C5320 Theoretical Concepts of NMR
Luk ´aˇs ˇZ´ıdek
December 11, 2018
ii
Contents
I Classical Introduction 1
1 Classical electromagnetism 3
1.1 Electric field, electric charge, electric dipole . . . . . . . . . . . . . . . . 3
1.2 Magnetic field and magnetic dipole . . . . . . . . . . . . . . . . . . . . . 4
1.3 Origin of the electric field . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Origin of the magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Current loop as a magnetic dipole . . . . . . . . . . . . . . . . . . . . . . 7
1.6 Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.7 Electrodynamics and magnetodynamics . . . . . . . . . . . . . . . . . . . 9
2 Diffusion 11
2.1 Translational diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Isotropic rotational diffusion . . . . . . . . . . . . . . . . . . . . . . . . . 12
3 Nuclear magnetic resonance 15
3.1 Nuclear magnetic moments in chemical substances . . . . . . . . . . . . . 15
3.1.1 Symmetric distribution . . . . . . . . . . . . . . . . . . . . . . . . 15
3.1.2 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.1.3 Coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.2 Chemical shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.2.1 Offset effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.2.2 Evolution of magnetization in B0 . . . . . . . . . . . . . . . . . . 25
4 Relaxation 27
4.1 Relaxation due to chemical shift anisotropy . . . . . . . . . . . . . . . . . 27
4.2 Loss of coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2.1 Time correlation function . . . . . . . . . . . . . . . . . . . . . . 32
4.3 Return to equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.4 Internal motions, structural changes . . . . . . . . . . . . . . . . . . . . . 40
4.5 Bloch equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
iii
iv CONTENTS
5 Signal acquisition and processing 43
5.1 NMR experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
5.1.1 Setting up the experiment . . . . . . . . . . . . . . . . . . . . . . 44
5.1.2 Quadrature detection . . . . . . . . . . . . . . . . . . . . . . . . . 45
5.1.3 Analog-digital conversion . . . . . . . . . . . . . . . . . . . . . . . 46
5.1.4 Signal averaging and signal-to-noise ratio . . . . . . . . . . . . . . 46
5.2 Fourier transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
5.2.1 Fourier transformation of an ideal NMR signal . . . . . . . . . . . 49
5.2.2 Properties of continuous Fourier transformation . . . . . . . . . . 53
5.2.3 Consequence of finite signal acquisition . . . . . . . . . . . . . . . 55
5.2.4 Discrete Fourier transformation . . . . . . . . . . . . . . . . . . . 57
5.2.5 Consequence of discrete signal acquisition . . . . . . . . . . . . . 59
5.3 Phase correction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
5.4 Zero filling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
5.5 Apodization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
II Quantum description 67
6 Spin 69
6.1 Wave function and state of the system . . . . . . . . . . . . . . . . . . . 69
6.2 Superposition and localization in space . . . . . . . . . . . . . . . . . . . 70
6.3 Operators and possible results of measurement . . . . . . . . . . . . . . . 71
6.4 Expected result of measurement . . . . . . . . . . . . . . . . . . . . . . . 72
6.5 Operators of position and momentum . . . . . . . . . . . . . . . . . . . . 74
6.5.1 Operator of momentum . . . . . . . . . . . . . . . . . . . . . . . 74
6.5.2 Operator of position . . . . . . . . . . . . . . . . . . . . . . . . . 74
6.5.3 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
6.6 Operator of energy and equation of motion . . . . . . . . . . . . . . . . . 76
6.7 Operator of angular momentum . . . . . . . . . . . . . . . . . . . . . . . 80
6.8 Relativistic quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . 83
6.8.1 Finding the matrices . . . . . . . . . . . . . . . . . . . . . . . . . 86
6.8.2 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
6.9 Hamiltonian of spin magnetic moment . . . . . . . . . . . . . . . . . . . 91
6.9.1 Operators of spin angular momentum . . . . . . . . . . . . . . . . 94
6.9.2 Eigenfunctions and eigenvalues of ˆIz . . . . . . . . . . . . . . . . 94
6.9.3 Eigenfunctions of ˆIx and ˆIy . . . . . . . . . . . . . . . . . . . . . 97
6.10 Real particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
6.11 Stationary states and energy level diagram . . . . . . . . . . . . . . . . . 99
6.12 Oscillatory states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
CONTENTS v
7 Mixed state of non-interacting spins 103
7.1 Mixed state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
7.2 Coherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
7.3 Basis sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
7.4 Liouville - von Neumann equation . . . . . . . . . . . . . . . . . . . . . . 106
7.5 Rotation in operator space . . . . . . . . . . . . . . . . . . . . . . . . . . 108
7.6 General strategy of analyzing NMR experiments . . . . . . . . . . . . . . 109
8 Chemical shift, NMR experiment 111
8.1 Operator of the observed quantity . . . . . . . . . . . . . . . . . . . . . . 111
8.2 Static field B0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
8.3 Radio-frequency field B1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
8.4 Hamiltonian of chemical shift . . . . . . . . . . . . . . . . . . . . . . . . 112
8.5 Secular approximation and averaging . . . . . . . . . . . . . . . . . . . . 113
8.6 Thermal equilibrium as the initial state . . . . . . . . . . . . . . . . . . . 114
8.7 Relaxation due to chemical shift anisotropy . . . . . . . . . . . . . . . . . 115
8.8 The one-pulse experiment . . . . . . . . . . . . . . . . . . . . . . . . . . 119
8.8.1 Excitation by radio wave pulses . . . . . . . . . . . . . . . . . . . 119
8.8.2 Evolution of chemical shift after excitation . . . . . . . . . . . . . 120
8.8.3 Spectrum and signal-to-noise ratio . . . . . . . . . . . . . . . . . 121
8.8.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
9 Product operators, dipolar coupling 125
9.1 Dipolar coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
9.2 Product operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
9.3 Liouville - von Neumann equation . . . . . . . . . . . . . . . . . . . . . . 128
9.4 Operator of the observed quantity . . . . . . . . . . . . . . . . . . . . . . 129
9.5 Hamiltonian of dipolar coupling . . . . . . . . . . . . . . . . . . . . . . . 129
9.6 Secular approximation and averaging . . . . . . . . . . . . . . . . . . . . 130
9.7 Dipole-dipole relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
10 2D spectroscopy, NOESY 137
10.1 Two-dimensional spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . 137
10.1.1 Thermal equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . 137
10.1.2 Evolution in the absence of dipolar coupling . . . . . . . . . . . . 139
10.2 NOESY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
11 J-coupling, spin echoes 145
11.1 Through-bond coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
11.2 Secular approximation and averaging . . . . . . . . . . . . . . . . . . . . 147
11.3 Relaxation due to the J-coupling . . . . . . . . . . . . . . . . . . . . . . 147
11.4 Spectroscopy in the presence of J-coupling . . . . . . . . . . . . . . . . . 148
vi CONTENTS
11.4.1 Thermal equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . 148
11.4.2 Evolution in the presence of J-coupling . . . . . . . . . . . . . . . 149
11.5 Spin echoes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
11.5.1 Free evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
11.5.2 Refocusing echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
11.5.3 Decoupling echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
11.5.4 Coupling echo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
12 INEPT, HSQC, APT 157
12.1 INEPT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
12.2 HSQC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
12.2.1 Decoupling trains . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
12.2.2 Benefits of HSQC . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
12.3 APT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
13 COSY 163
13.1 Homonucler correlation: COSY . . . . . . . . . . . . . . . . . . . . . . . 163
14 Strong coupling 167
14.1 Evolution with the strong coupling . . . . . . . . . . . . . . . . . . . . . 169
15 TOCSY 175
Part I
Classical Introduction
1
Chapter 1
Classical electromagnetism
Literature: Discussed in1
L2 and B11, with mathematical background in B4.
1.1 Electric field, electric charge, electric dipole
Objects having a property known as the electric charge (Q) experience forces (F) described
as the electric field. Since the force depends on both charge and field, a quantity
E = F/Q known as electric intensity has been introduced:
F = QE. (1.1)
Field lines are often used to visualize the fields: direction of the line shows the
direction of E, density of the lines describes the size of E (|E|). A homogeneous static
electric field is described by straight parallel field lines.
Two point electric charges of the same size and opposite sign (+Q and −Q) separated
by a distance r constitute an electric dipole. Electric dipoles in a homogeneous static
electric field experience a moment of force, or torque τ:
τ = r × F = r × QE = Qr × E = µe × E, (1.2)
where µe is the electric dipole moment.
τ = µe × E, (1.3)
is another possible definition of E.
1
The references consist of a letter specifying the textbook and a number specifying the section. The
letters refer to the following books: B, Brown: Essential mathematics for NMR and MRI spectroscopists,
Royal Society of Chemistry 2017; C, Cavanagh et al., Protein NMR spectroscopy, 2nd. ed.,
Academic Press 2006; K, Keeler, Understanding NMR spectroscopy, 2nd. ed., Wiley 2010; L Levitt:
Spin dynamics, 2nd. ed., Wiley 2008.
3
4 CHAPTER 1. CLASSICAL ELECTROMAGNETISM
Potential energy2 of the electric dipole can be calculated easily as a sum of potential energies of the individual
charges. Potential energy is defined as the work done by moving the charge from a position (1) to a reference position (0).
If we choose a coordinate system so that the z-axis is parallel with E, the dipole is in the yz-plane, and the origin is in the
middle of the distance r between the charges, then the force acts only in the z-direction (Fz = |F| = Q|E| for the positive
charge and Fz = −|F| = −Q|E| for the negative charge). Therefore, it is sufficient to follow only how the z-coordinates
of the charges change because changes of other coordinates do not change the energy. In the chosen coordinate system,
z-coordinates of the charges are always opposite, e.g., the reference value z−,0 of the negative charge is equal to minus the
reference z coordinate of the positive charge (z+,0). The natural choice of the reference position is that the z coordinates
are the same for both charges, which requires z+,0 = z−,0 = 0. Changing the z coordinate of the positive charge from
z+,0 = 0 to z+,1 ≡ z results in a work
Q|E|(z+,0 − z+,1) = −Q|E|z. (1.4)
Changing the z coordinate of the positive charge from z−,0 = 0 to z−,1 = −z+,1 results in a work
− Q|E|z−,0 − (z−,1) = −Q|E|z. (1.5)
Adding the works
E = −2Q|E|z = −2Q|E|
r
2
cos θ = −µe · E, (1.6)
where θ is the angle between E and µe.
Equivalently, the potential energy can be defined as the work done by the torque τ on µe when rotating it from
the reference orientation to the orientation described by the angle θ (between E and µe). The reference angle for
z+,0 = z−,0 = 0 is π/2, therefore,
E =
θˆ
π
2
|τ|dθ =
θˆ
π
2
|µe||E| sin θ dθ = −|µe||E| cos θ = −µe · E. (1.7)
Potential energy of an electric dipole is
E = −µe · E. (1.8)
1.2 Magnetic field and magnetic dipole
There is no ”magnetic charge”, but magnetic moments exist:
τ = µm × B, (1.9)
where µm is the magnetic dipole moment (because this course is about magnetic
resonance, we will write simple µ). This is the definition of the magnetic induction B as
a quantity describing magnetic field. As a consequence, potential energy of a magnetic
dipole can be derived as described by Eq. 1.7 for the electric dipole.
Potential energy of a magnetic moment µ is
E = −µ · B. (1.10)
2
Do not get confused: E (scalar) is the energy and E (vector) is electric intensity.
1.3. ORIGIN OF THE ELECTRIC FIELD 5
1.3 Origin of the electric field
The source of the electric field is the electric charge. The charge (i) feels (a surrounding)
field and (ii) makes (its own) field. Charge at rest is a source of a static electric field.
Parallel plates with homogeneous distribution of charges (a capacitor) are a source of a
homogeneous static electric field.
Force between charges is described by the Coulomb’s law. The force between two
charges is given by
F =
1
4π 0
Q1Q2
r2
r
|r|
, (1.11)
where 0 = 8.854187817 × 10−12
F m−1
is the vacuum electric permittivity.
Consequently, the electric intensity generated by a point charge is
E =
1
4π 0
Q
r2
r
|r|
. (1.12)
The electric intensity generated by a charge density ρ is
E =
1
4π 0
ˆ
V
dV
ρ
r2
r
|r|
(1.13)
Coulomb’s law implies that electric fields lines of a resting charge
1. are going out of the charge (diverge), i.e., the static electric field has a source (the
charge)
2. are not curved (do not have curl or rotation), i.e., the static electric field does not
circulate
This can been written mathematically in the form of Maxwell equations3
:
div E =
ρ
0
, (1.14)
rot E = 0. (1.15)
where div E is a scalar equal to ∂Ex
∂x
+ ∂Ey
∂y
+ ∂Ez
∂z
and rot E is a vector with the x, y, z
components equal to ∂Ez
∂y
− ∂Ey
∂z
, ∂Ex
∂z
− ∂Ez
∂x
, ∂Ey
∂x
− ∂Ex
∂y
, respectively. These expressions
can be written in a much more compact form, if we introduce a vector operator =
∂
∂x
, ∂
∂y
, ∂
∂z
. Using this formalism, the Maxwell equation have the form
3
The first equation is often written using electric induction D = 0E as div D = ρ.
6 CHAPTER 1. CLASSICAL ELECTROMAGNETISM
· E =
ρ
0
, (1.16)
× E = 0. (1.17)
1.4 Origin of the magnetic field
Electric charge at rest does not generate a magnetic field, but a moving charge does.
The magnetic force is a relativistic effect (consequence of the contraction of distances
in the direction of the motion, described by Lorentz transformation).4
Magnetic field of
a moving point charge is moving with the charge. Constant electric current generates
a stationary magnetic field. Constant electric current in an ideal solenoid generates a
homogeneous stationary magnetic field inside the solenoid.
Magnetic induction generated by a current density j (Biot-Savart law):
B =
1
4π 0c2
ˆ
V
dV
j
r2
×
r
|r|
=
µ0
4π
ˆ
V
dV
j
r2
×
r
|r|
(1.18)
Biot-Savart law implies that magnetic field lines of a constant current in a straight
wire
1. do not diverge, i.e., the static magnetic field does not have a source
2. make closed loops around the wire (have curl or rotation), i.e., the magnetic field
circulates around the wire
This can been written mathematically in the form of Maxwell equations5
:
· B = 0, (1.19)
× B = µ0j. (1.20)
4
A charge close to a very long straight wire which is uniformly charged experiences an electrical force
F⊥ in the direction perpendicular to the wire. If the charges in the wire move with a velocity v0 and
the charge close to the wire moves along the wire with a velocity v1, the perpendicular force changes to
F⊥(1− v0v1
c2 ), were c is the speed of light in vacuum. The modifying factor is clearly relativistic (B11.5).
5
The second equation is often written using magnetic intensity H = B/µ0 as × H = j.
1.5. CURRENT LOOP AS A MAGNETIC DIPOLE 7
1.5 Current loop as a magnetic dipole
Now we derive what is the magnetic dipole of a circular loop with an electric current.
The magnetic moment is defined by the torque τ it experiences in a magnetic field B (Eq. 1.9):
τ = µ × B, (1.21)
Therefore, we can calculate the magnetic moment of a current loop if we place it in a magnetic field B. Let us first
define the geometry of our setup. Let the axis z is the normal of the loop and let B is in the xz plane (⇒ By = 0). The
vector product in Eq. 1.9 then simplifies to
τx = µyBz, (1.22)
τy = µzBx − µxBz, (1.23)
τz = −µyBx. (1.24)
As the second step, we describe the electric current in the loop. The electric current is a motion of the electric
charge. We describe the current as a charge Q homogeneously distributed in a ring (loop) of a mass m which rotates
with a circumferential speed v. Then, each element of the loop of a infinitesimally small length dl = rdϕ contains the
same fraction of the charge dQ, moving with the velocity v. The direction of the vector v is tangent to the loop and the
amount of the charge per the length element is Q/2πr. The motion of the charge element dQ can be described, as any
circular motion, by the angular momentum
L = r × p = m(r × v), (1.25)
where r is the vector defining the position of the charge element dQ. In our geometry, r is radial and therefore always
perpendicular to v. Since both r and v are in the xy plane, L must have the same direction as the normal of the plane:
Lx = 0, (1.26)
Ly = 0, (1.27)
Lz = mrv. (1.28)
As the third step, we examine forces acting on dQ. The force acting on a moving charge in a magnetic field (the
Lorentz force) is equal to
F = Q(E + v × B), (1.29)
but we are now only interested in the magnetic component F = Q(v×B). The force acting on a single charge element
dQ is
dF = dQ(v × B) =
Q
2πr
dl(v × B) =
Q
2π
(v × B)dϕ. (1.30)
The key step in our derivation is the definition of the torque
τ = r × F = Qr × (v × B), (1.31)
which connects our analysis of the circular motion with the definition of µ (Eq. 1.9). The torque acting on a charge
element is
dτ = r × dF =
Q
2π
r × (v × B)dϕ =
Q
2π


