the ∆ axis for the
simple cubic lattice are those of a square, rather than those of a cube
and are the operations of C4v. The multiplication table for the elements
of the point group C4v which is appropriate for a reciprocal lattice point
∆ along the ˆx axis.
13.4. SIMPLE CUBIC LATTICE 355
Table 13.3: Character table for the group of the wave-vector ∆.†
Representation Basis Functions E C2
4 2C4 2iC2
4 2iC2
∆1 1, x, 2x2
− y2
− z2
1 1 1 1 1
∆2 y2
− z2
1 1 −1 1 −1
∆2 yz 1 1 −1 −1 1
∆1 yz(y2
− z2
) 1 1 1 −1 −1
∆5 y, z; xy, xz 2 −2 0 0 0
Class Operation Designation E α β γ δ ε ζ η
E x y z E E α β γ δ ε ζ η
C2
4 x −y −z α α E γ β ε δ η ζ
2C4
x
x
−z
z
y
−y
β
γ
β
γ
γ
β
α
E
E
α
ζ
η
η
ζ
ε
δ
δ
ε
2iC2
4
x
x
−y
y
z
−z
δ
ε
δ
ε
ε
δ
η
ζ
ζ
η
E
α
α
E
γ
β
β
γ
2iC2
x
x
−z
z
−y
y
ζ
η
ζ
η
η
ζ
δ
ε
ε
δ
β
γ
γ
β
E
α
α
E
†
The rule for using the multiplication table is that:
αβ = (x, −y, −z)(x, −z, y) = (x, −(−z), −(y)) = (x, z, −y) = γ
βδ = (x, −z, y)(x, −y, z) = (x, z, y) = η
where the right operator (β) designates the row and the left operator
(α) designates the column.
The character table (including basis functions) for the group of the
wave vector for ∆, where ∆ = (∆, 0, 0) is along ˆx, is given in Table 13.3.
In Table 13.3 the C4 rotation operation is along ˆx, the 2iC2
4 are along
ˆy, ˆz, and the 2iC2 are along {011}. The basis functions in the character
table can be found from inspection by taking linear combinations of
(x , ym
, zn
) following the discussion in Chapter 4. The basis functions
for the lower symmetry groups (such as the group of ∆) are related
to those of Oh by considering the basis functions of the point group
Oh as reducible representations of the subgroup ∆, and decomposing
these reducible representations into irreducible representations of the
group ∆. For example Γ+
25 of Oh is a reducible representation of C4v
356CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Table 13.4: Character table for the group of the wave vector Λ.
Character Table for the Λ Axis
Λ = C3v E 2C3 3iC2
Λ1 1 1 1
Λ2 1 1 −1
Λ3 2 −1 0
and reduction of Γ+
25 into irreducible representations of C4v yields the
compatibility relation
Γ+
25
Oh
→ ∆2 + ∆5
C4v
.
We note that yz is the longitudinal partner for ∆ = (∆, 0, 0) and corresponds
to the irreducible representation ∆2, while xy, xz are the transverse
partners corresponding to ∆5. The process of going from higher
to lower symmetry defines the compatibility relations between irreducible
representations of Oh and those of C4v.
Character tables for all the high symmetry points for k vectors in
the simple cubic lattice are given in this section. For example, the
symmetry group for a wave vector along the (111) axis or Λ axis is C3v.
(See Fig. 13.2 on p. 353). The character table for a Λ point along the
(111) direction with C3v point group symmetry is given in Table 13.4.
For a Λ point along the (111) direction, the 2C3 are along {111}, and the
3iC2 are along (1¯10), (10¯1), and (0¯11) directions. For the Λ point we
can do 3-fold rotations in both ± senses about ΓR. We can also do 180◦
rotations about 2-fold axes ΓM followed by inversion (see Fig. 13.2).
By ΓM we mean the wave vector to the center of an adjacent cube
edge. You will notice that a rotation by π about ΓM followed by
inversion does not leave Λ invariant. Only 3 of the “ΓM ” axes are
symmetry operations of the group; the other 3 such axes (like ΓM in
the diagram) are not symmetry operations. Therefore instead of the
symmetry operations 6iC2 which hold for the Γ and R points, the class
3iC2 for the group of the Λ point only has 3 symmetry elements.
In several instances, more than one high symmetry point in the
13.4. SIMPLE CUBIC LATTICE 357
Table 13.5: Group of the wave vector for points M and X.
M E 2C2
4 C2
4⊥ 2C4⊥ 2C2 i 2iC2
4 iC2
4⊥ 2iC4⊥ 2iC2
X E 2C2
4⊥ C2
4
2C4 2C2 i 2iC2
4⊥ iC2
4
2iC4 2iC2
M1, X1 1 1 1 1 1 1 1 1 1 1
M2, X2 1 1 1 −1 −1 1 1 1 −1 −1
M3, X3 1 −1 1 −1 1 1 −1 1 −1 1
M4, X4 1 −1 1 1 −1 1 −1 1 1 −1
M1, X1 1 1 1 1 1 −1 −1 −1 −1 −1
M2, X2 1 1 1 −1 −1 −1 −1 −1 1 1
M3, X3 1 −1 1 −1 1 −1 1 −1 1 −1
M4, X4 1 −1 1 1 −1 −1 1 −1 −1 1
M5, X5 2 0 −2 0 0 2 0 −2 0 0
M5, X5 2 0 −2 0 0 −2 0 2 0 0
Brillouin zone follows the same symmetry group–e.g., ∆ and T (see
Fig. 13.2). In considering point T, remember that any reciprocal lattice
point separated by a reciprocal lattice vector from T is an equally good
T point. The character table for the T-point is given in Table 13.3.
It can also happen that two symmetry points such as M and X
belong to the same point group D4h, but the symmetry operations for
the two groups of the wave vector can refer to different axes of rotation
as shown in Table 13.5. In the character Table 13.5, M and X both
belong to the same point group D4h, but the symmetry operations for
the 2 points can refer to different rotation axes. The notation C2
4 in
Table 13.5 refers to a 2-fold axis ΓX while 2C2
4⊥ refers to the two 2-fold
axes ⊥ to ΓX. These are in different classes because in one case X is
left exactly invariant, while in the other case X goes into an equivalent
X point. To put it in more physical terms, if X is not exactly on the
zone boundary but is at a ∆ point arbitrarily close, the C2
4 operation
still holds, while the 2C2
4⊥ does not.
Character tables for other high symmetry points in the Brillouin
zone for the simple cubic lattice (see Fig. 13.4) are given in Table 13.6
for the points Σ and S, and in Table 13.7 for point Z.
A word of explanation about the two C2
4 operations in group C2v
which are in different classes. Consider a π rotation about the kx axis;
Z goes into an equivalent Z point on the Brillouin zone boundary. For
symmetry operation iC2
4 about the kz axis, Z goes into itself identically,
358CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Table 13.6: Character table for the group of the wave vector for points
Σ and S
C2v
Σ, S E C2 iC2
4 iC2
Σ1 1 1 1 1
Σ2 1 1 −1 −1
Σ3 1 −1 −1 1
Σ4 1 −1 1 −1
Note that point S has extra symmetry because it is on the Brillouin
zone (B.Z.) boundary.
—
Table 13.7: Character table for the group of the wave vector for the
point Z
C2v
z x y
Z E C2
4 iC2
4 iC2
4⊥
Z1 1 1 1 1
Z2 1 1 −1 −1
Z3 1 −1 −1 1
Z4 1 −1 1 −1
13.5. HIGH SYMMETRY POINTS AND AXES 359
Figure 13.4: Brillouin zones for the (a) face-centered and (b) bodycentered
cubic lattices. Points and lines of high symmetry are indicated.
while for iC2
4 about the ky axis, Z goes into another Z point related to
it by a reciprocal lattice vector.
13.5 High Symmetry Points and Axes and
Their Character Tables for FCC and
BCC Structures
The group of the wave vector for arbitrary k is a subgroup of the group
of the wave vector k = 0, which displays the full point group symmetry
of the crystal. This situation applies to all crystal lattices, whether
they are cubic, hexagonal, etc. The various character tables given in
this section are for the B.C.C. (space group Im3m (O9
h) #229) and
F.C.C. (space group Fm3m (O5
h) #225) structures. The basis functions
are also given. The form of the basis functions is helpful in identifying
s, p and d states. The subgroups of the group of the wave vector at
k = 0 are called the small representations in contrast to the full
point group symmetry which is called the large representation.
360CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Table 13.8: Character Table for the Group of the wave vector for X
Group of X = (2π/a)(1, 0, 0)
Representation Basis E 2C2
4⊥
C2
4
2C2
4
2C2 i 2iC2
4⊥
iC2
4
2iC4 2iC2
X1 1, 2x2
− y2
− z2
1 1 1 1 1 1 1 1 1 1
X2 y2
− z2
1 1 1 −1 −1 1 1 1 −1 −1
X3 yz 1 −1 1 −1 1 1 −1 1 −1 1
X4 yz(y2
− z2
) 1 −1 1 1 −1 1 −1 1 1 −1
X5 xy, xz 2 0 −2 0 0 2 0 −2 0 0
X1 xyz(y2
− z2
) 1 1 1 1 1 −1 −1 −1 −1 −1
X2 xyz 1 1 1 −1 −1 −1 −1 −1 1 1
X3 x(y2
− z2
) 1 −1 1 −1 1 −1 1 −1 1 −1
X4 x 1 −1 1 1 −1 −1 1 −1 −1 1
X5 y, z 2 0 −2 0 0 −2 0 2 0 0
Table 13.9: Character table for group of the wave vector for L.
Group of L = (2π/a) 1
2
, 1
2
, 1
2
Representation Basis E 2C3 3C2 i 2iC3 3iC2
L1 1, xy + yz + xz 1 1 1 1 1 1
L2 yz(y2 − z2) + xy(x2 − y2) + xz(z2 − x2) 1 1 −1 1 1 -1
L3 2x2 − y2 − z2; y2 − z2 2 −1 0 2 −1 0
L1 x(y2 − z2) + y(z2 − x2) + z(x2 − y2) 1 1 1 −1 −1 −1
L2 x + y + z 1 1 −1 −1 −1 1
L3 y − z; 2x − y − z 2 −1 0 −2 1 0
Table 13.10: Character table for group of the wave vector for W.
Group of W = (2π/a)(1, 1
2
, 0)
Representation Basis E C2
4 2C2 2iC4 2iC2
4
W1 1, 2y2
− x2
− z2
1 1 1 1 1
W1 xz 1 1 1 −1 −1
W2 xyz 1 1 −1 1 −1
W2 y, z2
− x2
1 1 −1 −1 1
W3 xy, yz; x, z 2 −2 0 0 0
13.5. HIGH SYMMETRY POINTS AND AXES 361
Table 13.11: Character table for group of the wave vector for Σ.
Character Table, Group Σ = (2π/a)(x, x, 0)
Representation Basis E C2 iC2
4 iC2
Σ1 1, x + y 1 1 1 1
Σ2 z(x − y); z(x2
− y2
) 1 1 −1 −1
Σ3 z; z(x + y) 1 −1 −1 1
Σ4 x − y; x2
− y2
1 −1 1 −1
362CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Table 13.12: Character table for group of the wave vector for
G, K, U, D, Z, S.
Character Tables of G, K, U, D, Z, S
Z E C2
4 iC2
4 iC2
4⊥
Represen- G, K, U, S E C2 iC2
4 iC2
tation D E C2
4 iC2 iC2⊥
K1 1, x + y 1 1 1 1
K2 x(x − y), z(x2
− y2
) 1 1 −1 −1
K3 z, z(x + y) 1 −1 −1 1
K4 x − y; x2
− y2
1 −1 1 −1
G = 2π
a
(1
2
+ x, 1
2
− x, 0) (bcc) K = 2π
a
(3
4
, 3
4
, 0) (fcc)
U = 2π
a
(1, 1
4
.1
4
) (fcc) D = 2π
a
(1
2
, 1
2
, x) (bcc)
Z = 2π
a
(1, x, 0) (fcc) S = 2π
a
(1, x, x) (fcc)
For a given wave vector which is contained within the first Brillouin
zone for each of the simple cubic, FCC and BCC lattices, the character
table for the group of the wave vector at k = 0 is the same for all three
kinds of lattices. Having enumerated the symmetry operations for kvectors
in various cubic lattices, we are now ready to discuss the effect
of point group operations on the eigenfunctions of the Hamiltonian for
a solid with a periodic potential.
13.5. HIGH SYMMETRY POINTS AND AXES 363
Table 13.13: Character table for group of the wave vector for P.
Group P = (2π/a)(1
2
, 1
2
, 1
2
) for the BCC lattice
Representation Basis E 3C2
4 8C3 6iC4 6iC2
P1 1, xyz 1 1 1 1 1
P2 x4
(y2
− z2
) + y4
(z2
− x2
)+ 1 1 1 −1 −1
z4
(x2
− y2
)
P3 x2
− y2
, xyz(x2
− y2
) 2 2 −1 0 0
P4 x, y, z; xy; yz; zx 3 −1 0 −1 1
P5 z(x2
− y2
) 3 −1 0 1 −1
Table 13.14: Character table for group of the wave vector for N.
Group of N = (2π/a)(1
2
, 1
2
, 0) for the BCC lattice
Representation Basis E C2
4 C2 C2⊥ i iC2
4 iC2⊥ iC2
N1 1, xy, 3z2
− r2
1 1 1 1 1 1 1 1
N2 z(x − y) 1 −1 1 −1 1 −1 −1 1
N3 z(x + y) 1 −1 −1 1 1 −1 1 −1
N4 x2
− y2
1 1 −1 −1 1 1 −1 −1
N1 x + y 1 −1 1 −1 −1 1 1 −1
N2 z(x2
− y2
) 1 1 1 1 −1 −1 −1 −1
N3 z 1 1 −1 −1 −1 −1 1 1
N4 x − y 1 −1 −1 1 −1 1 −1 1
364CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Table 13.15: Character table for group of the wave vector for Λ and F.
F = (2π/a)(1
2
+ x, 1
2
− x, 1
2
− x); 0 ≤ x ≤ 1
2
Representation Basis E 2C3 3iC2
Λ1 1, x + y + z 1 1 1
Λ2 x(y2
− z2
) + y(z2
− x2
) + z(x2
− y2
) 1 1 −1
Λ3 2x − y − z, y − z 2 −1 0
13.6. GROUP OPERATIONS ON BLOCH FUNCTIONS 365
13.6 Effect of Translations and Point Group
Operations on Bloch Functions
We have considered the effect of the translation operator {ε|τ} on the
eigenfunctions for an electron in a periodic potential
ˆP{ε|τ}ψk(r) = eik·τ
ψk(r). (13.22)
The effect of a point group operation on this eigenfunction is
ˆP{Rα|0}ψk(r) = ˆP{Rα|0}eik·r
uk(r) (13.23)
in which we have written the eigenfunction in the Bloch form. Since
the effect of a point group operation on a function is equivalent to
preserving the form of the function and rotating the coordinate system
in the opposite sense, to maintain invariance of scalar products we
require
k · R−1
α (r) = Rα(k) · r. (13.24)
If we now define uRαk(r) ≡ uk(R−1
α r) and denote Rαk ≡ k , then we
have
ˆP{Rα|0}ψk(r) = eiRαk·r
uRαk(r) ≡ ψRαk(r) (13.25)
which we will now show to be of the Bloch form by operating with the
translation operator on ψRαk(r)
ˆP{ε|τ}ψRαk(r) = ˆP{ε|τ}[eiRαk·r
uk(R−1
α r)]
= eiRαk·(r+τ)
uk(R−1
α r + R−1
α τ). (13.26)
Because of the periodicity of uk(r) we have
uRαk(r + τ) = uk(R−1
α r + R−1
α τ) = uk(R−1
α r) ≡ uRαk(r) (13.27)
so that
ˆP{ε|τ}ψRαk(r) = eiRαk·τ
ψRαk(r) (13.28)
where uRαk(r) is periodic in the direct lattice. The eigenfunctions
ψRαk(r) thus forms basis functions for the Rαkth
irreducible representation
of the translation group T. As we saw in §13.3 on p. 349,
366CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
the set of distinct wave vectors in k-space which can be generated by
operating on one k vector by all the symmetry elements of the point
group g is called the “star of k” (see Fig. 13.1).
From the above arguments we have
ˆP{Rα|τ}ψk(r) = ˆP{ε|τ}
ˆP{Rα|0}ψk(r)
= ˆP{ε|τ}ψRαk(r)
= eiRαk·τ
ψRαk(r). (13.29)
Similarly we obtain
ˆP{Rβ|τ }ψRαk(r) = eiRβRαk·τ
ψRβRαk(r). (13.30)
Thus the set of eigenfunctions {ψRαk(r)} obtained by taking the star
of k spans the invariant subspace of the point group g since RβRα is
contained in g. If h is the order of the group g, there are h functions in
the set {ψRαk(r)}. All of these representations are completely specified
by k, but they are equally well specified by any of the k vectors in the
star of k. Since all the functions in the set {ψRαk(r)} correspond to the
same energy, we do not say that the function ψk(r) and ψRαk(r) are
degenerate. Instead we write {ψk(r)} for all the functions in the set
{ψRαk(r)} and consider the extra point group symmetry to yield the
relation E(k) = E(Rαk) for all Rα. In this way, we guarantee that the
energy E(k) will show the full point group symmetry of the reciprocal
lattice. Thus for the 2-dimensional square lattice, it is only necessary
to calculate E(k) explicitly for k points in 1/8 of the Brillouin zone
contained within the sector ΓΛRSX∆Γ (see Fig. 13.6).
We use the term “degeneracy” to describe states with exactly the
same energy and the same wave vector. Such degeneracies do in fact
occur because of symmetry restrictions at special high symmetry points
in the Brillouin zone and are called “essential” degeneracies. “Essential”
degeneracies occur only at high symmetry or special k points,
while accidental (“non-essential”) degeneracies occur at arbitrary k
points. “Special” high symmetry points in the Brillouin zone are those
for which
Rαk = k + K (13.31)
13.6. GROUP OPERATIONS ON BLOCH FUNCTIONS 367
  ¡
¢
£
¤
¥
.
Figure 13.5: One-eighth of
the simple cubic Brillouin
Zone.
where K is the reciprocal lattice vector including K = 0.
For these special points there are symmetry operations Rα which
obey Eq. 13.31 and these symmetry operators form the group of the
wave vector for wave vector k. We will denote this group by gk. When
gk is considered together with all the translation operations we denote
this group by Gk. The translations form a self-conjugate subgroup of
both Gk and the full space group G, and define a factor group Gk/T
and G/T for both of these cases. We label a wave function at one
of these special points in the Brillouin zone as ψi
kλ(r) where k is the
quantum number of the translations, i denotes a particular irreducible
representation of gk and λ is an index denoting the partners of (i). If
hk is the order of the group of the wave vector k, then if we operate on
an eigenfunction with symmetry elements Rk in gk, we obtain another
eigenfunction in the set {ψ
(i)
kλ(r)}
ˆP{Rk|0}ψ
(i)
kλ(r) =
µ
ψ
(i)
kµ(r)D(i)
({Rk|0})µλ (13.32)
where the sum is on partners µ. In this way the operations {Rk|0} in gk
will produce all partners of {ψ
(i)
kλ(r)}. If we have h symmetry elements
in g and if we now operate with one of these symmetry elements {Rβ|0}
which is not in gk we obtain other functions {ψ
(i )
Rβkλ(r)} corresponding
to the star of k and there will be (h/hk) such sets for all possible
i . In the cases where the symmetry operation yields Rαk = k + K,
then the eigenfunctions have essential degeneracies because we now
can have degenerate eigenfunctions with the same energy eigenvalue at
368CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
the same k vector (provided that the dimensionality of the irreducible
representation ≥ 2). These essential band degeneracies are lifted as we
move away from the high symmetry points to a general point in the
Brillouin zone. The rules governing the lifting of these degeneracies are
called compatibility relations, discussed in §13.7.
13.7 Compatibility Relations
To study these compatibility relations, let us follow some particular
energy band around the Brillouin zone and see how its symmetry type
and hence how its degeneracy changes. The problem of connectivity
(connecting energy bands as we move from one k point to a neighboring
k point with a different group of the wave vector) is exactly the same
type of problem as that occurring in crystal field splittings (Chapter 6)
as we go from a high symmetry situation to a perturbed situation of
lower symmetry.
As an illustration, consider the point group Oh in a simple cubic
lattice as we move along a (111) direction from Γ → Λ → R, from the
center of the Brillouin zone to the zone corner. At the Γ point (k = 0)
we have the full point group symmetry Oh. As we now go from a higher
point group symmetry Oh at Γ to a k vector along Λ, we go to a point
group of lower symmetry C3v. Since there are no 3-dimensional representations
in C3v, we know that the degeneracy of the Γ−
15, Γ−
25, Γ+
15, Γ+
25
levels will be at least partially lifted. We proceed as before to write
down the character table for the Λ point, and below it we will write
down the representations of the Γ point group, which we now treat as
reducible representations of the Λ point group. We then reduce out the
irreducible representations of the Λ point. This process is indicated in
the table below where we list the 10 irreducible representations of Oh
and indicate the irreducible representations of C3v therein contained.
This procedure gives a set of compatibility conditions.
13.7. COMPATIBILITY RELATIONS 369
irreducible
Λ E 2C3 3iC2 representations
Λ1 1 1 1
Λ2 1 1 −1
Λ3 2 −1 0
Γ+
1 1 1 1 Λ1
Γ+
2 1 1 −1 Λ2
Γ+
12 2 −1 0 Λ3
Γ+
15 3 0 −1 Λ2 + Λ3
Γ+
25 3 0 1 Λ1 + Λ3
Γ−
1 1 1 −1 Λ2
Γ−
2 1 1 1 Λ1
Γ−
12 2 −1 0 Λ3
Γ−
15 3 0 1 Λ1 + Λ3
Γ−
25 3 0 −1 Λ2 + Λ3
In a similar way, the compatibility relations for a simple cubic lattice
along the ∆ and Σ axes follow the progression from Γ to ∆ to X and also
from Γ to Σ to M as can be seen from Fig. 13.2. In going from ∆ → X
we go from C4v symmetry to D4h symmetry, since at the Brillouin
zone boundary, translation by a reciprocal lattice vector introduces
additional symmetries associated with a mirror plane. Similarly in
going from Σ → M we get 4 equivalent M points so that the symmetry
group goes from C2v to D4h.
Tables of compatibility relations are compiled in references such as
Miller and Love’s book. Compatibility relations for the simple cubic
lattice are summarized in Table 13.16.
As an example of using these compatibility relations, let us consider
what happens as we move away from the Γ point k = 0 on a 3-fold level
such as Γ+
25. There are many possibilities, as indicated below:
Γ+
25 → ∆2 + ∆5 → X3 + X5 (13.33)
Γ+
25 → Λ1 + Λ3 → R15 (13.34)
Γ+
25 → Σ1 + Σ2 + Σ3 → M1 + M5. (13.35)
Suppose that we want to find a set of compatible symmetries in
370CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Table 13.16: Compatibility Relations for the high symmetry points in
the simple cubic lattice.
Compatibility Relations Between Γ and ∆, Λ, Σ.
Γ+
1 Γ+
2 Γ+
12 Γ−
15 Γ+
25 Γ−
1 Γ−
2 Γ−
12 Γ+
15 Γ−
25
(100) ∆1 ∆2 ∆1∆2 ∆1∆5 ∆2 ∆5 ∆1 ∆2 ∆1 ∆2 ∆1 ∆5 ∆2∆5
(111) Λ1 Λ2 Λ3 Λ1Λ3 Λ1Λ3 Λ2 Λ1 Λ3 Λ2Λ3 Λ2Λ3
(110) Σ1 Σ4 Σ1Σ4 Σ1Σ3Σ4 Σ1Σ2Σ3 Σ2 Σ3 Σ2Σ3 Σ2Σ3Σ4 Σ1Σ2Σ4
Compatibility Relations Between X and ∆, Z, S
X1 X2 X3 X4 X5 X1 X2 X3 X4 X5
∆1 ∆2 ∆2 ∆1 ∆5 ∆1 ∆2 ∆2 ∆1 ∆5
Z1 Z1 Z4 Z4 Z3Z2 Z2 Z2 Z3 Z3 Z1Z4
S1 S4 S1 S4 S2S3 S2 S3 S2 S3 S1S4
Compatibility Relations Between M and Σ, Z, T
M1 M2 M3 M4 M1 M2 M3 M4 M5 M5
Σ1 Σ4 Σ1 Σ4 Σ2 Σ3 Σ2 Σ3 Σ2Σ3 Σ1Σ4
Z1 Z1 Z3 Z3 Z2 Z2 Z4 Z4 Z2Z4 Z1Z3
T1 T2 T2 T1 T1 T2 T2 T1 T5 T5
going around a circuit using the Brillouin zone (see Fig. 13.2).
Γ → Σ → M → Z → X → ∆ → Γ (13.36)
Then we must verify that when we arrive back at Γ we have the same
symmetry type as we started with. A set of such compatible symmetries
designates a whole band.
To go around one of these circuits, basis functions sometimes prove
very useful. Suppose that we are generating wave functions from the
tight binding point of view. Then we know that s-functions transform
like the identity representation so that a possible circuit would be Γ1 →
Λ1 → R1 → S1 → X1 → ∆1 → Γ1. If we have p-functions, the basis
functions are (x, y, z) and we can join up representations corresponding
to these basis functions, etc. The d-functions transform as (xy, xz, yz)
with Γ+
25 symmetry and (x2
+ ωy2
+ ω2
z2
), (x2
+ ω2
y2
+ ωz2
) where
ω = exp(2πi/3), corresponding to Γ+
12 symmetry. We must of course
always remember that the charge distribution of an s-electron in a
cubic crystal will exhibit the cubic symmetry of the crystal, and not
correspond to the full rotational symmetry of the free atomic state.
As an example of how compatibility relations are used in the labeling
of energy bands, we show the energy dispersion relation E(k) in
Fig. 13.6 for the high symmetry directions k100 and k111 for the simple
cubic structure. For the band with lower energy, we have the compat-
13.7. COMPATIBILITY RELATIONS 371
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¦ £
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Figure 13.6: Schematic diagram of energy bands illustrating compatibility
relations.
ibility relations Γ1 → ∆1 → X1 and Γ1 → Λ1 → R1. For the upper
band, a consistent set of compatibility relations is
Γ+
25 → ∆2 + ∆5, ∆2 → X2 and ∆5 → X5
Γ+
25 → Λ1 + Λ3, Λ1 → R1 and Λ3 → R12
In applying the compatibility relations as we approach the R point from
the Λ direction, we note that the R point has the same group of the
wave vector as k = 0 and the same subscript notation can be used to
label the R and Γ levels.
13.7.1 Irreducible Representations
For a symmorphic lattice, the irreducible representations for a given
point in the Brillouin zone are associated with the group of the wave
vector for that point.
For non-symmorphic lattices, the determination of the irreducible
representation can be more difficult, except at k = 0, where the irreducible
representation is the same as the point group formed by ignoring
the fractional translations of the space group. For each symmetry
axis leading away from k = 0, the character tables for those k points
can be obtained by selecting the appropriate point group table and by
372CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
multiplying the character for the symmetry operations that contains a
translation τ by a phase factor exp[ik · τ].
13.8 Selected Problems
1. (a) For the crystal structure shown below on the left, identify
the space group and list the symmetry elements.
(b) Identify the high symmetry points (and axes), and list the
group of the wave vector at these high symmetry points (and
axes).
(c) Using the space group identified in (a), explain the diagrams
for this space group as shown in the International Crystallography
Tables.
(d) Using the tables in (c), find the atom sites and site symmetries
for the structure shown in (a).
2. (a) For the crystal structure shown below on the left, list the
symmetry elements of the space group.
(b) Identify the high symmetry points (and axes), and list the
group of the wave vector at these high symmetry points (and
axes).
13.8. SELECTED PROBLEMS 373
3. (a) Show that in the diamond structure (§14.3.3) the product
of two symmetry operations involving translations τ yields
a symmetry element with no translations
{α|τ}{β|τ} = {γ|0}
where τ = (1, 1, 1)a/4. What is the physical significance of
this result?
(b) What are the symmetry operations of the group of the wave
vector for the diamond structure at k = 0? at the ∆ point?
at the L point?
(c) List the real space symmetry operations of the non-symmorphic
two-dimensional square space group p4gm (#12). Find the
group of the wave vector for the high symmetry points in
the space group p4gm and compare your results with those
for the symmorphic group p4mm (Fig. 13.1 of the notes).
4. (a) List the real space symmetry operations of the non-symmorphic
two-dimensional square space group p4gm (#12).
(b) Explain the diagrams and the point symmetry entries for
space group #12 (p4gm) in Fig. 12.20 which was taken from
the International Crystallography Tables.
(c) Find the group of the wave vector for the high symmetry
points in the space group p4gm and compare your results
with those for the symmorphic group p4mm (Fig. 13.1 of
the notes).
374CHAPTER 13. GROUP OF THE WAVE VECTOR AND BLOCH’S THEOREM
Chapter 14
Applications to Lattice
Vibrations
Our first application of the space groups to excitations in periodic solids
is in the area of lattice modes. Group theoretical techniques are important
for lattice dynamics in formulating the normal mode secular
determinant in block diagonal form, and symmetry is also important in
the area of selection rules for optical processes involving lattice modes
such as infrared and Raman activity. Transitions to lower symmetry
through either phase transitions or strain-induced effects may lead to
mode splittings. These mode splittings can be predicted using group
theoretical techniques and the changes in the infrared and Raman spectra
can be predicted.
14.1 Introduction
The atoms in a solid are in constant motion and give rise to lattice
vibrations which are very similar to the molecular vibrations which we
have discussed in Chapter 9. We discuss in this section the similarities
and differences between lattice modes and molecular vibrations.
Suppose that we have a solid with N atoms which crystallize into a
simple Bravais lattice with 1 atom/unit cell. For this system there are
3N degrees of freedom corresponding to 3 degrees of freedom/atom or 3
degrees of freedom/primitive unit cell. There are N allowed wave vector
375
376 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
states in the Brillouin zone which implies that there are 3 branches for
the phonon dispersion curves of a simple monatomic solid, each branch
containing solutions for N k-vectors. For the case of molecules, we
subtract three degrees of freedom corresponding to the uniform translation
of the molecule. In the solid, these uniform translational modes
correspond to the acoustic modes at k = 0, which are subject to the
constraint that ω2
acoustic ≡ 0. The three modes corresponding to the
rotations of the solid about the center of mass are not specifically con-
sidered.
We have found in Chapter 13 that the translational symmetry of
a crystal is conveniently handled by labeling the N irreducible representations
of the translation group by the N k vectors which are
accommodated in the 1st
Brillouin zone. So if we have a primitive
unit cell with 1 atom/unit cell, there are 3 vibrational modes for each
k value and together these 3 modes constitute the acoustic branches.
In particular, there are 3 acoustic vibrational modes for the k = 0
wave vector, which exhibits the full point group symmetry of the crystal;
these three acoustic modes correspond to the pure translational
modes which have zero frequency and zero restoring force.
We review here the phonon dispersion relations in a 1-dimensional
crystal with 1 atom/unit cell (see Fig. 14.1a) and with 2 atoms/unit
cell (see Fig. 14.1b) having masses m and M where m < M, and a is
the distance between adjacent atoms. For the acoustic branch at k = 0,
all atoms vibrate in phase with identical displacements u along the direction
of the atomic chain, thus corresponding to a pure translation
of the chain. The wave vector k distinguishes each normal mode of the
system by introducing a phase factor eika
between the displacements on
adjacent sites. For the case of one atom/unit cell, the lattice mode at
the zone boundary corresponds to atoms moving 90◦
out of phase with
respect to their neighbors. For the case of 2 atoms/unit cell, the size of
the unit cell is twice as large so that the size of the corresponding Brillouin
zone (B.Z.) is reduced by a factor of 2. The dispersion relations
and lattice modes in this case relate to those for one atom/unit cell by a
zone folding of the dispersion relation shown in Fig. 14.1a, thus leading
to Fig. 14.1b. Thus the optical mode at k = 0 has neighboring atoms
moving out of phase with respect to each other. The normal mode at
14.1. INTRODUCTION 377
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Figure 14.1: Phonon dispersion curves for a one-dimensional line of
atoms with (a) a single mass and (b) two different masses m and M.
the new B.Z. boundary k = π/2a thus corresponds to a mode where
one atom is at rest, while its neighbor is in motion.
In 3-dimensions, the phonon dispersion relations for Ge with the diamond
structure (with 2 atoms/unit cell) are plotted along high symmetry
directions in Fig. 14.2 and the dispersion relations are labeled
by the appropriate irreducible representations giving the symmetry of
the corresponding normal mode. The phonon dispersion relations for
germanium are determined from inelastic neutron scattering measurements
and are plotted as points in Fig. 14.2. At a general point k in the
Brillouin zone for the diamond structure, there are 3 acoustic branches
and 3 optical branches. However at certain high symmetry points and
along certain high symmetry directions, mode degeneracies occur as,
for example, along ΓL and ΓX. Group theory allows us to identify
the high symmetry points where degeneracies occur, which modes stick
together, which modes cross, and which modes show anti-crossings, to
be discussed further in this chapter.
The symmetry aspects of the lattice mode problem at k = 0 for simple
structures with 1 atom/unit cell are simply the uniform translation
of the solid and no group theory is required. However, group theory
is needed to deal with lattice modes away from k = 0. Furthermore
the lattice modes that are of interest in the current literature often
involve complicated crystal structures with many atoms/unit cell or
378 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
Figure 14.2: Phonon dispersion curves for Ge along certain high symmetry
axes in the Brillouin zone. The data at the Γ point are from
Raman scattering measurements and the data elsewhere in the zone
are from neutron scattering experiments.
14.2. LATTICE MODES RELATIVE TO MOLECULAR VIBRATIONS379
systems with reduced dimensionality; for such problems group theory
is a powerful tool for lattice mode classification and for the determination
of selection rules for infrared and Raman spectroscopy and for
phonon-assisted optical transitions more generally.
The general outline for discussing lattice modes in solids is:
1. Find the symmetry operations for the group of the wave vector
k = 0, the appropriate character table and irreducible represen-
tations.
2. Find χlattice modes = χatom sites ⊗ χvector. The meaning of this relation
is discussed below (item #3 in §14.2).
3. Find the irreducible representations of χlattice modes. The characters
for the lattice mode representation express the symmetry
types and degeneracies of the lattice modes.
4. Find the normal mode patterns.
5. Which modes are IR-active?
6. Which modes are Raman-active?
7. Are there any polarization effects?
8. Find the lattice modes at other points in the Brillouin zone.
9. Using the compatibility relations, connect up the lattice modes
at neighboring k points to form a phonon branch.
14.2 Unique Features of Lattice Modes Relative
to Molecular Vibrations
There are several aspects of the lattice mode problem in the solid phase
that differ from molecular vibrations in the gas phase (see §9.2 of Chapter
9 on p. 195):
380 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
1. The unit cell- In the lattice mode problem, we consider normal
modes for the atoms in a unit cell rather than for a molecule.
Since the symmetry is different for the various types of k-vectors
in the Brillouin zone, we must solve the lattice mode problem for
each distinct type of k-vector. On the other hand, for many experimental
studies of the lattice modes we use light as our probe.
Since the wavelength of light is long (λ ≈ 5000˚A) and the magnitude
of the corresponding k vector (k = 2π/λ) is very small
compared with Brillouin zone dimensions, our main interest is
in lattice modes at or near k = 0 (the Γ point). We therefore
emphasize the lattice modes for k = 0 in our discussions.
2. Equivalence- To find χatom sites for molecules, we consider the
action of a symmetry operator ˆPR on an atomic site and examine
the transformation matrix to see whether or not the site is transformed
into itself under the point symmetry operation ˆPR. In the
case of a crystal, however, we consider all points separated by a
lattice vector Rn as identical. Thus r → r + Rn is an identity
transformation for all Rn.
3. Counting of Lattice Modes-Phonon Branches- For the case
of molecules we have
χmolecular vibrations = χatom sites ⊗ χvector − χtranslations − χrotations
(14.1)
whereas for lattice modes, we simply write
χlattice modes = χatom sites ⊗ χvector. (14.2)
That is, we do not subtract χtranslations and χrotations in Eq. 14.2
for the following reasons. Each atom/unit cell has 3 degrees of
freedom, yielding a normal mode for each wave vector k in the
Brillouin zone. The collection of normal modes for a given degree
of freedom for all k vectors forms a phonon branch. Thus for
a structure with one atom/unit cell there are 3 phonon branches,
the acoustic branches. If there is more than 1 atom/unit cell,
then
#branches = (#atoms/unit cell) × 3 (14.3)
14.3. ZONE CENTER PHONON MODES 381
of which 3 are acoustic and the remainder are optic. The translational
degrees of freedom correspond to the trivial k = 0 solution
for the 3 acoustic branches which occur at ω = 0 and are smoothly
connected with non-trivial solutions as we move away from the Γ
point. Since the atoms in the solid are fixed in space there are no
rotational degrees of freedom to be subtracted.
We will now illustrate the application of group theory to the solution
of the lattice mode problem for several illustrative structures. First we
consider simple symmorphic structures. Then we consider some simple
non-symmorphic structures. Our initial examples will be for the k = 0
modes. This will be followed by a discussion of modes elsewhere in
the Brillouin zone. In this connection, a discussion of “zone-folding”
phenomena will be presented because of the connection between the
zone folding of normal modes at k = 0 to yield modes which can be
excited by optical techniques.
14.3 Zone Center Phonon Modes
In this section we consider the symmetries of zone center phonon modes
for some illustrative cases. The examples selected in this section are
chosen to demonstrate some important aspect of the lattice mode prob-
lem.
14.3.1 In the NaCl Structure
This very simple example is selected to illustrate how the symmetries
of the lattice modes are found. We take our “basic unit cell” to be the
primitive unit cell of either one of the interpenetrating fcc structures
(space group #225 (Fm3m) O5
h), so that each unit cell will contain an
Na atom and a Cl atom (see Fig. 14.3a). The primitive fcc unit cell is
shown in Fig. 14.3b and the primitive lattice vectors are indicated. The
larger cubic unit cell (Fig. 14.3a) contains 4 primitive unit cells with 4
Na and 4 Cl atoms (ions). The space group O5
h for the NaCl structure
is a symmorphic structure, and the group of the wave vector at k = 0
for the NaCl structure is Oh. Since the details of the translations do
382 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
(a) (b)
Figure 14.3: (a) The NaCl structure which is space group #225. (b)
The rhombohedral primitive cell of the fcc lattice which contains one
Na atom and one Cl atom.
not enter into the considerations of phonons at k = 0 for symmorphic
space groups, we need to consider only the point group operations for
Oh.
O(432) E 8C3 3C2 = 3C2
4 6C2 6C4
Γ1 A1 1 1 1 1 1
Γ2 A2 1 1 1 −1 −1
Γ12 (x2
− y2
, 3z2
− r2
) E 2 −1 2 0 0
Γ15
(Rx, Ry, Rz)
(x, y, z)
T1 3 0 −1 −1 1
Γ25 yz, zx, xy T2 3 0 −1 1 −1
Oh = O × i (m3m)
Under all symmetry operations of Oh each Na and Cl atom site is
transformed either into itself or into an equivalent atom site separated
by a lattice vector Rm. Thus,
χatom sites = 2A1g (14.4)
For Oh symmetry, χvector = T1u, so that at k = 0
χlattice modes = 2A1g ⊗ T1u = 2T1u. (14.5)
14.3. ZONE CENTER PHONON MODES 383
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Figure 14.4: In-phase and out-of-phase normal modes at k = 0 for
NaCl.
Thus both the acoustic branch and the optical branch at k = 0 have T1u
or Γ−
15 symmetry. The normal modes for the acoustic branches of the
NaCl structure have the two atoms moving in phase in the x, y, and
z directions, while for normal modes in the optical branches the two
atoms move out of phase in the x, y and z directions (see Fig. 14.4).
Since the electromagnetic interaction transforms as the vector (T1u),
the optic branch is infrared-active. The acoustic branch is not optically
excited because ω = 0 at k = 0. Since the optic branch for the NaCl
structure has odd parity, it is not Raman-active. As we move away
from the Γ point (k = 0), the appropriate symmetries can be found by
compatibility relations. For example along the (100) directions Γ−
15 →
∆1 + ∆5 in which ∆1 is the symmetry of the longitudinal mode and ∆5
that for the transverse mode. We will now give several other examples
of zone center modes in other structures and then return in §14.4.1 to
the discussion of non-zone-center modes for the NaCl structure.
14.3.2 In the Perovskite Structure
Let us now consider lattice modes in BaTiO3 (see Fig. 14.5), an example
of a crystal structure with slightly more complexity, but still corresponding
to a symmorphic space group. The focus of this section is
to illustrate the identification of the normal modes. For the perovskite
structure shown in Fig. 14.5, we have 5 atoms/unit cell and therefore
we have 15 degrees of freedom, giving rise to 3 acoustic branches and
12 optical branches. The point group of symmetry at k = 0 is Oh.
Consider the unit cell shown in Fig. 14.5. The Ba++
ions at the cube
384 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
Figure 14.5: The cubic
perovskite crystal
structure of barium
titanate,
containing one Ba,
one Ti and three O
atoms. The Ba2+
ions are at the cube
corners, O2−
ions at
the face centers, and
a Ti4+
ion at the
body center. The
space group is #221.
corners are shared by 8 neighboring unit cells, so that one Ba++
ion is
considered to be associated with the unit cell shown. Likewise the O−−
ions in the face centers are shared by two unit cells, so that 3O−−
ions
are treated in the unit cell shown. The Ti4+
ion at the cube center is
of course fully contained in the unit cell shown in Fig. 14.5.
Using the diagram in Fig. 14.5, we thus obtain for χatom sites
E 8C3 3C2
4 6C2 6C4 i 8iC3 3iC2
4 6iC2 6iC4
χatom sites 5 2 5 3 3 5 2 5 3 3
all Ba,Ti all Ba,Ti Ba,Ti all Ba,Ti all Ba,Ti Ba,Ti
one O one O one O one O
Looking at the character table for Oh we see that
χatom sites = 3A1g + Eg = 3Γ+
1 + Γ+
12. (14.6)
We note that the Ba2+
and Ti4+
ions each transform as A1g with the
three oxygens transforming as A1g + Eg = Γ+
1 + Γ+
12. In Oh symmetry
χvector = T1u = Γ−
15 (14.7)
14.3. ZONE CENTER PHONON MODES 385
so that
χlattice modes = (3A1g + Eg) ⊗ T1u = 3T1u + (Eg ⊗ T1u)
= 4T1u + T2u = 4Γ−
15 + Γ−
25.
(14.8)
Thus at k = 0 there are 5 distinct frequencies, including the acoustic
branch with Γ−
15 symmetry and ω = 0. Since the Ba2+
and Ti4+
ions
transform as A1g, we know that the Γ−
25 mode requires motion of the
oxygens. In the following we illustrate how the normal mode patterns
shown in Fig. 14.6 are obtained. Note the numbers assigned to the
oxygens in Fig. 14.6b.
From the character table for Oh we note that the characters for Cz
4
are different for the Γ−
15 and Γ−
25 modes, and for this reason Cz
4 is a
useful symmetry operation for finding the normal mode displacements.
First we consider the effect of Cz
4 on each of the 3 inequivalent oxygen
sites and on each of the 3 components of the vector; this consideration
is independent of the symmetry of the vibrational mode:
Cz
4



1
2
3


 =



2
1
3


 Cz
4



x
y
z


 =



y
−x
z


 . (14.9)
Finding the normal mode for the acoustic translational branch is trivial
(see Fig. 14.6a). The operations of Eq. 14.9 are now applied to find the
normal modes in Fig. 14.6b and e. For the Γ−
25 displacements, Fig. 14.6b
shows the motions for the z component of the mode. The partners are
found by cyclic operations on (x, y, z) and atom sites (1, 2, 3), as given
in Eq. 14.10. Then operation by Cz
4 yields
Cz
4



−x2 + x3
y1 − y3
−z1 + z2


 =



−y1 + y3
−x2 + x3
−z2 + z1


 =



0 −1 0
1 0 0
0 0 −1






−x2 + x3
y1 − y3
−z1 + z2



(14.10)
giving a character of −1 for Cz
4 in the Γ−
25 representation. Performing
representative operations on this normal mode will show that it provides
a proper basis function for the Γ−
25 irreducible representation in the
point group Oh.
386 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
Figure 14.6: Schematic diagram of lattice modes at k=0 for the BaTiO3
perovskite structure. (a) Γ−
15 acoustic mode; (b) Γ−
25 mode where only
2 of the 3 distinct oxygens move; (c) Γ−
15 mode with the Ti and Ba
vibrating against the oxygens. (d) Γ−
15 mode with the Ti4+
vibrating
against the Ba2+
and (e) Γ−
15 breathing mode of transverse oxygens vs.
longitudinal oxygens.
14.3. ZONE CENTER PHONON MODES 387
Now consider the Γ−
15 normal mode given in Fig. 14.6e. The displacements
shown in the diagram are for the z component of the mode.
To achieve no motion of the center of mass, the actual displacements
must be −z1 − z2 + 2z3 for the 3 oxygens at positions 1, 2 and 3. Using
cyclic permutations we obtain the 3 components of the mode given in
Eq. 14.11. Then action of Cz
4 yields
Cz
4



2x1 − x2 − x3
−y1 + 2y2 − y3
−z1 − z2 + 2z3


 =



2y2 − y1 − y3
x2 − 2x1 + x3
−z2 − z1 + 2z3



=



0 1 0
−1 0 0
0 0 1






2x1 − x2 − x3
−y1 + 2y2 − y3
−z1 − z2 + 2z3


 (14.11)
so that the character for this Γ−
15 mode is +1, in agreement with the
character for the Cz
4 operation in the Γ−
15,z irreducible representation
(see the character table for Oh). Operation with typical elements in
each class shows this mode provides a proper basis function for Γ−
15.
Clearly all the modes shown in Fig. 14.6 have partners x, y and z,
so that collectively they are all the normal modes for BaTiO3. Since
all modes at k = 0 have odd parity, none are Raman-active, noting
that for the Oh point group, Raman-active modes have A1g, Eg and T2g
symmetries. However, the 3T1u or 3Γ−
15 modes are infrared-active,
and can be excited when the E vector for the light is polarized in the
direction of the oscillating dipole moment indicated in Fig. 14.6.
14.3.3 Phonons in the Diamond Lattice: A NonSymmorphic
Structure
We now illustrate the mode symmetries for a non-symmorphic space
group with 2 atoms/unit cell (specifically we illustrate the lattice modes
of Ge or Si, which crystallize in the diamond structure). The two distinct
atoms per unit cell are indicated in Fig. 14.7 as light atoms and
dark atoms. We will take the “basic unit cell” for the diamond structure
to be the fcc primitive unit cell shown in Fig. 14.3b. Again, we are
most interested in the lattice modes at k = 0. The set of operations ˆPR
388 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
Figure 14.7: The zinc blende structure illustrating the two dissimilar
lattice sites. With identical atoms at the two sites, the diamond structure
results. The space group for the diamond lattice is Fd3m or #227
(O7
h).
that are relevant are the 48 operations of the Oh point group. In considering
the diamond structure, we think of the light atoms as being on
one sublattice and the dark atoms on another sublattice. Each of the
symmetry operators ˆPR of Td will leave each atom on the same sublattice.
However, the operations in Oh that are not in Td when combined
with a translation τ = a
4
(111) take each atom on one sublattice into the
other. This space group is non-symmorphic because the symmetry operations
of the group of the wave vector at k = 0 contains translations
τ = a
4
(111). The 48 operations and 10 classes for the diamond structure
at k = 0 are given below. The 24 operations requiring translations
τ are indicated by symmetry operations labeling the relevant classes.
In computing χatom sites, there are two kinds of lattice sites — one on
each of the fcc sublattices. Thus an atom is considered “to go into itself”
if it remains on its own sublattice and “not to go into itself” if it
switches sublattices under a symmetry operation ˆPR. Using this criterion
the results for χatom sites for the diamond structure are given below.
14.3. ZONE CENTER PHONON MODES 389
{E|0} {8C3|0} {3C2|0} {6C2|τ} {6C4|τ}
χa.s. 2 2 2 0 0
{i|τ} {8iC3|τ} {3iC2|τ} {6iC2|0} {6iC4|0}
χa.s. 0 0 0 2 2
Decomposition of χatom sites into irreducible representations of Oh leads
to χa.s. = A1g + A2u or Γ+
1 + Γ−
2 . Here Γ+
1 is even under inversion
and Γ−
2 which is odd under inversion, using the usual notation for irreducible
representations for solids. We note that the operation {i|τ}
interchanges sublattices 1 ↔ 2. We will also make use of this result
for χatom sites in discussing the electronic energy band structure of solids
crystallizing in the diamond structure.
To get the characters for the lattice vibrations, we then take χvector =
Γ−
15 = T1u and
χlattice modes = χa.s. ⊗χvector = (A1g +A2u)⊗T1u = T1u +T2g = Γ−
15 +Γ+
25
(14.12)
For each k value, there are 6 vibrational degrees of freedom with 2
atoms/unit cell. These break up into 2 triply degenerate modes at k =
0, one of which is even, the other odd under inversion. The odd mode
Γ−
15 is the acoustic mode, which at k = 0 is the pure translational mode.
The other mode is a Γ+
25 mode which is symmetric under inversion and
represents a breathing or optical mode. The optic mode is Ramanactive
but not infrared-active. Furthermore, the Raman-active mode is
observed only in the off-diagonal polarization Ei⊥Es for the incident
and scattered light.
Let us now illustrate a screw axis operation in the diamond structure
(see Fig. 14.7) and see how this operation is used in finding the normal
modes. Denoting the black atoms by 1 and the white atoms by 2,
consider the effect of {Cz
4 |τ} on atom sites
1
2
and on the vector
390 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS



x
y
z



{Cz
4 |τ}
1
2
=
2
1
{Cz
4 |τ}



x
y
z


 =



y
−x
z



(14.13)
Using these results we can then obtain the characters for the displacements
(R1 + R2) which has Γ−
15 symmetry and is identified with
the basic vibration of an fcc sublattice:
{Cz
4 |τ}



x1 + x2
y1 + y2
z1 + z2


 =



y2 + y1
−x2 − x1
z2 + z1


 =



0 1 0
−1 0 0
0 0 1






x1 + x2
y1 + y2
z1 + z2



(14.14)
yielding a character of +1 for {Cz
4 |τ}, in agreement with the character
for {Cz
4 |τ} in the Γ−
15 irreducible representation for the acoustic mode
translational branches of point group Oh. If all the symmetry operations
are then carried out, it is verified that R1 + R2 provides basis
functions for the Γ−
15 irreducible representation of Oh.
When the 2 fcc sublattices vibrate out of phase, their parity is reversed
and a mode with even parity (the Γ+
25 mode) is obtained
{Cz
4 |τ}



x1 − x2
y1 − y2
z1 − z2


 =



y2 − y1
−x2 + x1
z2 − z1


 =



0 −1 0
1 0 0
0 0 −1






x1 − x2
y1 − y2
z1 − z2



(14.15)
yielding a character of –1. This checks with the character for {Cz
4 |τ}
in the irreducible representation Γ+
25 for the point group Oh.
As we move away from k = 0 along the ∆ axis or the Λ axis,
the triply degenerate modes break up into longitudinal and transverse
branches. The symmetries for these branches can be found from the
compatibility relations (see §13.7 on p. 368). For example, as we move
away from k = 0 along the ∆ axis toward the X point (see Fig. 14.8),
we have the compatibility relations
Γ−
15 → ∆1 + ∆5
Γ+
25 → ∆2 + ∆5.
(14.16)
14.3. ZONE CENTER PHONON MODES 391
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¢ ¥¨!¤"¨#¨$¥%£¦&
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7 3
7 3
7 2
7986
Figure 14.8: Lattice modes along the ∆-axis for the diamond structure.
Group theory gives no information on the relative frequencies of the
Γ−
15 and Γ+
25 modes.
We finally note that the Raman tensor for the Raman-active Γ+
25
mode at the zone center transforms as Ei
xEs
y αxy(Γ+
25) plus cyclic permutations.
Thus, observation of this mode requires ( , ⊥) settings of
the incident and scattered polarizations, respectively.
14.3.4 Phonons in the Zincblende Structure
Closely related to the diamond structure is the zincblende structure
(space group F43m #216, T3
d ) where the two fcc sublattices are distinct.
This is the crystal structure for III-V semiconductor compounds such as
GaAs. For this case the Ga atoms (ions) would be on one fcc sublattice
and the As ions on the other fcc sublattice.
Since the sublattices are distinct, the group of the k-vector at k = 0
for the zincblende structure has only the operations of the point group
Td. In calculating χlattice modes we note that the vector in group Td
transforms as irreducible representation T2. Thus from the irreducible
representations contained in χatom sites
χatom sites = 2A1 = 2Γ1
392 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
we take the direct product of χatom sites with χvector to obtain
χlattice modes = 2A1 ⊗ T2 = 2T2 = 2Γ15. (14.17)
For the zincblende structure, the optic mode is both infrared-active and
Raman-active since the irreducible representation T2 for point group
Td corresponds to both T1u and T2g of the point group Oh. This correspondence
is apparent from the character tables for Td and Oh (see
Table 13.2 on p. 354).
14.4 Lattice Modes Away From k = 0
Modes at k = 0 can be observed by optical spectroscopy when superlattice
effects are present, giving rise to zone folding. Non-zone center
modes can also be observed in second-order Raman spectra (comprising
phonons with wave vectors +k and −k). Lattice modes at k = 0 are
routinely observed by inelastic neutron scattering techniques.
To construct phonon branches for the entire range of k vectors
within the first Brillouin zone, we must consider the general procedure
for finding the lattice modes at other high symmetry points and we
make use of compatibility relations to relate these solutions to related
solutions of neighboring k-points.
The procedure for finding lattice modes at k = 0 is outlined below:
1. Find the appropriate group of the wave vector at point k.
2. Find χatom sites and χvector for this group of the wave vector.
3. Within a unit cell
χlattice modes = χatom site ⊗ χvector (14.18)
Find the symmetry types and mode degeneracies of χlattice modes.
4. Introduce a phase factor relating unit cells with translation by τ:
P{ε|τ}Ψk(r) = eik·τ
Ψk(r) Bloch theorem (14.19)
5. Find lattice modes (including phase factor).
14.4. LATTICE MODES AWAY FROM K = 0 393
We illustrate these issues in terms of the NaCl structure which was
previously considered with regard to its normal modes at k = 0 (see
§14.3.1).
14.4.1 Phonons in NaCl at the X point k = π
a (100)
We use essentially the same steps to get normal modes at the X point
as we used for the normal modes in the Γ point (see §14.2). The group
of the wave vector at point X is given in the table in §13.4. We first
identify the symmetry operations of point group D4h. We then obtain
χatom sites for these symmetry operations.
We first review the situation for the Γ point:
Γ point E 8C3 3C2
4 6C2 6C4 i 8iC3 3iC2
4 6iC2 6iC4
χNa
atom sites 1 1 1 1 1 1 1 1 1 1
χCl
atom sites 1 1 1 1 1 1 1 1 1 1
Thus, we have χatom sites for the Na and Cl ions, and for χvector
χNa
a.s. = Γ1g = Γ+
1
χCl
a.s. = Γ1g = Γ+
1
χvector = Γ−
15
so that
χlattice vibrations = 2Γ+
1 ⊗ Γ−
15 = 2Γ−
15.
Similarly for the X point, we first find χatom sites for each type of
atom.
X point E 2C2
4⊥ C2
4 2C4 2C2 i 2iC2
4⊥ iC2
4 2iC4 2iC2
χNa
atom sites 1 1 1 1 1 1 1 1 1 1
χCl
atom sites 1 1 1 1 1 1 1 1 1 1
Thus, we obtain χatom sites, χvector, and χlattice vector at the X point:
χNa
a.s. = X1
χCl
a.s. = X1
χvector = X4 + X5,
394 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
where X4 corresponds to x and X5 corresponds to (y, z). We thus
obtain
χlattice vibrations = 2X1 ⊗ (X4 + X5) = 2X4 + 2X5.
Compatibility relations give Γ15 → ∆1 + ∆5 → X4 + X5.
The action of the translation operator on a basis function (normal
mode) yields
ˆP{ε|τ}u(r) = eik·τ
u(r) (14.20)
where k = π
a
ˆx at the X point under consideration. For Rn = aˆx we
obtain eik·τ
= eiπ
= −1 so that there is a π phase difference between
unit cells along ˆx. However, for Rn = aˆy or aˆz, we have eik·τ
= ei(0)
= 1,
and there is no phase difference along ˆy and ˆz.
The phase factor of Eq. 14.20 refers to the relative phase between
atoms in adjacent unit cells. The relative motion between atoms within
a unit cell was considered in §14.2. Thus the NaCl structure (#225)
has a set of 3 acoustic branches and 3 optical branches each having X4
and X5 symmetries at the X point, where
X4 → x
X5 → y, z
The normal modes for the 3 acoustic branches are shown in Fig. 14.9
in terms of the symmetry classifications X4 and X5 (2-fold) for the
longitudinal and transverse branches, respectively. The corresponding
normal modes for the 3 optical branches are shown in Fig. 14.10.
For rows of atoms in unit cells along the y and z directions, there
will be essentially zero phase difference (δ = π/N, where N ≈ 107
)
between molecules vibrating in the acoustic mode as we move in the
y and z directions. This is also true for the optical branches shown in
Fig. 14.10.
14.4.2 Phonons in BaTi3 at the X point
The modes in this case involve more than one atom of the same species
within the unit cell so that a few new aspects enter the lattice mode
problem in this case. The character table for the group of the wave
14.4. LATTICE MODES AWAY FROM K = 0 395
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¥43 576)8@9BA7C78D5 ¡ ¥EGFH
9BI ¡ 57PRQSIBPTSEGFU
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9BI ¡ 57PRQSIBPTS
VXW`Yba2cBd
Figure 14.9: Acoustic branch for the vibrational modes of NaCl showing
longitudinal and transverse modes at the X point (kx = π/a) in the
Brillouin zone for the X4 and X5 modes.
eGf eGf eGf
eGfeGf gih
eGf gih eGf g¦h eGf gih
gih gih
gih epf gih
g¦h
qbrts uwvyxt€)ƒ‚v…„†7‡
ˆ‰‘
h4’“•”)–D—B˜•™•–4“ f h
—Bd f “•eTfghdXeTgˆp‰i
—Bd f “•eTfghdXeTgˆ ‰i
jlknmo7pXq
Figure 14.10: Optic branch for the vibrational modes of NaCl showing
longitudinal and transverse modes at the X point (kx = π/a) in the
Brillouin zone for the X4 and X5 modes.
396 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
vector at the X point for BaTiO3 is the same as for NaCl. At the X
point, we compute χatom sites ≡ χa.s. using the symmetry operators for
the group of the wave vector at the X point making use of the notation
in Fig. 14.5.
X point E 2C2
4⊥ C2
4 2C4 2C2 i 2iC2
4⊥ iC2
4 2iC4 2iC2
χBa atom sites 1 1 1 1 1 1 1 1 1 1
χTi atom sites 1 1 1 1 1 1 1 1 1 1
χO3 atom sites 3 3 3 1 1 3 3 3 1 1
χBa atom sites = X1
χTi atom sites = X1
χO3 atom sites = 2X1 + X2
χvector = X4 + X5
(14.21)
where X4 corresponds to x, and X5 to (y, z). The symmetries of the
normal modes are found by taking the direct product of χatom sites ⊗
χvector
χBa lattice modes = X1 ⊗ (X4 + X5) = X4 + X5
χTi lattice modes = X1 ⊗ (X4 + X5) = X4 + X5
The Ba and Ti atoms form normal modes similar to NaCl with the
Ba moving along x (X4 symmetry) or along y or z (X5 symmetry) with
the Ti and O3 at rest, and likewise for the Ti atoms moving along the
x direction. The phase relations for atomic vibrations in adjacent unit
cells in the x direction have a phase factor eπi
= −1, while rows of
similar atoms in the y and z direction have no phase shift. For the
oxygens,
χO3 lattice modes = (2X1 + X2) ⊗ (X4 + X5) = 2X4 + X3 + 3X5 (14.22)
The mode patterns at the X point for BaTiO3 are given in Fig. 14.11.
14.4. LATTICE MODES AWAY FROM K = 0 397
Figure 14.11: Mode patterns models for the X point modes in BaTiO3.
The basis functions for each normal mode are indicated.
398 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
The mode symmetry and the normal mode displacements are verified
by the following considerations. Perusal of the X-point character
table shows that the symmetry types are uniquely specified by the operations
C4 , C2 and i. The effect of these operations on the coordinates
(x, y, z) and on the site locations are:
C4