v(r · B) − B (r · v)
=0


 dϕ =
Q
2π
(r · B)vdϕ. (1.32)
where a useful vector identity a × (b × c) = (a · c)b − (a · b)c helped us to simplify the equation because r ⊥ v).
Eq. 1.32 tells us that the torque has the same direction as the velocity v (v is the only vector on the right-hand side
because r · B is a scalar). In our coordinate frame, vx = −v sin ϕ, vy = v cos ϕ, vz = 0, and r · B = Bxr cos ϕ. Therefore,
we can calculate the components of the overall torque τ as
8 CHAPTER 1. CLASSICAL ELECTROMAGNETISM
τx = −
Qrv
2π
Bx
2πˆ
0
sin ϕ cos ϕdϕ = −
Qrv
4π
Bx
2πˆ
0
sin(2ϕ)dϕ = 0, (1.33)
τy =
Qrv
2π
Bx
2πˆ
0
cos2
ϕdϕ =
Qrv
4π
Bx
2πˆ
0
(1 + cos(2ϕ))dϕ =
Qrv
2
Bx, (1.34)
τz = 0. (1.35)
Comparison with Eqs. 1.22–1.24 immediately shows that
µx = 0, (1.36)
µy = 0, (1.37)
µz =
Qrv
2
(1.38)
and comparison with Eqs. 1.26–1.28 reveals that the magnetic dipole moment of the current loop is closely related
to the angular momentum L = r × mv:
µ =
Q
2m
L. (1.39)
We have derived that the magnetic moment of a current loop is proportional the
angular momentum of the circulating charge. The classical theory does not explain why
particles like electrons or nuclei have their own magnetic moments, even when they do
not move in circles (because the classical theory does not explain why such particles
have their own angular momenta). However, if we take the nuclear magnetic moment
as a fact (or if we obtain it using a better theory), the classical results are useful. It can
be shown that the magnetic moment is always proportional to the angular momentum6
,
but the proportionality constant is not always Q/2m; it is difficult to obtain for nuclei.
Magnetic dipolar moment µ is proportional to the angular momentum L
µ = γL, (1.40)
where γ is known as the magnetogyric ratio.
1.6 Precession
Magnetic dipole in a static homogeneous magnetic field does not adopt the energetically
most favored orientation (with the same direction of µ as B), but rotates around B
without changing the angle between µ and B. This motion on a cone is known as
precession.
6
A consequence of the rotational symmetry of space described mathematically by the Wigner-Eckart
theorem.
1.7. ELECTRODYNAMICS AND MAGNETODYNAMICS 9
This is not a result of quantum mechanics, but a classical effect – riding a bicycle is
based on the same effect.7
We can use another example of an object with an angular
momentum in the gravitational field, the spinning top, to derive the frequency of the
precession.8
The position of the center of the mass is given by vector r ( L), described by constant distance |r| and inclination
ϑ and by the changing azimuth ϕ. The gravitational force F points down, the torque is horizontal (τ = r × F). What is
the change of L?
dL
dt
= m
d(r × v)
dt
= m
dr
dt
× v + mr ×
dv
dt
= m (v × v)
0
+r × ma = r × F = τ. (1.41)
Rotation of L can be described using the angular frequency ω (its magnitude is the speed of the rotation in radians
per second and its direction is the axis of the rotation):
dL
dt
= ω × L. (1.42)
In the case of a magnetic field,
τ = µ × B = γL × B = −γB × L. (1.43)
Comparison with Eq. 1.42 immediately shows that ω = −γB.
Angular frequency of the precession of a magnetic dipolar momentum µ in a magnetic
field B is
ω = −γB. (1.44)
1.7 Electrodynamics and magnetodynamics
Similarly to the electric charge, the magnetic dipole (i) feels the surrounding magnetic
field and (ii) generates its own magnetic field. The magnetic field generated by a precessing
magnetic dipole is not stationary, it varies. To describe variable fields, the Maxwell
equations describing rotation must be modified9
:
× E = −
dB
dt
, (1.45)
× B =
1
c2
dE
dt
+ µ0j. (1.46)
Note that electric and magnetic fields are coupled in the dynamic equations. Not
only electric currents current, but also temporal variation of E induces circulation of B,
and circulation of E is possible if B varies. This has many important consequences: it
7
If you sit on a bike which does not move forward, gravity soon pulls you down to the ground. But
if the bike has a certain speed and you lean to one side, you do not fall down, you just turn a corner.
8
A qualitative discussion of precession using the spinning top and riding a bicycle is presented in
L2.4–L2.5.
9
The second equation can be written as × H = dD
dt + j.
10 CHAPTER 1. CLASSICAL ELECTROMAGNETISM
explains electromagnetic waves in vacuum and has numerous fundamental applications
in electrical engineering, including those used in NMR spectroscopy.
Eq. 1.45 shows us how the frequency of the precession motion can be measured. A
magnetic dipole in a magnetic field B0 generates a magnetic field B with the component
B0 constant and the component ⊥ B0 rotating around B0. If we place a loop of
wire next to the precessing dipole, with the axis of the loop perpendicular to the axis
of precession, the rotating component of B induces circulation of E which creates a
measurable oscillating electromotoric force (voltage) in the loop. As a consequence, an
oscillating electric current flows in the loop (L2.8).
We can use a simple example to analyze the induced voltage quantitatively. This voltage (the electromotoric force) is
an integral of the electric intensity along the detector loop. Stokes’ theorem (see B9) allows us to calculate such integral
from Eq. 1.45.
˛
L
Edl = −
ˆ
S
∂B
∂t
dS =
∂B
∂t
S, (1.47)
where S is the area of the loop. If the distance r of the magnetic moment from the detector is much larger than
the size of the loop, the magnetic induction of a field which is generated by a magnetic moment µ rotating in a plane
perpendicular to the detector loop and which crosses the loop (let us call it Bx) is10
Bx =
µ0
4π
2µx
r3
. (1.48)
As µ rotates with the angular frequency ω, µx = |µ| cos(ωt), and
∂Bx
∂t
= −
µ0
4π
2
r3
|µ|ω sin(ωt). (1.49)
Therefore, the oscillating induced voltage is
˛
L
Edl =
µ0
4π
2|µ|S
r3
ω sin(ωt). (1.50)
10
We describe the field generated by a magnetic moment in more detail later in Section 9.1 when we
analyze mutual interactions of magnetic moments of nuclei.
Chapter 2
Diffusion
2.1 Translational diffusion
Diffusion can be viewed as a result of collisions of the observed molecule with other molecules. Collisions change position of
the molecule is space (cause translation) and orientation of the molecule (cause rotation). Rotational diffusion is important
for NMR relaxation. Translational diffusion influences NMR experiments only if the magnetic filed is inhomogeneous.
Translational diffusion can be described as a random walk in a three-dimensional space, rotational diffusion can be
described as a random walk on a surface of a sphere. Although we are primarily interested in relaxation and we do not
discuss magnetic field inhomogeneity at this moment, we start our discussion with the random walk in a three-dimensional
space because the random walk on a surface of a sphere is just a special case of the general walk in three directions.
We start with several definitions. Let us assume that the position of our molecule is described by coordinates x, y, z
and its orientation is described by angles ϕ, ϑ, χ.
• Probability that the molecule is inside a cubic box of a volume ∆V = ∆x∆y∆z centered around x, y, z is
P(x, y, z, t, ∆x, ∆y, ∆z) =
x+ ∆x
2ˆ
x− ∆x
2
y+ ∆y
2ˆ
y− ∆y
2
z+ ∆z
2ˆ
z− ∆z
2
ρ(x, y, z, t)dxdydz,
where ρ(x, y, z, t) is probability density at x, y, z, corresponding to local concentration in a macroscopic picture.
If the box is small enough so that ρ(x, y, z, t) does not change significantly inside the box, the equation with the
triple integral can be simplified to
P(x, y, z, t, ∆x, ∆y, ∆z) = ρ(x, y, z, t)∆V.
• Probability that the molecule crosses one side of the box centered around x, y, z and jumps into the box centered
around x+∆x, y, z during a time interval δt is proportional to the area of the side between boxes centered around
x, y, z and around x + ∆x, y, z. This area is equal to ∆y∆z = ∆V/∆x and the probability of jumping from the
box centered around x, y, z to the box centered around x + ∆x, y, z can be written as
P(x → x + ∆x; x, y, z, t, ∆x, ∆y, ∆z, ∆t) = Φx→x+∆x∆y∆z = Φx→x+∆x∆V/∆x,
where Φx→x+∆x is the flux from the box centered around x, y, z to the box centered around x + ∆x, y, z (per
unit area). The corresponding probability density is
ρ(x → x + ∆x; x, y, z, t, ∆t) = P(x → x + ∆x; x, y, z, t, ∆x, ∆y, ∆z, ∆t)/∆V = Φx→x+∆x/∆x.
The probability of jumping to the box centered around x + ∆x, y, z is also proportional to the probability that
the molecule is inside the box centered around x, y, z (equal to ρ(x, y, z, t)∆V if the box is small enough). If the
probability of escaping the box is the same in all directions,
ρ(x → x + ∆x; x, y, z, t, ∆t) = ξρ(x, y, z, t),
11
12 CHAPTER 2. DIFFUSION
ρ(y → y + ∆y; x, y, z, t, ∆t) = ξρ(x, y, z, t),
ρ(z → z + ∆z; x, y, z, t, ∆t) = ξρ(x, y, z, t),
where ξ is a proportionality constant describing frequency of crossing a side of a box (per unit volume and including
the physical description of the collisions).
• The net flux in the x direction is given by
Φx = Φx→x+∆x − Φx+∆x→x = ξ∆x(ρ(x, y, z, t) − ρ(x + ∆x, y, z, t)) = −ξ∆x∆ρ = −ξ(∆x)2 ∂ρ
∂x
= −Dtr ∂ρ
∂x
,
where Dtr = ξ(∆x)2 is the translational diffusion coefficient.
• The net flux in all directions is
Φ = −Dtr
ρ,
which is the first Fick’s law.
• The continuity equation ˆ
V
∂ρ
∂t
dV +
‹
S
ΦdS = 0
states that any time change of probability that the molecule is in a volume V is due to the total flux through a
surface S enclosing the volume V (molecules are not created or annihilated). Using the divergence theorem,
∂ρ
∂t
+ · Φ =
∂ρ
∂t
+ · −Dtr
ρ ⇒
∂ρ
∂t
= Dtr 2
ρ,
which is the second Fick’s law.
• If the diffusion is not isotropic, the diffusion coefficient is replaced by a diffusion tensor. If we define a coordinate
frame so that the diffusion tensor is represented by a diagonal matrix with elements Dtr
xx, Dtr
yy, Dtr
zz, the second
Fick’s law has the following form:
∂ρ
∂t
= Dtr
xx
∂
∂x
∂ρ
∂x
+ Dtr
yy
∂
∂y
∂ρ
∂y
+ Dtr
zz
∂
∂z
∂ρ
∂z
= Dtr
xx
∂2
∂x2
+ Dtr
yy
∂2
∂y2
+ Dtr
zz
∂2
∂z2
ρ.
2.2 Isotropic rotational diffusion
Isotropic rotational diffusion can be viewed as random motions of a vector describing orientation of the molecule. Such
motions are equivalent to a random wandering of a point particle on a surface of a sphere with a unit diameter. In
order to describe such a random walk on a spherical surface, it is convenient to express the second Fick’s law in spherical
coordinates
∂ρ
∂t
=
Drot
r2 sin ϑ
∂
∂r
r2
sin ϑ
∂
∂r
+
∂
∂ϑ
sin ϑ
∂
∂ϑ
+
∂
∂ϕ
1
sin ϑ
∂
∂ϕ
ρ. (2.1)
Since r is constant and equal to unity,
∂ρ
∂t
=
Drot
sin ϑ
∂
∂ϑ
sin ϑ
∂
∂ϑ
+
∂
∂ϕ
1
sin ϑ
∂
∂ϕ
ρ. (2.2)
Using the substitution u = cos ϑ (and ∂u = − sin ϑ∂ϑ),
∂ρ
∂t
= Drot
(1 − u2
)
∂2
∂ϑ2
− 2u
∂
∂ϑ
+
1
1 − u2
∂2
∂ϕ2
ρ. (2.3)
Let us now try if time and space coordinates can be separated, i.e. if ρ can be expressed as a product ρ(ϑ, ϕ, t) =
f(ϑ, ϕ)g(t).
f
∂g
∂t
= gDrot
(1 − u2
)
∂2
∂ϑ2
− 2u
∂
∂ϑ
+
1
1 − u2
∂2
∂ϕ2
f. (2.4)
2.2. ISOTROPIC ROTATIONAL DIFFUSION 13
Dividing both sides of the equation by Drotρ = Drotfg,
1
Drot
1
g
∂g
∂t
=
1
f
(1 − u2
)
∂2
∂ϑ2
− 2u
∂
∂ϑ
+
1
1 − u2
∂2
∂ϕ2
f. (2.5)
If the separation of time and space coordinates is possible, i.e., if Eq. 2.5 is true for any t and any ϑ, ϕ independently,
both sides of the equation must be equal to the same constant (called λ bellow).
1
Drot
1
g
∂g
∂t
= λ (2.6)
1
f
(1 − u2
)
∂2
∂u2
− 2u
∂
∂u
+
1
1 − u2
∂2
∂ϕ2
= λ. (2.7)
Solution of the first equation is obviously
g(t) = g(0)eλDrot
t
, (2.8)
where λ is obtained by solving the second equation.
14 CHAPTER 2. DIFFUSION
Chapter 3
Nuclear magnetic resonance
Literature: A general introduction can be found in L2.6 and L2.7. A nice and detailed discussion, emphasizing
the importance of relaxation, is in Sz´antay et al.: Anthropic awareness, Elsevier 2015, Section 2.4. A useful review of
relevant statistical concepts is presented in B6. Chemical shift is introduced by Levitt in L3.7 and discussed in detail in
L9.1 (using a quantum approach, but the classical treatment can be obtained simply by using energy Ej instead of ˆHj
and magnetic moment µjk instead of γj
ˆIjk in Eqs. 9.11–9.14). A nice discussion of the offset effects (and more) can be
found in K4.
3.1 Nuclear magnetic moments in chemical substances
The classical theory does not explain the origin of the magnetic moment of some nuclei, but describes macroscopic effects
of the nuclear magnetic moments.
3.1.1 Symmetric distribution
Nuclei have permanent microscopic magnetic moments, but the macroscopic magnetic moment is induced only in the
magnetic field. This is the effect of symmetry. Outside a magnet, all orientations of the microscopic magnetic moments
have the same energy and are equally probable ⇒ the bulk magnetic moment is zero and the bulk magnetization M
(magnetic moment per unit volume) is zero (Fig. 3.1).
3.1.2 Polarization
In a static homogeneous magnetic field B0, the orientations of µ are no longer equally probable: the orientation of µ along
B0 is energetically most favored and the opposite orientation is least favored. The symmetry is broken in the direction
of B0, this direction is used to define the z axis of a coordinate system we work in. However, the state with all magnetic
moments in the energetically most favorable orientation is not most probable. Orienting all magnetic moments along
the magnetic field represents only one microstate. In contrast, there exist a large number of microstates with somewhat
higher energy. The correct balance between energy and probability is described by the Boltzmann distribution law, which
can be derived from purely statistical arguments. Thermodynamics thus helps us to describe the polarization along z
quantitatively.1
The average value of the z-component of µ is calculated as2
1
Thermodynamics also tells us that the energy of the whole (isolated) system must be conserved. Decreased
energy of magnetic moments is compensated by increased rotational kinetic energy of molecules
of the sample, coupled with the magnetic moments via magnetic fields of the tumbling molecules, as
discussed in the next chapter.
2
The integral represents summation (integration) over all possible orientations with respect to B0,
described by the inclination angles ϑ.
15
16 CHAPTER 3. NUCLEAR MAGNETIC RESONANCE
Figure 3.1: Distribution of magnetic moments in the absence of a magnetic field. Left, a schematic
representation of an NMR sample. Dots represent molecules, arrows represent magnetic moments (only
one magnetic moment per molecule is shown for the sake of simplicity, like e.g. in compressed 13
C16
O2).
Right, the molecules are superimposed to make the distribution of magnetic moments visible.
µeq
z =
πˆ
0
Peq
(ϑ)µz sin ϑdϑ =
πˆ
0
Peq
(ϑ)|µ| cos ϑ sin ϑdϑ, (3.1)
where ϑ is the inclination (angle between µ and axis z) and Peq(ϑ) is the probability of µ to be tilted by the angle
ϑ. If the magnetic dipoles are in a thermodynamic equilibrium, the angular distribution of the µ orientation is given by
the Boltzmann law3
Peq
(ϑ) =
e
−
E(ϑ)
kBT
π´
0
e
−
E(ϑ )
kBT
sin ϑ dϑ
, (3.3)
where T is the thermodynamic temperature, kB = 1.38064852 × 10−23 m2 kg s−2 K−1 is the Boltzmann constant,
and E = −|µ||B0| cos ϑ is the magnetic potential energy of the dipole. The distribution is axially symmetric, all values of
the azimuth ϕ are equally possible.
Using the substitutions
u = cos ϑ ⇒ du =
du
dϑ
dϑ =
d cos ϑ
dϑ
dϑ = − sin ϑdϑ (3.4)
and
w =
|µ||B0|
kBT
, (3.5)
3
Probability of a system to be in the state with the energy Ej at the temperature T is given by
Peq
(ϑ) =
e
−
Ej
kBT
Z
, (3.2)
where Z is sum of the e
−
Ek
kBT
terms of all possible states.
3.1. NUCLEAR MAGNETIC MOMENTS IN CHEMICAL SUBSTANCES 17
Peq
(ϑ) =
e
−
E(ϑ)
kBT
π´
0
e
−
E(ϑ )
kBT
sin ϑ dϑ
=
euw
−1´
1
−eu wdu
=
euw
1´
−1
eu wdu
=
euw
1
w
eu w 1
−1
=
w
ew − e−w
euw
= Peq
(u). (3.6)
Knowing the distribution, the average z-component of µ can be calculated
µeq
z =
πˆ
0
Peq
(ϑ)|µ| cos ϑ sin ϑdϑ =
1ˆ
−1
|µ|uPeq
(u)du =
|µ|w
ew − e−w
1ˆ
−1
ueuw
du. (3.7)
Using the chain rule,
µeq
z =
|µ|w
ew − e−w
1
w2
euw
(uw − 1)
1
−1
= |µ|
ew + e−w
w
−
ew − e−w
w2
= |µ|
ew + e−w
ew − e−w
−
1
w
= |µ| coth(w) −
1
w
.
(3.8)
The function coth(w) can be expanded as a Taylor series
coth(w) ≈
1
w
+
w
3
−
w3
45
+
2w5
945
− · · · ⇒ µz ≈ |µ|
w
3
−
w3
45
+
2w5
945
− · · · . (3.9)
At the room temperature, |µ||B0| kBT even in the strongest NMR magnets. Therefore, w is a very small number
and its high powers in the Taylor series can be neglected. In summary, the angular distribution can be approximated by
µeq
z =
1
3
|µ|2|B0|
kBT
, (3.10)
while
µeq
x = µeq
y = 0. (3.11)
The derived average magnetic moments allow us to calculate the bulk magnetization
of the NMR sample containing nuclei with µ:
Meq
x = 0 Meq
y = 0 Meq
z =
N
3
|µ|2
|B0|
kBT
, (3.12)
where N is the number of dipoles per unit volume.
In summary, dipoles are polarized in the static homogeneous magnetic fields. In
addition, all dipoles precess with the frequency ω = −γB0, but the precession cannot be
observed at the macroscopic level because the bulk magnetization is parallel with the
axis of precession (Fig. 3.2).
3.1.3 Coherence
In order to observe precession, we need to break the axial symmetry and introduce a
coherent motion of magnetic moments. This is achieved by applying another magnetic
field B1 ⊥ B0 and oscillating with the frequency close to (ideally equal to) γ|B0|/2π. In
NMR, sources of the oscillatory field are radio waves. Mathematically, such a radio field
can be decomposed into two components B+
radio and B−
radio rotating with the same angular
frequency but in opposite directions (ωradio and −ωradio, respectively). The component
18 CHAPTER 3. NUCLEAR MAGNETIC RESONANCE
M
B0
Figure 3.2: Distribution of magnetic moments in a homogeneous magnetic field B0. The red value
represents the bulk magnetization.
rotating in the same direction as the precessing dipoles (B−
radio ≡ B1 in this text) tilts
the magnetization vector M from the z direction, the other component can be neglected
as long as |B1| |B0|. This process represents a double rotation, the first rotation
is precession around the direction of B0, the second rotation around B1 is known as
nutation. The description can be simplified (the effect of the precession removed), if we
use B1 to define the x axis of our coordinate frame. As B1 rotates about B0 with an
angular frequency ωrot = −ωradio, we work in a rotating frame. In order to define the
direction of x in the rotating frame, we must also define the phase φrot.
The components of the field B1 rotating with the angular frequency −ωradio are in the laboratory frame
B1,x = |B1| cos(ωrott + φrot) = |B1| cos(−ωradiot + φrot), (3.13)
B1,y = |B1| sin(ωrott + φrot) = |B1| sin(−ωradiot + φrot), (3.14)
B1,z = 0 (3.15)
and in the rotating frame
B1,x = |B1| cos(φrot), (3.16)
B1,y = |B1| sin(φrot), (3.17)
B1,z = 0. (3.18)
Consequently, the rotation of magnetization is given by the angular frequency vector
3.1. NUCLEAR MAGNETIC MOMENTS IN CHEMICAL SUBSTANCES 19
M
B0
Bradio
Figure 3.3: Effect of the radio waves on the bulk magnetization. The thin green line shows oscillation
of the magnetic induction vector of the radio waves, the red trace shows evolution of the magnetization
during irradiation. The length of the irradiation period was chosen so that the magnetization is rotated
to the plane perpendicular to B0 (red arrow). Note that the |B0|/|Bradio| is much higher in a real
experiment.
20 CHAPTER 3. NUCLEAR MAGNETIC RESONANCE
ω = ω0 +ω1 = −γ(B0 +B1) =


0
0
−γ|B0|

+


−γ|B1| cos(−ωradiot + φrot)
−γ|B1| sin(−ωradiot + φrot)
0

 =


−γ|B1| cos(−ωradiot + φrot)
−γ|B1| sin(−ωradiot + φrot)
−γ|B0|

 (3.19)
in the laboratory frame, and by
ω = ω1 = −γB1 =


−γ|B1| cos(φrot)
−γ|B1| sin(φrot)
0

 (3.20)
in the coordinate frame rotating with the angular frequency ωrot = −ωradio = ω0.
What are the components of B1 in the rotating frame for different choices of φrot?
If φrot = 0, cos(0) = 1, sin(0) = 0, and
B1,x = |B1|, (3.21)
B1,y = 0, (3.22)
B1,z = 0. (3.23)
If φrot = π
2
, cos( π
2
) = 0, sin(π
2
) = 1, and
B1,x = 0, (3.24)
B1,y = |B1|, (3.25)
B1,z = 0. (3.26)
If φrot = π, cos(π) = −1, sin(π) = 0, and
B1,x = −|B1|, (3.27)
B1,y = 0, (3.28)
B1,z = 0, (3.29)
and so on.
The typical convention is to choose φrot = π for nuclei with γ > 0 and φrot = 0 for
nuclei with γ < 0. Then, the nutation frequency is ω1 = +γ|B1| (opposite convention to
the precession frequency!) for nuclei with γ > 0 and ω1 = −γ|B1| (the same convention
as the precession frequency) for nuclei with γ < 0.
If the radio waves are applied exactly for the time needed to rotate the magnetization
by 90 ◦
, they create a state with M perpendicular to B0. Such magnetization vector then
rotates with the precession frequency, also known as the Larmor frequency. Such rotation
corresponds to a coherent motion of nuclear dipoles polarized in the direction of M and
generates measurable electromotoric force in the detector coil (Fig. 3.4).
3.2 Chemical shift
The description of the motions of the bulk nuclear magnetization presented in the previous
section is simple but boring. What makes NMR useful for chemists and biologists
3.2. CHEMICAL SHIFT 21
M
B0
Figure 3.4: Distribution of magnetic moments after application of a 906◦ radiofrequency pulse.
is the fact that the energy of the magnetic moment of the observed nucleus is influenced
by magnetic fields associated with motions of nearby electrons. In order to understand
this effect, we need to describe the magnetic fields of moving electrons.
If a moving electron enters a homogeneous magnetic field, it experiences a Lorentz
force and moves in a circle in a plane perpendicular to the field (cyclotron motions).
Such an electron represents an electric current in a circular loop, and is a source of a
magnetic field induced by the homogeneous magnetic field. The homogeneous magnetic
field B0 in NMR spectrometers induces a similar motion of electrons in atoms, which
generates microscopic magnetic fields.
The observed nucleus feels the external magnetic field B0 slightly modified by the
microscopic fields of electrons.
If the electron distribution is spherically symmetric, with the observed nucleus in the
center (e.g. electrons in the 1s orbital of the hydrogen atom), the induced field of the
electrons decreases the effective magnetic field felt by the nucleus in the center. Since
the induced field of electrons Be is proportional to the inducing external field B0, the
effective field can be described as
B = B0 + Be = (1 + δ)B0. (3.30)
The constant δ is known as chemical shift and does not depend on the orientation
22 CHAPTER 3. NUCLEAR MAGNETIC RESONANCE
of the molecule in such a case4
. The precession frequency of the nucleus is equal5
to
(1 + δ)ω0.
Electron distribution is not spherically symmetric in most molecules. As a consequence,
the effective field depends on the orientation of the whole molecule and on mutual
orientations of atoms, defining the shapes of molecular orbitals6
. Therefore, the effective
field fluctuates as a result of rotational diffusion of the molecule and of internal motions
changing mutual positions of atoms. The induced field of electrons is still proportional
to the inducing external field B0, but the proportionality constants are different for each
combination of components of Be and B0 in the coordination frame used. Therefore, we
need six7
constants δjk to describe the effect of electrons:
Be,x = δxxB0,x + δxyB0,y + δxzB0,z (3.31)
Be,y = δyxB0,x + δyyB0,y + δyzB0,z (3.32)
Be,z = δzxB0,x + δzyB0,y + δzzB0,z (3.33)
Eqs. 3.31–3.33 can be written in more compact forms


Be,x
Be,y
Be,z

 =


δxx δxy δxz
δyx δyy δyz
δzx δzy δzz

 ·


B0,x
B0,y
B0,z

 (3.34)
or
Be = δ · B0, (3.35)
where δ is the chemical shift tensor.
It is always possible to find a coordinate system X, Y, Z known as the principal frame, where δ is represented
by a diagonal matrix. In such a system, we need only three constants (principal values of the chemical shift tensor):
δXX , δY Y , δZZ . However, three more parameters must be specified: three Euler angles (written as ϕ, ϑ, and χ in
this text) defining orientation of the coordinate system X, Y, Z in the laboratory coordinate system x, y, z. Note that
δXX , δY Y , δZZ are true constants because they do not change as the molecule tumbles in solution (but they may change
due to internal motions or chemical changes of the molecule). The orientation is completely described by the Euler angles.
The chemical shift tensor in its principal frame can be also written as a sum of three simple matrices, each multiplied
by one characteristic constant:


δXX 0 0
0 δY Y 0
0 0 δZZ

 = δi


1 0 0
0 1 0
0 0 1

 + δa


−1 0 0
0 −1 0
0 0 2

 + δr


1 0 0
0 −1 0
0 0 0

 , (3.36)
where
4
Instead of δ, a constant with the opposite sign defining the chemical shielding is sometimes used.
5
The value of δ in Eq. 3.30 describes how much the frequency of nuclei deviates from a hypothetical
frequency of free nuclei. Such a hypothetical frequency is difficult to measure. In practice, frequencies
of nuclei in certain, readily accessible chemical compounds are used instead of the frequencies of free
nuclei as the reference values of δ, as is described in Section 5.1.
6
The currents induced in orbitals of other atoms may decrease or increase (shield or deshield) the
effective magnetic field felt by the observed nucleus.
7
There are nine constants in Eqs. 3.31–3.33, but δxy = δyx, δxz = δzx, and δyz = δzy.
3.2. CHEMICAL SHIFT 23
δi =
1
3
Tr{δ} =
1
3
(δXX + δY Y + δZZ ) (3.37)
is the isotropic component of the chemical shift tensor,
δa =
1
3
∆δ =
1
6
(2δZZ − (δXX + δY Y )) (3.38)
is the axial component of the chemical shift tensor (∆δ is the chemical shift anisotropy), and
δr =
1
3
ηδ∆δ =
1
2
(δXX − δY Y ) (3.39)
is the rhombic component of the chemical shift tensor (ηδ is the asymmetry of the chemical shift tensor).
The chemical shift tensor written in its principle frame is relatively simple, but we need its description in the laboratory
coordinate frame. Changing the coordinate systems represents a rotation in a three-dimensional space. Equations
describing such a simple operation are relatively complicated. On the other hand, the equations simplify if B0 defines the
z axis of the coordinate frame (i.e., B0,z = B0 and B0,x = B0,y = 0):
Be = δiB0


0
0
1

+δaB0


3 sin ϑ cos ϑ cos ϕ
3 sin ϑ cos ϑ sin ϕ
3 cos2 ϑ − 1

+δrB0


−(2 cos2 χ − 1) sin ϑ cos ϑ cos ϕ + 2 sin χ cos χ sin ϑ sin ϕ
−(2 cos2 χ − 1) sin ϑ cos ϑ sin ϕ − 2 sin χ cos χ sin ϑ cos ϕ
+(2 cos2 χ − 1) sin2 ϑ

 . (3.40)
The first, isotropic contribution does not change upon rotation (it is a scalar). The second, axial contribution, is
insensitive to the rotation about the symmetry axis a, described by χ. Rotation of the chemical shift anisotropy tensor
from its principal frame to the laboratory frame can be also described by orientation of a in the laboratory frame:
δa


−1 0 0
0 −1 0
0 0 2

 −→ δa


3a2
x − 1 3axay 3axaz
3axay 3a2
y − 1 3ayaz
3axaz 3ayaz 3a2
z − 1

 , (3.41)
where ax = sin ϑ cos ϕ, ay = sin ϑ sin ϕ, and az = cos ϑ.
We have derived a not very simple equation (Eq. 3.40) describing how electrons
modify the external magnetic field. Do we really need it? When we analyze only the
(average) value of the precession frequency, it is sufficient to consider only the isotropic
component. The description of the effect of electrons then simplifies to Eq. 3.30, where δ
now represents δi of Eq. 3.40. When we analyze also the effect of stochastic motions, the
other terms become important as well. The correct quantitative analysis requires full
Eq. 3.40, but the basic principles can be discussed without using the rhombic component.
Therefore, we will use the axially symmetric approximation of Eq. 3.40 when we discuss
effects of molecular motions in Section 4.
3.2.1 Offset effects
The presence of electrons makes NMR a great method for chemical analysis. The measured
precession frequency depends not only on the type of nucleus (e.g. 1
H) but also on
the electronic environment: frequencies of protons in different chemical moieties differ
and can be used to identify chemical groups in organic molecules. But how the electrons
influence the physical description of the nuclear magnetization?
The effect of the isotropic component of the chemical shift on the precession frequency
is simply introducing a small correction constant 1 + δ modifying γ:
ω0 = −γB0 → ω0 = −γ(1 + δ)B0. (3.42)
24 CHAPTER 3. NUCLEAR MAGNETIC RESONANCE
The trouble is that the correction is different for each proton (or carbon etc.) in the
molecule. Therefore, the frequency of the radio waves can match ω0 = −γ(1 + δ)|B0|
only for one proton in the molecule. For example, if the radio wave resonate with the
frequency of the methyl proton in ethanol, it cannot resonate with the frequency of the
proton in the OH or CH2 group. In the rotating coordinate frame, only magnetization of
the methyl protons rotates about ω1 = γδ(methyl)B1. Magnetizations of other protons
rotate about other axes. Such rotations can be described by effective angular frequencies
ωeff = ω1 + Ω, (3.43)
where
Ω = ω0 − ωrot = ω0 − (−ωradio) = ω0 + ωradio (3.44)
is the angular frequency offset. As any vector in a 3D space, ωeff is characterized by
three parameters: magnitude ωeff, inclination ϑ, and azimuth ϕ.
The magnitude of the effective frequency is
ωeff = ω2
1 + Ω2. (3.45)
The inclination can be calculated from
tan ϑ =
ω1
Ω
. (3.46)
The azimuth is given by the phase of B1 (ϕ = ϕrot in a single-pulse experiment).
As a result of the chemical shift, only the magnetization of the nucleus with Ω = 0
(methyl protons in our case) rotates along the ”meridian” in the rotating coordinate
system. Magnetizations of other protons move in other circles. Therefore, if the radio
transmitter is switched off when the methyl magnetization is pointing horizontally (and
starts to rotate around the ”equator” with the precession frequency of methyl protons),
vectors of magnetizations of other protons point in different directions, and start to
precess on cones with different inclinations and with different initial phases. Such effects,
known as the offset effects, influence the measured signal.8
The discussed motion of the magnetization vector M is described by the following
equations
8
The result is the same as if apparent effective fields of the magnitude Beff = B2
1 + (Ω/γ)2 were
applied in the direction in the directions of ωeff . The apparent effective field Beff is often used to describe
the offset effects.
3.2. CHEMICAL SHIFT 25
dMx
dt
= −ΩMy + ω1 sin ϕMz, (3.47)
dMy
dt
= +ΩMx − ω1 cos ϕMz, (3.48)
dMz
dt
= −ω1 sin ϕMx + ω1 cos ϕMy, (3.49)
(3.50)
where ϕ is the azimuth of ωeff, which can be written in a compact form as
dM
dt
= ωeff × M. (3.51)
3.2.2 Evolution of magnetization in B0
Eqs. 3.47–3.49 are easy to solve in the absence of B1 (i.e., after turning off the radio waves):
dMx
dt
= −ΩMy (3.52)
dMy
dt
= ΩMx (3.53)
dMz
dt
= 0 (3.54)
The trick is to multiply the second equation by i and add it to the first equation or subtract it from the first equation.
d(Mx + iMy)
dt
= Ω(−My + iMx) = +iΩ(Mx + iMy) (3.55)
d(Mx − iMy)
dt
= Ω(−My − iMx) = −iΩ(Mx − iMy) (3.56)
Mx + iMy = C+e+iΩt
(3.57)
Mx − iMy = C−e−iΩt
(3.58)
where the integration constants C+ = Mx(0) + iMy(0) = M2
x(0) + M2
y (0)eφ0 and C− = Mx(0) − iMy(0) =
M2
x(0) + M2
y (0)e−φ0 are given by the initial phase φ0 of M in the coordinate system (in our case, t = 0 is defined by
switching off the radio waves):
Mx + iMy = M2
x(0) + M2
y (0)e+(iΩt+φ0)
= M2
x(0) + M2
y (0)(cos(Ωt + φ0) + i(sin(Ωt + φ0)) (3.59)
Mx − iMy = M2
x(0) + M2
y (0)e−(iΩt+φ0)
= M2
x(0) + M2
y (0)(cos(Ωt + φ0) − i(sin(Ωt + φ0)), (3.60)
Mx = M2
x(0) + M2
y (0) cos(Ωt + φ0) (3.61)
My = M2
x(0) + M2
y (0) sin(Ωt + φ0), (3.62)
26 CHAPTER 3. NUCLEAR MAGNETIC RESONANCE
where
tan φ0 =
My(0)
Mx(0)
. (3.63)
In order to obtain φ0 and M2
x(0) + M2
y (0), we must first solve Eqs. 3.47–3.49. This solution is not so easy, and
we look only at the result:
Mx(0) = M0 sin(ωeff τp) sin ϑ, (3.64)
My(0) = M0(1 − cos(ωeff τp)) sin ϑ cos ϑ, (3.65)
Mz(0) = M0(cos2
ϑ + cos(ωeff τp) sin2
ϑ), (3.66)
where M0 is the magnitude of the bulk magnetization in the thermodynamic equilibrium, τp is duration of irradiation
by the radio waves, and tan ϑ = ω1/Ω.
Chapter 4
Relaxation
Literature: A nice introduction is in K9.1 and K9.3, more details can be found in L19
and L20.1–L20.3.
4.1 Relaxation due to chemical shift anisotropy
The Boltzmann law allowed us to describe the state of the system in thermal equilibrium,
but it does not tell us how is the equilibrium reached. The processes leading to the
equilibrium states are known as relaxation. Relaxation takes places e.g. when the
sample is placed into a magnetic field inside the spectrometer or after excitation of the
sample by radio wave pulses.
Spontaneous emission is completely inefficient (due to low energy differences of spin
states). Relaxation in NMR is due to interactions with local fluctuating magnetic fields
in the molecule. One source of fluctuating fields is the anisotropy of chemical shift,
described by the axial and rhombic components of the chemical shift tensor. As the
molecule moves, the isotropic component of the chemical shift tensor does not change
because it is spherically symmetric. However, contributions to the local fields described
by the axial and rhombic components fluctuate even if the constants δa do not change
because the axial part of the chemical shift depends on the orientation of the molecule.
Here, we introduce the basic idea by analyzing the effects of fluctuating magnetic
fields in a classical manner.
4.2 Loss of coherence
Motion of a magnetic moment in a magnetic filed is described classically as (cf. Eq. 3.51)
dµ
dt
= ω × µ = −γB × µ, (4.1)
or for individual components:
27
28 CHAPTER 4. RELAXATION
dµx
dt
= ωyµz − ωzµy, (4.2)
dµy
dt
= ωzµx − ωxµz, (4.3)
dµz
dt
= ωxµy − ωyµx. (4.4)
Solving a set of three equations is not so easy. Therefore, we start with a simplified case. Remember what we learnt
when we tried to rotate the magnetization away from the z direction by magnetic fields perpendicular to B0, i.e., fields
with Bx and By components. Only Bx and By fields rotating with the frequency equal to the precession frequency of
individual magnetic moments (Larmor frequency) have the desired effect. Let us start our analysis by assuming that the
molecular motions are much slower than the Larmor frequency. Under such circumstances, the effects of Be,x and Be,y
can be neglected and the equations of motion simplify to
dµx
dt
= −ωzµy = γBzµy (4.5)
dµy
dt
= ωzµx = −γBzµx (4.6)
dµz
dt
= 0 (4.7)
Eqs. 4.5–4.7 are very similar to Eqs. 3.52–3.54, so we try the same approach and calculate
dµ+
dt
≡
d(µx + iµy)
dt
= iωz(µx + iµy) = −iγBz(µx + iµy) (4.8)
According to Eq. 3.40,
Bz = B0 + Be,z = B0(1 + δi + δa(3 cos2
ϑ − 1) + δr(2 cos2
χ − 1) sin2
ϑ). (4.9)
For the sake of simplicity, we assume that the chemical shift tensor is axially symmetric (δr = 0). Then, ωz can be
written as
ωz = −γ(B0 + Be,z) = −γB0(1 + δi) − γB0δa(3 cos2
ϑ − 1) = ω0 + bΘ , (4.10)
where
ω0 = −γB0(1 + δi) (4.11)
b = −2γB0δa (4.12)
Θ =
3 cos2 ϑ − 1
2
. (4.13)
This looks fine, but there is a catch here: Eq. 4.8 cannot be solved as easily as we solved 3.52–3.54 because ωz is
not constant but fluctuates in time. The value of ωz is not only changing, is changing differently for each molecule in
the sample and it is changing in a random, unpredictable way! Can we solve the equation of motion at all? The answer
is ”yes and no”. The equation of motion cannot be solved for an individual magnetic moment. However, we can take
advantage of statistics and solve the equation of motion for the total magnetization M+, given by the statistical ensemble
of magnetic moments.
We start by assuming that for a very short time ∆t, shorter than the time scale of molecular motions, the orientation
of the molecule does not change and Θ remains constant. We try to describe the evolution of µ+ in such small time
steps, assuming
∆µ+
∆t
≈
dµ+
dt
≈ i(ω0 + bΘ )µ+
(4.14)
If the initial value of µ+ is µ+
0 and if the values of ω0, b, Θ during the first time step are ω0,1, b1, Θ1, respectively,
the value of µ+ after the first time step is
4.2. LOSS OF COHERENCE 29
µ+
0
ibΘ1∆t
1
µ+
1
ibΘ2∆t
1
µ+
2
ibΘ3∆t
1
µ+
3
ibΘ4∆t
1
µ+
4 · · ·
ibΘk∆t
1
µ+
k
Figure 4.1: Evolution of magnetic moments due to longitudinal (parallel with B0) fluctuations of
magnetic fields. The symbols µ+
0 and µ+
k are connected by 2k
possible pathways composed of black and
green segments. Each black segment represents multiplication by one, each green segment represents
multiplication by ibΘj ∆t, where j ranges from 1 to k. The product of binomials in Eq. 4.21 is a sum
of 2k
terms. In order to obtain one term of the series, we walk along the corresponding pathway and
multiply all black and green numbers written above the individual steps. The pathway composed of
the black segments only gives the result of multiplication equal to one, the pathways containing just
one green segment give results of multiplication proportional to ∆t, the pathways containing two green
segments give results of multiplication proportional to (∆t)2
, etc. In order to get the complete product
in Eq. 4.21, we must walk through all possible pathways (all possible combinations of the segments)
and sum all results of the multiplication.
µ+
1 = µ+
0 + ∆µ+
1 = µ+
0 + i(ω0,1 + b1Θ1)∆tµ+
0 = [1 + i(ω0,1 + b1Θ1)∆t]µ+
0 . (4.15)
After the second step,
µ+
2 = µ+
1 + ∆µ+
2 = µ+
1 + i(ω0,2 + b2Θ2)∆tµ+
1 = [1 + i(ω0,2 + b2Θ2)∆t][1 + i(ω0,1 + b1Θ1)∆t]µ+
0 . (4.16)
After k steps,
µ+
k = [1 + i(ω0,k + bkΘk)∆t][1 + i(ω0,k−1 + bk−1Θk−1)∆t] · · · [1 + i(ω0,2 + b2Θ2)∆t][1 + i(ω0,1 + b1Θ1)∆t]µ+
0 . (4.17)
If the structure of the molecule does not change, the electron distribution is constant and the size and shape of
the chemical shift tensor described by δi and δa does not change in time. Then, ω0 and b are constant and the only
time-dependent parameter is Θ , fluctuating as the orientation of the molecule (described by ϑ) changes. The parameter
ω0 = −γB0(1 + δi) represents a constant frequency of coherent rotation under such circumstances. The coherent rotation
can be removed if we describe the evolution of µ+ in a coordinate frame rotating with the frequency ω0. The transformation
of µ+ to the rotating frame is given by
(µ+
)rot = µ+
e−iω0t
. (4.18)
We also need to express the derivative of (µ+)rot, which is done easily by applying the chain rule:
d(µ+)rot
dt
=
d(µ+e−iω0t)
dt
=
dµ+
dt
e−iω0t
− iω0µ+
e−iω0t
. (4.19)
Substituting dµ+/dt from Eq. 4.14 results in
d(µ+)rot
dt
= i(ω0 + bΘ )µ+
e−iω0t
− iω0µ+
e−iω0t
= ibΘ µ+
e−iω0t
= ibΘ (µ+
)rot. (4.20)
When compared with Eq. 4.14, we see that ω0 disappeared, which simplifies Eq. 4.17 to
(µ+
k )rot = [1 + ibΘk∆t][1 + ibΘk−1∆t] · · · [1 + ibΘ2∆t][1 + ibΘ1∆t](µ+
0 )rot. (4.21)
The process of calculating the product of brackets in Eq. 4.21 is shown schematically in Figure 4.1. The final product
is
(µ+
k )rot = [1+ib∆t(Θk+Θk−1+· · ·+Θ1)−b2
∆t2
(Θk(Θk−1+· · · Θ2 +Θ1)+· · ·+Θ2Θ1)−ib3
∆t3
(. . . )+· · · ](µ+
0 )rot. (4.22)
30 CHAPTER 4. RELAXATION
We can now return to the question how random fluctuations change µ+. Let us express the difference between µ+
after k and k − 1 steps:
∆(µ+
k )rot = (µ+
k )rot − (µ+
k−1)rot = [ib∆tΘk − b2
∆t2
Θk(Θk−1 + · · · + Θ1) − ib3
∆t3
(. . . ) + · · · ](µ+
0 )rot. (4.23)
Dividing both sides by ∆t
∆(µ+
k )rot
∆t
= [ibΘk − b2
∆tΘk(Θk−1 + · · · + Θ1) − ib3
∆t2
(. . . ) + · · · ](µ+
0 )rot (4.24)
and going back from ∆t to dt (neglecting terms with dt2, dt3, . . . , much smaller than dt),
d(µ+(tk))rot
dt
=

ibΘ (tk) − b2
tkˆ
0
Θ (tk)Θ (tk − tj)dtj

 (µ+
0 )rot. (4.25)
We see that calculating how fluctuations of Bz affect an individual magnetic moment in time tk requires knowledge of
the orientations of the molecule during the whole evolution (Θ (tk − tj)). However, we are not interested in the evolution
of a single magnetic moment, but in the evolution of the total magnetization M+. Total magnetization is given by the sum
of all magnetic moments (magnetic moments in all molecules). Therefore, we must average orientations of all molecules
in the sample. In other words, we should describe Θ using two indices, k and m, where k describes the time step and
m the orientation of the given molecule. Calculation of the evolution of M+ then should include summation of Θk,m for
all k and m, or integration over the angles describing orientations of the molecule in addition to the time integration.
As the magnetic moments move almost independently of the molecular motions, we can average Θ and µ+ separately.
In the case of the axially symmetric chemical shift tensor, the orientations of molecules are given by orientations of the
symmetry axes a of the chemical shift tensors of the observed nuclei in the molecules, described by the angles ϕ and
ϑ. In order to simplify averaging the orientations, we assume that all orientations are equally probable. This is a very
dangerous assumption. It does not introduce any error in this section, but leads to wrong results when we analyze the
effects of fluctuations of magnetic fields perpendicular to B0!
As the angle ϑ(t) is hidden in the function Θ (t) = (3 cos ϑ2 − 1)/2 in our equation, the ensemble averaging can be
written as1
d(M+(tk))rot
dt
=

ib
1
4π
2πˆ
0
dϕ
πˆ
0
Θ (tk) sin ϑdϑ − b2
tkˆ
0
dtj
1
4π
2πˆ
0
dϕ
πˆ
0
Θ (tk)Θ (tk − tj) sin ϑdϑ

 (M+
0 )rot, (4.26)
where ϕ ≡ ϕ(tk) and ϑ ≡ ϑ(tk).
In order to avoid writing too many integration signs, we mark the averaging simply by a horizontal bar above the
averaged function:
d(M+(tk))rot
dt
= −

ibΘ (tk) + b2
tkˆ
0
Θ (tk)Θ (tk − tj)dtj

 (M+
0 )rot. (4.27)
The average values of a2
z = cos2 ϑ, of a2
x = cos2 ϕ sin2 ϑ, and of a2
y = sin2 ϕ sin2 ϑ must be the same because none of
the directions x, y, z is preferred:
a2
x = a2
y = a2
z. (4.28)
Therefore,
1
Two integrals in the following equation represent calculation of an average of a function depending
on the orientation. Geometrically, it is summation of the values of the function for individual surface
elements (defined by inclination ϑ and azimuth ϕ) of a unit sphere, divided by the complete surface
of the sphere 4π. Note that the current orientation of each molecule at tk is described by ϑ(tk) and
ϕ(tk), the values ϑ(tj) hidden in the function Θ (tj) describe only history of each molecule. They are
somehow related to ϑ(tk) and ϕ(tk) and therefore treated as an unknown function of ϑ(tk) and ϕ(tk)
during the integration.
4.2. LOSS OF COHERENCE 31
a2
x + a2
y + a2
z = 1 ⇒ a2
x + a2
y + a2
z = 3a2
z ⇒ 3a2
z − 1 = (3 cos2 ϑ − 1) = 2Θ = 0 ⇒ Θ = 0. (4.29)
It explains why we did not neglect already the b2dt term – we would obtain zero on the right-hand side in the
rotating coordinate frame (this level of simplification would neglect the effects of fluctuations and describe just the
coherent motions).
We have derived that the equation describing the loss of coherence (resulting in a loss of transverse magnetization)
is
d(M+(tk))rot
dt
= −

b2
tkˆ
0
Θ (tk)Θ (tk − tj)dtj

 (M+
0 )rot, (4.30)
where the value of Θ (tk)Θ (tk − tj) is clearly defined statistically (by the averaging described above). Values of
Θ (tk)Θ (tk − tj) can be determined easily for two limit cases:
• tj = 0: If tj = 0, Θ (tk)Θ (tk − tj) = (Θ (tk))2, i.e., Θ (tk) and Θ (tk − tj) are completely correlated.
The average value of Θ (tk)2 is
Θ (tk)2 =
1
4
(3 cos2 ϑ − 1)2 =
1
16π
2πˆ
0
dϕ
πˆ
0
dϑ(sin ϑ)(3 cos2
ϑ − 1)2
=
1
5
. (4.31)
• tj → ∞: If the changes of orientation (molecular motions) are random, the correlation between Θ (tk) and
Θ (tk −tj) is lost for very long tj and they can be averaged separately: Θ (tk)Θ (tk − tj) = Θ (tk)·Θ (tk − tj).
But we know that average Θ (t) = 3 cos2 ϑ − 1 = 0. Therefore, Θ (tk)Θ (tk − tj) = 0 for tj → ∞.
If the motions are really stochastic, it does not matter when we start to measure
time. Therefore, we can describe the loss of coherence for any tk as
d(M+
)rot
dt
= −