1
2
3


 =



1
3
2


 C4



x
y
z


 =



x
−z
y



C2



1
2
3


 =



1
3
2


 C2



x
y
z


 =



−x
z
y



i



1
2
3


 =



1
2
3


 i



x
y
z


 =



−x
−y
−z



By carrying out the symmetry operations on the basis functions, we
verify that the matrix representations for each of the symmetry operations
have the correct characters for the X4 irreducible representation:
C4 (x1 + x2 + x3) = (x1 + x3 + x2) so that χ(C4 )
= +1
C2(x1 + x2 + x3) = −(x1 + x3 + x2) χ(C2)
= −1
i(x1 + x2 + x3) = −(x1 + x2 + x3) χ(i)
= −1
Applying the same approach to the normal mode displacements with
X5 symmetry we have:
C4
y1 + y2 + y3
z1 + z2 + z3
=
−z1 − z3 − z2
y1 + y3 + y2
=
0 −1
1 0
y1 + y2 + y3
z1 + z2 + z3
i
y1 + y2 + y3
z1 + z2 + z3
=
−1 0
0 −1
y1 + y2 + y3
z1 + z2 + z3
so that χ(C4 ) = 0, and χ(i) = −2, which are the correct characters
for the X5 irreducible representation. Finally for the X3 modes:
14.5. PHONONS IN TE AND QUARTZ 399
Table 14.1: Basis functions for the various irreducible representations
entering the lattice modes in BaTiO3.
Basis Functions Irreducible representation
x3 − x2 X3
y1 − y3
−z1 + z2
X5
2x1 − x2 − x3 X4
−y1 + 2y2 − y3
−z1 − z2 + 2z3
X5
x1 + x2 + x3 X4
y1 + y2 + y3
z1 + z2 + z3
X5
C4 (−x2 + x3) = (−x3 + x2) = −(−x2 + x3) → χ(C4 ) = −1
C2(−x2 + x3) = x3 − x2 = (−x2 + x3) → χ(C2) = +1
i(−x2 + x3) = −(−x2 + x3) → χ(i) = −1
These same calculations can be applied to the basis functions in Fig. 14.11
and their irreducible representations and the results are listed in Table
14.1.
The phase factors for oxygens separated by a lattice vector aˆx are
eπi
= −1 while the oxygens separated by a lattice vector aˆy or aˆz have
no phase difference (i.e., phase factor ≡ 1).
14.5 Phonons in Te and Quartz
In this section we discuss phonon modes for tellurium (with 3 atoms/unit
cell) and show how the lattice modes for this non-symmorphic structure
can be used to obtain the lattice modes for α-quartz (with 9 atoms/unit
cell) which has the same space group as Te.
400 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
14.5.1 Phonons in Tellurium: A Non-Symmorphic
Structure
The structure for Te (space groups P3121 , #152; P3221 , #154) is a
spiral non-symmorphic space group as shown in Fig. 14.12. There are
3 Te atoms/unit cell and these Te atoms are at levels 0, c/3 and 2c/3.
The structure for right-handed Te shows a right-handed screw when
viewed along +ˆz. When the atoms are arranged with the opposite
screw orientation, we have left-handed Te. Three-fold rotations about
the c axis must be combined with a translation τ = c
3
(001) to leave the
crystal invariant. The 3 two-fold symmetry axes normal to the 3-fold
axis do not require translations. The appropriate point group at k = 0
is D3 and the character table is shown below. Note that mirror planes
are not symmetry operations.
D3 (32) E 2C3 3C2
x2
+ y2
, z2
A1 1 1 1
Rz, z A2 1 1 −1
(xz, yz)
(x2
− y2
, xy)
(x, y)
(Rx, Ry)
E 2 −1 0
Following the same procedure as was used for the non-symmorphic
diamond structure (see §14.3.3), we find χatom sites by considering the
number of sites within the unit cell that remain invariant (or transform
into the identical site in a neighboring unit cell):
{E|0} 2{C3|τ} 3{C2 |0}
χatom sites 3 0 1 = A1 + E
To find the lattice vibrations, we note that the vector transforms as
A2 + E. This allows us to separate out the lattice modes in the zdirection
from those in the x − y plane. For the z-direction
χatom sites ⊗ χvector, z = (A1 + E) ⊗ A2 = A2 + E (14.23)
where the A2 mode corresponds to pure translations in the z direction at
k = 0. The phonon dispersion curves for tellurium have been measured
14.5. PHONONS IN TE AND QUARTZ 401
Figure 14.12: Model
for the Te crystal
structure.
402 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
Figure 14.13: Phonon modes for Te.
by inelastic neutron scattering and the results along the high symmetry
axes are shown in Fig. 14.13.
We show the normal modes with A2 and E symmetry in Fig. 14.14.
For the in-plane motion, the symmetries are obtained by computing:
χatom sites ⊗ χvector (x,y) = (A1 + E) ⊗ E = E + (A1 + A2 + E) (14.24)
The translational mode in the x, y directions transforms as E. The
in-plane modes at k = 0 are shown in Fig. 14.15. The A2 and E modes
are IR active, and the A1 and E modes are Raman-active.
Since the Te structure has a screw axis, right and left circularly
polarized light are of importance for optical experiments. For linear
polarization we consider the E vector for the light in terms of x, y, z
components. For circular polarization we take the linear combinations
(x + iy) and (x − iy). From the character table, we note that (x +
iy)(x − iy) = x2
+ y2
transforms as A1 and the dipole moment u is
14.5. PHONONS IN TE AND QUARTZ 403
 
 
 
 
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'(   ' ¤   '0)
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'(  3¢ ' ¤  4¢ ¤ '0)
5 ¤
Figure 14.14: Normal
modes for Te
for z-axis vibrations.
The A2 mode (a) is
a pure translational
mode along the zaxis.
The E mode
has displacements
along z which have
phase differences of
ω = exp(2πi/3) with
respect to one another.
One partner
of the E mode
is shown explicitly in
(b). For the other
partner, the displacements
correspond to
the interchange of
ω ↔ ω2
, yielding
the complex conjugate
(c.c.) of the
mode that is shown.
404 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
  
 
 
  
 
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§§
 ©¨
 ©¨ ¦
 ©¨
 ¨ ¦
"!$#&% ')(102 !$#&%
3 54 376 4
378 43 0 4
9 @#"A!CB 0@D 0@D 9 #")!EB 0@D 0@D
F ¢  G¨ F ¦  G¨ ¦ FIH P ¢  Q¨ P ¦  Q¨ ¦ PRH
P ¢   P ¦   PSHF ¢   F ¦   FIH
Figure 14.15: In-plane normal modes for Te. The A1 normal mode
(a) is a breathing mode, while the A2 mode (b) is a rocking mode
corresponding to rotations of the 3 tellurium atoms for each half cycle of
the vibration. The two E modes (c, d) can be described as a breathing
and a rocking mode with phase relations ω = exp(2πi/3) between each
of the atoms as indicated (with the complex conjugate partner in each
case obtained by the interchange of ω ↔ ω2
).
14.5. PHONONS IN TE AND QUARTZ 405
related to the polarizability tensor
↔
α by:



(ux + iuy)/
√
2
(ux − iuy)/
√
2
uz


 =



α11 α12 α13
α21 α22 α23
α31 α32 α33






(Ex + iEy)/
√
2
(Ex − iEy)/
√
2
Ez



(14.25)
so that the polarizability tensor for A1 modes will have the form:
↔
αA1
=



a 0 0
0 a 0
0 0 0



for in-plane motion with the Raman tensor having components (Ei
+Es
−+
Ei
−Es
+)α(A1). The polarizability tensor for the z-axis motion is
↔
αA1
=



0 0 0
0 0 0
0 0 b



with the Raman tensor having components Ei
zEs
zα(A1). Finally for
general A1 motion, the polarizability tensor is written as:
↔
αA1 =



a 0 0
0 a 0
0 0 b


 . (14.26)
To find the energy for aligning the dipole moment in an electric
field, we need to take the dot product of the dipole moment with the
electric field
E∗
· u = [Ex − iEy] /
√
2, (Ex + iEy) /
√
2, Ez ·



(ux + iuy)/
√
2
(ux − iuy)/
√
2
uz



so that
E∗
· u = (E−, E+, Ez) ·



u+
u−
uz



= E−u+ + E+u− + Ezuz = Exux + Eyuy + Ezuz = real quantity.
406 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
For the electromagnetic (infrared) interaction, the pertinent symmetries
are E+u−(E) + E−u+(E) for in-plane motion and Ezuz(A2) for z-axis
motion.
In considering the Raman effect, we find the energy of the Raman
interaction in terms of E∗
·
↔
α ·E which when properly symmetrized
becomes 1
2
E∗
·
↔
α ·E + E·
↔
α
∗
·E∗
. Thus for the Raman mode with A1
symmetry, the induced dipole u+ has the same sense of polarization as
the incident electric field. However, the energy involves E∗
i and Es or
alternatively E∗
s and Ei to yield 1
2
(Ei
+Es
− + Ei
−Es
+) which transforms
as (x + iy)(x − iy) = x2
+ y2
as desired for a basis function with A1
symmetry.
For Raman modes with E symmetry we can have a dipole moment
uz induced by E+, leading to the combination of electric fields E∗
z E+.
To have a symmetric polarizability tensor, we must also include the
term (E∗
z E+)∗
= E−Ez since the energy must be unchanged upon interchange
of electric fields E ↔ E∗
. Thus the polarizability and Raman
tensors must be of the form
↔
αE,1 =



0 0 0
0 0 r∗
r 0 0


 and
Ei
+Es
zα−(E) + Ei
−Es
zα+(E)
or Ei
zEs
+α−(E) + Ei
zEs
−α+(E)
(14.27)
The partner of this polarizability tensor with E symmetry will produce
the displacement uz from an electric field displacement E− yielding
↔
αE,2 =



0 0 r
0 0 0
0 r∗
0


 . (14.28)
The other lattice mode for Te with E symmetry (denoted here by E )
produces a dipole moment u+ from an electric field E−. This however
involves E−(E+)∗
= E2
− for the incident and scattered electric fields so
that the polarizability tensor in this case is
↔
αE ,1 =



0 s 0
0 0 0
0 0 0


 ; basis function x2
− (14.29)
14.5. PHONONS IN TE AND QUARTZ 407
and the corresponding partner is
↔
αE ,2 =



0 0 0
s∗
0 0
0 0 0


 ; basis function x2
+. (14.30)
The Raman tensor for the E mode has the form Ei
+Es
+α+(E)+Ei
−Es
−α−(E).
We can relate these partners of the E modes to the basis functions of
the character table for D3 by considering the basis functions for the
partners:
partner #1 :
1
2
(x − iy)2
= x2
−
partner #2 :
1
2
(x + iy)2
= x2
+
(14.31)
By taking the sums and differences of these partners we obtain
x2
+ + x2
− =
1
2
(x + iy)2
+
1
2
(x − iy)2
= (x2
− y2
)
x2
+ − x2
− =
1
2
(x + iy)2
−
1
2
(x − iy)2
= 2xy (14.32)
which form a set of partners listed in the character table for D3.
14.5.2 Phonons in the Non-Symmorphic α-Quartz
Structure
We will now examine the lattice modes of α-quartz (space group D4
3,
#152, P3121 for the right hand crystal or D5
3, #153, P3212 for the
left-hand crystal) in both its natural state and in the presence of an
applied uniaxial compressive force. We will use this as a means for
showing how lattice modes for crystals with several atoms per unit cell
can be built up from simpler units, in this case the tellurium structure
discussed in §14.5.1.
The spiral structure of α-quartz about the z-axis is shown in Fig. 14.16a
where each solid ball represents a SiO2 unit. This diagram is identical
to that for tellurium (see Fig. 14.12). The projection of the atoms onto
408 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
(a) (b)
Figure 14.16: Structure of (a) right-handed α-quartz and (b) the projection
of the atoms on the basal plane of α-quartz. Atoms #1, 4, 7
denote Si and the other numbers denote oxygen atoms.
the basal plane is shown in Fig. 14.16b. The Si atoms (1, 4 and 7)
occupy positions at levels 0, c/3, 2c/3 respectively (as for tellurium).
The oxygen atoms (9, 5, 3, 8, 6 and 2) occupy positions at levels c/9,
2c/9, 4c/9, 5c/9, 7c/9 and 8c/9 respectively. (These sites are of course
not occupied in tellurium.) Note that the space group of α-quartz is
D4
3. Figure 14.16 shows the right-handed tellurium structure.
There are 3 molecular SiO2 units per unit cell giving rise to 9 atoms
per unit cell or 27 lattice branches of which 24 are optic modes. By
examining the atom locations in Fig. 14.16b, we can determine the point
group symmetry of α-quartz. The z axis is a three-fold axis of rotation
when combined with the translation τ = (c/3)(001). In addition there
is a two-fold axis from the center to each of the silicon atoms. The
symmetry elements are the same as for tellurium discussed in §14.5.1.
In order to determine the normal modes of vibration we first find the
characters for the transformation of the atomic sites. It is convenient
to make use of the results for tellurium, noting that the silicon atoms
in quartz occupy the same sites as in tellurium. We thus obtain for the
modes in α-quartz at k = 0.
14.5. PHONONS IN TE AND QUARTZ 409
{E|0} 2{C3|τ} 3{C2 |0}
χSi atom sites 3 0 1 = A1 + E
χoxygen atom sites 6 0 0 = A1 + A2 + 2E
The lattice modes for the silicon are identical with those found previously
for Te, so that part of the problem is already finished. For the 6
oxygens we have:
χlattice modes, z = (A1 + A2 + 2E) ⊗ A2 for z motion
χlattice modes x,y = (A1 + A2 + 2E) ⊗ E for x, y motion
Carrying out the direct products we obtain:
χlattice modes, z = A2 + A1 + 2E for z motion
χlattice modes, x,y = 2A1 + 2A2 + 4E for x, y motion
(14.33)
where we note that for the D3 point group E ⊗ E = A1 + A2 + E.
The corresponding z-axis normal modes A2, A1, E and E for the
6 oxygens are shown in Fig. 14.17. The normal mode A2 is clearly a
uniform translation of the 6 oxygens while the A1 mode is a rocking of
the two oxygens on either side of a silicon atom (one going up, while
the other goes down). The two-fold E mode is derived from A2 by
introducing phases 1, ω, ω2
for each of the pairs of oxygens around a
silicon atom; the complex conjugate E mode is obtained from the one
that is illustrated by the substitution ω ↔ ω2
. Finally the E mode is
obtained from the A1 mode in a similar way as the E mode is obtained
from the A2 mode. In identifying the symmetry type for these normal
modes, we note the effect of symmetry operation C2.
We now combine the z motion for the silicons (symmetries A2 + E)
with the z motion for the oxygens (symmetries A1 +A2 +2E) to obtain
A1 +2A2 +3E for SiO2. The resulting normal mode patterns are shown
in Fig. 14.18. The z-axis translational mode for the 6 oxygens combine
either in-phase or out of phase to form the two normal modes with A2
symmetry. For the mode with A1 symmetry, the silicon atoms remain
stationary. Introducing the phases 1, ω, ω2
for each SiO2 group gives
the three E normal modes along the z-direction in α-quartz.
410 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
 
   
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Figure 14.17: Normal modes along the z-direction for the six oxygens
in the α-quartz crystal. The A2 mode is a uniform translation while
the A1 mode is a rocking of the oxygens around the Si. The E modes
are related to the A2 and A1 modes by combining the 1, ω, ω2
phases
with the translational and rocking motions.
14.5. PHONONS IN TE AND QUARTZ 411
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Figure 14.18: Normal modes along the z-direction for the three SiO2
groups in α-quartz. Here the motions of the Si atoms are combined
with those of the oxygens.
412 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
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Figure 14.19: Normal modes in the x − y plane for the six oxygens in
the α quartz crystal. In addition, the A1 tangential breathing mode,
the A2 radial breathing breathing mode, and the A2 rocking mode have
corresponding E modes, with phases 1, ω, ω2
for the three SiO2 units,
each having two partners related by ω ↔ ω2
. In the crystal, all modes
with the same symmetry are coupled so that the actual normal mode
is an admixture of the modes pictured here.
For the xy motion, the six oxygens form lattice modes with symmetries
2A1 + 2A2 + 4E and the normal mode patterns are shown in
Fig. 14.19. When we now combine the silicon (A1 + A2 + 2E) and oxygens
(2A1 + 2A2 + 4E) for the in-plane modes, we obtain symmetries
3A1 + 3A2 + 6E and the normal modes are shown in Fig. 14.19. For
the A1 breathing mode (a), all oxygens translate toward the center in
phase, whereas for the A2 mode (b) one oxygen of each SiO2 group
moves inward while the other moves outward. Another A1 mode arises
from each oxygen in a SiO2 pair moving circumferentially toward the
silicon atom, while an A2 mode is formed by a circumferential trans-
14.5. PHONONS IN TE AND QUARTZ 413
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Figure 14.20: The in-plane normal modes for α-quartz obtained by
superposition of the normal modes for the oxygens and the silicons.
Corresponding to each of the one-dimensional modes shown here are
two-dimensional E modes with phases 1, ω, ω2
for the three SiO2 units,
with the two partners related by ω ↔ ω2
.
lational rocking motion of the oxygens. Each of these 4 modes has
corresponding E modes with each SiO2 group assigned phases 1, ω, ω2
and its complex conjugate. For the A1 breathing mode in Fig. 14.20,
we can have the silicons moving either in-phase or out of phase with
respect to the oxygens, while for the A2 mode in Fig. 14.20, the silicons
remain at rest. For the circumferential translational A2 motion, the silicons
can go either in phase or they can move out of phase with respect
to the oxygens. Finally the rocking motion of the oxygens with respect
to static silicons forms an A1 lattice mode. Associated with each A1
and A2 in-plane mode, are E modes with phases 1, ω, ω2
as before. By
superposing the normal modes of the oxygens and the silicons for the
in-plane and c-axis modes, we can obtain all the normal modes (27 of
them) for α-quartz.
414 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
The infrared-active modes have A2 and E symmetries while the
Raman-active modes have A1 and E symmetries. The polarizability
tensor for the A1 and E modes is of the same form as given for the case
of tellurium.
Since the E modes are infrared-active, polarization fields lift the
two-fold degeneracy of every E mode giving rise to the so-called LOTO
splitting of the doubly degenerate Raman modes.
14.5.3 Effect of Uniaxial Stress on Phonons
In general, an external perturbation, when applied to a crystal, reduces
the symmetry of the crystal. The fundamental principle used to deduce
this lower symmetry is called the Curie principle which states that
only those symmetry operations are allowed which are common to both
the unperturbed system and to the perturbation itself. This condition
restricts the new symmetry group to a subgroup common to the original
group.
When a homogeneous uniaxial compression is applied to a crystal,
the resulting strain is described by a symmetric tensor of the second
rank. The strain tensor can be represented by a triaxial ellipsoid which
has at least D2h point group symmetry; if two of its major modes are
equal, the ellipsoid acquires rotational symmetry about the third, and
the point group symmetry is D∞h, whereas, if all three axes are equal
it becomes a sphere with three dimensional continuous rotation and
reflection symmetry. In order to determine the symmetry operations
of the strained crystal it is necessary to know the orientation of the
strain ellipsoid relative to the crystallographic axes. An alternative
procedure is to treat the stress itself as the imposed condition and find
the symmetry elements common to the unstrained crystal and to the
symmetry of the stress tensor.
Using the symmetry properties of the stress tensor is particularly
simple when the external perturbation is a uniaxial compression. In
this case the stress ellipsoid has D∞h point group symmetry and can
be conveniently represented by a right circular cylinder with its center
coinciding with the center of the crystal and its axis of revolution along
the direction of the force. The symmetry operations common to the
unstrained crystal and to the cylinder representing the stress can then
14.5. PHONONS IN TE AND QUARTZ 415
be easily determined by inspection.
As an illustrative case, consider the point group D3, the point group
of α-quartz. The symmetry operations of D3 are a three-fold axis of
rotation along the z axis and three two-fold axes perpendicular to the
z axis, one of which is taken to be the x axis. If the force, F, is applied
along the z direction, all of the operations of the group are common
to the symmetry of the stress and hence the symmetry remains D3. If,
however, the force is applied along the x direction, the only remaining
symmetry operation is C2. Similarly, if the force is applied along the y
axis, the only remaining symmetry operation is again the two-fold axis
of rotation along the x-axis and the symmetry is reduced to the point
group C2. If the force is in a direction other than along z or parallel
or perpendicular to a two-fold axis, the crystal symmetry is reduced to
C1.
Once the reduced symmetry of the crystal in the presence of the
external perturbation is determined, the correlation between the irreducible
representations of the two groups can be obtained. From such
a correlation, the removal of the degeneracy of a particular energy level
can be immediately deduced.
C2 (2) E C2
x2
, y2
, z2
, xy Rz, z A 1 1
xz, yz
x, y
Rx, Ry
B 1 −1
Representations of D3 A1 1 1 A
A2 1 −1 B
E 2 0 A + B
This group theoretical analysis thus predicts that the Raman lines of
E symmetry should split and the Raman inactive A2 mode in D3 symmetry
should become Raman-active in C2 symmetry. We note that the
basis functions that are used for C2 are x, y, z while for D3, the combinations
(x+iy, x−iy, z) are used. The form of the polarizability tensors
for the Raman-active modes in D3 and C2 point group symmetries are
given in Fig. 14.21.
416 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
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quartz.
14.6. LATTICE MODES IN HIGH TC RELATED MATERIALS 417
14.6 Lattice Modes in High Tc Related Ma-
terials
As an example of complex crystal structures of current interest, let us
consider the lattice modes of high Tc related materials.
14.6.1 The K2NiF4 Structure
Let us now consider lattice modes in K2NiF4 structure. This is the
structure for one of the high Tc related materials La2CuO4. The space
group is I4/mmm (D17
4h) #139. The structure is given by Wyckoff
(Vol. 3, p. 68) as:
Cu: (2a) (000), (1/2,1/2,1/2)
La: (4e) ±(00u; 1/2, 1/2, u + 1/2)
O(1): (4c) (0, 1/2, 0) (1/2, 0, 0), (1/2, 0, 1/2), (0, 1/2, 1/2)
O(2): (4e) ±(00u; 1/2, 1/2, u + 1/2)
This example is slightly more complex than examples given earlier in
the chapter, but still corresponds to a symmorphic space group.
Using the diagram in Fig. 14.22 and considering which atoms remain
unchanged under the symmetry operations of D4h or are transformed
into sites separated by a lattice vector, we thus obtain for χatom sites:
D4h E C2 = C2
4 2C4 2C2 2C2 i S2 2S4 2σv 2σv
χNi2 atom sites 2 2 2 2 2 2 2 2 2 2 ⇒ 2A1g
χK4 atom sites 4 4 4 0 0 0 0 0 4 4 ⇒ 2A1g + 2A2u
χF4 (4e) sites 4 4 4 0 0 0 0 0 4 4 ⇒ 2A1g + 2A2u
χF4 (4c) sites 4 4 2 2 0 0 0 2 2 4 ⇒ 2A1g + A2u + B1u
The character table for D4h shown below
418 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
Figure 14.22: The K2NiF4 crystal
structure, or equivalently the
La2CuO4 structure, where La↔K,
Cu↔Ni, and O↔F.
14.6. LATTICE MODES IN HIGH TC RELATED MATERIALS 419
D4h E C2 = C2
4 2C4 2C2 2C2 i S2 2S4 2σv 2σv
A1g 1 1 1 1 1 1 1 1 1 1 (x2
+ y2
), z2
A2g 1 1 1 –1 –1 1 1 1 –1 –1 Rz
B1g 1 1 –1 1 –1 1 1 –1 1 –1 x2
− y2
B2g 1 1 –1 –1 1 1 1 –1 –1 1 xy
Eg 2 –2 0 0 0 2 –2 0 0 0 (xz, yz)
A1u 1 1 1 1 1 –1 –1 –1 –1 –1 zRz
A2u 1 1 1 –1 –1 –1 –1 –1 1 1 z
B1u 1 1 –1 1 –1 –1 –1 1 –1 1 xyz
B2u 1 1 –1 –1 1 –1 –1 1 1 –1 xyzRz
Eu 2 –2 0 0 0 –2 2 0 0 0 (x, y)
gives the decomposition of χatom sites as well as the irreducible representations
for the vector:
z → A2u and (x, y) → Eu
Thus, taking the direct product χatom sites ⊗ χvector gives
Ni2 2A2u + 2Eu
K4 2A1g + 2Eg + 2A2u + 2Eu
F4 (4e) 2A1g + 2Eg + 2A2u + 2Eu
F4 (4c) A1g + B1g + 2Eg + 2A2u + 2Eu
(K2NiF4)2 5A1g + B1g + 6Eg + 8A2u + 8Eu
Of these the A2u and Eu modes are IR active (14 frequencies), two are
acoustic modes, while the A1g, B1g and Eg constitute the Raman-active
modes (12 frequencies). The B1g, Eg and Eu modes are in-plane modes
and the A2u mode is a c-axis mode. The A1g mode may be either an
in-plane or a c-axis mode. Experimentally these mode symmetries can
all be determined by a suitable sequence of settings of the polarizers.
14.6.2 Phonons in the YBa2Cu3O6 Structure
Let us now consider lattice modes in the YBa2Cu3O6 structure. This
is the structure for one of the high Tc related materials where there is
420 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
Figure 14.23: Model for the crystal structure of (a) YBa2Cu3O6 (space
group is P4/mmm–D1
4h); (b) YBa2Cu3O7 (space group is Pmmm–
D1
2h); and (c) YBa2Cu3O6.5.
14.6. LATTICE MODES IN HIGH TC RELATED MATERIALS 421
no oxygen in the chains, and the material is not conducting. The space
group is the symmorphic space group P4/mmm (D1
4h) #123.
Using the diagram in Fig. 14.23 and considering which atoms remain
unchanged under the symmetry operations of D4h or are transformed
into sites separated by a lattice vector, we thus obtain for χatom sites
D4h E C2 = C2
4 2C4 2C2 2C2 i S2 2S4 2σv 2σv
χY atom sites 1 1 1 1 1 1 1 1 1 1 ⇒ A1g
χBa2 atom sites 2 2 2 0 0 0 0 0 0 2 ⇒ A1g + A2u
χCu3 atom sites 3 3 3 1 1 1 1 1 3 3 ⇒ 2A1g + A2u
χO6 atom sites 6 6 2 1 1 1 1 1 6 2 ⇒ 2A1g + 2A2u + B1g + B1u
Looking at the character table for D4h we see that
z → A2u and (x, y) → Eu
Thus, taking the direct product χatom sites ⊗ χvector gives
Y A2u + Eu
Ba2 A1g + Eg + A2u + Eu
Cu3 A1g + Eg + 2A2u + 2Eu
O6 2A1g + B1g + 3Eg + 2A2u + B2u + 3Eu
YBa2Cu3O6 4A1g + B1g + 5Eg + 6A2u + B2u + 7Eu
so that there are 2 acoustic branch modes, 11 IR active modes and 10
Raman-active modes. The in-plane and c-axis polarization of the various
modes can be found from the basis functions listed in the character
table.
14.6.3 In The YBa2Cu3O7 Structure
Let us now consider lattice modes in the YBa2Cu3O7 structure. This
is the structure for one of the high Tc related materials where the chain
layers contain 1 copper and 1 oxygen atom. The space group for the
YBa2Cu3O7 structure is Pmmm (D1
2h) #47. This example has less
symmetry and is slightly more complex, but still corresponds to a symmorphic
space group.
Using the diagram in Fig. 14.24 and considering which atoms remain
unchanged under the symmetry operations of D2h or are transformed
422 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
Figure 14.24: Model for the
orthorhombic YBa2Cu3O7 crystal
structure (Pmmm (D1
2h) #47).
14.6. LATTICE MODES IN HIGH TC RELATED MATERIALS 423
into sites separated by a lattice vector, we thus obtain for χatom sites
D2h E C2 C2 C2 i S2 σh σv σv
χY atom sites 1 1 1 1 1 1 1 1 ⇒ A1g
χBa2 atom sites 2 2 0 0 0 0 2 2 ⇒ A1g + B1u
χCu3 atom sites 3 3 1 1 1 1 3 3 ⇒ 2A1g + B1u
χO7 atom sites 7 7 1 1 1 0 7 7 ⇒ 4A1g + 3B1u
The character table for D2h is given by:
D2h E Cz
2 Cy
2 Cx
2 i Sz
2 Sy
2 Sx
2
A1g 1 1 1 1 1 1 1 1 x2
, y2
, z2
B1g 1 1 –1 –1 1 1 –1 –1 xy
B2g 1 –1 1 –1 1 –1 1 –1 xz
B3g 1 –1 –1 1 1 –1 –1 1 yz
A1u 1 1 1 1 –1 –1 –1 –1 xyz
B1u 1 1 –1 –1 –1 –1 1 1 z
B2u 1 –1 1 –1 –1 1 –1 1 y
B3u 1 –1 –1 1 –1 1 1 –1 x
Thus in D2h symmetry we see that
x → B3u, y → B2u and z → B1u
Thus, taking the direct product χatom sites ⊗ χvector gives
Y B1u + B2u + B3u
Ba2 A1g + B2g + B3g + B1u + B2u + B3u
Cu3 A1g + B2g + B3g + 2B1u + 2B2u + 2B3u
O7 3A1g + 3B2g + 3B3g + 4B1u + 4B2u + 4B3u
YBa2Cu3O7 5A1g + 5B2g + 5B3g + 8B1u + 8B2u + 8B3u
The number of IR modes in this case is 21 (3 acoustic modes). The
IR selection rules are that 7 modes are seen with each polarization.
The number of Raman-active modes is 15. Their polarizations can be
obtained from the basis functions given in the character table and show
the 5 modes are seen with , polarization, 5 modes with x, z and 5
modes with y, z polarization.
424 CHAPTER 14. APPLICATIONS TO LATTICE VIBRATIONS
14.7 Selected Problems
1. Using the results of Problem # 1:
(a) Find the number of lattice modes at k = 0. What are their
symmetries and what are their mode degeneracies?
(b) What are the normal mode displacements for each of these
lattice modes?
(c) What modes are IR active, Raman active? What are the
polarizations of the Raman active modes?
2. Consider the crystal structure in the diagram for Nb3Sn, a prototype
superconductor with the A–15 (or β–W) structure used for
high field superconducting magnet applications.
(a) How many lattice modes are there at k = 0, what are their
symmetries and what are their degeneracies?
(b) What are the normal mode displacements for each of these
lattice modes?
(c) Which modes are IR active, Raman active? What are the
polarizations of the Raman-active modes?
3. Tin oxide (SnO2 with space group #136) is an important electronic
material.
(a) Find the lattice modes at k = 0, their symmetries, degeneracies
and the normal mode patterns.
(b) Indicate the IR-activity and Raman activity of these modes.
4. Bromine forms a molecular crystal.
(a) What is the appropriate space group?
(b) Find the lattice modes at k = 0, their symmetries, degeneracies
and the normal mode patterns.
(c) Indicate the IR-activity and Raman activity of these modes.
Chapter 15
Use of Standard Reference
Texts
15.1 Introduction
In Chapter 14 we discussed the lattice modes for a number of crystals
assuming that the crystal structure and the space group are known. In
many research situations, the researcher must first identify the space
group and the pertinent group of the wave vector, before solving for the
lattice modes or for the electronic structure. In this chapter we discuss
the procedure to be used in such cases. This procedure involves use of
3 standard reference sources:
1. Wyckoff’s books which give the atom locations for hundreds of
crystal structures:
R.W.G. Wyckoff, Crystal Structures,
QD951.W977 1963 (7 volumes).
2. The International Tables for X-ray Crystallography through which
one can identify the space group from the atom locations:
International Tables for X-ray Crystallography
QD 945.I61 1965 (4 volumes)
3. The Character Tables for the group of the wave vector for each
unique k vector for each of the 230 space groups:
425
426 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
Miller and Love, “Irreducible Representations of Space Groups”,
QA171.5 .M651
In this chapter we comment on the use of each of these standard reference
texts.
15.2 Determination of the Crystal Struc-
ture
The standard determinations of crystal structures are carried out using
diffraction techniques, either x-ray or neutron diffraction. The elastically
scattered beams give rise to a series of diffraction peaks which
can be indexed according to the points in the reciprocal lattice. The
scattering intensities can be calculated using the form factors listed in
the “International Tables for X-Ray Crystallography”. The results of
many such structural determinations for specific materials are listed in
the series of books by Wyckoff (R.W.G. Wyckoff, Crystal Structures,
QD 951.W977 1963 (7 volumes)).
We illustrate the use of Wyckoff’s books to find the crystal structure
of a particular material. We choose graphite for the illustrative
material. For the crystal structure of graphite see p. 26–28 in Volume 1
of Wyckoff. The information to be extracted from Wyckoff concerns
the number of allotropic structures, the site symmetries of the atoms
in each of the structures and the space group designations. Wyckoff
generally gives you most of the information you need. For the case of
graphite, there are 3 crystal structures: ordinary hexagonal graphite,
puckered graphite and rhombohedral graphite, each of which is listed
in Wyckoff.
First we have ordinary graphite with 4 atoms/unit cell as shown in
Fig. 15.1, containing two a-atoms at (0,0,0) and (0,0,1/2) denoted by
dark circles and two b-atoms at (1/3,2/3,0) and (2/3,1/3,1/2) denoted
by open circles. The lattice constants a0 = 2.456˚A and c0 = 6.696˚A are
indicated in Fig. 15.1 and the layer stacking is ABAB . . .. To do the
various symmetry operations it is sometimes convenient to take the
origin at (0,0,1/4).
The second form that is listed is puckered graphite where the two
15.2. DETERMINATION OF THE CRYSTAL STRUCTURE 427
Figure 15.1: Crystal structure of
hexagonal graphite.
a atoms are translated in the z direction by u and the two b atoms are
translated by v in the z direction. This makes the a and b carbon atoms
slightly non-planar relative to ordinary graphite, which is a sheet-like
material.
In the rhombohedral form we have 3 different layer planes per unit
cell, so that the repeat distance is (3/2)(6.696)˚A=10.044˚A in the c
direction, with the 3rd
layer having an atom over the (2/3, 1/3, 1/2)
atom but none over the (0, 0, 1/2) atom. This gives 6 atoms in a
hexagonal unit cell shown in Fig. 15.2 with ABCABC . . . layer stacking.
It is not always easy to see the stacking arrangements from the
diagrams in Wyckoff. Sometimes you must make additional pictures,
like the projection of the planar stacking for the 3 layers shown in
Fig. 15.3. This stacking is equivalent to a rhombohedral unit cell with
lattice constants a0 = 3.635˚A and α = 39◦
30 and two atoms at (u, u, u)
and (¯u, ¯u, ¯u) with u ≈ 1/6 shown by the parallelepiped in the figure.
The space group is specified in terms of the rhombohedral unit cell
containing 2 carbon atoms.
Once we know the atom locations we can then use the International
Tables for X-Ray Crystallography to identify the space group. In many
cases the summary in Wyckoff actually gives the space group assign-
428 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
Figure 15.2: Rhombohedral
graphite showing ABC stacking of
the individual sheets.
Figure 15.3: Top view of a three layered projection for rhombohedral
graphite.
15.3. DETERMINATION OF THE SPACE GROUP 429
ment.
15.3 Determination of the Space Group
The International Tables for X-Ray Crystallography helps with the determination
of the space group and the symmetry operations of the
space group (International Tables for X-ray Crystallography; QD 945.I61
1965 (4 volumes)). These volumes deal with space groups in general
but do not refer to specific materials, which is the central theme of
Wyckoff’s books. In some cases Wyckoff gives the space group designation,
and if so, this will greatly simplify our work. If the space
group designation is known, it should be an easy matter to match up
the atom site locations in Wyckoff with the atom site locations listed
under the correct space group in the International Tables. For the case
of graphite we have to find three different space groups for each of the
three allotropic forms.
For ordinary graphite, the space group is D4
6h (P63/m 2/m 2/c,
#194). The designation D6h refers to the point group at k = 0 and the
superscript 4 refers to a space group index based on this point group.
The full symmetry listing is P63/m 2/m 2/c and an abbreviated form is
P63/mmc. Since Wyckoff does not list the space group designation we
must consult the International Tables for X-ray Crystallography where
the tables are arranged according to space group designation (e.g., D4
6h).
If Wyckoff doesn’t list the space group designation, then it is necessary
to use the index of the International Tables (p. 552) under the title
“hexagonal systems”. Under “hexagonal systems” there are 27 space
groups. What needs to be done at this point is to see which of these
27 space groups is the right one. The method of identification involves
matching up the atom site locations in the International Tables with the
one in Wyckoff. Matching up the point group symmetry operations is
often a big help in eliminating almost all of the irrelevant space groups
within a given system, such as the hexagonal system. The stereograph
shown in Fig. 15.4 can be used to deduce the point group symmetry
operations for the space group #194.
Once this matching up is done correctly, the space group assignment
with the following atom site assignments is found:
430 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
Figure 15.4: Site locations for space group #194.
15.3. DETERMINATION OF THE SPACE GROUP 431
# of desig- site International Tables Wyckoff
atoms nation symmetry atom locations atom locations
2 b ¯6m2 (0,0,1/4) (0,0,3/4) (0,0,0) (0,0,1/2)
2 c ¯6m2 (1/3,2/3,1/4) (2/3,1/3,3/4) (1/3,2/3,0) (2/3,1/3,1/2)
space group#194; D4
6h hexagonal P63/mmc; P63/m 2/m 2/c; 6/mmm
The International Tables for X-ray Crystallography tell us that sites b
and c each have ¯6m2 point group symmetry. The Wyckoff atom site
locations differ from the International Tables atom site locations by
a translation of the unit cell by (c/4)(001). Once we have the correct
space group, we record the space group number which is #194, because
we will need this number to use Miller and Love which is discussed in
§15.4, since Miller and Love only designates each space group by its
number. By the way, space group #194 also describes the hexagonal
close packed structure.
The expanded crystallographic notation P63/m 2/m 2/c for space
group #194 means that we have a primitive (P) lattice, the highest
symmetry axis is a 6-fold screw axis along the c-direction corresponding
to a translation (3/6) of the length of the c-axis unit vector. Mirror
planes pass through the c-axis. Perpendicular to the main symmetry
directions are three two-fold axes with mirror planes through these axes
(2/m), and involving no translations. Also perpendicular to the main
symmetry axis is a set of 2-fold axes with a glide plane or translation in
the c-direction (2/c). For the ranking system used to list the ordering
of the mirror planes, the simple mirror plane is listed first, followed by
glide planes perpendicular to the a, b, and c directions, followed by a net
glide (n-glide) τ = 1/2(a + b); and finally the diamond glide (d-glide)
τ = 1/4(a + b) : m > a > b > c > n > d.
The match-up between the site locations and the space group is
made as follows. We first identify the symmetry operations of the
graphite lattice ({ε|0}, {C2|τ}, 2{C3|0}, 2{C6|τ}, 3{C2 |0}, 3{C2 |τ}, and
all operations compounded with inversion). These are the point group
operations of D6h. There are four space groups in the hexagonal system
with D6h symmetry. We can immediately rule out the space group
#191 (P63/m 2/m 2/m) D1
6h which is symmorphic, and we can rule
out the space group #192 (P63/m 2/c 2/c) D2
6h which has two kinds of
non-symmorphic translations. Group #193 D3
6h is ruled out because it
does not contain the proper site locations. Group #194 (see Fig. 15.4)
however contains all the proper symmetry operations and site locations.
432 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
In the puckered graphite structure the a and b atoms are no longer
co-planar so that the white lattice is slightly displaced along the z
direction from the dark lattice in the diagram shown in Fig. 15.1. In
comparing the site symmetries between Wyckoff and the International
Crystallography Tables we see that a match is found with group #186
(C4
6v) another hexagonal group with P63mc symmetry.
Combining the information in Wyckoff and the International Tables
we prepare a table for the site locations of puckered graphite for space
group #186; C4
6v hexagonal; P63mc; 6mm:
# of designation site International Tables Comments
atoms in international symmetry and Wyckoff
tables
2 a 3m (0,0, ¯w) (0,0,1/2+ ¯w) In Wyckoff (International Tables)
listed as: ¯w is u(z)
(0,0,u) (0,0,1/2+u)
2 b 3m (1/3,2/3,w) (2/3,1/3,1/2+w) In Wyckoff (International Tables)
listed as: w is v(z)
(1/3,2/3,v) (2/3,1/3,1/2+v)
The site symmetries for the a and b sites are given as 3m in the International
Tables. We note that puckered graphite has lower symmetry
than ordinary graphite because the value of z for the site locations
is arbitrary. Wyckoff tells us that u = v; in fact Wyckoff selects a
coordinate system where u = 0 and v is small (< 0.05).
Since no space group information is provided for rhombohedral
graphite in Wyckoff, we must again use the index in the International
Tables (p. 552) under the heading of “Trigonal Groups”. The space
group that matches the site locations in Wyckoff is D5
3d #166. This
space group is described in terms of rhombohedral axes (p. 272 of the
International Tables) and also hexagonal axes (p. 273 of the International
Tables). In terms of the rhombohedral axes we find the correct
site symmetries as:
# Designation Site Symmetry Site Locations
2 c 3m (x, x, x); (¯x, ¯x, ¯x); Wyckoff gives x ≈ 1/6
space group D5
3d #166
The stereographs for the hexagonal and rhombohedral systems for space
group #166 are shown in Fig. 15.5. Also listed in the International Tables
are the site locations in the hexagonal system.
15.3. DETERMINATION OF THE SPACE GROUP 433
Figure 15.5: Stereographs for the hexagonal and rhombohedral systems
for space group #166.
434 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
# designation site International Tables
symmetry site locations
6 c 3m (0, 0, z), (0, 0, ¯z), (1/3, 2/3, 2/3+z)
(1/3, 2/3, 2/3+¯z)
(2/3, 1/3, 1/3+z), (2/3, 1/3, 1/3+¯z)
Agreement between the Wyckoff listing and the listing in the International
Tables is obtained by taking z = 1/6, and ¯z = −1/6, and by
translating the coordinate system by (c/6)(0, 0, 1).
15.4 Finding Character Tables for all Groups
of the Wave Vector
To find all the pertinent group theory information for a space group
we use Miller and Love. This book contains character tables for all
groups of the wave vectors for every space group. This reference is a
big computer printout. Neither Miller and Love nor the International
Tables refer to specific materials – these two books only refer to the
space group which describes specific materials.
To find the character tables for the various points in the Brillouin
zone we use Miller and Love. Let us take the case of ordinary graphite
(see §15.2 for the atomic site designations) which has space group #194
(p. 356-8 Miller and Love). The first character table listed is for the Γ
point (k = 0) and is shown in Fig. 15.6.
The arrangement of the classes and irreducible representations is
opposite to that used in Tinkham. The classes are listed on the lefthand
column and follow the notation on p. 124 of Miller and Love which
is reproduced in Fig. 15.7. The symmetry elements numbered n ≥49
are for double group operations, to be discussed later in the course.
The double group representations are 7±
, 8±
, 9±
and we will not be
concerned with these further for discussing lattice modes of graphite.
The representations with a +(−) sign are even (odd) under inversion
so that we need only be concerned with the first 6±
representations.
Applying the notation for the classes (see Fig. 15.7) in Miller and
Love to Tinkham’s tables, we make the following correspondence between
the classes:
15.4. CHARACTER TABLES FOR GROUPS OF THE WAVE VECTOR435
Figure 15.6: Miller and Love character table for the group of the wave
vector at k = 0 for space group #194.
Figure 15.7: Miller and Love notation for the group elements for the
rhombohedral and hexagonal systems.
436 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
Tinkham Miller & Love Tinkham Miller & Love
E 1 i ⊗ E 13
→ C2 4 i ⊗ C2 16
2C3 3,5 i ⊗ 2C3 15,17
→ 2C6 2,6 i ⊗ 2C6 14,18
3C2 7,9,11 i ⊗ 3C2 19,21,23
→ 3C2 8,10,12 i ⊗ 3C2 20,22,24
in which indicators are used to distinguish operations with no translations
(no arrows) and with translations τ = c
2
(0, 0, 1) (with arrows).
The operations without arrows are all in point group D3d = D3 ⊗ i.
The operations with translations are in D6h = D6 ⊗ i but not in D3d.
Thus graphite crystallizes in a non-symmorphic space group.
The correspondence between irreducible representations in Tinkham
and Miller and Love is given by the following table:
A1g → 1+
A1u → 1−
A2g → 2+
A2u → 2−
B1g → 3+
B1u → 3−
B2g → 4+
B2u → 4−
E1g → 6+
E1u → 6−
E2g → 5+
E2u → 5−
With these identifications we can construct a character table from
Miller and Love that looks like the character tables in Tinkham for
D6 (622), with D6h = D6 ⊗ i. (T=Tinkham, M+L=Miller and Love).
The resulting table is
M+L → 1 4,1 3;5 2,1;6,1 7;9;11 8,1;10,1;12,1
D6 (622) ↓ {E|0} {C2|τ} 2{C3|0} 2{C6|τ} 3{C2 |0} 3{C2 |τ}
x2 + y2, z2 1 A1 1 1 1 1 1 1
Rz, z 2 A2 1 1 1 1 −1 −1
3 B1 1 −1 1 −1 1 −1
4 B2 1 −1 1 −1 −1 1
(xz, yz)(x, y)(Rx, Ry) 6 E1 2 −2 −1 1 0 0
(x2 − y2, xy) 5 E2 2 2 −1 −1 0 0
15.5. PHONONS IN GRAPHITE 437
Figure 15.8: Miller and Love character tables for puckered graphite,
space group #186.
For puckered graphite we must use space group #186 (p. 338 in
Miller and Love). We rewrite the character table for k = 0 from Miller
and Love (given in Fig. 15.8) in the Tinkham notation, and go through
the same steps as were taken for ordinary graphite corresponding to
space group #194.
In the case of rhombohedral graphite, we need to consider the
high symmetry points of the rhombohedral Brillouin zone, shown in
Fig. 15.9. The Miller and Love character table (p. 304) for k = 0 rhombohedral
graphite (space group #166) is given in Fig. 15.10. The double
group irreducible representations in this case are 4±
, 5±
, 6±
and need
not be considered here further.
15.5 Phonons in Graphite
Now that we have reviewed the use of the Standard Reference Materials
we will find the phonon modes in graphite: first for ordinary hexagonal
graphite, and then for puckered graphite and for rhombohedral
graphite.
438 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
Figure 15.9: High symmetry points in the rhombohedral Brillouin zone
taken from Miller and Love.
Figure 15.10: Miller and Love character table (p. 304) for space group
#166 at k = 0.
15.5. PHONONS IN GRAPHITE 439
 ¢¡¡£
  ¡£
¤
¤ ¤
¥
Figure 15.11: Schematic showing the symmetry operations in graphite.
O labels the origin, which for convenience is taken halfway between the
upper and lower planes of the unit cell.
15.5.1 Phonons in Ordinary Hexagonal Graphite
To find χatom sites we use a diagram showing the symmetry operations
in graphite (see Fig. 15.11). The hexagonal Brillouin zone is shown in
Fig. 15.12.
From Fig. 15.11, we compile χatom sites = χa.s.
{E|0} {C2|τ} 2{C3|0} 2{C6|τ} 3{C2 |0} 3{C2 |τ} {i|0} {iC2|τ} 2{iC3|0} 2{iC6|τ} 3{iC2 |0}3{iC2 |τ}
χa.s. 4 0 4 0 0 4 0 4 0 4 4 0
Next we find the irreducible representations in χa.s. and χvector:
χa.s. = 2A1g + 2B2u (15.1)
χvector = A2u + E1u (15.2)
The direct product then yields
χa.s. ⊗ χvector = (2A1g + 2B2u) ⊗ (A2u + E1u)
and can be decomposed into
χin−plane lattice modes = [(2A1g + 2B2u) ⊗ E1u]
= 2E1u + 2E2g
χz−axis lattice modes = [(2A1g + 2B2u) ⊗ A2u]
= 2A2u + 2B1g
440 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
Figure 15.12: High symmetry points in the hexagonal Brillouin zone
taken from Miller and Love.
15.5. PHONONS IN GRAPHITE 441
The lattice mode diagrams are given in Fig. 15.13 for the vibrational
modes. In addition there are in-plane translations (E1u) and z-axis
translations A2u (not shown).
15.5.2 Phonons in Puckered Graphite
To find the lattice modes at k = 0 for puckered graphite we use the
diagram in Fig. 15.14 and the character table below:
Point Group C6v M+L → 1 4,1 3;5 2,1;6,1 19;21;23 20,1;22,1;24,1
↓ Tinkham {E|0} {C2|τ} 2{C3|0} 2{C6|τ} 3{σd|0} 3{σv|τ}
x2
+ y2
, z2
z 1 A1 1 1 1 1 1 1
Rz 2 A2 1 1 1 1 −1 −1
3 B1 1 −1 1 −1 −1 1
4 B2 1 −1 1 −1 1 −1
(xz, yz)(x, y)(Rx, Ry) 6 E1 2 −2 −1 1 0 0
(x2
− y2
, xy) 5 E2 2 2 −1 −1 0 0
We also have included in the character table the basis functions for C6v.
We find χatom sites for the two a atoms and for the two b atoms of
puckered graphite using Fig. 15.14:
{E|0} {C2|τ} 2{C3|0} 2{C6|τ} 3{σd|0} 3{σv|τ}
χa.s. a−atoms 2 0 2 0 2 0 A1 + B2
χa.s. b−atoms 2 0 2 0 2 0 A1 + B2
χa.s. for both the a- and b-atoms transforms as A1 + B2; also the vector
transforms as A1 + E1. Thus we obtain for the lattice modes for
puckered graphite
χin−plane lattice modes = 2(A1 + B2) ⊗ E1 = 2E1 + 2E2
χz−axis lattice modes = 2(A1 + B2) ⊗ A1 = 2A1 + 2B2
The lattice mode patterns are similar to those for ordinary graphite
(see Fig. 15.13). Furthermore the A1 and E1 modes are infrared-active,
the E2 modes are Raman-active and the B2 modes are silent. Thus
puckered graphite would yield IR and Raman spectra similar to that
observed for ordinary graphite. For the two hexagonal forms of graphite
the high symmetry points in the Brillouin zone are given in Fig. 15.12
(from Miller and Love, p. 131). For each of these high symmetry points,
the group of the wave vector is given in Miller and Love.
Finally we discuss the lattice modes for rhombohedral graphite. The
character table from Tinkham corresponding to the the group of the
442 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
Figure 15.13: Schematic diagram
showing the lattice mode displacements
for hexagonal graphite.
15.5. PHONONS IN GRAPHITE 443
 
¡
¡ ¡
¢¤£
¢¦¥
Figure 15.14: Schematic diagram
showing the symmetry operations
in puckered graphite.
wave vector at k = 0 is written below (D3d = D3 ⊗ i):
1 3, 5 7, 9, 11 13 15, 17 19, 21, 23
D3 (32) E 2C3 3C2 i 2iC3 3iC2
x2 + y2, z2 A1 1 1 1 1 1 1
Rz, z A2 1 1 −1 1 1 −1
(xz, yz)
(x2 − y2, xy)
(x, y)
(Rx, Ry)
E 2 −1 0 2 −1 0
where the correspondence in notation is:
Miller and Love Tinkham Miller and Love Tinkham
1 E 1+
A1g
3, 5 2C3 2+
A2g
7, 9, 11 3C2 3+
Eg
13 i 1−
A1u
15, 17 2iC3 2−
A2u
19, 21, 23 3iC2 3−
Eu
Referring to Figs. 15.2 and 15.3 for the rhombohedral structure and
layer stacking, we note that the operations E and C3 take each carbon
atom into an equivalent carbon atom (i.e., either into itself or
displaced by a lattice vector), while operations C2 and i interchange
444 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
the two inequivalent atoms/unit cell. We also note that iC2 also takes
each carbon into an equivalent site. We thus obtain for χatom sites = χa.s.
for rhombohedral graphite:
E 2C3 3C2 i 2iC3 3iC2
χa.s. 2 2 0 0 0 2 = A1g + A2u
For D3 symmetry
χvector = A2u + Eu (15.3)
so that taking the direct product χa.s. ⊗ χvector we obtain
χin−plane lattice modes = Eu ⊗ (A1g + A2u) = Eu + Eg (15.4)
χz−axis lattice modes = A2u ⊗ (A1g + A2u) = A2u + A1g. (15.5)
After identifying the A2u and Eu modes with z-axis and in-plane translations,
we identify the two modes with A1g and Eg symmetries as
optical modes at k = 0. Both of these modes are Raman-active and
infrared-inactive and execute the normal optical type motions.
15.6 Selected Problems
1. The electronic energy band structure of graphite near the Fermi
level is of particular interest along the KH edge of the Brillouin
zone (see Fig. 15.12).
(a) “Translate” the K-point character table for graphite from
Miller and Love in terms of the appropriate point group
character table in Tinkham.
(b) Find χatom sites at the K-point for the 4 atoms in the unit cell
of graphite. Give the K point irreducible representations
contained in χatom sites.
2. Find the space groups and give the appropriate site locations for
the atomic constituents from the International Crystallography
Tables for the following crystalline solids:
15.6. SELECTED PROBLEMS 445
(a) Solid molecular crystalline bromine.
(b) Crystalline SnO2.
(c) Crystalline Nb3Sn.
446 CHAPTER 15. USE OF STANDARD REFERENCE TEXTS
Chapter 16
Calculation of the Electronic
Energy Levels in a Cubic
Crystal
We now apply the space groups to the electronic dispersion relations in
crystalline materials. In this chapter we will use group theory to find
symmetrized plane wave solutions to the nearly free electron dispersion
relations in crystalline materials.
16.1 Introduction
Suppose that we wish to calculate the electronic energy levels of a
solid from a specified potential. There are many standard techniques
available for this purpose based on plane waves, such as the orthogonalized
plane wave method (OPW), the augmented plane wave method
(APW); these are discussed in some detail in the sequence of solid state
courses. In all cases these techniques utilize the space symmetry of
the crystal. Because of the relative importance of the electronic energy
bands at high symmetry points and along high symmetry axes
for the interpretation of experimental data, these model calculations
exploit the simplifications which result from the application of group
theory.
To illustrate how group theory is utilized in these calculations, we
447
448 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
will consider explicitly the energy bands of the nearly free electron
model. If there were no periodic potential, the energy eigenvalues would
be the free electron energies
E(k ) =
¯h2
k 2
2m
V (r) = 0 (16.1)
and the free electron eigenfunctions would be
ψk (r) =
1
√
Ω
eik ·r
(16.2)
where k is a wave vector in the extended Brillouin zone.
The presence of a weak periodic potential imposes the symmetry of
the crystal on the “empty lattice” energy bands. From a group theoretical
point of view, the free electron energy bands correspond
to the symmetry of the full rotation group and the finite periodic
potential serves to lower the symmetry, as for example
to Oh symmetry for a simple cubic crystal. Thus, the introduction of
a periodic potential results in a situation similar to the crystal field
problem which we have by now encountered in several contexts.
We consider the empty lattice energy bands in the reduced zone by
writing the wave vector k in the extended zone scheme as:
k = k + K (16.3)
where k is a reduced wave vector in the 1st
Brillouin zone and K is 2π
times the reciprocal lattice vector to obtain
E(k + K) =
¯h2
2m
(k + K) · (k + K) (16.4)
where
K =
2π
a
(n1, n2, n3), and ni = integer. (16.5)
We use the subscript K on the energy eigenvalues to denote the pertinent
K vector when using the wave vector k within the first Brillouin
zone. If we write k in dimensionless units
ξ =
ka
2π
(16.6)
16.1. INTRODUCTION 449
Figure 16.1: Free-electron bands in a face centered cubic structure.
The labels of the high symmetry points in the fcc structure are given
in Fig. 16.8(b). The band degeneracies are indicated on the diagram.
we obtain
EK(k) =
¯h2
2m
2π
a
2
(ξ1 + n1)2
+ (ξ2 + n2)2
+ (ξ3 + n3)2
. (16.7)
The empty lattice energy bands for the fcc cubic structure are shown
in Fig. 16.1 at the high symmetry points and along the high symmetry
directions indicated by the Brillouin zone for the fcc lattice [see
Fig. 16.8(b)]. The energy bands are labeled by the symmetries of the
irreducible representations appropriate to the group of the wave vector
corresponding to the space group. Group theory provides us with the
symmetry designations and with the level degeneracies. In §16.2, we
treat the symmetry designations and mode degeneracies for the simple
cubic lattice at k = 0, and in §16.3 and §16.4 at other symmetry points
in the Brillouin zone.
450 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
In the reduced zone scheme, the wave functions for the plane wave
become the Bloch functions
ψk (r) =
1
√
Ω
eik ·r
=
1
√
Ω
eik·r
eiK·r
(16.8)
where the periodic part of the Bloch function is written as
uk(r) = eiK·r
. (16.9)
According to Bloch’s theorem, the effect of the translation operator is
to introduce a phase factor
{ε|t}ψk(r) = eik·t
ψk(r) (16.10)
in which t is a lattice vector such as Rn.
In calculating the electronic energy bands in the nearly free electron
approximation, we recognize that the main effect of the periodic
potential is to lift the degeneracy of EK(k). At certain high symmetry
points or axes and at the Brillouin zone boundary, the degeneracy in
many cases is not fully lifted because of the degeneracy of some of the
pertinent irreducible representations. Group theory tells us the form
of the interactions, the symmetry of the levels and their degeneracies.
For each of the high symmetry points in the Brillouin zone, different
symmetry operations will be applicable, depending on the appropriate
group of the wave vector for the k point under consideration.
16.2 Plane Wave Solutions at k = 0
The highest symmetry point in the Brillouin zone is of course the Γ
point (k = 0) and we will therefore illustrate the application of group
theoretical considerations to the energy bands at the Γ point first. Setting
k = 0 in Eq. (16.7) for EK(k) we obtain
EK(0) =
¯h2
2m
2π
a
2
n2
1 + n2
2 + n2
3 =
¯h2
2m
2π
a
2
N2
, (16.11)
where
N2
= n2
1 + n2
2 + n2
3. (16.12)
16.2. PLANE WAVE SOLUTIONS AT K = 0 451
Corresponding to each reciprocal lattice vector K, a value for EK(0)
is obtained. For most K vectors, these energies are degenerate. We
will now enumerate the degeneracy of the first few levels, starting with
K = 0 and n1 = n2 = n3 = 0. We then find which irreducible representations
for Oh are contained in each degenerate state. If then a periodic
potential is applied, the degeneracy of some of these levels will be lifted.
Group theory provides a powerful tool for specifying how these degeneracies
are lifted. In Table 16.1 we give the energy, the degeneracy and
the set of K vectors that yield each of the five lowest energy eigenvalues
EK(0). The example that we explicitly work out here is for the simple
cubic lattice (space group #221).
At K = 0 we have ψk(r) = (1/
√
Ω)eik·r
. For a general K vector,
(n1, n2, n3) there will in general be a multiplicity of states with the
same energy. We now show how to choose a properly symmetrized
combination of plane waves which transform as irreducible representations
of the group of the wave vector at k = 0, and therefore bring
the Hamiltonian into block diagonal form. In the presence of a weak
cubic periodic potential, the degeneracy of states which transform as
different irreducible representations will be lifted.
By calculating χequivalence = χatom sites we can specify which plane
waves are transformed into one another by the elements of the group of
the wave vector at the Γ point (k = 0). From χequivalence we can find the
irreducible representations of Oh which correspond to the degenerate
empty lattice state and we can furthermore find the appropriate linear
combination of plane wave states which correspond to a particular
irreducible representation of Oh.
To calculate χequivalence we use the diagram in Fig. 16.2 which shows
the cubic symmetry operations of point group Oh. The character table
for Oh symmetry is given in Table 16.2, where the column on the right
gives the familiar solid state notation for the irreducible representations
of Oh. Computation of χequivalence is identical with the calculation of
χatom sites. If a given plane wave goes into itself under the symmetry
operations of Oh a contribution of 1 is made to the character; otherwise
no contribution is made. Using these definitions, we form χatom sites and
the characters for the various plane waves are given in Table 16.3, where
the various plane wave states are denoted by one of the reciprocal lat-
452 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
Table 16.1: Listing of the energy, degeneracy and the list of K vectors
for the five lowest energy levels for the simple cubic lattice at k = 0.
(i) E{000}(0) = 0 degeneracy=1 K{000} = 2π
a
(0,0,0) N2
= 0
(ii) E{100}(0) = ¯h2
2m
2π
a
2
degeneracy=6 K{100} = 2π
a



(1, 0, 0)
(¯1, 0, 0)
(0, 1, 0)
(0, ¯1, 0)
(0, 0, 1)
(0, 0, ¯1)
N2
= 1
Plane Wave States:
e± 2πix
a , e± 2πiy
a , e± 2πiz
a
(iii) E{110}(0) = 2 ¯h2
2m
2π
a
2
degeneracy=12 K{110} = 2π
a



(1, 1, 0)
(¯1, 1, 0)
(1, 0, 1)
(¯1, 0, 1)
(0, 1, 1)
(0, ¯1, 1)
(1, ¯1, 0)
(¯1, ¯1, 0)
(1, 0, ¯1)
(¯1, 0, ¯1)
(0, 1, ¯1)
(0, ¯1, ¯1)
N2
= 2
(iv) E{111}(0) = 3 ¯h2
2m
2π
a
2
degeneracy=8 K{111} = 2π
a



(1, 1, 1)
(1, ¯1, 1)
(1, 1, ¯1)
(¯1, 1, 1)
(¯1, ¯1, 1)
(1, ¯1, ¯1)
(¯1, 1, ¯1)
(¯1, ¯1, ¯1)
N2
= 3
(v) E{200}(0) = 4 ¯h2
2m
2π
a
2
degeneracy=6 K{200} = 2π
a



(2, 0, 0)
(¯2, 0, 0)
(0, 2, 0)
(0, ¯2, 0)
(0, 0, 2)
(0, 0, ¯2)
N2
= 4
16.2. PLANE WAVE SOLUTIONS AT K = 0 453
Figure 16.2: Diagram of Cubic
Symmetry Operations
Table 16.2: The character table for Oh symmetry.
O (432) E 8C3 3C2 = 3C2
4 6C2 6C4
A1 1 1 1 1 1 Γ1
A2 1 1 1 −1 −1 Γ2
(x2
− y2
, 3z2
− r2
) E 2 −1 2 0 0 Γ12
(Rx, Ry, Rz)
(x, y, z)
T1 3 0 −1 −1 1 Γ15
(xy, yz, zx) T2 3 0 −1 1 −1 Γ25
Oh = O ⊗ i (m3m)
454 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
Table 16.3: Characters for the equivalence representation for the set of
plane wave states labeled by {K}.
K E 3C2
4 6C2 8C3 6C4 i 3iC2
4 6iC2 8iC3 6iC4
{0,0,0} 1 1 1 1 1 1 1 1 1 1 Γ+
1
{1,0,0} 6 2 0 0 2 0 4 2 0 0 Γ+
1 + Γ+
12 + Γ−
15
{1,1,0} 12 0 2 0 0 0 4 2 0 0 Γ+
1 + Γ+
12 + Γ−
15 + Γ+
25 + Γ−
25
{1,1,1} 8 0 0 2 0 0 0 4 0 0 Γ+
1 + Γ−
2 + Γ−
15 + Γ+
25
{2,0,0} 6 2 0 0 2 0 4 2 0 0 Γ+
1 + Γ+
12 + Γ−
15
tice vectors which describe each of these states using the notation of
Table 16.1. The reducible representations for χequivalence for the various
plane wave states in the simple cubic lattice are decomposed into irreducible
representations of Oh and the results are given on the right
hand side of Table 16.3.
Once we know the irreducible representations of Oh that are contained
in each of the degenerate levels of the simple cubic empty lattice,
we can find appropriate linear combinations of these plane wave states
which will then transform as the desired irreducible representations of
Oh. When a cubic periodic potential is now applied, the degeneracy of
these empty lattice states will be lifted in accordance with the decomposition
of the reducible representations of χequivalence into the irreducible
representations of Oh. Thus the proper linear combinations of the plane
wave states will bring the secular equation of the nearly free electron
model energy bands into block diagonal form. As an example of how
this works, let us list the six appropriate linear combinations for the
{1,0,0} set of reciprocal lattice vectors exp(±2πix/a), exp(±2πiy/a),
and exp(±2πiz/a) which will bring the secular equation into block diagonal
form:
1
√
6
[(1, 0, 0) + (¯1, 0, 0) + (0, 1, 0) + (0, ¯1, 0) + (0, 0, 1) + (0, 0, ¯1)] → Γ+
1
1√
6
[(1, 0, 0) + (¯1, 0, 0) + ω(0, 1, 0) + ω(0, ¯1, 0) + ω2
(0, 0, 1) + ω2
(0, 0, ¯1)]
1√
6
[(1, 0, 0) + (¯1, 0, 0) + ω2
(0, 1, 0) + ω2
(0, ¯1, 0) + ω(0, 0, 1) + ω(0, 0, ¯1)]
→ Γ+
12
16.2. PLANE WAVE SOLUTIONS AT K = 0 455
1
i
√
2
[(1, 0, 0) − (¯1, 0, 0)]
1
i
√
2
[(0, 1, 0) − (0, ¯1, 0)]
1
i
√
2
[(0, 0, 1) − (0, 0, ¯1)]



→ Γ−
15,
(16.13)
in which we have used (1,0,0) to denote exp(2πix/a) and similarly for
the other plane waves. Substituting
1
2
[(1, 0, 0) + (¯1, 0, 0)] = cos(2πx/a)
1
2i
[(1, 0, 0) − (¯1, 0, 0)] = sin(2πx/a)
(16.14)
we obtain the following linear combinations of symmetrized plane waves
from Eq. (16.13):
2
√
6
cos
2πx
a
+ cos
2πy
a
+ cos
2πz
a
→ Γ+
1
2√
6
cos 2πx
a
+ ω cos 2πy
a
+ ω2
cos 2πz
a
2√
6
cos 2πx
a
+ ω2
cos 2πy
a
+ ω cos 2πz
a



→ Γ+
12
√
2 sin 2πx
a√
2 sin 2πy
a√
2 sin 2πz
a



→ Γ−
15
(16.15)
The linear combinations of plane wave states given in Eq. (16.15) transform
as irreducible representations of Oh, and bring the secular equation
for E(k = 0) into block diagonal form. For example, using the 6
combinations of plane wave states given in Eq. (16.15), we bring the
(6 × 6) secular equation for K = {1, 0, 0} into a (1 × 1), a (2 × 2)
and a (3 × 3) block, with no coupling between the blocks. Since there
are 3 distinct energy levels, each corresponding to a different symmetry
type, the introduction of a weak periodic potential will, in general, split
the 6–fold level into 3 levels with degeneracies 1 (Γ+
1 ), 2 (Γ+
12) and 3
(Γ−
15). This procedure is used to simplify the evaluation of E(k) and
ψk(r) in first-order degenerate perturbation theory. Referring to Table
16.1, Eq. (16.15) gives the symmetrized wave functions for the six
K{100} vectors. The corresponding analysis can be done for the twelve
456 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
K{110} vectors for the third lowest energy level, etc. The results for
E(k) for the empty lattice for the simple cubic group #221 are shown
in Fig. 16.3.
The results obtained for the simple cubic lattice can be extended to
other cubic lattices. The space group numbers for common cubic crystals
are as follows: simple cubic (#221), fcc (#225), diamond (#227),
bcc (#229). For the fcc lattice the (n1, n2, n3) integers are all even or
all odd so that the allowed K vectors are {000}, {111}, {200}, etc. For
the bcc lattice, the integers (n1 + n2 + n3) must all sum to an even
number, so that we can have reciprocal lattice K vectors {000}, {110},
{200}, etc. Thus Table 16.1 can be used together with an analysis such
as given in this section to obtain the proper linear combination of plane
waves for the pertinent K vectors for the various cubic groups.
To complete the discussion of the use of group theory for the solution
of the electronic states of the empty lattice (or more generally the
nearly free electron) model, we will next consider the construction of
the symmetrized plane wave states E(k) as we move away from k = 0.
16.3 Symmetrized Plane Wave Solutions
at the ∆ Point
As an example of a non-zero k vector, let us consider E(k) as we move
from Γ(k = 0) toward point X (k = π
a
(100)). For intermediate points
along the (100) direction (labeled ∆ in Fig. 16.4), the appropriate point
group of the wave vector is C4v, with character table:
Below the character table (Table 16.4) for point group C4v, are
listed the characters for the three irreducible representations of K =
{100}(π/a) corresponding to the k = 0 solution and Oh symmetry. We
consider these as reducible representations of point group C4v. The decomposition
of these three reducible representations in C4v point group
symmetry is indicated on the right of Table 16.4. This decomposition
yields the compatibility relations (see §13.7):
Γ+
1 → A1 = ∆1
16.3. SYMMETRIZED PLANE WAVES AT ∆ 457
(a) (b)
Figure 16.3: Diagram of the empty lattice energy levels along (a) Γ−X
and (b) Γ − L for the simple cubic lattice #221.
458 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
 ¢¡
 ¤£
 ¦¥
§
¨
©




 
Figure 16.4: Brillouin zone for a
simple cubic lattice.
Table 16.4: Character table for the point group C4v and other relevant
information.
C4v (4mm) E C2 2C4 2σv 2σd
x2
+ y2
, z2
z A1 1 1 1 1 1 ∆1
Rz A2 1 1 1 −1 −1 ∆1
x2
− y2
B1 1 1 −1 1 −1 ∆2
xy B2 1 1 −1 −1 1 ∆2
(xz, yz)
(x, y)
(Rx, Ry)
E 2 −2 0 0 0 Γ5
A1g (Oh) Γ+
1 1 1 1 1 1 A1
Eg (Oh) Γ+
12 2 2 0 2 0 A1 + B1
T1u (Oh) Γ−
15 3 –1 1 1 1 A1 + E
16.4. PLANE WAVE SOLUTIONS AT THE X POINT 459
Γ+
12 → A1 + B1 = ∆1 + ∆2
Γ−
15 → A1 + E = ∆1 + ∆5. (16.16)
In the above character table, the main symmetry axis is the x axis, so
that the basis functions that should be used require the transformation:
x → y, y → z, z → x. The symmetry axis σv = iC100
2 denotes
the mirror planes y = 0 and z = 0, while σd = iC100
2 denotes the diagonal
(011) planes, with all symmetry operations referring to reciprocal
space, since we are considering the group of the wave vector at a ∆
point. The results of Eq. (16.16) are of course in agreement with the
compatibility relations given in §13.7 for the simple cubic structure.
Compatibility relations of this type can be used to obtain the degeneracies
and symmetries for all the levels at the ∆ point, starting from
the plane wave solution at k = 0. A similar approach can be used to
obtain the symmetries and degeneracies as we move away from k = 0
in other directions. For an arbitrary crystal structure we have to use
Miller and Love to construct the compatibility relations using the tables
for the group of the wave vector given in this reference.
16.4 Plane Wave Solutions at the X Point
As we move in the Brillouin zone from a point of high symmetry to a
point of lower symmetry, the solution using the compatibility relations
discussed in §16.3 is unique. On the other hand, when going from
a point of lower symmetry to one of higher symmetry, the solution
from the compatibility relations is not unique, and we must then go
back to consideration of the equivalence transformation. An example
of this situation occurs when we go from the ∆ point to the X-point
(D4h symmetry), which has higher symmetry than the ∆ point (C4v
symmetry). The appropriate character table for the X point is D4h =
D4 ⊗ i for the group of the wave vector shown in Table 16.5. At the
X-point, the nearly free electron solutions for the simple cubic lattice
given by Eq. (16.7) become:
E k =
π
a
ˆx =
¯h2
2m
2π
a
2
1
2
+ n1
2
+ n2
2 + n2
3 . (16.17)
460 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
Table 16.5: Character table for the point group D4, showing the solid
state notation in the right hand column.
D4 (422) E C2 = C2
4 2C4 2C2 2C2
x2
+ y2
, z2
A1 1 1 1 1 1 X1
Rz, z A2 1 1 1 −1 −1 X4
x2
− y2
B1 1 1 −1 1 −1 X2
xy B2 1 1 −1 −1 1 X3
(xz, yz)
(x, y)
(Rx, Ry)
E 2 −2 0 0 0 X5
The lowest energy level at the X-point is
E1 k =
π
a
ˆx =
¯h2
2m
2π
a
2
1
4
. (16.18)
The pertinent plane waves which contribute to the energy level in
Eq. (16.18) correspond to K vectors:
K = (0, 0, 0)
K =
2π
a
(¯1, 0, 0).
We will now find χequivalence for these plane waves, using the symmetry
operations in Fig. 16.5 and in the character table for D4h in which
we use the transformation x → y, y → z, z → x to obtain the proper
X-point. We note that K = (0, 0, 0) yields a plane wave e
π
a
ix
while
K = 2π
a
(¯1, 0, 0) yields a plane wave e(π
a
ix− 2π
a
ix) = e− π
a
ix
and both have
energies E1 = ¯h2
2m
2π
a
2
1
4
. The plane waves denoted by K = (0, 0, 0)
and K = 2π
a
(¯1, 0, 0) form partners of a reducible representation:
E C2 2C4 2C2 2C2 i iC2 2iC4 2iC2 2iC2
exp(±πix/a) 2 2 2 0 0 0 0 0 2 2 A1g + A2u
A1g + A2u = X+
1 + X−
4 . (16.19)
16.4. PLANE WAVE SOLUTIONS AT THE X POINT 461
 
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Figure 16.5: Diagram of a square
showing the two-fold axes normal to
the principal C4 symmetry axis.
We thus obtain irreducible representations with X+
1 and X−
4 symmetries
for the lowest X-point level so that a periodic potential will split
the degeneracy of these levels at the X-point. In this case the level
separations becomes 2|VK| where K = 2π
a
(¯1, 0, 0). The appropriate
linear combination of plane waves corresponding to the X+
1 and X−
4
irreducible representations are:
X+
1 symmetry : (0, 0, 0) + (¯1, 0, 0) → 2 cos
π
a
x
X−
4 symmetry : (0, 0, 0) − (¯1, 0, 0) → 2 sin
π
a
x. (16.20)
and each of the X+
1 and X−
4 levels is non-degenerate.
Referring to Eq. (16.17), the next lowest energy level at the X point
is:
E2 k =
π
a
ˆx =
¯h2
2m
2π
a
2
5
4
. (16.21)
The 8 pertinent plane waves for this energy level correspond to the K
vectors
K =
2π
a
(0, 1, 0),
2π
a
(0, ¯1, 0),
2π
a
(0, 0, 1),
2π
a
(0, 0, ¯1)
462 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
K =
2π
a
(¯1, 1, 0),
2π
a
(¯1, ¯1, 0),
2π
a
(¯1, 0, 1),
2π
a
(¯1, 0, ¯1).
More explicitly, the 8 plane waves corresponding to these K vectors
are:
exp πix
a
+ 2πiy
a
, exp πix
a
− 2πiy
a
,
exp πix
a
+ 2πiz
a
, exp πix
a
− 2πiz
a
,
exp −πix
a
+ 2πiy
a
, exp −πix
a
− 2πiy
a
,
exp −πix
a
+ 2πiz
a
, exp −πix
a
− 2πiz
a
(16.22)
To find χequivalence for the 8 plane waves of Eq. (16.22) we use the character
table for D4h and Fig. 16.5. The results for several pertinent plane
wave combinations are given below:
E C2 2C4 2C2 2C2 i iC2 2iC4 2iC2 2iC2
Eq. (16.22) 8 0 0 0 0 0 0 0 4 0
exp(±2πiy/a)
exp (±2πiz/a)
4 0 0 2 0 0 4 0 2 0
The reducible representation for the 8 plane waves given by Eq. (16.22)
yields the following X point irreducible representations
X+
1 + X+
2 + X−
5 + X−
4 + X−
3 + X+
5 . (16.23)
The same result can be obtained by considering the e± πix
a functions as
common factors of the e± 2πiy
a and e± 2πiz
a functions. The χequivalence for
the four e± 2πiy
a and e± 2πiz
a plane waves is also tabulated above. The
e± πix
a functions transform as X+
1 +X−
4 (see above), and the 4 functions
e± 2πiy
a and e± 2πiz
a transform as X+
1 + X+
2 + X−
5 . If we now take the
direct product, we obtain:
(X+
1 +X−
4 )⊗(X+
1 +X+
2 +X−
5 ) = X+
1 +X+
2 +X−
5 +X−
4 +X−
3 +X+
5 (16.24)
in agreement with the result of Eq. (16.23). The proper linear combination
of the eight plane waves which transform as irreducible representations
of the D4h point symmetry group for the second lowest X
point level is found from the K vectors given below:
X+
1 : (0, 1, 0) + (0, ¯1, 0) + (0, 0, 1) + (0, 0, ¯1) + (¯1, 1, 0) + (¯1, ¯1, 0) + (¯1, 0, 1) + (¯1, 0, 1)
X−
4 : (0, 1, 0) + (0, ¯1, 0) + (0, 0, 1) + (0, 0, ¯1) − (¯1, 1, 0) − (¯1, ¯1, 0) − (¯1, 0, 1) − (¯1, 0, 1)
16.4. PLANE WAVE SOLUTIONS AT THE X POINT 463
X+
2 : (0, 1, 0) − (0, 0, 1) + (0, ¯1, 0) − (0, 0, ¯1) + (¯1, 1, 0) − (¯1, 0, 1) + (¯1, ¯1, 0) − (¯1, 0, ¯1)
X−
3 : (0, 1, 0) − (0, 0, 1) + (0, ¯1, 0) − (0, 0, ¯1) − (¯1, 1, 0) + (¯1, 0, 1) − (¯1, ¯1, 0) + (¯1, 0, ¯1)
X−
5 :
(0, 1, 0) − (0, ¯1, 0) + (¯1, 1, 0) − (¯1, ¯1, 0)
(0, 0, 1) − (0, 0, ¯1) + (¯1, 0, 1) − (¯1, 0, ¯1)
2 partners
X+
5 :
(0, 1, 0) − (0, ¯1, 0) − (¯1, 1, 0) + (¯1, ¯1, 0)
(0, 0, 1) − (0, 0, ¯1) − (¯1, 0, 1) + (¯1, 0, ¯1)
2 partners (16.25)
in which the plane waves are denoted by their corresponding K vectors.
We note that the wave vector K = 2π
a
(0, 1, 0) gives rise to a plane wave
exp[πix
a
+ 2πiy
a
]. Likewise the wave vector K = 2π
a
(¯1, 1, 0) gives rise to a
plane wave exp[πix
a
− 2πix
a
+ 2πiy
a
]. Using this notation we find that the
appropriate combinations of plane waves corresponding to Eq. (16.25)
are:
X+
1 : cos
πx
a
cos
2πy
a
+ cos
2πz
a
X−
4 : sin
πx
a
cos
2πy
a
+ cos
2πz
a
X+
2 : cos
πx
a
cos
2πy
a
− cos
2πz
a
X−
3 : sin
πx
a
cos
2πy
a
− cos
2πz
a
X−
5 :
cos πx
a
sin 2πy
a
cos πx
a
sin 2πz
a
2 partners
X+
5 :
sin πx
a
sin 2πy
a
sin πx
a
sin 2πz
a
2 partners (16.26)
A summary of the energy levels and symmetries along Γ − X and
Γ − R for the simple cubic lattice is given in Fig. 16.3. A similar
procedure is used to find the degeneracies and the symmetrized linear
combination of plane waves for any of the energy levels at each of the
high symmetry points in the Brillouin zone. We show for example
results in Fig. 16.3b also for the empty lattice bands along Γ − R. The
corresponding results can be obtained by this same procedure for the
fcc and bcc lattices as well (see Figs. 16.1 and 16.6).
In the following section we will consider the effect of non-symmorphic
operations on plane waves.
464 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
Figure 16.6: Diagram of empty lattice
energy levels along Γ − H for
the bcc lattice. The irreducible representations
are indicated.
16.5. EFFECT OF GLIDE PLANES AND SCREW AXES 465
Figure 16.7: Brillouin zone for a
rectangular lattice (such as p2mg
(#14))
16.5 Effect of Glide Planes and Screw Axes
The main effect of non-symmorphic operations connected with glide
planes and screw axes is to cause energy bands to stick together along
some of the high symmetry points and axes in the Brillouin zone. We
first illustrate this feature using the 2D space group p2mg (#14) which
has a 2-fold axis, mirror planes normal to the x axis at x = 1
4
a and
x = 3
4
a, and a glide plane g parallel to the x axis for a distance a
2
. In
addition, group p2mg has inversion symmetry. Suppose that X(x, y)
is a solution to Schr¨odinger’s equation at the X point kX = π
a
(1, 0)
(see Fig. 16.7). This degeneracy at the zone boundary in the twodimensional
non-symmorphic space group p2mg is also found in many
of the common three-dimensional non-symmorphic groups.
In the two-dimensional case for the space group p2mg, the mirror
glide operation g implies
gX(x, y) = X(x +
1
2
a, −y) (16.27)
while inversion i implies
iX(x, y) = X(−x, −y). (16.28)
466 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
The mirror plane m at x = a/4 implies
mX(x, y) = X(−x +
1
2
a, y) (16.29)
so that
gX(x, y) = m iX(x, y) (16.30)
where m denotes reflection in a mirror plane and i denotes inversion.
Since i2
X(x, y) = X(x, y) and m2
X(x, y) = X(x, y), we would expect
from Eq. (16.30) that
g2
X(x, y) = X(x, y). (16.31)
But direct application of the glide operation twice yields
g2
X(x, y) = X(x + a, y) = eikxa
X(x, y) = eπi
X(x, y) = −X(x, y)
(16.32)
which contradicts Eq. (16.31). This contradiction is resolved by having
the solutions ±X(x, y) stick together at the X point.
In fact, if we employ time reversal symmetry (to be discussed in
Chapter 21), we can show that bands ±ΦZ(x, y) stick together along the
entire Brillouin zone edge for all Z points, i.e., (π/a, ky) (see Fig. 16.7).
Thus in addition to the degeneracies imposed by the group of the wave
vector, other symmetry relations can in some cases cause energy bands
to stick together at high symmetry points and axes.
The same situation also arises for the 3D space groups. Some common
examples where energy bands stick together are on the hexagonal
face of the hexagonal close packed structure (space group #194, see
Brillouin zone in Fig. 16.8a), and the square face in the diamond structure
(#227) for which the Brillouin zone is given in Fig. 16.8b. For
the case of the hexagonal close packed structure, there is only a single
translation τ = c
2
(001) connected with non-symmorphic operations in
space group #194. The character table for the group of the wave vector
at the A point shows that the bands stick together, i.e., there are no
non-degenerate levels at the A point. To illustrate this point, we give
in Table 16.6 the character tables for the Γ point and the A point for
space group #194.
For the case of the diamond structure (space group #227), Miller
and Love shows that there are 3 different translations (a/4)(110), (a/4)(011),
16.5. EFFECT OF GLIDE PLANES AND SCREW AXES 467
Table 16.6: Character tables from Miller and Love for the Γ point
and the A point for the hexagonal close packed structure (hcp) (space
group #194). Note the degeneracy of all the irreducible representations
at point A in the Brillouin zone.
468 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
Figure 16.8: Brillouin zone for (a) the hexagonal close pack structure,
D4
6h, #194 and (b) the fcc structure (e.g., diamond #227).
and (a/4)(101). The reason why these translations differ from those
used in this section is the selection of a different origin for the unit cell.
In Miller and Love the origin is selected to lie halfway between the two
inequivalent lattice points. We can take the equivalent lattice points
as a white atom and a black atom for discussion purposes. Miller and
Love selects the origin at a
8
(111) or at a
8
(¯1¯1¯1), so that the inversion
operation takes the white sublattice into a black sublattice, and vice
versa. In contrast, we have taken the origin to coincide with the origin
of the white sublattice so that in this case the space group operation for
inversion contains a translation by τ = (a/4)(111) and is denoted by
{i|τ}. In Table 16.7 we show the character tables appropriate for the
diamond structure at the Γ point and at the X point using the origin
at (000).
In all the E(k) diagrams for the diamond structure (see Fig. 16.9 for
E(k) for Ge), we see that all the bands stick together at the X point,
all being either 2-fold or 4-fold degenerate, as seen in the character
table for the X point in Table 16.7. The plane wave basis functions
for the irreducible representations X1, X2, X3 and X4 for the diamond
structure are listed in Table 16.8.
Because of the non-symmorphic features of the diamond structure,
the energy bands at the X point behave differently from the bands
16.5. EFFECT OF GLIDE PLANES AND SCREW AXES 469
Table 16.7: Character table for the diamond structure (space group
#227; Fd3m) at the Γ point and at the X point. If we take the origin
of the white sublattice at (000), then the site location of the black
sublattice is at (a/4)(111).
470 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
Figure 16.9: Energy band structure for germanium. Note that the
bands stick together at the X point.
16.5. EFFECT OF GLIDE PLANES AND SCREW AXES 471
Table 16.8: Plane wave basis functions for the group of the wave vector
for the X-point [2π
a
(100)] for the non-symmorphic diamond structure.
Representation Function
X1 x11 = cos 2π
a
x
x12 = sin 2π
a
x
X2 x21 = cos 2π
a
x[cos 4π
a
y − cos 4π
a
z]
x22 = sin 2π
a
x[cos 4π
a
y − cos 4π
a
z]
X3 x31 = sin 4π
a
(y + z)[cos 2π
a
x + sin 2π
a
x]
x32 = sin 4π
a
(y − z)[cos 2π
a
x − sin 2π
a
x]
X4 x41 = sin 4π
a
(y − z)[cos 2π
a
x + sin 2π
a
x]
x42 = sin 4π
a
(y + z)[cos 2π
a
x − sin 2π
a
x]
at high symmetry points where “essential” degeneracies occur. For
the case of essential degeneracies, the energy bands E(k) come into
the Brillouin zone with zero slope. For the X point in the diamond
structure, the E(k) dispersion relations with X1 and X2 symmetries in
general have a non-zero slope, but the slopes are equal and opposite for
the two levels (X1 and X2) that stick together. The physical reason for
this behavior is that the x-ray structure factor for the Bragg reflection
associated with the X point in the Brillouin zone for the diamond
structure vanishes and thus no energy discontinuity in E(k) is expected,
nor is it observed upon small variation of kx relative to the X point.
Explicitly the structure factor at the X point for the diamond structure
is:
i
eiKX ·ri
= 1 + ei 4π
a
(100)·a
4
(111)
= 1 − 1 ≡ 0 (16.33)
where the sum is over the two inequivalent atom sites in the unit cell.
The vanishing of this structure factor for the reciprocal lattice vector
KX = (4π/a)(100) associated with the X point implies that there is
no Fourier component of the periodic potential to split the degeneracy
caused by having two atoms per unit cell and thus the energy bands
at the X-point stick together. In fact, the structure factor in diamond
vanishes for all points on the square face of the fcc Brillouin zone (see
Fig. 16.8(b)), and we have energy bands sticking together across the
472 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
entire square face.
For the L-point in the Brillouin zone, the levels do not stick together
because the structure factor at the L-point does not vanish:
i
eiKL·ri
= 1 + ei 2π
a
(111)·a
4
(111)
= 1 − i = 0. (16.34)
Thus for the non-symmorphic diamond structure, some high symmetry
points behave normally (such as the L point), while for other points
(such as the X point) the energy bands stick together.
To show how the energy bands at the X-point stick together, consider
the operations of the inversion symmetry operator {i|τ} on the
basis functions for the X-point listed in Table 16.8. Similar results can
be obtained by considering other operations in the point group Oh (and
not in Td), that is by considering symmetry operations involving the
translation operation τ = (a/4)(111). treat the effect of {i|τ} on the
various functions of (x, y, z) in Table 16.8, consider first the action of
{i|τ} on the coordinates:
{i|τ}



x
y
z


 =



−x + (a/4)
−y + (a/4)
−z + (a/4)


 . (16.35)
Then using the trigonometric identity:
cos(α + β)=cos α cos β − sin α sin β
sin(α + β)=sin α cos β + cos α sin β
(16.36)
we obtain for the effect of {i|τ} on the various trigonometric functions
in Table 16.8:
{i|τ} cos(2π
a
x) = cos(2π
a
(−x) + π
2
)=sin(2π
a
x)
{i|τ} sin(2π
a
x) = sin(2π
a
(−x) + π
2
)=cos(2π
a
x)
{i|τ} cos(4π
a
y) = cos(4π
a
(−y) + π)=− cos(4π
a
y)
{i|τ} sin(4π
a
y) = sin(4π
a
(−y) + π)=sin(4π
a
y)
{i|τ} sin(4π
a
(y + z)) = sin(4π
a
(−y − z) + 2π)=− sin(4π
a
(y + z))
{i|τ} sin(4π
a
(y − z)) = sin(4π
a
(−y + z))=− sin(4π
a
(y − z)).
(16.37)
16.5. EFFECT OF GLIDE PLANES AND SCREW AXES 473
Thus we obtain
{i|τ}
x11
x12
=
cos(2π
a
(−x) + π
2
)
sin(2π
a
(−x) + π
2
)
=
sin(2π
a
x)
cos(2π
a
x)
=
x12
x11
(16.38)
and we see that the effect of {i|τ} is to interchange x11 ↔ x12. Similarly
the effect of {i|τ} on x12 and x22 is
{i|τ}
x21
x22
=
− sin(2π
a
(x))[cos(4π
a
y) − cos(4π
a
z)]
− cos(2π
a
(x))[cos(4π
a
y) − cos(4π
a
z)]
=
−x22
−x21
(16.39)
so that {i|τ} in this case interchanges the functions and reverses their
signs x21 ↔ −x22. Correspondingly, the other symmetry operations
involving translations also interchange the basis functions for the X1
and X2 irreducible representations. The physical meaning of this phenomenon
is that the energy bands EX1 (k) and EX2 (k) go right through
the X point without interruption in the extended zone scheme, except
for an interchange in the symmetry designations of their basis functions
in crossing the X point, consistent with the E(k) diagram for Ge where
bands with X1 symmetry are seen.
In contrast, the effect of {i|τ} on the x31 and x32 basis functions:
{i|τ}
x31
x32
=
− sin(4π
a
(y + z))[sin(2π
a
x) + cos(2π
a
x)]
− sin(4π
a
(y − z))[sin(2π
a
x) − cos(2π
a
x)]
=
−x31
x32
(16.40)
does not interchange x31 and x32. Thus the X3 level comes into the X
point with zero slope. The behavior for the X4 levels is similar
{i|τ}
x41
x42
=
− sin(4π
a
(y − z))[sin(2π
a
x) + cos(2π
a
x)]
− sin(4π
a
(y + z))[sin(2π
a
x) − cos(2π
a
x)]
=
−x41
x42
(16.41)
so that the X3 and X4 levels behave like doubly degenerate levels.
Equations 16.38 – 16.41 show that the character χ({i|τ}) vanishes at the
X point for the X1, X2, X3 and X4 levels, consistent with the character
table for the diamond X-point given in Table 16.7. These results also
explain the behavior of the energy bands for Ge at the X-point shown in
Fig. 16.9. The non-degenerate ∆1 and ∆2 energy bands going into the
X point stick together and interchange their symmetry designations
474 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
on crossing the X point, while the doubly degenerate ∆5 levels go
into a doubly-degenerate X4 level with zero slope at the Brillouin zone
boundary. The doubly-degenerate X5 levels are split by the spin-orbit
interaction into ∆6 and ∆7 levels, and when spin-orbit interaction is
taken into account all the levels at the X-point have X5 symmetry and
show sticking-together properties.
16.6 Selected Problems
1. (a) For the simple cubic lattice find the proper linear combinations
of plane waves for the twelve (110) plane wave states
at k = 0 which transform as irreducible representations of
the Oh point group.
(b) As we move away from k = 0, find the plane wave eigenfunctions
which transform according to ∆1 and ∆5 and are
compatible with the eigenfunctions for the Γ−
15 level at k = 0.
(c) Repeat part (b) for the case of Γ−
12 → ∆1 + ∆2.
2. (a) Considering the empty lattice model for the 2D triangular
lattice (space group # 17 p6mm), find the symmetries of the
two lowest energy states at the Γ point (k = 0).
(b) Find the linear combination of plane waves that transform
according to the irreducible representations in part (a).
(c) Repeat (a) and (b) for the lowest energy state at the M
point shown in the diagram below.
 ¢¡¡£
 ¢¡£
¤
¥
¦
¥
§¨
3. Using the empty lattice, find the energy eigenvalues, degeneracies
and symmetry types for the two electronic levels of lowest energy
for the fcc lattice at the L point.
16.6. SELECTED PROBLEMS 475
(a) Find the appropriate linear combinations of plane waves
which provide basis functions for the two lowest L-point electronic
states for the fcc lattice.
(b) Which states of the lower and upper energy levels in (a) are
coupled by optical dipole transitions?
(c) Repeat parts (a, b, c) for the two lowest X point energy
levels for the fcc empty lattice (i.e., the X1, X4 and X1, X3,
X5 levels).
(d) Compare your results to those for the simple cubic lattice.
476 CHAPTER 16. ENERGY LEVELS IN CUBIC CRYSTALS
Chapter 17
Energy Band Models Based
on Symmetry
Chapter 16 addressed the general application of space groups to the
one-electron energy bands in a periodic solid. This chapter deals with
some specific models that make extensive use of the crystal symmetry.
17.1 Introduction
Just from the symmetry properties of a particular crystal, a good deal
can be deduced concerning the energy bands of a solid. Our study of the
group of the wave vector illustrates that questions such as degeneracy
and connectivity are answered by group theory alone. It is not necessary
to solve Schr¨odinger’s equations explicitly to find the degeneracies
and connectivity relations for En(k).
There are two interpolation or extrapolation techniques for energy
band dispersion relations that are based on symmetry and provide the
form of En(k) without actual solution of Schr¨odinger’s equation. These
methods are useful as interpolation schemes for experimental data or
for band calculations that are carried out with great care at a few
high symmetry points in the Brillouin zone. These methods are called
k · p perturbation theory [extrapolation or Taylor’s series expansion of
E(k)] and the Slater–Koster Fourier expansion [interpolation or Fourier
series expansion of E(k)]. Both of these methods are discussed in this
477
478CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
Chapter.
The particular example that is used here to illustrate k · p perturbation
theory is the electronic structure for a material with simple
cubic symmetry. This discussion is readily extended to the electronic
structure of semiconductors that crystallize in the diamond structure
(e.g., silicon). The valence and conduction bands for these semiconductors
are formed from hybridized s- and p-bands. For the diamond
structure, the s- and p-functions in the Oh point group (at k = 0)
transform as the Γ+
1 and Γ−
15 irreducible representations, respectively.
In the diamond structure there are 2 atoms/unit cell and χatom sites at
k = 0 transforms as Γ+
1 + Γ−
2 or (A1g + A2u). Thus we must consider
8 bands in discussing the valence and conduction bands formed by sand
p-bands. These bands have symmetries
χa.s. ⊗ χs−functions (Γ+
1 + Γ−
2 ) ⊗ Γ+
1 = Γ+
1 + Γ−
2
χa.s. ⊗ χp−functions (Γ+
1 + Γ−
2 ) ⊗ Γ−
15 = Γ−
15 + Γ+
25
(17.1)
We identify the Γ+
1 and Γ+
25 bands as the bonding s- and p-bands and
the Γ−
2 and Γ−
15 bands as antibonding s- and p-bands. The reason why
the bonding p-band has Γ+
25 symmetry follows from the direct product
Γ−
2 ⊗ Γ−
15 = Γ+
25 in Eq. (17.1).
Our discussion starts with a review of k · p perturbation theory in
general. An example of k · p perturbation theory for a non-degenerate
level is then given. This is followed by an example of degenerate secondorder
k · p perturbation theory which is appropriate to the p-bonding
and antibonding levels in the diamond structure. In all of these cases,
group theory tells us which are the non-vanishing matrix elements and
which matrix elements are equal to each other.
17.2 k · p Perturbation Theory
An electron in a periodic potential obeys the one-electron Hamiltonian:
p2
2m
+ V (r) ψn,k(r) = En(k)ψn,k(r) (17.2)
where the eigenfunctions of the Hamiltonian are the Bloch functions
ψn,k(r) = eik·r
un,k(r) (17.3)
17.2. K · P PERTURBATION THEORY 479
and n is the band index.
Substitution of ψn,k(r) into Schr¨odinger’s equation gives an equation
for the periodic function un,k(r)


p2
2m
+ V (r) +
¯hk · p
m
+
¯h2
k2
2m

 un,k(r) = En(k) un,k(r). (17.4)
In the spirit of the (k ·p) method we assume that En(k) is known at
point k = k0 either from experimental information or from direct solution
of Schr¨odinger’s equation for some model potential V (r). Assume
the band in question has symmetry Γi so that the function un,k0
(r)
transforms as the irreducible representation Γi. Then we have
Hk0
u
(Γi)
n,k0
= εn(k0) u
(Γi)
n,k0
(17.5)
where
Hk0
=
p2
2m
+ V (r) +
¯hk0 · p
m
(17.6)
and
εn(k0) = En(k0) −
¯h2
k2
0
2m
. (17.7)
If εn(k0) and un,k0
(r) are specified at k0, the k · p method prescribes
the development of the periodic un,k0
(r) functions under variation of k.
At point k = k0 + κ, the eigenvalue problem becomes:
Hk0+κun,k0+κ(r) = Hk0
+ ¯hκ·p
m
un,k0+κ(r)
= εn k0 + κ un,k0+κ(r).
(17.8)
In the spirit of the usual k · p perturbation theory κ is small so that
the perturbation Hamiltonian is taken as H = ¯hκ · p/m and the energy
eigenvalue at the displaced k vector εn(k0 + κ) is given by Eq. (17.7),
and En(k0) is given by Eq. (17.2). We will illustrate this method first
for a non-degenerate band (a Γ±
1 band for the simple cubic lattice) and
then for a degenerate band (a Γ±
15 band for the simple cubic lattice).
480CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
17.3 k · p perturbation theory for a nondegenerate
Γ±
1 Band–Simple Cubic Lat-
tice
Suppose the energy of the Γ±
1 band at k = 0 is established by the
identification of an optical transition and measurement of its resonant
photon energy. The unperturbed wave function at k = 0 is uΓ1
n,0(r) and
its eigenvalue from Eq. (17.7) is ε(Γ1)
n (0) = E(Γ1)
n (0) since k0 = 0. Then
ε(Γ1)
n (κ) = E(Γ1)
n (0) + uΓ1
n,0|H |uΓ1
n,0 +
n =n
uΓ1
n,0|H |uΓi
n ,0 uΓi
n ,0|H |uΓ1
n,0
EΓ1
n (0) − EΓi
n (0)
(17.9)
where the sum is over states n which have symmetries Γi.
Now H = ¯hκ · p/m transforms like a vector since H is proportional
to the vector p, which pertains to the electronic system. If we expand
the eigenfunctions and eigenvalues of Eq. (17.9) about the Γ point (k =
0), then H transforms according to the irreducible representation Γ−
15
in Oh symmetry. In the spirit of k ·p perturbation theory, the vector k0
determines the point symmetry group that is used to classify the wave
functions and eigenvalues for the perturbation Hamiltonian.
For the k·p expansion about the Γ point, the linear term in k which
arises in 1st
order perturbation theory vanishes since u
Γ+
1
n,0|H |u
Γ+
1
n,0
transforms according to the direct product Γ+
1 ⊗ Γ−
15 ⊗ Γ+
1 = Γ−
15 which
does not contain Γ+
1 . The same result is obtained using arguments
relevant to the oddness and evenness of the functions which enter the
matrix elements of Eq. (17.9). At other points in the Brillouin zone,
the k · p expansion may contain linear k terms.
Now let us look at the terms uΓi
n ,0|H |u
Γ+
1
n,0 that arise in 2nd
order
perturbation theory. The product H u
Γ+
1
n,0 transforms as Γ−
15 ⊗ Γ+
1 = Γ−
15
so that Γi must be of Γ−
15 symmetry if a non-vanishing matrix is to
17.3. K · P PERTURBATION THEORY IN SC LATTICE 481
result. We thus obtain:
εΓ+
1
n (κ) = EΓ+
1
n (0) + Σn =n (Γ−
15)
u
Γ+
1
n,0|H |u
Γ−
15
n ,0 u
Γ−
15
n ,0|H |u
Γ+
1
n,0
E
Γ+
1
n (0) − E
Γ−
15
n (0)
+ . . .
(17.10)
and a corresponding relation is obtained for the non-degenerate Γ−
2
level. Thus we see that by using group theory, our k · p expansion is
greatly simplified, since it is only the Γ−
15 levels that couple to the Γ+
1
level by k · p perturbation theory in Eq. (17.10). These statements
are completely independent of the explicit wave functions which enter
the problem, but depend only on the symmetry. Further simplifications
result from the observation that for cubic symmetry the matrix
elements u
Γ+
1
n,0|H |u
Γ−
15
n ,0 can all be expressed in terms of a single matrix
element, if u
Γ−
15
n ,0 is identified with specific basis functions such as
p-functions (denoted by x, y, z for brevity) and u
Γ+
1
n,0 with an s-function
(denoted by 1 for brevity). Thus for the Oh group, the selection rules
(see §7.5) give
(1|px|x) = (1|py|y) = (1|pz|z) (17.11)
and all other cross terms of the form (1|px|y) vanish. This result that
the matrix elements of p in Oh symmetry have only one independent
matrix element also follows from the discussion in Chapter 11. Combining
these results with ε
Γ+
1
n (κ) = E
Γ+
1
n (κ) − ¯h2
κ2
2m
we get
EΓ+
1
n (κ) = EΓ+
1
n (0) +
¯h2
κ2
2m
+
¯h2
κ2
m2
n =n
|(1|px|x)|2
E
Γ+
1
n (0) − E
Γ−
15
n (0)
(17.12)
where the sum is over all states n with Γ−
15 symmetry. A similar expansion
formula is applicable to E
Γ−
2
n (k) which corresponds to the conduction
antibonding s-band in the diamond structure. Equation 17.12
is sometimes written in the form
EΓ+
1
n (κ) = EΓ+
1
n (0) +
¯h2
κ2
2m∗
n
(17.13)
482CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
where the effective mass parameter m∗
n is related to band couplings
through the momentum matrix element:
m
m∗
n
= 1 +
2
m n =n
|(1|px|x)|2
E
Γ+
1
n (0) − E
Γ−
15
n (0)
(17.14)
in which the sum over n is restricted to states with Γ−
15 symmetry.
Consistent with Eq. (17.12), the effective mass m∗
n is related to the
band curvature by the relation
∂2
E
Γ+
1
n (κ)
∂κ2
=
¯h2
m∗
n
. (17.15)
Thus m∗
n is proportional to the inverse of the band curvature. If the
curvature is large, the effective mass is small and conversely, if the
bands are “flat” (essentially k-independent), the effective masses are
large. Thus the k · p expansion for a non-degenerate band in a cubic
crystal leads to an isotropic parabolic dependence of En(k) on k which
looks just like the free electron dispersion relation except that m is
replaced by m∗
.
For the case of a single band with Γ−
15 symmetry, the formula for
the effective mass [Eq. (17.14)] becomes
m
m∗
n
= 1 +
2
m
|(1|px|x)|2
εg
(17.16)
which is useful for estimating effective masses, provided that we know
the magnitude of the matrix element and the band gap εg. On the
other hand, if m∗
and εg are known experimentally, then Eq. (17.16) is
useful for evaluating |(1|px|x)|2
. This is, in fact, the most common use
of Eq. (17.16). The words matrix element or oscillator strength
typically refer to the momentum matrix element (un,k|px|un ,k) when
discussing the optical properties of solids.
The treatment given here for the non-degenerate bands is easily
carried over to treating the k · p expansion about some other high symmetry
point in the Brillouin zone. For arbitrary points in the Brillouin
zone, the diagonal term arising from 1st
order perturbation theory does
17.4. TWO BAND MODEL IN PERTURBATION THEORY 483
not vanish. Also the matrix element (u
Γ±
i
n,k0
|pα|u
Γj
n,k0
) need not be the
same for each component α = x, y, z, and for the general case six independent
matrix elements would be expected. For example, along the ∆
and Λ axes, the matrix element for momentum to the high symmetry
axis is not equal to the components ⊥ to the axis, and there are two
independent matrix elements along each of the ∆ and Λ axes. These
two directions are called longitudinal ( to the axis) and transverse
(⊥ to the axis), and lead to longitudinal and transverse effective mass
components away from the Γ point.
17.4 Two Band Model—Degenerate FirstOrder
Perturbation Theory
One of the simplest applications of k · p perturbation theory is to twoband
models for solids. These models are applicable to describe the
energy E(k) about a point k0 for one of two bands that are strongly
coupled to each other and are weakly coupled to all other bands. The
strongly coupled set is called the nearly degenerate set (NDS) and, if
need be, the weakly coupled bands can always be treated in perturbation
theory after the problem of the strongly interacting bands is
solved. Simple extensions of the 2-band model are made to handle 3
strongly coupled bands, such as the valence band of silicon, germanium
and related semiconductors or even 4 strongly coupled bands as occur
in graphite.
The eigenvalue problem to be solved is


p2
2m
+ V (r) +
¯hk0 · p
m
+
¯hκ · p
m

 un,k0+κ(r) = εn k0 + κ un,k0+κ(r)
(17.17)
in which εn(k0) is related to the solution of Schr¨odinger’s equation
En(k0) by Eq. (17.7).
Let n = i, j be the two bands that are nearly degenerate. Using first
order degenerate perturbation theory, the secular equation is written
484CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
as
i j
i
j
i|H0 + H |i − ε
j|H0 + H |i
i|H0 + H |j
j|H0 + H |j − ε
= 0
(17.18)
in which we have explicitly written i and j to label the rows and
columns. Equation 17.18 is exact within the 2-band model, i.e., all
the coupling occurs between the nearly degenerate set and no coupling
is made to bands outside this set. For most cases where the 2-band
model is applied (e.g., bismuth, InSb, PbTe), the unperturbed wave
functions un,k0
(r) are invariant under inversion. Then because of the
oddness of H = ¯hκ · p/m, the matrix elements vanish
i|H |i = j|H |j = 0. (17.19)
Also since the “band edge” wave functions un,k0
(r) are constructed to
diagonalize the Hamiltonian
H0un,k0
(r) = εn(k0)un,k0
(r) (17.20)
there are no off-diagonal matrix elements of H0 or
i|H0|j = 0 for i = j. (17.21)
We then write
i|H0|i = E0
i and j|H0|j = E0
j (17.22)
where for n = i, j
E0
n = En(k0) −
¯h2
k2
0
2m
. (17.23)
In this notation the secular equation can be written as
E0
i − ε (¯h/m)κ · i|p|j
(¯h/m)κ · j|p|i E0
j − ε
= 0 (17.24)
where i|p|j = 0 for the 2-band model. The secular equation implied
by Eq. (17.24) is equivalent to the quadratic equation
ε2
− ε E0
i + E0
j + E0
i E0
j −
¯h2
m2
κ · i|p|j j|p|i · κ = 0. (17.25)
17.4. TWO BAND MODEL IN PERTURBATION THEORY 485
 ¢¡
£¥¤¥¦ §©¨§¤¥¨
¨! "¨#¤"¨ Figure 17.1: Two strongly
coupled mirror bands separated
by an energy εg at the
band extremum.
We write the symmetric tensor
↔
p2
ij coupling the 2 bands as:
↔
p2
ij= i|p|j j|p|i (17.26)
where i and j in the matrix elements refer to the band edge wave
functions un,k0
(r) and n = i, j. The solution to the quadratic equation
[Eq. (17.25)] yields
ε(κ) =
E0
i + E0
j
2
±
1
2
(E0
i − E0
j )2 +
4¯h2
m2
κ·
↔
p2
ij ·κ. (17.27)
We choose our zero of energy symmetrically such that
E0
i = εg/2 E0
j = −εg/2 (17.28)
to obtain the two-band model result (see Fig. 17.1):
ε(κ) = ±
1
2
ε2
g +
4¯h2
m2
κ·
↔
p2
ij ·κ (17.29)
which at κ = 0 reduces properly to ε(0) = ±1
2
εg.
Equation 17.29 gives a non-parabolic dependence of E upon
κ. For strongly coupled bands, the 2-band model is characterized by
its non-parabolicity. In the approximation that there is no coupling to
bands outside the non-degenerate set, these bands are strictly mirror
bands—one band is described by an E(κ) relation given by the ⊕ sign;
the other by the identical relation with the sign.
486CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
For cubic materials there is only one independent matrix element
↔
p2
ij= i|pα|j j|pα|i ≡ p2
ij α = x, y, z (17.30)
and the
↔
p2
ij tensor assumes the form
↔
p2
ij=



p2
ij 0 0
0 p2
ij 0
0 0 p2
ij


 . (17.31)
In applying the two-band model to cubic symmetry, the degeneracy
of the Γ+
25 valence bands or the Γ−
15 conduction bands is ignored. The
2-band model formula then becomes
ε(κ) = ±
1
2
ε2
g +
4¯h2
κ2p2
ij
m2
where κ2
= κ2
x + κ2
y + κ2
z. (17.32)
In this form Eq. (17.32) is called the Kane 2-band model. The generalization
of Eq. (17.32) to non-cubic materials is usually called the
Lax 2-band model and in the case of bismuth the
↔
p2
ij tensor has the
following form
↔
p2
ij=



p2
xx 0 0
0 p2
yy p2
yz
0 p2
yz p2
zz


 (17.33)
where the x axis is a binary axis ⊥ to the mirror plane in bismuth
(space group R3m, #166), and the matrix elements of Eq. (17.33) have
4 independent components.
We now show that for small κ we recover the parabolic ε(κ) relations.
For example, for the Kane 2-band model a Taylor’s series
expansion of Eq. (17.32) yields
ε(κ) = ±
1
2
ε2
g +
4¯h2
κ2p2
ij
m2
= ±
εg
2
1 +
4¯h2
κ2
p2
ij
ε2
gm2
1
2
(17.34)
which to order κ4
becomes:
ε(κ) = ±
εg
2
+
¯h2
κ2
p2
ij
εgm2
−
¯h4
κ4
p4
ij
ε3
gm4
+ . . . (17.35)
17.4. TWO BAND MODEL IN PERTURBATION THEORY 487
where ε(κ) is given by Eq. (17.7), and the momentum matrix elements
are given by
p2
ij = |(1|px|x)|2
(17.36)
and the bandgap at the band extrema by En(k0) − En (k0) = ±εg.
If the power series expansion in Eq. (17.35) is rapidly convergent
(either because κ is small or the bands are not that strongly coupled–
i.e., p2
ij is not too large), then the expansion through terms in κ4
is
useful. We note that within the two-band model the square root formula
of Eq. (17.34) is exact and is the one that is not restricted to small κ
or small p2
ij. It is valid so long as the 2-band model itself is valid.
Some interesting consequences arise from these non-parabolic features
of the dispersion relations. For example, the effective mass (or
band curvature) is energy or κ-dependent. Consider the expression
which follows from Eq. (17.35):
En(k0 + κ)
¯h2
|k0 + κ|2
2m
±
εg
2
+
¯h2
κ2
p2
ij
εgm2
−
¯h4
κ4
p4
ij
ε3
gm4
. (17.37)
Take k0 = 0, so that
∂2
E
∂κ2
=
¯h2
m
±
2¯h2
p2
ij
εgm2
−
12κ2
¯h4
p4
ij
ε3
gm4
≡
¯h2
m∗
. (17.38)
From this equation we see that the curvature ∂2
E/∂κ2
is κ dependent.
In fact as we more further from the band extrema, the band curvature
decreases, the bands become more flat and the effective mass increases.
This result is also seen from the definition of m∗
[Eq. (17.38)]
m
m∗
= 1 ±
2
m
p2
ij
εg
−
12¯h2
κ2
p4
ij
ε3
gm3
. (17.39)
Another way to see that the masses become heavier as we move
higher into the band (away from k0) is to work with the square root
formula Eq. (17.34):
ε = ±
1
2
ε2
g +
4¯h2
κ2p2
ij
m2
. (17.40)
488CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
Squaring Eq. (17.40) and rewriting this equation we obtain:
(2ε − εg)(2ε + εg) =
4¯h2
κ2
p2
ij
m2
(17.41)
(2ε − εg) =
4¯h2
κ2
p2
ij
m2(2ε + εg)
. (17.42)
For κ = 0 we have ε = εg/2, and we then write an expression for ε(κ):
ε =
εg
2
+
2¯h2
κ2
p2
ij
m2(2ε + εg)
=
εg
2
+
¯h2
κ2
p2
ij
m2(ε + εg
2
)
. (17.43)
Therefore we obtain the non-parabolic two-band model relation
E(κ) =
εg
2
+
¯h2
κ2
2m
1 +
2p2
ij
m(ε + εg
2
)
(17.44)
which is to be compared with the result for simple non-degenerate bands
[Eq. (17.12)]:
Ei(κ) = Ei(0) +
¯h2
κ2
2m
1 +
2p2
ij
mεg
. (17.45)
Equation (17.44) shows that for the non-parabolic 2-band model, the
effective mass at the band edge is given by
m
m∗
= 1 +
2p2
ij
mεg
. (17.46)
and the effective mass becomes heavier as we move away from k0 and
as we move up into the band. The magnitude of the k or energy dependence
of the effective mass is very important in narrow gap materials
such as bismuth. At the band edge, the effective mass parameter for
electrons in Bi is ∼ 0.001m0 whereas at the Fermi level m∗
∼ 0.008m0.
The number of electron carriers in Bi is only 1017
/cm3
. Since the density
of states for simple bands has a dependence ρ(E) ∼ m∗3
2 E
1
2 , we can
expect a large increase in the density of states with increasing energy
in a non-parabolic band. Since bismuth has relatively low symmetry,
the tensorial nature of the effective mass tensor must be considered and
17.5. DEGENERATE K · P PERTURBATION THEORY 489
the dispersion relations for the coupled bands at the L point in bismuth
are generally written as
ε(κ) = ±
1
2
ε2
g + 2¯h2
εg
κ·
↔
α ·κ
m
(17.47)
in which
↔
α is a reciprocal effective mass tensor.
17.5 Degenerate k·p Perturbation Theory
For many cubic crystals it is common to have triply-degenerate energy
bands at k = 0. Such bands are of great importance in the transport
properties of semiconductors such as silicon, germanium, and the III-V
compounds. The analysis of experiments such as cyclotron resonance
in the valence band of these semiconductors depends upon degenerate
second-order k·p perturbation theory which is discussed in this section.
Second-order degenerate k · p perturbation theory becomes a good
deal more complicated than the simpler applications of perturbation
theory discussed in §17.2–§17.4. Group theory thus provides a valuable
tool for the solution of practical problems. As an example, we will
consider how the degeneracy is lifted as we move away from k = 0 for
a Γ−
15 level; a similar analysis applies for the Γ+
25 level.
Suppose that we set up the secular equation for a Γ−
15 level using
degenerate perturbation theory
x y z
x
y
z
(x|H |x) − ε
(y|H |x)
(z|H |x)
(x|H |y)
(y|H |y) − ε
(z|H |y)
(x|H |z)
(y|H |z)
(z|H |z) − ε
= 0
(17.48)
where the x, y and z symbols denote the (x, y, z) partners of the basis
functions in the Γ−
15 irreducible representation derived from atomic pfunctions
and the diagonal matrix elements for H0 are set equal to zero
at the band extremum, such as the top of the valence band. We notice
that since H = ¯hk · p/m, then H transforms like the Γ−
15 irreducible
representation and (Γ−
15|H |Γ−
15) = 0 since
Γ−
15 ⊗ Γ−
15 = Γ+
1 + Γ+
12 + Γ+
15 + Γ+
25 (17.49)
490CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
 
 
¡¢ £¥¤§¦
¨
Figure 17.2: NDS ≡ nearly degenerate
set. We use Roman letter subscripts
for levels within the NDS
and Greek indices for levels outside
of the NDS.
or more simply since H is odd under inversion each matrix element
in Eq. (17.48) vanishes because of parity considerations. Since each of
the matrix elements of Eq. (17.48) vanishes, the degeneracy of the Γ−
15
level is not lifted in first-order degenerate perturbation theory; thus we
must use second-order degenerate perturbation theory to lift this level
degeneracy. We show below the derivation of the form of the matrix
elements for the off-diagonal matrix elements in Eq. (17.48 showing
that the vanishing Hmn is replaced by
Hmn → Hmn +
α
HmαHαn
E
(0)
m − E
(0)
n
. (17.50)
We will see below that the states given in Eq. (17.49 will serve as the
intermediate states α which arise in second-order perturbation theory.
In applying second-order degenerate perturbation theory, we assume
that we have a degenerate (or nearly degenerate) set of levels–
abbreviated NDS. We assume that the states inside the NDS are strongly
coupled and those outside the NDS are only weakly coupled to states
within the NDS. (See Fig. 17.2)
The wave function for a state is now written in terms of the unperturbed
wave functions and the distinction is made as to whether we are
dealing with a state inside or outside of the NDS. If we now expand
the wavefunction ψn in terms of the unperturbed band edge states, we
obtain:
ψn =
n
anψ(0)
n +
α
aαψ(0)
α (17.51)
17.5. DEGENERATE K · P PERTURBATION THEORY 491
where ψ(0)
n and ψ(0)
α are, respectively, the unperturbed wavefunctions
inside (n) and outside (α) of the nearly degenerate set. Substitution
into Schr¨odinger’s equation yields
Hψn = Eψn =
n
an(E0
n + H )ψ(0)
n +
α
aα(E(0)
α + H )ψ(0)
α . (17.52)
We multiply the left hand side of Eq. (17.52) by ψ(0)∗
m0
and integrate
over all space, making use of the orthogonality theorem ψ(0)∗
m ψ(0)
n dr =
δmn to obtain the iterative relation between the expansion coefficients
(Brillouin-Wigner Perturbation Theory):
[E − E(0)
m ]am = amHmm +
n =m
an Hmn +
α
aαHmα (17.53)
where the sum over n denotes coupling to states in the NDS and the
sum over α denotes coupling to states outside the NDS. A similar procedure
also leads to a similar equation for levels outside the NDS:
[E − E(0)
α ]aα = aαHαα +
n
anHαn +
β=α
aβHαβ. (17.54)
We now substitute Eq. (17.54) for the coefficients aα outside the NDS
in Eq. (17.53) to obtain:
[E − E(0)
m ]am = amHmm + n =m an Hmn
+ α
Hmα
E−E
(0)
α
n anHαn + aαHαα + β aβHαβ .
(17.55)
If we neglect terms in Eq. (17.55) which couple states outside the NDS
to other states outside the NDS, we obtain:
am(E(0)
m − E) +
n
anHmn +
n
an
α
HmαHαn
E
(0)
m − E
(0)
α
= 0 (17.56)
in which the first sum is over all n without restriction, and for E in the
denominator of the second-order perturbation term in Eq. (17.55) we
replace E by E(0)
m in the spirit of perturbation theory. Equation 17.56
then implies the secular equation
n
n=1
an (E(0)
m − E)δmn + Hmn +
α
HmαHαn
E
(0)
m − E
(0)
α
= 0 (17.57)
492CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
which yields an n × n secular equation with each matrix element given
by
Hmn +
α
HmαHαn
E
(0)
m − E
(0)
α
. (17.58)
In degenerate k · p perturbation theory we found that Hmn = 0
for a Γ−
15 level, and it was for this precise reason that we had to go
to degenerate second order perturbation theory. In this case, each
state in the NDS couples to other states in the NDS only through an
intermediate state outside of the NDS.
In second-order degenerate perturbation theory Eq. (17.49) shows
us that for a 3-fold Γ−
15 level k · p degenerate perturbation theory will
involve only states of Γ+
1 , Γ+
12, Γ+
15, or Γ+
25 symmetry as intermediate
states. In our discussion of non-degenerate k·p perturbation theory (see
§17.3) we found that there was only one independent matrix element
of p coupling a Γ+
1 state to a Γ−
15 state. We include below a useful table
of matrix elements of p between states of different symmetries for Γ
point levels in cubic crystals. These matrix elements are found using
the basis functions for each of the irreducible representations of Oh
given in Table 16.2. Table 17.1 lists the non-vanishing matrix elements
appearing in the k · p perturbation theory for electronic energy bands
with cubic Oh symmetry.
For the matrix element A2 in Table 17.1 we note with the help
of Table 13.2 that the pertinent basis functions are Γ−
2 = xyz and
Γ+
25,x = yz. For A4 we note that the basis function Γ−
25,z = z(x2
− y2
)
gives C2Γ−
25,z = −Γ−
25,z where C2 is a rotation of π around the (011) axis.
For A5 we use as basis functions: Γ−
15,x = x and Γ+
15,x = yz(z2
−y2
) which
is odd under the interchange y ↔ z. For A6 we use as basis functions:
Γ+
25,x = yz and Γ−
15,x = x, where A6 = (Γ±
15,x|py|Γ25,z). For A7 we use as
basis functions: Γ+
25,x = yz; Γ−
25,x = x(y2
− z2
); Γ−
25,z = z(x2
− y2
).
Let us make a few comments on this table of matrix elements for
H . Since H is odd, only states of opposite parity are coupled. For
each of the 7 symmetry type couplings given in the table, there is only
one independent matrix element. Likewise the coupling between the
Γ+
12 and Γ−
15 representations involve 2 × 3 × 3 = 18 matrix elements but
there is only one independent matrix element:
(x|px|f1) = (x|px|f2) = ω(y|py|f1) = ω2
(y|py|f2) = ω2
(z|pz|f1) = ω(z|pz|f2)
17.5. DEGENERATE K · P PERTURBATION THEORY 493
Table 17.1: Matrix Element Table for H = ¯hk · p/m in cubic Oh Sym-
metry
(Γ±
1 |H |Γ15,α) = A1
¯h
m
kα A1 = (Γ±
1 |px|Γ15,x)
(Γ±
2 |H |Γ25,α) = A2
¯h
m
kα A2 = (Γ±
2 |px|Γ25,x)
(Γ±
12,1|H |Γ15,x) = A3
¯h
m
kx
(Γ±
12,1|H |Γ15,y) = A3
¯h
m
kyω2
(Γ±
12,1|H |Γ15,z) = A3
¯h
m
kzω



A3 = (f±
1 |px|Γ15,x)
f1 = f∗
2 = x2 + ωy2 + ω2z2
(Γ±
12,2|H |Γ15,x) = A∗
3
¯h
m
kx ω = exp(2πi/3)
(Γ±
12,2|H |Γ15,y) = A∗
3
¯h
m
kyω
(Γ±
12,2|H |Γ15,z) = A∗
3
¯h
m
kzω2
(Γ±
12,1|H |Γ25,x) = A4
¯h
m
kx A4 = (f1|px|Γ25,x)
(Γ±
12,1|H |Γ25,y) = A4
¯h
m
kyω2 f1 = f∗
2 = x2 + ωy2 + ω2z2
(Γ±
12,1|H |Γ25,z) = A4
¯h
m
kzω
(Γ±
12,2|H |Γ25,x) = A∗
4
¯h
m
kx
(Γ±
12,2|H |Γ25,y) = A∗
4
¯h
m
kyω
(Γ±
12,2|H |Γ25,z) = A∗
4
¯h
m
kzω2



(Γ±
15,x|H |Γ15,x) = 0
(Γ±
15,x|H |Γ15,y) = −A5
¯h
m
kz
(Γ±
15,x|H |Γ15,z) = A5
¯h
m
ky
A5 = (Γ±
15,y|px|Γ15,z)



(Γ±
15,y|H |Γ15,x) = A5
¯h
m
kz
(Γ±
15,y|H |Γ15,y) = 0
(Γ±
15,y|H |Γ15,z) = −A5
¯h
m
kx



(Γ±
15,z|H |Γ15,x) = −A5
¯h
m
ky
(Γ±
15,z|H |Γ15,y) = A5
¯h
m
kx
(Γ±
15,z|H |Γ15,z) = 0



(Γ±
15,x|H |Γ25,x) = 0
(Γ±
15,x|H |Γ25,y) = A6
¯h
m
kz
(Γ±
15,x|H |Γ25,z) = A6
¯h
m
ky
A6 = (Γ±
15,x|py|Γ25,z)



(Γ±
15,y|H |Γ25,x) = A6
¯h
m
kz
(Γ±
15,y|H |Γ25,y) = 0
(Γ±
15,y|H |Γ25,z) = A6
¯h
m
kx



(Γ±
15,z|H |Γ25,x) = A6
¯h
m
ky
(Γ±
15,z|H |Γ25,y) = A6
¯h
m
kx
(Γ±
15,z|H |Γ25,z) = 0



(Γ±
25,x|H |Γ25,x) = 0
(Γ±
25,x|H |Γ25,y) = −A7
¯h
m
kz
(Γ±
25,x|H |Γ25,z) = A7
¯h
m
ky
A7 = (Γ±
25,x|py|Γ25,z)



(Γ±
25,y|H |Γ25,x) = A7
¯h
m
kz
(Γ±
25,y|H |Γ25,y) = 0
(Γ±
25,y|H |Γ25,z) = −A7
¯h
m
kx