b2
∞ˆ
0
Θ (0)Θ (t)dt

 (M+
)rot, (4.32)
which resembles a first-order chemical kinetics with the rate constant
R0 = b2
∞ˆ
0
Θ (0)Θ (t)dt. (4.33)
In order to calculate the value of the rate constant R0, we must be able to avaluate
the averaged term Θ (0)Θ (t), known as the time correlation function. Note that
statistics play the key role here: the whole analysis relies on the fact that although
the product Θ (0)Θ (t) changes randomly, the value of the time correlation function is
defined statistically. If the structure of the molecule does not change (rigid body rotational
diffusion), which is the case we analyze, the analytical form of Θ (0)Θ (t) can be
derived.
32 CHAPTER 4. RELAXATION
4.2.1 Time correlation function
Analysis of the isotropic rotational diffusion in Section 2.2 allows us to calculate the time correlation function Θ (0)Θ (t)
for this type of diffusion (with a spherical symmetry). The ensemble-averaged product of randomly changing (3 cos2 ϑ(t)−
1)/2, evaluated for a time difference t, can be expressed as
3
2
cos2 ϑ(0) −
1
2
3
2
cos2 ϑ(t) −
1
2
(4.34)
=
2πˆ
0
dϕ(0)
πˆ
0
sin ϑ(0)dϑ(0)ρ0
2πˆ
0
dϕ(t)
πˆ
0
sin ϑ(t)dϑ(t)
3
2
cos2
ϑ(0) −
1
2
3
2
cos2
ϑ(t) −
1
2
G(ϑ(0), ϕ(0)|ϑ(t), ϕ(t)),
(4.35)
where ρ0 is the probability density of the original orientation described by ϑ(0) and ϕ(0), and G(ϑ(0), ϕ(0)|ϑ(t), ϕ(t))
is the conditional probability density or propagator (also known as the Green function) describing what is the chance to
find an orientation given by ϑ(t), ϕ(t) at time t, if the orientation at t = 0 was given by ϑ(0), ϕ(0).
If the molecule is present in an isotropic environment2, ρ0 plays a role of a normalization constant and can be
calculated easily from the condition that the overall probability of finding the molecule in any orientation is equal to one:
2πˆ
0
dϕ(0)
πˆ
0
sin ϑ(0)dϑ(0)ρ0 = 4πρ0 = 1 ⇒ ρ0 =
1
4π
. (4.36)
Evaluation of G(ϑ(0), ϕ(0)|ϑ(t), ϕ(t)) requires to solve the diffusion equation Eq. 2.7. We again express G as a
product of time-dependent and time-independent functions g(t)P(ϑ). The function g(t) is defined by Eq. 2.6, the function
P(ϑ) is a simplified version of function f(ϑ, ϕ) from Eq. 2.7. Since our correlation correlation function does not depend
on ϕ, we can assume ∂P/∂ϕ = 0 and further simplify Eq. 2.7 to
(1 − u2
)
d2
du2
− 2u
d
du
P = λP, (4.37)
(1 − u2
)
d2P
du2
− 2u
dP
du
− λP = 0. (4.38)
We expand P in a Taylor series
P =
∞
k=0
akuk
, ak =
1
k!
dkP(0)
duk
, (4.39)
calculate its first and second derivatives
dP
du
=
∞
k=0
kakuk−1
, (4.40)
d2P
du2
=
∞
k=0
k(k − 1)akuk−2
, (4.41)
and substitute them into Eq. 4.38
(1 − u2
)
∞
k=0
k(k − 1)akuk−2
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
− 2u = 0 (4.42)
∞
k=0
k(k − 1)akuk−2
−
∞
k=0
k(k − 1)akuk
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
= 0. (4.43)
Note that the first two terms of the first sum are equal to zero (the first term includes multiplication by k = 0 and
the second term includes multiplication by k − 1 = 0 for k = 1). Therefore, we can start summation from k = 2.
2
Note that in the isotropic environment, where all orientations of the molecule are equally probable,
the diffusion can be very anisotropic if the shape of the molecule greatly differs from a sphere.
4.2. LOSS OF COHERENCE 33
∞
k=2
k(k − 1)akuk−2
−
∞
k=0
k(k − 1)akuk
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
= 0. (4.44)
We shift the index in the first sum by two to get the first sum expressed in the same power of u as the other sums
∞
k=0
(k + 2)(k + 1)ak+2uk
−
∞
k=0
k(k − 1)akuk
− 2
∞
k=0
kakuk
− λ
∞
k=0
akuk
= 0 (4.45)
∞
k=0
((k + 2)(k + 1)ak+2 − (k(k − 1) + 2k + λ)ak) uk
=
∞
k=0
((k + 2)(k + 1)ak+2 − (k(k + 1) + λ)ak) uk
= 0. (4.46)
This equation is true only if all terms in the sum are equal to zero
(k + 2)(k + 1)ak+2 − (k(k + 1) + λ)ak = 0, (4.47)
which gives us a recurrence formula relating ak+2 and ak:
ak+2 =
k(k + 1) + λ
(k + 2)(k + 1)
ak = 0. (4.48)
We can use the recurrence formula to express the Taylor series in terms of a0 and a1:
P = a0 1 +
0 · 1 + λ
1 · 2
u2
+
0 · 1 + λ
1 · 2
·
2 · 3 + λ
3 · 4
u4
+ . . . + a1 u +
1 · 2 + λ
2 · 3
u3
+
1 · 2 + λ
2 · 3
·
3 · 4 + λ
4 · 5
u5
+ . . . = 0.
(4.49)
What is the value of λ? Note that ak+2 = 0 for each λ = −k(k + 1), which terminates one of the series in large
parentheses, while the other series grows to infinity (for u = 0). To keep P finite, the coefficient before the large parentheses
in the unterminated series must be set to zero. It tells us that we can find a possible solution for each even or odd k if
a1 = 0 or a00, respectively.
k = 0 a1 = 0 P = P0 = 1 λ = −k(k + 1) = 0 (4.50)
k = 1 a0 = 0 G = P1 = u = cos ϑ λ = −k(k + 1) = −2 (4.51)
k = 2 a1 = 0 G = P2 =
3u2 − 1
2
=
3 cos2 ϑ − 1
2
λ = −k(k + 1) = −6 (4.52)
k = 3 a0 = 0 G = P3 =
5u3 − 3u
2
=
5 cos3 ϑ − 3 cos ϑ
2
λ = −k(k + 1) = −12 (4.53)
... (4.54)
The value of a0 or a1 preceding the terminated series was chosen so that Pk(u = 1) = Pk(ϑ = 0) = 1.
Which of the possible solutions is the correct one? It can be shown easily that
1ˆ
−1
Pk(u)Pk (u)du =
πˆ
0
Pk(ϑ)Pk (ϑ)dϑ =
2
2k + 1
δkk , (4.55)
i.e., is equal to zero for each k = k (Pk are orthogonal). As we are going to use G = g(t)P(ϑ) to calculate a
correlation function for functions having the same form as the solutions for k = 2 and as the calculation of the correlation
function includes the same integration as in Eq. 4.55, it is clear that the only solution which gives us non-zero correlation
function is that for k = 2, i.e. P2. Our function G is therefore given by
G = g0
3 cos2 ϑ − 1
2
e−6Drot
t
. (4.56)
Still, we need to evaluate the factor g0. This value must be chosen so that we fulfill the following conditions:
34 CHAPTER 4. RELAXATION
2πˆ
0
dϕ
πˆ
0
sin ϑdϑG = 1 (4.57)
and
G(t = 0) = δ(θ − θ0), (4.58)
where δ(θ − θ0) is a so-called Dirac delta function, defined as
∞ˆ
−∞
f(x)δ(x − x0) = f(x0). (4.59)
The second condition says that ϑ must have its original value for t = 0. This is fulfilled for g0 proportional to
(3 cos2 ϑ0 − 1)/2:
g0 = c0
3
2
cos2
ϑ(0) −
1
2
. (4.60)
We can re-write our original definition of the correlation function with the evaluated G function and in a somewhat
simplified form (omitting integration over ϕ and ϕ0):
3
2
cos2 ϑ(0) −
1
2
3
2
cos2 ϑ(t) −
1
2
=
1ˆ
−1
du0ρ0c0
1ˆ
−1
du
(3u2
0 − 1)2
4
(3u2 − 1)2
4
e−6Drot
t
, (4.61)
where ρ0 can be evaluated from the normalization condition
1ˆ
−1
du0ρ0 = 2ρ0 = 1 ⇒ ρ0 =
1
2
(4.62)
and
c0 from
1ˆ
−1
du0
1ˆ
−1
c0
3u2
0 − 1
2
3u2 − 1
2
δ(u − u0)du =
1ˆ
−1
du0c0
(3u2
0 − 1)2
4
=
2
5
c0 ⇒ c0 =
5
2
. (4.63)
Finally, the correlation function can be calculated
3
2
cos2 ϑ(0) −
1
2
3
2
cos2 ϑ(t) −
1
2
=
5
4
1ˆ
−1
du0
1ˆ
−1
du
(3u2
0 − 1)2
4
(3u2 − 1)2
4
e−6Drot
t
=
1
5
e−6Drot
t
. (4.64)
We have derived that the time correlation function for spherically symmetric rotational
diffusion is a single exponential function. Analytical solutions are also available
(but more difficult to derive) for axially symmetric and asymmetric rotational diffusion,
with the time correlation function in a form of three and five exponential functions,
respectively.
For spherically symmetric rotational diffusion, described by a mono-exponential function
characterized by the rotational correlation time τc,
R0 = b2
∞ˆ
0
1
5
e−t/τc
dt =
b2
5
τc =
b2
5
1
6Drot
, (4.65)
4.3. RETURN TO EQUILIBRIUM 35
where Drot
is the rotational diffusion coefficient, given by the Stokes’ law
kBT
8πη(T)r3
, (4.66)
where r is the radius of the spherical particle, T is the temperature, and η(T) is the
dynamic viscosity of the solvent, strongly dependent on the temperature.3
4.3 Return to equilibrium
After introducing the correlation function, we can repeat the analysis using the same
simplifications (rigid molecule, isotropic liquid), but taking the transverse (perpendicular)
field fluctuations into account.
dµx
dt
= ωyµz − ωzµy (4.68)
dµy
dt
= ωzµx − ωxµz (4.69)
dµz
dt
= ωxµy − ωyµx (4.70)
Expressing ωx as bΘ⊥ cos ϕ and ωy as bΘ⊥ sin ϕ, where
b = −2γB0δa (4.71)
Θ⊥
=
3
2
sin ϑ cos ϑ, (4.72)
gives
dµx
dt
= (bΘ⊥
sin ϕ)µz − (ω0 + bΘ )µy (4.73)
dµy
dt
= (ω0 + bΘ )µx − (bΘ⊥
cos ϕ)µz (4.74)
dµz
dt
= (bΘ⊥
cos ϕ)µy − (bΘ⊥
sin ϕ)µx, (4.75)
Introducing µ+ = µx + iµy and µ− = µx − iµy results in
dµ+
dt
= −ibΘ⊥
eiϕ
µz + i(ω0 + bΘ )µ+
(4.76)
dµ−
dt
= ibΘ⊥
e−iϕ
µz − i(ω0 + bΘ )µ−
(4.77)
dµz
dt
=
i
2
bΘ⊥
e−iϕ
µ+
− eiϕ
µ−
, (4.78)
3
Dynamic viscosity of water can be approximated by
η(T) = η0 × 10T0/(T −T1)
, (4.67)
where η0 = 2.414×10−5
kg m−1
s−1
, T0 = 247.8 K, and T1 = 140 K (Al-Shemmeri, T., 2012. Engineering
Fluid Mechanics. Ventus Publishing ApS. pp. 1718.).
36 CHAPTER 4. RELAXATION
In a coordinate frame rotating with ω0,
d(µ+)rot
dt
= −ibΘ⊥
ei(ϕ−ω0t)
µz + ibΘ (µ+
)rot (4.79)
d(µ−)rot
dt
= ibΘ⊥
e−i(ϕ−ω0t)
µz − ibΘ (µ−
)rot (4.80)
dµz
dt
=
i
2
bΘ⊥
e−i(ϕ−ω0t)
(µ+
)rot − ei(ϕ−ω0t)
(µ−
)rot , (4.81)
Note that now the transformation to the rotating frame did not remove ω0 completely, it survived in the exponential
terms.
Again, the set of differential equations cannot be solved because Θ , Θ⊥, and ϕ fluctuate in time, but we can analyze
the evolution in time steps short enough to keep Θ , Θ⊥, and ϕ constant.
µ+
1 = µ+
0 + ∆µ+
1 = [1 + i(ω0 + bΘ1)∆t]µ+
0 − ibΘ⊥
1 ∆tei(ϕ1−ω0t1)
µz,0 (4.82)
µ−
1 = µ−
0 + ∆µ−
1 = [1 − i(ω0 + bΘ1)∆t]µ−
0 + ibΘ⊥
1 ∆te−i(ϕ1−ω0t1)
µz,0 (4.83)
µz,1 = µz,0 + ∆µz,1 = µz,0 −
i
2
bΘ⊥
1 ∆te−i(ϕ1−ω0t1)
µ+
0 +
i
2
bΘ⊥
1 ∆tei(ϕ1−ω0t1)
µ−
0 . (4.84)
The µ+, µ−, and µz,0 are now coupled which makes the step-by-step analysis much more complicated. Instead of
writing the equations, we just draw a picture (Figure 4.2) similar to Fig. 4.1. Derivation of the values of relaxation rates
follows the procedure described for the parallel fluctuations (Eqs. 4.21–4.26). As the number of possible pathways in
Fig. 4.2 is very high, already the list of the terms proportional to ∆t and ∆t2 is very long. Fortunately, we are not
interested in evolution of magnetic moments in individual molecules, described in Fig. 4.2. The values of Θ1, Θ⊥
1 , ϕ1,
etc. are different for each molecule and we are interested in what we get after averaging results of multiplications for all
molecules (all possible orientations). In order to avoid writing the long expressions for magnetic moments of individual
molecules, we skip steps corresponding to Eqs. 4.21–4.25 and jump directly to the calculation of the evolution of total
magnetization (corresponding to Eq. 4.26).
Let us start with the terms proportional to ∆t, which give us the imaginary term proportional to b when calculating
dM+/dt (and dM−/dt, dMz/dt). We have already seen that the average of Θ (the green segment) is zero. The terms
containing Θ⊥ (red and blue segments) contain the exponential expression with the phase including ϕ. If the azimuth ϕ
is random4, the ”red” and ”blue” terms average to zero.
Let us now turn to the terms proportional to ∆t2, which give us the time integral multiplied by b2 when calculating
dM+/dt (and dM−/dt, dMz/dt). The pathways containing two red segments or two blue segments correspond to ∆t2
terms with a random phase in the exponent (random sums of ϕj − ω0tj). When averaged for all orientations, such phases
tend to zero. The ∆t2 terms do not average to zero only in two cases: (i) if the pathway contains two green segments
(effect of longitudinal fluctuations described above) or (ii) if the pathway contains a combination of one red and one blue
segment. The former case is obvious, but the latter one is more subtle.
We can distinguish two combinations of one red and one blue segment:
1
2
b2
∆t2
Θ⊥
k ei(ϕk−ω0tk)
Θ⊥
j e−i(ϕj −ω0tj )
=
1
2
b2
∆t2
Θ⊥
k Θ⊥
j ei(ϕk−ϕj −ω0(tk−tj ))
(4.85)
(with −ω0(tk − tj) in the exponent) and
1
2
b2
∆t2
Θ⊥
k e−i(ϕk−ω0tk)
Θ⊥
j ei(ϕj −ω0tj )
=
1
2
b2
∆t2
Θ⊥
k Θ⊥
j ei(−ϕk+ϕj +ω0(tk−tj ))
(4.86)
(with −ω0(tk − tj) in the exponent). As discussed in Section 4.2, we can replace tk by zero and tj by t because the
molecular motions are random:
1
2
b2
∆t2
Θ⊥
(0)Θ⊥
(t)ei(−(ϕ(t)−ϕ(0))+ω0t))
(4.87)
(with +ω0t in the exponent) and
1
2
b2
∆t2
Θ⊥
(0)Θ⊥
(t)ei(+(ϕ(0)−ϕ(t))+ω0t))
(4.88)
4
Note that this is true even in the presence of B0 and in molecules aligned along the direction of B0,
for example in liquid crystals oriented by the magnetic field.
4.3. RETURN TO EQUILIBRIUM 37
µ+
0
ibΘ1∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
1
∆te
−i(ϕ1−ω0t1)
µ+
1
ibΘ2∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
2
∆te
−i(ϕ2−ω0t2)
µ+
2
ibΘ3∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
3
∆te
−i(ϕ3−ω0t3)
µ+
3
ibΘ4∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
4
∆te
−i(ϕ4−ω0t4)
µ+
4 · · ·
ibΘk∆t
1
e
e
e
e
e
e
e
e
e
e
e
e
e
−i
b
2
Θ
⊥
k
∆te
−i(ϕk
−ω0tk
)
µ+
k
µz,0
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
1∆tei(ϕ1−ω0t1)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
1
∆te
−i(ϕ1−ω0t1)
µz,1
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
2∆tei(ϕ2−ω0t2)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
2
∆te
−i(ϕ2−ω0t2)
µz,2
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
3∆tei(ϕ3−ω0t3)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
3
∆te
−i(ϕ3−ω0t3)
µz,3
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
4∆tei(ϕ4−ω0t4)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
4
∆te
−i(ϕ4−ω0t4)
µz,4 · · ·
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
−ibΘ⊥
k∆tei(ϕk−ω0tk)
e
e
e
e
e
e
e
e
e
e
e
e
e
ibΘ
⊥
k
∆te
−i(ϕk
−ω0tk
)
µz,k
µ−
0
−ibΘ1∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
1∆tei(ϕ1−ω0t1)
µ−
1
−ibΘ2∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
2∆tei(ϕ2−ω0t2)
µ−
2
−ibΘ3∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
3∆tei(ϕ3−ω0t3)
µ−
3
−ibΘ4∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
4∆tei(ϕ4−ω0t4)
µ−
4 · · ·
−ibΘk∆t
1¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
ib
2Θ⊥
k∆tei(ϕk−ω0tk)
µ−
k
Figure 4.2: Evolution of magnetic moments due to longitudinal (parallel) and transverse (perpendicular)
fluctuations of magnetic fields. The meaning of the diagram is the same as in Fig. 4.1, but
additional segments (red and blue) interconnect µ+
j , µ−
j , and µz,j., substantially increasing the number
of possible pathways. The pathway composed of the black segments only gives the result of multiplication
equal to one, the pathways containing just one segment of a different color give results of
multiplication proportional to ∆t, the pathways containing two segments of a color different than black
segments give results of multiplication proportional to (∆t)2
, etc.
38 CHAPTER 4. RELAXATION
(with −ω0t in the exponent).
In both cases, the phase is not randomly distributed for different orientations only if ϕ(0) − ϕ(t) is similar to ω0t.
The average value of Θ⊥(0)2 is 3/10:
Θ⊥(t)2 =
9
4
cos2 ϑ sin2 ϑ =
9
16π
2πˆ
0
dϕ
πˆ
0
dϑ(sin3
ϑ cos2
ϑ) =
3
10
(4.89)
for any t.
The Mz component of magnetization is given by the average of the µz components at tk. In order to get to µz,k
through paths giving terms proportional to ∆t2, we must start at µz,0 and pass one blue segment and one red segment
in Figure 4.2. Eqs. 4.87 and 4.88 mathematically describe that orientations of magnetic moments are redistributed if the
molecular motions (described by the azimuth ϕ) accidently resonate for a short time with the frequencies ω0t and −ω0t.
Then the magnetic energy of the magnetic moments is exchanged with the rotational kinetic energy of the molecules.
This energy exchange must be taken into account when we average magnetic moments of individual molecules to calculate
Mz. Let us call the total rotational energy of molecules Erot
0 . The exchange of the magnetic energy Eµ of a magnetic
moment µ with a small amount of rotational energy of molecules ∆Erot can be described as
Erot
0 → Erot
0 + ∆Erot
+ Eµ. (4.90)
The molecular motions have much more degrees of freedom (both directions of rotational axes and rates of rotation
vary) than the magnetic moments (size is fixed, only orientation changes). We can therefore assume that the exchange perturbs
distribution of the magnetic moments, but the rotating molecules stay very close to the termodynamic equilibrium.
At the equilibrium, the probability to find a molecule with the rotational kinetic energy Erot
0 + ∆Erot is proportional to
(Boltzmann law) to
e−∆Erot
≈ 1 − ∆Erot
. (4.91)
The conservation of energy requires
Erot
0 + ∆Erot
+ Eµ = Erot
0 , (4.92)
showing that ∆Erot = −Eµ. Consequently, the population of molecules with the given rotational energy is proportional
to 1 − ∆Erot = 1 + Eµ. According to Eq. 3.3, the probability of finding a magnetic moment in the orientation
described by a given u = cos ϑµ is
Peq
(u) =
w
ew − e−w
euw
≈
w
1 − w − 1 + w
(1 + uw) =
1
2
(1 + uw). (4.93)
Consequently, Eµ = −uw = 1 − 2Peq(u) and the probability to find a molecule with the rotational kinetic energy
Erot
0 + ∆Erot is proportional to
1 − ∆Erot
= 1 + Eµ = 2 − 2Peq
(u) = 2(1 − Peq
(u)), (4.94)
where the factor of two can be absorbed to the normalization constant.
We have derived that the averaged values of µz are weigted by 1 − Peq(u). How does it affect calculation of Mz? In
the expression µz − Peq(u)µz, µz in the first term is not weighted by anything and its average (multiplied by the number
of magnetic moments per unit volume) is equal to Mz. The average value of the second term has been already calculated
in Eqs. 3.7–3.12. It represents the quilibrium value of the magnetization, Meq. Therefore, averaging of µz results in
Mz − Meq, usually abbreviated as ∆Mz.
Using the same arguments as in Section 4.2,
d∆Mz
dt
= −

 1
2
b2
∞ˆ
0
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt +
1
2
b2
∞ˆ
0
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt

 ∆Mz (4.95)
The relaxation rate R1 for Mz, known as longitudinal relaxation rate in the literature, is the real part of the expression
in the parentheses
R1 = b2



∞ˆ
0
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt +
∞ˆ
0
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt



(4.96)
4.3. RETURN TO EQUILIBRIUM 39
If the fluctuations are random and their statistical properties do not change in time, they are stationary: the current
orientation of the molecule is correlated with the orientation in the past in the same manner as it is correlated with the
orientation in the future. Therefore,
∞ˆ
0
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt =
1
2


∞ˆ
0
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt +
0ˆ
−∞
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt


(4.97)
=
1
2
∞ˆ
−∞
Θ⊥(0)Θ⊥(t)e−i(ϕ(t)−ϕ(0))eiω0t
dt. (4.98)
∞ˆ
0
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt =
1
2


∞ˆ
0
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt +
0ˆ
−∞
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt


(4.99)
=
1
2
∞ˆ
−∞
Θ⊥(0)Θ⊥(t)ei(ϕ(t)−ϕ(0))e−iω0t
dt. (4.100)
In isotropic solutions, the motions of molecules are very little affected by magnetic fields. Therefore, the choice
of the z axes is arbitrary form the point of the view of the molecule (not of the magnetic moment!). Therefore, the
terms with Θ⊥ can be replaced by those with Θ , multiplied by 3/2 to match the difference between Θ (0)2 = 1/5 with
Θ⊥(0)2 = 3/10:
1
2
∞ˆ
−∞
Θ⊥(0)Θ⊥(t)e i(ϕ(t)−ϕ(0))e±iω0t
dt =
3
4
∞ˆ
−∞
Θ (0)Θ (t)e±iω0t
dt. (4.101)
Real parts of the integrals in the right-hand side of Eq. 4.101 are known as spectral density functions J(ω). Note
that the integral in Eq. 4.32 in Section 4.2 can be also included in the definition of the spectral density function if we
replace ω0 by zero:
∞ˆ
0
Θ (0)Θ (t)dt =
1
2


∞ˆ
0
Θ (0)Θ (t)dt +
0ˆ
−∞
Θ (0)Θ (t)dt

 =
1
2
∞ˆ
−∞
Θ (0)Θ (t)e0
dt =
1
2
J(0). (4.102)
The relaxation rate R1 can be therefore written as
R1 =
3
4
b2 1
2
J(ω0) +
1
2
J(−ω0) ≈
3
4
b2
J(ω0). (4.103)
Let us now turn to M+. Its value is given by the average of µ+ components at tk. The analysis of Figure 4.2 showed
that the relaxation-relevant ∆t2 terms are obtained by walking through paths containing two green segments (leading to
Eq. 4.32 in Section 4.2) or one blue segment and one red segment (which gave us Eq. 4.95 in this section). In any case,
we must start at µ+
0 . As we have already analyzed both relevant types of pathways in Figure 4.2, we can directly write
the result
R2 =
1
2
b2
J(0) +
3
8
b2
J(−ω0). (4.104)
Similarly, relaxation of M− is given by
R2 =
1
2
b2
J(0) +
3
8
b2
J(ω0). (4.105)
40 CHAPTER 4. RELAXATION
The longitudinal relaxation rate R1, describing the decay of ∆Mz due to the chemical
shift anisotropy in randomly reorienting molecules, and the transverse relaxation
rate R2, describing the decay of transverse magnetization M+ (or M−) are given by
R1 =
3
4
b2
J(ω0), (4.106)
R2 =
1
2
b2
J(0) +
3
8
b2
J(ω0). (4.107)
4.4 Internal motions, structural changes
So far, we analyzed only the rigid body motions of molecules, assuming that the structures
of molecules are rigid. What happens if the structure of the molecule changes? Let
us first assume that the structural changes are random internal motions which change
orientation of the chemical shift tensor relative to the orientation of the whole molecule,
but do not affect its size or shape. Then, Eq. 4.21 can be still used and R0 is still given
by Eq. 4.33, but the correlation function is not mono-exponential even if the rotational
diffusion of the molecule is spherically symmetric. The internal motions contribute to
the dynamics together with the rotational diffusion, and in a way that is very difficult
to describe exactly. Yet, useful qualitative conclusions can be made.
• If the internal motions are much faster than rotational diffusion, correlation between
Θ (tk) and Θ (tj) is lost much faster. The faster the correlation decays,
the lower is the result of integration. The internal motions faster than rotational
diffusion always decrease the value of R0 (make relaxation slower). Amplitude and
rate of the fast internal motions can be estimated using approximative approaches.
• If the internal motions are much slower than rotational diffusion, the rate of decay
of the correlation function is given by the faster contribution, i.e., by the rotational
diffusion. The internal motions slower than rotational diffusion do not change the
value of R0. Amplitude and rate of the fast internal motions cannot be measured
if the motions do not change size or shape of the diffusion tensor.
If the structural changes alter size and/or shape of the chemical shift tensor,5
parameters
ω0j and bj in Eq. 4.17 vary and cannot be treated as constants. E.g., the
parameter ω0j is not absorbed into the frequency of the rotating coordinate frame and
terms ω0(tk)ω0(tk − tj) contribute to R0 even if a(tk)a(tk − tj) decays much slower than
Θ (tk)Θ (tk − tj).
5
Examples of such changes are internal motions changing torsion angles and therefore distribution
of electrons, or chemical changes (e.g. dissociation of protons) with similar effects.
4.5. BLOCH EQUATIONS 41
• Internal motions or chemical processes changing size and/or shape of the chemical
shift tensor may have a dramatic effect on relaxation even if their frequency is much
slower than the rotational diffusion of the molecule. If the molecule is present in
two inter-converting states (e.g. in two conformations or in a protonated and
deprotonated state), the strongest effect is observed if the differences between
the chemical shift tensors of the states are large and if the frequency of switching
between the states is similar to the difference in γB0δa of the states. Such processes
are known as chemical or conformational exchange and increase the value of R0
and consequently R2.
4.5 Bloch equations
The effects of relaxation can be included in the equations describing evolution of the
bulk magnetization (Eqs. 3.47–3.49). The obtained set of equations, known as Bloch
equations, provides a general macroscopic description of NMR for proton and similar
nuclei.
dMx
dt
= −R2Mx − ΩMy + ω1 sin ϕMz, (4.108)
dMy
dt
= +ΩMx − R2My − ω1 cos ϕMz, (4.109)
dMz
dt
= −ω1 sin ϕMx + ω1 cos ϕMy − R1(Mz − Meq
z ). (4.110)
(4.111)
How does the incorporation of the relaxation effects influence the solution of the Bloch equations? In the single-pulse
experiment, irradiation by the radio waves is usually short and the relaxation can be neglected. Therefore the values
Mx(0) and My(0) obtained by solving Eqs. 3.47–3.49 can be used as the initial conditions for solving the Bloch eqations.
The evolution of magnetization during the relatively long period of signal acquisition in the absence of B1 (i.e., after
turning off the radio waves), when the relaxation cannot be neglected, is given by
dMx
dt
= −R2Mx − ΩMy, (4.112)
dMy
dt
= ΩMx − R2My, (4.113)
dMz
dt
= 0. (4.114)
The same trick can be applied as when solving Eqs. 3.52–3.54:
d(Mx + iMy)
dt
= Ω(−My + iMx) = (+iΩ − R2)(Mx + iMy), (4.115)
d(Mx − iMy)
dt
= Ω(−My − iMx) = (−iΩ − R2)(Mx − iMy), (4.116)
42 CHAPTER 4. RELAXATION
Mx + iMy = C+e(+iΩ−R2)t
= Mx(0) + My(0)e−R2t
e(+iΩ+φ0)t
, (4.117)
Mx − iMy = C−e(−iΩ−R2)t
= Mx(0) + My(0)e−R2t
e(−iΩ+φ0)t
, (4.118)
Mx = Mx(0) + My(0)e−R2t
cos(Ωt + φ0) (4.119)
My = Mx(0) + My(0)e−R2t
sin(Ωt + φ0). (4.120)
As a result of relaxation, the detected NMR signal does not oscillate as a cosine
(or sine) function with a constant amplitude, but decays exponentially, with the rate
constant of the decay equal to R2. Such a signal is usually described as the free induction
decay (FID).
Chapter 5
Signal acquisition and processing
Literature: Function of an NMR spectrometer is nicely described in L4, K13, or
C3.1. More details are provided in B23. Experimental setup is discussed in C3.8.2.
Signal averaging is described in L5.2, quadrature detection in L5.7 and LA.5, K13.6,
and C3.2.3, Fourier transformation is introduced in K5.1–K5.3.1 and L5.8.1.–L5.8.3,
and treated moro thoroughly in B8 and C3.3.1. Phase correction is described nicely in
K5.3.2–K5.3.4 and discussed also in C3.3.2.3 and L5.8.4–L5.8.5, zero filling is discussed
in C3.3.2.1 and K5.5, and apodization is explained in K5.4 and C3.3.2.2.
5.1 NMR experiment
The real NMR experiment closely resembles FM radio broadcast. The mega-hertz radio
frequency ωradio plays the role of the carrier frequency, and is frequency-modulated by
the offset, which usually falls in the range of kilo-hertz audio frequencies. In the same
fashion, the carrier frequency of the FM broadcast is modulated by the audio frequency
of the transmitted signal (voice, music). Like when listening to the radio, we need to
know the carrier frequency to tune the receiver, but its value is not interesting. The
interesting information about the chemical environment is hidden in the audio-frequency
offset. Note, however, that the numerical value of Ω is arbitrary as it depends on the
actual choice of the carrier frequency. What can be interpreted unambiguously, is the
constant δ, given just by the electron density. But in practice, the absolute value of δ
is extremely difficult to obtain because the reference δ = 0 represents nuclei with no
electrons – definitely not a sample we are used to produce in our labs. Therefore, more
accessible references (precession frequencies ωref of stable chemical compounds) are used
instead of the vacuum frequency. The value of δ is than defined as (ω − ωref)/ωref and
usually presented in the units of ppm.
43
44 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
5.1.1 Setting up the experiment
• Temperature control and calibration. Temperature affects molecular motions and
chemical shits, it should be controlled carefully to obtain reproducible spectra and
to analyze them quantitatively. The sample temperature is controlled by a flow of
pre-heated/cooled air or nitrogen gas. The exact temperature inside the sample is
not so easy to measure. Usually, spectra of compounds with known temperature
dependence of chemical shifts are recorded (e.g. methanol). The temperature is
obtained by comparing a difference of two well defined chemical shifts (of methyl
and hydroxyl protons in the case of methanol) with its values reported for various
temperatures. Purity of the standard samples is a critical issue.
• Field-frequency lock. The external magnetic field should be stationary. It is
achieved by a feedback system known as field-frequency lock. A deuterated compound
(usually heavy water or other deuterated solvent) is added to the sample
and the deuterium frequency is measured continually and kept constant by adjusting
electric current in an auxiliary electromagnet. The lock parameters for the
particular deuterium compound used are selected and the deuterium spectrometer
is switched on before the measurement.
• Shimming. The external magnetic field should be also homogeneous. The inhomogeneities
caused e.g. by the presence of the sample are compensated by adjusting
electric current in a set of correction coils called shims. This is usually at least
partially automated.
• Tuning. Each radio-frequency circuit in the probe consists of a receiver coil and
two adjustable capacitors. The capacitors should be adjusted for each sample.
The tuning capacitor of the capacitance CT and the coil of the inductance L
make an LC circuit, acting as a resonator. Adjusting the value of CT defines
the resonant frequency, which should be equal to the precession frequency of the
measured nucleus ω0. If we neglect the second capacitor, the resonant frequency
is ω = 1/
√
LCT. The second, matching capacitor of the capacitance CM is used
to adjust the impedance of the resonator. The radio waves do not travel from
the transmitter to the coil through air but through co-axial cables. In order to
have minimum of the wave reflected back to the transmitter, the impedance of the
resonator should match the input impedance Zin.
The impedance of the coil circuit is given by
Zc =
1
1
ZM
+ 1
ZT+ZL+R
=
1
iωCM + 1
1
iωCT
+ iωL + R
.
In order to tune the circuit, CT and CM must be adjusted simultaneously to get
(i) Zc = Zin and (ii) ω = ω0.
5.1. NMR EXPERIMENT 45
• Calibration of pulse duration. The magnitude of B1 cannot be set directly. Therefore,
the duration of irradiation rotating M by 360 ◦
at the given strength of radio
waves is searched for empirically. This duration is equal to 2π/ω1 and can be used
to calculate ω1 or |B1| = ω1/γ. As |B1| is proportional to the square root of power
P, durations of pulses of radio waves of other strengths need not be calibrated,
but can be recalculated.
Power is measured in the units of Watt, but the relative power is usually expressed on a logarithmic scale in decibells
(dB). One Bell represents a ten-fold attenuation of power
log10
P2
P1
= attenuation/B. (5.1)
Consequently,
10 log10
P2
P1
= attenuation/dB (5.2)
and
20 log10
P2
2
P2
1
= 20 log10
|B1|2
2
|B1|2
1
= 10 log10
|B1|2
|B1|1
= attenuation/dB. (5.3)
5.1.2 Quadrature detection
Precession of the magnetization vector in the sample induces a signal oscillating with
the same frequency (Larmor frequency ω0) in the coil of the NMR probe. The signal
generated in the coil and amplified in the preamplifier is split into two channels. The
signal in each channel is mixed with a reference wave supplied by the radio-frequency
synthesizer. The reference waves have the same frequency ωref in both channels, but their
phases are shifted by 90 ◦
. Let us assume that the signal oscillates as a cosine function
cos(ω0t) and that the reference wave in the first channel is a cosine wave cos(ωreft) and
that the reference wave in the second channel is a sine wave − sin(ωreft).
Mathematically, the procedure can be described as
cos(ω0t) →
1
2
cos(ω0t) → 1
2
cos(ω0t) cos(ωref t)
1
2
cos(ω0t) → −1
2
cos(ω0t) sin(ωref t)
(5.4)
Basic trigonometric identities show that the result of mixing in the first channel is a sum of a high-frequency cosine
wave cos((ω0 + ωref )t) and a low-frequency cosine wave cos((ω0 − ωref )t), while the result of mixing in the second channel
is a difference of the corresponding sine waves:
1
2
cos(ω0t) cos(ωref t) =
1
4
cos((ω0 + ωref )t) +
1
4
cos((ω0 − ωref )t), (5.5)
−
1
2
cos(ω0t) sin(ωref t) = −
1
4
sin((ω0 + ωref )t) +
1
4
sin((ω0 − ωref )t). (5.6)
The high-frequency waves are filtered out by a low-pass filter, resulting in signals oscillating with a low frequency
ω0 − ωref . If ωref = −ωradio, then ω0 − ωref = Ω. The procedure, similar to the demodulation in an ordinary radio
receiver, thus produces audio signals in both channels
cos(ω0t) →
1
2
cos(ω0t) → 1
2
cos(ω0t) cos(ωref t) → 1
4
cos(Ωt)
1
2
cos(ω0t) → −1
2
cos(ω0t) sin(ωref t) → 1
4
sin(Ωt)
(5.7)
It is convenient to treat the signals in the individual channels as a real and imaginary
component of a single complex number, denoted y(t) in this text:
46 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
y(t) =
1
4
cos(Ωt) + i
1
4
sin(Ωt) =
1
4
eiΩt
. (5.8)
5.1.3 Analog-digital conversion
The output of the quadrature receiver is converted to a digital form. Therefore, the
information obtained from an NMR experiment is a set of complex numbers describing
the signal intensities at the time points t ∈ {0, ∆t, 2∆t, · · · , (N − 1)∆t}.
5.1.4 Signal averaging and signal-to-noise ratio
The NMR signal induced by precession of the magnetization vector is very weak, comparable
to the noise, generated mostly by random motions of electrons in the receiver coil.
Therefore, the NMR experiments are usually repeating several times, adding the signal
together. If the experiment is repeated in the same manner N-times, the evolution of the
magnetization vector is identical in all repetitions (magnetization is evolving coherently),
and the sum of the signals from the individual measurements, called transients, is simply
Ny(t). However, the absolute size of the signal is not important, what really matters
is the signal-to-noise ratio. Therefore, it is also important how noise accumulates when
adding signals of separate measurements.
The noise n(t) is random and so its average1 n(t) = 0. The size of the noise is typically defined by the root-meansquare
n(t)2 . Sum of the noise from N independent experiments is
(n1(t) + n2(t) + · · · + nN (t))2
. (5.9)
Because the random motions of electrons in the individual experiments are not correlated (are independent), all
terms like 2n1(t)n2(t) are equal to zero. Therefore, calculation of the square in Eq. 5.9 simplifies to
(n1(t) + n2(t) + · · · + nN (t))2
= n1(t)2 + n2(t)2 + · · · + nN (t)2 . (5.10)
We can also assume that the root-mean-square is the same in all experiments, and write it as n(t)2 . The sum of
the noise can be then calculated as
N n(t)2 =
√
N n(t)2 . (5.11)
We can now calculate the signal-to-noise ratio as
Ny(t)
√
N n(t)2
=
√
N
y(t)
n(t)2
. (5.12)
The signal-to-noise ratio is proportional to the square root of the number of summed
transients.
1
To avoid writing the integrals defining averaging, we indicate the time average by the angled brack-
ets.
5.2. FOURIER TRANSFORMATION 47
5.2 Fourier transformation
The effect of electrons (chemical shift) makes NMR signal much more interesting but also
much more complicated. Oscillation of the voltage induced in the receiver coil is not described
by a cosine function, but represents a superposition (sum) of several cosine curves
(phase-shifted and dumped). It is practically impossible to get the frequencies of the
individual cosine functions just by looking at the recorded interferograms. Fortunately,
the signal acquired as a function of time can be converted into a frequency dependence
using a straightforward mathematical procedure, known as Fourier transformation.
It might be useful to present the basic idea of the Fourier transformation in a pictorial
form before we describe details of Fourier transformation by mathematical equations.
The oscillating red dots in Figure 5.1 represent an NMR signal defined by one frequency
ν. Let us assume that the signal oscillates as a cosine function but we do not know the
frequency. We generate a testing set of cosine functions of different known frequencies fj
(blue curves in Figure 5.1) and we multiply each blue testing function by the red signal.
The resulting product is plotted as magenta dots in Figure 5.1. Then we sum the values
of the magenta points for each testing frequency getting one number (the sum) for each
blue function. Finally, we plot these numbers (the sums) as the function of the testing
frequency. How does the plot looks like? If the testing frequency differs from ν, the
magenta dots oscillate around zero and their sum is close to zero (slightly positive or
negative, depending on how many points were summed). But if we are lucky and the
testing frequency matches ν (f3 in Figure 5.1), the result is always positive (we always
multiply two positive numbers or two negative numbers). The sum is then also positive,
the larger the more points are summed. Therefore, the sum for the matching frequency
is much higher than the other sums, making a positive peak in the final green plot (the
dependence on fj). The final plot represents a frequency spectrum and the position of
the peak immediately identifies the value of the unknown frequency. If the NMR signal
is composed of two frequencies, the red dots oscillate in a wild interference patterns,
not allowing to get the frequency simply by measuring the period of the oscillation.
However, the individual components (if they are sufficiently different) just make several
peaks in the final green plot and their frequencies can be easily obtained by reading the
positions of the peaks.
Let us now try to describe the Fourier transformation in a bit more mathematical manner. We start with a special
case of a signal which can be described by a sum of cosine functions with frequencies that are integer multiples of some
small frequency increment ∆ω. All such cosine functions must have the same value at time t and t + 2π/∆ω: the whole
signal periodic with the period 2π/∆ω. If we record such a signal using quadrature detection, we obtain
y(t) =
∞
k=−∞
Akeiωkt
=
∞
k=−∞
Akeik∆ωt
. (5.13)
The mentioned periodicity allows us to determine Ak by calculating the integrals
2π
∆ωˆ
0
y(t)e−iωj t
dt =
∞
j=−∞
Aj
2π
∆ωˆ
0
ei(k−j)∆ωt
dt =
2π
∆ω
Ak (5.14)
48 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
f1 f2 f3 f4 f5
ν = f3
t t
ν = ?
f5
Σ = 0
f4
Σ = 0
f3
Σ = 50
f2
Σ = 0
f1
Σ = 0
Figure 5.1: The basic idea of Fourier transformation.
(All integrated functions are periodic and their integrals are therefore equal to zero with the exception of the case
when k = j, which is a constant function).
The same result is obtained for any integration limits which differ by 2π/∆ω, e.g.
+ π
∆ωˆ
− π
∆ω
y(t)e−iωj t
dt =
∞
j=−∞
Aj
+ π
∆ωˆ
− π
∆ω
ei(k−j)∆ωt
dt =
2π
∆ω
Ak (5.15)
We can now continue in two different directions. We can describe the signal as it is actually measured, not as a
continuous function of time, but as a discrete series of points sampled in time increments ∆t. Then, the integral in
Eq. 5.14 is replaced by summation of a finite number of measured signal points:
Yk =
N−1
j=0
yje−ik∆ωj∆t
∆t, (5.16)
where Yk = 2π
∆ω
Ak. As the time and frequency are treated in the same manner, we can also define the inverse
operation
yj =
N−1
k=0
Ykeik∆ωj∆t
∆ω. (5.17)
This way of the signal analysis, discussed in more details in Section 5.2.4, handles the signal as it is measured in
reality. It is also instructive to follow the other direction and to increase the period 2π/∆ω by decreasing ∆ω. The series
of ωk becomes a continuous variable ω and π/∆ω → ∞ if ∆ω → 0. The sum in Eq. 5.13 is replaced by the integral
y(t) =
1
2π
∞ˆ
−∞
Y (ω)eiωt
dω (5.18)
5.2. FOURIER TRANSFORMATION 49
and the integral in Eq. 5.15 becomes
Y (ω) =
∞ˆ
−∞
y(t)e−iωt
dt. (5.19)
If we apply Eq. 5.19 to a function y(t) and Eq. 5.18 to the obtained result, we should get back the function y(t).
Such a double transformation can be written as
y(t) =
1
2π
∞ˆ
−∞
Y (ω)eiωt
dω =
1
2π
∞ˆ
−∞
eiωt
dω
∞ˆ
−∞
y(t )e−iωt
dt =
∞ˆ
−∞
y(t )dt
1
2π
∞ˆ
−∞
eiω(t−t )
dω. (5.20)
This requires the second integral to be equal to 2π for t = t and to zero for t = t. Therefore, the integral can be
used to define the delta function
δ(t − t ) =
1
2π
∞ˆ
−∞
eiω(t−t )
dω. (5.21)
An alternative definition
Y (ω) =
1
√
2π
∞ˆ
−∞
y(t)e−iωt
dt, (5.22)
y(t) =
1
√
2π
∞ˆ
−∞
Y (ω)eiωt
dω. (5.23)
is equally acceptable.
Although the actual NMR signal is not recorded and processed in a continuous
manner, the idealized continuous Fourier transformation helps to understand the fundamental
relation between the shapes of FID and frequency spectra and reveals important
features of signal processing. Therefore, we discuss the continuous Fourier transformation
before we proceed to the discrete analysis.
5.2.1 Fourier transformation of an ideal NMR signal
An ”ideal signal” (see Figure 5.2) has the form y(t) = 0 for t ≤ 0 and y(t) = Ae−R2t
eiΩt
for t ≥ 0, where A can be a complex number (complex amplitude), including the real
amplitude |A| and the initial phase φ0:
A = |A|eφ0
. (5.24)
Y (ω) =
∞ˆ
−∞
y(t)e−iωt
dt =
∞ˆ
0
Ae(i(Ω−ω)−R2)t
dt =
−A
i(Ω − ω) − R2
= A
1
R2 − i(Ω − ω)
R2 + i(Ω − ω)
R2 + i(Ω − ω)
= A
R2 + i(Ω − ω)
R2
2 + (Ω − ω)2
(5.25)
50 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω
1/R22R2
{Y(ω)}
ω
Ω
Figure 5.2: Ideal signal detected with a quadrature detection (top) and its Fourier transform (bottom).
5.2. FOURIER TRANSFORMATION 51
Fourier transform of the ”ideal” signal is
Y (ω) =
∞ˆ
−∞
Ae−R2t
eiΩt
e−iωt
dt = A
R2
R2
2 + (Ω − ω)2
+ iA
Ω − ω
R2
2 + (Ω − ω)2
(5.26)
If φ0 = 0, the blue term, known as the absorption line is a real function ( {Y (ω)})
having a shape of the Lorentz curve (see Figure 5.2). The shape of the absorption line
is given2
by the relaxation rate R2:
• Peak height ∝ 1/R2 (Y = Ymax at ω = Ω ⇒ Ymax = Y (Ω) = A/R2)
• Linewidth at the half-height = 2R2 (Y = Ymax/2 at Ω − ω = ±R2)
The red term, the dispersion line, is purely imaginary ( {Y (ω)}) if φ0 = 0. Such
shape is less convenient in real spectra containing several lines because the broad wings
of the dispersion line distort the shape of the neighbouring lines (see Figure 5.2).
Figure 5.3 documents that Fourier transformation allows us to immediately determine
several Larmor frequencies in spectra even if the signal in the time domain (FID) is very
difficult to interpret, and that the real (absorption) part of the complex spectrum is
much better for such purpose.
The discussed transformation of a continuous signal is extremely useful for understanding
the relation between evolution of the magnetization vector and shape of the
peaks observed in the frequency spectra. But in reality, the signal is finite (tmax < ∞)
and discrete (∆t > 0):
• t ∈ {0, ∆t, 2∆t, · · · , (N − 1)∆t} y(t) ∈ {y0, y1, y2, · · · , yN−1}
• ω ∈ {0, ∆ω, 2∆ω, · · · , (N − 1)∆ω} Y (t) ∈ {Y0, Y1, Y2, · · · , YN−1}
The seemingly marginal difference between ideal and real (finite and discrete) signal
has several practical consequences, discussed below.
Figures 5.4 and 5.5 document the advantage of recording the signal with the quadrature
detection, as a complex number. If we take only the signal from the first channel,
oscillating as the cosine function if φ = 0, and stored as the real part if the quadrature
detection is used (Figure 5.4), and perform the Fourier transformation, we get a spectrum
with two peaks with the frequency offsets Ω and −Ω. Such a spectrum does not
tell us if the actual Larmor frequecy is ω0 = ωradio − Ω or ω0 = ωradio + Ω. If we use the
signal from the second channel only, oscillating as the sine function if φ = 0 (Figure 5.5),
a spectrum with two peaks is obtained again, the only difference is that the peaks have
2
In practice, it is also affected by inhomogeneities of the static magnetic field, increasing the apparent
value of R2. This effect is known as inhomogeneous broadening.
52 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω1
Ω2
Ω3
{Y(ω)}
ω
Ω1
Ω2 Ω3
Figure 5.3: Signal (top) and frequency spectrum (bottom) with three Larmor frequencies.
5.2. FOURIER TRANSFORMATION 53
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω−Ω
{Y(ω)}
ω
Ω−Ω
Figure 5.4: A signal detected in the first (”real”) channel (top) and its Fourier transform (bottom).
opposite phase (i.e., their phases differ by 180 ◦
). But if we combine both signals, the
false peaks at −Ω disappear because they have opposite signs and cancel each other in
the sum of the spectra.
5.2.2 Properties of continuous Fourier transformation
The continuous Fourier transformation has several important properties:
• Parseval’s theorem
∞´
−∞
|y(t)|2
dt = 1
2π
∞´
−∞
|Y (ω)|2
dω
A conservation law, documents that the signal energy (information content) is
preserved by the Fourier transformation.
• Linearity
∞´
−∞
(y(t) + z(t))e−iωt
dt = Y (ω) + Z(ω)
54 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω
−Ω
{Y(ω)}
ω
Ω−Ω
Figure 5.5: A signal detected in the second (”imaginary”) channel (top) and its Fourier transform
(bottom).
5.2. FOURIER TRANSFORMATION 55
It documents that a sum of periodic functions (difficult to be distinguished in the
time domain) can be converted to a sum of resonance peaks (easily distinguishable
in the frequency domain if the resonance frequencies differ).
• Convolution
∞´
−∞
(y(t) · z(t))e−iωt
dt =
∞´
−∞
Y (ω)Z(ω − ω )dω
It provides mathematical description of apodization (Section 5.5)
• Time shift
∞´
−∞
y(t − t0)e−iωt
dt = Y (ω)e−iωt0
It shows that time delays result in frequency-dependent phase shifts in the frequency
domain (Section 5.3)
• Frequency modulation
∞´
−∞
y(t)eiω0t
e−iωt
dt = Y (ω − ω0)
It shows that the apparent frequencies can be shifted after acquisition.
• Causality
∞´
−∞
y(t)e−iωt
dt =
∞´
0
y(t)e−iωt
dt
It says that no signal is present before the radio-wave pulse (this is why we can
start integration at t = 0 or t = −∞, y(t) = 0 for t < 0). This provides an extra
piece of information allowing us to reconstruct the imaginary part of the signal
from the real one and vice versa (Figure 5.6).
The mentioned consequence of causality is rather subtle. As mentioned above, the NMR signal is recorded in two
channels, as a real and imaginary part of a complex number. It is because Fourier transformation of a cosine (or sine)
function gives a symmetric (or antisymmetric) spectrum with two frequency peaks and thus does not allow us to distinguish
frequencies higher than the carrier frequency from those lower than the carrier frequency. Once we have the transformed
complex signal in the frequency domain, we can ask whether we need both its parts (real and imaginary). It looks like
we do because the inverse Fourier transformation of just the real (imaginary) part produces a symmetric (antisymmetric)
picture in the time domain (the second row in Figure 5.6). But the causality tells us that this is not a problem because
we know that there is no signal left from the zero time – the symmetry does not bother us because we know that we can
reconstruct the time signal simply by discarding the left half of the inverse Fourier image (the third row in Figure 5.6).
The time signal reconstructed from the real part of the frequency spectrum only, can be then Fourier transformed to
provide the missing imaginary part of the frequency spectrum.
5.2.3 Consequence of finite signal acquisition
In reality, the acquisition of signal stops at a finite time tmax:
Y (ω) =
tmaxˆ
0
Ae(i(Ω−ω)−R2)t
dt = A
1 − e−R2tmax
ei(Ω−ω)tmax
R2 − i(Ω − ω)
. (5.27)
It has some undesirable consequences:
Leakage: Part of the signal is lost, peak height Y (Ω) < A/R2.
56 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
{Y(ω)}
ω
Ω
{Y(ω)}
ω
1
2π
∞´
−∞
{Y(ω)}eiωt
dω
t
1
2π
∞´
−∞
{Y(ω)}eiωt
dω
t
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
Figure 5.6: Causality of NMR signal. If we take a frequency spectrum, discard its imaginary part (the
first row), and perform the inverse Fourier transformation, we do not get the original signal (starting
at t = 0), but a set of symmetric (real part) and antisymmetric (imaginary part) functions predicting
non-zero signal before t = 0 (the second row). However, we can apply our knowledge that no signal was
present before t = 0 and multiply the left half of the predicted signal by zero. This recovers the actual
signal (the third row). Fourier transformation of this signal provides both real and inmaginary parts of
the spectrum, as shown in Figure 5.2.
5.2. FOURIER TRANSFORMATION 57
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω
{Y(ω)}
ω
Ω
{Y(ω)}
ω
Figure 5.7: Effect of finite acqusition in the limit R2 → 0.
Truncation artifacts: For R2 → 0,
Y (ω) =
tmaxˆ
0
Ae(i(Ω−ω))t
dt = A
1 − ei(Ω−ω)tmax
−i(Ω − ω)
= A
sin(Ω − ω)tmax
Ω − ω
+iA
1 − cos(Ω − ω)tmax
Ω − ω
.
(5.28)
If the acquisition is stopped before the signal relaxes completely, artifacts (baseline
oscillation) appear. In the limit of no relaxation, the real part of the Fourier-transformed
signal does not have a pure absorption shape (Lorentz curve), but has a shape of the
sin(Ω − ω)tmax/(Ω − ω)tmax function (sinc function).
5.2.4 Discrete Fourier transformation
In reality, the acquired signal is finite (tmax < ∞) and discrete (∆t > 0):
• t ∈ {0, ∆t, 2∆t, · · · , (N − 1)∆t} y(t) ∈ {y0, y1, y2, · · · , yN−1}
58 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
• ω ∈ {0, ∆ω, 2∆ω, · · · , (N − 1)∆ω} Y (t) ∈ {Y0, Y1, Y2, · · · , YN−1}
We may try to define the discrete Fourier transform as
Yk =
N−1
j=0
yje−ik∆ωj∆t
∆t =
N−1
j=0
yje−i2π∆f∆tkj
∆t, (5.29)
yj =
N−1
k=0
Ykeik∆ωj∆t
∆t =
N−1
k=0
Ykei2π∆f∆tkj
∆f. (5.30)
However, there is a catch here. It turns out that ∆t and ∆f are not independent, but closely related. The transformation
can be written in a matrix form as







Y0
Y1
Y2
...
YN−1







=







F0,0 F0,1 F0,2 . . . F0,N−1
F1,0 F1,1 F1,2 . . . F1,N−1
F2,0 F2,1 F2,2 . . . F2,N−1
...
...
...
...
...
FN−1,0 FN−1,1 FN−1,2 . . . FN−1,N−1







ˆF







y0
y1
y2
...
yN−1







∆t, (5.31)
where the elements of the matrix ˆF are Fjk = e−i2π∆f∆t·k·j.
Let us now try to transform Yk back to the time domain:







y0
y1
y2
...
yN−1







=









F−1
0,0 F−1
0,1 F−1
0,2 . . . F−1
0,N−1
F−1
1,0 F−1
1,1 F−1
1,2 . . . F−1
1,N−1
F−1
2,0 F−1
2,1 F−1
2,2 . . . F−1
2,N−1
...
...
...
...
...
F−1
N−1,0 F−1
N−1,1 F−1
N−1,2 . . . F−1
N−1,N−1









ˆF −1







Y0
Y1
Y2
...
YN−1







∆f, (5.32)
where the elements of the matrix ˆF−1 are F−1
jk = e+i2π∆f∆t·k·j. Substituting from Eq. 5.31,







y0
y1
y2
...
yN−1







=









F−1
0,0 F−1
0,1 F−1
0,2 . . . F−1
0,N−1
F−1
1,0 F−1
1,1 F−1
1,2 . . . F−1
1,N−1
F−1
2,0 F−1
2,1 F−1
2,2 . . . F−1
2,N−1
...
...
...
...
...
F−1
N−1,0 F−1
N−1,1 F−1
N−1,2 . . . F−1
N−1,N−1
















F0,0 F0,1 F0,2 . . . F0,N−1
F1,0 F1,1 F1,2 . . . F1,N−1
F2,0 F2,1 F2,2 . . . F2,N−1
...
...
...
...
...
FN−1,0 FN−1,1 FN−1,2 . . . FN−1,N−1














y0
y1
y2
...
yN−1







∆f∆t.
(5.33)
In order to get the original signal, the product of the transformation matrices, ˆF−1 ˆF multiplied by ∆f∆t, must be
a unit matrix:









F−1
0,0 F−1
0,1 F−1
0,2 . . . F−1
0,N−1
F−1
1,0 F−1
1,1 F−1
1,2 . . . F−1
1,N−1
F−1
2,0 F−1
2,1 F−1
2,2 . . . F−1
2,N−1
...
...
...
...
...
F−1
N−1,0 F−1
N−1,1 F−1
N−1,2 . . . F−1
N−1,N−1
















F0,0 F0,1 F0,2 . . . F0,N−1
F1,0 F1,1 F1,2 . . . F1,N−1
F2,0 F2,1 F2,2 . . . F2,N−1
...
...
...
...
...
FN−1,0 FN−1,1 FN−1,2 . . . FN−1,N−1