(Γ±
25,z|H |Γ25,x) = −A7
¯h
m
ky
(Γ±
25,z|H |Γ25,y) = A7
¯h
m
kx
(Γ±
25,z|H |Γ25,z) = 0
+ denotes even and − denotes odd states under inversion.
494CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
and all others vanish. Here we write
f1 = x2
+ ωy2
+ ω2
z2
f2 = x2
+ ω2
y2
+ ωz2 (17.59)
as the basis functions for the Γ+
12 representation. For Γ+
25 symmetry we
can take our basis functions as



yz
zx
xy
which in the table are denoted by



(Γ+
25,x)
(Γ+
25,y)
(Γ+
25,z)
The three Γ+
25 basis functions are derived from 3 of the 5 atomic d
functions, the other two being Γ+
12 functions. Using these results for
the matrix elements, the secular equation Eq. (17.48) can be written
as a function of kx, ky and kz to yield the dispersion relations for the
degenerate Γ−
15 bands as we move away from the Γ point k = 0 in the
Brillouin zone.
Since Γ−
15 ⊗ Γ−
15 = Γ+
1 + Γ+
12 + Γ+
15 + Γ+
25, and from Eq. (17.57), the
secular equation [Eq. (??)] for the Γ−
15 levels involves the following sums:
F =
¯h2
m2
Γ+
1 (n )
|(x|px|1)|2
E
Γ−
15
n (0) − E
Γ+
1
n (0)
G =
¯h2
m2
Γ+
12(n )
|(x|px|f1)|2
E
Γ−
15
n (0) − E
Γ+
12
n (0)
H1 =
¯h2
m2
Γ+
25
|(x|py|xy)|2
E
Γ−
15
n (0) − E
Γ+
25
n (0)
H2 =
¯h2
m2
Γ+
15
|(x|py|xy(x2
− y2
))|2
E
Γ−
15
n (0) − E
Γ+
15
n (0)
. (17.60)
We are now ready to solve the secular equation [Eq. (??)] using
Eq. (17.57 to include the various terms which occur in second-order
degenerate perturbation theory. Let us consider the diagonal entries
first, as for example the xx entry. We can go from an initial Γ−
15,x state
to the same final state through an intermediate Γ+
1 state which brings
17.5. DEGENERATE K · P PERTURBATION THEORY 495
down a k2
x term through the F term in Eq. (17.60). We can also couple
the initial Γ−
15 state to itself through an intermediate Γ+
12,1 or Γ+
12,2 state,
in either case bringing down a k2
x term through the G contribution–so
far we have Fk2
x + 2Gk2
x. We can also go from a Γ−
15,x state and back
again through a Γ+
25,y or Γ+
25,z state to give a (k2
y+k2
z )H1 contribution and
also through a Γ+
15 state to give a (k2
y + k2
z )H2 contribution. Therefore
on the diagonal xx entry we get
Lk2
x + M(k2
y + k2
z ) where L = F + 2G and M = H1 + H2. (17.61)
From this discussion we obtain the results for other diagonal entries yy
and zz, using a cyclic permutation of indices.
Now let us consider an off-diagonal entry such as (x|H |y), where
we start with an initial Γ−
15,x state and go to a final Γ−
15,y state. This
can be done through either of four intermediate states:
intermediate state Γ+
1 gives kxkyF
intermediate state Γ+
12 gives (ω2
+ ω)kxkyG = −kxkyG
intermediate state Γ+
15 gives −kxkyH2
intermediate state Γ+
25 gives kxkyH1
Therefore we get Nkxky = (F − G + H1 − H2)kxky for the total xy
entry. Using the same procedure we calculate the other four independent
entries to the secular equation. Collecting terms we have the final
result for the secular equation for the Γ−
15 degenerate p-band:
0 =
Lk2
x + M(k2
y + k2
z) − ε(k) Nkxky Nkxkz
Nkxky Lk2
y + M(k2
z + k2
x) − ε(k) Nkykz
Nkxkz Nkykz Lk2
z + M(k2
x + k2
y) − ε(k)
(17.62)
The secular equation [Eq. (17.62)] is greatly simplified along the high
symmetry directions. For a [100] axis, ky = kz = 0, and kx = κ, then
Eq. (17.62) reduces to
Lκ2
− ε(κ) 0 0
0 Mκ2
− ε(κ) 0
0 0 Mκ2
− ε(κ)
= 0 (17.63)
496CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
which has the roots
ε(κ) = Lκ2
ε(κ) = Mκ2
twice.
(17.64)
The result in Eq. (17.64) must be consistent with the compatibility
relations about the k = 0 Γ-point whereby
Γ+
15 = ∆1 + ∆5 (17.65)
in which the ∆1 level is non-degenerate and the ∆5 level is doubly
degenerate.
Along a Λ [111] axis, kx = ky = kz = κ and the general secular
equation of Eq. (17.62) simplifies into
(L + 2M)κ2
− ε(κ) Nκ2
Nκ2
Nκ2
(L + 2M)κ2
− ε(κ) Nκ2
Nκ2
Nκ2
(L + 2M)κ2
− ε(κ)
= 0
(17.66)
which can readily be diagonalized to give
ε(κ) =
L + 2M + 2N
3
κ2
once (Λ2 level)
ε(κ) =
L + 2M − N
3
κ2
twice (Λ3 level) (17.67)
where the Λ2 level is non-degenerate and the Λ3 level is doubly degen-
erate.
The secular equation for a general κ point is more difficult to solve,
but it can still be done in closed form by solving a cubic equation.
In practice, the problem is actually simplified by including the effects
of the electron spin (see Chapter 19). For each partner of the Γ−
15
levels we get a spin up state and a spin down state so that the secular
equation is now a (6×6) equation. However, we will see that spin-orbit
interaction simplifies the problem somewhat and the secular equation
can be solved analytically.
The band parameters L, M and N, which enter the secular equation
[Eq. (17.62)], express the strength of the coupling of the Γ−
15 levels to the
various other levels. In practice, these quantities are determined from
17.6. NON-DEGENERATE K · P PERTURBATION THEORY 497
experimental data. The cyclotron resonance experiment carried out
along various high symmetry directions provides accurate values for the
band curvatures and hence for the quantities L, M and N (Dresselhaus,
Kip and Kittel, Phys. Rev. 98, 386 (1955)). In the spirit of the k ·
p perturbation theory, solution of the secular equation provides the
most general form allowed by symmetry for E(k) about k = 0. The
solution reduces to the proper form along the high symmetry directions,
∆, Λ and Σ. However, group theory cannot provide information about
the magnitude of these coefficients. These magnitudes are most easily
obtained from experimental data.
The k · p method has also been used to obtain the energy bands
throughout the Brillouin zone for such semiconductors as silicon and
germanium (Cardona and Pollack, Phys. Rev. 142, 530 (1966)). In the
k·p approach of Cardona and Pollack, 7 other bands outside this “nearly
degenerate set” of 8 (Γ+
1 , Γ−
2 , Γ−
15, Γ+
25) are allowed to couple to these
eight bands. The secular equation (a 15 × 15 determinantal equation
in this case) is then constructed in standard first-order perturbation
theory.
New features in the problem arise in going from points of lower
symmetry to points of higher symmetry. Along the Λ or (111) axis, the
k · p expansion will connect a Λ point to an L point. The k · p method
has been made to work well and has been used for the analysis of many
experiments.
17.6 Non-Degenerate k·p Perturbation Theory
at a ∆ Point
The momentum operator in the k·p Hamiltonian transforms as a vector.
For the group of the wave vector at a ∆ point (C4v point group), the
vector transforms as ∆1 + ∆5 for the longitudinal component x and
for the transverse components y, z, respectively. The conduction band
extrema for Si are located at the six equivalent (∆, 0, 0) locations, where
∆ is 85% of the distance from Γ to X. This level has Γ−
2 symmetry
at k = 0 and ∆2 symmetry as we move away from k = 0 (see the
compatibility relations in §13.7).
498CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
In first-order perturbation theory we have a non-vanishing contribution
along kx of the form (∆2|px|∆2) since ∆1 ⊗ ∆2 = ∆2. Thus,
there is in general a linear k term for E(k) in the longitudinal direction.
At the band extremum this matrix element however vanishes (not
by symmetry but because of the band extremum).
The second-order contributions are as follows. The longitudinal
terms (∆2|∆1|∆j) require that the intermediate state ∆j transforms
as ∆2 according to the compatibility relations, or else the matrix element
vanishes. States with ∆2 symmetry at a ∆ point arise from
Γ+
25, Γ−
2 and Γ−
12 states at k = 0. For transverse terms, the matrix
element (∆2|∆5|∆j) requires the intermediate state ∆j to transform as
∆5. States with ∆5 symmetry arise from Γ±
25 levels at k = 0.
Since the basis function for ∆2 is yz (see character tables in the
notes), the vector component ∆5,y couples to the z component of the
intermediate state with symmetry ∆5,z while the vector component ∆5,z
couples to the y component of the intermediate state with symmetry
∆5,y.
We thus obtain for E(k) about the band extremum at k0 using the
matrix element (∆2 |∆5,y|∆5,z):
E(k) = E(k0) +
¯h2
k2
x
2m∗
+
¯h2
(k2
y + k2
z )
2m∗
t
(17.68)
in agreement with the expression used in all solid state physics courses.
These arguments can be extended to other points in the Brillouin
zone, and to 2-band and 3-band models for materials with cubic symmetry.
The k · p perturbation theory can also be extended to crystals
with other symmetries.
17.7 Optical Matrix Elements
The Hamiltonian in the presence of electromagnetic fields can be written
as
H =
1
2m
(p −
e
c
A)2
+ V (17.69)
17.8. FOURIER EXPANSION OF ENERGY BANDS 499
Then the proper form of the Hamiltonian for an electron in a solid in
the presence of an optical field is
H =
(p − e/cA)2
2m
+ V (r) =
p2
2m
+ V (r) −
e
mc
A · p +
e2
A2
2mc2
(17.70)
in which A is the vector potential due to the optical fields, V(r) is the
periodic potential. Thus, the one-electron Hamiltonian without optical
fields is
H0 =
p2
2m
+ V (r) (17.71)
and the optical perturbation terms are
H = −
e
mc
A · p +
e2
A2
2mc2
. (17.72)
The momentum matrix elements v|p|c which determine the strength
of optical transitions also govern the magnitudes of the effective mass
components. The coupling of the valence and conduction bands through
the optical fields depends on the matrix element for the coupling to the
electromagnetic field
H ∼= −
e
mc
p · A. (17.73)
With regard to the spatial dependence of the vector potential we can
write
A = A0 exp[i(K · r − ωt)] (17.74)
where for a loss-less medium K = ˜nω/c = 2π˜n/λ is a slowly varying
function of r since 2π˜n/λ is much smaller than typical wave vectors in
solids. Here ˜n, ω, and λ are respectively the real part of the index of
refraction, the optical frequency, and the wavelength of light.
The relation between the momentum matrix element v|p|c and the
effective mass components are discussed in the previous section.
17.8 Fourier Expansion of Energy Bands—
Slater–Koster Method
This technique provides the most general form for the energy bands
throughout the Brillouin zone which is consistent with the crystal symmetry.
Like the k · p method, it is an approach whereby the energy
500CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
bands can be determined from experimental data without recourse to a
definite energy band model or to a specific crystal potential. The original
work on this method was done by Slater and Koster to provide an
interpolation formula for calculating energy bands at high symmetry
points in the Brillouin zone (Slater and Koster, Phys. Rev. 94, 1498
(1954)) and the method was later applied to silicon and germanium
(Phys. Rev. 160, 649 (1967)). We will illustrate the method here for
a simple cubic lattice (ref. “The Optical Properties of Solids”, Proceedings
of the International School of Physics, Enrico Fermi course
XXXIV, p. 202-8).
Because of the periodicity of the lattice, the energy bands En(k) are
periodic in the extended zone
En(k + K) = En(k) (17.75)
where K is 2π times a reciprocal lattice vector so that K · Rm = 2π
integer.
The energy bands En(k) are furthermore continuous across a zone
boundary and they approach this boundary with zero slope (giving
the electrons zero velocity at a zone boundary). We make use of this
periodicity as follows. Suppose that we have a function V (r) which
is periodic in the 3-dimensional lattice. This function can be Fourier
expanded in the reciprocal lattice
V (r) =
K
v(K)eiK·r
(17.76)
in which the summation is over all reciprocal lattice vectors. In the extended
zone scheme, the energy En(k) is periodic in a 3-dimensional
space defined by the reciprocal lattice vectors. Therefore it is possible
to Fourier expand En(k) in a space “reciprocal” to the reciprocal
lattice, i.e., in the direct lattice, to obtain:
En(k) =
d
εn(d)eik·d
(17.77)
where d = Rm are Bravais lattice vectors and εn(d) can be interpreted
as an overlap integral in the tight binding approximation. Crystal symmetry
restricts the number of independent expansion coefficients εn(d).
17.8. FOURIER EXPANSION OF ENERGY BANDS 501
Provided that the Fourier series of Eq. (17.77) is rapidly convergent, it
is possible to describe En(k) in terms of a small number of expansion
parameters εn(d) which can, in principle, be determined by experiment.
For example, let us consider a non-degenerate, isolated s-band in
a simple cubic crystal. Such a band has Γ1 symmetry and is invariant
under the point group operations of the cubic group. The Fourier
expansion would then take the form of the tight binding functions:
En(k) = εn(0) + εn(1) cos akx + cos aky + cos akz
+ εn(2) cos a(ky + kz) + cos a(ky − kz) + cos a(kz + kx)
+ cos a(kz − kx) + cos a(kx + ky) + cos a(kx − ky)
+ εn(3) cos a(kx + ky + kz) + cos a(kx − ky − kz)
+ cos a(−kx + ky − kz) + cos a(−kx − ky + kz) + . . .(17.78)
where
d = 0 is the zeroth neighbor at a(0,0,0)
d = 1 is the nearest neighbor at a(1,0,0)
d = 2 is the next nearest neighbor at a(1,1,0)
d = 3 is the next-next nearest neighbor at a(1,1,1), etc.
In the tight binding approximation, the expansion coefficients appear
as overlap integrals and transfer integrals of various kinds. Thus, the
tight binding form is written to satisfy crystal symmetry and is of the
Slater–Koster form.
Now for energy bands of practical interest, we will not have isolated
non-degenerate bands, but rather coupled bands of some sort. We can
express the problem for n coupled bands in terms of an (n × n) secular
equation of the form
| i|H|j − En(k)δij| = 0. (17.79)
502CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
In Eq. (17.79) the indices i and j denote Bloch wave functions which
diagonalize the Hamiltonian
H =
p2
2m
+ V (r) (17.80)
and are labeled by the wave vector k. The matrix elements i|H|j
thus constitute a k-dependent matrix. But at each k point these matrix
elements are invariant under the symmetry operations of the group of
the wave vector at k. The Hamiltonian at k = 0 has Γ+
1 symmetry
just like its eigenvalues En(k). This matrix is also periodic in the
reciprocal lattice in the extended zone scheme and therefore can be
Fourier expanded.
The expansion is carried out in terms of a complete set of basis
matrices which are taken as angular momentum matrices. For example,
a (2×2) Hamiltonian including the electron spin (i.e., the double group
representations Γ±
6 or Γ±
7 to be discussed in Chapter 19) would be
expanded in terms of 4 basis matrices 1, Sx, Sy and Sz, representing
the angular momentum matrices for spin 1
2
. A (3 × 3) Hamiltonian is
expanded in terms of the 9 linearly independent basis matrices which
span this space, namely, 1, Sx, Sy, Sz, S2
x, S2
y , {Sz, Sy}, {Sz, Sx}
and {Sx, Sy}, in which 1 is a (3 × 3) unit matrix, Sx, Sy, Sz are
angular momentum matrices for spin 1, and {Si, Sj} denotes the anticommutator
for matrices Si and Sj. Under the point group operations
of the group of the wave vector, the angular momentum matrices Si
transform as an axial vector–i.e., at k = 0, Si transforms as Γ+
15, while
the matrix Hamiltonian still is required to be invariant. Therefore, it is
necessary to take products of symmetrized combinations of the n basis
matrices with appropriate symmetrized combinations of the Fourier
expansion functions so that an invariant matrix Hamiltonian results.
The (n×n) matrix Hamiltonian which is denoted by DΓ1 (k) can be
Fourier expanded in terms of these basis function matrices in the form
DΓ1 (k) =
d
αd,Γj
CΓj
(d) · SΓj
(17.81)
which is a generalization of Eq. (17.77). In Eq. (17.81), SΓj
denotes a
collection of basis matrices which transforms as Γj, and these sym-
17.8. FOURIER EXPANSION OF ENERGY BANDS 503
Table 17.2: Symmetrized products of angular momenta for the cubic
group
Order Representation Notation Symmetrized
Products
0 Γ+
1 SΓ+
1
(0) 1
1 Γ+
15 Sx
Γ+
15
(1) Sx
2 Γ+
12 S
(1)
Γ+
12
(2) S2
x + ωS2
y + ω2
S2
z
Γ+
25 S
(x)
Γ+
25
(2) {Sy, Sz}
3 Γ+
2 SΓ+
2
(3) SxSySz + SxSzSy
Γ+
15 S
(x)
Γ+
15
(3) S3
x
Γ+
25 S
(x)
Γ+
15
(3) {Sx, (S2
y − S2
z )}
metrized products of angular momentum matrices are given in Table
17.2. The distance d denotes the order of the expansion in Eq. (17.81)
and corresponds to the distance of neighbors in the Fourier expansion in
the tight binding sense. The angular momentum matrices in Table 17.2
are given by:
Sx =



0 0 0
0 0 i
0 −i 0


 Sy =



0 0 −i
0 0 0
i 0 0


 Sz =



0 i 0
−i 0 0
0 0 0



(17.82)
Products of the dimensionless angular momentum matrices Si are listed
in Table 17.2, using an abbreviated notation. For example, S
(x)
Γ+
15
(1)
denotes the x component of a 3 component vector Sx, Sy, Sz and all
three components would appear in Eq. (17.81). Similarly, S
(i)
Γ+
12
(2) is a
2 component vector with partners
S2
x + ωS2
y + ω2
S2
z
and
S2
x + ω2
S2
y + ωS2
z
504CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
and only one of the partners is listed in the table. In Table 17.2 several
other 3 component matrices are found such as S
(α)
Γ+
25
(2) where the x
component is the anti-commutator {Sy, Sz} and the y and z components
of S
(α)
Γ+
25
(2) are found by cyclic permutation of the indices x, y, z. It is
worth mentioning that all of the S matrices in Eq. (17.81) are 3 × 3
matrices which are found explicitly by carrying out the indicated matrix
operations. For example:
{Sy, Sz} = SySz+SzSy =



0 0 0
0 0 0
0 −1 0


+



0 0 0
0 0 −1
0 0 0


 =



0 0 0
0 0 −1
0 −1 0


 .
(17.83)
Also useful for carrying out matrix operations are the definitions:
Sx =
¯h
i
y
∂
∂z
− z
∂
∂y
(17.84)
so that
Sx



x
y
z


 =
¯h
i



0
−z
y


 . (17.85)
Another point worth mentioning about Table 17.2 concerns the
terms that do not appear. For example, in second-order we could have
terms like S2
x + S2
y + S2
z but this matrix is just the unit matrix which
has already been listed in the table. Similarly, the commutators [Sy, Sz]
which enter in second-order are matrices that have already appeared in
first-order as iSx.
We give below the nine basis matrices that span the (3×3) matrices
for spin 1.
SΓ+
1
=



1 0 0
0 1 0
0 0 1


 (17.86)
S
(1)
Γ+
12
=



−1 0 0
0 1 + ω2
0
0 0 1 + ω


 =



−1 0 0
0 −ω 0
0 0 −ω2


 (17.87)
17.8. FOURIER EXPANSION OF ENERGY BANDS 505
S
(2)
Γ+
12
=



−1 0 0
0 1 + ω 0
0 0 1 + ω2


 (17.88)
S
(x)
Γ+
15
=



0 0 0
0 0 i
0 −i 0


 (17.89)
S
(y)
Γ+
15
=



0 0 −i
0 0 0
i 0 0


 (17.90)
S
(z)
Γ+
15
=



0 i 0
−i 0 0
0 0 0


 (17.91)
S
(x)
Γ+
25
=



0 0 0
0 0 1
0 1 0


 (17.92)
S
(y)
Γ+
25
=



0 0 1
0 0 0
1 0 0


 (17.93)
S
(z)
Γ+
25
=



0 1 0
1 0 0
0 0 0


 (17.94)
Any arbitrary (3 × 3) matrix can be written as a linear combination of
these nine matrices.
Table 17.2 however was constructed to be more general that to
describe interacting p-bands in a 3 × 3 matrix formulation. The table
can equally well be used to form the appropriate 16 basis matrices which
are needed to deal with interacting s and p bands such as would arise
in semiconductor physics. Such interacting s and p bands give rise to
a 4 × 4 matrix Hamiltonian and therefore 16 basis matrices are needed
to span the space for the secular equation in this case.
Now let use return to the Fourier expansion of Eq. (17.81). For
each neighbor distance |d| there are several lattice vectors that enter,
just as in the plane wave problem of Chapter 16 where we considered
506CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
sets of K vectors of equal magnitude. The terms in Eq. (17.81) can be
labeled by their symmetry types so that the sum on d breaks up into
a sum on the magnitude |d| and on the symmetry type Γj occurring
at distance d. The linear combinations of the exponential functions
exp(ik · d) which transform as the pertinent irreducible representations
of the cubic group are given in Table 17.3 out through 3rd
nearest neighbor
distances. Once again, if a representation is one-dimensional, the
basis function itself is given. For the two-dimensional representations,
only one of the functions is listed, the partner being the complex conjugate
of the listed function. For the three-dimensional representations,
only the x-component is listed; the partners are easily found by cyclic
permutations of the indices.
The combinations of plane waves and basis functions that enter
the Fourier expansion of Eq. (17.81) are the scalar products of these
symmetrized Fourier functions CΓj
(d) and the basis functions SΓj
(d).
This means that for the 2-dimensional representations, we write
C
(1)
Γ+
12
S
(1)
Γ+
12
∗
+ C
(2)
Γ+
12
S
(2)
Γ+
12
∗
(17.95)
where the second term is the complex conjugate of the first so that
the sum is real. For the 3-dimensional representations we write for the
scalar product
Cx
Sx
+ Cy
Sy
+ Cz
Sz
. (17.96)
Finally, the Fourier expansion parameters αd,Γj
are just numbers that
give the magnitude of all the terms which enter the Fourier expansion.
These coefficients are often evaluated from experimental data.
Now suppose that we are going to do a Fourier expansion for pbands.
These have Γ−
15 symmetry. We ask what symmetry types can
we have in the coupling between p-bands—clearly only the symmetries
that enter into the direct product
Γ−
15 ⊗ Γ−
15 = Γ+
1 + Γ+
12 + Γ+
15 + Γ+
25. (17.97)
We will now indicate the terms which contribute at each neighbor distance
to Eq. (17.81).
17.8. FOURIER EXPANSION OF ENERGY BANDS 507
Table 17.3: Symmetrized Fourier functions for a simple cubic lattice.
d Repr. Notation Symmetrized Fourier functions
a(0, 0, 0) Γ+
1 CΓ+
1
(000) 1
a(1, 0, 0) Γ+
1 CΓ+
1
(100) cos akx + cos aky + cos akz
Γ+
12 C
(1)
Γ+
12
(100) cos akx + ω cos aky + ω2
cos akz
Γ−
15 C
(x)
Γ−
15
(100) sin akx
a(1, 1, 0) Γ+
1 CΓ+
1
(110) cos a(ky +kz)+cos a(ky −kz)+cos a(kz +kx)
+cos a(kz −kx)+cos a(kx +ky)+cos a(kx −
ky)
Γ+
12 C
(1)
Γ+
12
(110) [cos a(ky + kz) + cos a(ky − kz)]
+ ω[cos a(kz + kx) + cos a(kz − kx)]
+ ω2
[cos a(kx + ky) + cos a(kx − ky)]
Γ−
15 C
(x)
Γ−
15
(110) sin a(kx + ky) + sin a(kx − ky)
+ sin a(kx + kz) + sin a(kx − kz)
Γ−
25 C
(x)
Γ−
25
(110) sin a(kx + ky) + sin a(kx − ky)
− sin a(kx + kz) − sin a(kx − kz)
Γ+
25 C
(x)
Γ+
25
(110) cos a(ky + kz) − cos a(ky − kz)
a(1, 1, 1) Γ+
1 CΓ+
1
(111) cos a(kx + ky + kz) + cos a(kx − ky − kz)
+cos a(−kx +ky −kz)+cos a(−kx −ky +kz)
Γ−
2 CΓ−
2
(111) sin a(kx + ky + kz) + sin a(kx − ky − kz)
sin a(−kx + ky − kz) + sin a(−kx − ky + kz)
Γ−
15 C
(x)
Γ−
15
(111) sin a(kx + ky + kz) + sin a(kx − ky − kz)
−sin a(−kx +ky −kz)−sin a(−kx −ky +kz)
Γ+
25 C
(x)
Γ+
25
(111) cos a(kx + ky + kz) + cos a(kx − ky − kz)
−cos a(−kx +ky −kz)−cos a(−kx −ky +kz)
where ω = exp(2πi/3) and a is the lattice constant.
508CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
17.8.1 Contributions at d = 0:
From Table 17.3 we can have only Γ+
1 symmetry at d = 0 for which the
basis matrix is 


1 0 0
0 1 0
0 0 1


 (17.98)
and the symmetrical Fourier function is 1, so that the net contribution
to Eq. (17.81) is
α0,Γ+
1



1 0 0
0 1 0
0 0 1


 . (17.99)
17.8.2 Contributions at d = 1:
For Γ+
1 symmetry the contribution is in analogy to Eq. (17.99)
α1,Γ+
1
CΓ+
1
(100)



1 0 0
0 1 0
0 0 1


 (17.100)
while for Γ+
12 symmetry, the contribution is:
α1,Γ+
12
C
(1)
Γ+
12



ω + ω2
0 0
0 1 + ω2
0
0 0 1 + ω


+α1,Γ+
12
C
(2)
Γ+
12



ω + ω2
0 0
0 1 + ω 0
0 0 1 + ω2



(17.101)
where we have used the relation S
(1)
Γ12
= S2
x + ωS2
y + ω2
S2
z to obtain the
appropriate matrices. We also use the relations 1 + ω + ω2
= 0 for the
cube roots of unity to simplify Eq. (17.101). We note that both terms
in Eq. (17.101) have the same expansion parameter α1,Γ+
12
.
These are all the contributions for d = 1. The symmetry type Γ−
15
does not enter into this sum since there are no basis matrices with
symmetries Γ−
15 for d = 1 (see Table 17.2). This symmetry would
however enter into treating the interaction between s and p bands.
Therefore, we get no off-diagonal terms until we go to second-neighbor
distances. This should not be surprising to us since this is exactly
what happens in the k · p treatment of p bands. In fact, the Fourier
17.8. FOURIER EXPANSION OF ENERGY BANDS 509
expansion technique contains in it a k · p expansion for every
point in the Brillouin zone.
17.8.3 Contributions at d = 2:
At the second-neighbor distance Table 17.3 yields contributions from
Γ+
1 , Γ+
12 and Γ+
25 symmetries. These contributions at d = 2 are:
Γ+
1 symmetry
α2,Γ+
1
CΓ+
1
(110)



1 0 0
0 1 0
0 0 1


 (17.102)
Γ+
12 symmetry
α2,Γ+
12


C
(1)
Γ+
12
(110)



−1 0 0
0 −ω 0
0 0 −ω2


 + c.c.


 (17.103)
Γ+
25 symmetry
α2,Γ+
25






0 C
(z)
Γ+
25
(110) C
(y)
Γ+
25
(110)
C
(z)
Γ+
25
(110) 0 C
(x)
Γ+
25
(110)
C
(y)
Γ+
25
(110) C
(x)
Γ+
25
(110) 0






(17.104)
Terms with Γ−
15 and Γ−
25 symmetries in Table 17.3 do not enter because
there are no basis matrices with these symmetries.
17.8.4 Contributions at d = 3:
Symmetries Γ+
1 and Γ+
25 contribute and these are written down as above.
To get the matrix Hamiltonian we add up contributions from Eqs. (17.99–
17.104). There are 6 parameters αd,Γj
that enter into the Fourier expansion
through second-neighbor terms (d = 0, 1, 2). The Γ+
1 representation
at d = 0 contributes to the (1,1) position in the secular equation
a term in α0,Γ+
1
and at d = 1 contributes a term α1,Γ+
1
(cos akx +cos aky +
cos akz) in which the two coefficients α0,Γ+
1
and α1,Γ+
1
will have different
numerical values. The other entries into the (3 × 3) matrix are found
510CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
similarly. The resulting (3 × 3) matrix Hamiltonian is then diagonalized
and the eigenvalues are the En(k) we are looking for. This En(k)
properly expresses the crystal symmetry at all points in the Brillouin
zone.
It is instructive to write out in detail this matrix Hamiltonian along
the (100), (110) and (111) directions and to verify that all connectivity
relations and symmetry requirements are automatically satisfied. It is
directly shown that near k = 0, the Hamiltonian of Eq. (17.81) is of the
k · p form previously derived. As stated above, the Fourier expansion
approach contains the k · p form for all expansion points k0 in the
Brillouin zone.
17.8.5 Other Degenerate Levels
The Fourier expansion can also be applied to the two-fold Γ+
12 levels in
cubic symmetry arising from d-bands, or to Γ±
12 levels more generally.
Of particular interest is application of the Slater–Koster method to
coupled s and p-bands as has been done for silicon and germanium,
both of which crystallize in the diamond structure. In the case of
coupled s and p bands, the 3 × 3 expansion in §17.8 and the s-band
expansion in §17.3 are coupled with the Fourier terms from Table 17.3
having symmetries Γi ⊗ Γ−
15. We give an outline in this section for
setting up the secular equation to solve the Fourier expansion for these
two interesting cases.
The four 2 × 2 matrices that are used as basis matrices for Fourier
expanding the Γ±
12 levels are implied by Γ±
12 ⊗ Γ±
12 = Γ+
1 + Γ+
2 + Γ+
12:
for Γ+
1 symmetry
SΓ+
1
=
1 0
0 1
(17.105)
for Γ+
2 symmetry
SΓ+
2
=
1 0
0 −1
(17.106)
for Γ+
12 symmetry
SΓ+
12,1
=
0 1
0 0
(17.107)
17.8. FOURIER EXPANSION OF ENERGY BANDS 511
where the partner of SΓ+
12,1
is the Hermitian transpose
SΓ+
12,2
= S∗
Γ+
12,1
= S†
Γ+
12,1
=
0 0
1 0
. (17.108)
Using these matrices we see that
SΓ+
12,1
S†
Γ+
12,1
+ SΓ+
12,2
S†
Γ+
12,2
=
1 0
0 1
= SΓ+
1
(17.109)
and
SΓ+
12,1
S†
Γ+
12,1
− SΓ+
12,2
S†
Γ+
12,2
=
1 0
0 −1
= SΓ+
2
. (17.110)
The dispersion relation of E(k) for a band with Γ+
12 symmetry at k = 0
can then be Fourier expanded throughout the Brillouin zone in terms
of the basis functions in Eqs. (17.105–17.108) as:
EΓ±
12
(k) = d αd,Γ+
1
CΓ+
1
(d)
1 0
0 1
+ d αd,Γ+
2
CΓ+
2
(d)
1 0
0 −1
+ d αd,Γ+
12
C
(1)
Γ±
12
(d)
0 1
0 0
+ d αd,Γ+
12
C
(2)
Γ±
12
(d)
0 0
1 0
(17.111)
where C
(2)
Γ±
12
(d) = C
(1)∗
Γ±
12
(d) and the CΓ±
i
(d) functions are found in Table
17.3.
For the case of interacting s (Γ+
1 ) and p (Γ−
15) bands, the interaction
terms have Γ+
1 ⊗ Γ−
15 = Γ−
15 symmetry so the 4 × 4 expansion matrices
must be supplemented by the matrices
Sx
Γ−
15
=





0 1 0 0
1 0 0 0
0 0 0 0
0 0 0 0





(17.112)
and the two partners
Sy
Γ−
15
=





0 0 1 0
0 0 0 0
1 0 0 0
0 0 0 0





Sz
Γ−
15
=





0 0 0 1
0 0 0 0
0 0 0 0
1 0 0 0





(17.113)
512CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
The complete treatment of the Fourier expansion for the 8 coupled s
and p bonding and anti-bonding bands in the non-symmorphic diamond
structure has been presented in Phys. Rev. 160, 649 (1967) and was
used to describe the Si and Ge bands throughout the Brillouin zone.
The same basic treatment without the s bands was used to treat the
lattice dynamics for these structures in Int. J. of Quantum Chemistry,
Vol IIs, 333 (1968).
17.8.6 Summary
A discussion of the use of k · p perturbation theory extrapolation approach
and the use of the Slater–Koster interpolation approach is based
on the following considerations. If the available experimental data are
limited to one small region in the Brillouin zone and that is all that is
known, then k·p perturbation theory is adequate to describe E(k). This
is often the case in practice for semiconductors. If, however, the available
experimental data relates to several points in the Brillouin zone,
then the Slater–Koster approach is more appropriate. Experiments on
various parts of the Fermi surface usually require knowledge of E(k)
over several regions of the Brillouin zone. Although such experiments
might seem to yield unrelated information about the energy bands, the
Slater–Koster approach is useful for interrelating such experiments.
In Chapter 20, the effective mass Hamiltonian will be considered in
the presence of a magnetic field, taking into account the spin on the
electron. In this case we form the following symmetrized combinations
of wave vectors:
Γ+
1 → k2
x + k2
y + k2
z
Γ+
12 → k2
x + ωk2
y + ω2
k2
z , cc
Γ+
25 → ({ky, kz}, {kz, kx}, {kx, ky})
Γ+
15 → ([ky, kz], [kz, kx], [kx, ky])
(17.114)
treating the wave vector as an operator. This formulation of the effective
mass equation is used to yield the effective g-factor for an electron
in a periodic solid.
17.9. SELECTED PROBLEMS 513
17.9 Selected Problems
1. (a) Using k · p perturbation theory, find the form of the E(k)
relation near the L-point in the Brillouin zone for a face centered
cubic lattice arising from the lowest levels with L1 and
L2 symmetry that are doubly degenerate in the free electron
model. Which of the non-vanishing k · p matrix elements at
the L-point are equal to each other by symmetry?
(b) Using the Slater–Koster technique, find the form for E(k)
for the lowest two levels for a face centered cubic lattice.
(c) Expand your results for (b) about the L-point in a Taylor
expansion.
(d) Compare your results in (c) to those in (a).
(e) Using k · p perturbation theory, find the form of E(k) for a
non-degenerate band with W1 symmetry about the W point
in the fcc lattice.
2. (a) Using k · p perturbation theory, find the form of the secular
equation for the valence band of Si with Γ+
25 symmetry.
(b) Which intermediate states couple to the Γ+
25 valence band
states in second-order k · p perturbation theory?
(c) Which matrix elements (listed in Table 17.1) enter the secular
equation in (a)?
(d) Write the secular equation for the Γ+
25 valence bands that is
analogous to Eq. (17.59) for the Γ−
15 band?
(e) What result is obtained along a Λ (111) axis?
3. (a) Write down the matrices for Sx, Sy and Sz for angular momentum
3/2. Products of these and the (4 × 4) unit matrix
form the 16 matrix basis functions which span the vector
space for the (4 × 4) Slater–Koster secular equation for coupled
s and p bands for a simple cubic lattice.
(b) Returning to the Slater–Koster (3 × 3) secular determinant
for a simple cubic lattice given in the class notes, write the
514CHAPTER 17. ENERGY BAND MODELS BASED ON SYMMETRY
explicit expression for this matrix along a (100) direction.
Show that the proper k ·p Hamiltonian is obtained at the X
point.
Chapter 18
Application of Group Theory
to Valley-Orbit Interactions
in Semiconductors
In this chapter, we shall discuss the application of group theory to the
impurity problem of a multi-valley semiconductor, such as occurs in the
donor carrier pockets in silicon and germanium. In the case of silicon
the lowest conduction bands occur at the 6 equivalent (∆,0,0) points
where (∆ =0.85 on a scale where the Γ point is at the origin and the X
point is at 1). In the case of germanium, the conduction band minima
occur at the L points so that the Fermi surface for electrons consists
of eight equivalent half-ellipsoids of revolution (4 full ellipsoids). Other
cases where valley-orbit interactions are important are multi-valleyed
semiconductors, such as PbTe or Te, where the conduction and valence
band extrema are both away from k = 0.
18.1 Introduction
Group theory tells us that the maximum degeneracy that energy levels
or vibrational states can have with cubic symmetry is 3-fold degeneracy.
Cubic symmetry is imposed on the problem of donor doping of
a semiconductor through the valley-orbit interaction which causes a
partial lifting of the n-fold degeneracy of an n-valley semiconductor.
515
516CHAPTER 18. APPLICATION TO VALLEY-ORBIT INTERACTIONS
In this presentation we show how group theory prescribes the partial
lifting of this n-fold degeneracy. This effect is important in describing
the ground state energy of a donor-doped n-valley semiconductor.
Our discussion of the application of group theory to the classification
of the symmetries of the impurity levels in a degenerate semiconductor
proceeds with the following outline:
1. Review of the one-electron Hamiltonian and the effective mass
Hamiltonian for a donor impurity in a semiconductor yielding
hydrogenic impurity levels for a single-valley semiconductor.
2. Discussion of the impurity states for multivalley semiconductors
in the effective mass approximation.
3. Discussion of the valley-orbit interaction. In this application we
consider a situation where the lower symmetry group is not a
subgroup of the higher symmetry group.
18.2 Background
In this section we review the one-electron Hamiltonian, effective mass
approximation and the hydrogenic impurity problem for a single-valley
semiconductor. We write the one-electron Hamiltonian for an electron
in a crystal which experiences a perturbation potential U(r) due to an
impurity:
p2
2m
+ V (r) + U(r) Ψ(r) = EΨ(r) (18.1)
in which V (r) is the periodic potential. In the effective mass approximation,
the perturbing potential due to an impurity is taken as
U(r) = −e2
/(εr) where ε is the dielectric constant. This problem is
usually solved in terms of the effective mass theorem to obtain
p2
2m∗
αβ
+ U(r) fj(r) = (E − E0
j )fj(r) (18.2)
where m∗
αβ is the effective mass tensor for electrons in the conduction
band about the band extremum at energy E0
j , and fj(r) is the effective
mass wave function. We thus note that by replacing the periodic
18.2. BACKGROUND 517
potential V (r) by an effective mass tensor, we have lost most of the
symmetry information contained in the original periodic potential.
This symmetry information is restored by introducing the valley-orbit
interaction, as in §18.3 and §18.4.
The simplest case for an impurity in a semiconductor is that for a
shallow impurity level described by hydrogenic impurity states in a nondegenerate
conduction band, as for example a Si atom substituted for a
Ga atom in GaAs, a direct gap semiconductor with the conduction band
extremum at the Γ point (k = 0). To satisfy the bonding requirements
in this case, one electron becomes available for conduction and a donor
state is formed. The effective mass equation in this case becomes
p2
2m∗
−
e2
εr
f(r) = (E − E0
j )f(r) (18.3)
where U(r) = −e2
/(εr) is the screened Coulomb potential for the donor
electron, ε is the low frequency dielectric constant, and the donor energies
are measured from the band edge E0
j . This screened Coulomb
potential is expected to be a good approximation for r at a sufficiently
large distance from the impurity site, so that ε can be considered
to be independent of r.
The solutions to this hydrogenic problem are the hydrogenic levels
En − E0
j = −
e2
2εa∗
0n2
n = 1, 2, . . . (18.4)
where the effective Bohr radius is
a∗
0 =
ε¯h2
m∗e2
. (18.5)
Since En − E0
j ∼ m∗
/ε2
, we have shallow donor levels located below
the band extrema. These levels are shallow because of the large value
of ε and the small value of m∗
, a common occurrence in many of the
high mobility semiconductors.
Group theoretical considerations enter in the following way. For
many III–V compound semiconductors, the valence and conduction
band extrema are at k = 0 so that the effective mass Hamiltonian
518CHAPTER 18. APPLICATION TO VALLEY-ORBIT INTERACTIONS
has full rotational symmetry. Since the hydrogenic impurity is embedded
in a crystal with a periodic potential, the crystal symmetry (i.e.,
Td point group symmetry) will perturb the hydrogenic levels and cause
a splitting of various degenerate levels:
s levels → Γ1 (no splitting)
p levels → Γ15 (no splitting)
d levels → Γ12 + Γ15 (splitting occurs)
f levels → Γ2 + Γ15 + Γ25 (splitting occurs)
In principle, if a multiplet has the same symmetry as an s or p level then
an interaction can occur giving rise to an admixture of states of similar
symmetries. In practice, the splittings are very small in magnitude and
the effects of the crystal field are generally unimportant for shallow
donor levels in single valley semiconductors.
18.3 Impurity States for Multivalley Semi-
conductors
Group theory plays a more important role in the determination of impurity
states in multi-valley semiconductors than for the simple hydrogenic
case. A common example of a multi-valley impurity state is As in
Si (or in Ge). In Si there are six equivalent valleys while for Ge there are
four equivalent valleys. The multi-valley aspect of the problem results
in two departures from the simple hydrogenic series.
The first of these departures is associated with the fact that the
constant energy surfaces in this case are ellipsoids rather than spheres.
In this case we write Schr¨odinger’s equation for a single valley in the
effective mass approximation as:
p2
x + p2
y
2mt
+
p2
z
2m
−
e2
εr
= E f(r) (18.6)
in which mt is the transverse mass component, m is the longitudinal
mass component, and the energy E is measured from the energy band
extremum. The appropriate symmetry group for the effective mass
18.3. IMPURITY STATES FOR MULTIVALLEY SEMICONDUCTORS519
equation given by Eq. (18.6) is D∞h rather than the full rotation group
which applies to the hydrogenic impurity levels. This form for the effective
mass Hamiltonian follows from the fact that the constant energy
surfaces are ellipsoids of revolution, which in turn is a consequence of
the selection rules for the k · p Hamiltonian at a ∆ point (group of
the wave vector C4v) in the case of Si, and at an L point (group of the
wave vector D3d) in the case of Ge. The anisotropy of the kinetic energy
terms corresponds to the anisotropy of the effective mass tensor. For example
in the case of silicon m /m0 = 0.98 (heavy mass), mt/m0 = 0.19
(light mass). This anisotropy in the kinetic energy terms results in a
splitting of the impurity levels with angular momentum greater than
1, in accordance with the irreducible representations of D∞h. For example,
in D∞h symmetry we have the following correspondence with
angular momentum states:
s states → Σ+
g = A1g
p states → Σ+
u + πu = A2u + E1u
d states → ∆g + πg + Σ+
g = A1g + E1g + E2g.
We note that s and d states are even (g) and p states are odd (u) under
inversion in accordance with the character table for D∞h.
D∞h (∞/mm) E 2Cφ C2 i 2iCφ iC2
x2 + y2, z2 A1g(Σ+
g ) 1 1 1 1 1 1
A1u(Σ−
u ) 1 1 1 −1 −1 −1
Rz A2g(Σ−
g ) 1 1 −1 1 1 −1
z A2u(Σ+
u ) 1 1 −1 −1 −1 1
(xz, yz) (Rx, Ry) E1g(Πg) 2 2 cos φ 0 2 2 cos φ 0
(x, y) E1u(Πu) 2 2 cos φ 0 −2 −2 cos φ 0
(x2 − y2, xy) E2g(∆g) 2 2 cos 2φ 0 2 2 cos 2φ 0
E2u(∆g) 2 2 cos 2φ 0 −2 −2 cos 2φ 0
Thus a 2p level with angular momentum = 1 splits into a two-fold
2p±1
level and a non-degenerate 2p0
level in which the superscripts
denote the m values. Furthermore in D∞h symmetry, the splitting
of d-levels gives rise to the same irreducible representation (Σ+
g ) that
describes the s-levels, and consequently a mixing of these levels occurs.
Referring back to the effective mass equation [Eq. (18.6)], we note
that this equation cannot be solved exactly if m = mt. Thus, the
520CHAPTER 18. APPLICATION TO VALLEY-ORBIT INTERACTIONS
donor impurity levels in these indirect gap semiconductors must be
deduced from some approximate technique such as a variational calculation
or using perturbation theory. The effective mass approximation
itself works very well for these p-states because |ψp|2
for p states vanishes
for r = 0; consequently, for r values small enough for central cell
corrections to be significant, the wave function has a small amplitude
and thus small r values do not contribute significantly to the expectation
value of the energy for p-states.
18.4 The Valley-Orbit Interaction
The second departure from the hydrogenic series in a multi-valley semiconductor
is one that relates closely to group theory. This effect is most
important for s-states, particularly for the 1s state.
For s-states, a sizable contribution to the expectation value for the
energy is made by the perturbing potential for small r. The physical
picture of a spherically symmetric potential U(r) for small r cannot
fully apply because the tetrahedral bonding must become important
for |r| ≤ a, i.e., within the unit cell dimension. This tetrahedral crystal
field which is important within the central cell lifts the spherical symmetry
of an isolated atom. Thus we need to consider corrections to the
effective mass equation due to the tetrahedral crystal field. This tetrahedral
crystal field term is called Hvalley−orbit, the valley-orbit effective
Hamiltonian which couples equivalent conduction band extrema in the
various valleys.
Therefore to find the wave functions for the donor states in a multivalley
semiconductor, we must find linear combinations of wave functions
from each of the valleys that transform as irreducible representations
of the crystal field about the impurity ion. For example, in silicon,
the symmetrized linear combination of valley wave functions is in the
form
ψγ
(r) =
6
j=1
Aγ
j fj(r)uj,kj
0
(r)eikj
0·r
(18.7)
in which ψγ
(r) denotes one of 6 possible linear combinations. The index
j is the valley index and fj(r) is the envelope effective mass wave
18.4. THE VALLEY-ORBIT INTERACTION 521
 
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Figure 18.1: Constant energy ellipsoids
of the conduction-band minima
of silicon along {100} direc-
tions.
Table 18.1: Character table for the point group Td.
Td (¯43m) E 8C3 3C2 6σd 6S4
A1 1 1 1 1 1 Γ1
A2 1 1 1 −1 −1 Γ2
E 2 −1 2 0 0 Γ12
(Rx, Ry, Rz) T1 3 0 −1 −1 1 Γ25
(x, y, z) T2 3 0 −1 1 −1 Γ15
function, while uj,kj
0
(r) is the periodic part of the Bloch function in
which kj
0 is the wave vector to the band minimum of valley j. The 6
equivalent valleys along the (100) axes for silicon are shown in Fig. 18.1.
The indices j which label the various ellipsoids or valleys in Fig. 18.1
correspond to the indices j of Eq. (18.7). The local symmetry close
to the impurity center is Td, reflecting the tetrahedral bonding at the
impurity site. The character table for the Td point group is shown in
Table 18.1. The diagram which is useful for finding which valleys are
invariant under the symmetry operations of Td is given in Fig. 18.2. We
ask for the number of valleys which remain invariant under the various
symmetry operations of Td. This is equivalent to finding χatom sites or
χvalley sites, which forms a reducible representation of group Td. From
Figure 18.2, we immediately see that the characters for χvalley sites are
522CHAPTER 18. APPLICATION TO VALLEY-ORBIT INTERACTIONS
Figure 18.2: The regular tetrahedron
inscribed inside a cube, illustrating
the locations and symmetry
operations of the six valleys.
E 8C3 3C2 6σd 6S4
χvalley sites 6 0 2 2 0 = Γ1 + Γ12 + Γ15
and that the irreducible representations contained in χvalley sites are Γ1 +
Γ12+Γ15. To find the splitting of a level we must take the direct product
of the symmetry of the level with χvalley sites, provided that the level
itself transforms as an irreducible representation of group Td:
χlevel ⊗ χvalley sites. (18.8)
Since χlevel for s-states transforms as Γ1, the level splitting for s-states
is just χvalley sites:
—–Γ15
—–Γ12
—–Γ1
The appropriate linear combination of valley functions corresponding
to each of these irreducible representations is [using the notation
18.4. THE VALLEY-ORBIT INTERACTION 523
from Eq. (18.7)]:
A
(Γ1)
j = 1√
6
(1, 1, 1, 1, 1, 1)
A
(Γ12,1)
j = 1√
6
(1, 1, ω, ω, ω2
, ω2
)
A
(Γ12,2)
j = 1√
6
(1, 1, ω2
, ω2
, ω, ω)



A
(Γ15,1)
j = 1√
2
(1, −1, 0, 0, 0, 0)
A
(Γ15,2)
j = 1√
2
(0, 0, 1, −1, 0, 0)
A
(Γ15,3)
j = 1√
2
(0, 0, 0, 0, 1, −1)



(18.9)
in which each of the six components refers to one of the valleys. The
totally symmetric linear combination Γ1 is a non-degenerate level, while
the Γ12 basis functions have 2 partners which are given by f1 = x2
+
ωy2
+ω2
z2
and f2 = f∗
1 and the Γ15 basis functions have three partners
(x, y, z).
The analysis for the p-levels is more complicated because the plevels
in D∞h do not transform as irreducible representations of group
Td. The p-level in group D∞h transforms as a vector, with A2u and
E1u symmetries for the longitudinal and transverse components, respectively.
Since Td does not form a subgroup of D∞h we write the
vector for group Td as
χvector = χlongitudinal + χtransverse (18.10)
where χvector = Γ15. We treat the longitudinal component of the vector
as forming a σ-bond and the transverse component as forming a π-bond
so that χlongitudinal = Γ1 and χtransverse = Γ15 − Γ1, where we note that:
Γ15 ⊗(Γ1 +Γ12 +Γ15) = Γ15 +(Γ15 +Γ25)+(Γ1 +Γ12 +Γ15 +Γ25). (18.11)
We thus obtain:
χ2p0
= χvalley sites ⊗ Γ1 = Γ1 + Γ12 + Γ15 for m = 0 (18.12)
χ2p±
= χvalley sites⊗(Γ15−Γ1) = 2Γ15+2Γ25 for m = ±1 (18.13)
for group Td. If we perform high resolution spectroscopy experiments
for the donor impurity levels, we would observe transitions between the
various 1s multiplets to the various 2p-multiplets.
524CHAPTER 18. APPLICATION TO VALLEY-ORBIT INTERACTIONS
In addition to spectroscopic studies of impurity states, these donor
states for multi-valley semiconductors have been studied by the ENDOR
technique (G. Feher, Phys. Rev. 114, 1219 (1959) for work on Si).
Here the nuclear resonance of the 29
Si atoms is observed. The random
distribution of the 29
Si sites with respect to the donor impurity sites is
used to study the spatial dependence of the donor wavefunction, and
to determine the location in k-space of the conduction band extrema.
References for the group theoretical analysis of the valley-orbit splitting
are W. Kohn and J.M. Luttinger, Phys. Rev. 97, 1721 (1955) and
Phys. Rev. 98, 915 (1955). Experimental evidence for the splitting of
the degeneracy of the 1s donor levels in silicon is provided by infrared
absorption studies: R.L. Aggarwal and A.K. Ramdas, Phys. Rev. 140,
A1246 (1965) and V.J. Tekippe, H.R. Chandrasekhar, P. Fisher and
A.K. Ramdas, Phys. Rev. B6, 2348 (1972). An experimental trace for
the excitation spectrum of phosphorus impurities in silicon is shown in
Fig. 18.3 for several sample temperatures. The interpretation of this
spectrum follows from the energy level diagram in Fig 18.4.
It is of interest that the valley orbit splitting effect is only important
for the 1s levels. For the higher levels, the tetrahedral site location of
the impurity atom becomes less important as the Bohr orbit for the
impurity level
a∗
Bohr =
ε¯h2
m∗e2
n2
(18.14)
increases, where n is the principal quantum number for the donor impurity
level.
18.4. THE VALLEY-ORBIT INTERACTION 525
 
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Figure 18.3: Excitation spectrum of phosphorus donors in silicon. The
donor concentration ND ∼ 5×1015
/cm3
. Various donor level transitions
to valley-orbit split levels are indicated. The labels for the final state of
the optical transitions are in accordance with the symmetries of point
group Td.
526CHAPTER 18. APPLICATION TO VALLEY-ORBIT INTERACTIONS
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Figure 18.4: Energy-level scheme for transitions from the valley-orbit
split 1s multiplet of states to the 2p0, 2p± levels. The irreducible representations
for the various valley-orbit split levels in Td symmetry are
indicated. The conduction band edge (C.B.) is also indicated.
18.5. SELECTED PROBLEMS 527
18.5 Selected Problems
1. Find the symmetries and appropriate linear combination of valley
functions for the 1s and 2p donor levels for germanium (conduction
band minima at the L-point in the Brillouin zone), including
the effect of valley-orbit interaction. Indicate the transitions expected
in the far infrared spectra for these low temperature donor
level states.
528CHAPTER 18. APPLICATION TO VALLEY-ORBIT INTERACTIONS
Chapter 19
Spin Orbit Interaction in
Solids and Double Groups
The discussion of angular momentum and the rotation group has thus
far been limited to integral values of the angular momentum. The inclusion
of half integral angular momentum states requires the introduction
of “double groups”, which is the focus of this chapter.
19.1 Introduction
The spin angular momentum of an electron is half integral or Sz =
¯h/2. Furthermore, associated with each electron is a magnetic moment
µB = −|e|¯h/(2mc) = 0.927 × 10−20
erg/gauss. The magnetic moment
and spin angular momentum for the free electron are related by
µ = −
|e|
mc
S = −
|e|
mc
¯h
2
S
|S|
(19.1)
and are oppositely directed. This relation between the spin angular
momentum and the magnetic moment gives rise to an interaction, the
spin-orbit interaction, which is important in describing the electronic
structure of crystalline materials. In this section we briefly review this
interaction and then in the following sections of this Chapter, we consider
the group theoretical consequences of the half-integral spin and
the spin-orbit interaction.
529
530 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
An electron in an atom sees a magnetic field because of its own
orbital motion and consequently gives rise to the spin-orbit interaction
whereby this internal magnetic field tends to line up its magnetic
moment along the magnetic field: HSO = −µ · H. Substitution for
H = −(v/c) × E and µ = −|e|/(mc)S together with a factor of 1/2 to
make the result correct relativistically yields
HSO =
1
2m2c2
( V × p) · S. (19.2)
For an atom the corresponding expression is written as
HSO atom = ξ(r)L · S (19.3)
since ∆V ∼ r/r3
and where L is the orbital angular momentum. A
detailed discussion of the spin-orbit interaction is found in standard
quantum mechanics text books.
This spin-orbit interaction gives rise to a spin-orbit splitting of
atomic levels corresponding to different j values. As an example, consider
an atomic p state ( = 1). Writing the total angular momentum
J = L + S (19.4)
where L and S are, respectively, the orbital angular momentum operator
and the spin angular momentum operator, we obtain
J · J = (L + S) · (L + S) = L · L + S · S + (L · S + S · L) (19.5)
in which the operators L and S commute. Since L and S are coupled
through the spin-orbit interaction, m and ms are no longer good
quantum numbers, though and s remain good quantum numbers. To
find the magnitude of the spin-orbit interaction in Eq. (19.2), we need
to take the matrix elements of HSO in the |j, , s, mj representation.
Using Eq. (19.5) for the operators J, L and S, we obtain the diagonal
matrix element of J · J
j(j + 1) = ( + 1) + s(s + 1) + 2 L · S /¯h2
(19.6)
so that the expectation value of L · S in the |j, , s, mj representation
becomes:
L · S =
¯h2
2
[j(j + 1) − ( + 1) − s(s + 1)]. (19.7)
19.1. INTRODUCTION 531
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Figure 19.1: Splitting of the p levels
as a result of the spin-orbit interaction
into a 4-fold j = 3/2 level and
a 2-fold j = 1/2 level.
For p states, we have = 1, and s = 1/2 so that j = 3/2 or 1/2
L · S = ¯h2
/2 for j = 3/2
L · S = −¯h2
for j = 1/2.
(19.8)
Thus the spin-orbit interaction introduces a splitting between the j =
3/2 and j = 1/2 angular momentum states of the p-levels, as is indicated
in Fig. 19.1. From the expression for the expectation value of
L · S , we note that the degeneracy of an s-state is unaffected by the
spin-orbit interaction. On the other hand, a d-state is split up into
a D5/2 (6-fold degenerate) and a D3/2 (4-fold degenerate). Thus, the
spin-orbit interaction does not lift all the degeneracy of atomic states.
To lift the remaining degeneracy, it is necessary to lower the symmetry
further, for example, by the application of a magnetic field.
The magnitude of the spin-orbit interaction in atomic physics depends
also on the expectation value of ξ(r). For example,
n, j, , s, mj|H SO|n, j, , s, mj = j, , s, mj|L·S|j, , s, mj
∞
0
R∗
n ξ(r)Rn dr
(19.9)
where Rn (the radial part of the wave function) has an r dependence.
The magnitude of the integral in Eq. (19.9) increases rapidly with increasing
atomic number Z, approximately as Z3
or Z4
.
The physical reason behind the strong Z dependence of HSO is
that atoms with high Z have more electrons to generate larger internal
532 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Table 19.1: Spin-orbit interaction energies for some important semi-
conductors.
Semiconductor Atomic Number Γ-point Splitting
Diamond Z =6 ∆ = 0.006 eV
Silicon Z =14 ∆ = 0.044 eV
Germanium Z =32 ∆ = 0.29 eV
InSb
In
Sb
Z = 49
Z = 51
∆ = 0.9 eV
H fields and more electrons with magnetic moments to experience the
interaction with these magnetic fields. References for tabulation of the
spin-orbit splittings are:
1. C.E. Moore – Atomic Energy Levels (National Bureau of Standards,
Circular #467), Vol. 1 (1949), Vol. 2 (1952) and Vol. 3
(1958). These references give the measured spectroscopic levels
for any atom in a large number of excited configurations. The
lowest Z values are in Vol. 1, the highest in Vol. 3.
2. F. Herman and S. Skillman – Atomic Structure Calculations (PrenticeHall,
Inc. 1963). This reference gives the most complete listing of
the calculated atomic levels.
For most atomic species that are important in semiconducting materials,
the spin-orbit interaction plays a significant role. Some typical
values for the spin-orbit splitting in semiconductors are shown in Table
19.1. The listing in Table 19.1 gives the Γ-point splittings. We will
see that in crystalline solids the spin-orbit splittings are k-dependent.
For example, at the L-point, the spin-orbit splittings are typically about
2/3 of the Γ-point value.
The one-electron Hamiltonian for a solid including spin-orbit interaction
is
H =
p2
2m
+ V (r)
H0
+
1
2m2c2
( V × p) · S
H SO
. (19.10)
19.1. INTRODUCTION 533
(a) (b)
Figure 19.2: Energy versus dimensionless wave vector for a few highsymmetry
directions in germanium. (a) The spin-orbit interaction has
been neglected. (b) The spin-orbit interaction has been included and
the bands are labeled by the double group representations.
When the spin-orbit interaction is included, the wave functions consist
of a spatial part and a spin part. This means that the irreducible
representations that classify the states in a solid must depend on the
spin angular momentum. To show the effect of the k-dependence of the
spin-orbit interaction on the energy bands of a semiconductor, consider
the energy bands for germanium shown in Fig. 19.2(a) along the ∆(100)
axis, Λ(111) axis and Σ(110) axes for no spin-orbit interaction. Here
we show the four bonding and the four anti-bonding s and p bands.
This picture is to be compared with the energy bands with spin-orbit
interaction shown in Fig. 19.2.
We note that the Fermi level is between the top of the highest valence
band (the Γ25 or Γ+
25 band) and the bottom of the lowest conduc-
534 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
tion band (the L1 or L+
1 band). We note that the energy band extrema
for the more common semiconductors usually occur at high symmetry
points.
We further note that the inclusion of the spin-orbit interaction has
two major effects on the energy band structure affecting both the level
degeneracies and the labeling of the energy bands. Note that the (L±
4 +
L±
5 ) and (Λ4 +Λ5) are Kramers-degenerate doublet states, which means
that these bands stick together at high symmetry points and along high
symmetry directions, because of time reversal symmetry to be discussed
in Chapter 21. The Γ+
7 band which lies below the Γ+
8 valence band in
Fig. 19.2(b) is called the split-off band, and the separation between
the Γ+
7 and the Γ+
8 bands is the Γ-point spin-orbit splitting given in
Table 19.1.
19.2 Crystal Double Groups
Figure 19.2(b) shows energy bands that are labeled by irreducible representations
of the double group for the diamond structure. Double
groups come into play when we are dealing with the electron spin,
whereby half-integral angular momentum states are introduced. In this
section we discuss the double group irreducible representations which
arise when the electron spin is introduced.
The character tables for states of half-integral angular momentum
are constructed from the same basic formula as we used in Chapter 6 for
finding the characters for a rotation by an angle α in the full rotation
group:
χj(α) =
sin(j + 1/2)α
sin(α/2)
. (19.11)
Not only is Eq. (19.11) valid for integral j (as we have discussed in
Chapter 6) but the formula is also valid for j equal to half-integral
angular momentum states. We will now discuss the special issues that
must be considered for the case of half-integral spin.
We note that Eq. (19.11) behaves differently under the transformation
α → (α+2π) depending on whether j is an integral or half-integral
angular momentum state. This difference in behavior is responsible for
the name of double groups when j is allowed to assume half-integral
19.2. CRYSTAL DOUBLE GROUPS 535
values. We first consider how rotation by α+2π is related to a rotation
by α:
χj(α + 2π) =
sin(j + 1/2)(α + 2π)
sin α+2π
2
=
sin(j + 1/2)α · cos(j + 1/2)2π
sin(α/2) · cos π
(19.12)
since sin(j + 1/2)2π = 0 whether j is an integer or a half-integer. For
integral values of j, cos(j + 1/2)2π = −1 while for half-integral values
of j, cos(j + 1/2)2π = +1. Therefore we have the important relation
χj(α + 2π) = χj(α)(−1)2j
(19.13)
which implies that for integral j, a rotation by α, α ± 2π, α ± 4π,
etc, yields identical characters (integral values of j correspond to odddimensional
representations of the full rotation group), the dimensionality
being given by 2j + 1. For half-integral values of j, corresponding
to the even-dimensional representations of the rotation group, we have
χj(α ± 2π) = −χj(α)
χj(α ± 4π) = +χj(α) (19.14)
so that rotation by 4π is needed to yield the same character for χj(α).
The need to rotate by 4π (rather than by 2π) to generate the identity
operation leads to the concept of double groups which is the main theme
of this chapter.
Although the concept of double groups goes back to 1929 (H.A.
Bethe, Ann. der Phys. 3, 133 (1929)), experimental evidence that wave
functions for Fermions are periodic in 4π and not 2π was not available
until 1975 (S.A. Werner, R. Colella, A.W. Overhauser and C.F. Eagen,
Phys. Rev. Lett. 16, 1053 (1975)) when an ingenious experiment was
carried out to measure the phase shift of a neutron due to its precession
in a magnetic field. The experiment utilizes a neutron interferometer
and determines the phase shift of the neutron as it travels along path
AC, where it sees a magnetic field Bgap as opposed to path AB where
there is no magnetic field, as shown in Fig. 19.3(a). The phase shift
measured by counters C2 and C3 shows an interference pattern that is
periodic, as shown in Fig. 19.3(b), implying a magnetic field precession
with a periodicity of 4π.
536 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
(a) (b)
Figure 19.3: (a) A schematic diagram of the neutron interferometer
used to establish the phase of the electron wave function along the
path AC along which the neutrons are in a magnetic field B (0 to 500
G) for a distance (2 cm), while the path AB has no magnetic field.
(b) The periodic interference pattern as a function of magnetic field,
showing a periodicity of 4π, is presented on the right.
To account for this behavior of the wave function, it is convenient to
introduce a new group element in dealing with symmetry properties
of crystals for which half-integral values of the angular momentum
arise as, for example, through the introduction of the electron spin.
Let R denote a rotation by 2π, and now let us assume that R = ±E
or equivalently R2
= E, since the rotation by 4π leaves the characters
for the full rotation group invariant for both integral and half-integral j
values. Suppose that the elements of the crystal group without the electron
spin are E, A2, A3, . . . , Ah. Then with spin we have twice as many
group elements. That is, we have the same h elements that we had
before the spin on the electron was considered, plus h new elements
of the form RAi. Just as the matrix representation for the identity
operator E is the unit matrix 1 and for RE it is ±1, the matrix representation
for Ai is D(Γj)
(Ai) and for RAi it is ±D(Γj)
(Ai), depending
upon whether the representation Γj is related by compatibility relations
to even- or odd-dimensional representations of the full rotation
group. The introduction of this symmetry element R leads to no difficulties
with the quantum mechanical description of the problem, since
the wave functions ψ and −ψ describe the same physical problem and
19.3. DOUBLE GROUP PROPERTIES 537
the matrices ±D(Γj)(Ai)
each produce the same linear combination of
the basis functions.
Because of the introduction of the symmetry element R, the point
groups of the crystal have twice as many elements as before. These
point groups also have more classes, but not exactly twice as many
classes because some of the elements RAi are in the same classes as
other elements Ak. For example, according to Eq. (19.11), when j
assumes half-integral values, then we have for a C2 operation
χj(π) =
sin(j + 1/2)π
sin(π/2)
= 0 (19.15)
and
χj(π ± 2π) =
sin(j + 1/2)(π ± 2π)
sin π±2π
2
=
0
−1
= 0. (19.16)
As presented in §19.3, for some classes of two-fold axes, the elements
RC2 and C2 are, in fact, in the same class.
19.3 Double Group Properties
We will now state some properties of the even-dimensional representations
of the full rotation group and of double groups corresponding
to the half-integral angular momentum states. These properties are
given here without proof. More complete treatments can be found, for
example, in Heine’s book on group theory.
AUTHOR :Heine, Volker
TITLE :Group theory in quantum mechanics; an introduction to its present usage.
PUBLISHED :Oxford, New York, Pergamon Press [1960]
PHYSICAL DESC :468 p. illus.
SERIES :International series of monographs on pure and applied mathematics; v.9
QC174.5.H468 1960C
We list below four important rules for the properties of double
groups.
538 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
1. If a set of symmetry operations {Ak} forms a class in the original
point group, then {Ak} and the corresponding symmetry operations
for the double group {RAk} form 2 different classes in the
double group, except in the case noted below under heading 2.
2. The exceptions to property 1 are classes of rotations by π, if, and
only if, there is another two-fold axis ⊥ to the 2-fold axis under
consideration. In this case only, are C2 and RC2 in the same
class.
3. Any irreducible representation of the original group is also an
irreducible representation of the double group, with the same set
of characters.
4. In addition to the irreducible representations described in property
3, there must be additional double group representations,
so that we have as many irreducible representations as there are
classes. For these additional representations, the characters for
the class RCk are found from the characters of class Ck according
to the relation χ(RCk) = −χ(Ck). In the special case where
property 2 applies and {Ak} and {RAk} are in the same class,
then
χ(Ck) = +χ(RCk) = −χ(RCk) = 0 (19.17)
where χ(Ck) = +χ(RCk) since both types of symmetry operations
are in same class. The relation χ(Ck) = −χ(RCk) follows because
the signs of the wavefunctions change as a result of the symmetry
operation RCk. Therefore, for classes obeying property 2, it is
always the case that χ(C2)=0.
We can now write down the characters for double group representations
and relate these results to the spin-orbit interaction. In a solid,
without spin-orbit coupling
H0 =
p2
2m
+ V (r). (19.18)
Now if we include the electron spin, but still neglect the spin-orbit
interaction, the Bloch functions in the simplest case can be written as
ψ+
nk = eik·r
unk(r)α
19.3. DOUBLE GROUP PROPERTIES 539
ψ−
nk = eik·r
unk(r)β (19.19)
where α, β are the spin up and spin down eigenfunctions for spin 1
2
,
and n, k denote the band index and wave number, respectively, and
for a single electron with Sz = ±1
2
. Without spin-orbit coupling, each
state is doubly degenerate and is an eigenstate of Sz. If the spin-orbit
interaction is included, then the states are no longer eigenstates of Sz
and the wave function becomes some linear combination of the states
given by Eq. (19.19):
ψnk = aψ+
nk + bψ−
nk. (19.20)
The group theoretical way to describe these states is in terms of
the direct product Γi ⊗ D1/2 of the irreducible representations of the
spatial wave functions Γi with the irreducible representations of the
spin function which we will denote by D1/2.
To illustrate how we write the characters for D1/2, let us consider
cubic crystals with an O symmetry point group. (The results for Oh
are immediately obtained from O by taking the direct product O ⊗ i).
From the above, the classes of the double group for O are E, R, (3C2
4 +
3RC2
4 ), 6C4, 6RC4, (6C2 + 6RC2), 8C3, 8RC3. Having listed the classes
(8 in this case), we can now find the characters for D1/2 by the formula
χj(α) =
sin(j + 1/2)α
sin(α/2)
=
sin α
sin(α/2)
(19.21)
since j = 1/2. For the Full Rotational Symmetry group, the characters
for a rotation by α for the double O point group are:
α χ1
2
(α) χ1
2
(Rα)
0 α
α/2
= 2 −2
π 0 0
π
2
sin π
2
sin π
4
= 1
1√
2
=
√
2 −
√
2
π
3
sin 2π
3
sin π
3
=
√
3
2√
3
2
= 1 −1
This procedure for finding the characters for the spinor D1/2 is general
and can be done for any point group.
540 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Now we will write down the complete character table for the double
group O. In O itself, there are 24 elements, and therefore in the
double group derived from O there are 24 × 2 = 48 elements. There
are 8 classes and therefore 8 irreducible representations. From the ordinary
representations for point group O, we already have 5 irreducible
representations (see Table 3.32 in Chapter 3). Using rule 2 for the
character tables of double group representations, we have the following
condition for the dimensionality of the three additional double group
representations (Γ6, Γ7, Γ8) that are not present in the original group
i
2
i = h (19.22)
12
+ 12
+ 22
+ 32
+ 32
+ 2
6 + 2
7 + 2
8 = 48, (19.23)
yielding the following restriction on the dimensionalities of the double
group irreducible representations:
2
6 + 2
7 + 2
8 = 24. (19.24)
This allows us to fill in many of the entries in the double group character
table (Table 19.2). For example, we can not have any 5-dimensional representations,
because then 2
j = 25 > 24 for a 5-dimensional irreducible
representation. Among 1, 2, 3 and 4-dimensional irreducible representations,
the only combination we can make to satisfy Eq. (19.24) is:
22
+ 22
+ 42
= 24. (19.25)
We already have identified a 2-dimensional irreducible representation
of the double group, namely the “spinor” D1/2. We see immediately
that D1/2 obeys all the orthogonality relations, and the characters for
D1/2 can be added to the character table.
In Table 19.2 we have also filled in zeros for the characters for all the
C2 classes in the extra representations Γ+
6 , Γ+
7 and Γ+
8 . Using orthogonality
and normalization conditions which follow from the wonderful
orthogonality theorem on character, it is quite easy to complete this
character table. To get the Γ+
7 representation we have to consider the
following:
19.3. DOUBLE GROUP PROPERTIES 541
Table 19.2: Worksheet for the double group characters for the group
O.
E R 3C2
4 + 3RC2
4 6C4 6RC4 6C2 + 6RC2 8C3 8RC3
Γ1 1 1 1 1 1 1 1 1
Γ2 1 1 1 −1 −1 −1 1 1
Γ12 2 2 2 0 0 0 −1 −1
Γ15 3 3 −1 1 1 −1 0 0
Γ25 3 3 −1 −1 −1 1 0 0
Γ+
6 2 −2 0
√
2 −
√
2 0 1 −1
Γ+
7 2 −2 0 0
Γ+
8 4 −4 0 0
E 8C3 6C4
Γ+
6 2 1
√
2
Γ+
7 2 x y
and orthogonality requires 4 + 8x + 6
√
2y = 0 which is satisfied for
x = ±1, and y = −
√
2. Having filled in those entries it is easy to get
the 4-dimensional representation:
E 8C3 6C4
Γ+
6 2 1
√
2
Γ+
7 2 1 −
√
2
Γ+
8 4 x y
Orthogonality now requires: 8 + 8x ±
√
2y = 0 which is satisfied for
x = −1, y = 0. So now we have the whole character table, as shown
in Table 19.3.
In practice, we don’t have to construct these character tables because
the double group character tables have already been tabulated,
e.g., Koster’s article in Vol. 5 of the Seitz–Turnbull series, or the book on
the “Properties of the 32 Point Groups” by Koster, Dimmock, Wheeler
and Statz (QD911 .K86), or Miller and Love.
We will now apply the double group characters to a cubic crystal
with Oh symmetry at the Γ point, k = 0. The spin functions α and β
542 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Table 19.3: Double group character table for the group O.
O E R 3C2
4 + 3RC2
4 6C4 6RC4 6C2 + 6RC2 8C3 8RC3
Γ1 1 1 1 1 1 1 1 1
Γ2 1 1 1 −1 −1 −1 1 1
Γ12 2 2 2 0 0 0 −1 −1
Γ15 3 3 −1 1 1 −1 0 0
Γ25 3 3 −1 −1 −1 1 0 0
Γ6 2 −2 0
√
2 −
√
2 0 1 −1
Γ7 2 −2 0 −
√
2
√
2 0 1 −1
Γ8 4 −4 0 0 0 0 −1 1
Table 19.4: Direct products Γi ⊗ Γ+
6 for Oh symmetry.
Γ+
1 ⊗ Γ+
6 = Γ+
6 Γ−
1 ⊗ Γ+
6 = Γ−
6
Γ+
2 ⊗ Γ+
6 = Γ+
7 Γ−
2 ⊗ Γ+
6 = Γ−
7
Γ+
12 ⊗ Γ+
6 = Γ+
8 Γ−
12 ⊗ Γ+
6 = Γ−
8
Γ+
15 ⊗ Γ+
6 = Γ+
6 + Γ+
8 Γ−
15 ⊗ Γ+
6 = Γ−
6 + Γ−
8
Γ+
25 ⊗ Γ+
6 = Γ+
7 + Γ+
8 Γ−
25 ⊗ Γ+
6 = Γ−
7 + Γ−
8
transform as the partners of the irreducible representation D1/2 which
is written as Γ+
6 . Now we see that the appropriate double group representations
(which must be used when the effects of the electron spin are
included) are obtained by taking the direct product of the irreducible
representation Γi with the spinor (D1/2) as shown in Table 19.4.
When the spin-orbit interaction is introduced into the description
of the electronic structure, then the energy bands are labeled by double
group irreducible representations (e.g., Γ±
6 , Γ±
7 and Γ±
8 for the Oh group
at k = 0). Table 19.4 shows that the 1-dimensional representations
without spin-orbit interaction Γ±
1 and Γ±
2 all become doubly degenerate.
This result is independent of the symmetry group. When the
spin-orbit interaction is introduced, all formerly non-degenerate levels
therefore become double degenerate. (This effect is called the Kramers
degeneracy.)
19.3. DOUBLE GROUP PROPERTIES 543
In the case of the Oh group, the 2-fold levels Γ±
12 become 4-fold
degenerate when spin is included as is shown in Table 19.4. However
something different happens for the triply degenerate Γ±
15 and Γ±
25
states. These states would become 6-fold degenerate with spin, but the
spin-orbit interaction partly lifts this degeneracy so that these 6-fold
levels split into a 2-fold and a 4-fold level, just as in the atomic case.
Group theory does not tell us the ordering of these levels, nor the magnitude
of the splitting, but it does give the symmetry of the levels. By
including the spin-orbit interaction in dealing with the valence band
of a semiconductor like germanium, the 6-fold level can be partially
diagonalized; the (6 × 6) k · p effective Hamiltonian breaks up into a
(2 × 2) block and a (4 × 4) block.
Figures 19.2(a) and (b) show the effect of the spin-orbit interaction
on the energy bands of germanium. We note that the magnitudes of
the spin-orbit splittings are k dependent. These effects are largest
at k = 0, moderately large along the (111) direction (Λ) and at the
L-point, but much smaller along the (100) direction (∆) and at the
X-point. Group theory doesn’t provide information on these relative
magnitudes.
As was mentioned above, the spin-orbit interaction effects tend to be
very important in the III–V compound semiconductors. Since in
this case the two atoms in the unit cell correspond to different species,
the appropriate point group at k = 0 is Td and the bonding and antibonding
bands both have symmetries Γ1 and Γ15. The general picture
of the energy bands for the III–V compounds is qualitatively similar to
that given in Figs. 19.2(a) and (b) except for a generally larger spinorbit
splitting.
Another important class of semiconductors where the spin-orbit interaction
is important is the narrow gap lead salts (e.g., PbTe). Since
Pb has a high atomic number it is necessary to give a more exact theory
for the spin-orbit interaction in this case including relativistic correction
terms. However, the group theoretical considerations given here
apply equally well. Setting up a secular equation when the spin-orbit
interaction is large, as for example in the lead salts, is treated in J.B.
Conklin, L.E. Johnson and G.W. Pratt, Jr., Phys. Rev. 137, A1282
(1965).
544 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
In writing down the double group irreducible representations, we
see that a particular representation may be associated with various
single group representations. For example, the direct products in Table
19.4 show that the Γ+
7 irreducible double group representation could
be associated with either a Γ+
2 , aΓ+
15 or a Γ+
25 irreducible single group
representation. In dealing with basis functions in the double group representations
it is often useful to know which single group representation
corresponds to a particular double group representation. The standard
notation used for this association is for example Γ+
8 (Γ+
12), in which the
appropriate single group representation is put in parenthesis, indicating
that Γ+
8 arises from the direct product Γ+
12 ⊗ Γ+
6 rather than from one
of the other possibilities listed in Table 19.4.
19.4 Crystal Field Splitting Including SpinOrbit
Coupling
In our treatment of crystal field splittings in solids in chapter 6 we
ignored the spin-orbit coupling. The treatment in Chapter 6 thus provided
a first approximation for describing the crystal field levels for the
impurity ions in a host lattice. To improve on this, we consider in this
chapter the effect of the spin-orbit interaction which will allow us to
treat crystal field splittings in host lattices with rare earth ions, and
also to obtain a better approximation to the crystal field splittings for
3d transition metal ions discussed in Chapter 6.
The introduction of a transition-metal ion in an atomic d-state into
an octahedral crystal field gives rise to crystal field splittings as shown
in Fig. 19.4 (see §6.4). For a single d-electron, s = 1/2 and the
appropriate double group representation for the spinor is Γ+
6 . Thus
when the spin-orbit interaction is included in the crystal field problem,
the d-levels are further split. Thus the 2-fold crystal field level in cubic
symmetry transforms as
Γ+
12 ⊗ Γ+
6 = Γ+
8 (19.26)
and the 3-fold crystal field level in Oh symmetry is split according to
Γ+
25 ⊗ Γ+
6 = Γ+
7 + Γ+
8 . (19.27)
19.4. CRYSTAL FIELD INCLUDING SPIN-ORBIT 545
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Figure 19.4: Schematic diagram of the crystal field splitting of a 2
D
state, followed by further splitting by the spin-orbit interaction. This
model is appropriate for a 3d transition metal ion in a crystal with Oh
symmetry. The degeneracy of each of the levels is indicated by the
parentheses. Also shown in this figure are the labels for the crystal
field levels associated with each of the Γ+
8 levels in the absence of the
spin-orbit interaction. Below the crystal field splitting diagram, the
form of the crystal field Hamiltonian is indicated on the left, in the
absence of the spin-orbit interaction, and on the right when the spinorbit
interaction is included.
546 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
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Figure 19.5: Schematic diagram of the spin-orbit splitting of a 2
D level
and of the subsequent crystal field splittings of these levels in a cubic
field for an ion with a spin-orbit interaction that is large compared to
the crystal field splittings. The degeneracy of each level is shown in
parentheses.
19.4. CRYSTAL FIELD INCLUDING SPIN-ORBIT 547
Table 19.5: Decomposition of double group representations for a d
band.
E R 3C2
4 + 3RC2
4 6C4 6RC4 6C2 + 6RC2 8C3 8RC3
χ(2
D5/2) 6 −6 0 −
√
2
√
2 0 0 0
χ(2
D3/2) 4 −4 0 0 0 0 −1 1
In Eqs. (19.26) and (19.27) Γ+
12 and Γ+
25 denote the spatial wave-functions
and Γ+
6 denotes the spin wave-function. Here we see that the Eg level
does not split further but the T2g level splits into a 2-fold and a 4-fold
level.
For the 2
D state of the 3d transition-metal ion in a host crystal,
we use Fig. 19.4 (where the number of states is given in parentheses).
The analysis in Fig. 19.4 is valid only if the crystal field interaction
is large compared with the spin-orbit splitting. This situation
describes the iron-group transition metal ions.
When we move down the periodic table to the palladium group
(4d) and the platinum group (5d), the spin-orbit interaction is large
compared with the crystal field. In this case, we consider first the
spin-orbit splitting of the free ion state as the major perturbation (see
Fig. 19.5). We now have to consider the effect of the crystal field on
these levels. To compute the characters for the full rotation group we
use the formula
χj(α) =
sin(j + 1/2)α
sin(α/2)
. (19.28)
We then find the characters for the 2
D5/2 and 2
D3/2 states to see how
they split in the cubic field (see Table 19.5). Using Table 19.5 we see
immediately that
χ2D5/2
→ Γ7 + Γ8 (19.29)
χ2D3/2
→ Γ8 (19.30)
as indicated in Fig. 19.5. The symmetries in Figs. 19.4 and 19.5 for
the levels in the presence of both the spin-orbit interaction and the
cubic field of the crystalline solid are Γ+
7 + 2Γ+
8 in both cases. However
in Fig. 19.5, the crystal field splittings are small compared with the
spin-orbit splittings in contrast to the case in Fig. 19.4.
548 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Let us consider another example of crystal field levels that show
some new features. Consider the levels of the holmium ion Ho3+
in a
cubic field for which the atomic configuration is 4f10
5s2
5p6
so that by
Hund’s rule the ground state after the spin-orbit interaction is included
becomes s = 2, l = 6, j = 8 denoted by the spectroscopic notation 5
I8.
Since j = 8 is an integer, application of the formula
χj(α) =
sin(j + 1/2)α
sin(α/2)
(19.31)
gives only ordinary irreducible representations, even though the electron
spin is included. We thus get for the characters for the ground
state 5
I8:
E 3C2
4 6C4 6C2 8C3
χ( 5
I8) 17 1 1 1 −1
χ( 4
I15/2) 16 0 0 0 −1
Decomposition of the χ(5
I8) level into irreducible representations of Oh
yields
χ( 5
I8) → Γ1 + 2Γ12 + 2Γ15 + 2Γ25 (19.32)
where there are 7 levels for 17 states.
Finding the crystal field splittings for a 17-fold level would be a
very difficult problem without group theory. As another example, let
us consider the erbium ion Er3+
in a host crystal. This ion is the basis
for recent applications of amplification capabilities in optical fibers.
We consider the level splitting for the rare earth ion Er3+
in a
4f11
5s2
p6
which gives a 4
I15/2 ground state. The characters for the
j = 15/2 state are given in the above table and the splitting of these
states in a cubic Oh field is also included in the Table above. The
j = 15/2 state splits in cubic symmetry into:
χ( 4
I15/2) → Γ6 + Γ7 + 3Γ8.
In dealing with the crystal field problem, we often encounter a situation
where a perturbation is applied to lower the crystal symmetry.
19.4. CRYSTAL FIELD INCLUDING SPIN-ORBIT 549
In such cases we follow the procedure which we have used many times
before – the irreducible representation of the high symmetry group is
treated as a reducible representation for the lower symmetry group and
we look for the irreducible representations contained therein. The only
difference in including the spin-orbit interaction is the use of double
groups for all point groups – both for the high symmetry and the low
symmetry groups. It is the case that the single group irreducible representations
in a group of higher symmetry will always go into single
group irreducible representations of the lower symmetry group. For
example, the level Γ8 in point group O goes into Γ4 + Γ5 + Γ6 in point
group D3, when the symmetry is lowered.
In considering optical transitions between crystal field states which
are described by either single or double group representations, the electromagnetic
interaction Hamiltonian will in all cases transform as the
vector within the single group representations. Thus, we can write
Γ−
15 ⊗ Γ−
7 = Γ+
7 + Γ+
8 (19.33)
for the coupling of the electromagnetic field to the conduction band of
germanium at k = 0. Thus, single group states are optically coupled to
other single group states and double group states are optically coupled
to other double group states,
Whereas the wave function for a single electron transforms as D1/2
(or Γ+
6 for Oh symmetry), a two-electron wavefunction transforms as
the direct product D1/2 ⊗ D1/2. For Oh symmetry, we have
Γ+
6 ⊗ Γ+
6 = Γ+
1 + Γ+
15 (19.34)
where Γ+
1 is the singlet s = 0 state and the Γ+
15 corresponds to the
triplet s = 1 level. We note that in both cases the levels have integral
values of spin angular momentum and thus the state transforms as
a single group irreducible representation. Finally we note that for a
D3/2 p-state in full rotational symmetry, the double group irreducible
representation in cubic symmetry is Γ−
8 and no further splitting occurs.
550 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
19.5 Comment on the Use of the Koster
et al. Reference for Double Groups
An important group theory reference book on the 32 point groups and
their double groups is “Properties of the Thirty-Two Point Groups”, by
G.F. Koster, J.O. Dimmock, R.G. Wheeler and H. Statz (QD911 .K86).
The book is especially valuable for its many useful tables including
tables for double groups. In this reference you will find for each of the
32 point groups:
1. a character table including the double group representations (see
e.g., Table 19.6 for groups O and Td);
2. a table giving the decomposition of the direct product of any 2
irreducible representations (an example of such a table is given in
Table 19.7);
3. tables of coupling coefficients for the product of any two basis
functions (which are explained in Table 19.8);
4. compatibility tables with other point groups (Table 19.10);
5. compatibility tables with the Full Rotation Group (Table 19.11).
We will illustrate some examples of these tables for the group O,
which is tabulated together with Td (p. 88–101 in Koster’s book). The
first table we reproduce is (Table 19.6). Note the listing of the double
group representations and basis functions for the double group representations.
The basis functions for Γ4(Γ15) are Sx, Sy, Sz which refer
to the three Cartesian components of the angular momentum (integral
values of angular momentum). The basis functions for the Γ6 and
Γ8 irreducible representations are written in the basic form
φ(j, mj) for the angular momentum and all the mj partners are
listed. Koster et al. use the notation ¯E for R and they use ¯C3 for
RC3. The meaning of the time inversion entries will be explained in
Chapter 21 where time inversion symmetry is discussed.
Table 19.7 for groups O and Td gives the decomposition of the direct
product of any irreducible representation labeling a column with one
19.5. USE OF THE KOSTER ET AL. REFERENCE 551
Table 19.6: Character Table and Basis Functions for the Groups O and
Td
O E ¯E 8C3 8 ¯C3
3C2
3 ¯C2
6C4 6 ¯C4
6C2
6 ¯C2
Td E ¯E 8C3 8 ¯C3
3C2
3 ¯C2
6S4 6 ¯S4
6σd
6¯σd
Time
Inv.
Bases
for O
Bases
for Td
Γ1 1 1 1 1 1 1 1 1 a R R or xyz
Γ2 1 1 1 1 1 -1 -1 -1 a xyz SxSySz
Γ3(Γ12) 2 2 -1 -1 2 0 0 0 a (2z2 −x2 −y2),√
3(x2 − y2)
(2z2 −x2 −y2),√
3(x2 − y2)
Γ4(Γ15) 3 3 0 0 -1 1 1 -1 a Sx, Sy, Sz Sx, Sy, Sz
Γ5(Γ25) 3 3 0 0 -1 -1 -1 1 a yz, xz, xy x, y, z
Γ6 2 -2 1 -1 0
√
2 -
√
2 0 c φ(1/2, −1/2),
φ(1/2, 1/2)
φ(1/2, −1/2),
φ(1/2, 1/2)
Γ7 2 -2 1 -1 0 -
√
2
√
2 0 c Γ6 ⊗ Γ2 Γ6 ⊗ Γ2
Γ8 4 -4 -1 1 0 0 0 0 c φ(3/2, −3/2),
φ(3/2, −1/2),
φ(3/2, 1/2),
φ(3/2, 3/2)
φ(3/2, −3/2),
φ(3/2, −1/2),
φ(3/2, 1/2),
φ(3/2, 3/2)
Table 19.7: Table of direct products of irreducible representations for
the groups O and Td
Γ1 Γ2 Γ3 Γ4 Γ5 Γ6 Γ7 Γ8
Γ1 Γ2 Γ3 Γ4 Γ5 Γ6 Γ7 Γ8 Γ1
Γ1 Γ3 Γ5 Γ4 Γ7 Γ6 Γ8 Γ2
Γ1 + Γ2 + Γ3 Γ4 + Γ5 Γ4 + Γ5 Γ8 Γ8 Γ6 + Γ7 + Γ8 Γ3
Γ1 + Γ3 + Γ4 + Γ5 Γ2 + Γ3 + Γ4 + Γ5 Γ6 + Γ8 Γ7 + Γ8 Γ6 + Γ7 + 2Γ8 Γ4
Γ1 + Γ3 + Γ4 + Γ5 Γ7 + Γ8 Γ6 + Γ8 Γ6 + Γ7 + 2Γ8 Γ5
Γ1 + Γ4 Γ2 + Γ5 Γ3 + Γ4 + Γ5 Γ6
Γ1 + Γ4 Γ3 + Γ4 + Γ5 Γ7
Γ1 + Γ2 + Γ3
+2Γ4 + 2Γ5
Γ8
labeling a row, which is entered in the matrix position of their intersection.
In our experience with Koster et al., there are a few typographical
errors in the tables, so beware!
The extensive tables of coupling coefficients are perhaps the most
useful tables in Koster et al. These tables give the basis functions for
the irreducible representations obtained by taking the direct product
of two irreducible representations. We illustrate in Table 19.8 the basis
functions obtained by taking the direct product of each of the two
partners of the Γ12 representation (denoted by Koster et al. as u3
1 and
u3
2) with each of the 3 partners of the Γ15 representation (denoted as
v4
x, v4
y, v4
z ) to yield 3 partners with Γ15 symmetry (denoted by ψ4
x, ψ4
y, ψ4
z )
552 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Table 19.8: Coupling coefficients for the indicated basis functions for
single group O.
u3
1v4
x u3
1v4
y u3
1v4
z u3
2v4
x u3
2v4
y u3
2v4
z
ψ4
x −1/2 0 0
√
3/2 0 0
ψ4
y 0 −1/2 0 0 −
√
3/2 0
ψ4
z 0 0 1 0 0 0
ψ5
yz −
√
3/2 0 0 −1/2 0 0
ψ5
xz 0
√
3/2 0 0 −1/2 0
ψ5
xy 0 0 0 0 0 1
and 3 partners with Γ25 symmetry (denoted by ψ5
yz, ψ5
zx, ψ5
xy). This is
table 83 on p. 91 of Koster et al. From Table 19.8 we see that the
appropriate linear combinations for the ψ4
and ψ5
functions are:
ψ4
x = −(1/2)u3
1v4
x + (
√
3/2)u3
2v4
x
ψ4
y = −(1/2)u3
1v4
y − (
√
3/2)u3
2v4
y
ψ4
z = u3
1v4
z
ψ5
yz = −(
√
3/2)u3
1v4
x − (1/2)u3
2v4
x
ψ5
xz = (
√
3/2)u3
1v4
y − (1/2)u3
2v4
y
ψ5
xy = u3
2v4
z
Note that the basis functions for the ψ4
and ψ5
functions depend on
the choice of basis functions for u and v. Journal articles often use the
notation
Γ15 ⊗ Γ12 = Γ15 + Γ25 (19.35)
where Γ4 ↔ Γ15 and Γ3 ↔ Γ12. Thus taking the direct product between
irreducible representations Γ3 and Γ4 in O or Td symmetries yields:
Γ4 ⊗ Γ3 = Γ4 + Γ5 (19.36)
where Γ5 ↔ Γ25.
We next illustrate the use of a typical coupling coefficient table
relevant to the introduction of spin into the electronic energy level
19.5. USE OF THE KOSTER ET AL. REFERENCE 553
Table 19.9: Coupling coefficient tables for the indicated basis functions
for double group Oh.
u4
xv6
−1/2 u4
xv6
1/2 u4
yv6
−1/2 u4
yv6
1/2 u4
zv6
−1/2 u4
zv6
1/2
ψ6
−1/2 0 −i/
√
3 0 −1/
√
3 i/
√
3 0
ψ6
1/2 −i/
√
3 0 1/
√
3 0 0 −i/
√
3
ψ8
−3/2 i/
√
2 0 1/
√
2 0 0 0
ψ8
−1/2 0 i/
√
6 0 1/
√
6 i
√
2/
√
3 0
ψ8
1/2 −i/
√
6 0 1/
√
6 0 0 i
√
2/
√
3
ψ8
3/2 0 −i/
√
2 0 1/
√
2 0 0
problem. In this case we need to take a direct product of Γ+
6 with
a single group representation, where Γ+
6 is the representation for the
spinor (D1/2). For example, for a p-level Γ−
15 ⊗ Γ+
6 = Γ−
6 + Γ−
8 and
the appropriate coupling coefficient table is Table 19.9 (in Koster et al.
Table 83, p. 92).
Table 19.9 gives us the following relations between basis functions:
ψ6
−1/2 =
1
2
, −
1
2
= −(i/
√
3)(u4
x − iu4
y) ↑ +(i/
√
3)u4
z ↓
ψ6
1/2 =
1
2
,
1
2
= −(i/
√
3)(u4
x + iu4
y) ↓ −(i/
√
3)u4
z ↑
ψ8
−3/2 =
3
2
, −
3
2
= (i/
√
2)(u4
x − iu4
y) ↓
ψ8
−1/2 =
3
2
, −
1
2
= (i/
√
6)(u4
x − iu4
y) ↑ +(i
√
2/
√
3)u4
z ↓
ψ8
1/2 =
3
2
,
1
2
= −(i/
√
6)(u4
x − iu4
y) ↓ +(i
√
2/
√
3)u4
z ↑
ψ8
3/2 =
3
2
,
3
2
= −(i/
√
2)(u4
x + iu4
y) ↑ (19.37)
where we note that Γ−
15 = Γ4 and v6
1/2 =↑ and v6
−1/2 =↓. The relations
in Eq. (19.37) give the transformation of basis functions in the
| sm ms representation to the |j smj representation, appropriate to
energy bands for which the spin-orbit interaction is included. These
554 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Table 19.10: Compatibility Table
Td O Γ1 Γ2 Γ3 Γ4 Γ5 Γ6 Γ7 Γ8
T T Γ1 Γ1 Γ2 + Γ3 Γ4 Γ4 Γ5 Γ5 Γ6 + Γ7
D2d D4 Γ1 Γ3 Γ1 + Γ3 Γ2 + Γ5 Γ4 + Γ5 Γ6 Γ7 Γ6 + Γ7
C3v; E(w) D3 Γ1 Γ2 Γ3 Γ2 + Γ3 Γ1 + Γ3 Γ4 Γ4 Γ4 + Γ5 + Γ6
S4 : H(z) C4 : H(z) : E(z) Γ1 Γ1 Γ2 + Γ3 Γ1 + Γ2 + Γ3 Γ1 + Γ2 + Γ3 Γ4 + Γ5 Γ4 + Γ5 Γ5 + Γ6 + Γ7 +
C2v : E(z) Γ1 Γ3 Γ1 + Γ3 Γ2 + Γ3 + Γ4 Γ1 + Γ2 + Γ4 Γ5 Γ5 2Γ5
Cs : E(v) : H(v) C2 : E(v) : H(v) Γ1 Γ2 Γ1 + Γ2 Γ1 + 2Γ2 2Γ1 + Γ2 Γ3 + Γ4 Γ3 + Γ4 2Γ3 + 2Γ4
linear combinations are basically the Clebsch-Gordan coefficients
in quantum mechanics. We make use of Eq. (19.37) in the next section
when we discuss the introduction of spin and spin-orbit interaction into
the plane wave relations of the energy eigenvalues of the empty lattice.
Table 19.10 gives the point groups that are subgroups of Groups
Td and O, and the decomposition of the irreducible representations of
Td and O into the irreducible representations of the lower symmetry
group. Note in Table 19.10 that E refers to the electric field and H to
the magnetic field. The table can be used for many applications such
as finding the resulting symmetries under crystal field splittings as for
example Oh → D3.
The notation for each of the irreducible representations is consistent
with that given in the character tables of Koster’s book. The decomposition
of the irreducible representations of the full rotation group into
irreducible representations of groups O and Td is given in Tables 19.11
and 19.12. Note that all the irreducible representations of the full rotation
group are listed, with the ± sign denoting the parity (even or
odd under inversion) and the subscript giving the angular momentum
quantum number (j), so that the dimensionality of the irreducible representation
D±
j is (2j + 1).
19.6 Plane Wave Basis Functions for Double
Group Representations
In Chapter 16 we discussed the nearly free electron approximation for
the energy bands in crystalline solids, neglecting the electron spin. In
this case the electron wave functions were expressed in terms of symmetrized
linear combinations of plane waves transforming according to
irreducible representations of the group of the wave vector.
19.6. PLANE WAVE FUNCTIONS FOR DOUBLE GROUPS 555
Table 19.11: Full Rotation Group Compatibility Table for the Group
O
S D+
0 Γ1
P D−
1 Γ4
D D+
2 Γ3 + Γ5
F D−
3 Γ2 + Γ4 + Γ5
G D+
4 Γ1 + Γ3 + Γ4 + Γ5
H D−
5 Γ3 + 2Γ4 + Γ5
I D+
6 Γ1 + Γ2 + Γ3 + Γ4 + 2Γ5
D±
1/2 Γ6
D±
3/2 Γ8
D±
5/2 Γ7 + Γ8
D±
7/2 Γ6 + Γ7 + Γ8
D±
9/2 Γ6 + 2Γ8
D±
11/2 Γ6 + Γ7 + 2Γ8
D±
13/2 Γ6 + 2Γ7 + 2Γ8
D±
15/2 Γ6 + Γ7 + 3Γ8
556 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Table 19.12: Full Rotation Group Compatibility Table for the Group
Td
D+
0 Γ1 D−
0 Γ2
D+
1 Γ4 D−
1 Γ5
D+
2 Γ3 + Γ5 D−
2 Γ3 + Γ4
D+
3 Γ2 + Γ4 + Γ5 D−
3 Γ1 + Γ4 + Γ5
D+
4 Γ1 + Γ3 + Γ4 + Γ5 D−
4 Γ2 + Γ3 + Γ4 + Γ5
D+
5 Γ3 + 2Γ4 + Γ5 D−
5 Γ3 + Γ4 + 2Γ5
D+
6 Γ1 + Γ2 + Γ3 + Γ4 + 2Γ5 D−
6 Γ1 + Γ2 + Γ3 + 2Γ4 + Γ5
D+
1/2 Γ6 D−
1/2 Γ7
D+
3/2 Γ8 D−
3/2 Γ8
D+
5/2 Γ7 + Γ8 D−
5/2 Γ6 + Γ8
D+
7/2 Γ6 + Γ7 + Γ8 D−
7/2 Γ6 + Γ7 + Γ8
D+
9/2 Γ6 + 2Γ8 D−
9/2 Γ7 + 2Γ8
D+
11/2 Γ6 + Γ7 + 2Γ8 D−
11/2 Γ6 + Γ7 + 2Γ8
D+
13/2 Γ6 + 2Γ7 + 2Γ8 D−
13/2 2Γ6 + Γ7 + 2Γ8
19.6. PLANE WAVE FUNCTIONS FOR DOUBLE GROUPS 557
In this section, we give an explicit example for Oh symmetry for
the corresponding situation where the spin of the electron is included
and the wave functions are described in terms of the double group
irreducible representations.
It is relatively simple to include the effect of the electron spin for the
irreducible representations Γ±
1 and Γ±
2 because there are no splittings
induced by the spin-orbit coupling. Thus the basis functions in this case
are simple product functions given by Γ±
6 = Γ±
1 ⊗Γ+
6 and Γ±
7 = Γ±
2 ⊗Γ+
6
or more explicitly
ΨΓ±
6
(K) = ψΓ±
1
(K)
α
β
ΨΓ±
7
(K) = ψΓ±
2
(K)
α
β
(19.38)
in which the ψΓ±
1
(K) and ψΓ±
2
(K) denote symmetrized plane wave combinations
considered in Chapter 16, ignoring the effect of the electron
spin, while α and β denote spin up and spin down functions, respectively,
which form partners of the Γ+
6 double group irreducible repre-
sentation.
For the degenerate plane wave combinations such as those with Γ±
12,
Γ±
15 and Γ±
25 symmetries, one method to find an appropriate set of basis
functions when the electron spin is included is to use the Koster tables
discussed in §19.5. For example, basis functions for the four partners
for Γ±
8 = Γ±
3 ⊗ Γ+
6 can be found in the following Koster table:
u3
1v6
−1/2 u3
1v6
+1/2 u3
2v6
−1/2 u3
2v6
+1/2
ψ8
−3/2 0 0 0 1
ψ8
−1/2 1 0 0 0
ψ8
+1/2 0 −1 0 0
ψ8
+3/2 0 0 −1 0
where the Koster functions u3
1, u3
2 for Γ3 are related to the ψΓ12 of Table
17.3 on p. 507 by:
u3
1 ∝ 3z2
− r2
∼ [ωψΓ+
12
+ ω2
ψ∗
Γ+
12
]
u3
2 ∝
√
3(x2
− y2
) ∼ [ωψΓ+
12
− ω2
ψ∗
Γ+
12
]
(19.39)
558 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
and
v6
+1/2 ∝ α
v6
−1/2 ∝ β.
(19.40)
Thus the application of Koster’s table gives:
ΨΓ±
8
(K) = 1√
2






√
3(x2
− y2
)α
(3z2
− r2
)β
−(3z2
− r2
)α
−
√
3(x2
− y2
)β






(19.41)
A more symmetric set of basis functions for Γ±
8 = Γ±
12 ⊗ Γ+
6 is
ΨΓ±
8
(K) = 1√
2








[ω2
ψ∗
Γ±
12
(K) + ωψΓ±
12
(K)]α
−i[ω2
ψ∗
Γ±
12
(K) − ωψΓ±
12
(K)]β
i[ω2
ψ∗
Γ±
12
(K) − ωψΓ±
12
(K)]α
−[ω2
ψ∗
Γ±
12
(K) + ωψΓ±
12
(K)]β








(19.42)
in which ψΓ+
12
(K) = x2
+ ωy2
+ ω2
z2
.
Since the 3-dimensional levels Γ±
15 and Γ±
25 split under the spin-orbit
interaction
Γ±
15 ⊗ D1/2 = Γ±
6 + Γ±
8
Γ±
25 ⊗ D1/2 = Γ±
7 + Γ±
8
the basis functions for these levels are somewhat more complicated. In
these cases we can use the following tables from Koster’s Table 83 on
coupling coefficients (see Table 19.9):
u4
xv6
−1/2 u4
xv6
+1/2 u4
yv6
−1/2 u4
yv6
+1/2 u4
zv6
−1/2 u4
zv6
+1/2
ψ6
−1/2 0 −i/
√
3 0 −1/
√
3 i/
√
3 0
ψ6
+1/2 −i/
√
3 0 1/
√
3 0 0 −i/
√
3
ψ8
−3/2 i/
√
2 0 1/
√
2 0 0 0
ψ8
−1/2 0 i/
√
6 0 1/
√
6 i
√
2/
√
3 0
ψ8
+1/2 −i/
√
6 0 1/
√
6 0 0 i
√
2/
√
3
ψ8
+3/2 0 −i/
√
2 0 1/
√
2 0 0
19.6. PLANE WAVE FUNCTIONS FOR DOUBLE GROUPS 559
and
u5
xv6
−1/2 u5
xv6
+1/2 u5
yv6
−1/2 u5
yv6
+1/2 u5
zv6
−1/2 u5
zv6
+1/2
ψ7
−1/2 0 −i/
√
3 0 −1/
√
3 i/
√
3 0
ψ7
+1/2 −i/
√
3 0 1/
√
3 0 0 −i/
√
3
ψ8
−3/2 −i/
√
6 0 1/
√
6 0 0 i
√
2/
√
3
ψ8
−1/2 0 i/
√
2 0 −1/
√
2 0 0
ψ8
+1/2 −i/
√
2 0 −1/
√
2 0 0 0
ψ8
+3/2 0 i/
√
6 0 1/
√
6 i
√
2/
√
3 0
from which we obtain for the two-fold levels:
ΨΓ±
6
(K) = 1√
3








− i ψx
Γ±
15
(K) − iψy
Γ±
15
(K) α + iψz
Γ±
15
(K)β
− i ψx
Γ±
15
(K) + iψy
Γ±
15
(K) β − iψz
Γ±
15
(K)α








ΨΓ±
7
(K) = 1√
3






− i ψx
Γ±
25
(K) − iψy
Γ±
25
(K) α + iψz
Γ±
25
(K)β
− i ψx
Γ±
25
(K) + iψy
Γ±
25
(K) β − iψz
Γ±
25
(K)α






(19.43)
The corresponding 4-fold levels are also found from the same two Koster
tables:
ΨΓ±
8
(K) = 1√
6



















i
√
3 ψx
Γ±
15
(K) − iψy
Γ±
15
(K) β
i ψx
Γ±
15
(K) − iψy
Γ±
15
(K) α + 2iψz
Γ±
15
(K)β
− i ψx
Γ±
15
(K) + iψy
Γ±
15
(K) β + 2iψz
Γ±
15
(K)α
− i
√
3 ψx
Γ±
15
(K) + iψy
Γ±
15
(K) α



















(19.44)
560 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
and
ΨΓ±
8
(K) =
1
√
6



















− i ψx
Γ±
25
(K) + iψy
Γ±
25
(K) β + 2iψz
Γ±
25
(K)α
i
√
3 ψx
Γ±
25
(K) + iψy
Γ±
25
(K) α
− i
√
3 ψx
Γ±
25
(K) − iψy
Γ±
25
(K) β
i ψx
Γ±
25
(K) − iψy
Γ±
25
(K) α + 2iψz
Γ±
25
(K)β



















(19.45)
in which the ψΓi
(K) for the single group are the plane wave combinations
obtained in Chapter 16 for the plane wave solutions to the electronic
problem, and α, β are, respectively, the spin up and spin down
spin states. The basis functions for Γ±
8 (Γ±
15) are given in Eq. (19.44)
and the basis functions for Γ±
8 (Γ±
25) are given in Eq. (19.45).
If no special precautions are taken, the various partners of the Γ±
8
representation will differ from one another, depending on whether they
come from a Γ±
3 , a Γ±
4 or a Γ±
5 orbital state. In all cases, the sets of partners
for the Γ±
8 representation will be related to one another through a
unitary transformation. However, for some applications it is desirable
that the unitary matrix be the identity or unit matrix so that all partners
for a given irreducible representation look the same, independent
of origin. Koster’s tables generally require use of a unitary transformation
(other than the unitary matrix) to relate the various Γ±
8 basis
functions. Despite this possible undesirable feature, the completeness
of coverage in Koster’s book make it a very useful resource for research
problems in this topic.
To treat the plane wave states at other points in the Brillouin zone,
we again use the direct product approach. For the orbital wave functions
we use the procedures outlined in Chapter 16 to get the appropriate
symmetrized combinations of plane waves for each of the high
symmetry points and axes in the Brillouin zone using the character
tables for the group of the wave vector at these symmetry points. For
the same group of the wave vector, we use the appropriate Koster table
for coupling coefficients which give the basis functions for the direct
19.7. USE OF REFERENCE BOOKS 561
product between the orbital plane wave states and the spinors. In this
way the E(k) relations in the nearly free electron approximation can
be found, including the effect of the electron spin and the spin-orbit
interaction.
19.7 Use of Reference Books to Find the
Group of the Wave Vector for NonSymmorphic
Groups
In Chapters 16 and 17 we discussed the form of the E(k) relations for
symmorphic space groups, neglecting the spin and the spin-orbit interaction.
In the case of non-symmorphic space groups we found in §16.5
that bands are often required to stick together at certain high symmetry
points on the Brillouin zone boundary where the structure factor
vanishes. In §16.5 it was explicitly shown that for the diamond structure
the non-degenerate ∆1 and ∆2 levels come into the X point with
equal and opposite non-zero slopes, so that in the extended Brillouin
zone, the E(k) curves pass through the X point continuously together
with all their derivatives, as they interchange their symmetry designations.
The physical basis for bands sticking together in this way is that
if the structure factor vanishes, it is as if there was no Brillouin zone
boundary and the energy eigenvalues continue through the symmetry
point without interruption.
In this section, we consider the corresponding situation including
the electron spin and the spin-orbit interaction. Explicitly we illustrate
the sticking together of energy bands in terms of space group #194 for
the hexagonal close packed structure. Another objective of this section
is to illustrate the use of Koster’s tables for double group irreducible
representations. Space group #194 was previously discussed in §15.3,
§15.4 and §15.5 in relation to lattice modes in graphite. In the case
of lattice modes we only made use of the single group representations.
Mention of space group #194 was also made in §16.5 in connection
with bands sticking together at the zone boundary in cases where the
structure factor vanishes for non-symmorphic groups.
Consider the character table of space group #194 in Miller and Love
562 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Table 19.13: Miller and Love character table for the group of the wave
vector at k = 0 (the Γ point) for space group #194 P63/mmc, which
is applicable to the hexagonal close packed structure and to graphite.
for the Γ point shown in Table 19.13. The related point group symmetry
at the Γ point is D6h. The double group character table for D6h from
Koster et al. is given in Table 19.14. We then make the identification
of the symmetry elements and classes between Miller and Love and
Koster. On top of the labels of the classes given in Koster’s character
table (Table 19.14) we have included the Miller and Love symmetry
operations of the space group #194 including the translation operation
τ = (0, 0, c/2) for the classes which have translations. For example,
the symmetry operation {C4|τ} in Table 19.14 is identified with the
operation ‘4, 1’ in Miller and Love where ‘4’ refers to the C4 rotation
and ‘,1’ refers to the translation τ.
In the character table for the Γ point (corresponding to the D6h
point group) from Miller and Love (Table 19.13), we note that half
of the operations in #194 have translations τ. The operations in D6h
that are also in D3d ≡ D3 ⊗ i have no translations while those in
19.7. USE OF REFERENCE BOOKS 563
Figure 19.6: Brillouin zone for a hexagonal Bravais lattice.
D6h ≡ D6 ⊗ i and not in D3d do have the translation τ = (c/2)(001).
Irreducible representations 6±
, 7±
and 8±
in Miller and Love are double
group representations and correspond to Γ±
7 , Γ±
8 and Γ±
9 in Table 19.14
from Koster. The Miller and Love notation for the irreducible representations
for the group of the wave vector at k = 0 are included in
Table 19.14 on the left margin.
Comparing Miller and Love with Koster identifies B (which appears
in Table 19.13) with B =
√
3. In Table 19.14, we note the mixing of
symmetry elements ‘3’ and ‘53’ in the same class, where element ‘53’
is a double group symmetry operation related to symmetry element ‘5’
compounded with ¯E (rotation by 2π). Note that the representations Γ±
1
through Γ±
6 are even under multiplication by ¯E, while representations
Γ±
7 through Γ±
9 are odd, as required for double groups.
As we move away from the Γ point in the kz direction, the symmetry
is lowered and the appropriate group of the wave vector is that
for a ∆ point. Miller and Love give the character table for the group
of the wave vector at the ∆ point which is reproduced in Table 19.15.
The corresponding point group is C6v which has the classes:
E, ¯E,
C2
¯C2
, 2C3, 2 ¯C3, 2C6, 2 ¯C6,
3σd
3¯σd
,
3σv
3¯σv
.
564 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Table 19.14: Character Table and Basis Functions for the Group D6h
from Koster.
19.7. USE OF REFERENCE BOOKS 565
Table 19.15: Miller and Love character table for the group of the wave
vector at the ∆ point for space group #194.
These classes are identified with the listing given in the character table
from Miller and Love, Table 19.15, by comparison with the character
table for the Γ point (Table 19.14). The character table for the double
point group C6v from Koster is given in Table 19.16 and the correspondence
is explicitly given between symmetry operations in Miller and
Love and in Koster’s table. Again B =
√
3 relates the characters in
Miller and Love with those in Koster. All characters corresponding to
symmetry operations containing τ must be multiplied by a phase factor
α = exp[ickz] which is indicated in Table 19.15 as T.
We note that elements ‘3’ and ‘53’ are in class 2C3 as are elements
‘5’ and ‘51’ in class 2 ¯C3. Similarly elements ‘6,1’ and ‘50,1’ are in
class 2{C6|τ} as are elements ‘2,1’ and ‘54,1’ in class 2{ ¯C6|τ}. All
characters with operations containing τ must be multiplied by a phase
factor α = exp[ickz], indicated by arrows on the bottom of Table 19.16.
From Tables 19.14 and 19.16 we can write down compatibility relations
between the Γ point and the ∆ point representations (see Table 19.17).
In the limit k → 0 the phase factors α → 0, so that the compatibility
relations are satisfied as ∆ → Γ.
In §16.5 we discussed the phenomenon of bands sticking together
for the case where the electron spin and the spin-orbit interaction was
neglected. A typical high symmetry point in the Brillouin zone where
energy bands for the non-symmorphic hexagonal close packed lattice
stick together is the A point (see Fig. 19.7). Reference: Phys. Rev.
140, A401 (1965). The character table for the A point taken from Miller
566 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Table 19.16: Character Table and Basis Functions for the Group C6v
from Koster.
19.7. USE OF REFERENCE BOOKS 567
Table 19.17: Compatibility relations between irreducible representations
at Γ and ∆ for space group #194 using both Koster and Miller
and Love notations.
Γ point reps. ∆ point reps. Γ point reps. ∆ point reps.
Koster M&L M&L Koster Koster M&L M&L Koster
Γ+
1 1+
→ 1 ∆1 Γ−
1 1−
→ 2 ∆2
Γ+
2 2+
→ 2 ∆2 Γ−
2 2−
→ 1 ∆1
Γ+
3 3+
→ 4 ∆3 Γ−
3 3−
→ 3 ∆3
Γ+
4 4+
→ 3 ∆4 Γ−
4 4−
→ 4 ∆4
Γ+
5 6+
→ 6 ∆5 Γ−
5 6−
→ 6 ∆6
Γ+
6 5+
→ 5 ∆6 Γ−
6 5−
→ 5 ∆5
Γ+
7 8+
→ 7 ∆7 Γ−
7 8−
→ 7 ∆7
Γ+
8 7+
→ 8 ∆8 Γ−
8 7−
→ 8 ∆8
Γ+
9 9+
→ 9 ∆9 Γ−
9 9−
→ 9 ∆9
and Love illustrates the sticking together of bands at the Brillouin zone
boundary (see Table 19.18).
At the A point we have 6 irreducible representations, 3 of which are
ordinary irreducible representations and 3 of which are double group
representations. There are only 6 classes with non-vanishing characters
(see Table 19.19 which lists the characters for all the symmetry operations
at the A point). The compatibility relations below
(A) 1 2 3 4 5 6
↓ ↓ ↓ ↓ ↓ ↓
(∆) (1+4) (2+3) (5+6) 9 9 (7+8)
show that in the vicinity of the A point we have band crossings for
all the single group bands with A1, A2 and A3 symmetry. These band
crossings are based on the compatibility relations and the non-vanishing
of the pz matrix element for the first-order k · p perturbation is shown
in Fig. 19.7. Since the structure factor vanishes at A, the energy bands
pass through the A point without interruption and merely change their
568 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Table 19.18: Character table for point A in space group #194 from
Miller and Love.
Table 19.19: Character table for the group of the wave vector at the
point A adapted from character tables in Koster.
19.7. USE OF REFERENCE BOOKS 569
symmetry designations at the A point, as for example ∆1 → A1 → ∆4.
Bands for doubly degenerate double group irreducible representations
∆7 and ∆8 stick together as an A6 band at the A point. At the A point
(kz = π/c) the phase factor exp[i(c/2)kz], associated with the symmetry
operations containing τ such as {C6|τ}, becomes eiπ/2
= i. Energy
bands with double group representations A4 and A5 have complex characters
and are complex conjugates of each other. In Chapter 21 we will
see that such bands stick together because of time reversal symmetry.
Thus two ∆9 levels come into the A point to form A4 + A5 levels and
leave the A point with the same ∆9 symmetry (see Fig. 19.7). The pz
matrix element couples the A4 and A5 levels (see Table 19.19) and since
the A4 and A5 levels are degenerate by time reversal symmetry there is
once again a finite slope of the ∆9 levels as they approach the A-point.
We note that for the non-symmorphic groups at points in the Brillouin
zone where the structure factor vanishes (such as the A point for group
#194), the various components of the momenta need not transform as
irreducible representations of the non-symmorphic group.
570 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
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Figure 19.7: Energy band splittings near the A point for bands going
through the A point without interruption because of the vanishing
structure factor at the A point. Note that A4, A5, A6, ∆7, ∆8 and ∆9
are double group representations. The A4 and A5 levels stick together
because of time reversal symmetry discussed in Chapter 21.
19.8. SELECTED PROBLEMS 571
19.8 Selected Problems
1. Consider an Er3+
rare earth ion entering an insulating ionic crystal
in a position with point group symmetry D4h.
(a) Find the double group irreducible representations of the crystal
field (D4h point group symmetry) corresponding to the
ground state configuration for the free ion. Compare with
the crystal field splitting that would occur for icosahedral
point group symmetry Ih.
(b) Use Charlotte Moore’s tables to identify the lowest energy
optical transitions that can be induced from the ground state
level of the free Er3+
ion. Find the corresponding optical
transitions for the Er3+
ion in a crystal field with Ih point
group symmetry.
(c) Repeat part (b) for the case of D4h point group symmetry.
(d) What changes in the spectra (part c) are expected to occur
if a stress is applied in the z-direction? In the x-direction?
(e) Now suppose that a Dy3+
rare earth ion is introduced in the
same lattice instead of the Er3+
ion. What are the symmetry
types for levels to which optical transitions can be
induced from a multiplet corresponding to the ground state
level of the free ion. (Use Hund’s rule to obtain the ground
state energies.) Work the problem only for the D4h point
group symmetry. Comment on the expected difference in
the optical spectrum for the Dy3+
and the Er3+
ions in part
(c).
572 CHAPTER 19. SPIN ORBIT INTERACTION IN SOLIDS
Chapter 20
Application of Double
Groups to Energy Bands
with Spin
In this chapter we apply the group theoretical background for the electron
spin and the spin-orbit interaction (which is discussed in Chapter
19) to the treatment of energy band models for solids (which are
discussed in Chapters 16 and 17). By including the spin-orbit interaction
we can also discuss the effective g-factor, which together with the
effective mass tensor, characterizes the properties of a semiconductor
in a magnetic field.
20.1 Introduction
The one-electron Hamiltonian including spin-orbit interaction is written
as:
H =
p2
2m
+ V (r) +
¯h
4m2c2
( V × p) · σ (20.1)
where σ is the dimensionless spin operator. The first two terms of
Eq. (20.1) denote the kinetic energy and periodic potential of the oneelectron
Hamiltonian in a simple periodic potential V (r), and the third
573
574CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
term denotes the spin-orbit interaction HSO
HSO =
¯h
4m2c2
( V × p) · σ (20.2)
where H = H0+HSO. The Hamiltonian [Eq. (20.1)] is appropriate when
the spin-orbit splittings are significant compared with typical energy
gaps. The presence of the spin operator σ in the spin-orbit term HSO
requires the use of spin-dependent wave functions with double group
symmetry designations for the energy bands. Since the magnitude of
the spin-orbit interaction is comparable to energy band gaps for many
important electronic materials, it is important to consider the spin-orbit
interaction when carrying out energy band calculations.
Thus explicit band calculations of E(k) with spin-orbit interaction
have been carried out using all the standard techniques for energy
band calculations. Quite independent of the particular calculational
technique that is used, group theoretical techniques are introduced to
classify the states and to bring the secular equation into block diagonal
form. To illustrate these points we consider explicitly the use of
group theory (i.e., double groups as discussed in Chapter 19) to treat
the energy bands for the empty lattice (nearly free electron approximation)
and for k · p perturbation theory. These examples are designed to
provide some experience with the handling of double groups.
20.2 The k·p Perturbation with Spin-Orbit
Interaction
Schr¨odinger’s equation including the spin-orbit interaction can be written
as:
p2
2m
+ V (r) +
¯h
4m2c2
( V × p) · σ ψnk(r) = En(k)ψnk(r) (20.3)
in which the Bloch functions ψnk(r) for HSO become spinors ψnk↑(r)
and ψnk↓(r) rather than simple wave functions. These spinor basis
functions can be written in more expanded notation as
ψnk↑(r) = eik·r
unk↑(r)
20.2. K · P PERTURBATION WITH SPIN-ORBIT 575
ψnk↓(r) = eik·r
unk↓(r) (20.4)
where the arrow on ψnk↑(r) means that the state is generally spin up
or the expectation value of σz in this state is positive, and the down
arrow gives a negative expectation value for σz so that
ψnk↑|σz|ψnk↑ > 0
ψnk↓|σz|ψnk↓ < 0.
(20.5)
The Bloch states are only pure spin up or spin down states when the
spin-orbit interaction is neglected (HSO ≡ 0). The spin-orbit interaction
mixes the spin-up and spin-down partners, and, as was discussed in
Chapter 19 for the atomic case, the |j, , s, mj representation becomes
the appropriate irreducible representation for the spin-orbit coupled
system rather than the | , s, m , ms representation.
Let us focus our attention on one of the spinor unk(r) functions
(either of the components which diagonalize the Schr¨odinger equation
[Eq. (20.3)]. Using k · p perturbation theory, the corresponding differential
equation for unk(r) is
p2
2m
+ V (r) + ¯h
4m2c2 ( V × p) · σ unk(r) +
¯hk
m
· p + ¯h
4mc2 σ × V unk(r) =
En(k) − ¯h2
k2
2m
unk(r)
(20.6)
in which we have made use of the vector identities:
(A × B) · C = (B × C) · A = (C × A) · B, (20.7)
or more explicitly
( V × p) · σeik·r
unk(r) = (σ × V ) · p eik·r
unk(r) (20.8)
and
peik·r
unk(r) = eik·r
¯hkunk(r) + punk(r) . (20.9)
576CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
If we identify terms in Eq. (20.6) with an unperturbed Hamiltonian
H0 and a perturbation Hamiltonian Hk·p
we obtain
H0 =
p2
2m
+ V (r) +
¯h
4m2c2
( V × p) · σ (20.10)
and
Hk·p
=
¯hk
m
· p +
¯h
4mc2
σ × V (20.11)
so that Rayleigh-Schr¨odinger perturbation theory for energy bands near
k = 0 yields the following expression for the non-degenerate state Γi
[see Eq. (17.4)]
EΓi
n (k) = EΓi
n (0) + (uΓi
n,0|H |uΓi
n,0) +
n =n
(uΓi
n,0|H |u
Γj
n ,0)(u
Γj
n ,0|H |uΓi
n,0)
EΓi
n (0) − E
Γj
n (0)
(20.12)
in which the unperturbed functions uΓi
n,0 are evaluated at k = 0 (the
expansion point for the k ·p perturbation) and Γj labels the irreducible
representations for bands n . The sum in Eq. (20.12) is over states
Γj that couple to state Γi through the k · p perturbation Hamiltonian
given by Eq. (20.11). We note that Eq. (20.12) has the same form as
the corresponding expression without spin-orbit interaction [Eq. (17.4)]
except that in Eq. (20.12):
1. The unperturbed Hamiltonian yielding the energy eigenvalues at
k = 0 explicitly contains a spin-orbit term.
2. The k · p perturbation explicitly contains the spin operator and a
spin-orbit term.
3. The irreducible representations Γi and Γj are both double group
representations.
For example, if the spin-orbit interaction is neglected for a crystal with
Oh symmetry, then a non-degenerate Γ+
1 state is coupled by the k · p
perturbation Hamiltonian only to a Γ−
15 intermediate state. When the
20.2. K · P PERTURBATION WITH SPIN-ORBIT 577
spin-orbit interaction is included, the Γ+
1 and Γ−
15 states become double
group states:
Γ+
1 → Γ+
6
Γ−
15 → Γ−
6 + Γ−
8
(20.13)
so that with the spin-orbit interaction a Γ+
6 band will couple to bands
with Γ−
6 and Γ−
8 symmetries. We note that bands with Γ−
8 symmetry
can arise from single-group bands with Γ−
12, Γ−
15 and Γ−
25 symmetries. In
this sense the spin-orbit interaction gives more possibilities for immediate
states.
In treating k · p perturbation theory (see Chapter 17), we have 3
possibilities: non-degenerate levels, degenerate (or nearly degenerate)
levels that are treated in first-order degenerate perturbation theory, and
degenerate levels that are treated in second-order degenerate perturbation
theory. In all three of these cases, we use group theory to determine
which are the non-vanishing matrix elements of a vector operator taken
between double group states, and which of the non-vanishing matrix elements
are equal to each other. More explicitly, for the case of a crystal
with Oh symmetry, all the Γi and Γj representations have either Γ±
6 , Γ±
7
and Γ±
8 symmetry since the spatial part of the wavefunctions transform
according to one of the five ordinary irreducible representations and the
direct product of an ordinary irreducible representation with D+
6 yields
one of the double group representations. By inspection, we find that
for the Oh group all the irreducible representations Γi are at least 2fold
degenerate. But this degeneracy is maintained for all k values and
is lifted only by the application of an external (or internal) magnetic
field. This 2-fold degeneracy, know as the Kramers degeneracy is
generally found in the absence of a magnetic field. We therefore look
for this degeneracy when working practical problems, because it greatly
reduces the labor in dealing with problems involving spin). Because of
this Kramers degeneracy, we can effectively use non-degenerate perturbation
theory to deal with 2-fold levels such as the Γ±
6 and Γ±
7
levels in many applications.
Group theory can be used to greatly simplify the k · p expansion for
one of the Γ±
6 or Γ±
7 levels. For example, take Γi = Γ+
6 and note that
578CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
the momentum operator including the spin-orbit interaction
P = p +
¯h
4mc2
σ × V (20.14)
transforms like the Γ−
15 irreducible representation. The momentum operator
P transforms as Γ−
15 whether or not the spin-orbit interaction
is included, since p is a vector and so is (σ × V ), both being radial
vectors. Since Γ+
6 ⊗ Γ−
15 = Γ−
6 + Γ−
8 and Γ+
6 are orthogonal to Γ−
6 and
Γ−
8 , we have no linear k term in the k·p expansion of Eq. (20.12). In the
quadratic term we can only have intermediate states with Γ−
6 and Γ−
8
symmetry. Again we can use group theory to show relations between
the various non-vanishing matrix elements of p, and as before, only a
very small number of matrix elements are independent. To study these
matrix elements we will need to look at the basis functions for the double
group irreducible representations, which are discussed in the next
section, §20.3.
20.3 Basis Functions for Double Group Rep-
resentations
We will use the notation for single electron spin states:
↑= spin up =
1
0
↓= spin down =
0
1
.
(20.15)
Operation by the Pauli spin matrices σx, σy and σz
σx =
0 1
1 0
σy =
0 −i
i 0
(20.16)
σz =
1 0
0 −1
20.3. BASIS FUNCTIONS FOR DOUBLE GROUP REPRESENTATIONS579
on the pure spin up and spin down states yields
σx ↑ = ↓
−iσy ↑ = ↓
σz ↑ = ↑ (20.17)
σx ↓ = ↑
−iσy ↓ = − ↑
σz ↓ = − ↓
The Pauli spin matrices σx, σy, σz together with the (2×2) unit matrix
1 =
1 0
0 1
(20.18)
span a 2 × 2 space, so that every matrix can be expressed in terms of
these four matrices, 1, σx, −iσy, σz. Also the raising and lowering
operators are defined by
σ± = σx ± iσy (20.19)
so that
1
2
σ− ↑=↓ and
1
2
σ+ ↓=↑ . (20.20)
One set of basis functions for Γ+
6 is the pair ↑, ↓ which form partners
for Γ+
6 . Any other pair can be found from multiplication of this pair by
another basis function for Γ+
1 , since Γ+
6 = Γ+
1 ⊗ Γ+
6 . We will see below
how very different-looking basis functions can be used for Γ+
6 depending
on the single group representation with which Γ+
6 is connected, such as
a Γ+
1 or a Γ+
15 state. Thus, it is convenient to label the basis functions for
any double group representation with the single group representation
from which it comes. Thus the pair ↑, ↓ would be associated with a
Γ+
6 (Γ+
1 ) state.
To understand this problem better, consider the Γ+
8 (Γ+
15) state which
comes from the direct product Γ+
15 ⊗Γ+
6 = Γ+
6 +Γ+
8 . For the Γ+
15 state we
may take the basis functions Lx, Ly, Lz (angular momentum components).
Then the 6 functions Lx ↑, Lx ↓, Ly ↑, Ly ↓, Lz ↑, Lz ↓ make up
basis functions for the combined Γ+
6 and Γ+
8 representations, assuming
580CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
no spin-orbit interaction. However, when the spin-orbit interaction is
included, we must now find the correct linear combinations of the above
6 functions so that two of these transform as Γ+
6 and four transform
as Γ+
8 . The correct linear combinations are found by identifying
these basis functions which arise in the energy band problem with basis
functions that occur in the angular momentum problem. If the group
theory problem is solved for the angular momentum functions, then the
same solution can be applied to the energy band eigenfunctions more
generally. This approach was taken in writing down basis functions in
§19.6.
20.4 Basis Functions for j = 3/2 and 1/2
States
For the angular momentum functions in the | sm ms representation,
the 6 eigenfunctions correspond to the orbital states = 1, m =
1, 0, −1 and the spin states s = 1/2, ms = 1/2, −1/2. The transformations
we are looking for will transform these states into j = 3/2, mj =
3/2, 1/2, −1/2, −3/2 and j = 1/2, mj = 1/2, −1/2. The matrices
which carry out these transformations generate what are known as the
Clebsch-Gordan coefficients.
Tables of Clebsch-Gordan coefficients are found in quantum mechanics
and group theory books for many of the useful combinations of
spin and orbital angular momentum that occur in practical problems.
A set that is appropriate for = 1, s = 1/2 is given below for a Γ+
8
double group state derived from a Γ+
15 single group state (see §19.6)
|j, mj State Basis Function
|3
2
, 3
2
ξ1 = 1√
2
(Lx + iLy) ↑
|3
2
, 1
2
ξ2 = 1√
6
[(Lx + iLy) ↓ +2Lz ↑]
|3
2
, −1
2
ξ3 = 1√
6
[(Lx − iLy) ↑ +2Lz ↓]
|3
2
, −3
2
ξ4 = 1√
2
(Lx − iLy) ↓
(20.21)
in which we have used the fundamental relations for raising operators
L+| , m = ( − m )( + m + 1) | , m + 1
20.4. BASIS FUNCTIONS FOR J = 3/2 AND 1/2 STATES 581
J+|j, mj = (j − mj)(j + mj + 1) |j, mj + 1 . (20.22)
We further note that the state |j = 3/2, mj = −3/2 is identical with
the state for = 1, s = 1/2 and |m = −1, ms = −1/2 . Therefore,
we start with the j = 3/2, mj = −3/2 state and apply the raising
operator to obtain the other states:
J+ j = 3/2, mj = −3/2 =
3
2
+
3
2
3
2
−
3
2
+ 1 j = 3/2, mj = −1/2
= (L+ + S+) m = −1, ms = −1/2 = (1 + 1)(1 − 1 + 1) m = 0, ms = −1/2
+
1
2
+
1
2
1
2
−
1
2
+ 1 m = −1, ms = 1/2 .
Collecting terms, we obtain
j =
3
2
, mj = −
1
2
=
√
2
3
m = 0, ms = −
1
2
+
1
√
3
m = −1, ms =
1
2
.
(20.23)
We make the identification:
m = +1 →
1
√
2
(Lx + iLy)
m = 0 → Lz
m = −1 →
1
√
2
(Lx − iLy)
ms =
1
2
→ ↑
ms = −
1
2
→ ↓ .
from which we obtain the basis functions
|j, mj State Basis Function
|3
2
, −3
2
1√
2
(Lx − iLy) ↓
|3
2
, −1
2
1√
6
[(Lx − iLy) ↑ +2Lz ↓].
(20.24)
Similarly, operation of J+ on the state |j = 3/2, mj − 1/2 results in a
state |j = 3/2, mj = 1/2 and operation of L+ + S+ on the corresponding
functions of |m = 0, ms = −1/2 and |m = −1, ms = 1/2 results
582CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
in states |m = 0, ms = 1/2 and |m = +1, ms = −1/2 . In this way
we obtain all the basis functions for Γ+
8 (Γ+
15) given in Eq. (20.21).
We will now proceed to obtain the basis functions for Γ+
6 (Γ+
15) which
are
|j, mj State Basis Function
|1
2
, 1
2
λ1 = 1√
3
[(Lx + iLy) ↓ −Lz ↑]
|1
2
, −1
2
λ2 = 1√
3
[−(Lx − iLy) ↑ +Lz ↓].
(20.25)
The notation “ξi” was used in Eq. (20.21) to denote the four Γ+
8 (Γ+
15)
basis functions for j = 3/2 and “λi” for the two Γ+
6 (Γ+
15) basis functions
for j = 1/2. This notation “ξi” and “λi” is arbitrary and not standard
in the literature.
To obtain the Γ+
6 (Γ+
15) basis functions we note that the appropriate
(m , ms) quantum numbers corresponding to j = 1/2 and mj = ±1/2
are
m = 0 ms = ±
1
2
m = 1 ms = −
1
2
m = −1 ms = +
1
2
so that the corresponding basis functions are completely specified by
making them orthogonal to the |j = 3/2, mj = +1/2 and |j = 3/2, mj =
−1/2 states. For example, the function orthogonal to
2
3
m = 0, ms = −
1
2
+
1
√
3
m = −1, ms = +
1
2
(20.26)
is the function
1
√
3
m = 0, ms = −
1
2
−
2
3
m = −1, ms = +
1
2
(20.27)
which yields the basis functions for the |j = 1/2, mj = −1/2 state:
1
√
3
Lz ↓ −(Lx − iLy) ↑ . (20.28)
20.5. BASIS FUNCTIONS FOR OTHER Γ+
8 STATES 583
Similarly the basis function for the |j = 1/2, mj = +1/2 state can be
found by application of the raising operators J+ and (L+ + S+) to the
|j = 1/2, mj = −1/2 state, or alternatively by requiring orthogonality
to the |j = 3/2, mj = +1/2 state. Applying the raising operator to
the state
1
√
3
m = 0, ms = −
1
2
−
2
3
m = −1, ms = +
1
2
(20.29)
yields:
1
√
3
m = 0, ms = +
1
2
−
2
3
m = +1, ms = −
1
2
=
1
√
3
[(Lx+iLy) ↓ −Lz ↑]
(20.30)
which is seen to be orthogonal to
1
√
6
[(Lx + iLy) ↓ +2Lz ↑]. (20.31)
In finding the basis functions for Γ+
8 (Γ+
15) we have made use of the
symmetry properties of the angular momentum operators. As far as the
symmetry properties are concerned, it makes no difference whether L is
an angular momentum function or an energy band wave function with
Γ+
15 symmetry. This concept allows us to write down wave functions
with Γ+
8 symmetry derived from other single group states, and tables of
these results are given in §19.6, taken from Koster’s book, where tables
of coupling coefficients for all point groups are given.
20.5 Basis Functions for Other Γ+
8 States
Basis functions for the Γ±
8 state derived from Γ−
8 (Γ−
15), Γ+
8 (Γ+
25), Γ−
8 (Γ−
25),
etc. can be found from Γ+
8 (Γ+
15) and Γ+
6 (Γ+
15). All we have to do is to
replace
Lx, Ly, Lz → x, y, z
in the Eqs. (20.21) of §20.4 to obtain the basis functions for Γ−
8 (Γ−
15).
Likewise to obtain Γ+
8 (Γ+
25), we have to replace
584CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
Lx, Ly, Lz → εx, εy, εz ,
where εx = yz, εy = zx, εz = xy. By using this prescription, the basis
functions for Γ±
8 will be of the same form for all symmetry-related
partners, whether the basis functions are derived from a Γ±
15 or a Γ±
25
single group representation. This correspondence is a highly desirable
feature for working practical problems.
We note that the Γ+
8 (Γ+
12) representation can also be produced by
considering the electron spin for a Γ+
12 spinless level: Γ+
6 ⊗ Γ+
12 = Γ+
8 .
We can always make a set of 4 basis functions for this representation
out of f1 ↑, f1 ↓, f2 ↑, f2 ↓ where f1 = x2
+ ωy2
+ ω2
z2
, f2 = f∗
1 and
ω = exp(2πi/3). This makes up a perfectly good representation, but
the partners look very different from those of Γ+
8 (Γ+
15) or Γ+
8 (Γ+
25). We
can however make a unitary transformation of these 4 functions so that
they look like the Γ+
8 (Γ+
15) set.
We can make use of these double group basis functions in many
ways. For example, these basis functions are used to determine the nonvanishing
k · p matrix elements (uΓi
n,0|H |u
Γj
n,0) in Eq. (20.12). These
basis functions also determine which of the non-vanishing matrix elements
are equal to each other for a given group of the wave vector.
One technique that can be used to determine the number of nonvanishing
matrix elements in cases involving multidimensional representations
is as follows. If the relevant matrix element is of the form
(Γi|Γinteraction|Γj) then the number of independent matrix elements
is the number of times the identity representation (Γ+
1 ) is contained
in the triple direct product Γi ⊗ Γinteraction ⊗ Γj. For example,
the direct product of the matrix element (Γ+
1 |Γ−
15|Γ−
15) is:
Γ+
1 ⊗ Γ−
15 ⊗ Γ−
15 = Γ+
1 + Γ+
12 + Γ+
15 + Γ+
25.
and since all non-vanishing matrix elements must be invariant under
all symmetry operations of the group, only the Γ+
1 term leads to a nonvanishing
matrix element. Of the 9 possible combinations of partners,
there is only one independent non-vanishing matrix element.
For the case of double groups, the matrix element (Γ+
6 |Γ−
15|Γ−
6 ) has
2 × 3 × 2 = 12 possible combinations. Now Γ+
6 ⊗ Γ−
15 ⊗ Γ−
6 = Γ+
1 +
Γ+
15 + Γ+
12 + Γ+
15 + Γ+
25, so that once again there is only one independent
matrix element. Finally, for the case (Γ+
6 |Γ−
15|Γ−
8 ) there are 24 possible
20.6. E(K) INCLUDING SPIN-ORBIT INTERACTION 585
combinations. The direct product Γ+
6 ⊗ Γ−
15 ⊗ Γ−
8 = Γ+
1 + Γ+
2 + Γ+
12 +
2Γ+
15 + 2Γ+
25, and once again there is one independent matrix element.
Furthermore, if Γ−
6 and Γ−
8 are both related through a Γ−
15 interaction
term, then the same independent matrix element applies to both
(Γ+
6 |Γ−
15|Γ−
6 ) and (Γ+
6 |Γ−
15|Γ−
8 ). We illustrate this simplifying feature in
§20.6.
20.6 E(k) for a Non-Degenerate Band Including
Spin-Orbit Interaction
We are now ready to give the form of E(k) for a non-degenerate band
using non-degenerate k ·p perturbation theory [see Eq. (20.12)] by considering
the form of the k · p matrix elements implied by group theory.
As an example, consider a non-degenerate Γ+
6 band. We take as basis
functions for the Γ+
6 state:
Γ+
6 :
1 ↑
1 ↓
. (20.32)
Within the framework of k ·p perturbation theory, the Γ+
6 state couples
only to Γ−
6 and Γ−
8 since Γ+
6 ⊗Γ−
15 = Γ−
6 +Γ−
8 . For the Γ−
6 and Γ−
8 states
we use the basis functions derived from Eq. (20.21) and Eq. (20.25), so
that for Γ+
6 (Γ+
15) we write
|j, mj State Basis Function
|1
2
, 1
2
( 1√
3
)[(x + iy) ↓ −z ↑]
|1
2
, −1
2
( 1√
3
)[−(x − iy) ↑ +z ↓]
(20.33)
and for Γ+
8 (Γ+
15) we write
|j, mj State Basis Function
|3
2
, 3
2
( 1√
2
)(x + iy) ↑
|3
2
, 1
2
( 1√
6
)[(x + iy) ↓ +2z ↑]
|3
2
, −1
2
( 1√
6
)[(x − iy) ↑ +2z ↓]
|3
2
, −3
2
( 1√
2
)(x − iy) ↓
. (20.34)
586CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
We can read off the basis functions relating the |j, mj representation
and the | sm ms representation for the Γ−
6 (j = 1/2) and Γ−
8 (j = 3/2)
states that are derived from the Γ−
15 level directly from Eqs. (20.33) and
(20.34). The x, y and z in Eqs. (20.33) and (20.34) refer to the three
partners of the Γ+
15 state. For this case there are no non-vanishing
matrix elements in Eq. (20.12) in first-order perturbation theory. In
second-order, the non-vanishing terms are:
1 ↑ |Px|
1
√
2
(x + iy) ↑ =
1
√
2
(1|Px|x)
1 ↑ |Py|
1
√
2
(x + iy) ↑ =
i
√
2
(1|Py|y)
1 ↑ |Pz|
1
√
6
{(x + iy) ↓ +2z ↑} =
2
√
6
(1|Pz|z)
1 ↑ |Px|
1
√
6
{(x − iy) ↑ +2z ↓} =
1
√
6
(1|Px|x)
1 ↑ |Py|
1
√
6
{(x − iy) ↑ +2z ↓} = −
i
√
6
(1|Py|y)
1 ↑ |Pz|
1
√
3
{(x + iy) ↓ −z ↑} = −
1
√
3
(1|Pz|z)
1 ↑ |Px|
1
√
3
{(−x + iy) ↑ +z ↓} = −
1
√
3
(1|Px|x)
1 ↑ |Py|
1
√
3
{(−x + iy) ↑ +z ↓} =
i
√
3
(1|Py|y).(20.35)
Summing up the second-order terms and utilizing the equality
(1|Px|x) = (1|Py|y) = (1|Pz|z) (20.36)
we obtain
EΓ+
6 (k) = EΓ+
6 (0) +
¯h2
|(1|Px|x)|2
m2Eg
1
3
k2
x +
1
3
k2
y +
1
3
k2
z
+
¯h2
|(1|Px|x)|2
m2(Eg + ∆)
1
2
k2
x +
1
2
k2
y +
2
3
k2
z +
1
6
k2
x +
1
6
k2
y
= EΓ+
6 (0) +
¯h2
k2
m2
|(1|Px|x)|2 1
3Eg
+
2
3(Eg + ∆)
(20.37)
20.7. E(K) FOR DEGENERATE BANDS 587
 ¢¡
£
¤¦¥§
¤©¨
¤ ¥ Figure 20.1: Energy versus k at
the Γ point showing the effect of
the spin-orbit interaction in splitting
the p-level. The relevant bands
are labeled by the double group rep-
resentations.
where Eg and Eg + ∆ are defined in Fig. 20.1.
20.7 E(k) for Degenerate Bands Including
Spin-Orbit Interaction
In dealing with k ·p perturbation theory for degenerate states we again
use basis functions such as are given by Eqs. (20.21) and (20.25) to classify
the degenerate states. For example, instead of the (3× 3) secular
equation for p-bands (Γ−
15 symmetry) that was discussed in § 17.5, inclusion
of the spin-orbit interaction leads to solution of a (6×6) secular
equation. This (6×6) equation assumes block diagonal form containing
a (4 × 4) and a (2 × 2) block, because the spin functions transform as
D1/2 or Γ+
6 and because
Γ+
6 ⊗ Γ−
15 = Γ−
6 + Γ−
8 (20.38)
where Γ−
6 corresponds to a j = 1/2 state and Γ−
8 to a j = 3/2 state
(see Fig. 20.1).
An important application of degenerate k · p perturbation theory
including the effects of spin-orbit interaction is to the valence band of
the group IV and III-V compound semiconductors. A description of
E(k) for the valence band is needed to construct the constant energy
surfaces for holes in these semiconductors. This method is useful for
588CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
analysis of cyclotron resonance measurements on holes in group IV and
III-V semiconductors.
One way to solve for the energy levels of the valence band of a
group IV semiconductor about the valence band maximum k = 0 (Γ+
25
single group level) is to start with the (6 × 6) matrix labeled by the
double group basis functions. The secular equation is constructed by
considering
H = H0 + Hk·p
(20.39)
in which the matrix elements for Hk·p
vanish in first-order. Therefore
in degenerate second-order perturbation theory we must replace each
matrix element i|H |j by
i|H |j +
α
i|H |α α|H |j
Ei − Eα
(20.40)
in which H denotes the k ·p perturbation Hamiltonian (see §17.5), and
i, j, α all denote double group irreducible representations. In this case
we obtain the appropriate basis functions for the Γ+
7 and Γ+
8 states from
the combination that we previously derived using the raising operator
J+ = L+ + S+ [see Eqs. (20.21) and (20.25)]. Thus for the Γ+
7 (Γ+
25)
states, the basis functions are
|j, mj State Basis Function
|1
2
, 1
2
µ1 = 1√
3
[(εx + iεy) ↓ −εz ↑]
|1
2
, 1
2
µ2 = 1√
3
[−(εx − iεy) ↑ +εz ↓]
(20.41)
and for the Γ+
8 (Γ+
25) states, the basis functions are
|j, mj State Basis Function
|3
2
, 3
2
ν1 = 1√
2
(εx + iεy) ↑
|3
2
, 1
2
ν2 = 1√
6
[(εx + iεy) ↓ +2εz ↑]
|3
2
, −1
2
ν3 = 1√
6
[(εx − iεy) ↑ +2εz ↓]
|3
2
, −3
2
ν4 = 1√
2
(εx − iεy) ↓
(20.42)
in which the states Γ+
7 and Γ+
8 are labeled by |j, mj and the components
of the function εi relate to x, y, z partners according to
εx=yz
εy=zx
εz=xy.
(20.43)
20.7. E(K) FOR DEGENERATE BANDS 589
In solving for E(k) for the valence band of a semiconductor such as
germanium we use the unperturbed and perturbed Hamiltonians given
by Eqs. (20.10) and (20.11), respectively. The states used to solve the
eigenvalue problem are labeled by the wave functions that diagonalize
the “unperturbed” Hamiltonian H0 of Eq. (20.10). Since Hk·p
transforms
as Γ−
15 and Γ−
15 ⊗Γ+
7 = Γ−
7 +Γ−
8 , Hk·p
does not couple band Γ+
7 to
band Γ+
7 . This result follows more easily just from parity arguments.
A solution to the (6 × 6) secular equation involves explicit computation
of matrix elements as was done for the spinless case in §17.5. For
brevity, we will not include a detailed evaluation of all the matrix elements,
but we will instead just summarize the results. For the Γ+
7 (Γ+
25)
level, E(k) assumes the form
E(Γ+
7 ) = k2 ¯h2
2m
+ 4C1 +
4
3
C2 + C3 (20.44)
where
C1 =
¯h2
m2