∆f∆t =







1 0 0 . . . 0
0 1 0 . . . 0
0 0 1 . . . 0
...
...
...
...
...
0 0 0 . . . 1







.
(5.34)
According to the matrix multiplication rule, the jl-element of the product ˆF−1 ˆF is given by
N−1
k=0
e−i2π∆f∆t(jk−kl)
∆t. (5.35)
5.2. FOURIER TRANSFORMATION 59
Clearly, the exponential terms in the sums representing the diagonal elements (j = l) are equal to e−i2π∆f∆t(jk−kj)∆t =
e0 = 1. Therefore, the diagonal elements (sums of N terms e0 = 1) are equal to N. Obviously, we need to set N∆f∆t = 1
to get the elements of the product ˆF−1 ˆF equal to one.
What about the off-diagonal elements? For N∆f∆t = 1, the elements of ˆF−1 ˆF are equal to
N−1
k=0
e−i 2π
N
(j−l)k
∆t. (5.36)
The complex numbers in the sum can be visualized as points in the Gauss plane (plane of complex numbers) with
the phase of 2πk(l − j)/N. Let us assume that N is an integer power of two (N = 2n, a typical choice in discrete Fourier
transform). Then all numbers in the series are symmetrically distributed in the Gauss plane. As a consequence, their sum
is equal to zero (they cancel each other). We can therefore conclude that setting N∆f∆t = 1 ensures that the product
ˆF−1 ˆF is a unit matrix.
The consequences of the requirement ∆f∆t = 1/N are:
• spectral width N∆f = 1/∆t, it is defined by the choice of the time increment
• digital resolution ∆f = 1/N∆t, it is defined by the choice of the maximum acquisition
time
Possible definitions of the discrete Fourier transform with a correct normalization (so
that ∆f∆t = 1/N) are
Yk =
N−1
j=0
yje−i 2π
N
kj
yj =
1
N
N−1
k=0
Ykei2π
N
kj
(5.37)
or
Yk =
1
√
N
N−1
j=0
yje−i 2π
N
kj
yj =
1
√
N
N−1
k=0
Ykei 2π
N
kj
. (5.38)
5.2.5 Consequence of discrete signal acquisition
The ”ideal” NMR signal converted to the digital form
yj = Ae−R2j∆t
ei2πνj∆t
(5.39)
has a Fourier transform
Yk =
N−1
j=0
Ae−R2j∆t
ei2πνj∆t
e−i 2π
N
kj
∆t. (5.40)
The summation formula
N−1
j=0
zj
=
1 − zN
1 − z
(5.41)
60 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
helps us to evaluate the sum. For the sake of simplicity, let us assume that the carrier
frequency is chosen so that the peak is in the middle of the spectrum
ν =
1
2
N∆f =
1
2∆t
. (5.42)
Then, z and zN
in the summation formula are
z = e−R2∆t
ei2π(1
2
− k
N ) = e−R2∆t
1−R2∆t
eiπ
−1
e−i2π k
N = −(1 − R2∆t)e−i2π k
N , (5.43)
zN
= e−R2N∆t
eiπ(N−2k)
. (5.44)
Therefore,
Yk = A∆t
1 − e−R2N∆t
eiπ(N−2k)
1 + (1 − R2∆t)e−i2π k
N
. (5.45)
• The signal is discrete ⇒ the spectral width is limited ∆t > 0 ⇒ N∆f = 1/∆t < ∞
The consequences of the discrete sampling are:
Aliasing: If we add a value of N∆f to the frequency which was originally in the
middle of the frequency spectrum (1
2
N∆f = 1
2∆t
), the second exponent in Eq. 5.40
changes from iπj to i3πj, i.e. by one period (2π), and the transformed signal (the
spectrum) does not change. In general, a peak of the real frequency ν+N∆f (outside the
spectral width) appears at the apparent frequency ν in the spectrum (Nyquist theorem:
frequencies ν and ν + 1/∆t cannot be distinguished).
Offset: Peak height of the continuous Fourier transform Y (f) = A/R2 and offset
of the continuous Fourier transform Y (±∞) = 0. Peak height of the discrete Fourier
transform.
YN
2
= A∆t
1 − e−R2N∆t
R2∆t
→ A/R2 (5.46)
for N∆t → ∞, but offset of the discrete Fourier transform
Y0 = A∆t
1 − e−R2N∆t
eiNπ
2 − R2∆t
→
1
2
A∆t =
1
2
y0∆t (5.47)
for N∆t → ∞ and ∆t → 0. The offset of discrete Fourier transform is non-zero,
equal to half of the intensity of the signal at the first time point y(0) if the signal was
acquired sufficiently long to relax completely (N∆t 1/R2).
Loss of causality: The algorithm of the discrete Fourier transform assumes that the
signal is periodic. This contradicts the causality theorem: a periodic function cannot be
equal to zero for t < 0 and different from zero t > 0. The causality must be introduced
5.3. PHASE CORRECTION 61
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
Ω1 = Ω2 − 2π/∆t
{Y(ω)}
ω
Ω1 = Ω2 − 2π/∆t
Figure 5.8: Aliasing. If the signal is acquired in discrete time intervals (dots in the top plots), the
signals with frequencies different by an integer multiple of 2π/∆t, shown by solid (Ω1) and dotted (Ω2)
lines, cannot be distinguished. Both signals give a peak with the same frequency in the spectrum. This
frequency is equal to Ω1 and to Ω2 − 2π/∆t, where 2π/∆t is the width of the spectrum.
in a sort of artificial manner. After recording N time points, another N zeros should be
added to the signal3
(see Section 5.4).
5.3 Phase correction
So-far, we ignored the effect of the initial phase φ0 and analyzed Fourier transforms of
NMR signals consisting of a collection of (damped) cosine functions, with zero initial
phase. In reality, the signal has a non-zero phase, difficult to predict
3
In practice, the zeros are added after the last point of the measured signal, not before the first one,
as one may expect based on the fact that signal should be equal to zero for t < 0.
62 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
{Y(ω)}
ω
{Y(ω)}
ω
Figure 5.9: A signal with the initial phase of 60 ◦
(top) provides distorted spectra (bottom), unless a
phase correction is applied.
y(t) = Ae−R2t
eiΩ(t+t0)
= |A|e−R2t
eiΩ(t+t0)+φ0
. (5.48)
The phase has a dramatic impact on the result of the Fourier transformation. Real
and imaginary parts are mixtures of absorption and dispersion functions. If we plot the
real part as a spectrum, it looks really ugly for a non-zero phase.
For a single frequency, the phase correction is possible (multiplication by the function
e−(iΩt0+φ0)
, where t0 and φ0 are found empirically):
|A|e−R2t
eiΩ(t+t0)+φ0
e−(iΩt0+φ0)
= |A|e−R2t
eiΩt
. (5.49)
In practice, phase corrections are applied also to signal with more frequencies –
multiplication by a function e−i(ϑ0+ϑ1ω)
, where ϑ0 and ϑ1 are zero-order and first-order
phase corrections, respectively (we try to find ϑ0 and ϑ1 giving the best-looking spectra).
5.4. ZERO FILLING 63
Note that phase correction is always necessary, but only approximative corrections are
possible for a signal with multiple frequencies!
5.4 Zero filling
Routinely, a sequence of NZ zeros is appended to the recorded signal, mimicking data
obtained at time points N∆t to (N + NZ − 1)∆t:
0, ∆t, 2∆t, · · · , (N − 1)∆t
y0, y1, y2, · · · , yN−1
↓
0, ∆t, 2∆t, · · · , (N − 1)∆t, N∆t, (N + 1)∆t, · · · , (N + NZ − 1)∆t
y0, y1, y2, · · · , yN−1, 0, 0, · · · , 0
(5.50)
This may look like a completely artificial procedure, but there are several practical
reasons to do it.
1. The very fast computational algorithm of calculating Fourier transform, known as
Cooley–Tukey FFT, requires the number of time points to be an integer power of
2. If the number of collected time points N is not a power of 2, NZ zeros are added
to the data prior to Fourier transformation so that N + NZ is an integer power of
2.
2. In order to obtain a spectrum with the full content of information by discrete
Fourier transformation, the collected data must be extended by a factor of 2 by
zero-filling. As discussed in Section 5.2.5, this operation reintroduces causality and
the full information content of N experimental complex points (i.e., N points of
the real part and N points of the imaginary part, together 2N bits of information)
is encoded in the spectrum (i.e., in the real part of the Fourier transform, which
now consists of 2N frequency points because we artificially increased the maximum
time from N − 1 to 2N − 1 and therefore narrowed the frequency sampling step
∆f from 1/N∆t to 1/2N∆t).
3. The digital resolution ∆ν, given by 1/(N∆t), can be improved (narrowed) to
1/((N + NZ)∆t) by zero-filling. In this manner, the visual appearance of spectra
can be improved by interpolation between data points. Note, however, that adding
more than N zeros does not improve the informational content of the spectrum.
Although the digital resolution is improved, the real resolution is the same, zerofilling
does not help to resolve frequencies that differ less than 1/(N∆t)!
64 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
5.5 Apodization
The NMR signal is very often multiplied by a so-called window function prior to Fourier
transformation.4
This process is known as apodization. The goal is to
1. improve resolution. As the resolution is given by 1/(N∆t), resolution is improved
if the signal is multiplied by a window function that amplifies the late data points.
2. improve sensitivity. Due to the relaxation, signal of data acquired at later time
points is lower, but the noise is the same. Therefore, the late time points decrease
the signal-to-noise ratio. The sensitivity can be improved by discarding or
attenuating the late time points.
3. suppress truncation artifacts. We have seen that oscillations of the baseline appear
if the data acquisition stops before the signal relaxes to zero (i.e., to the noise
level). The desired effect of relaxation can be mimicked by a window function that
smoothly converges to zero at N∆t.
Obviously, the three listed goals are in conflict, and only a compromise can been
reached. There is no ”best apodization”. The choice of the optimal window function
depends on the actual needs.
The simplest window function is a rectangle: multiplying the signal by a rectangular
function equal to 1 for j∆t ≤ m∆t and to 0 for j∆t > m∆t represents discarding data
recorded for times longer than m∆t. It is a very useful way of improving signal-to-noise
ratio if the signal relaxed before m∆t. Otherwise, it produces severe truncation artifacts.
The highest signal-to-noise ratio is provided by a matched filter window function.
The matched filter has the shape of the envelope of the signal. The matched filter for
our ideal signal is e−R2j∆t
. The price paid for the signal-to-noise improvement is a lower
resolution: Multiplying e−R2t
eiΩ∆t
by e−R2t
obviously doubles the linewidth, given by
the decay rate, which is now 2R2.
The best balance between resolution and truncation artifacts for an allowed extra line broadening λ is obtained with
the Dolph–Chebyshev window, defined as
1
√
N
N−1
k=0
cos 2(N − 1) arccos
cos(πk/N)
cos(πλ∆t/2)
cosh 2(N − 1)arccosh 1
cos(πλ∆t/2)
ei 2π
N
kj
, (5.51)
which is, however, not used in practice due to its very complex form.
Instead, sine-bell windows sinp 2π−φ
N
j + φ are used routinely, usually with the phase
φ = π/2 (i.e., cosine function) and with the power p = 1 or p = 2.
4
The mathematical expression describing the Fourier-transformed product of two functions, signal
and window in our case, is given by the convolution theorem, presented in Section 5.2.2.
Summary of conventions
As mentioned in the preceding chapters, some of the conventions used in practical NMR
spectroscopy do not follow directly from the physical analysis of NMR.
• The phase of the rotating coordinated system is chosen to be π for nuclei with
γ > 0 and 0 for nuclei with γ < 0. As a consequence, ω1 does not depend on the
sign of γ. The exciting irradiation by radio waves, defining the rotating coordinate
system, always rotates the magnetization vector about the +x axis, from the z to
the −y direction (after 90 ◦
rotation), etc.
• The frequency axes of NMR spectra is plotted from left to right for nuclei with
γ > 0, but from right to left for nuclei with γ < 0. As the nuclei with γ > 0 have
negative ω0 = −γB0, peaks of γ > 0 nuclei precessing faster, i.e., having a higher
positive chemical shift δ, are shifted to the left in the spectra. The nuclei with
γ < 0 have positive ω0 = −γB0, but peaks of γ < 0 nuclei precessing faster, i.e.,
having a higher positive chemical shift δ, are also shifted to the left in the spectra
because the spectrum is plotted in the opposite direction. The position of the peak
in the spectrum does not depend on the sign of γ, it is only given by the value of
its chemical shift, which increases from right to left.
• The zero value of the chemical shift is not given by the absence of shielding by
electrons, but by the chosen reference compound. The zero value of the frequency
offset Ω is given by the value of −ωradio, chosen by the operator in each NMR
experiment. If the spectrum is plotted in Hertz, the position of zero is defined just
by the experimental setup. If the spectrum is plotted in ppm and the zero value
was correctly calculated based on a comparison with the reference compound, the
position of zero and all values of chemical shifts are independent of the experimental
conditions and unambiguously defined for the given chemical compound. Always
report position of peaks in ppm!
65
66 CHAPTER 5. SIGNAL ACQUISITION AND PROCESSING
Part II
Quantum description
67
Chapter 6
Spin
Literature: This chapter starts with a brief review of quantum mechanics. Textbooks
covering this topic represent the best source of information. Brown presents in B9 a useful
review of classical mechanics, usually missing in the quantum mechanics textbooks
(assuming that students learnt the classical mechanics earlier, which is true in the case
of students of physics, but not so often in the case of chemistry or biology students),
and reviews quantum mechanics in B13, B15, and B16. B1–B5 provides overview of
the relevant mathematical tools. NMR books also provide some introduction. Keeler reviews
quantum mechanics in very understandable fashion, using the concept of spin from
the very beginning (K3.2 and K6). Levitt proceeds more like us (L6–7). A condensed
summary is presented in C2.1 (short, rigorous, but not a good start for a novice). Introduction
to the special theory of relativity can be found in B10, but relativistic quantum
mechanics is not discussed in the literature recommended for this course or in general
physical chemistry textbooks (despite the important role of spin in chemistry). Therefore,
more background information is presented here than in the other chapters. NMR
can be correctly described if the spin is introduced ad hoc. The purpose of Section 6.8
is to show how the spin emerges naturally. Origin of nuclear magnetism is touched in
L1.3 and L1.4. Quantum mechanics of spin angular momentum is reviewed in K6, L7,
and L10.
6.1 Wave function and state of the system
We postulate that the state of the system is completely described by a wave function.
• Newton mechanics: coordinates and moments of all particles describe all properties
of the current state and all future states
• Quantum mechanics: wave function describes all properties of the current state
and all future states
69
70 CHAPTER 6. SPIN
Quantum mechanics is postulated, not derived. It can be only tested experimentally.
Introduced because Newton mechanics did not described experiments correctly.
Example – two-slit (Young) experiment:
• Question: Particles or waves?
• Answer: Particles, but with probabilities added like waves
(Complex) probability amplitude: Ψ = Ceiφ
(Real) probability density: ρ = Ψ∗
Ψ = |Ψ|2
= |C|2
Probability of finding single particle in volume L3
:
L´
0
L´
0
L´
0
Ψ∗
Ψdxdydz
Wave function of a free particle moving in direction x (coordinate frame can be
always chosen so that x is the direction of motion of a free particle):
Ψ = Cei2π( x
λ
− t
T
)
= Ce
i
(px−Et)
, (6.1)
where h = 2π is the Planck’s constant, p = mv is momentum (along x), and E is
(kinetic) energy.
Note that Ψ corresponds to a monochromatic wave with period equal to h/E, wavelength
equal to h/p, and a complex amplitude C (it may contain a phase factor eiφ
).
Calculating ”square”: real number c2 = cc, complex number |c|2 = cc∗, real vector |v|2 = v · v = v1v1 + v2v2 + · · · ,
complex vector |v|2 = v† · v = v∗
1 v1 + v∗
2 v2 + · · · , (continuous) function
´ b
a f∗(x)f(x)dx (function can be viewed as a
vector of infinite number of infinitely ”dense” elements – summation → integration).
Dirac’s notation: |v , |f is a vector v or function f, respectively:
v|v = v†
· v =
N
j=1
v∗
j vj, (6.2)
f|f =
∞ˆ
−∞
f∗
(x)f(x)dx. (6.3)
6.2 Superposition and localization in space
Note that a monochromatic wave function describes exactly what is p of the particle,
but does not say anything about position of the particle because ρ = Ψ∗
Ψ = |C| is the
same for any x (distribution of probability is constant from x = −∞ to x = ∞). Wave
function describing a particle (more) localized in space can be obtained by superposition
of monochromatic waves.
Ψ(x, t) = c1 Ae
i
(p1x−E1t)
ψ1
+c2 Ae
i
(p2x−E2t)
ψ2
+ · · · (6.4)
We postulate that if possible states of our system are described by wave functions
ψ1, ψ2, . . . , their linear combination also describes a possible state of the system.
6.3. OPERATORS AND POSSIBLE RESULTS OF MEASUREMENT 71
Note that monochromatic waves are orthogonal:
∞ˆ
−∞
A∗
e− i (p1x−E1t)
Ae
i (p2x−E2t)
dx = |A|2
e
i (E1−E2)t
∞ˆ
−∞
e
i (p1−p2)x
dx =
|A|2
e
i (E1−E2)t
∞ˆ
−∞
cos
(p1 − p2)x
dx + i|A|2
e
i (E1−E2)t
∞ˆ
−∞
sin
(p1 − p2)x
dx = 0 (6.5)
unless p1 = p2 (positive and negative parts of sine and cosine functions cancel each other during integration, with
the exception of cos 0 = 1).
Values of A can be also normalized to give the result of Eq. 6.5 equal to 1 if p1 = p2 and E1 = E2. It follows from the
property of the Fourier transform that in such a case |A|2 = 1/h if we integrate over a single coordinate (or |A|2 = 1/h3
if we integrate over three coordinates etc.).
In the language of algebra, the complete set of normalized monochromatic waves constitutes orthonormal basis for
wave functions, in a similar way as unit vectors ı, , k are the orthonormal basis for all vectors in the Cartesian coordinate
system x, y, z.
Also, Ψ can be normalized based on the condition
∞ˆ
−∞
Ψ∗
Ψdx = P = 1 (6.6)
(if a particle exists, it must be somewhere). It requires
∞ˆ
−∞
(c∗
1c1 + c∗
2c2 + · · · )dx = 1. (6.7)
6.3 Operators and possible results of measurement
We postulate that any measurable property is represented by an operator (acting
on the wave function) and that result of a measurement must be one of eigenvalues
of the operator.
We postulated that the wave function contains a complete information about the system, but how can we extract this
information from the wave function? For example, how can we get the value of a momentum of a free particle described
by Eq. 6.4? Calculation of ∂Ψ/∂x gives us a clue:
∂Ψ
∂x
= c1
∂
∂x
e
i (p1x−E1t)
+ c2
∂
∂x
e
i (p2x−E2t)
+ · · · =
i
p1c1e
i (p1x−E1t)
+
i
p2c2e
i (p2x−E2t)
+ · · · (6.8)
It implies that
− i
∂
∂x
e
i (p1x−E1t)
= p1e
i (p1x−E1t)
, −i
∂
∂x
e
i (p2x−E2t)
= p2e
i (p2x−E2t)
, . . . (6.9)
We see that
1. calculation of the partial derivative of any monochromatic wave and multiplying the result by −i gives us the
same wave just multiplied by a constant. In general, the instruction to calculate the partial derivative and multiply
the result by −i is known as operator. If application of the operator to a function gives the same function, only
multiplied by a constant, the function is called eigenfunction of the operator and the constant is called eigenvalue
of the operator.
2. the eigenvalues are well-defined, measurable physical quantities – possible values of the momentum along x.
3. the eigenvalues can be obtained by applying the operator to the eigenfunction and multiplying the result by the
complex conjugate of the eigenfunction:
72 CHAPTER 6. SPIN
p1 = e− i (p1x−E1t)
−i
∂
∂x
e
i (p1x−E1t)
= e− i (p1x−E1t)
p1e
i (p1x−E1t)
= p1 e− i (p1x−E1t)
e
i (p1x−E1t)
=1
. (6.10)
We usually write operators with ”hats”, like ˆA. Writing ˆAΨ means ”take function
Ψ and modify it as described by ˆA”. It is not a multiplication: ˆAΨ = ˆA · Ψ, ˆA is not a
number but an instruction what to do with Ψ!
Recipe to calculate possible results of a measurement:
1. Identify the operator representing what you measure ( ˆA)
2. Find all eigenfunctions |ψ1 , |ψ2 , . . . of the operator and use them as an orthonormal
basis for Ψ: Ψ = c1|ψ1 + c2|ψ2 , . . .
3. Calculate individual eigenvalues Aj as
ψj| ˆAψj = ψj|Aj · ψj = Aj ψj|ψj
=1
= Aj. (6.11)
The first equality in step 3 follows from the definition of eigenfunctions, then Aj
is just a (real) number and can be factored out of the brackets (representing integration
or summation) as described by the second equality, and the last equality reflects
orthonormality of |ψj .
6.4 Expected result of measurement
Eq. 6.11 tells us what are the possible results of a measurement, but it does not say which
value is actually measured. We can only calculate probabilities of getting individual
eigenvalues and predict the expected result of the measurement.
We postulate that the expected result of measuring a quantity A represented by an
operator ˆA in a state of the system described by a wave function Ψ is
A = Ψ| ˆA|Ψ . (6.12)
There are three ways how to do the calculation described by Eq. 6.12:
1. Express Ψ, calculate its complex conjugate Ψ∗
≡ Ψ|, calculate ˆAΨ ≡ | ˆAΨ , and
in the manner of Eq. 6.3
A = Ψ| ˆA|Ψ ≡ Ψ|( ˆAΨ) =
∞ˆ
−∞
· · · Ψ∗
(x, . . . ) ˆAΨ(x, . . . )dx . . . . (6.13)
6.4. EXPECTED RESULT OF MEASUREMENT 73
Three dots in Eq. 6.13 tell us that for anything else that a single free particle (with
zero spin) we integrate over all degrees of freedom, not just over x.
2. Find eigenfunctions ψ1, ψ2, . . . of ˆA and write Ψ as their linear combination Ψ =
c1ψ1 + c2ψ2 + · · · (use the eigenfunctions as an orthonormal basis for Ψ). Due to
the orthonormality of the basis functions, the result of Eq. 6.13 is A = c∗
1c1A1 +
c∗
2c2A2 +· · · , where A1, A2, . . . are eigenvalues of ˆA. We see that A is a weighted
average of eigenvalues Aj with the weights equal to the squares of the coefficients
(c∗
j cj = |cj|2
). The same result is obtained if we calculate
A = c∗
1 c∗
2 · · ·



A1 0 · · ·
0 A2 · · ·
...
...
...






c1
c2
...


 . (6.14)
We see that we can replace (i) operators by two-dimensional diagonal matrices,
with eigenvalues forming the diagonal, and (ii) wave functions by one-dimensional
matrices (known as state vectors) composed of the coefficients cj. Eq. 6.14 shows
calculation of the expected results of the measurement of A using matrix representation
of operators and wave functions. Matrix representation is a big simplification
because it allows us to calculate A without knowing how the operator ˆA
and its eigenfunctions look like! We just need the eigenvalues and coefficients cj.
This simplification is paid by the fact that the right coefficients are defined by the
right choice of the basis.
3. Write Ψ as a linear combination of basis functions ψ1, ψ1, . . . (not necessarily
eigenfunctions of ˆA)
Ψ = c1ψ1 + c2ψ2 + · · · (6.15)
Build a two-dimensional matrix ˆP from the products of coefficients c ∗
j ck:
ˆP =



c1c ∗
1 c1c ∗
2 · · ·
c2c ∗
1 c2c ∗
2 · · ·
...
...
...


 . (6.16)
Multiply the matrix ˆP by a matrix1 ˆA representing the operator ˆA in the basis
1
How can we get a matrix representation of an operator with eigenfunctions different from the basis?
The complete set of N functions defines an abstract N-dimensional space (N = ∞ for free particles!).
The wave function Ψ is represented by a vector in this space built from coefficients c1, c2, . . . , as
described by Eq. 6.15, and a change of the basis is described as a rotation in this space. The same
rotation describes how the matrix representing the operator ˆA changes upon changing the basis. Note
that the matrix is not diagonal if the basis functions are not eigenfunctions of ˆA.
74 CHAPTER 6. SPIN
ψ1, ψ1, . . . . The sum of the diagonal elements (called trace) of the resulting matrix
ˆP ˆA is equal to the expected value A
A = Tr{ ˆP ˆA }. (6.17)
Why should we use such a bizarre way of calculating the expected value of A
when it can be calculated easily from Eq. 6.14? The answer is that Eq. 6.17
is more general. We can use the same basis for operators with different sets of
eigenfunctions.
6.5 Operators of position and momentum
We need to find operators in order to describe measurable quantities. Let us start with
the most fundamental quantities, position of a particle x and momentum p = mv.
6.5.1 Operator of momentum
We have already obtained the operator of momentum of a particle moving in the x
direction when calculating ∂Ψ/∂x (Eq. 6.9). If a particle moves in a general direction,
operators of components of the momentum tensor are derived in the same manner.
ˆpx ≡ −i
∂
∂x
, (6.18)
ˆpy ≡ −i
∂
∂y
, (6.19)
ˆpz ≡ −i
∂
∂z
. (6.20)
6.5.2 Operator of position
The wave function Ψ(x, t) defined by Eq. 6.4 is a function of the position of the particle,
not of the momentum (it is a sum of contributions of all possible momenta). If we
define basis as a set of functions ψj = Ψ(xj, t) for all possible positions xj, operator
of position is simply multiplication by the value of the coordinate describing the given
position. Operators of the y and z are defined in the same manner.
ˆx ≡ x · ˆy ≡ y · ˆz ≡ z · (6.21)
To see how the operator acts, write Ψ∗(x, t) and xΨ(x, t) as the set of functions Ψ(xj, t) for all possible positions xj:
6.5. OPERATORS OF POSITION AND MOMENTUM 75
xΨ(x, t) =







x1c1e
i (p1x1−E1t)
+ x1c2e
i (p2x1−E2t)
+ x1c3e
i (p3x1−E3t)
+ · · ·
x2c1e
i (p1x2−E1t)
+ x2c2e
i (p2x2−E2t)
+ x2c3e
i (p3x2−E3t)
+ · · ·
x3c1e
i (p1x3−E1t)
+ x3c2e
i (p2x3−E2t)
+ x3c3e
i (p3x3−E3t)
+ · · ·
...







=





ψ1
ψ2
ψ3
...





. (6.22)
If the position of the particle is e.g. x2,
Ψ(x2, t) =






0
c1e
i (p1x2−E1t)
+ c2e
i (p2x2−E2t)
+ c3e
i (p3x2−E3t)
+ · · ·
0
...






=





0
ψ2
0
...





(6.23)
and x · Ψ(x, t) for x = x2 is
x2Ψ(x2, t) =






0
x2 c1e
i (p1x2−E1t)
+ c2e
i (p2x2−E2t)
+ c3e
i (p3x2−E3t)
+ · · ·
0
...






=





0
x2ψ2
0
...