Γ−
8 (Γ−
12)
| Γ+
7 |Px|Γ−
8 |2
E0 − E
+
Γ−
8 (Γ−
25)
| Γ+
7 |Px|Γ−
8 |2
E0 − E



C2 =
¯h2
m2
Γ−
8 (Γ−
15)
| Γ+
7 |Px|Γ−
8 |2
E0 − E
C3 =
¯h2
m2
Γ−
7 (Γ−
2 )
| Γ+
7 |Pz|Γ−
7 |2
E0 − E
(20.45)
in which
P = p +
¯h
4m2c2
(σ × V ) (20.46)
and E is an intermediate state. Since bands with Γ−
12 and Γ−
25 symmetries
do not lie close to the valence band Γ+
25 in germanium we would
expect C1 to be much smaller than C2 or C3.
The solution for the Γ+
8 level is a good deal more complicated and
yields the result
E[Γ+
8 (Γ+
25)] = Ak2
± B2k4 + C2(k2
xk2
y + k2
yk2
z + k2
z k2
x) (20.47)
590CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
where
A = ¯h2
2m
+ 2
3
E1 + 2E2 + E3 + 5E4 + 1
2
E5
B2
= 4
9
E2
1 + 4E2
2 + 16E2
4 + 1
4
E2
5 − 8
3
E1E2 + 16
3
E1E4
−2
3
E1E5 − 16E2E4 + 2E2E5 − 4E4E5
C2
= − 9
16
E2
5 + 16E1E2 − 32E1E4 + E1E5 − 9E2E5 + 18E4E5
(20.48)
and where
E1 =
¯h2
m2
Γ−
6 (Γ−
15)
| Γ+
8 |Px|Γ−
6 |2
E0 − E
E2 =
¯h2
m2
Γ−
7 (Γ−
2 )
| Γ+
8 |Px|Γ−
7 |2
E0 − E
E3 =
¯h2
m2
Γ−
8 (Γ−
15)
| Γ+
8 (Γ+
25)|Pz|Γ−
8 (Γ−
15) |2
E0 − E
E4 =
¯h2
m2
Γ−
8 (Γ−
25)
| Γ+
8 (Γ+
25)|Pz|Γ−
8 (Γ−
25) |2
E0 − E
E5 =
¯h2
m2
Γ−
8 (Γ−
12)
| Γ+
8 (Γ+
25)|Pz|Γ−
8 (Γ−
12) |2
E0 − E
. (20.49)
In Eqs. (20.49), E4 and E5 are expected to be small. Because of the
E0 − E denominator that enters second-order degenerate perturbation
theory, the most important contributions to k · p perturbation theory
come from bands lying close in energy to the E0 level, which in this
case are the Γ-point valence band energy extrema. For germanium the
levels lying close to the Fermi level have Γ+
25, Γ+
1 , Γ−
2 and Γ−
15 symmetries
(see Fig. 19.2) so that only the double group states derived from these
states will contribute significantly to the sums in Eqs. (20.49). The
far-lying levels only contribute small correction terms.
Although the spin-orbit perturbation term contained in H0 in Eq. (20.10)
does not depend on k, the resulting energy bands show a k-dependent
spin-orbit splitting. For example, in Figure 19.2 we note that the spinorbit
splitting of the Γ+
8 (Γ+
25) level is ∆ =0.29 eV at the Γ point while
along the Λ axis, the splitting is only about 2/3 this value and remains
20.8. EFFECTIVE G–FACTOR 591
constant over most of the Λ axis. For the corresponding levels along
the ∆ or (100) direction, the spin-orbit splitting is very much smaller.
When the spin-orbit interaction is weak, it is convenient to deal with
this interaction in perturbation theory. We note that the spin-orbit
interaction can be written in diagonal form using the |j, mj representation.
Therefore instead of writing the wavefunctions for the unperturbed
problem in the | , s, m , ms representation, it is convenient to
use the |j, mj representation for the whole perturbation theory problem.
A classic work on spin-orbit interaction in solids is R.J. Elliott,
Phys. Rev. 96, 266 (1954). An application to k · p perturbation theory
is found in Phys. Rev. 98, 368 (1955).
20.8 Effective g–Factor
One of the important applications of double groups in solid state physics
is to the treatment of the effective g–factor. In calculating the effective
g-factor (geff), we employ k · p perturbation theory with spin, and show
that in a magnetic field B, new terms arise in the one-electron Hamiltonian.
Some of these new terms have the symmetry of an axial vector
(e.g., the magnetic moment µeff), giving rise to an interaction µeff · B.
We review first the origin of the effective g–factor in solid state physics
and show the important role of group theory in the evaluation of the
pertinent matrix elements. In this problem we consider 3 perturbations
1. spin-orbit interaction
2. k · p perturbation
3. perturbation by a magnetic field.
We will see that the effective one-electron Hamiltonian for an electron
in a solid in an applied magnetic field can be written as
Heff =
1
2m∗
αβ
p −
e
c
A
2
− geffµBmsB (20.50)
which implies that in effective mass theory, the periodic potential is
replaced by both an effective mass tensor and an effective g-factor.
592CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
Just as the effective mass of an electron can differ greatly from the free
electron value, so can the effective g–factor differ greatly from the free
electron value of 2. To see how this comes about, let us consider energy
bands about a band extrema in a crystal with Oh symmetry. The
discussion given here follows closely that given for k · p perturbation
theory in Chapter 17.
Every entry in the secular equation for the k · p Hamiltonian is of
the following form since there are no entries in first-order that couple
the degenerate states:
¯h2
k2
2m
δn,n +
n
n|H |n n |H |n
En − En
(20.51)
where n denotes the sum over states outside the nearly degenerate
set (NDS) and where we are assuming that every member in the NDS
is of approximately the same energy, like the situation for degenerate
p-bands or of strongly coupled s and p bands. The k · p perturbation
Hamiltonian is H = ¯h
m
k · p for the spinless problem or H = ¯h
m
k · P
for the problem with spin, where P = p + ¯h
4mc2 σ × V . With this
identification of H we can rewrite the entries to the secular equation
as
αβ
Dnn αβkαkβ =
αβ
kαkβ
¯h2
2m
δnn δαβ +
¯h2
m2
n
n|Pα|n n |Pβ|n
E
(0)
n − E
(0)
n
(20.52)
where αβ denotes a sum on components of the k vectors, and n denotes
a sum over members outside the NDS, and where Dnn αβ denotes
the term in curly brackets, and depends on the band indices n, n . The
eigenvalues are found by solving the secular equation
n


αβ
Dnn αβkαkβ − Eδnn

 fn = 0. (20.53)
Equation 20.53 is the eigenvalue problem in zero magnetic field. The
same form for the secular equation also applies when B = 0. This
equation symbolically represents the problem with spin if the fn functions
are taken to transform as irreducible representations of the crystal
double group and the P vectors are chosen so that they include the
spin-orbit interaction P = p + ¯h
4mc2 (σ × V ).
20.8. EFFECTIVE G–FACTOR 593
In an external magnetic field we replace the operator p → p− e
c
A in
the Hamiltonian and from this it follows generally that in Eq. (20.53)
we must replace
¯hk →
¯h
i
−
e
c
A (20.54)
when a magnetic field is applied. The relation Eq. (20.54) is called
the Kohn–Luttinger transcription and is widely used in the solution
of magnetic field problems in semiconductor physics. As a result of
Eq. (20.54), k in a magnetic field becomes a non-commuting operator,
rather than just a simple commuting operator. Let us, for example,
select a gauge for the vector potential
Ax = −By
Ay = 0
Az = 0
(20.55)
so that B = Bˆz,
¯hkx =
¯h
i
∂
∂x
+
e
c
By (20.56)
¯hky =
¯h
i
∂
∂y
(20.57)
which results in the commutation relation
[kx, ky] =
ieB
¯hc
. (20.58)
The commutation relation [Eq. (20.58)] tells us that the amount by
which the operators kx and ky fail to commute is proportional to B.
We note that all other pairs of wave vector components, such as [kx, kz]
etc., commute. Since the order of operators is important in a magnetic
field, we will need to rewrite the secular equation [Eq. (20.53)] when
B =0 in terms of a symmetric and an antisymmetric part:
Dnn αβkαkβ =
1
2
DS
nn αβ {kα, kβ}
anticommutator
+
1
2
DA
nn αβ [kα, kβ]
commutator
(20.59)
where the symmetric part is
DS
nn αβ =
1
2
[Dnn αβ + Dnn βα] (20.60)
594CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
and the antisymmetric part is
DA
nn αβ =
1
2
[Dnn αβ − Dnn βα] . (20.61)
Thus the symmetric part can be written explicitly as
DS
nn αβ =
¯h2
2m
δnn δαβ+
¯h2
2m2
n
n|Pα|n n |Pβ|n + n|Pβ|n n |Pα|n
En(0) − En (0)
(20.62)
and gives the effective mass tensor through the relation
1
m∗
αβ
=
∂2
En
¯h2
∂kα∂kβ
. (20.63)
Since the electron spin is included, the states in Eq. (20.62) are labeled
by irreducible representations of the double groups and P is a function
of σ, as seen in Eq. (20.11).
The antisymmetric part DA
nn αβ is from the above definition:
DA
nn αβ =
¯h2
2m2
n
n|Pα|n n |Pβ|n − n|Pβ|n n |Pα|n
En(0) − En (0)
. (20.64)
In the case of a spinless electron in a cubic crystal, DA
nn αβ would
vanish identically because there is only one independent momentum
matrix element in cubic Oh symmetry in the absence of a magnetic
field. If now we also include the electron spin and the double group
representations, these arguments do not apply and we will find that
DA
nn αβ does not generally vanish and in fact contributes strongly to the
effective g-factor.
By way of comparison, the zero magnetic field eigenvalue problem
is
n


αβ
Dnn αβkαkβ − Eδnn

 fn = 0 (20.65)
and the magnetic field eigenvalue problem then becomes
n



αβ
1
2
DS
nn αβ{kα, kβ} + DA
nn αβ[kα, kβ] − µBσ · B − Eδnn



fn = 0
(20.66)
20.8. EFFECTIVE G–FACTOR 595
where µB is the Bohr magneton
µB = −
|e|¯h
2mc
and σ = 2S/¯h. The term DS
nn αβ gives rise to a replacement of the
periodic potential by an effective mass tensor. In computing m∗
αβ we
ordinarily neglect the difference between p and P.
In the presence of a magnetic field, the wavevectors k are operators
which act on the effective mass wave functions fn . From Eq. (20.58) we
see that the components of the wave vector operator do not commute,
so that
[kα, kβ] =
ieBγ
¯hc
(20.67)
and the commutator in Eq. (20.67) vanishes in zero magnetic field, as
it should. Here the α, β, γ directions form a right-handed coordinate
system. The term DA
nn αβ vanishes if there is no spin. The commutator
[kα, kβ] transforms as an axial vector. Because of the form of DA
nn αβ
given in Eq. (20.64), we see that DA
nn αβ also transforms as an axial
vector. Therefore the term DA
nn αβ has the same symmetry properties
as −µBσ and gives rise to an effective magnetic moment different from
the free electron value of the Bohr magneton µB. If we now write
[kx, ky] =
ieBz
¯hc
= iBz
e¯h
2mc
2m
¯h2 = iµBBz
2m
¯h2 (20.68)
then
DA
nn xy[kx, ky] =
iBz
m
µB
n
n|Px|n n |Py|n − n|Py|n n |Px|n
En(0) − En (0)
(20.69)
so that the effective magnetic moment of an electron in a crystal is
µ∗
αβ = |µB| δαβ +
i
m n
n|Pα|n n |Pβ|n − n|Pβ|n n |Pα|n
En(0) − En (0)
(20.70)
where the effective g–factor is related to µ∗
αβ by geff αβ = 2µ∗
αβ/µB.
We recall that the energy levels of a free electron in a magnetic field
are
Ems = gµBmsB (20.71)
596CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
so that for spin 1/2, the spin splitting of the levels is 2µBB. In a
crystalline solid, the spin splitting becomes 2µ∗
B.
For comparison we include the formula for the effective mass tensor
component
1
m∗
αβ
=
δαβ
m
+
1
m2
n
n|Pα|n n |Pβ|n + n|Pβ|n n |Pα|n
En(0) − En (0)
(20.72)
in which
P = p +
¯h
4mc2
σ × V. (20.73)
Thus an electron in a magnetic field and in a periodic potential acts
as if the periodic potential can be replaced by letting m → m∗
αβ and
µB → µ∗
αβ. Thus, symbolically we would write an effective Hamiltonian
as
Heff =
1
2m∗
p −
e
c
A
2
− µ∗
σ · B (20.74)
where
µ∗
= µBgeff/2. (20.75)
In deriving the formula for the effective g–factor above, we did not pay
much attention to whether P was merely the momentum operator p or
the more complete quantity including the spin-orbit interaction
p +
¯h
4mc2
(σ × V ).
It turns out that it is not very important whether we distinguish between
matrix elements of p and of P since the matrix element of
¯h
4mc2
(σ × V )
is generally quite small. However, what is important, and even crucial,
is that we consider the states n, n , n as states characterized by the
irreducible representations of the crystal double groups.
Let us illustrate how we would proceed to calculate an effective g–
factor for a typical semiconductor. Let us consider the effective g-factor
for germanium at the Γ point (k = 0). In Fig. 20.2 we let Eg denote the
20.8. EFFECTIVE G–FACTOR 597
 
¡£¢
¤¦¥§
¤¦¥¨
¤©§
Figure 20.2: Level ordering at the
Γ point in Ge.
energy gap between the conduction band and the uppermost valence
band, and ∆ denote the spin-orbit splitting of the valence band. In
germanium Eg ∼ 0.8eV and ∆ ∼ 0.3eV. We will assume in this simple
example that these are the only bands to be included in carrying out
the sum on n .
To evaluate µ∗
and m∗
in Eqs. (20.70) and (20.72) we use the basis
function discussed in §20.4 to find the non-vanishing matrix elements
of ¯hk · p/m. We write the basis functions for Γ+
8 (Γ+
25) and Γ+
7 (Γ+
25) in a
symbolic form from Eqs. (20.41) and (20.42) so that we can make use
of all the group theory ideas that were discussed in §17.5 in connection
with the corresponding problem without spin. This approximation is
valid if ∆ Eg and each double group level can be clearly identified
with the single group level from which it originates. Otherwise the Γ+
8
levels mix appreciably with one another and all matrix elements must
be evaluated in the double group representation directly, so that the
numerical estimates obtained here would have to be revised.
Take the basis functions for the Γ−
7 state to be (γ−
↑, γ−
↓) where
γ−
is a basis function for the Γ−
2 representation. Now let us evaluate
the matrix elements that go into Eq. (20.70) for µ∗
. For example, we
obtain
γ−
↑ |px|
3
2
,
3
2
= γ−
↑ |px|
1
√
2
(εx + iεy) ↑ (20.76)
598CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
using the basis functions for Γ+
8 (Γ+
25)
|j, mj State Basis Function
|3
2
, 3
2
ν1 = 1√
2
(εx + iεy) ↑
|3
2
, 1
2
ν2 = 1√
6
[(εx + iεy) ↓ +2εz ↑]
|3
2
, −1
2
ν3 = 1√
6
[(εx − iεy) ↑ +2εz ↓]
|3
2
, −3
2
ν4 = 1√
2
(εx − iεy) ↓
(20.77)
From §17.5 we have (Γ±
2 |H |Γ25,α) = A2¯hkα/m where A2 = (Γ±
2 |px|Γ25,x)
is the only independent matrix element connecting these symmetry
types, where we note that the basis function for Γ2− symmetry is xyz.
Using the basis functions for Γ+
8 (Γ+
25) given by Eq. (20.77) we obtain
γ−
↑ |px|
3
2
,
3
2
=
1
√
2
A2
γ−
↑ |px|
3
2
,
1
2
= 0
γ−
↑ |px|
3
2
, −
1
2
=
1
√
6
A2
γ−
↑ |px|
3
2
, −
3
2
= 0
where we consider the ortho-normality of both the spin and orbital
states. For the py matrix, the same procedure gives
γ−
↑ |py|
3
2
,
3
2
=
i
√
2
A2
γ−
↑ |py|
3
2
,
1
2
= 0
γ−
↑ |py|
3
2
, −
1
2
= −
i
√
6
A2
γ−
↑ |py|
3
2
, −
3
2
= 0.
20.8. EFFECTIVE G–FACTOR 599
To find the contribution to µ∗
/µB, we sum Eq. (20.70) over the four
Γ+
8 levels we obtain
i γ−
↑ |px|νi νi|py|γ−
↑ − γ−
↑ |py|νi νi|px|γ−
↑ /Eg
= 1
Eg
A2√
2
−
iA∗
2√
2
+ A2√
6
iA∗
2√
6
− iA2√
2
A∗
2√
2
− −iA2√
6
A∗
2√
6
= |A2|2
Eg
−2i
3
.
(20.78)
We thus obtain for the contribution from the Γ+
8 (Γ+
25) levels to µ∗
µB
a
value of
i
m
−
2i
3
|A2|2
Eg
=
2|A2|2
3mEg
. (20.79)
Let us now find the contribution to µ∗
/µB from the spin-orbit split-off
bands. Here we use the basis functions for Γ+
7 (Γ+
25)
|j, mj State Basis Function
|1
2
, 1
2
µ1 = 1√
3
[(εx + iεy) ↓ −εz ↑]
|1
2
, −1
2
µ2 = 1√
3
[−(εx − iεy) ↑ +εz ↓]
(20.80)
so that the matrix elements for px and py become
γ−
↑ |px|
1
2
,
1
2
= 0
γ−
↑ |px|
1
2
, −
1
2
= −
1
√
3
A2
γ−
↑ |py|
1
2
,
1
2
= 0
γ−
↑ |py|
1
2
, −
1
2
=
i
√
3
A2.
We thus obtain the contribution of
i
m(Eg + ∆)
2i
3
|A2|2
= −
2
3
|A2|2
m(Eg + ∆)
. (20.81)
to µ∗
/µB in Eq. (20.70) from the Γ+
7 (Γ+
25) levels. Adding up the two
contributions from Eqs. (20.79) and (20.81) we finally obtain
µ∗
µB orbital
= −
2|A2|2
3m
1
Eg + ∆
−
1
Eg
+ 1 (20.82)
600CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
where +1 is the free electron contribution.
We can now evaluate |A2|2
in terms of the conduction band effective
mass using the symmetric contribution DS
nn αβ and for this term we can
use the relation
m
m∗
= 1 +
2
m n
| γ−
↑ |px|n |2
EΓ2
(0) − En(0)
. (20.83)
Evaluating the matrix elements in Eq. (20.83) we thus obtain
m
m∗
= 1 +
2
m
|A2|2
2Eg
+
|A2|2
6Eg
+
|A2|2
3(Eg + ∆)
≈
2
3m
|A2|2 2
Eg
+
1
Eg + ∆
(20.84)
where the free electron term of unity is usually small compared to other
terms in the sum in Eq. (20.84) and can be neglected in many cases.
Neglecting this term, we now substitute for |A2|2
in terms of m∗
to
obtain
geff =
2µ∗
µB
= 2 −
2m
m∗
∆
3Eg + 2∆
. (20.85)
In the limit, ∆ → 0, then g → 2 in agreement with the results for the
free electron g-factor. In the limit ∆ Eg
geff → 2 −
m
m∗
(20.86)
which implies geff → − m
m∗ for very light masses.
For germanium, for which m∗
/m ∼ 0.12, ∆ ∼ 0.3eV, and Eg ∼
0.8eV, the effective g–factor mostly cancels the free electron contribu-
tion:
geff = 2 1 −
1
0.12
0.3
3(0.8) + 2(0.3)
= 2 1 −
1
1.2
1
3
. (20.87)
For InSb, the spin-orbit splitting is large compared with the direct band
gap m∗
/m ∼ 0.013, ∆ ∼ 0.9eV, and Eg ∼ 0.2eV
geff ∼ 2 1 −
1
0.013
0.9
3(0.2) + 2(0.9)
∼ 2(1 − 28) −54 (20.88)
leading to the picture for InSb shown in Figure 20.3. In InSb, the spin
splitting is almost as large as the Landau level separation. However,
20.8. EFFECTIVE G–FACTOR 601
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67
6
6
6
7
7
7
Figure 20.3: Landau levels in InSb showing the spin splitting resulting
from the large negative effective g-factor.
the geff has the opposite sign as compared with free electron spin g–
value, where we note that because of the negative sign of the charge on
the electron and on the Bohr magneton, the free electron spin state of
lowest energy is aligned antiparallel to the applied field. Sometimes it
is convenient to define the spin effective mass by the relation
µ∗
µB
=
m
m∗
s
(20.89)
where m∗
s denotes spin effective mass, so that geff = 2m/m∗
s.
In general, the spin and orbital effective masses will not be the same.
If they are, the Landau level spacing is equal to the spacing between
spin levels. The physical reason why these masses are not expected to
be equal is that the orbital mass is determined by a momentum matrix
element (which transforms as a radial vector). Since the spin mass
depends on the coupling between bands through an operator which
transforms as an axial vector, different bands are coupled for the 2
cases.
In treating cyclotron resonance transitions, the transitions are spin
conserving and the g–factors usually cancel out. They are, however,
important for interband Landau level transitions even though the transitions
are spin conserving, since the g–factors in the valence and conduction
bands can be different. Thus spin up and spin down transitions
can occur at different energies. The effective g–factors are directly observed
in spin resonance experiments which occur between the same
602CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
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6798
6798
6798
Figure 20.4: Strict 2-band model where the Landau level separation is
equal to the spin splitting as for the case of a free electron gas. This
limit applies quite well to the L-point Landau levels in bismuth.
Landau level but involve a spin flip.
Of interest also is the case where the spin effective mass and the
orbital effective mass are equal. In a strict 2-band model this must be
the case. For bismuth the strongly coupled 2-band model is approximately
valid and m∗
s m∗
(see Figure 20.4). Landau level separations
equal to the spin splitting also occur for the free electron magnetic energy
levels. However, for band electrons, the Landau level separations
are proportional to the inverse cyclotron effective mass rather than the
inverse free electron mass.
For high mobility (low effective mass) materials with a small spinorbit
interaction, the Landau level separation is large compared with
the spin splitting (see Fig. 20.3). On the other hand, some high mobility
narrow gap semiconductors with a large spin-orbit interaction can have
spin splittings larger than the Landau level separations; such a situation
gives rise to interesting phenomena at high magnetic fields.
References for Effective g–Factors
Luttinger, Phys. Rev. 102, 1030 (1956).
Yafet, Vol. 14 of Seitz-Turnbull Solid State Physics Series.
Roth, Phys. Rev. 118, 1534 (1960).
Cohen and Blount, Phil. Mag. 5, 115 (1960).
Wolff, J. Phys. Chem. Solids 25, 1057 (1964).
20.9. SELECTED PROBLEMS 603
20.9 Selected Problems
1. Bismuth is a semimetal which crystallizes in a rhombohedral
structure (see Wyckoff vol. 1 p. 32) which is a slight distortion
from a simple cubic structure. These distortions result in the
semimetallic behavior for Bi with small electron pockets about
the L points and a small hole pocket around the T point in the
Brillouin zone. See the diagram below for the Bi band structure.
As this distortion becomes small, the T-point becomes degenerate
with the L-point.
604CHAPTER 20. APPLICATION TO ENERGY BANDS WITH SPIN
(a) Using double group k · p perturbation theory, find the form
of E(k) for the L point occupied conduction bands. How
many L point electron pockets are there? (Note that in
the cubic limit where the distortion vanishes, the L and T
points of bismuth all become equivalent “L points” for a fcc
Brillouin zone.) In the bismuth band structure shown below,
the conduction band has Ls= L5 + L6 symmetry while the
filled valence band strongly coupled to Ls has La = L7 + L8
symmetry, where Ls and La are, respectively, symmetric and
antisymmetric under time reversal symmetry.
(b) Between which energy bands and where in the Brillouin zone
do the far infrared transitions occur?
(c) Between which energy bands and where in the Brillouin zone
do allowed optical transitions in the visible range occur?
Chapter 21
Time Reversal Symmetry
In this chapter we consider the properties of the time reversal operator
for the case of no spin and when the spin-orbit interaction is included.
The effect of time reversal symmetry on the energy dispersion relations
is then considered, first for the case of no spin and then including the
spin-orbit interaction.
In high energy physics, arguments regarding time inversion were essential
in providing guidance for the development of a theory for the
fundamental particles. The CPT invariance in particle physics deals
with charge conjugation (C) which is the reversal of the sign of the
electrical charge, parity (P) which is spatial inversion, and time inversion
(T).
21.1 The Time Reversal Operator
Knowledge of the state of a system at any instant of time t and the
deterministic laws of physics are sufficient to determine the state of
the system both into the future and into the past. If ψ(r, t) specifies
the time evolution of state ψ(r, 0), then ψ(r, −t) is called the timereversed
conjugate of ψ(r, t). The time-reversed conjugate state is
achieved by running the system backwards in time or reversing all the
velocities (or momenta) of the system.
The time evolution of a state is governed by Schr¨odinger’s equation
605
606 CHAPTER 21. TIME REVERSAL SYMMETRY
(one of the deterministic laws of physics)
i¯h
∂ψ
∂t
= Hψ (21.1)
which is satisfied by a time-dependent wave function of the form
ψ(r, t) = e
−iHt
¯h ψ(r, 0) (21.2)
where ˜T ≡ exp [−iHt/¯h] is the time evolution operator. Under time
reversal t → −t we note that ψ → ψ∗
so that
ˆTψ(r, t) = ψ(r, −t) = ψ∗
(r, t). (21.3)
In the following section, we derive some of the important properties of
ˆT.
21.2 Properties of the Time Reversal Op-
erator
The important properties of the time reversal operator include:
1. commutation: [ ˆT, H] = 0
Because of energy conservation, the time reversal operator ˆT
commutes with the Hamiltonian ˆTH = H ˆT. Since ˆT commutes
with the Hamiltonian, eigenstates of the time reversal operator
are also eigenstates of the Hamiltonian.
2. anti-linear: ˆTi = −i ˆT
From Schr¨odinger’s equation (Eq. 21.1), it is seen that the reversal
of time corresponds to a change of i → −i, which implies that
ˆTi = −i ˆT. We call an operator anti-linear if its operation on
a complex number yields the complex conjugate of the number
rather than the number itself ˆTa = a∗ ˆT.
3. action on wave functions: ˆTψ = ψ∗ ˆT
Since ˆTψ = ψ∗ ˆT, the action of ˆT on a scalar product is
ˆT(ψ, φ) = φ∗
(r)ψ(r)d3
r ˆT = (ψ, φ)∗ ˆT (21.4)
21.2. TIME REVERSAL OPERATOR 607
4. In the case of no spin ˆT = ˆK where ˆK is the complex conjugation
operator. With spin, we show below that ˆT = ˆKσy where σy is
the Pauli spin operator,
σy =
0 −i
i 0
.
We will see below that both ˆT and ˆK are anti-unitary operators.
From Schr¨odinger’s equation (no spin), the effect of ˆT on p is to
reverse p (time goes backward) and ˆT leaves V (r) invariant, so
that indeed H is invariant under ˆT; and furthermore ˆT = ˆK for
the case of no spin. When spin is included, however, the Hamiltonian
H must still be invariant under ˆT. We note that ˆTp = −p
and ˆTL = −L (orbital angular momentum). We likewise require
that ˆTS = −S where S = spin angular momentum. If these
requirements are imposed, we show below that the H is still invariant
under ˆT (i.e., H commutes with ˆT) when the spin-orbit
interaction is included:
H =
p2
2m
+ V (r) +
¯h
4m2c2
σ · ( V × p). (21.5)
We note that ˆK[σx, σy, σz] = [σx, −σy, σz] when the spin components
are written in terms of the Pauli matrices
σx =
0 1
1 0
σy =
0 −i
i 0
σz =
1 0
0 −1
(21.6)
since only the Pauli matrix σy contains i. Thus ˆK by itself is not
sufficient to describe the time reversal operation on the Hamiltonian
H (Eq. 21.5) when the spin-orbit interaction is included. We
will see below that the product ˆKσy can describe time reversal of
H.
608 CHAPTER 21. TIME REVERSAL SYMMETRY
Let us now consider the effect of ˆKσy on the spin matrices ˆKσy[σx, σy, σz].
We note that
σyσx = −σxσy so that ˆKσyσx = − ˆKσxσy = −σx
ˆKσy
σyσz = −σzσy so that ˆKσyσz = − ˆKσzσy = −σz
ˆKσy
ˆKσyσy = −σy
ˆKσy since, from above ˆKσy = −σy
ˆK.
Thus we obtain
ˆKσyσ = −σ ˆKσy
so that the operator ˆKσy transforms σ (or S) into −σ (or −S).
Clearly σy does not act on any of the other terms in the Hamiltonian.
We note that ˆK cannot be written in matrix form.
Since ˆK ˆK = ˆK2
= 1, we can write the important relation ˆT =
ˆKσy which implies ˆK ˆT = σy = unitary operator. Thus σ†
yσ−1
y =
1 and since σ2
y = σyσy = 1 we have σ†
y = σy and σ†2
y = 1, where
the symbol † is used to denote the adjoint of an operator.
5. In the case of no spin ˆT2
= 1, since ˆK2
= 1 and ˆT = ˆK. With
spin we will now show that ˆT2
= −1. Since ˆT = ˆKσy when the
effect of the electron spin is included,
ˆT2
= ( ˆKσy)( ˆKσy) = −(σy
ˆK)( ˆKσy) = −σy
ˆK2
σy = −σyσy = −1.
More generally if we write ˆK ˆT = U = unitary operator (not
necessarily σy), we can then show that ˆT2
= ±1. Since two
consecutive operations by ˆT on a state ψ must produce the same
physical state ψ, we have ˆT2
= C1 where C is a phase factor eiφ
of unit magnitude. Since ˆK2
= 1, we can write
ˆK2 ˆT = ˆT = ˆKU = U∗ ˆK (21.7)
ˆT2
= ˆKU ˆKU = U∗ ˆK2
U = U∗
U = C1 (21.8)
We show below that C = ±1. Making use of the unitary property
U†
U = UU†
= 1, we obtain by writing U∗
= U∗
UU†
= CU†
,
U∗
= CU†
= C ˜U∗ (21.9)
21.2. TIME REVERSAL OPERATOR 609
Taking the transpose of both sides of Eq. 21.9 yields
˜U∗ = U†
= CU∗
= C(C ˜U∗) = C2
U†
or C2
= 1 and C = ±1.
(21.10)
We thus obtain either ˆT2
= +1 or ˆT2
= −1.
6. Operators H, r, V (r) are even under time reversal ˆT; operators
p, L, σ are odd under ˆT. Operators are either even or odd under
time reversal. We can think of spin angular momentum classically
as due to a current loop in a plane ⊥ to the z-axis. Time reversal
causes the current to flow in the opposite direction.
7. ˆT and ˆK are anti-unitary operators, as shown below.
In this subsection we show that ˆT ˆT†
= −1 and ˆK ˆK†
= −1, which is
valid whether or not the spin is considered explicitly. The properties of
the inverse of ˆT and ˆK are readily found. Since ˆK2
= 1, then ˆK ˆK = 1
and ˆK−1
= ˆK. If for the case where the spin is treated explicitly
ˆT2
= −1, then ˆT ˆT = −1 and ˆT−1
= − ˆT; ˆT = ˆKσy for the case of spin.
For the spinless case, ˆT2
= 1 and ˆT−1
= ˆT.
Since complex conjugation changes i → −i, we can write ˆK†
= − ˆK
so that ˆK is anti-unitary.
We now use this result to show that both ˆT and ˆK are anti-unitary.
This is the most important property of ˆT from the point of view of
group theory. Since ˆK = ˆT in the absence of spin, and since ˆK is antiunitary,
it follows that ˆT is anti-unitary in this case. However, when
spin is included, ˆT = ˆKσy and
σy = ˆK ˆT
σ†
y = ˆT† ˆK†
.
(21.11)
Since σy is a unitary operator, thus ˆT† ˆK† ˆK ˆT = 1 but since ˆK† ˆK = −1
it follows that ˆT† ˆT = −1, showing that ˆT is also anti-unitary.
Furthermore ˆK and ˆT behave differently from all the operators that
we have thus far encountered in group theory, such as the point group
operations (rotations, improper rotations, mirror planes, inversion and
R= rotation of 2π for spin problems). Thus in considering symmetry
610 CHAPTER 21. TIME REVERSAL SYMMETRY
operations in group theory, we treat all the unitary operators separately
by use of character tables and all the associated apparatus, and then we
treat time reversal symmetry as an additional symmetry constraint.
We will see in Chapter 22 how time reversal symmetry enters
directly as a symmetry element for magnetic point groups.
We discuss first in §21.3 and §21.4 the general effect of ˆT on the
form of E(k) for the case of electronic bands (a) neglecting spin and
(b) including spin. After that, we will consider the question of degeneracies
imposed on energy levels by time reversal symmetry (the
Herring Rules).
21.3 The Effect of ˆT on E(k), Neglecting
Spin
If for the moment we neglect spin, then the time reversal operation
acting on a solution of Schr¨odinger’s equation yields
ˆTψ(r) = ψ∗
(r). (21.12)
Since the Hamiltonian commutes with ˆT, then both ψ(r) and ψ∗
(r)
satisfy Schr¨odinger’s equation for the same energy eigenvalue, so that
a two-fold degeneracy occurs. We will now show that time reversal
symmetry leads to two symmetry properties for the energy eigenvalues
for Bloch states: the evenness of the energy eigenvalues E(k) = E(−k),
and the zero slope of En(k) at the Brillouin zone boundaries.
The effect of the translation operation on a Bloch state is
ψk(r + Rn) = eik·Rn
ψk(r) (21.13)
and the effect of time reversal is
ˆTψk(r) = ψ∗
k(r). (21.14)
We can write the following relation for the complex conjugate of Bloch’s
theorem
ψ∗
k(r + Rn) = e−ik·Rn
ψ∗
k(r) (21.15)
21.3. EFFECT OF ˆT ON E(K) 611
and we can also rewrite Eq. 21.15 in terms of k → −k as
ψ∗
−k(r + Rn) = eik·Rn
ψ∗
−k(r) (21.16)
which upon comparing Eqs. 21.13, 21.15 and 21.16 implies that for
non-degenerate levels the time reversal operator transforms k → −k
ˆTψk(r) = ψ−k(r) = ψ∗
k(r). (21.17)
If the level is doubly degenerate and ψk(r) and φk(r) are the corresponding
eigenstates, then if ˆTψk(r) = φk(r) = ψ−k(r), no additional
degeneracy is required by time reversal symmetry. Time reversal symmetry
thus implies that for a spinless system
En(k) = En(−k) (21.18)
and the energy is an even function of wave vector k whether or not
there is inversion symmetry.
Using this result (Eq. 21.18) and the E(k) = E(k + K) periodicity
in k space, we obtain:
E


K
2
− δk

 = E

−
K
2
+ δk

 = E


K
2
+ δk

 (21.19)
where δk is an infinitesimal distance to the Brillouin zone boundary.
Thus referring to Fig. 21.1, E(k) comes into the zone boundary with
zero slope for both the lower and upper branches of the solutions in
Fig. 21.1. For the case where there is degeneracy at the zone boundary,
the upper and lower bands will have equal and opposite slopes.
We have been using the symmetry properties in Eqs. 21.18 and
21.19 throughout our solid state physics courses. In the most familiar
cases, E(k) depends on k2
. Figure 21.1 taken from Kittel illustrates
the symmetry properties of Eqs. 21.18 and 21.19 for a simple parabolic
band at k = 0.
Let us now consider the consequences of these ideas from a group
theoretical point of view, and enumerate Herring’s rules. If ψ(r) belongs
to the irreducible representation D, then ˆTψ(r) = ψ∗
(r) will transform
612 CHAPTER 21. TIME REVERSAL SYMMETRY
 
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Figure 21.1: Simple E(k) diagram from Kittel for a spinless electron
illustrating both E(k) = E(−k) and the zero slope of E(k) at the
Brillouin zone boundary.
21.3. EFFECT OF ˆT ON E(K) 613
Table 21.1: Character table for point group C4.
C4 (4) E C2 C4 C3
4 Time reversal
x2
+ y2
, z2
Rz, z A 1 1 1 1 (a)
x2
− y2
, xy B 1 1 −1 −1 (a)
(xz, yz)
(x, y)
(Rx, Ry)
E
1
1
−1
−1
i
−i
−i
i
(b)
(b)
according to D∗
which consists of the complex conjugate of all the
matrices in D.
We can distinguish three different possibilities in the case of no
spin:
(a) All of the matrices in the representation D can be written as
real matrices. In this case, the time reversal operator leaves the
representation D invariant and no additional degeneracies in E(k)
result.
(b) If the representations D and D∗
cannot be brought into equivalence
by a unitary transformation, there is a doubling of the
degeneracy of such levels due to time reversal symmetry. Then
the representations D and D∗
are said to form a time reversal
symmetry pair and these levels will stick together.
(c) If the representations D and D∗
can be made equivalent under
a suitable unitary transformation, but the matrices in this representation
cannot be made real, then the time reversal symmetry
also requires a doubling of the degeneracy of D and the bands
will stick together.
To illustrate these possibilities, consider the point group C4 (see
Table 21.1). Here irreducible representations A and B are of type (a)
above and each of these representations correspond to non-degenerate
energy levels. However, the two representations labeled E are complex
conjugates of each other and are of type (b) since there is no unitary
transformation that can bring them into equivalence. Thus because of
614 CHAPTER 21. TIME REVERSAL SYMMETRY
the time reversal symmetry requirement, representation E corresponds
to a doubly degenerate level. This is an example where time reversal
symmetry gives rise to an additional degeneracy.
The time reversal partners are treated as different representations
when applying the following rules on character:
1. The number of irreducible representations is equal to the number
of classes.
2. i
2
i = h.
Using the character table for the group of the wave vector, we can
distinguish which of the 3 cases apply for a given irreducible representation
using the Herring test (ref. C. Herring, Phys. Rev. 52, 361
(1937)). Let Q0 be an element in the space group which transforms k
into −k. Then Q2
0 is an element in the group of the wave vector k and
all elements in the group of the wave vector are elements of Q2
0. If the
inversion operator i is contained in the group of the wave vector k, then
all the elements Q0 are in the group of the wave vector k. If i is not an
element of the group of the wave vector k, then the elements Q0 may
or may not be an element in the group of the wave vector. Let h equal
the number of elements Q0. The Herring space group test is then
Q0
χ(Q2
0) = h case (a)
= 0 case (b)
= −h case (c)
where χ is the character for a representation of the group of the wave
vector k. These tests can be used to decide whether or not time reversal
symmetry introduces any additional degeneracies to this representation.
Information on the Herring test is contained for every one of the 32
point groups in the character tables in Koster’s book.
To apply the Herring test to the point group C4, and consider the
group of the wave vector for k = 0. Then all four symmetry operations
take k → −k since k = 0. Furthermore, E2
= E, C2
2 = E, C2
4 = C2 and
(C3
4 )2
= C2 so that for representations A and B
Q0
χ(Q2
0) = 1 + 1 + 1 + 1 = 4 (21.20)
21.4. INCLUDING THE SPIN-ORBIT INTERACTION 615
from which we conclude that A and B correspond to case (a), in agreement
with Koster’s tables.
On the other hand, for each representation under E,
Q0
χ(Q2
0) = 1 + 1 + (−1) + (−1) = 0 (21.21)
from which we conclude that representations E correspond to case (b).
Therefore the two irreducible representations under E correspond to
the same energy and the corresponding E(k) will stick together. The
two representations under E are called time reversal conjugate rep-
resentations.
21.4 The Effect of ˆT on E(k), Including
the Spin-Orbit Interaction
When the spin-orbit interaction is included, then the Bloch functions
transform as irreducible representations of the double group. The
degeneracy of the energy levels is different from the spinless situation,
and in particular every level is at least doubly degenerate.
When the spin-orbit interaction is included, ˆT = ˆKσy and not only
do we have k → −k, but we also have σ → −σ under time reversal
symmetry. This is written schematically as:
ˆTψn,k↑(r) = ψn,−k↓(r) (21.22)
so that the time reversal conjugate states are
En↑(k) = En↓(−k) (21.23)
and
En↓(k) = En↑(−k). (21.24)
If inversion symmetry exists as well,
En(k) = En(−k) (21.25)
then
En↑(k) = En↑(−k) and En↓(k) = En↓(−k) (21.26)
616 CHAPTER 21. TIME REVERSAL SYMMETRY
making En↑(k) and En↓(k) degenerate. In more detail, since ˆT = ˆKσy
and since
σy ↑ =
0 −i
i 0
1
0
= i
0
1
= i ↓
σy ↓ =
0 −i
i 0
0
1
= −i
1
0
= −i ↑
we obtain
ˆTψn,k↑(r) = ˆTeik·r
un,k↑
1
0
= e−ik·r
iu∗
n,k↑
0
1
= e−ik·r
un,−k↓
0
1
(21.27)
which is a Bloch state for wave vector −k and spin ↓. Likewise
ˆTψn,k↓(r) = ˆTeik·r
un,k↓
0
1
= e−ik·r
−iu∗
n,k↓
1
0
= e−ik·r
un,−k↑
1
0
(21.28)
which is a Bloch state for wave vector −k and spin ↑ in which we have
written
iu∗
n,k↑ = un,−k↓
and
−iu∗
n,k↓ = un,−k↑.
For a general point in the Brillouin zone, and in the absence of
spin-orbit coupling but including the spin on the electron, the energy
levels have a necessary 2-fold spin degeneracy and also exhibit the
property E(k) = E(−k), whether or not there is inversion symmetry.
This is illustrated in Fig. 21.2(a). When the spin-orbit interaction is
turned on and there is inversion symmetry then we get the situation
illustrated in Fig. 21.2(b) where the 2-fold degeneracy remains. However,
if there is no inversion symmetry, then the only relationships that
remain are those of Eqs. 21.23 and 21.24 shown in Fig. 21.2(c), and the
Kramers degeneracy results in E↑(k) = E↓(−k) and E↓(k) = E↑(−k).
The role of inversion symmetry is also important for the E(k) relations
for degenerate bands. This is illustrated in Fig. 21.3 for degenerate
bands near k = 0. We take as examples: (a) diamond for which the
21.4. INCLUDING THE SPIN-ORBIT INTERACTION 617
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Figure 21.2: Schematic example of Kramers degeneracy in a crystal in
the case of: (a) no spin-orbit interaction where each level is doubly degenerate
(↑, ↓), (b) both spin-orbit interaction and inversion symmetry
are present and the levels are doubly degenerate, (c) spin-orbit interaction
and no spatial inversion symmetry where the relations 21.23 and
21.24 apply.
spin-orbit interaction can be neglected and all levels are doubly degenerate
at a general point in the Brillouin zone, (c) InSb or GaAs which
have Td symmetry (lacking inversion) so that relations 21.23 and 21.24
apply and the two-fold Kramers degeneracy is lifted, (b) Ge or Si which
have Oh symmetry (including inversion) and the two-fold Kramers degeneracy
is retained at a general point in the Brillouin zone.
We give in Table 21.2 the Herring rules (see §21.3) whether or not
the spin-orbit interaction is included. When the spin-orbit interaction
Table 21.2: Summary of rules regarding degeneracies and time reversal.
Case Relation between
D and D∗
FrobeniusSchur
test
Spinless
electron
Half-integral
spin electron
Case (a) D and D∗
are equivalent
to the same real irreducible
representation
R χ(Q2
0) = h No extra
degeneracy
Doubled
degeneracy
Case (b) D and D∗
are
inequivalent
R χ(Q2
0) = 0 Doubled
degeneracy
Doubled
degeneracy
Case (c) D and D∗
are equivalent
to each other but not to
a real representation
χ(Q2
0) = −h Double
degeneracy
No extra
degeneracy
618 CHAPTER 21. TIME REVERSAL SYMMETRY
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Figure 21.3: Schematic examples of energy bands E(k) in diamond, Ge
and GaAs near k = 0. (a) Without spin-orbit coupling, each band in
diamond has a two-fold spin degeneracy. (b) Splitting by spin-orbit coupling
in Ge, with each band remaining doubly degenerate. (c) Splitting
of the valence bands by the spin-orbit coupling in GaAs. The magnitudes
of the splittings are not to scale.
21.4. INCLUDING THE SPIN-ORBIT INTERACTION 619
is included, there are also three cases which can be distinguished. When
the time reversal operator ˆT acts on a spin dependent wavefunction ψ
which transforms according to an irreducible representation D, then we
have three possibilities:
(a) If the representation D is real, or can be transformed by a unitary
transformation into a set of real matrices, then the action of ˆT
on these matrices will yield the same set of matrices. To achieve
the required additional degeneracy, we must have D occur twice.
(b) If representations D and D∗
cannot be brought into equivalence
by a unitary transformation, then the corresponding levels must
stick together in pairs to satisfy the time reversal degeneracy re-
quirement.
(c) If representations D and D∗
can be brought into equivalence but
neither can be made all real, then no additional degeneracy need
be introduced and both make up the time reversal degenerate
pair.
These results are summarized in Table 21.2 for both the case of
no spin and when spin-orbit interaction is included. We now illustrate
these rules with two cases:
1. the double group representations of the point group C4 (symmor-
phic)
2. the double group representation at the L point in Ge (or Si)
where the levels are degenerate by time reversal symmetry (non-
symmorphic)
For the first illustration, we give the character table for the double
group C4 taken from Koster et al. in Table 21.3. We note that the
Koster table contains an entry for time inversion, which summarizes the
results discussed in §21.1 for the spinless bands. Inspection of this character
table shows that the double group representations involve the 4th
roots of unity (as shown below) and obey the relation χ(Ai) = −χ( ¯Ai)
for each of the pairs of symmetry operations Ai and ¯Ai. Note that
the character table originally given in Koster has some misprints with
620 CHAPTER 21. TIME REVERSAL SYMMETRY
Table 21.3: Character table for C4
C4 E ¯E C4
¯C4 C2
¯C2 C−1
4
¯C−1
4 Time
Inv.
Bases
for
C4
Γ1 1 1 1 1 1 1 1 1 a z or Sz
Γ2 1 1 −1 −1 1 1 −1 −1 a xy
Γ3 1 1 i i −1 −1 −i −i b −i(x + iy) or
−(Sx + iSy)
Γ4 1 1 −i −i −1 −1 i i b i(x − iy) or
(Sx − iSy)
Γ5 1 −1 ω −ω i −i −ω3
ω3
b φ(1/2, 1/2)
Γ6 1 −1 −ω3
ω3
−i i ω −ω b φ(1/2, −1/2)
Γ7 1 −1 −ω ω i −i ω3
−ω3
b φ(3/2, −3/2)
Γ8 1 −1 ω3
−ω3
−i i −ω ω b φ(3/2, 3/2)
regard to χ(C−1
4 ) = −χ( ¯C−1
4 ), which are corrected in Table 21.3. This
character table shows that the characters for the Γ5 and Γ6 irreducible
representations are time reversal degenerate pairs, and likewise for the
Γ7 and Γ8 irreducible representations:
E ¯E C4
¯C4 C2
¯C2 C−1
4
¯C−1
4
Γ5: ω0
ω4
ω ω5
ω2
ω6
ω7
ω3
Γ6: ω0
ω4
ω7
ω3
ω6
ω2
ω ω5
Γ7 ω0
ω4
ω5
ω ω2
ω6
ω3
ω7
Γ8: ω0
ω4
ω3
ω7
ω6
ω2
ω5
ω
Application of the Frobenius–Schur test for Γ5 yields:
χ(Q2
0) = (1)(−1) + (1)(−1) − ω2
− ω2
+ 1 + 1 − ω6
− ω6
= −1 − 1 − i − i + 1 + 1 + i + i = 0
(21.29)
where we note that for the double group representations we consider
the character χ(Q0
¯Q0) in the Frobenius–Schur test. We thus find that
the representations Γ6, Γ7 and Γ8 are also of the b type with respect to
time reversal symmetry and this information is also given in Table 21.3.
21.4. INCLUDING THE SPIN-ORBIT INTERACTION 621
Table 21.4: Character Table and Basis Functions for the Group D3d
D3d E ¯E 2C3 2 ¯C2 3C2 3 ¯C2 I ¯I 2S6 2 ¯S6 3σd 3¯σd Time
Inv.
Bases
L+
1 Γ+
1 1 1 1 1 1 1 1 1 1 1 1 1 a R
L+
2 Γ+
2 1 1 1 1 −1 −1 1 1 1 1 −1 −1 a Sx
L+
3 Γ+
3 2 2 −1 −1 0 0 2 2 −1 −1 0 0 a (Sx − iSy),
−(Sx + iSy)
L−
1 Γ−
1 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 a zSz
L−
2 Γ−
2 1 1 1 1 −1 −1 −1 −1 −1 −1 1 1 a z
L−
3 Γ−
3 2 2 −1 −1 0 0 −2 −2 1 1 0 0 a (x − iy),
−(x + iy)
L+
6 Γ+
4 2 −2 1 −1 0 0 2 −2 1 −1 0 0 c φ(1/2, −1/2)
L+
4 Γ+
5 1 −1 −1 1 i −i 1 −1 −1 1 i −i b φ(3/2, −3/2)
−iφ(3/2, 3/2)
L+
5 Γ+
6 1 −1 −1 1 −i i 1 −1 −1 1 −i i b − (φ(3/2, 3/2)
−iφ(3/2, −3/2))
L−
6 Γ−
4 2 −2 1 −1 0 0 −2 2 −1 1 0 0 c Γ+
4 × Γ−
1
L−
4 Γ−
5 1 −1 −1 1 i −i −1 1 1 −1 −i i b Γ+
5 × Γ−
1
L−
5 Γ−
6 1 −1 −1 1 −i i −1 1 1 −1 i −i b Γ+
6 × Γ−
1
For the L-point levels in Ge, see the E(k) diagram in Fig. 19.2b
for the case where the spin-orbit interaction is included. The character
table appropriate to the L-point is given in Table 21.4. The designation
for the L-point representations have been added on the left column of
Koster’s table.
For a Λ point, the operations E, 2C3 and 3C2 take k → −k. For the
L-point, all operations are of the Q0 type, so that for the representations
L1, L2 and L3, we have Σχ(Q2
0) = 12, yielding representations of type
a, in agreement with the character table for D3d (Table 21.4).
For the double group representation L+
6 we obtain
L+
6 = Σχ(Q2
0) = −4 − 2 + 0 − 4 − 2 + 0 = −12 type (c) (21.30)
where again we write Q0
¯Q0 for Q2
0. For the double group representation
L+
4 the Frobenius–Schur test yields:
L+
4 : Σχ(Q2
0) = −1 − 2 + 3 − 1 − 2 + 3 = 0 type (b) (21.31)
Likewise L+
5 is of type b. Since L+
4 and L+
5 are complex conjugate representations,
L+
4 and L+
5 form time reversal degenerate pairs. Similarly,
L−
4 and L−
5 are type b representations and form time reversal degenerate
pairs (see Fig. 19.2b).
622 CHAPTER 21. TIME REVERSAL SYMMETRY
With this discussion of time reversal symmetry, we have explained
all the entries to the character tables, and have explained why because
of time reversal symmetry certain bands stick together on the E(k) diagrams.
In the following Chapter we see how the time reversal operator
becomes a symmetry element in magnetic point groups.
21.5 Selected Problems
1. Consider the space group D4
6h (#194) which we discussed in connection
with the lattice modes for graphite. We will now concern
ourselves with the electronic structure. Since the Fermi surfaces
are located close to the HK axes in the Brillouin Zone it is important
to work with the group of the wave vector at points H,
K and P (see diagram).
(a) Using Miller and Love, and Koster et al., give the character
table including double groups for the group of the wave
vector at point K. Classify each of the irreducible representations
according to whether they behave as a, b or c under
time reversal symmetry.
(b) Find the compatibility relations as we move away from K
toward H.
Chapter 22
Magnetic Groups
If atoms at each lattice site, can be represented as a charge distribution
ρ(r) with no particular spin symmetry (paramagnetic or diamagnetic),
the ordinary space groups are used. If, however, we have ordered arrangements
of spins, then the time reversal operator (which reverses
the spin direction) can be combined with other group elements to form
elements of a new type of symmetry group. Groups in which the time
reversal operator forms group elements are called magnetic space
groups and the corresponding point groups are called magnetic point
groups.
In this chapter we present some of the essential properties of magnetic
space groups and give some examples of interest to solid state
physics.
22.1 Introduction
When magnetically ordered phases are taken into account, the magnetic
unit cell is often larger than the chemical unit cell, as for example
in an antiferromagnetic system. Additional symmetry elements are
introduced (see §22.2), and as a result many more point groups and
space groups are possible (see §22.3).
There are, in fact, 122 (58 + 2×32) magnetic point groups (rather
than 32), and 1651 (1191 + 2×230) magnetic space groups (rather
than 230), and 36 (22 + 14) magnetic Bravais lattices rather than 14.
623
624 CHAPTER 22. MAGNETIC GROUPS
The magnetic Bravais lattices which are important for describing antiferromagnetic
structures are shown in Fig. 22.1(b), and for comparison
the 14 ordinary Bravais lattices are also shown in Fig. 22.1(a), and are
further explained below. We will confine our discussion in this chapter
to magnetic single groups (not double groups), and we shall only
discuss magnetic point groups.
22.2 Types of Elements
Magnetic groups have symmetry elements corresponding to unitary operators
(denoted by Ai) and anti-elements Mk = ˆTAk corresponding to
anti-unitary operators, where ˆT is the anti-unitary time reversal operator
(see Chapter 21). We show in Fig. 22.2(a) a one-dimensional lattice
in which ˆT when combined with a translation is a symmetry operation.
However, by displacing the non-magnetic white atoms in Fig. 22.2(b)
relative to Fig. 22.2(a), we see that ˆT is no longer a symmetry operation.
If we neglect spin, then ˆT = ˆK where ˆK is the complex conjugation
operator (see Chapter 21), and ˆT2
= ˆK2
= 1. The product of two unitary
elements Ai or of two anti-unitary elements Mk yields a unitary
element, while the product of a unitary element Ai with an anti-unitary
element Mk yields an anti-unitary element:
AiAi = Ai
AiMk = Mk
MkAi = Mk
Mk Mk = Ai (22.1)
To satisfy these relations, group properties and the rearrangement theorem,
there must be equal numbers of elements of the type Ai and
of the type Mk in a magnetic point group.
22.3 Types of Magnetic Point Groups
In classifying the magnetic point groups we must consider three types
of point groups:
22.3. TYPES OF MAGNETIC POINT GROUPS 625
(a) (b)
Figure 22.1: (a) The 14 ordinary Bravais lattices and (b) the 22 additional
magnetic Bravais lattices. The open circles represent the time
reversed sites.
626 CHAPTER 22. MAGNETIC GROUPS
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Figure 22.2: Diagram showing a one-dimensional lattice where: (a) the
operation ˆT is combined with the translation symmetry operation, (b)
ˆT is not a symmetry operation even if combined with translations.
(a) 32 ordinary point groups G where ˆT is not an element
(b) 32 ordinary point groups G ⊗ ˆT. In these magnetic point groups,
all elements Ai of G are contained together with all elements
ˆTAi.
(c) 58 point groups G in which half of the elements are {Ai} and half
are {Mk} where Mk = ˆTAk and the {Ai, Ak} form an ordinary
point group G . Also {Ai} is a subgroup of G .
Summing the number of types (a), (b), and (c) we obtain (32+32+
58) = 122 magnetic point groups. Case (a) can apply to non-magnetic
materials and some ferromagnetic materials. Case (b) can apply to
some antiferromagnetic materials. Case (c) can apply to magnetic materials
with a variety of spin orderings.
We list in Table 22.1 (from Tinkham) the 58 magnetic point groups
of type (c) and denoted by G; also included in the table are the 32
ordinary point groups of type (a) which are denoted by G . The 32
point groups of type (b), obtained from those in type (a) as G ⊗ ˆT,
are not listed. The magnetic groups of type (c) are related to elements
of a group G and a subgroup Hr and are denoted by G (Hr). The
22.4. 58 MAGNETIC POINT GROUPS 627
appropriate group G contains the symmetry elements {Ai, Ak} while
the subgroup Hr of G only has elements {Ai}.
22.4 Properties of the 58 Magnetic Point
Groups {Ai, Mk}
We list below some of the properties of the magnetic point groups [type
(c)] that contain both unitary and anti-unitary symmetry elements, Ai
and Mk = ˆTAk, respectively. We denote a typical magnetic point group
of this category by G = {Ai, Mk}.
1. ˆT is not an element in the magnetic point group G (since the
identity element is one of the elements of {Ai} but not of {Ak}).
2. Ai and Ak are distinct, so that no element in the set {Ai} is also
in {Ak} where {Mk} = { ˆTAk}. (If there were one Aj in common,
then we could have ˆTAj in {Mk} and A−1
j in {Ai}, which on
multiplication ˆTAjA−1
j implies that ˆT is in G, in contradiction
with property (1)).
3. G ≡ {Ai, Ak} is one of the 32 ordinary point groups.
4. The set Hr = {Ai} forms an invariant unitary subgroup of G.
This subgroup is self-conjugate because conjugation of an element
in Ai with any element in {Mk} written as MkAiM−1
k yields an
element in {Ai} as a result of Eq. (22.1), and likewise the conjugation
AiMkA−1
i yields an element in {Mk}.
5. The number of unitary operators Ai= the number of anti-unitary
operators Mk, to satisfy the multiplication rules in Eq. 22.1 and
the group properties of G.
6. {Ai} is the only coset of Hr in G and {Ak} is the only coset of
Hr in G .
7. Since Hr and G are groups, and properties (5) and (6) apply,
then G is a group of the form
628 CHAPTER 22. MAGNETIC GROUPS
Table 22.1: The magnetic point groups of type (a) and type (c).
22.4. 58 MAGNETIC POINT GROUPS 629
Table 22.1: CONTINUED: The magnetic point groups of type (a) and
type (c).
630 CHAPTER 22. MAGNETIC GROUPS
Table 22.2: Character Table for Group C2h
C2h (2/m) E C2 σh i
x2
, y2
, z2
, xy Rz Ag 1 1 1 1
z Au 1 1 −1 −1
xz, yz Rx, Ry Bg 1 −1 −1 1
x, y Bu 1 −1 1 −1
G = Hr + ˆT(G − Hr) (22.2)
8. From property (7), we see that the procedure for finding magnetic
point groups is to start with one of the 32 point groups G and find
all invariant subgroups of index 2. Denoting each such subgroup
by Hr we can form
Gr = Hr + ˆT(G − Hr). (22.3)
We denote each magnetic group Gr thus formed by G (Hr) in which
the relevant G and Hr for each Gr are listed. This notation is used in
Table 22.1 and the various G (Hr) can be found in Table 22.1.
To illustrate the elements of magnetic point groups, consider the 4
entries under C2h in Table 22.1. We list below the symmetry elements
of each of the C2h(Hr) magnetic point groups.
C2h(C2h) : E, C2, i, iC2 (iC2 = σh)
C2h(C2) : E, C2, ˆTi, ˆTiC2
C2h(Ci) : E, i, ˆTC2, ˆTiC2
C2h(C1h) : E, iC2, ˆTi, ˆTC2 (22.4)
in which the magnetic point group C2h(C2h) is of type (a), and the
other three are of type (c). Not listed is the magnetic space group
C2h ⊗ ˆT of type (b) which contains the eight symmetry elements {Ai} =
{E, C2, i, iC2} and { ˆT ⊗Ai} = { ˆT, ˆTC2, ˆTi, ˆTiC2}. The character table
for the ordinary point group C2h is given as Table 22.2.
22.5. EXAMPLES OF MAGNETIC STRUCTURES 631
Table 22.3: Character Table for Group D2
D2 (222) E Cz
2 Cy
2 Cx
2
x2
, y2
, z2
A1 1 1 1 1
xy Rz, z B1 1 1 −1 −1
xz Ry, y B2 1 −1 1 −1
yz Rx, x B3 1 −1 −1 1
D2h = D2 ⊗ i
We note that the time reversal operator of ˆT reverses the sign of a
spin, while the inversion operator i leaves a spin invariant (since
the angular momentum L is even under inversion while r and p are
each odd).
22.5 Examples of Magnetic Structures
22.5.1 Orthorhombic Ferromagnetic Unit Cell with
D2h(C2h) Symmetry
The notation D2h (C2h) for a magnetic point group denotes a point
group D2h from which the subgroup (C2h) forms the set of symmetry
elements {Ai} and the remaining symmetry elements of G are of the
form {Ak} where the elements Mk in G are of the form Mk = ˆTAk. We
note from Table 22.1 that D2h (C2h) corresponds to a ferromagnetic
structure such as the one shown in Fig. 22.3. In the paramagnetic
state, the proper symmetry group for this structure in D2h.
The symmetry operations for D2h = D2 ⊗ i are: E, C2x, C2y, C2z,
i, iC2x, iC2y, iC2z (see Table 22.3). It is immediately seen that the
subgroup of D2h which leaves the spin invariant consists of the elements
{Ai} = E, C2z, i, iC2z, since both orbital and spin angular momentum
are invariant under inversion. These four elements form the group
C2h = C2 ⊗ i, noting again that the spin angular momentum S is even
under inversion. The remaining elements of D2h reverse the spins, so
that the time reversal operator ˆT is needed to keep all the spins ferro-
632 CHAPTER 22. MAGNETIC GROUPS
 
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Figure 22.3: Magnetic spin
arrangement in D2h(C2h) for
an orthorhombic ferromagnetic
system.
22.5. EXAMPLES OF MAGNETIC STRUCTURES 633
magnetically aligned. We therefore obtain {Mk} = ˆTC2x, ˆTC2y, ˆTiC2x
and ˆTiC2y for the remaining symmetry elements of D2h (C2h). The appropriate
Bravais lattice in this case is the orthorhombic Bravais lattice
#4 for the non-magnetic groups [see Fig. 22.1(a)].
22.5.2 Antiferromagnets with the Rutile Structure
The antiferromagnets MnF2, FeF2 and CoF2 crystallize in the rutile
structure shown in Fig. 22.4. The open circles are the F ions while
the shaded circles with spins denote the magnetic cations. The point
group for this structure in the paramagnetic state is D4h = D4 ⊗ i.
In the antiferromagnetic state, each unit cell has one spin up and one
spin down cation. The chemical and magnetic unit cells contain the
atoms shown in Fig. 22.4. The space group symmetry operations for
D4h pertinent to the rutile structure are the 16 operations listed below:
1. {E|0} 9. {i|0}
2. {C2|0} 10. {σh|0} = {C2|0}{i|0}
3. {C2ξ|0} 11. {σdξ|0} = {C2ξ|0}{i|0}
4. {C2ν|0} 12. {σdν|0} = {C2ν|0}{i|0}
5. {C4|τ0} 13. {S−1
4 |τ0} = {C4|τ0}{i|0}
6. {C−1
4 |τ0} 14. {S4|τ0} = {C−1
4 |τ0}{i|0}
7. {C2x|τ0} 15. {σvx|τ0} = {C2x|τ0}{i|0}
8. {C2y|τ0} 16. {σvy|τ0} = {C2y|τ0}{i|0}
(22.5)
where the axes ξ = (110) and ν = (1¯10) denote two-fold axes and
the translation τ0 = 1
2
(a1 + a2 + a3) is to the body center of the unit
cell (see Fig. 22.4). The point group D4h corresponding to these space
group operations is found by setting τ0 = 0. The character table for
D4 is given in Table 22.4 where D4h = D4 ⊗ i. Thus the operations in
Eq. (22.5) correspond to the space group for the chemical unit cell.
The unitary subgroup that forms the symmetry group for antiferromagnetic
MF2 (M=magnetic cation) consists of the 4 elements of the
symmetry group D2 {E|0}, {C2|0}, {C2x|τ0}, {C2y|τ0} and 4 additional
elements formed by combining these with inversion. These 8 elements
constitute {Ai} which corresponds to the group D2h = D2 ⊗ i (see Table
22.3). Note that the operations C2x and C2y invert the spins. The
634 CHAPTER 22. MAGNETIC GROUPS
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§
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Figure 22.4: The common antiferromagnets
MnF2, FeF2 and
CoF2 crystallize in the rutile
structure with |a1| = |a2| =
a; |a3| = c; c = a. The
diagram shows the magnetic
point group D4h (D2d) which
describes the antiferromagnetic
spin alignment.
Table 22.4: Character Table for Group D4 (422)
D4 (422) E C2 = C2
4 2C4 2C2 2C2
x2
+ y2
, z2
A1 1 1 1 1 1
Rz, z A2 1 1 1 −1 −1
x2
− y2
B1 1 1 −1 1 −1
xy B2 1 1 −1 −1 1
(xz, yz)
(x, y)
(Rx, Ry)
E 2 −2 0 0 0
D4h = D4 ⊗ i
22.5. EXAMPLES OF MAGNETIC STRUCTURES 635
appropriate Bravais lattice for MnF2 is the tetragonal Bravais lattice
PI for the magnetic groups (see Fig. 22.1).
If we ignore the fluorine anions, the chemical unit cell would be half
as large containing only one magnetic cation. The magnetic unit cell
would then be twice as large as the chemical unit cell. Nevertheless
the magnetic point group for the magnetic antiferromagnetic system
remains D4h(D2h).
22.5.3 The Magnetic States of EuSe
Because the nearest and next-nearest exchange constants are of approximately
equal magnitude and of opposite sign, EuSe exhibits several
different magnetic phases, depending on the magnetic field and
temperature variables. In Figs. 22.5a, 22.5b, 22.5c we see, respectively,
the spin arrangement for the antiferromagnetic (AF-II) two spin (↑↓)
phase, the ferrimagnetic three spin (↑↑↓) phase, and the antiferromagnetic
(AF-I) four spin (↑↑↓↓) phase.
A ferromagnetic phase is also found upon application of a high applied
magnetic field. In all 4 magnetically ordered phases, the spins
in a given (1¯11) plane are parallel to each other and are oriented
along the [011] direction. The resulting magnetic space group has
very low symmetry. For the AF-II phase, the symmetry elements are:
{E|0}, {i|0}, ˆT{E|τ0}, ˆT{i|τ0} in which the vector τ0 takes the spins
from one sublattice to the other
τ0 =
1
4
(ax, 0, az) (22.6)
Thus the magnetic point group is S2 ⊗ ˆT.
If, however, the spins were oriented instead along [1¯11] and [¯11¯1]
directions in alternate (111) planes, then the magnetic symmetry of
the group increases and is C3 ⊗ ˆT. Thus the spin direction is important
in determining the magnetic point group and the magnetic space group.
We note that the number of sublattices (1, 2, 3, or 4) is also important
in determining the symmetry operations in the magnetic space groups.
For some cases it is useful to ignore the spin directions and just to
consider each atom on a given sublattice as a colored atom. Such
636 CHAPTER 22. MAGNETIC GROUPS
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¨6BCD(§EFEHG IP©¢Q9RD(4SFG 5T"8$D©¢')(@¨A1§143§
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UXV
Figure 22.5: Magnetic structure of EuSe in (a) the AF-II phase (↑↓),
(b) the ferrimagnetic phase (↑↑↓), and (c) the antiferromagnetic AF-I
phase (↑↑↓↓).
22.6. SELECTED PROBLEMS 637
groups are called color groups. Color groups have more symmetry
than the magnetic groups.
22.6 Selected Problems
1. Suppose that we have a magnetic compound MX (where M is
the magnetic species) that crystallizes in the zincblende structure.
Suppose that at the Ne´el temperature the magnetic species
undergo a magnetic phase transition to an antiferromagnetic two
sublattice phase such that by treating the M↑ and M↓ as different
species, the magnetic crystal is described by the chalcopyrite
structure. Find the change in the Raman spectra associated with
this magnetic phase transition from the zincblende to the chalcopyrite
structures.
In the prototype chalcopyrite structure, shown
on the right for ZnGeP2, the lattice is compressed
slightly along the vertical direction
and the phosphorus atoms are slightly displaced
from the positions they would have in
the zincblende structure.
638 CHAPTER 22. MAGNETIC GROUPS
Chapter 23
Symmetry Considerations of
Fullerene Molecules and
Carbon Nanotubes
Many of the special properties that fullerenes exhibit are directly related
to the very high symmetry of the C60 molecule, where the 60
equivalent carbon atoms are at the vertices of a truncated icosahedron.
The regular truncated icosahedron is obtained from the regular icosahedron
by passing planes normal to each of the six five-fold axes passing
through the center of the icosahedron so that the edges of the pentagonal
faces thus formed are equal in length to the edges of the hexagonal
faces. Figure 23.1 shows this soccer-ball configuration for C60. This
structure is thought to have been constructed by Leonardo da Vinci in
about 1500, and Fig. 23.1 shows the location of the carbon atoms at
the vertices of the truncated icosahedron. The first application of the
icosahedral group to molecules was by Tisza in 1933.
In this chapter the group theory for the icosahedron is reviewed,
and mathematical tables are given for simple applications of the icosahedral
group symmetry to the vibrational and electronic states of the
icosahedral fullerenes. The effect of lowering the icosahedral symmetry
is discussed in terms of the vibrational and electronic states. Symmetry
considerations related to the isotopic abundances of the 12
C and 13
C
nuclei are discussed. The space group symmetries appropriate to several
crystalline phases of C60 are reviewed. The symmetry properties of
639
640 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.1: Character tablea,b,c
for the point group Ih.
R E 12C5 12C2
5 20C3 15C2 i 12S3
10 12S10 20S3 15σv
Ag 1 +1 +1 +1 +1 +1 +1 +1 +1 +1
F1g 3 +τc
1−τ 0 −1 +3 +τ 1 − τ 0 −1
F2g 3 1−τ +τ 0 −1 +3 1 − τ +τ 0 −1
Gg 4 −1 −1 +1 0 +4 −1 −1 +1 0
Hg 5 0 0 −1 +1 +5 0 0 −1 +1
Au 1 +1 +1 +1 +1 −1 −1 −1 −1 −1
F1u 3 +τ 1−τ 0 −1 −3 −τ τ − 1 0 +1
F2u 3 1−τ +τ 0 −1 −3 τ − 1 −τ 0 +1
Gu 4 −1 −1 +1 0 –4 +1 +1 −1 0
Hu 5 0 0 −1 +1 –5 0 0 +1 −1
a
Note: the symmetry operations about the 5-fold axes are in two different classes,
labeled 12C5 and 12C2
5 in the character table. Then iC5 = S−1
10 and iC−1
5 = S10
are in the classes labeled 12S3
10 and 12S10, respectively. Also iC2 = σv.
b
See Table 23.2 for a complete listing of the basis functions for the Ih point group
in terms of spherical harmonics.
c
In this table τ ≡ (1 +
√
5)/2.
symmorphic and non-symmorphic carbon nanotubes are also discussed.
23.1 Icosahedral Symmetry Operations
The truncated icosahedron (see Fig. 23.1) has 12 pentagonal faces, 20
hexagonal faces, 60 vertices and 90 edges. The 120 symmetry operations
for the icosahedral point group are listed in the character table
given in Table 23.1, where they are grouped into 10 classes, as indicated
in this character table. These classes are the identity operator which
is in a class by itself, the 12 primary five-fold rotations (12 C5 and
12 C2
5 ) going through the centers of the pentagonal faces, the 20 secondary
3-fold rotations going through the centers of the 20 hexagonal
faces, and the 30 secondary 2-fold rotations going through the 30 edges
joining two adjacent hexagons. Each of these symmetry operations is
compounded with the inversion operation. Also listed in the character
23.1. ICOSAHEDRAL SYMMETRY OPERATIONS 641
C2
C5
C3
Ih
C2
C3C5
C60
5-fold axis 3-fold axis
2-fold axis σ
Figure 23.1: Symmetry operations of the regular truncated icosahedron.
(a) The 5-fold axis, (b) the 3-fold axis, (c) the 2-fold axis, and (d) a
composite of the symmetry operations of the point group Ih.
642 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
table are the 10 irreducible representations of the point group Ih.
The C60 molecule has carbon atoms at the 60 equivalent vertices of
a truncated icosahedron for which the lengths of the pentagonal edges
are slightly longer (a5 = 1.46 ˚A) than the bond lengths shared by
two hexagons (a6 = 1.40 ˚A). If we take this difference in bond length
into account, then the C60 cage forms a truncated icosahedron, but
not a regular truncated icosahedron where all bond lengths would be
equal. Nevertheless, from a symmetry point of view, the truncated
icosahedron describes the C60 molecule group symmetry, as does the
regular truncated icosahedron.
Since the truncated icosahedron is close in shape to a sphere, it is
suggestive to relate the basis functions of the icosahedron to those of
the sphere, namely the spherical harmonics. Basis functions for each
irreducible representation and each partner for group Ih are listed in
Table 23.2 in terms of spherical harmonics Y ,m with minimal values.
Many physical problems dealing with fullerenes, such as the electronic
states or vibrational modes, are treated in terms of spherical harmonics
which are basis functions for the full rotational group. The spherical
harmonics therefore form reducible representations of the Ih point group
for > 2, and irreducible representations for = 0, 1, 2. Odd and even
integers , respectively, correspond to odd and even representations of
the group Ih. The basis functions for group Ih in an l-dimensional
manifold are obtained by solving the eigenvalue problem for irreducible
tensor operations. The decomposition of the spherical harmonics into
irreducible representations of the point group I is given in Table 23.3
for both integral and half integral values of the angular momentum J,
where groups I and Ih are related by Ih = I⊗i. The integral values of J
are pertinent to the vibrational spectra (§23.2), while the half integral
J values are also needed to describe the electronic states when electron
spin is included in the wave function (§23.3).
Five-fold symmetry is not often found in solid state physics, because
it is impossible to construct a Bravais lattice based on five-fold symmetry
(see §12.4). Thus fullerenes in the solid state crystallize into solids
of lower point group symmetries, such as the fcc lattice (e.g., C60 at
room temperature) or the hcp lattice (e.g., some phases of C70). Nevertheless,
the local point group symmetry of the individual fullerene
molecules is very important because they crystallize into highly molec-
23.1. ICOSAHEDRAL SYMMETRY OPERATIONS 643
Table 23.2: Basis functions for each of the irreducible representations
R of point group Ih expressed as spherical harmonics Yl,m. For the
multi-dimensional representations, the basis functions for each partner
are listed.
R Basis function
Ag Y0,0; and
√
7
5 (Y6,5 − Y6,−5) +
√
11
5 Y6,0
F1g



−
√
3/5Y6,−6 +
√
66/10Y6,−1 +
√
22/10Y6,4
1/2(Y6,5 + Y6,−5)√
3/5Y6,6 +
√
66/10Y6,1 −
√
22/10Y6,−4
F2g



− 28/125Y8,−8 + 39/500Y8,−3 + 143/250Y8,2 − 63/500Y8,7
1/2(Y8,5 + Y8,−5)
− 28/125Y8,8 − 39/500Y8,3 + 143/250Y8,−2 + 63/500Y8,−7
Gg



8/15Y4,−4 + 7/15Y4,1
1/15Y4,−3 + 14/15Y4,2
− 1/15Y4,3 + 14/15Y4,−2
8/15Y4,4 − 7/15Y4,−1
Hg



Y2,−2
Y2,−1
Y2,0
Y2,1
Y2,2
Au −
√
5·7·11·13
250 (Y15,15 + Y15,−15) −
√
2·3·5·11·29
250 (Y15,10 − Y15,−10)
+
√
23·29
50 (Y15,5 + Y15,−5)
F1u



Y1,−1
Y1,0
Y1,1
F2u



− 2/5Y3,−3 + 3/5Y3,2)
Y3,0
2/5Y3,3 + 3/5Y3,−2)
Gu



3/5Y3,−3 + 2/5Y3,2
Y3,−1
Y3,1
− 3/5Y3,3 + 2/5Y3,−2
Hu



3/10Y5,−4 + 7/10Y5,1
3/5Y5,−3 − 2/5Y5,2
1/2(Y5,5 + Y5,−5)
3/5Y5,3 + 2/5Y5,−2
− 3/10Y5,4 + 7/10Y5,−1
644 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.3: Decomposition of angular momenta basis functions in the
full rotation group labeled by J into irreducible representations of the
double group of I. Both integral and half-integral angular momentum
basis functions are included.a
J Γ1 Γ2 Γ3 Γ4 Γ5 J Γ6 Γ7 Γ8 Γ9
A F1 F2 G H
0 (S) 1 1
2 1
1 (P) 1 3
2 1
2 (D) 1 5
2 1
3 (F) 1 1 7
2 1 1
4 (G) 1 1 9
2 1 1
5 (H) 1 1 1 11
2 1 1 1
6 (I) 1 1 1 1 13
2 1 1 1 1
7 (K) 1 1 1 1 15
2 1 2
8 (L) 1 1 2 17
2 1 1 2
9 (M) 1 1 2 1 19
2 1 1 1 2
10 (N) 1 1 1 1 2 21
2 1 2 2
11 (O) 2 1 1 2 23
2 1 1 2 2
12 (Q) 1 1 1 2 2 25
2 1 1 1 3
13 (R) 1 2 2 2 27
2 1 2 3
14 (T) 1 1 2 3 29
2 1 1 2 3
15 (U) 1 2 2 2 2 31
2 2 1 2 3
16 (V) 1 2 1 2 3 33
2 1 1 3 3
17 (W) 2 2 2 3 35
2 1 1 2 4
18 (X) 1 1 2 3 3 37
2 1 2 2 4
19 (Y) 2 2 2 4 39
2 1 1 3 4
20 (Z) 1 2 2 2 4 41
2 2 1 3 4
21 1 3 2 3 3 43
2 2 2 3 4
22 1 2 2 3 4 45
2 1 1 3 5
23 2 3 3 4 47
2 1 2 3 5
24 1 2 2 4 4 49
2 2 2 3 5
25 1 3 3 3 4 51
2 2 1 4 5
26 1 3 2 3 5 53
2 2 2 4 5
27 1 3 3 4 4 55
2 2 2 3 6
28 1 2 3 4 5 57
2 1 2 4 6
29 3 3 4 5 59
2 2 2 4 6
30 2 3 3 4 5 61
2 3 2 4 6
31 1 4 3 4 5 63
2 2 2 5 6
32 1 3 3 4 6 65
2 2 2 4 7
33 1 3 4 4 5 67
2 2 3 4 7
34 1 3 3 4 6 69
2 2 2 5 7
35 1 4 4 4 6 71
2 3 2 5 7
36 2 4 3 5 6 72
2 3 3 5 7
...
...
...
...
...
...
...
...
...
...
...
a
As an example in using the table, Γ(J = 5) = Γ2(F1) + Γ3(F2) + Γ5(H).
23.1. ICOSAHEDRAL SYMMETRY OPERATIONS 645
Table 23.4: Characters χa.s.
for the equivalence transformation of various
atomic sites in icosahedral Ih symmetry.a
The corresponding irreducible
representations of group Ih are listed in Table 23.5.
Cluster E C5 C2
5 C3 C2 i S3
10 S10 S6 σv
# elements 1 12 12 20 15 1 12 12 20 15
X12 12 2 2 0 0 0 0 0 0 4
C20 20 0 0 2 0 0 0 0 0 4
X30 30 0 0 0 2 0 0 0 0 4
C60 60 0 0 0 0 0 0 0 0 4
C80 80 0 0 2 0 0 0 0 0 8
C140 140 0 0 2 0 –a
–a
–a
–a
–a
C180 180 0 0 0 0 0 0 0 0 4
C240 240 0 0 0 0 0 0 0 0 8
a
Since C140 lacks inversion symmetry, entries in the table are made only for the
classes that are pertinent to point group I.
ular solids in which the electronic and vibrational states are closely
related to those of the free molecule. Therefore we summarize in this
chapter the group theoretical considerations that are involved in finding
the symmetries and degeneracies of the vibrational and electronic
states of the C60 molecule, with some discussion also given to higher
mass fullerenes.
To describe the symmetry properties of the vibrational modes and
of the electronic levels, it is necessary to find the equivalence transformation
for the carbon atoms in the molecule. The characters for
the equivalence transformation χa.s.
for the 60 equivalent carbon atom
sites (a.s.) for the C60 molecule in icosahedral Ih symmetry are given
in Table 23.4. Also listed in Table 23.4 are the characters for the
equivalence transformation for the 12 five-fold axes, the 20 three-fold
axes and the 30 two-fold axes which form classes of the icosahedral
Ih group. The decomposition of the reducible representations of Table
23.4 into their irreducible constituents is given in Table 23.5, which
directly gives the number of π-orbitals for each irreducible representation.
For example, if guest species X are attached to each 5-fold axis at
an equal distance from the center of the icosahedron to yield a molecule
646 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.5: Irreducible representations contained in the various χa.s.
in
the icosahedral group Ih given in Table 23.4.a
Cluster χa.s.
X12
Ag, Hg,
F1u, F2u
C20
Ag, Gg, Hg,
F1u, F2u, Gu
X30
Ag, Gg, 2Hg,
F1u, F2u, Gu, Hu
C60
Ag, F1g, F2g, 2Gg, 3Hg,
2F1u, 2F2u, 2Gu, 2Hu
C80
2Ag, F1g, F2g, 3Gg, 4Hg,
3F1u, 3F2u, 3Gu, 2Hu
C140 3A, 7F1, 7F2, 10G, 11H
C180
2Ag, 5F1g, 5F2g, 6Gg, 8Hg
Au, 4F1u, 5F2u, 6Gu, 7Hu
C240
3Ag, 5F1g, 5F2g, 8Gg, 11Hg,
Au, 7F1u, 7F2u, 8Gu, 9Hu
a
Since C140 lacks inversion symmetry, the irreducible representations for χa.s.
refer
to the group I.
23.2. SYMMETRY OF VIBRATIONAL MODES 647
X12C60, then the full icosahedral Ih symmetry is preserved. Table 23.4
also lists χa.s.
for a few higher icosahedral fullerenes which are specified
by CnC
with nC = 20(m2
+ n2
+ mn). Thus C60 corresponds to
(m, n) = (1, 1), C80 has (n, m) = (2, 0), C140 has (n, m) = (2, 1), and
C240 has (n, m) = (2, 2). We note that fullerenes with either m = 0, or
n = 0, or those with m = n have inversion symmetry and therefore are
described by group Ih. Other icosahedral fullerenes with m = n = 0
lack inversion symmetry (e.g., C140 and C180) and are described by the
point group I. The decomposition of the equivalence transformation
Γa.s.
which is a reducible representation of the group Ih (or I) into its
irreducible constituents is given in Table 23.5 for every entry in Table
23.4, and the even and odd constituents are listed on separate lines.
23.2 Symmetry of Vibrational Modes
In this section we review the symmetries and degeneracies of the vibrational
modes for the C60 molecule. There are 180 degrees of freedom
(3 × 60) for each C60 molecule. Three of these degrees of freedom correspond
to translations and three to rotation, leaving 174 vibrational
degrees of freedom. Since icosahedral symmetry gives rise to a large
number of degenerate modes, only 46 distinct mode frequencies are expected
for the C60 molecule. The number of distinct modes Nω for C60
and other icosahedral configurations is given in Table 23.6. The results
given in Table 23.6 follow directly from group theoretical arguments,
using the entries given in Table 23.4 for the characters for the equivalence
transformation χa.s.
(C60) for the 60 equivalent carbon atoms in
C60. Taking the direct product of χa.s.
(C60) with the characters for the
vector (which transforms according to the irreducible representation
F1u), as given in Table 23.1, and subtracting off the irreducible representations
for pure rotations (F1g) and pure translations (F1u), yields
the irreducible representations for the vibrational modes of C60. The
symmetries of the resulting vibrational modes are listed in Table 23.6,
where the multiplicities for each symmetry type are given. For example,
X12 in Ih symmetry has Nω = 8 distinct modes, and the symmetry
types that are found include Ag + Gg + 2Hg + F1u + F2u + Gu + Hu
where 2Hg indicates that there are two distinct mode frequencies cor-
648 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.6: Symmetry properties of vibrational modes in molecules
with icosahedral symmetry.
Cluster Nω
a
Ag
b
F1g F2g Gg Hg
c
Au F1u
d
F2u Gu Hu
X12 8 1 1 2 1 1 1 1
X20 14 1 1 2 3 1 2 2 2
X30 22 1 1 2 3 4 2 3 3 3
C60 46 2 3 4 6 8 1 4 5 6 7
X12C60 56 3 4 4 7 10 1 6 6 7 8
C80 62 3 6 5 8 10 1 5 6 8 10
C140
e
110 7 20 20 28 35 – – – – –
C240 190 7 17 17 24 31 5 17 19 24 29
a
The number of distinct mode frequencies in icosahedral symmetry is denoted by
Nω.
b
Raman-active mode is seen only in , polarization.
c
Raman-active mode is seen in both , and , ⊥ polarizations.
d
Infrared-active mode symmetry.
e
In I symmetry, there are no gerade or ungerade modes. In this case the F1 modes
are IR-active and the A and H modes are Raman-active.
23.2. SYMMETRY OF VIBRATIONAL MODES 649
Figure 23.2: First-order infrared
(A) and Raman (B)
spectra for C60 taken with
low incident optical power
levels (<50 /mm2
)
responding to (5 × 2) = 10 normal modes. We note that for the C60
molecule, every irreducible representation is contained at least once. In
carrying out the direct product χa.s.
⊗χF1u the entries in Table 23.7 for
the decomposition of direct products for the point group I are useful,
noting that Ih = I ⊗ i.
The Raman-active modes have Ag and Hg symmetry (corresponding
to the basis functions for all symmetrical quadratic forms, and the
antisymmetric F1g does not contribute to the Raman scattering). The
infrared-active modes have F1u symmetry (the linear forms associated
with a vector). One can see from the basis functions listed in Table 23.2
that the symmetry of the Raman tensor allows , scattering for Ag
modes, and both , and , ⊥ scattering for Hg modes, where the directions
and ⊥ refer to the polarization directions of the incident
and scattered photon electric fields. For example, , ⊥ implies that the
polarizations for the incident and scattered beams are orthogonal. Table
23.6 shows that of the 46 distinct vibrational mode frequencies for
C60, only four are infrared active with symmetry F1u, and only 10 are
Raman active (two with Ag symmetry and eight with Hg symmetry),
while the remaining 32 modes are silent in the first-order infrared and
Raman spectra. The experimental observation of the infrared and Raman
spectra for C60 is shown in Fig. 23.2. Many of these silent modes
can, however, be observed by inelastic neutron scattering, electron energy
loss spectroscopy, as vibronic sidebands on the photoluminescence
spectra, and most sensitively in the higher-order infrared and Raman
spectra, because of the different selection rules governing these higher-
650 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.7: Decomposition of direct productsa
in the icosahedral point
group I.
R Γ1 (A) Γ2 (F1) Γ3 (F2) Γ4 (G) Γ5 (H)
Γ1 (A) Γ1 Γ2 Γ3 Γ4 Γ5
Γ2 (F1) Γ2



Γ1
Γ2
Γ5
Γ4
Γ5



Γ3
Γ4
Γ5



Γ2
Γ3
Γ4
Γ5
Γ3 (F2) Γ3
Γ4
Γ5



Γ1
Γ3
Γ5



Γ2
Γ4
Γ5



Γ2
Γ3
Γ4
Γ5
Γ4 (G) Γ4



Γ3
Γ4
Γ5



Γ2
Γ4
Γ5



Γ1
Γ2
Γ3
Γ4
Γ5



Γ2
Γ3
Γ4
2Γ5
Γ5 (H) Γ5



Γ2
Γ3
Γ4
Γ5



Γ2
Γ3
Γ4
Γ5



Γ2
Γ3
Γ4
2Γ5



Γ1
Γ2
Γ3
2Γ4
2Γ5
Γ6 Γ6
Γ6
Γ8
Γ9
Γ7
Γ9
Γ8
Γ9
Γ7 Γ7 Γ9
Γ6
Γ8
Γ6
Γ9
Γ8
Γ9
Γ8 Γ8



Γ6
Γ8
Γ9



Γ7
Γ8
Γ9
Γ8
2Γ9



Γ6
Γ7
Γ8
2Γ9
Γ9 Γ9



Γ7
Γ8
2Γ9



Γ6
Γ8
2Γ9



Γ6
Γ7
2Γ8
2Γ9



Γ6
Γ7
2Γ8
3Γ9
a
As an example of using this table, the direct product Γ4 ⊗ Γ2 = Γ3 + Γ4 + Γ5.
23.2. SYMMETRY OF VIBRATIONAL MODES 651
order processes. The observation that the Raman spectra for C60 remain
essentially unchanged for isolated C60 molecules in solution and
in a crystalline film is indicative of the very weak coupling between the
C60 molecules in the solid.
Various possible attachments could be made to the C60 molecule
without lowering its symmetry (e.g., by attaching 12 equivalent guest
species (X) along the 12 five-fold axes, or 20 guest species along the 20
three-fold axes, or 30 guest species along the two-fold axes). To preserve
the Ih symmetry, all the equivalent sites must be occupied. However,
if only some of these sites are occupied, the symmetry is lowered (see
§23.4). Table 23.4 gives the characters χa.s.
for the equivalence transformation
for special arrangements of guest species that preserve the Ih or
I symmetries, and these guest species may be attached through doping
or a chemical reaction. The corresponding vibrational modes associated
with such guest species are included in Table 23.6, both separately and
in combination with the C60 molecule, as for example X12C60, where
we might imagine a guest atom to be located at the center of each
pentagonal face, as occurs in the alkali metal coated Li12C60. Any detailed
solution to the normal mode problem will involve solutions of
a dynamical matrix in which tangential and radial modes having the
same symmetry will mix.
Vibrational modes that are silent in the first-order spectrum can,
however, contribute to the second- and higher-order Raman and infrared
spectra. Anharmonic terms in the potential couple the normal
mode solutions of the harmonic potential approximation, giving rise
to overtones (nωi) and combination modes (ωi ± ωj), many of which
are observable in the second-order spectra. Group theory requires that
the direct product of the second-order combination modes Γi ⊗ Γj (see
Table 23.7) must contain the irreducible representations F1u to be observable
in the second-order infrared spectrum, and Ag or Hg to be
observable in the second-order Raman spectrum. By parity considerations
alone, overtones (or harmonics) can be observed in the Raman
spectrum, but are not symmetry-allowed in the second-order infrared
spectrum, since all second-order overtones have even parity. Because of
the highly molecular nature of crystalline C60, the second-order infrared
and Raman spectra are especially strong in crystalline films. Whereas
only about 10 strong features are seen experimentally in the first-order
652 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Raman spectrum, over 100 features are resolved and identified in the
second-order Raman spectrum. The observation of a multitude of sharp
lines in the higher-order infrared and Raman spectra is a unique aspect
of the spectroscopy of highly molecular solids. For typical crystals, dispersion
effects in the solid state broaden the higher-order spectra, so
that detailed features commonly observed in the gas phase spectra can
no longer be resolved in the crystalline phase. Nevertheless, analysis
of the second-order infrared and Raman spectra of fcc C60 provides a
powerful method for the determination of the silent modes of C60.
Although the icosahedral C80, C140, and C240 icosahedral molecules
have not yet been studied by Raman or infrared spectroscopy, the symmetry
analysis for these molecules is included in Table 23.6. The corresponding
symmetry analysis for fullerene molecules of lower symmetry
is given in §23.4.
23.3 Symmetry for Electronic States
Symmetry considerations are also important for describing the electronic
states of fullerenes and their related crystalline solids. In this section
we consider a simple description of the electronic states of fullerene
molecules. The basic concepts presented in this section can be extended
to the electronic states in crystalline solids and to carbon nanotubules.
To treat the electronic energy levels of a fullerene molecule it is necessary
to consider a many-electron system with the point group symmetry
appropriate to the fullerene. From a group theoretical standpoint,
treatment of the electronic states for the neutral C60 molecule or the
charged C±n
60 molecular ion requires consideration of both integral and
half-integral angular momentum states. To describe the half-integral
states, it is necessary to use the double group based on the point group
Ih. The character table for the double group I is given in Table 23.8.
The double group of Ih is found from that for I by taking the direct
product group I ⊗ i, and the irreducible representations and characters
for the double group are obtained by taking appropriate direct
products of the characters in Table 23.8 with those of the inversion
group Ci consisting of two elements (the identity and the inversion operator)
and having two irreducible representations (1, 1) and (1, −1).
23.3. SYMMETRY FOR ELECTRONIC STATES 653
Table 23.8: Character tablea,b
for the double point group I.
R E E C5 C5 C2
5 C
2
5 C3 C3 C2
# elements 1 1 12 12 12 12 20 20 30
Γ1 (A) 1 +1 +1 +1 +1 +1 +1 +1 +1
Γ2 (F1) 3 +3 +τ +τ 1−τ 1−τ 0 0 −1
Γ3 (F2) 3 +3 1−τ 1−τ +τ +τ 0 0 −1
Γ4 (G) 4 +4 −1 −1 −1 −1 +1 +1 0
Γ5 (H) 5 +5 0 0 0 0 −1 −1 +1
Γ6 2 −2 +τ −τ −(1 − τ) 1 − τ +1 −1 0
Γ7 2 −2 1 − τ −(1 − τ) −τ +τ +1 −1 0
Γ8 4 −4 +1 −1 −1 +1 −1 +1 0
Γ9 6 −6 −1 +1 +1 −1 0 0 0
a
Note: C5 and C−1
5 are in different classes, labeled 12C5 and 12C2
5 in the character
table. The class E represents a rotation by 2π and classes Cn represent rotations
by 2π/n between 2π and 4π. In this table τ = (1 +
√
5)/2, while −1/τ = 1 − τ, and
τ2
= 1 + τ.
b
The basis functions for Γ1 − Γ5 are given in Table 23.2 and for the double group
irreducible representations Γ6 − Γ9 are given in Table 23.9.
Table 23.9: Basis functions for the double group irreducible representations
R of point group I expressed as half-integer spherical harmonics
φj,nj
.
R Basis function
Γ6 φ1/2,−1/2, φ1/2,1/2
Γ7
( (7/10)φ7/2,−3/2 − (3/10)φ7/2,7/2)
( (7/10)φ7/2,3/2 + (3/10)φ7/2,−7/2)
Γ8 φ3/2,−3/2, φ3/2,−1/2, φ3/2,1/2, φ3/2,3/2
Γ9



φ5/2,−5/2
φ5/2,−3/2
φ5/2,−1/2
φ5/2,1/2
φ5/2,3/2
φ5/2,5/2
654 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Basis functions for each of the irreducible representations of the double
group of I are also listed in Table 23.9. Spin states enter both in the
application of the Pauli principle and in considering the effect of the
spin-orbit interaction. Though the spin-orbit interaction of carbon is
small, it has been determined in graphite by means of detailed electron
spin resonance (ESR) measurements.
Each C60 molecule with icosahedral symmetry can be considered to
have 60 × 3 = 180 σ electrons making bonds along the surface of the
icosahedron and 60 π electrons with higher-lying energy levels than the
σ-bonds for a given angular momentum state. On a graphene sheet
(denoting a single layer of the graphite crystal), the bonding π states
have nearest-neighbor orbitals parallel to one another, and the antibonding
states have antiparallel orbitals. More generally, for fullerenes
with nC carbon atoms, the molecular electronic problem involves nC π
electrons.
The electronic levels for the π electrons for a fullerene molecule can
be found by starting with a spherical approximation where spherical
harmonics can be used to specify the electronic wave functions according
to their angular momentum quantum numbers. As stated above,
for > 2 these spherical harmonics form reducible representations for
the icosahedral group symmetry. By lowering the symmetry from full
rotational symmetry to icosahedral symmetry (see Table 23.3), the irreducible
representations of the icosahedral group are found. In general,
the bonding σ levels will lie well below the Fermi level in energy and
are not as important for determining the electronic properties as the π
electrons.
To obtain the symmetries for the 60 π electrons for C60 we focus
our attention on the 60 bonding π electrons whose energies lie close to
the Fermi level. Assigning angular momentum quantum numbers to
this electron gas, we see from the Pauli principle that 60 π electrons
will completely fill angular momentum states up through = 4, leaving
10 electrons in the = 5 level which can accommodate a total of 22
electrons. In Table 23.10 we list the number of electrons that can
be accommodated in each angular momentum state , as well as the
splitting of the angular momentum states in the icosahedral field.
Table 23.10 thus shows that the = 4 level is totally filled by the
C50 molecule or by nC = 50. The filled states in icosahedral symmetry
23.3. SYMMETRY FOR ELECTRONIC STATES 655
Table 23.10: Filled shell configurations for fullerene molecules.a
electrons/state nC HOMO in Ih symmetry
0 2 2 a2
g
1 6 8 f6
1u
2 10 18 h10
g
3 14
24
26
32
f6
2u
g8
u
(f6
2ug8
u)
4 18
40
42
50
g8
g
h10
g
(g8
gh10
g )
5 22
56
60
62
66
72
f6
1u or f6
2u
h10
u
f6
1uf6
2u
(f6
1uh10
u ) or (f6
2uh10
u )
(f6
1uf6
2uh10
u )
6 26
74
78
80
82
84
88
90
92
96
98
a2
g
f6
1g
g8
g or (a2
gf6
1g)
h10
g or (a2
gg8
g)
(a2
gh10
g )
(f6
1gg8
g)
(a2
gf6
1gg8
g) or (f6
1gh10
g )
(a2
gg8
gh10
g )
(f6
1gg8
gh10
g )
(a2
gf6
1gg8
gh10
g )
...
...
...
...
a
The angular momentum quantum number for a spherical shell of π electrons is
denoted by , while nC denotes the number of π electrons for fullerenes with closed
shell (1
Ag) ground state configurations in icosahedral symmetry. The last column
gives the symmetries of all the levels of the value corresponding to the highest
occupied molecular orbital (HOMO). The superscript on the symmetry label indicates
the total spin and orbital degeneracy of the level. All of the listed levels are
assumed to be occupied.
656 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
for = 4 are labeled by the irreducible representations g8
g and h10
g to
accommodate a total of 18 electrons. On filling the = 4 level, possible
ground states occur when either the gg level is filled with 8 electrons
at nC = 40 or when the hg level is filled with 10 electrons at nC = 42,
or when the complete shell = 4 is filled (i.e., g8
gh10
g ) at nC = 50.
Following the same line of reasoning, the 22-fold degenerate = 5 level
in full rotational symmetry will be filled by C72 which splits into the
irreducible representations Hu + F1u + F2u of the icosahedral group
with filled shell occupations for these levels of 10, 6, and 6 electrons,
respectively. Ten electrons in the = 5 angular momentum states of
C60 are sufficient to completely occupy the hu level, leaving the f1u and
f2u levels completely empty, so that the highest occupied molecular
orbital (HOMO) corresponds to the hu level and the lowest unoccupied
molecular orbital (LUMO) corresponds to the f1u level, in agreement
with H¨uckel calculations for the one-electron molecular orbitals. It
should be noted that for H¨uckel calculations the next lowest unoccupied
molecular orbital is not an f2u level but rather an f1g level, associated
with the angular momentum state = 6. The reason why an = 6
derived level becomes lower than an = 5 derived level is due to the
form of the atomic potential. In fact, the C60 molecule has sufficiently
large icosahedral splittings so that some of the = 6 states lie lower
than the highest = 5 state, so taht the lowest = 6 state becomes
occupied before the = 5 shell is completely filled. Such level crossings
occur even closer to the HOMO level as nC increases.
Not only C60, but also other higher mass fullerenes, have icosahedral
symmetry. As discussed previously, all icosahedral fullerenes can be
specified by CnC
where nC = 20(n2
+ nm + m2
). Using the same
arguments as for C60, the angular momentum states and electronic
configurations for the nC π electrons in these larger fullerenes (up to
nC = 780) can be found (see Table 23.11). In this table, the symmetry
of each icosahedral fullerene is given. Fullerenes with (n, m) values such
that n = 0, m = 0 or m = n have Ih symmetry (including the inversion
operation), while other entries have I symmetry, lacking inversion. Also
listed in this table is max, the maximum angular momentum state that
is occupied, from which ntot, the maximum number of electrons needed
23.3. SYMMETRY FOR ELECTRONIC STATES 657
Table 23.11: Symmetries and configurations of the π-electrons for icosahedral
fullerenes.
CnC max
a
ntot
b
nv
c
Config.d
JHund Ih (or I) Symmetriese
C20 3 32 2 . . . f2
4 Gg, Hg
C60 5 72 10 . . . g18
h10
0 Ag
C80 6 96 8 . . . h22
i8
16 Ag, 2F1g, F2g, 2Gg, 3Hg
C140 8 162 12 . . . k30
l12
24 A, 2F1, 2F2, 4G, 4H
C180 9 200 18 . . . l34
m18
0 Ag
C240 10 242 40 . . . m38
n40
20 Ag, 2F1g, 2F2g, 2Gg, 4Hg
C260 11 288 18 . . . n42
o18
36 2A, 4F1, 3F2, 5G, 6H
C320 12 338 32 . . . o46
q32
72 . . .
C380 13 392 42 . . . q50
r42
96 . . .
C420 14 450 28 . . . r54
t28
0 A
C500 15 512 50 . . . t58
u50
120 . . .
C540 16 578 28 . . . u62
v28
56 . . .
C560 16 578 48 . . . u62
v48
144 . . .
C620 17 648 42 . . . v66
w42
112 . . .
C720 18 722 72 . . . w70
x72
36 2Ag, 4F1g, 3F2g, 5Gg, 6Hg
C740 19 800 18 . . . x74
y18
180 . . .
C780 19 800 58 . . . x74
y58
200 . . .
C860 20 882 60 . . . y78
z60
220 . . .
C960 21 968 78 . . . z82
a78
144 . . .
C980 22 1058 12 . . . a86
b12
192 . . .
...
...
...
...
...
...
...
a
max represents the maximum value of the angular momentum for the HOMO
level.
b
ntot = 2( max + 1)2
is the number of electrons needed for a filled shell.
c
nv represents the number of electrons in the HOMO level.
d
The notation for labeling the angular momentum states is consistent with Table
23.3.
e
The splittings of the Hund’s rule ground state JHund in icosahedral symmetry.
658 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
to fill a spherical shell can be calculated according to
ntot = 2( max + 1)2
. (23.1)
The number of valence electrons in the HOMO level nv is included in
Table 23.11, which also lists the full electronic configuration of the π
electrons, the JHund value for the ground state configuration according
to Hund’s rule and the icosahedral symmetries for the valence electron
states. Then by decomposing these angular momentum states into
irreducible representations of the icosahedral group (using Table 23.3
and the notation for the electronic configurations in Table 23.3), the
symmetry designation for the ground state energy levels (according to
Hund’s rule) in icosahedral symmetry can be found (see Table 23.11).
For example, the icosahedral C80 molecule has a sufficient number of
π electrons to fill the = 5 level, with 8 electrons available for filling
states in the = 6 level. H¨uckel calculations for this molecule suggest
that the f1u and f1g levels are completely filled, and the hu level is
partially filled with 8 electrons.
There are, in general, many Pauli-allowed states that one can obtain
from the spherical molecule configurations listed in Table 23.11. For
example, the hypothetical C20 icosahedral molecule with the s2
p6
d10
f2
(or simply f2
) configuration has Pauli-allowed states with S = 0,
L = 0, 2, 4, 6 and with S = 1, L = 1, 3, 5. The Hund’s rule ground
state is the JHund = 4 state that comes from S = 1, L = 5. The
symmetries of these Hund’s rule ground states are listed in the column
labeled JHund in Table 23.11 together with the decomposition of these
states of the JHund reducible representation into the appropriate irreducible
representations of the icosahedral group. If the perturbation
to the spherical symmetry by the icosahedral potentials is small, and
Hund’s rule applies, then the ground state will be as listed. If, however,
the icosahedral perturbation is large compared with the electron
correlation and exchange energies, then the icosahedral splitting must
be considered first before the electrons are assigned to the spherical
symmetry angular momentum states.
23.4. GOING FROM HIGHER TO LOWER SYMMETRY 659
23.4 Going from Higher to Lower Sym-
metry
A lowering of the symmetry from full icosahedral symmetry occurs
in a variety of fullerene-derived structure-property relations. One example
of symmetry lowering results from elongation of the icosahedral
shape of the C60 fullerene molecules to a rugby-ball shape for
the C70 molecule, as discussed below. Another example involves the
introduction of chirality into the fullerene molecule. A third example
is found in many chemical or photochemical reactions which add
side-groups at various sites, and with various symmetries. In these
cases the symmetry-lowering effect is specific to the side-groups that
are added. A fourth example is the introduction of fullerenes into a
crystal lattice. Since no Bravais lattice with five-fold symmetry is possible,
symmetry-lowering must occur in this case. As a fifth example,
carbon nanotubes can be considered to be related to fullerenes through
a symmetry-lowering process (see §23.6).
23.4.1 Symmetry Considerations for C70
The most common fullerene which has lower than icosahedral symmetry
is C70, and the structure and properties of this fullerene have also
been studied in some detail. The C70 molecule can be constructed from
C60 by appropriately bisecting the C60 molecule normal to a five-fold
axis, rotating one hemisphere relative to the other by 36◦
(thereby losing
inversion symmetry), then adding a ring of 5 hexagons around the
equator (or belt), and finally reassembling these three constituents (see
Fig. 23.3). The elongation of the icosahedral C60 in this way to yield
C70 results in a lowering of the symmetry of the molecule from Ih to
D5h. The point group D5h does not have inversion symmetry, but does
have a mirror plane normal to the five-fold axis. In contrast, group
Ih has inversion symmetry but no mirror plane. If a second ring of 5
hexagons is added around the equator, we then obtain a C80 molecule
with D5d symmetry, which is symmetric under inversion but has no σh
mirror plane. The character tables for the point groups D5h and D5d
are given in Tables 23.12 and Tables 23.13, respectively, and the corre-
660 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
1
2
3
4
5
1
2
3
4 5
7 6
8
Figure 23.3: Geometry of the C70
molecule with D5h symmetry. In
the C70 cluster, there are five inequivalent
atomic sites (1–5) and
eight kinds of bonds (boxed num-
bers).
sponding basis functions are given in Table 23.14. The corresponding
compatibility relations for the irreducible representations of the point
group Ih in going to lower symmetry groups (I, Th, D5d, D5, and C1h)
are provided in Table 23.15. Since the group D5d has inversion symmetry,
the irreducible representations of Ih form reducible representations
of D5d, so that the compatibility relations between the two groups are
easily written. For the group D5h which has a mirror plane but no inversion
symmetry, one must use the lower symmetry icosahedral group
I for relating the icosahedral irreducible representations to those in D5,
which is a subgroup of I. The compatibility relations for I → D5 are
also included in Table 23.15, in addition to compatibility tables for
groups Ih → Th and I → C1h.
In treating the electronic levels and vibrational modes for the C70
molecule, we can either go from full rotational symmetry (see Table
23.3) to D5h symmetry in analogy to §23.3, or we can first go from
full rotational symmetry to I symmetry, and then treating D5 as a
subgroup of I, go from I to D5 in the sense of perturbation theory.
Referring to Table 23.16, which shows the decomposition of the various
angular momentum states into irreducible representations of point
group I and then to group D5 (or directly from = 5 to point group
D5), we obtain
Γ =5 → H + F1 + F2 → A1 + 2A2 + 2E1 + 2E2 (23.2)
23.4. GOING FROM HIGHER TO LOWER SYMMETRY 661
Table 23.12: Character table for point group D5h.
R E 2C5 2C2
5 5C2 σh 2S5 2S3
5 5σv
A1 +1 +1 +1 +1 +1 +1 +1 +1
A2 +1 +1 +1 −1 +1 +1 +1 −1
E1 +2 τ − 1 −τ 0 +2 τ − 1 −τ 0
E2 +2 −τ τ − 1 0 +2 −τ τ − 1 0
A1 +1 +1 +1 +1 −1 −1 −1 −1
A2 +1 +1 +1 −1 −1 −1 −1 +1
E1 +2 τ − 1 −τ 0 −2 1 − τ τ 0
E2 +2 −τ τ − 1 0 −2 τ 1 − τ 0
In the table τ ≡ (1 +
√
5)/2 and σv = C2 ⊗ σh.
Table 23.13: Character table for point group D5d.
R E 2C5 2C2
5 5C2 i 2S−1
10
a
2S10 5σd
A1g +1 +1 +1 +1 +1 +1 +1 +1
A2g +1 +1 +1 −1 +1 +1 +1 −1
E1g +2 τ − 1 −τ 0 +2 τ − 1 −τ 0
E2g +2 −τ τ − 1 0 +2 −τ τ − 1 0
A1u +1 +1 +1 +1 −1 −1 −1 −1
A2u +1 +1 +1 −1 −1 −1 −1 +1
E1u +2 τ − 1 −τ 0 −2 1−τ +τ 0
E2u +2 −τ τ − 1 0 −2 +τ 1−τ 0
a
Note: In this table τ ≡ (1+
√
5)/2, while iC5 = S−1
10 and iC2
5 = S10. Also iC2 = σd
662 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.14: Basis functions for the irreducible representations of
groups D5h and D5d.
D5h D5d Basis functions
A1 A1g x2
+ y2
, z2
A2 A2g Rz
E1 E1u (x, y), (xz2
, yz2
), [x(x2
+ y2
), y(x2
+ y2
)]
E2 E2g (x2
− y2
, xy), [y(3x2
− y2
), x(x2
− 3y2
)]
A1 A1u –
A2 A2u z, z3
, z(x2
+ y2
)
E1 E1g (Rx, Ry), (xz, yz)
E2 E2u [xyz, z(x2
− y2
)]
Table 23.15: Compatibility relations between the icosahedral groups,
Ih, I, and several point groups of lower symmetry.
Ih I Th D5d D5 C1h
Ag A Ag A1g A1 A1
F1g F1 Tg A2g + E1g A2 + E1 A1 + 2A2
F2g F2 Tg A2g + E2g A2 + E2 A1 + 2A2
Gg G Ag + Tg E1g + E2g E1 + E2 2A1 + 2A2
Hg H Eg + Tg A1g + E1g + E2g A1 + E1 + E2 3A1 + 2A2
Au A Au Au A1 A2
F1u F1 Tu A2u + E1u A2 + E1 2A1 + A2
F2u F2 Tu A2u + E2u A2 + E2 2A1 + A2
Gu G Au + Tu E1u + E2u E1 + E2 2A1 + 2A2
Hu H Eu + Tu A1u + E1u + E2u A1 + E1 + E2 3A1 + 2A2
23.4. GOING FROM HIGHER TO LOWER SYMMETRY 663
Table 23.16: Decomposition of spherical angular momentum states labeled
by (for ≤ 10) into irreducible representations of lower symmetry
groups.a
Ih Th D5d D5h C1h
0 Ag Ag A1g A1 A1
1 F1u Tu
A2u
E1u
A2
E1
2A1
A2
2 Hg
Eg
Tg