. (6.24)
We see that multiplication of Ψ(x2, t) = ψ2 by x2 results in x2ψ2, i.e., ψ2 is an eigenfunction of the operator ˆx = x·
and x2 is the corresponding eigenvalue.
Note that multiplication by pj does not work in the same way! We could multiply ψ2 by x2 because ψ2 does not
depend on any other value of the x coordinate. However, ψ2 depends on all possible values of p. On the other hand,
partial derivative gave us each monochromatic wave multiplied by its value of p and ensured that the monochromatic
waves acted as eigenfunctions.
6.5.3 Commutators
If we apply two operators subsequently to the same wave function, order of the operators
sometimes does not matter.
For example, ˆxˆpyΨ = ˆpy ˆxΨ (ˆx and ˆpy commute). It means that x and py can be measured independently at the
same time. However, sometimes the order of operators makes a difference. For example
ˆxˆpxΨ = −i x
∂Ψ
∂x
(6.25)
but
ˆpx ˆxΨ = −i
∂(xΨ)
∂x
= −i Ψ − i x
∂Ψ
∂x
. (6.26)
The difference is known as the commutator and is written as ˆxˆpx − ˆpx ˆx = [ˆx, ˆpx]. A non-zero commutator tells
us that ˆx and ˆpx are not independent and cannot be measured exactly at the same time. Analysis of the action of the
operators shows reveals the basic commutation relations:
• Commutators of operators of a coordinate and the momentum component in the
same direction are equal to i (i.e., multiplication of Ψ by the factor i )
• All other position and coordinate operators commute.
Written in a mathematically compact form,
[ˆrj, ˆpk] = i δj,k [ˆrj, ˆrk] = [ˆrj, ˆpk] = 0, (6.27)
76 CHAPTER 6. SPIN
where j and k are x, y, or z, rj is the x, y, or z component of the position vector
r = (rx, ry, rz) ≡ (x, y, z), pk is the x, y, or z component of the momentum vector
p = (px, py, pz), and δj,k = 1 for j = k and δj,k = 0 for j = k.
The described commutator relations follow from the way how we defined Ψ in Eq. 6.4.
However, we can also use Eq. 6.27 as the fundamental definition and Eq. 6.4 as its
consequence:
We postulate that operators of position and momentum obey the relations
[ˆrj, ˆpk] = i δj,k [ˆrj, ˆrk] = [ˆpj, ˆpk] = 0. (6.28)
Note that we only postulate relations between operators. Other choices are possible
and correct as long as Eq. 6.27 holds.
6.6 Operator of energy and equation of motion
We obtained the operator of momentum by calculating ∂Ψ/∂x. What happens if we calculate ∂Ψ/∂t?
∂Ψ
∂t
= c1
∂
∂t
e
i (p1x−E1t)
+ c2
∂
∂t
e
i (p2x−E2t)
+ · · · = −
i
E1c1e
i (p1x−E1t)
−
i
E2c2e
i (p2x−E2t)
− · · · (6.29)
and consequently
i
∂
∂t
e
i (p1x−E1t)
= E1e
i (p1x−E1t)
, i
∂
∂t
e
i (p2x−E2t)
= E2e
i (p2x−E2t)
, . . . (6.30)
1. First, we obtain the operator of energy from Eq. 6.30, in analogy to Eq. 6.9.
2. The second achievement is Eq. 6.29 itself. Energy of free particles is just the kinetic energy (by definition).
Therefore, all energies Ej in the right-hand side of Eq. 6.29 can be written as
Ej =
mv2
j
2
=
p2
j
2m
, (6.31)
resulting in
∂Ψ
∂t
= −
i p2
1
2m
c1e
i (p1x−E1t)
+
p2
2
2m
c2e
i (p2x−E2t)
+ · · · . (6.32)
But an equation with the p2
j terms can be also obtained by calculating
1
2m
∂2Ψ
∂x2
=
1
2m
∂
∂x
∂Ψ
∂x
= −
1
2
p2
1
2m
c1e
i (p1x−E1t)
+
p2
2
2m
c2e
i (p2x−E2t)
+ · · · . (6.33)
Comparison of Eqs. 6.32 and 6.33 gives us the equation of motion
i
∂Ψ
∂t
= −
2
2m
∂2Ψ
∂x2
+ · · · . (6.34)
If we extend our analysis to particles experiencing a time-independent potential energy Epot(x, y, z), the energy
will be given by
Ej =
p2
j
2m
+ Epot (6.35)
6.6. OPERATOR OF ENERGY AND EQUATION OF MOTION 77
where pj is now the absolute value of a momentum vector pj (we have to consider all three direction x, y, z because
particles change direction of motion in the presence of a potential). The time derivative of Ψ is now
∂Ψ
∂t
= −
i p2
1
2m
c1e
i (p1r−E1t)
+
p2
2
2m
c2e
i (p2r−E2t)
+ · · · −
i
Epot(r)Ψ (6.36)
and
p2
1
2m
c1e
i (p1r−E1t)
+
p2
2
2m
c2e
i (p2r−E2t)
+ · · · = −
2
2m
∂2Ψ
∂x2
+
∂2Ψ
∂x2
+
∂2Ψ
∂x2
. (6.37)
Substituting Eq. 6.37 into Eq. 6.36 gives us the famous Schr¨odinger equation
i
∂Ψ
∂t
= −
2
2m
∂2
∂x2
+
∂2
∂x2
+
∂2
∂x2
+ Epot(x, y, z)
ˆH
Ψ. (6.38)
The sum of kinetic and potential energy is known as Hamiltonian in the classical
mechanics and the same term is used for the operator ˆH.
In our case, the Hamiltonian is expressed in terms of linear momentum p = mv. In general, the canonical (or
generalized) momentum should be used. The canonical momentum is defined by the Lagrange mechanics. Motions of
objects can be described by the least action principle (nicely described in The Feynman lectures on physics, Vol. 2,
Chapter 19), which can be formulated as
d
dt
∂L
∂ ˙qj
=
∂L
∂qj
, (6.39)
where qj are generalized coordinates, the dot represents time derivative, and L is a scalar function of qj and ˙qj,
known as Lagrangian. For example, for a free particle moving in one direction (x) in a field described by a potential
energy Epot(x),
d
dt
∂L
∂v
=
∂L
∂x
. (6.40)
What L(x, v) gives the correct equation of motion? The equation of motion has the form
ma = m
dv
dt
= F = −
∂Epot(x)
∂x
. (6.41)
Since
∂Ekin(v)
∂v
=
∂ 1
2
mv2
∂v
= mv ⇒
d
dt
∂Ekin(v)
∂v
= m
dv
dt
= ma, (6.42)
and
∂Ekin(v)
∂x
= 0,
∂Epot(x)
∂v
= 0, (6.43)
we see that L(x, v) can be taken as a difference of the kinetic energy (depending on v, but not on x) and the potential
energy (depending on x, but not on v)
ma =
d(mv)
dt
=
d
dt
∂(Ekin(v) − Epot(x))
∂v
=
d
dt
∂L
∂v
= F = −
∂(Ekin(v) − Epot(x))
∂x
=
∂L
∂x
. (6.44)
Hamiltonian and Lagrangian are related by the Legendre transform2
2
Legendre transform has a simple graphical representation. If we plot a function of a variable x,
e.g. f(x), slope at a certain value of x = x0 is equal to s(x0) = (∂f/∂x)x0
. A tangent line y(x0)
touching the plotted f for x = x0 is described by the slope s(x0) and intercept a(x0) as y = a+s(x0)x0.
The value of the intercept for all possible values of x0 can be expressed as a function of the slope
a(s(x0)) = y(x0) − s(x0)x0 = f(x0) − s(x0)x0 (y and f are equal at x0 because they touch each other).
If we identify x with q, f with L, and −a with H, we get Eq. 6.45 for a one-dimensional case (i = 1).
78 CHAPTER 6. SPIN
H(qj, pj) + L(qj, ˙qj) =
j
(pj · ˙qj), (6.45)
where
pj =
∂L
∂ ˙qj
(6.46)
is the canonical (generalized) momentum. In our example, pj = px = p is the linear momentum mv and the
Hamiltonian is pv − L = mv2 − 1
2
mv2 + Epot = 1
2
mv2 + Epot. The whole procedure may seem to be unnecessarily
complicated, but it becomes useful when we analyze motions of magnetic particles in magnetic fields.
Derivation of the Hamiltonian (classical or quantum) for magnetic particles in magnetic fields is much more demanding
because the canonical momentum is no longer identical with the linear momentum. We start our analysis by searching for
a classical Lagrangian describing motion of a charged particle in a magnetic field. We know that the Lagrangian should
give us the Lorentz force
F = Q(E + v × B). (6.47)
The information about E and B can be extracted from the following Maxwell equations
· B = 0 (6.48)
× E = −
∂B
∂t
, (6.49)
but we have to employ our knowledge of vector algebra to handle the divergence in Eq. 6.48 and the curl in Eq. 6.49.
It may be unclear at the beginning why we go this way, but the purpose becomes evident when we combine the obtained
expressions.
First, we notice that a · (a × b) = 0 for any vectors a and b because a × b ⊥ a. As a consequence, we can replace B by
a curl (rotation) of some vector A because · ( × A) = 0 as required by Eq. 6.48. The vector A is known as the vector
potential. The first step gives us a new definition of B
B = × A (6.50)
which can be inserted into Eq. 6.47
F = Q(E + v × B) = Q(E + v × ( × A)). (6.51)
Using the identity a × (b × c) = b(a · c) − (a · b)c,
F = Q(E + v × B) = Q(E + v × ( × A)) = Q(E + (v · A) − (v · )A). (6.52)
Second, we use our new definition of B and rewrite Eq. 6.49 as
0 =
∂B
∂t
+ × E = ×
∂A
∂t
+ × E = ×
∂A
∂t
+ E . (6.53)
Third, we notice that that for any vector a and constant c, a × (ca) = 0 because a ca. As a consequence, we can
replace (∂A/∂t + E) by a gradient of some scalar V because × ( (∂A/∂t + E)) = × (− V ) = 0 as required by
Eq. 6.49. The scalar V is the well-known electric potential and allows us to express E as
E = −
∂A
∂t
− V. (6.54)
which can be also inserted into Eq. 6.47
F = Q(E + v × B) = Q −
∂A
∂t
− V + (v · A) − (v · )A . (6.55)
Finally, we notice that
dA
dt
=
∂A
∂t
+
∂A
∂x
dx
dt
+
∂A
∂y
dy
dt
+
∂A
∂z
dz
dt
=
∂A
∂t
+ v · A ⇒
∂A
∂t
=
dA
dt
− v · A, (6.56)
6.6. OPERATOR OF ENERGY AND EQUATION OF MOTION 79
which shows that v · A in Eq. 6.55 can be can be included into dA/dt
F = Q(E + v × B) = Q −
∂A
∂t
− V + (v · A) − (v · )A = Q −
dA
dt
− V + (v · A) . (6.57)
Let us now try to write L as
L = Ekin − Eel + Emagn =
1
2
mv2
− QV + Emagn, (6.58)
where Eel is a typical potential energy dependent on position but not on speed, and Emagn can depend on both
position and speed.
For this Lagrangian,
∂L
∂x
=
∂Eel
∂x
+
∂Emagn
∂x
= −Q
∂V
∂x
+
∂Emagn
∂x
(6.59)
d
dt
∂L
∂vx
=
d
dt
∂Ekin
∂vx
+
∂Emagn
∂vx
= max +
d
dt
∂Emagn
∂vx
. (6.60)
If we use Emagn = Qv · A, Eqs. 6.59 and 6.60 give us
max = −Q
dAx
dt
−
∂V
∂x
+
∂(v · A)
∂x
(6.61)
and a sum with similar y- and z-components is equal to the Lorentz force
ma = F = Q −
dA
dt
− V + (v · A) = Q(E + v × B). (6.62)
We have found that our Lagrangian has the form
L =
1
2
mv2
− QV + Q(v · A). (6.63)
According to Eq. 6.46, the canonical momentum has the following components
px =
∂L
∂vx
= mvx + QAx py =
∂L
∂vy
= mvy + QAy pz =
∂L
∂vz
= mvy + QAz. (6.64)
The Hamiltonian can be obtained as usually by the Legendre transform
H =
j=x,y,z
pjvj − L = p · v − L. (6.65)
In order to express H as a function of p, we express v as (p − QA)/m:
H =
2p · (p − QA) − (p − QA)2 − 2Q(p − QA) · A
2m
+ QV =
(p − QA)2
2m
+ QV. (6.66)
The association of Hamiltonian (energy operator) with the time derivative makes it
essential for analysis of dynamics of systems in quantum mechanics:
We postulate that evolution of a system in time is given by the Hamiltonian:
i
∂Ψ
∂t
= ˆHΨ. (6.67)
Eq.6.67 can be also written for matrix representation of Ψ and ˆH. If eigenfunctions of
ˆH are used as a basis (Ψ = c1(t)ψ1 +c2(t)ψ2 +· · · ), the time-independent eigenfunctions
80 CHAPTER 6. SPIN
ψj can be factored out from ∂Ψ/∂t (left-hand side) and Ψ (right-hand side), and canceled,
giving
i
d
dt



c1
c2
...


 =



E1 0 · · ·
0 E2 · · ·
...
...
...






c1
c2
...


 , (6.68)
which is simply a set of independent differential equations
dcj
dt
= −i
Ej
cj ⇒ cj = aje−i
Ej
t
, (6.69)
where the (possibly complex) integration constant aj is given by the value of cj at t = 0.
Note that the coefficients cj evolve, but the products c∗
j cj = |aj|2
do not change in
time. Each product c∗
j cj describes the probability that the system is in the state with
the energy equal to the eigenvalue Ej, described by an eigenfunction ψj.
• States corresponding to the eigenfunctions of the Hamiltonian are stationary (do
not vary in time).
• Only stationary states can be described by the energy level diagram.
6.7 Operator of angular momentum
In order to understand NMR experiments, we also need to describe rotation in space.
The fundamental quantity related to the rotation is angular momentum. In a search for
its operator, we start from what we know, position and momentum operators. We use
classical physics and just replace the values of coordinates and momentum components
by their operators.
Classical definition of the vector of angular momentum L is
L = r × p. (6.70)
The vector product represents the following set of equations:
Lx = rypz − rzpy, (6.71)
Ly = rzpx − rxpz, (6.72)
Lz = rxpy − rypx. (6.73)
Going to the operators
6.7. OPERATOR OF ANGULAR MOMENTUM 81
ˆLx = ˆry ˆpz − ˆrz ˆpy = −i y
∂
∂z
+ i z
∂
∂y
, (6.74)
ˆLy = ˆrz ˆpx − ˆrx ˆpz = −i z
∂
∂x
+ i x
∂
∂z
, (6.75)
ˆLz = ˆrx ˆpy − ˆry ˆpx = −i x
∂
∂y
+ i y
∂
∂x
, (6.76)
ˆL2
= ˆL2
x + ˆL2
y + ˆL2
z. (6.77)
It follows from Eq. 6.27 that
[ˆLx, ˆLy] = i ˆLz, (6.78)
[ˆLy, ˆLz] = i ˆLx, (6.79)
[ˆLz, ˆLx] = i ˆLy, (6.80)
but
[ˆL2
, ˆLx] = [ˆL2
, ˆLy] = [ˆL2
, ˆLz] = 0. (6.81)
• Two components of angular momentum cannot be measured exactly at the same
time
• Eqs. 6.78–6.81 can be used as a definition of angular momentum operators if the
position and momentum operators are not available.
Let us find eigenvalues Lz,j and eigenfunctions ψj of ˆLz. In spherical coordinates (r, ϑ, ϕ), ψj = Q(r, ϑ)Rj(ϕ) and
ˆLz = −i ∂
∂ϕ
Eigenvalues and eigenfunctions are defined by
ˆLzψj = Lz,jψj, (6.82)
−i
∂(QRj)
∂ϕ
= Lz,j(QRj), (6.83)
−i Q
dRj
dϕ
= Lz,jQRj, (6.84)
−i
d ln Rj
dϕ
= Lz,j, (6.85)
Rj = ei
Lz,j
ϕ
. (6.86)
Since ψj(ϕ) = ψj(ϕ + 2π),
• value of the z-component of the angular momentum must be an integer multiple of
82 CHAPTER 6. SPIN
There is a close relation between the angular momentum operators and description of rotation in quantum mechanics.
Rotation about an axis given by the angular frequency vector ω
dr
dt
= ω × r, (6.87)
or more explicitly
drx
dt
= ωyrz − ωzry, (6.88)
dry
dt
= ωzrx − ωxrz, (6.89)
drz
dt
= ωxry − ωyrx. (6.90)
If a coordinate frame is chosen so that ω = (0, 0, ω)
drx
dt
= −ωry, (6.91)
dry
dt
= ωrx, (6.92)
drz
dt
= 0. (6.93)
We already know that such a set of equation can be solved easily: multiply the second equation by i and add it to
the first equation or subtract it from the first equation.
d(rx + iry)
dt
= ω(−ry + irx) = +iω(rx + iry), (6.94)
d(rx − iry)
dt
= ω(−ry − irx) = −iω(rx − iry), (6.95)
rx + iry = C+e+iωt
, (6.96)
rx − iry = C−e−iωt
, (6.97)
where the integration constants C+ = rx(0) + iry(0) = reiφ0 and C− = rx(0) − iry(0) = re−iφ0 are given by the
initial phase φ0 of r in the coordinate system:
rx + iry = re+i(ωt+φ0)
= r(cos(ωt + φ0) + i(sin(ωt + φ0)), (6.98)
rx − iry = re−i(ωt+φ0)
= r(cos(ωt + φ0) − i(sin(ωt + φ0)). (6.99)
The angle of rotation ϕ is obviously given by ωt + φ0.
rx + iry = re+iϕ
= r(cos(ϕ) + i(sin(ϕ)), (6.100)
rx − iry = re−iϕ
= r(cos(ϕ) − i(sin(ϕ)). (6.101)
Comparison with Eq. 6.86 documents the relation between ˆLz and rotation:
• Eigenfunction of ˆLz describes rotation about z.
6.8. RELATIVISTIC QUANTUM MECHANICS 83
Knowing the operator of the angular momentum, we can easily define the operators of the orbital magnetic moment.
A moving charged particle can be viewed as an electric current. Classical definition of the magnetic moment of a
charged particle travelling in a circular path (orbit) is
µ =
Q
2
(r × v) =
Q
2m
(r × p) =
Q
2m
L = γL, (6.102)
where Q is the charge of the particle, m is the mass of the particle, v is the velocity of the particle, and γ is known
as the magnetogyric ratio (constant).3
Therefore, we can write the operators
ˆµx = γ ˆLx ˆµy = γ ˆLy ˆµz = γ ˆLz ˆµ2
= γ2 ˆL2
. (6.103)
Finally, we can define the operator of energy (Hamiltonian) of a magnetic moment in a magnetic field. Classically,
the energy of a magnetic moment µ in a magnetic field of induction B is E = −µ · B. Accordingly, the Hamiltonian of
the interactions of an orbital magnetic moment with a magnetic field is
ˆH = −Bx ˆµx − By ˆµy − Bz ˆµz = −γ (Bx ˆµx + By ˆµy + Bz ˆµz) = −
Q
2m
Bx
ˆLx + By
ˆLy + Bz
ˆLz . (6.104)
6.8 Relativistic quantum mechanics
The angular momentum discussed in Section 6.7 is associated with the change of direction
of a moving particle. However, the theory discussed so-far does not explain
the experimental observation that even point-like particles moving along straight lines
possess a well defined angular momentum, so-called spin.
The origin of the spin is relativistic. The Schr¨odinger equation is not relativistic and
does not describe the spin. In order to describe spin, we need a relativistic theory, i.e.,
a theory which in agreement with two fundamental postulates of the special theory of
relativity:
• The laws of physics are invariant (i.e. identical) in all inertial systems (nonaccelerating
frames of reference).
• The speed of light in a vacuum is the same for all observers, regardless of the
motion of the light source.
According to the special theory of relativity, time is slower and mass increases at a speed v close to the speed of light
(in vacuum) c, and energy is closely related to the mass:
t =
t0
1 − v2/c2
m =
m0
1 − v2/c2
Et = mc2
=
m0c2
1 − v2/c2
, (6.105)
where m0 is the rest mass, m0c2 is the rest energy, t0 is the proper time (i.e., mass, energy, and time in the coordinate
frame moving with the particle), and Et is the total energy. The first equation can be used to express dt2
t2
=
dt2
0
1 − v2/c2
=
m2c4dt2
0
m2c4 − m2c2v2
, (6.106)
where numerator and denominator were multiplied by E2
t = m2c4 in the second step. Eqs. 6.105 show that t0/t =
m0/m. Therefore
3
The term gyromagnetic ratio is also used.
84 CHAPTER 6. SPIN
dt2
=
m2
0c4dt2
m2c4 − m2c2v2
, (6.107)
m2
0c4
= m2
c4
− (mcvx)2
− (mcvy)2
− (mcvz)2
, (6.108)
m2
0c4
= E2
t − c2
p2
x − c2
p2
y − c2
p2
z. (6.109)
We see that the special theory of relativity requires that the quantity m2
0c4 − E2
t + c2p2
x + c2p2
y + c2p2
z is equal to
zero. Let us look for an operator which represents the quantity m2
0c4 − E2
t + c2p2
x + c2p2
y + c2p2
z. We know that for a
monochromatic wave function
ψ = e
i (pxx+pyy+pzz−Ett)
, (6.110)
partial derivatives of ψ serve as operators of energy and momentum:
i
∂ψ
∂x
= −pxψ i
∂ψ
∂y
= −pyψ i
∂ψ
∂z
= −pzψ i
∂ψ
∂t
= Etψ. (6.111)
Therefore, the operator of m2
0c4 − E2
t + c2p2
x + c2p2
y + c2p2
z should have a form
2 ∂2
∂t2
− c2 2 ∂2
∂z2
− c2 2 ∂2
∂x2
− c2 2 ∂2
∂y2
+ (m0c2
)2
. (6.112)
Let us look for equation(s) of motion leading to such operator. As this problem is not easy to solve, we will proceed
step by step. Let us first assume that particles do not move, i.e., p = 0. Then, Eq. 6.109 simplifies to
m2
0c4
− E2
t = 0, (6.113)
which can be written as
(m0c2
+ Et)(m0c2
− Et) = 0, (6.114)
Using the operator of energy,
2 ∂2ψ
∂t2
+ (m0c2
)2
ψ = m2
0c4
− E2
t ψ = 0 (6.115)
if ψ is an eigenfunction of the energy operator. The operator of m2
0c4 − E2
t (let us call it ˆO2) can be obtained by a
subsequent application of operators ˆO+ and ˆO− in the following equations of motion:
i
∂
∂t
− m0c2
ψ = ˆO+
ψ = 0, (6.116)
−i
∂
∂t
− m0c2
ψ = ˆO−
ψ = 0. (6.117)
The operators ˆO− and ˆO+ can be viewed as ”square roots” of ˆO2:
ˆO2
ψ ≡ 2 ∂2ψ
∂t2
+ (m0c2
)2
ψ = ˆO+ ˆO−
ψ = i
∂
∂t
− m0c2
−i
∂
∂t
− m0c2
ψ = 0. (6.118)
What are the eigenfuctions? One solution is the monochromatic wave described by Eq. 6.110 (with px = py = pz = 0):
i
∂
∂t
− m0c2
−i
∂
∂t
− m0c2
e
i (−Ett)
= i
∂
∂t
− m0c2
−Et − m0c2
e
i (−Ett)
= −Et − m0c2
i
∂
∂t
− m0c2
e
i (−Ett)
= −Et − m0c2
Et − m0c2
e
i (−Ett)
= m2
0c4
− E2
t e
i (−Ett)
= 0.
(6.119)
But the complex conjugate of the monochromatic wave described by Eq. 6.110 is another possible solution:
i
∂
∂t
− m0c2
−i
∂
∂t
− m0c2
e
i (Ett)
= i
∂
∂t
− m0c2
Et − m0c2
e
i (Ett)
6.8. RELATIVISTIC QUANTUM MECHANICS 85
= Et − m0c2
i
∂
∂t
− m0c2
e
i (Ett)
= Et − m0c2
−Et − m0c2
e
i (Ett)
= m2
0c4
− E2
t e
i (Ett)
= 0. (6.120)
The second eigenfunction can be interpreted as a particle with a positive energy moving backwards in time, or as an
antiparticle moving forward in time.
Let us now turn our attention to particles that can move (p = 0). For the most interesting particles as electron or
quarks, the operator ˆO2 should have the form described by Eq. 6.112
2 ∂2
∂t2
− c2 2 ∂2
∂z2
− c2 2 ∂2
∂x2
− c2 2 ∂2
∂y2
+ (m0c2
)2
ψ. (6.121)
Let us try to find ”square roots” of the operator ˆO2 for a particle with a momentum p. In Eq. 6.118, ˆO+ and ˆO+
were complex conjugates. A similar choice for a particle with a momentum p, i.e.,
ˆO+
ψ = i
∂
∂t
+ ic
∂
∂x
+ ic
∂
∂y
+ ic
∂
∂z
− m0c2
ψ (6.122)
ˆO−
ψ = −i
∂
∂t
− ic
∂
∂x
− ic
∂
∂y
− ic
∂
∂z
− m0c2
ψ (6.123)
gives
ˆO− ˆO+ψ = ˆO2ψ = 2 ∂2
ψ
∂t2 +c 2 ∂ψ
∂t
∂ψ
∂x
+c 2 ∂ψ
∂t
∂ψ
∂y
+c 2 ∂ψ
∂t
∂ψ
∂z
−im0c2 ∂ψ
∂t
+c 2 ∂ψ
∂x
∂ψ
∂t
+ 2 ∂2
ψ
∂x2 +c 2 ∂ψ
∂x
∂ψ
∂y
+c 2 ∂ψ
∂x
∂ψ
∂z
−im0c2 ∂ψ
∂x
+c 2 ∂ψ
∂y
∂ψ
∂t
+c 2 ∂ψ
∂y
∂ψ
∂x
+ 2 ∂2
ψ
∂y2 +c 2 ∂ψ
∂y
∂ψ
∂z
−im0c2 ∂ψ
∂y
+c 2 ∂ψ
∂z
∂ψ
∂t
+c 2 ∂ψ
∂z
∂ψ
∂x
+c 2 ∂ψ
∂z
∂ψ
∂y
+ 2 ∂2
ψ
∂z2 −im0c2 ∂ψ
∂z
+im0c2 ∂ψ
∂t
+im0c2 ∂ψ
∂x
+im0c2 ∂ψ
∂y
+im0c2 ∂ψ
∂z
+(m0c2)2ψ
(6.124)
with the correct five square terms along the ”diagonal”, but also with additional twenty unwanted mixed terms.
As the second trial, let us try (na¨ıvely) to get rid of the unwanted mixed terms by introducing coefficients γj that
hopefully cancel them:
ˆO+
ψ = i
∂
∂t
γ0 + ic
∂
∂x
γ1 + ic
∂
∂y
γ2 + ic
∂
∂z
γ3 − m0c2
ψ
(6.125)
ˆO−
ψ = −i
∂
∂t
γ0 − ic
∂
∂x
γ1 − ic
∂
∂y
γ2 − ic
∂
∂z
γ3 − m0c2
ψ.
(6.126)
Then,
ˆO− ˆO+ψ = ˆO2ψ = γ2
0
2 ∂2
ψ
∂t2 +γ0γ1c 2 ∂ψ
∂t
∂ψ
∂x
+γ0γ2c 2 ∂ψ
∂t
∂ψ
∂y
+γ0γ3c 2 ∂ψ
∂t
∂ψ
∂z
−iγ0m0c2 ∂ψ
∂t
+γ1γ0c 2 ∂ψ
∂x
∂ψ
∂t
+γ2
1
2 ∂2
ψ
∂x2 +γ1γ2c 2 ∂ψ
∂x
∂ψ
∂y
+γ1γ3c 2 ∂ψ
∂x
∂ψ
∂z
−iγ1m0c2 ∂ψ
∂x
+γ2γ0c 2 ∂ψ
∂y
∂ψ
∂t
+γ2γ1c 2 ∂ψ
∂y
∂ψ
∂x
+γ2
2
2 ∂2
ψ
∂y2 +γ2γ3c 2 ∂ψ
∂y
∂ψ
∂z
−iγ2m0c2 ∂ψ
∂y
+γ3γ0c 2 ∂ψ
∂z
∂ψ
∂t
+γ3γ1c 2 ∂ψ
∂z
∂ψ
∂x
+γ3γ2c 2 ∂ψ
∂z
∂ψ
∂y
+γ2
3
2 ∂2
ψ
∂z2 −iγ3m0c2 ∂ψ
∂z
+iγ0m0c2 ∂ψ
∂t
+iγ1m0c2 ∂ψ
∂x
+iγ2m0c2 ∂ψ
∂y
+iγ3m0c2 ∂ψ
∂z
+(m0c2)2ψ.
(6.127)
Obviously, the terms with −iγjm0c2 cancel each other, which removes six unwanted terms. Can we also remove the
remaining dozen of unwanted mixed derivative terms? In order to do it, we need the following conditions to be fulfilled:
γ2
0 = 1 (6.128)
γ2
1 = −1 (6.129)
γ2
2 = −1 (6.130)
γ2
3 = −1 (6.131)
γjγk + γkγj = 0 for j = k. (6.132)
These conditions are clearly in conflict. The first four condition require γj to be ±1 or ±i, but the last condition
requires them to be zero. There are no complex numbers that allow us to get the correct operator ˆO2. However, there
are mathematical objects, that can fulfil the listed conditions simultaneously. Such objects are matrices.
86 CHAPTER 6. SPIN
6.8.1 Finding the matrices
Let us replace the coefficients γj in Eqs. 6.125–6.125 by matrices4 ˆγj:
ˆO+
Ψ = i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0 (6.133)
ˆO−
Ψ = −i
∂
∂t
ˆγ0
− ic
∂
∂x
ˆγ1
− ic
∂
∂y
ˆγ2
− ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = 0. (6.134)
We need a set of four matrices ˆγj with the following properties:
ˆγ0
· ˆγ0
= 1, (6.135)
ˆγ1
· ˆγ1
= −1 ˆγ2
· ˆγ2
= −1 ˆγ3
· ˆγ3
= −1 (6.136)
and
ˆγj
· ˆγk
+ ˆγk
· ˆγj
= 0 for j = k. (6.137)
In addition, there is a physical restriction. We know that the operator of energy (Hamiltonian) is
ˆH = i
∂
∂t
(6.138)
We can get the Dirac Hamiltonian by multiplying Eq. 6.133 by ˆγ0 from left:
i
∂
∂t
ˆ1Ψ = −ic
∂
∂x
ˆγ0
· ˆγ1
− ic
∂
∂y
ˆγ0
· ˆγ2
− ic
∂
∂z
ˆγ0
· ˆγ3
+ m0c2
ˆγ0
Ψ = 0. (6.139)
Operator of any measurable quantity must be Hermitian ( ψ| ˆOψ = ˆOψ|ψ ) in order to give real values of the
measured value. Since the terms in the Hamiltonian are proportional to ˆγ0 or to ˆγ0 · ˆγj, all these matrices must be
Hermitian (the elements in the j-th row and k-th column must be equal to the complex conjugates of the elements in the
k-th row and j-th column for each j and k.).
We have a certain liberty in choosing the matrices. A matrix equation is nothing else than a set of equations. One of
the matrices can be always chosen to be diagonal. Let us assume that ˆγ0 is diagonal.5 How should the diagonal elements
of ˆγ0 look like? In order to fulfill Eq. 6.135, the elements must be +1 or −1.
Another requirement follows from a general property of matrix multiplication: Trace of the matrix product ˆA · ˆB is
the same as that of ˆB · ˆA. Let us assume that ˆA = ˆγj and ˆB = ˆγ0 · ˆγj. Then,
Tr{ˆγj
· ˆγ0
· ˆγj
} = Tr{ˆγj
· ˆγj
· ˆγ0
}. (6.140)
But Eq. 6.137 tells us that ˆγ0 · ˆγj = −ˆγj · ˆγ0. Therefore, the left-hand side of Eq. 6.140 can be written as Tr{ˆγj ·
(−ˆγj) · ˆγ0}, resulting in
− Tr{ˆγj
· ˆγj
· ˆγ0
} = Tr{ˆγj
· ˆγj
· ˆγ0
}, (6.141)
and using Eq. 6.137
Tr{ˆγ0
} = −Tr{ˆγ0
}. (6.142)
It can be true only if the trace is equal to zero. Consequently, the diagonal of ˆγ0 must contain the same number of
+1 and −1 elements. It also tells us that the dimension of the ˆγj matrices must be even. Can they be two-dimensional?
4
In relativistic quantum mechanics, these matrices can be treated as four components of a fourvector.
There are two types of four-vectors (contravariant and covariant) which transform differently.
There is a convention to distinguish these two types by writing components of covariant vectors with
lower indices and components of contravariant vectors with upper indices. To keep this convention, we
label the gamma matrices with upper indices, do not confuse them with power!
5
This is a good choice because it results in a diagonal matrix representing the Hamiltonian, which
is convenient.
6.8. RELATIVISTIC QUANTUM MECHANICS 87
No, for the following reason. The four ˆγj matrices must be linearly independent, and it is impossible to find four
linearly independent 2 × 2 matrices so that all fulfill Eq. 6.137.6
Is it possible to find four-dimensional ˆγj matrices? Yes. We start by choosing
ˆγ0
=