A1g
E1g
E2g



A1
E1
E2
3A1
2A2
3
F2u
Gu
Au
2Tu



A2u
E1u
2E2u



A2
E1
2E2
4A1
3A2
4
Gg
Hg



Ag
Eg
2Tg



A1g
2E1g
2E2g



A1
2E1
2E2
5A1
4A2
5



F1u
F2u
Hu
Eu
3Tu



A1u
2A2u
2E1u
2E2u



A1
2A2
2E1
2E2
6A1
5A2
6



Ag
F1g
Gg
Hg



2Ag
Eg
3Tg



2A1g
A2g
3E1g
2E2g



2A1
A2
3E1
2E2
7A1
6A2
7



F1u
F2u
Gu
Hu



Au
Eu
4Tu



A1u
2A2u
3E1u
3E2u



A1
2A2
3E1
3E2
8A1
7A2
8



F2g
Gg
2Hg



Ag
2Eg
4Tg



2A1g
A2g
3E1g
4E2g



2A1
A2
3E1
4E2
9A1
8A2
9



F1u
F2u
2Gu
Hu
5Tu



2Au
Eu
5Tu



A1u
2A2u
4E1u
4E2u



A1
2A2
4E1
4E2
10A1
9A2
10



Ag
F1g
F2g
Gg
2Hg



2Ag
2Eg
5Tg



3A1g
2A2g
4E1g
4E2g



3A1
2A2
4E1
4E2
11A1
10A2
a
Note that D5d = D5 ⊗ i and D5h = D5 ⊗ σh
664 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.17: Characters of atomic sites χa.s.
for D5h of relevance to the
C70 moleculea,b
.
χa.s.
(D5h) E 2C5 2C2
5 5C2 σh 2S5 2S3
5 5σv
C10(cap0
) 10 0 0 0 0 0 0 2
C20(off-belt) 20 0 0 0 0 0 0 0
C10(belt) 10 0 0 0 10 0 0 0
a
See text for a discussion of C10(cap0
), C20(off-belt), and C10(belt). The same
building blocks listed in this table are found in C50, C70, C90, etc.
b
The irreducible representations for each χa.s.
in this table are given in Table 23.18.
Table 23.18: Irreducible representations of atomic sites χa.s.
for D5h of
relevance to the C70 moleculea
.
χa.s.
(D5h) Irreducible representations
C10(cap0
) A1 + A2 + E1 + E1 + E2 + E2
C20(off-belt) A1 + A2 + A1 + A2 + 2E1 + 2E1 + 2E2 + 2E2
C10(belt) A1 + A2 + 2E1 + 2E2
a
The characters for the equivalence transformation for these sets of carbon atoms
are given in Table 23.17.
where the irreducible representations of group I go into irreducible
representations of group D5
H → A1 + E1 + E2
F1 → A2 + E1
F2 → A2 + E2
(23.3)
using the results of Table 23.15. Since = 5 corresponds to states that
are odd under reflection in the mirror plane, the proper states in D5h
symmetry for = 5 A1 + 2A2 + 2E1 + 2E2 , as given in Table 23.16.
Symmetry considerations also play a major role in classifying the
normal modes of the C70 molecule. To find the symmetries of the
normal mode vibrations of C70, we first find the symmetries for the
transformation of the 70 carbon atoms denoted by χa.s.
(C70) for the
23.4. GOING FROM HIGHER TO LOWER SYMMETRY 665
Table 23.19: Characters of atomic sites χa.s.
for D5d of relevance to the
D5d isomer of the C80 moleculea
.
χa.s.
(D5d) E 2C5 2C2
5 5C2 i 2S−1
10 2S10 5σd
C10 (cap0
) 10 0 0 0 0 0 0 2
C20 (cap1
)
C20 (off−belt)
C20 (belt)
20
20
20
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
a
For an explanation about the notation for the atom sites, see text. Irreducible
representations contained in χa.s.
are given in Table 23.20.
Table 23.20: Irreducible representations of atomic sites χa.s.
for D5d of
relevance to the D5d isomer of the C80 moleculea
.
χa.s.
(D5d) Irreducible representations
C10 (cap0
)
A1g + A2u
+E1g + E1u + E2g + E2u
C20 (cap1
)
C20 (off−belt)
C20 (belt)



A1g + A1u + A2g + A2u
+2E1g + 2E1u + 2E2g + 2E2u
a
The characters for χa.s.
in this table are given in Table 23.19.
666 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.21: Symmetries of molecular vibrational modesa,b
for groups
of carbon atoms with D5h symmetry.
D5h A1 A2 E1 E2 A1 A2 E1 E2
Caxial
10 (cap0
) 1 0 1 1 0 1 1 1
Cradial
10 (cap0
) 2 2 2 1 2 2 2 1
Caxial
20 (off-belt) 2 2 2 1 2 2 2 1
Cradial
20 (off-belt) 2 2 2 1 2 2 2 1
Caxial
10 (belt) 1 0 1 1 0 1 1 1
Cradial
10 (belt) 2 2 2 1 2 2 2 1
Caxial
70+20j
4
+
j
1
+
j
6
+
2j
6
+
2j
3
+
j
4
+
j
8
+
2j
8
+
2j
Cradial
70+20j
8
+
2j
8
+
2j
15
+
4j
16
+
4j
6
+
2j
6
+
2j
11
+
4j
12
+
4j
a
One A2 mode corresponding to translations of the center of mass of the free
molecule along the 5-fold axis and one A2 mode corresponding to rotations of the
free molecule about the 5-fold axis have been subtracted.
b
One E1 mode corresponding to translations of the center of mass of the free
molecule normal to the 5-fold axis and one E1 mode corresponding to rotations of
the free molecule about axes normal to the 5-fold axis have been subtracted.
23.4. GOING FROM HIGHER TO LOWER SYMMETRY 667
Table 23.22: Symmetries of molecular vibrational modesa,b
for groups
of carbon atoms with D5d symmetry.
D5d A1g A2g E1g E2g A1u A2u E1u E2u
Caxial
10 (cap0
) 1 0 1 1 0 1 1 1
Cradial
10 (cap0
) 2 2 2 1 2 2 2 1
Caxial
20 (off-belt) 2 2 2 1 2 2 2 1
Cradial
20 (off-belt) 2 2 2 1 2 2 2 1
Caxial
20 (belt) 1 0 1 1 0 1 1 1
Cradial
20 (belt) 2 2 2 1 2 2 2 1
Caxial
80+20j
5
+
j
2
+
j
8
+
2j
8
+
2j
3
+
j
4
+
j
8
+
2j
8
+
2j
Cradial
80+20j
8
+
2j
8
+
2j
15
+
4j
16
+
4j
8
+
2j
8
+
2j
15
+
4j
16
+
4j
a
One A2u mode corresponding to translations of the center of mass of the free
molecule along the 5-fold axis and one A2g mode corresponding to rotations of the
free molecule about the 5-fold axis have been subtracted.
b
One E1u mode corresponding to translations of the center of mass of the free
molecule normal to the 5-fold axis and one E1g mode corresponding to rotations of
the free molecule about axes normal to the 5-fold axis have been subtracted.
668 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
point group D5h where “a.s.” refers to atom sites (as in Table 23.5
for icosahedral symmetry). Because of the large number of degrees of
freedom in fullerenes, it is advantageous to break up the 70 atoms in C70
into subunits which themselves transform as a subgroup of D5h. This
approach allows us to build up large fullerene molecules by summing
over these building blocks. The equivalence transformation (χa.s.
) for
each of the building blocks can be written down by inspection.
The characters for the equivalence transformation for χa.s.
for these
subgroup building blocks, which are expressed in terms of sets of atoms
normal to the 5-fold axis, are listed in Table 23.17. The symmetry operations
of the group transform the atoms within each of these subgroups
into one another. The χa.s.
entries in Table 23.17 under the various
symmetry operations denote the number of carbon atoms that remain
invariant under the various classes of symmetry operations. The set
C10(cap0
) denotes the 5 carbon atoms around the two pentagons (10
atoms in total) through which the 5-fold axis passes. Another 10 carbon
atoms that are nearest neighbors to the 10 atoms on the axial
pentagons transform in the same way as the set C10(cap0
). The set
C10(belt) refers to the 10 equatorial atoms in the 5 hexagons on the
equator that form a subgroup. There are also two sets of 20 carbon
atoms on hexagon double bonds, labeled C20(off-belt), that form another
subgroup. The characters for the equivalence transformation for
C70 are found by summing the contributions from the various layers
appropriately:
χa.s.
(C70) = 2χa.s.
[C10(cap0
)] + 2χa.s.
[C20(off−belt)] + χa.s.
[C10(belt)].
(23.4)
From Tables 23.17 and 23.21 and from . (23.4), we then obtain the
irreducible representations of D5h contained in the equivalence transformation
for C70 as a whole:
χa.s.
(C70) = 5A1 + 3A2 + 2A1 + 4A4 + 8E1 + 8E2 + 6E1 + 6E2 . (23.5)
If instead of C70, we were to consider an isomer of C90 with D5h symmetry,
the same procedure as in Eqs. (23.4) and (23.5) would be used,
except that an additional χa.s.
(off-belt) would be added to Eq. (23.4).
The same building block approach could be used to describe C80 or
C100 isomers with D5d symmetry using Tables 23.19 and 23.20.
23.4. GOING FROM HIGHER TO LOWER SYMMETRY 669
Table 23.23: The D5h irreducible representations (R) together with the
number of distinct eigenvalues (Nω) and the corresponding degeneracies
g of the normal modes of the C70 molecule. The symbols Nbelt
ω and Ncap
ω
denote the number of distinct eigenvalues associated with the “belt”
and “cap” modes, respectively, for each irreducible representation.
R Nbelt
ω Ncap
ω Nω g
A1 2 10 12 1
A2 2 7 9 1
E1 4 17 21 2
E2 4 18 22 2
A1 1 8 9 1
A2 1 9 10 1
E1 2 17 19 2
E2 2 18 20 2
The symmetries of the molecular vibrations χm.v.
(see Table 23.23)
are then found using the relation
χm.v.
(C70) = χa.s.
(C70) ⊗ χvector
− χtranslations
− χrotations
(23.6)
in which the direct product is denoted by ⊗ and the irreducible representations
for the χvector
, χtranslation
, and χrotation
for group D5h are
given by
χvector
= A2 + E1
χtranslation
= A2 + E1
χrotation
= A2 + E1 .
(23.7)
Table 23.21 lists the number of axial and radial molecular vibrations
associated with each of the layers of carbon atoms of C70. This division
into axial and radial molecular modes is only approximate but often
gives a good description of the physics of the molecular vibrations. The
terms axial and transverse refer to modes associated with motions along
and perpendicular to the 5-fold axis, respectively. Also included in
Table 23.21 are the total number of axial and radial modes for C70+20j,
which contains summaries of the mode symmetries for C70, C90, etc.
From Table 23.21 and Eq. (23.5), we see that the number of distinct
mode frequencies for C70 is 122. It is sometimes useful to consider these
670 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
modes as being approximately divided into cap modes and belt modes.
Thus the 122 modes for C70 are classified as 104 cap modes (corresponding
to the 60 carbon atoms of the two hemispheres of C60) and 18 are
belt modes. This division into cap and belt modes becomes more important
in the limit of carbon nanotubes which are discussed elsewhere.
The symmetries and degeneracies of the distinct mode frequencies for
C70 are given in Table 23.23.
Among the modes given in Table 23.23, those that transform according
to the A1, E2 or E1 irreducible representations are Raman active,
with the A1 modes being observed only in the ( , ) polarization geometry
and the E1 mode observed in the ( , ⊥) polarization. The E2
symmetry mode is seen in both polarization geometries. The modes
with A2 and E1 symmetries are infrared active.
23.4.2 Symmetry Considerations for Higher Mass
Fullerenes
Similar arguments can be made to classify the symmetries of the molecular
vibrations of the rugby-shaped C80 (see Fig. 23.3) which follows
symmetry group D5d. Table 23.19 lists the characters for the equivalence
transformations for groups of carbon atoms comprising the C80
isomer with D5d symmetry. Each of these equivalence transformations
forms a reducible representation of D5d and the decomposition of χa.s.
into irreducible representations of D5d is given in Table 23.20. The
vibrations associated with these groups of atoms are found using a
variant of Eq. (23.6), and the classification of the vibrational modes
into irreducible representations of D5d is given in Table 23.22. Finally
in Table 23.24, we give the number of distinct eigenfrequencies for the
C80 isomer with D5d symmetry, listed according to their symmetry type,
and again distinguishing between the cap and belt modes. It should be
noted that the building block approach using point group D5h can be
used to obtain χa.s.
and χm.v.
for C50, C70, C90, etc., and using group D5d
the corresponding information can simply be found for D5d isomers of
C80, C100, etc. The building block approach provides a simple method
for constructing the dynamical matrix of large fullerenes molecules or
for treating their electronic structure when using explicit potentials.
23.5. SYMMETRY FOR ISOTOPIC EFFECTS 671
Table 23.24: The D5d irreducible representations (R) together with the
number of distinct eigenvalues (Nω) and the corresponding degeneracies
g of the normal modes of the C80 molecule. The symbols Nbelt
ω and Ncap
ω
denote the number of distinct eigenvalues associated with the “belt”
and “cap” modes, respectively, for each irreducible representation.
R Nbelt
ω Ncap
ω Nω g
A1g 3 10 13 1
A2g 2 8 10 1
E1g 3 20 23 2
E2g 2 22 24 2
A1u 2 9 11 1
A2u 3 9 12 1
E1u 3 20 23 2
E2u 2 22 24 2
23.5 Symmetry Considerations for Isotopic
Effects
Carbon has two stable isotopes: 12
C which is 98.892% abundant and
has a molecular weight of 12.011 and a zero nuclear spin, and 13
C with
atomic weight 13.003, a natural abundance of 1.108%, and a nuclear
spin of 1
2
. It is the nuclear spin of the 13
C isotope that is exploited in
the NMR experiments on fullerenes.
Though small in abundance, the 13
C isotope occurs on approximately
half of the C60 molecules synthesized using the natural abundance
of carbon isotopes as shown in Table 23.25. The probability pm
for m isotopic substitutions to occur on an nC atom fullerene CnC
is
given by
pm(CnC
) =
nC
m
xm
(1 − x)nC−m
, (23.8)
where x is the fractional abundance of the isotope and the binomial
coefficient appearing in Eq. (23.8) is given by
nC
m
=
nC!
(nC − m)!m!
. (23.9)
672 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
For the natural abundance of carbon isotopes (x = 0.01108), we obtain
the results for the C60 and C70 molecules listed in Table 23.25 and shown
graphically in Fig. 23.4. Less than 1% of the C60 and C70 fullerenes
have more than three 13
C isotopes per fullerene. Also included in the
table are the corresponding results for an isotopic enrichment to 5%
and 10% 13
C. The results in Table 23.25 show that as x increases (and
also as nC increases in CnC
), the peak in the distribution moves to
larger m values and the distribution gets broader. These conclusions
follow from binomial statistics, where the average ¯m and the standard
deviation ∆m ≡ (m − ¯m)2 1/2
give
¯m = nCx (23.10)
and
∆m = nCx(1 − x), (23.11)
respectively. For example, for a fullerene with x = 0.05, then Eqs. (23.10)
and (23.11) yield ¯m = 3 and ∆m = 1.7. Thus as x increases, so does ¯m
and ∆m, thereby accounting for the broader distribution with increasing
x. In general, for isotopically enriched samples, the distribution
pm(CnC
) is sufficiently shifted and broadened so that graphical displays
are desirable, such as shown in Fig. 23.4.
The results of Table 23.25 have important consequences for both
symmetry considerations and the rotational levels of fullerene molecules.
The molecules containing one or more 13
C atoms show much lower symmetry
than that of the full Ih point group. In fact, the singly 13
C substituted
molecule 13
C1
12
C59 has only one symmetry operation, a single
reflection plane (point group C1h, see Table 23.26); two or more substitutions
generally show no symmetry, i.e., they belong to point group
C1 and have no symmetry-imposed degeneracies, which implies that
all levels (electronic, vibrational, rotational, etc.) are non-degenerate
and every state is both IR and Raman active. Group theory predicts
this symmetry-lowering, but the intensity of IR and Raman lines do
not change much upon addition of 13
C isotopes. Inactive modes before
the isotopic symmetry lowering effect remain mostly inactive, and the
optically-active modes still show strong intensity.
The isotopic distribution has unique consequences with regard to
the rotational levels of the C60 molecule and hence also regarding the
23.5. SYMMETRY FOR ISOTOPIC EFFECTS 673
Table 23.25: The probability pm(CnC
) of 13
C occurring among C60 and
C70 fullerenes.a
x m pm(C60) pm(C70)
0.0001 0 0.9940 0.9930
0.0005 0 0.9704 0.9656
0.001 0 0.9417 0.9324
0.005 0 0.7403 0.7041
0.01 0 0.5472 0.4948
0.01108 0 0.5125 0.4584
0.01108 1 0.3445 0.3595
0.01108 2 0.1139 0.1390
0.01108 3 0.0247 0.0353
0.01108 4 0.0039 0.0066
0.05 0 0.0461 0.0276
0.05 1 0.1455 0.1016
0.05 2 0.2259 0.1845
0.05 3 0.2298 0.2201
0.05 4 0.1724 0.1941
0.05 5 0.1016 0.1348
0.05 6 0.0490 0.0769
0.10 0 0.0018 0.00063
0.10 1 0.0120 0.00487
0.10 2 0.0393 0.01868
0.10 3 0.0844 0.04705
0.10 4 0.1336 0.08756
0.10 5 0.1662 0.12843
0.10 6 0.1693 0.15459
0.10 7 0.1451 0.15704
a
Here x is the isotopic abundance, m is the number of 13
C per fullerene, and
pm(CnC
) is the probability a fullerene CnC
has m 13
C atoms. Although the table
would normally be used for small concentrations of 13
C in 12
C, the same probabilities
as given in the table apply to: 1.108% 12
C in 98.892% 13
C; or to 5% 12
C in
95% 13
C; or to 10% 12
C in 90% 13
C.
674 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
0 10 20 30
m
0.0
0.1
0.2
0.3
0.4
0.5
Pm(C60)
x=0.01108
0.05
0.10
0.15
0.50
Figure 23.4: Plot of the m dependence
of pm(C60) the distribution
of C60 molecules with m 13
C
atoms for various concentrations of
x from 5% to 50% in steps of 5%.
A plot of pm(C60) for x equal to the
natural abundance is also shown.
Table 23.26: Character table for point group C1h.
R E σh basis functions
A1 1 1 1; x; y; x2
; etc.
A2 1 –1 z; xz; yz; etc.
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 675
librational states of the corresponding solid. Other experiments sensitive
to the isotopic abundance include NMR measurements and studies
of the isotope effect regarding the superconducting transition temperature
Tc.
23.6 Symmetry Properties of Carbon Nan-
otubes
Although the symmetry of the 2D graphene layer (a single honeycomb
layer plane of the graphite structure) is greatly lowered in the 1D nanotube,
the single-walled nanotubes have interesting symmetry properties
that lead to nontrivial physical effects, namely a necessary degeneracy
at the Fermi surface for certain geometries.
23.6.1 Relation between Carbon Nanotubes and
Fullerenes
In this section we consider first two simple examples of single-walled
carbon nanotubes based on the C60 fullerene. The concept of a singlewalled
nanotube is then generalized to specify the idealized structure
of single-walled nanotubes in general.
In analogy to a C60 molecule, we can specify a single-walled C60derived
nanotube by bisecting a C60 molecule at the equator and joining
the two resulting hemispheres with a cylindrical tube one monolayer
thick and with the same diameter as C60. If the C60 molecule is bisected
normal to a fivefold axis, the “armchair” nanotube shown in Fig. 23.5(a)
is formed. If the C60 molecule is bisected normal to a threefold axis,
the “zigzag” nanotube in Fig. 23.5(b) is formed. Armchair and zigzag
carbon nanotubes of larger diameter, and having correspondingly larger
caps, are described below.
Figures 23.6 (a) and (b) show the only nanotube types that have the
σh symmetry operation. Other nanotubes, such as (c), belong to a nonsymmorphic
translational group which only has pure spiral symmetry
operations. From the shape of the cross sections of the nanotubes shown
in Fig. 23.5 and listed in Table 23.27, carbon nanotubes with high sym-
676 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
(b)
(a)
(c)
Figure 23.5: By rolling up a
graphene sheet (a single layer from
a 3D graphite crystal) as a cylinder
and capping each end of the
cylinder with half of a fullerene
molecule, a “fullerene-derived carbon
nanotube,” one atomic layer in
thickness, is formed. Shown here is
a schematic theoretical model for a
single-wall carbon nanotube with
the nanotube axis normal to: (a)
the θ = 30◦
direction (an “armchair”
nanotube), (b) the θ = 0◦
direction (a “zigzag” nanotube),
and (c) a general direction
−→
OB (see
Fig. 23.8) with 0 < θ < 30◦
(a “chiral”
nanotube). The actual nanotubes
shown in the figure correspond
to (n, m) values of: (a) (5,
5), (b) (9, 0), and (c) (10, 5).
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 677
(a) (b)
(d)(c)
(5,5)
(10,0)(9,0)
(6,6)
i
i
i
i Figure 23.6: Symmetry of armchair
(a,b) and zigzag (c,d)
tubules with odd (a,c) and even
(b,d) numbers of the unit cell
around the circumferential direction:
(a) (5,5) armchair, (b)
(6,6) armchair (c) (9,0) zigzag
and (d) (10,0) zigzag tubes.
Here we show the inversion cen-
ters.
metry are called (a) armchair and (b) zigzag nanotubes, respectively,
and are described by symmorphic space groups. All nanotubes, except
for the armchair and zigzag nanotubes, are called chiral nanotubes.1
The classification of carbon nanotubes is listed in Table 23.27.
In addition to the armchair and zigzag nanotubes, a large number of
chiral carbon nanotubes can be formed with a screw axis along the axis
of the nanotube and with a variety of hemispherical caps. These carbon
nanotubes can be specified mathematically in terms of the nanotube
diameter dt and chiral angle θ, which are shown in Fig. 23.8, where the
chiral vector Ch
Ch = nˆa1 + mˆa2 (23.12)
is shown, as well as the basic translation vector T for the nanotube,
which is discussed below. In Fig. 23.8, the chiral vector Ch connects
two crystallographically equivalent sites A and A on a two-dimensional
(2D) graphene sheet where a carbon atom is located at each vertex of
the honeycomb structure. The construction in Fig. 23.8 shows the chiral
1
The name for the chiral nanotube comes from the designation of spiral symmetry
by ‘axial chirality’ in chemistry. Axial chirality is commonly discussed in
connection with optical activity.
678 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
NR
hC
C
MT
T
O R
Figure 23.7: The vector NR =
(ψ|τ)N
is shown on the cylindrical
surface. After rotating by
2π around the tube M times,
the vector NR reaches a lattice
point equivalent to point O,
but separated from O by MT.
In the figure we show the case
Ch = (4, 2) where M = 6.
angle θ of the nanotube with respect to the zigzag direction (θ = 0)
and the unit vectors ˆa1 and ˆa2 of the hexagonal honeycomb lattice.
The armchair nanotube [Fig. 23.5(a)] corresponds to θ = 30◦
on this
construction. An ensemble of possible chiral vectors can be specified
by Eq. (23.12) in terms of pairs of integers (n, m) and this ensemble is
shown in Fig. 23.8(b). Each pair of integers (n, m) defines a different
way of rolling up the graphene sheet to form a carbon nanotube. We
now show how the construction in Fig. 23.8(a) specifies the geometry
of the carbon nanotube.
The cylinder connecting the two hemispherical caps of Fig. 23.5 is
formed by superimposing the two ends OA of the vector Ch. The cylinder
joint is made by joining the line AB to the parallel line OB in
Fig. 23.8(a), where lines OB and AB are perpendicular to the vector
Ch at each end. distortion of bond angles other than distortions caused
by the cylindrical curvature of the nanotube. Differences in chiral angle
θ and in the nanotube diameter dt give rise to differences in the
properties of the various carbon nanotubes. In the (n, m) notation for
specifying the chiral vector Ch in Eq. (23.12), the vectors (n, 0) denote
zigzag nanotubes and the vectors (n, n) denote armchair nanotubes,
and the larger the value of n, the larger the nanotube diameter. Both
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 679
a1
a2
(0,0) (1,0) (2,0) (3,0) (4,0) (5,0) (6,0) (7,0) (8,0) (9,0)
(1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (7,1) (8,1)
(2,2) (3,2) (4,2) (5,2) (6,2) (7,2) (8,2)
(3,3) (4,3) (5,3) (6,3) (7,3) (8,3)
(4,4) (5,4) (6,4) (7,4)
(5,5) (6,5)
(6,6) (6,7)
:metal :semiconductor
(10,0) (11,0) (12,0)
(9,1) (10,1)
(9,2)
(11,1)
(10,2) (11,2)
(8,3)
(8,4) (9,4)
(7,5) (8,5)
48
(10,4)
(9,5)
zigzag
armchair
1 3 17 24
3 20
10 19 56
7 18
(9,3)
5 17 43 92
1 1 13 37 80
15 32 87
(7,7)
(b)
y
x
C h
a1
a2
(a)
Ψ
T
τ
θ
y
x
O
A
B
B
Figure 23.8: (a) The chiral vec-
tor
−→
OA or Ch = nˆa1 +mˆa2 is defined
on the honeycomb lattice
of carbon atoms by unit vectors
ˆa1 and ˆa2 and the chiral angle
θ is defined with respect to the
zigzag axis. Along the zigzag
axis θ = 0◦
. Also shown are
the lattice vector
−→
OB= T of the
1D nanotube unit cell and the
rotation angle ψ and the translation
τ which constitute the
basic symmetry operation R =
(ψ|τ) for the carbon nanotube.
The diagram is constructed for
(n, m) = (4, 2). (b) Chiral vectors
specified by the pairs of integers
(n, m) for carbon nanotubes,
including zigzag, armchair,
and chiral nanotubes. Below
each pair of integers (n, m)
is listed the number of distinct
caps that can be joined continuously
to the carbon nanotube
denoted by (n, m). The encircled
dots denote metallic nanotubes
while the small dots are
for semiconducting nanotubes.
680 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
the (n, 0) and (n, n) nanotubes have especially high symmetry, corresponding
to symmorphic space groups, as discussed in §23.6.4, and
exhibit a mirror symmetry plane normal to the nanotube axis. All other
vectors (n, m) correspond to chiral nanotubes. Since both right- and
left-handed chirality are possible for chiral nanotubes, it is expected
that chiral nanotubes are optically active to either right or left circularly
polarized light. In terms of the integers (n, m), the nanotube
diameter dt is given by
dt = Ch/π =
√
3aC−C(m2
+ mn + n2
)1/2
/π (23.13)
where aC−C is the nearest-neighbor C–C distance (1.421 ˚A in graphite),
Ch is the length of the chiral vector Ch, and the chiral angle θ is given
by
θ = tan−1
[
√
3m/(m + 2n)]. (23.14)
For example, a zigzag nanotube (θ = 0◦
) specified by (9, 0) has a theoretical
nanotube diameter of dt = 9
√
3aC−C/π = 7.05 ˚A, while an
armchair nanotube specified by (5,5) has dt = 15aC−C/π = 6.83 ˚A,
both derived from hemispherical caps for the C60 molecule and assuming
an average aC−C = 1.43 ˚A appropriate for C60. If the graphite value
of aC−C = 1.421 ˚A is used, slightly smaller values for dt are obtained.
Substitution of (n, m) = (5, 5) into Eq. (23.14) yields θ = 30◦
while
substitution of (n, m) = (9, 0) and (0, 9) yields θ = 0◦
and 60◦
, respectively.
The nanotubes (0, 9) and (9, 0) are equivalent, because of the
sixfold symmetry of the graphene layer. Because of the point group
symmetry of the honeycomb lattice, several different integers (n, m)
give rise to equivalent nanotubes. To define each nanotube uniquely,
we restrict ourselves to consideration of nanotubes arising from the 30◦
wedge of the 2D Bravais lattice shown in Fig. 23.8(b). Because of the
small diameter of a carbon nanotube (∼10 ˚A) and the large length-todiameter
ratio (> 104
), carbon nanotubes provide an important system
for studying one-dimensional physics, both theoretically and experi-
mentally.
Many of the carbon nanotubes that were observed experimentally
in early work were multilayered, consisting of capped concentric cylinders
separated by ∼ 3.5 ˚A. In a formal sense, each of the constituent
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 681
cylinders can be specified by the chiral vector Ch in terms of the indices
(n, m) of Eq. (23.12), or equivalently by the nanotube diameter
dt and chiral angle θ. Because of the different numbers of carbon atoms
around the various coaxial cylinders, it is not possible to achieve the
ABAB... interlayer graphite stacking in carbon nanotubes. Thus, an
interlayer spacing closer to that of turbostratic graphite (3.44 ˚A) is expected,
subject to the quantized nature of the (n, m) integers, which
determine Ch.
23.6.2 Specification of Lattice Vectors in Real Space
To specify the symmetry properties of carbon nanotubes as 1D systems,
it is necessary to define the lattice vector T along the nanotube
axis and normal to the chiral vector Ch defined by Eq. (23.12) and
Fig. 23.8(a). The vector T thus defines the unit cell of the 1D nanotube.
The length T of the translation vector T corresponds to the
first lattice point of the 2D graphene sheet through which the vector T
passes. From Fig. 23.8(a) and these definitions, we see that the translation
vector T of a general chiral nanotube as a function of n and m,
can be written as:
T = [(2m + n)a1 − (2n + m)a2]/dR (23.15)
with a length
T =
√
3Ch/dR (23.16)
where the length Ch is given by Eq. (23.13), d is the highest common
divisor of (n, m), and
dR =
d if n − m is not a multiple of 3d
3d if n − m is a multiple of 3d.
(23.17)
Thus for the (5, 5) armchair nanotube dR = 3d = 15, while for the
(9, 0) zigzag nanotube dR = d = 9. The relation between the translation
vector T and the symmetry operations on carbon nanotubes is discussed
below and in §23.6.3. As a simple example, T =
√
3aC−C for a (5, 5)
armchair nanotube and T = 3aC−C for a (9, 0) zigzag nanotube, where
aC−C is the nearest-neighbor carbon–carbon distance. We note that the
682 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
length T is greatly reduced when (n, m) have a common divisor and
when (n − m) is a multiple of 3.
Having specified the length T of the smallest translation vector
for the 1D carbon nanotube, it is useful to determine the number of
hexagons, N, per unit cell of a chiral nanotube specified by integers
(n, m) having a highest common divisor of d. From the size of the unit
cell of the 1D carbon nanotube defined by the orthogonal vectors T and
Ch, the number N is given by
N =
2(m2
+ n2
+ nm)
dR
(23.18)
where dR is given by Eq. (23.17) and we note that each hexagon contains
two carbon atoms. As an example, application of Eq. (23.18) to the
(5, 5) and (9, 0) nanotubes yields values of 10 and 18, respectively, for
N. We will see below that these unit cells of the 1D nanotube contain,
respectively, five and nine unit cells of the 2D graphene lattice, each 2D
unit cell containing two hexagons of the honeycomb lattice. This multiplicity
is used in the application of zone-folding techniques to obtain
the electronic and phonon dispersion relations for carbon nanotubes.
Referring to Fig. 23.8(a), we see that the basic space group symmetry
operation of a chiral fiber consists of a rotation by an angle ψ combined
with a translation τ, and this space group symmetry operation
is denoted by R = (ψ|τ) and corresponds to the vector R = pa1 + qa2
shown in Fig. 23.9. The physical significance of the vector R is that
the projection of R on the chiral vector Ch gives the angle ψ scaled by
Ch/2π, while the projection of R on T gives the translation vector τ of
the basic symmetry operation of the 1D space group. The integer pair
(p, q) which determines R is found using the relation
mp − nq = d (23.19)
subject to the conditions q < m/d and p < n/d. Taking the indicated
scalar product R·T in Fig. 23.9 we obtain the expressions for the length
of τ
τ = Td/N (23.20)
where d is the highest common divisor of (n, m), T is the magnitude
of the lattice vector T, and N is the number of hexagons per 1D unit
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 683
N
d R
R
T
τ
Ch
X
Figure 23.9: Relation between
the fundamental symmetry vector
R = pa1 + qa2 of the 1D
unit cell and the two vectors
that specify the carbon nanotube
(n, m): the chiral vector
Ch and translation vector T.
The projections of R on the Ch
and T axes, respectively, yield
ψ and τ (see text). After (N/d)
translations, R reaches a lattice
point B (see text). The vertical
dashed lines divide the vector
Ch into d sectors.
cell, given by Eq. (23.18). For the armchair and zigzag nanotubes,
Eq. (23.20) yields τ = T/2 and τ =
√
3T/2, respectively.
Regarding the angle of rotation, the scalar product R · Ch yields
ψ = 2π
Ω
Nd
+
λ
d
(23.21)
where λ = 0, 1, . . . , d − 1 and
Ω = [p(m + 2n) + q(n + 2m)](d/dR) (23.22)
in which the integers (p, q) determine the vector R (see Fig. 23.9). Thus
(p, q) denotes the coordinates reached when the symmetry operation
(ψ|τ) acts on an atom at (0,0), i.e., (ψ|τ)(0, 0) = (p, q).
If (ψ|τ) is a symmetry operation for the nanotube, then (ψ|τ)2
,
(ψ|τ)3
, . . . (ψ|τ)N/d
are all distinct symmetry operations where (ψ|τ)N/d
=
E is the identity operation, bringing the lattice point O to an equivalent
lattice point B , where
−→
OB = (N/d)R = (1/d)[ΩCh + T d]. (23.23)
684 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.27: Values for characterization parametersa
for selected carbon
nanotubes labeled by (n, m).
(n, m) d dR dt (˚A) L/a0 T/a0 N ψ/2π τ/a0 Ω/d
(4, 2) 2 2 4.14 5
√
21 28 5/28
√
21/14 5
(5, 5) 5 15 6.78
√
75 1 10 1/10 1/2 1
(9, 0) 9 9 7.05 9
√
3 18 −1/18
√
3/2 −1
(6, 5) 1 1 7.47
√
91
√
273 182 149/182 3/364 149
(7, 4) 1 3 7.55
√
93
√
31 62 17/62 1/
√
124 17
(8, 3) 1 1 7.72
√
97
√
291 194 71/194 3/388 71
(10, 0) 10 10 7.83 10
√
3 20 −1/20
√
3/2 −1
(6, 6) 6 18 8.14
√
108 1 12 1/12 1/2 1
(10, 5) 5 5 10.36
√
175
√
21 70 1/14 3/28 5
(20, 5) 5 15 17.95
√
525
√
7 70 3/70 1/(
√
28) 3
(30, 15) 15 15 31.09
√
1575
√
21 210 1/42 3/28 5
...
...
...
...
...
...
...
...
...
...
(n, n) n 3n
√
3na/π
√
3n 1 2n 1/2n 1/2 1
(n, 0) n n na/π n
√
3 2n −1/2n
√
3/2 −1
a
Ω is given by Eq. (23.22); T is given by Eq. (23.16) with a0 =
√
3aC−C; L is the
length of the chiral vector πdt given by Eq. (23.13); N is given by Eq. (23.18); ψ is
given by Eq. (23.21); τ is given by Eq. (23.20); Ω/d is the number of complete 2π
revolutions about the nanotube axis for one cycle.
Referring to Fig. 23.9, we see that after N/d symmetry operations (ψ|τ),
the vector (N/d)R along the zigzag direction reaches a lattice point,
which we denote by B . Correspondingly, after N/d symmetry operations
(ψ|τ)N/d
, the translation (N/d)τ yields one translation of the
lattice vector T of the nanotube and (Ω/d2
) revolutions of 2π around
the nanotube axis. Although Ω/d is an integer, Ω/d2
need not be. In
Table 23.27 we list the characteristic parameters of carbon nanotubes
specified by (n, m), including d, the highest common divisor of n and
m, and the related quantity dR given by Eq. (23.17). Also listed in
Table 23.27 are the nanotube diameter dt in units of ˚A, the translation
repeat distance T of the 1D lattice in units of the lattice constant
a0 =
√
3aC−C for a 2D graphene sheet, the number N of hexagons per
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 685
unit cell of the 1D nanotube, the rotation angle ψ expressed in units of
Ch/2π, and the translation τ of the symmetry operation R = (ψ|τ) expressed
in units of a0. The length of the chiral vector L (L ≡ Ch = πdt)
is listed in Table 23.27 in units of a0. For cases where n and m have a
common divisor d, the angle of rotation ψ is defined modulo 2π/d instead
of 2π, so that the group CN/Ω has N/d elements. To illustrate use
of Table 23.27 we refer to the (4, 2) nanotube illustrated in Fig. 23.8(a),
which has N = 28, d = dR = 2, and T = 14τ =
√
21a0 and also
Ch =
√
28a0. Thus for the (4, 2) nanotube, while (N/d)ψ = (5/2)(2π)
corresponding to 5π rotations around the nanotube axis, (N/d) translations
R reach a lattice point.
As a second example, we consider the case of the nanotube specified
by (7, 4). For this nanotube, there are no common divisors, so d = 1,
but since n−m = 3, we obtain dR = 3. Solution of Eq. (23.19) for (p, q)
gives p = 2 and q = 1, which yields the rotation angle (in units of 2π)
Nψ = Ω/d = 17(2π), where N = 62. Thus, after 62 operations (ψ|τ)N
,
the origin O is transformed into a new lattice point a distance T from O
along the T axis, after having completed 17 rotations of 2π around the
nanotube axis. For the armchair nanotube (5, 5), the highest common
divisor is 5, and since n − m = 0, we have dR = 3 × 5 = 15, yielding
N = 10 and Nψ = 2π. Regarding translations, use of Eq. (23.20) yields
τ = a0/2 = T/2. Finally, we give the example of the smallest zigzag
nanotube (0, 9), for which d = dR = 9 and N = 18. Using p = 1 and
q = 0 and N = 18, we obtain Nψ = 2π and τ =
√
3a0/2 = T/2.
23.6.3 Symmetry for Symmorphic Carbon Nan-
otubes
In this section we consider the symmetry properties of the highly symmetric
armchair and zigzag carbon nanotubes, which can be described
by symmorphic groups. We then summarize the symmetry operations
for the general chiral nanotube in §23.6.4 which is described by a nonsymmorphic
space group.
For symmorphic groups, the translations and rotations are decoupled
from each other, and we can treat the rotations simply by point
group operations. Since symmorphic groups generally have higher sym-
686 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
Table 23.28: Character table for point group D(2j+1).
R E 2C1
φj
a
2C2
φj
. . . 2Cj
φj
(2j + 1)C2
A1 1 1 1 . . . 1 1
A2 1 1 1 . . . 1 −1
E1 2 2 cos φj 2 cos 2φj . . . 2 cos jφj 0
E2 2 2 cos 2φj 2 cos 4φj . . . 2 cos 2jφj 0
...
...
...
...
...
...
...
Ej 2 2 cos jφj 2 cos 2jφj . . . 2 cos j2
φj 0
a
Where φj = 2π/(2j + 1).
metry than the nonsymmorphic groups, it might be thought that group
theory plays a greater role in specifying the dispersion relations for electrons
and phonons for symmorphic space groups and in discussing their
selection rules. It is shown in §23.6.4 that group theoretical considerations
are also very important for the nonsymmorphic groups, leading
to important and simple classifications of their dispersion relations.
In discussing the symmetry of carbon nanotubes, it is assumed that
the nanotube length is much larger than its diameter, so that the nanotube
caps can be neglected when discussing the electronic and lattice
properties of the nanotubes. Hence, the structure of the infinitely long
armchair nanotube (n = m) or zigzag nanotube (m = 0) is described
by the symmetry groups Dnh or Dnd for even or odd n, respectively,
since inversion is an element of Dnd only for odd n and is an element
of Dnh only for even n. Character tables for groups D5h and D5d are
given in §17.7 under Tables 3.38 and 3.37, respectively. We note that
group D5d has inversion symmetry and is a subgroup of group Ih, so
that compatibility relations can be specified between the lower symmetry
group D5d and Ih, as given in Table 23.15. In contrast, D5h is
not a subgroup of Ih, although D5 is a subgroup of I. Character
tables for D6h and D6d are readily available in standard group theory
texts. For larger diameter nanotubes, appropriate character tables can
be constructed from the generalized character tables for the Dn group
given in Table 23.28 (for odd n = 2j + 1) and in Table 23.29 (for even
n = 2j), and the basis functions are listed in Table 23.30. These tables
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 687
Table 23.29: Character Table for Group D(2j)
R E C2 2C1
φj
a
2C2
φj
. . . 2Cj−1
φj
(2j)C2 (2j)C2
A1 1 1 1 1 . . . 1 1 1
A2 1 1 1 1 . . . 1 −1 −1
B1 1 −1 1 1 . . . 1 1 −1
B2 1 −1 1 1 . . . 1 −1 1
E1 2 −2 2 cos φj 2 cos 2φj . . . 2 cos(j − 1)φj 0 0
E2 2 2 2 cos 2φj 2 cos 4φj . . . 2 cos 2(j − 1)φj 0 0
...
...
...
...
...
...
...
...
...
Ej−1 2 (−1)j−1
2 2 cos(j − 1)φj 2 cos 2(j − 1)φj . . . 2 cos(j − 1)2
φj 0 0
a
Where φj = 2π/(2j).
Table 23.30: Basis Functions for Groups D(2j) and D(2j+1)
Basis functions D(2j) D(2j+1) CN/Ω
(x2
+ y2
, z2
) A1 A1 A
z A2 A1 A
Rz A2 A2 A
(xz, yz)
(x, y)
(Rx, Ry)
E1 E1 E1
(x2
− y2
, xy) E2 E2 E2
...
...
...
are adapted from the familiar character table for the semi-infinite group
D∞h.
The character table for group Dnd for odd integers n = 2j + 1 is
constructed from Table 23.28 for group Dn [or group (n2) in the international
notation] by taking the direct product Dn ⊗ i where i is the
two element inversion group containing E and i. In addition to the
identity class E, the classes of Dn in Table 23.28 constitute the nth
roots of unity where ±φj rotations belong to the same class, and in addition
there is a class (2j +1)C2 of n twofold axes at right angles to the
main symmetry axis Cφj
. Thus, Table 23.28 yields the character tables
688 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
for Dnd, or D5d, D7d, . . . for symmorphic nanotubes with odd numbers
of unit cells around the circumference [(5, 5), (7, 7), .. armchair nanotubes,
and (9, 0), (11, 0), ... zigzag nanotubes]. Likewise, the character
table for Dnh, or D6h, D8h, . . . for even n is found from Table 23.29
by taking the direct product Dn ⊗ i = Dnh. Table 23.29 shows two
additional classes for group D2j because rotation by π about the main
symmetry axis is in a class by itself. Also the 2j twofold axes nC2 form
a class and are in a plane normal to the main symmetry axis Cφj
, while
the nC2 dihedral axes, which are bisectors of the nC2 axes, also form
a class for group Dn when n is an even integer. Correspondingly, there
are two additional one-dimensional representations (B1 and B2) in D2j,
since the number of irreducible representations equals the number of
classes. Table 23.30 lists the irreducible representations and basis functions
for the various 1D nanotube groups and is helpful for indicating
the symmetries that are infrared active (transform as the vector x, y, z)
and Raman active (transform as the symmetric quadratic forms).
23.6.4 Symmetry for Nonsymmorphic Carbon Nan-
otubes
The symmetry groups for carbon nanotubes can be either symmorphic
(as for the special case of the armchair and zigzag nanotubes) or nonsymmorphic
for the general case of chiral nanotubes. For chiral nanotubes
the chiral angle in Eq. (23.14) is in the range 0 < θ < 30◦
and the
space group operations (ψ|τ) given by Eqs. (23.20) and (23.21) involve
both rotations and translations, as discussed in §23.6.2. Figure 23.9
shows the symmetry vector R which determines the space group operation
(ψ|τ) for any carbon nanotube specified by (n, m). From the
symmetry operation R = (ψ|τ) for nanotube (n, m), the symmetry
group of the chiral nanotube can be determined. If the nanotube is
considered as an infinite molecule, then the set of all operations Rj
, for
any integer j, also constitutes symmetry operators of the group of the
nanotube. Thus from a symmetry standpoint, a carbon nanotube is
a one-dimensional crystal with a translation vector T along the cylinder
axis and a small number of carbon hexagons associated with the
circumferential direction.
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 689
Table 23.31: The character table for the group CN/Ω for chiral nanotubes,
where N and Ω have no common divisor, corresponding to
(n, m) having no common divisor.
CN/Ω E C1
C2
· · · C · · · CN−1
A 1 1 1 · · · 1 · · · 1
B 1 –1 1 · · · (−1) · · · –1
E1
1
1 ∗
2
∗2 · · · ∗ · · ·
N−1
∗(N−1)
E2
1
1
2
∗2
4
∗4 · · ·
2
∗2 · · ·
2(N−1)
∗2(N−1)
...
...
...
...
...
...
...
...
EN
2 −1
1
1
N
2 −1
∗ N
2 −1
2( N
2 −1)
∗2( N
2 −1)
· · ·
( N
2 −1)
∗ ( N
2 −1)
· · ·
(N−1)( N
2 −1)
∗(N−1)( N
2 −1)
a
The complex number is e2πiΩ/N
.
The symmetry groups for the chiral nanotubes are Abelian groups.
All Abelian groups have a phase factor , such that all h symmetry
elements of the group commute, and are obtained from any symmetry
element by multiplication of by itself an appropriate number of times,
such that h
= E, the identity element. To specify the phase factors for
the Abelian group, we introduce the quantity Ω of Eq. (23.22), where
Ω/d is interpreted as the number of 2π rotations which occur after N/d
rotations of ψ. The phase factor for the Abelian group for a carbon
nanotube then becomes = exp(2πiΩ/N) for the case where (n, m)
have no common divisors (i.e., d = 1). If Ω = 1 and d = 1, then
the symmetry vector R in Fig. 23.9 reaches a lattice point after a 2π
rotation. Many of the actual nanotubes with d = 1 have large values
for Ω; for example, for the (6, 5) nanotube, Ω = 149, while for the (7, 4)
nanotube Ω = 17, so that many 2π rotations around the nanotube axis
are needed to reach a lattice point of the 1D lattice.
The character table for the Abelian group of a carbon nanotube
is given in Table 23.31 for the case where (n, m) have no common divisors
and is labeled by CN/Ω. The number of group elements is N,
690 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
all symmetry elements commute with each other, and each symmetry
operation is in a class by itself. The irreducible representation A in
Table 23.31 corresponds to a 2π rotation, while representation B corresponds
to a rotation by π. Except for the irreducible representations
A and B, all other irreducible representations are doubly degenerate.
The E representations correspond to two levels which stick together
by time reversal symmetry, for which the corresponding eigenvectors
are related to one another by complex conjugation. Since N can be
quite large, there can be a large number of symmetry operations in the
group CN/Ω, all of which can be represented in terms of a phase factor
= exp(iψ) = exp(2πiΩ/N) and the irreducible representations are related
to the Nth roots of unity. The number of classes and irreducible
representations of group CN/Ω is thus equal to N, counting each of the
partners of the E representations as distinct. Various basis functions
for group CN/Ω that are useful for determining the infrared and Raman
activity of the chiral nanotubes are listed in Table 23.30. We note that
the irreducible representations A and E1 are infrared active and A, E1,
and E2 are Raman active.
For nanotubes where (n, m) have a common divisor d, then after N
translations τ, a length Td is produced, and Ω unit cells of area (ChT)
are generated. For d = 1 the character table for the Abelian group in
Table 23.31 contains 1/d as many elements, so that the order of the
Abelian group becomes N/d. This case is discussed further below.
The space group for a chiral nanotube specified by (n, m) is given
by the direct product of two Abelian groups
C = Cd ⊗ CNd/Ω (23.24)
where d is the highest common divisor of Ω/d and d, while d is the
highest common divisor of (n, m). The symmetry elements in group
Cd which is of order d include
Cd = {E, Cd , C2
d , ..., Cd −1
d }, (23.25)
and, correspondingly, the symmetry elements in group CNd/Ω include
CNd/Ω = {E, CNd/Ω, C2
Nd/Ω, ..., C
(N/d )−1
Nd/Ω }, (23.26)
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 691
and group CNd/Ω is of order N/d . The irreducible representations of
the groups Cd and CNd/Ω in Eq. (23.24) are given in Table 23.31 and
are appropriate roots of unity. For the two-dimensional En irreducible
representations, the characters of the symmetry operations Cd and
CNd/Ω are given by
χEn (Cd ) =
ei2πn/d
e−i2πn/d (23.27)
where 1 ≤ n = d and
χEn (CNd/Ω) =
ei2πnΩ/Nd
e−i2πnΩ/Nd
.
(23.28)
For the chiral nanotubes a vector transforms according to the basis
functions for the A and E1 irreducible representations, whereas
quadratic terms in the coordinates form basis functions for the A, E1,
and E2 irreducible representations, as shown in the Table 23.30.
Referring to Fig. 23.8(a) for the (4, 2) nanotube, we have ψ =
2π(5/28) and Ω/d = 5, so that N rotations produce a total rotation
of 2π(5). Since d = 2, N translations produce a distance 2T, so that
NR is the length of the diagonal of a rectangle that contains Ω = 10
times the area of the 1D nanotube unit cell. The first lattice point is,
however, reached at (N/d)R = [(5/2)Ch, T] of the 1D nanotube lattice.
The number of elements in the Abelian group CNd/Ω is N/d, which is
14 in this case. The value of d which is the highest common divisor of d
and Ω/d is d = 1, so that Cd is the identity group containing only one
symmetry element. This example shows that for those (n, m) having a
common divisor d, a lattice point is reached after (N/d)R translations,
so that Ω/d2
is not necessarily a multiple of 2π.
We now give some examples to show that if either d > 1 or n−m =
3r (so that dR = 3d), the 1D nanotube unit cell is reduced in size,
and the number of R vectors to reach a lattice point is reduced. For
example, nanotube (7, 5) has a diameter of 8.18 ˚A, N = 218 symmetry
operations, and 91 translations R are needed to reach a lattice point.
Since the integers (7, 5) have no common divisor, and since n−m = 3r,
the unit cell is large. If we now consider a nanotube with (n, m) = (8, 5)
with a diameter of 8.89 ˚A, nearly 10% larger than the diameter for the
692 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
(7, 5) nanotube, we note that since n−m is a multiple of 3, then N = 86
for the (8, 5) nanotube, where N is much smaller than for the (7, 5)
nanotube. Likewise, for the (8, 5) nanotube, the length T = 16.1 ˚A is
much shorter than T = 44.5 ˚A for the (7, 5) nanotube, and a lattice
point is reached after 53 translations. An even smaller unit cell is
obtained for the nanotube corresponding to (n, m) = (10, 5), despite
the larger nanotube diameter of 10.36 ˚A. For the (10, 5) nanotube, N
is only 70, and a lattice point is reached after only five translations R.
Although the number of symmetry operations of the nonsymmorphic
groups tends to increase with nanotube diameter, those nanotubes for
which either n − m = 3r, where r is an integer, or (n, m) contains a
common divisor d, the size of the unit cell and therefore the number of
symmetry operations is reduced by factors of 3 and d, respectively.
The symmorphic groups corresponding to the armchair and zigzag
nanotubes have relatively small unit cells with T = a0 and T =
√
3a0,
respectively, where a0 = 2.46 ˚A=
√
3aC−C. The basic rotation angle ψ
for both the (n, n) armchair nanotube and the (n, 0) zigzag nanotube
is ψ = 2π/n.
23.6.5 Reciprocal Lattice Vectors
To express the dispersion relations for electrons and phonons in carbon
nanotubes, it is necessary to specify the basis vectors in reciprocal space
Ki which relate to those in real space Rj by
Rj · Ki = 2πδij. (23.29)
The lattice vectors and the unit cells in real space are discussed in
§23.6.2. In Cartesian coordinates we can write the two real space lattice
vectors as
Ch = (a0/2dR)
√
3(n + m), (n − m) ,
T = (3a0/2dR) − (n − m)/
√
3, (n + m)
(23.30)
where dR is defined by Eq. (23.17) and the length a0 is the lattice
constant of the 2D graphene unit cell, a0 = aC−C
√
3. From Eq. (23.30),
23.6. SYMMETRY PROPERTIES OF CARBON NANOTUBES 693
ky
kx
1b’
b22
b’
a2
2a
b2
’
b2
a1 b1
kx
ky
(b)
y
x
(a)
11a a
Γ
K
Μ
’
1
Μ K
x
y
a1’
Γ
2a’
2a’
1b
’b
Figure 23.10: Real space unit cell
and Brillouin zone for (a) armchair
and (b) zigzag nanotubes (dotted
lines). Those for a 2D graphene
sheet (dashed lines) are shown for
comparison.
it is easy to see that the lengths of Ch and T are in agreement with
Eqs. (23.13) and (23.16). Then using Eqs. (23.29) and (23.30), we
can write the corresponding reciprocal lattice vectors K1 and K2 in
Cartesian coordinates as
K1 = (2π/Na0)
√
3(n + m), (n − m) ,
K2 = (2π/Na0) − (n − m)/
√
3, (n + m) .
(23.31)
For the case of the armchair and zigzag nanotubes, it is convenient to
choose a real space rectangular unit cell in accordance with Fig. 23.10.
The area of each real space unit cell for both the armchair and zigzag
nanotubes contains two hexagons or four carbon atoms. It should be
noted that the real space unit cell defined by the vectors T and Ch is
n times larger than the real space unit cells shown in Fig. 23.10. Also
shown in this figure are the real space unit cells for a 2D graphene layer.
It should be noted that the real space unit cell for the zigzag nanotubes
follows from the definition given for the unit cells for chiral nanotubes
694 CHAPTER 23. FULLERENES AND CARBON NANOTUBES
[Fig. 23.8(a)], but the unit cell for the armchair nanotube is specially
selected for convenience.
Having specified the real space unit cells in Fig. 23.10, the corresponding
unit cells in reciprocal space (or Brillouin zones) are determined
by Eqs. (23.29) and (23.31) and are shown in Fig. 23.10 in
comparison to the reciprocal space unit cell for a 2D graphene sheet.
We note that for both the armchair and zigzag nanotubes, the 1D reciprocal
space unit cells shown in Fig. 23.10 are half as large as those
for the 2D graphene sheet. On the graphene sheet, the real space lattice
vectors Ch and T form a rectangular unit cell for the zigzag or
armchair nanotubes. This unit cell has an area N times larger than the
area of the corresponding primitive cell in the graphene sheet, where N
is given by Eq. (23.18), spans the circumference of the cylinder of the
nanotube.
The one-dimensional Brillouin zone of the chiral nanotube is a segment
along the vector K2 of Eq. (23.31). The extended Brillouin zone
for the nanotube is a collection of N wave vector segments of length
|K2|, each separated from the next segment by the vector K1. Thus,
by zone folding the N wave vector segments of the 2D dispersion relations
of the graphene layer back to the first Brillouin zone of the 1D
nanotube, the 1D dispersion relations for the N electron energy bands
(or phonon branches) of the nanotube are obtained. For the special
cases of the armchair and zigzag nanotubes, the real space unit cells in
Fig. 23.10 correspond to four carbon atoms and two hexagons, while
the reciprocal space unit cells also contain two wave vectors. Therefore
each dispersion relation for the (n, n) armchair nanotubes and the
(n, 0) zigzag nanotubes is zone folded n = N/2 times in the extended
Brillouin zone to match the 1D Brillouin zone of the nanotube.
23.7 Suggested Problems
1. A regular dodecahedron has 12 regular pentagonal faces, 20 vertices
and 30 edges.
(a) What are the symmetry classes for the regular dodecahe-
dron?
23.7. SUGGESTED PROBLEMS 695
(b) Is this symmetry group isomorphic to that for the regular
icosahedron? Explain.
(c) Show (by finding the characters of the rotation group) that
the d-level for a transition metal impurity in a quasicrystal
with Ih point symmetry is not split by the icosahedral crystal
field.
2. Suppose that we stretch the B12H12 molecule pictured in Fig. 3.3
along one of the 5-fold axes.
(a) What are the resulting symmetry elements on the stretched
molecule?
(b) What is the appropriate point group?
(c) Consider the Gu and Hg irreducible representations of group
Ih as reducible representation of the lower symmetry group
of a 4-fold level transforming as Gu and a 5-fold level transforming
as Hg in the Ih group. Assuming the basis functions
given in the character table, give the corresponding basis
functions for the levels in the stretched molecule.