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 (6.143)
(the diagonal must contain two +1 elements and two −1 elements, their order is arbitrary, but predetermines forms
of the other matrices).
Being diagonal, ˆγ0 is of course Hermitian. The ˆγ0 · ˆγj products




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 ·





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
γj
3,1 γj
3,2 γj
3,3 γj
3,4
γj
4,1 γj
4,2 γj
4,3 γj
4,4





=





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
−γj
3,1 −γj
3,2 −γj
3,3 −γj
3,4
−γj
4,1 −γj
4,2 −γj
4,3 −γj
4,4





(6.144)
must be also Hermitian, i.e.,





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
−γj
3,1 −γj
3,2 −γj
3,3 −γj
3,4
−γj
4,1 −γj
4,2 −γj
4,3 −γj
4,4





=





(γj
1,1)∗ (γj
2,1)∗ −(γj
3,1)∗ −(γj
4,1)∗
(γj
1,2)∗ (γj
2,2)∗ −(γj
3,2)∗ −(γj
4,2)∗
(γj
1,3)∗ (γj
2,3)∗ −(γj
3,3)∗ −(γj
4,3)∗
(γj
1,4)∗ (γj
2,4)∗ −(γj
3,4)∗ −(γj
4,4)∗





. (6.145)
At the same time, Eq. 6.137 requires ˆγ0 · ˆγj = −ˆγj · ˆγ0





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
−γj
3,1 −γj
3,2 −γj
3,3 −γj
3,4
−γj
4,1 −γj
4,2 −γj
4,3 −γj
4,4





= −





γj
1,1 γj
1,2 γj
1,3 γj
1,4
γj
2,1 γj
2,2 γj
2,3 γj
2,4
γj
3,1 γj
3,2 γj
3,3 γj
3,4
γj
4,1 γj
4,2 γj
4,3 γj
4,4





·




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 =





−γj
1,1 −γj
1,2 γj
1,3 γj
1,4
−γj
2,1 −γj
2,2 γj
2,3 γj
2,4
−γj
3,1 −γj
3,2 γj
3,3 γj
3,4
−γj
4,1 −γj
4,2 γj
4,3 γj
4,4





, (6.146)
which is possible only if the red elements are equal to zero. Eq. 6.145 shows that the blue elements form two adjoint
2 × 2 matrices for each j > 0:
ˆγj
=





0 0 γj
1,3 γj
1,4
0 0 γj
2,3 γj
2,4
γj
3,1 γj
3,2 0 0
γj
4,1 γj
4,2 0 0





=





0 0 γj
1,3 γj
1,4
0 0 γj
2,3 γj
2,4
−(γj
1,3)∗ −(γj
2,3)∗ 0 0
−(γj
1,4)∗ −(γj
2,4)∗ 0 0





=
ˆ0 ˆσj
−(ˆσj)† ˆ0
. (6.147)
Now we use Eqs. 6.136 and 6.137 to find the actual forms of three ˆσj (and consequently ˆγj) matrices for j > 0.
Eq. 6.136 requires
ˆ0 ˆσj
−(ˆσj)† ˆ0
·
ˆ0 ˆσj
−(ˆσj)† ˆ0
=
−ˆσj · (ˆσj)† ˆ0
ˆ0 −(ˆσj)† · ˆσj = −
ˆ1 ˆ0
ˆ0 ˆ1
, (6.148)
6
If the ˆγj
matrices are linearly independent, they can be used as a basis. If they constitute a
basis, there must exist a linear combination of ˆγj
giving any 2 × 2 matrix, e.g., the unit matrix ˆ1:
ˆ1 = c0ˆγ0
+ c1ˆγ1
+ c2ˆγ2
+ c3ˆγ3
. Let us now multiply this equation by ˆγ0
from left (and use Eq. 6.135)
ˆγ0
= c0
ˆ1 + c1ˆγ0
· ˆγ1
+ c2ˆγ0
· ˆγ2
+ c3ˆγ0
· ˆγ3
,
then from right
ˆγ0
= c0
ˆ1 + c1ˆγ1
· ˆγ0
+ c2ˆγ2
· ˆγ0
+ c3ˆγ3
· ˆγ0
,
and sum both equations. If the matrices fulfill Eq. 6.137, the result must be 2ˆγ0
= 2c0
ˆ1, but this cannot
be true for our choice of ˆγ0
=
1 0
0 −1
.
88 CHAPTER 6. SPIN
i.e.,
ˆσj
· (ˆσj
)†
= (ˆσj
)†
· ˆσj
= ˆ1 (6.149)
Eq. 6.149 is obviously true if the ˆσj matrices are Hermitian (ˆσj = (ˆσj)†), i.e. σj
m,n = (σj
n,m)∗. It implies that the
ˆσj matrices have the following form:
ˆσj
=
aj cj
c∗
j bj
, (6.150)
where aj and bj are real, and cj is complex. Eq. 6.149 can be then written as
ˆσj
· (ˆσj
)†
= ˆσj
· ˆσj
=
aj cj
c∗
j bj
·
aj cj
c∗
j bj
=
a2
j + |cj|2 (aj + bj)cj
(aj + bj)c∗
j b2
j + |cj|2 =
1 0
0 1
. (6.151)
The off-diagonal terms of the product matrix must be equal to zero, which is true if aj = −bj or |cj| = 0. In the
former case, matrices ˆσj can be written as
ˆσj
=
1 − |cj|2 cj
c∗
j − 1 − |cj|2 , (6.152)
in the latter case, there are only two possibilities how to construct the ˆσj matrix:
ˆσj
=
1 0
0 1
or ˆσj
=
1 0
0 −1
(6.153)
(note that |cj|2 = 0 ⇒ a2
j = b2
j = 1.) Eq. 6.137 shows that the second option is correct. Eq. 6.137 requires
ˆ0 ˆσj
−(ˆσj)† ˆ0
·
ˆ0 ˆσk
−(ˆσk)† ˆ0
+
ˆ0 ˆσk
−(ˆσk)† ˆ0
·
ˆ0 ˆσj
−(ˆσj)† ˆ0
= −
ˆσj · (ˆσk)† + ˆσk · (ˆσj)† ˆ0
ˆ0 (ˆσj)† · ˆσk + (ˆσk)† · ˆσj =
ˆ0 ˆ0
ˆ0 ˆ0
,
(6.154)
therefore no ˆσj can be a unit matrix.
As Eq. 6.153 unambiguously defines one sigma matrix (let us call it ˆσ3), the other two (ˆσ1 and ˆσ2) are given by
Eq. 6.152. According to Eq. 6.137,
1 0
0 −1
·
1 − |cj|2 cj
c∗
j − 1 − |cj|2 +
1 − |cj|2 cj
c∗
j − 1 − |cj|2 ·
1 0
0 −1
=
2 1 − |cj|2 0
0 −2 1 − |cj|2
=
0 0
0 0
,
(6.155)
showing that |cj|2 = 1 and the diagonal elements of ˆσ1 and ˆσ2 are equal to zero. Therefore, these equations can be
written as
ˆσ1
=
0 eiφ1
e−iφ1 0
ˆσ2
=
0 eiφ2
e−iφ2 0
(6.156)
According to Eq. 6.137,
0 eiφ1
e−iφ1 0
·
0 eiφ2
e−iφ2 0
+
0 eiφ2
e−iφ2 0
·
0 eiφ1
e−iφ1 0
=
0 ei(φ1−φ2) + e−i(φ1−φ2)
e−i(φ1−φ2) + ei(φ1−φ2) 0
=
0 2 cos (φ1 − φ2)
2 cos (φ1 − φ2) 0
=
0 0
0 0
. (6.157)
The off-diagonal elements of the sum of the matrix products are equal to zero if the phases differ by π/2. Choosing
φ1 = 0, the set of three sigma matrices is
ˆσ1
=
0 1
1 0
ˆσ2
=
0 −i
i 0
ˆσ3
=
1 0
0 −1
(6.158)
and the set of the four gamma matrices is
ˆγ0
=




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 ˆγ1
=




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 ˆγ2
=




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0



 ˆγ3
=




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 . (6.159)
6.8. RELATIVISTIC QUANTUM MECHANICS 89
6.8.2 Dirac equation
With the help of the ˆγj matrices, we can modify our definition of ˆO+ and ˆO− to get the correct operator ˆO2:



i
∂
∂t




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 + ic
∂
∂z




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 + ic
∂
∂x




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 + ic
∂
∂y




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0




−m0c2




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1












ψ1
ψ2
ψ3
ψ4



 = ˆO+
Ψ = 0,
(6.160)



−i
∂
∂t




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 − ic
∂
∂z




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 − ic
∂
∂x




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 − ic
∂
∂y




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0




−m0c2




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1












ψ1
ψ2
ψ3
ψ4



 = ˆO−
Ψ = 0.
(6.161)
Introducing matrices means that we do not have a single equation of motion, but a set of four equations for four
coupled wave functions. The complete wave function Ψ is therefore a vector consisting of four components. The operators
ˆO+ and ˆO− consist of partial derivative operators summarized in Eq. 6.111, and Eq. 6.111 also shows that a monochromatic
wave ψ = e
i (pxx+pyy+pzz−Ett)
is eigenfunction of the partial derivative operators, with the eigenvalues equal to
Et, px, py, pz. Also note that the 2 × 2 sub-matrices, which form the ˆγj matrices, always appear with the opposite sign
on the first and last two lines, except for the unit matrix associated with the m0c2 term. It is therefore useful to use
a complex conjugate of the aforementioned monochromatic wave as eigenfuction on the last two lines, in order to get
eigenvalues with opposite signs. Possible solutions of the Dirac equation can be than assumed to have a form
Ψ =




u1ψ
u2ψ
v1ψ∗
v2ψ∗



 , (6.162)
where u1, u2, v1, v2 are coefficients to be determined. The Dirac equation7 can be written as
7
Actually, two equations, one for ˆO+ and another one for ˆO−.
90 CHAPTER 6. SPIN



i
∂
∂t




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 + ic
∂
∂z




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 + ic
∂
∂x




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 + ic
∂
∂y




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0




−m0c2




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1












u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = ˆO+
Ψ = 0,
(6.163)



−i
∂
∂t




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 − ic
∂
∂z




0 0 1 0
0 0 0 −1
−1 0 0 0
0 1 0 0



 − ic
∂
∂x




0 0 0 1
0 0 1 0
0 −1 0 0
−1 0 0 0



 − ic
∂
∂y




0 0 0 −i
0 0 i 0
0 i 0 0
−i 0 0 0




−m0c2




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1












u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = ˆO−
Ψ = 0,
(6.164)
or shortly
i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1




u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = ˆO+
Ψ = 0
(6.165)
−i
∂
∂t
ˆγ0
− ic
∂
∂x
ˆγ1
− ic
∂
∂y
ˆγ2
− ic
∂
∂z
ˆγ3
− m0c2ˆ1




u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = ˆO−
Ψ = 0.
(6.166)
For our wavefunctions,
ˆO+
Ψ =




Etu1ψ + cpxv2ψ∗ − icpyv2ψ∗ + cpzv1ψ∗ − m0c2u1ψ
Etu2ψ + cpxv1ψ∗ + icpyv1ψ∗ − cpzv2ψ∗ − m0c2u2ψ
Etv1ψ∗ + cpxu2ψ∗ − icpyu2ψ∗ + cpzu1ψ∗ − m0c2v1ψ∗
Etv2ψ∗ + cpxu1ψ∗ + icpyu1ψ∗ − cpzu2ψ∗ − m0c2v2ψ∗



 = 0 (6.167)
and
ˆO2
Ψ = ˆO− ˆO+
Ψ =




E2
t − c2p2 − (m0c2)2
E2
t − c2p2 − (m0c2)2
E2
t − c2p2 − (m0c2)2
E2
t − c2p2 − (m0c2)2








u1ψ
u2ψ
v1ψ∗
v2ψ∗



 = (E2
t − c2
p2
− (m0c2
)2
)Ψ (6.168)
in agreement with Eq.6.109.
Eq. 6.167 can be also used to find four explicit solutions of the Dirac equation, treating u1, u2, v1, v2 as unknown
variables to be determined. The solutions are found by setting one of the coefficients u1, u2, v1, v2 to zero, and calculating
the other coefficients so that the following normalization condition is fulfilled
Lˆ
0
Lˆ
0
Lˆ
0
Ψ∗
Ψdxdydz = 1 (6.169)
(other normalizations could be used as well). The solutions have the following form.
6.9. HAMILTONIAN OF SPIN MAGNETIC MOMENT 91
Ψ1 =
Et + m0c2
2EtL3





ψ
0
cpz
Et+m0c2 ψ∗
c(px+ipy)
Et+m0c2 ψ∗





, Ψ2 =
Et + m0c2
2EtL3





0
ψ
c(px−ipy)
Et+m0c2 ψ∗
−cpz
Et+m0c2 ψ∗





,
Ψ3 =
Et + m0c2
2EtL3





cpz
Et+m0c2 ψ
c(px+ipy)
Et+m0c2 ψ
ψ∗
0





, Ψ4 =
Et + m0c2
2EtL3





c(px−ipy)
Et+m0c2 ψ
−cpz
Et+m0c2 ψ
0
ψ∗





, (6.170)
where ψ = e
i (pxx+pyy+pzz−Ett)
.
Eq. 6.133
i
∂
∂t
ˆγ0
+ ic
∂
∂x
ˆγ1
+ ic
∂
∂y
ˆγ2
+ ic
∂
∂z
ˆγ3
− m0c2ˆ1 Ψ = ˆO+
Ψ = 0 (6.171)
is known as the Dirac equation. When postulated by Dirac, Eq. 6.133 naturally
explained the behavior of particles with spin number 1/2 and predicted existence of
antiparticles, discovered a few years later.
How is the Dirac equation related to the Schr¨odinger equation? We came to the Schr¨odinger equation using the
relation E = p2/2m (energy of a free particle, i.e., kinetic energy), which is only an approximation for low speeds,
obtained by neglecting the E2 term (E2 (m0c2)2 for v2 c2) in Eq. 6.109:
(m0c2
)2
= (m0c2
+ E)2
− c2
p2
= (m0c2
)2
+ 2E(m0c2
) + E2
− c2
p2
≈ (m0c2
)2
+ 2E(m0c2
) − c2
p2
⇒ E =
p2
2m0
. (6.172)
6.9 Hamiltonian of spin magnetic moment
Our goal is to find Hamiltonian for a relativistic charged particle in a magnetic field. When we compare the classical
Hamiltonian of a particle in an electromagnetic (Eq. 6.66) with the classical Hamiltonian of a free particle H = (p)2/(2m)
outside the field, we see that the presence of an electromagnetic field requires the following modifications:
H → H − QV p → p − QA, (6.173)
Accordingly, the operators of energy and momentum in the quantum description change to
i
∂
∂t
→ i
∂
∂t
− QV − i
∂
∂x
→ −i
∂
∂x
− QAx − i
∂
∂y
→ −i
∂
∂y
− QAy − i
∂
∂z
→ −i
∂
∂z
− QAz. (6.174)
This modifies Eq. 6.139 to
i
∂
∂t
− QV ˆ1Ψ = −c i
∂
∂x
+ QAx ˆγ0
ˆγ1
− c i
∂
∂y
+ QAy ˆγ0
ˆγ2
− c i
∂
∂z
+ QAz ˆγ0
ˆγ3
+ m0c2
ˆγ0
Ψ
(6.175)
In order to obtain the Hamiltonian describing energy of our particle in a magnetic field, we apply the operator
(i ∂/∂t − QV ) twice
92 CHAPTER 6. SPIN
i
∂
∂t
− QV i
∂
∂t
− QV ˆ1Ψ = i
∂
∂t
− QV
2
Ψ
= c2
i
∂
∂x
+ QAx
2
ˆγ0
ˆγ1
ˆγ0
ˆγ1
+ c2
i
∂
∂y
+ QAy
2
ˆγ0
ˆγ2
ˆγ0
ˆγ2
+ c2
i
∂
∂z
+ QAz
2
ˆγ0
ˆγ3
ˆγ0
ˆγ3
+ m2
0c4
ˆγ0
ˆγ0
Ψ
−m0c3
i
∂
∂x
+ QAx ˆγ0
ˆγ1
ˆγ0
+ i
∂
∂y
+ QAy ˆγ0
ˆγ2
ˆγ0
+ i
∂
∂z
+ QAz ˆγ0
ˆγ3
ˆγ0
Ψ
−m0c3
i
∂
∂x
+ QAx ˆγ0
ˆγ0
ˆγ1
+ i
∂
∂y
+ QAy ˆγ0
ˆγ0
ˆγ2
+ i
∂
∂z
+ QAz ˆγ0
ˆγ0
ˆγ3
Ψ
+c2
i
∂
∂x
+ QAx i
∂
∂y
+ QAy ˆγ0
ˆγ1
ˆγ0
ˆγ2
+ i
∂
∂y
+ QAy i
∂
∂x
+ QAx ˆγ0
ˆγ2
ˆγ0
ˆγ1
Ψ
+c2
i
∂
∂y
+ QAy i
∂
∂z
+ QAz ˆγ0
ˆγ2
ˆγ0
ˆγ3
+ i
∂
∂z
+ QAz i
∂
∂y
+ QAy ˆγ0
ˆγ3
ˆγ0
ˆγ2
Ψ
+c2
i
∂
∂z
+ QAz i
∂
∂x
+ QAx ˆγ0
ˆγ3
ˆγ0
ˆγ1
+ i
∂
∂x
+ QAx i
∂
∂z
+ QAz ˆγ0
ˆγ1
ˆγ0
ˆγ3
Ψ = 0. (6.176)
We use the properties of the gamma matrices (Eqs. 6.135–6.137) to simplify the equation. In particular, we invert of
the order of matrices in the products
ˆγ0
ˆγj
ˆγ0
= −(ˆγ0
ˆγ0
)ˆγj
= ˆγj
, (6.177)
ˆγ0
ˆγj
ˆγ0
ˆγj
= −(ˆγ0
ˆγ0
)(ˆγj
ˆγj
) = −(ˆ1)(−ˆ1) = ˆ1, (6.178)
ˆγ0
ˆγj
ˆγ0
ˆγk
= −(ˆγ0
ˆγ0
)(ˆγj
ˆγk
) = −(ˆ1)(ˆγj
ˆγk
) = −ˆγj
ˆγk
= ˆγk
ˆγj
(6.179)
and obtain
i
∂
∂t
− QV
2
ˆ1Ψ = c2
i
∂
∂x
+ QAx
2
ˆ1 + c2
i
∂
∂y
+ QAy
2
ˆ1 + c2
i
∂
∂z
+ QAz
2
ˆ1 + m2
0c4ˆ1 Ψ
−m0c3
i
∂
∂x
+ QAx ˆγ1
+ i
∂
∂y
+ QAy ˆγ2
+ i
∂
∂z
+ QAz ˆγ3
Ψ
+m0c3
i
∂
∂x
+ QAx ˆγ1
+ i
∂
∂y
+ QAy ˆγ2
+ i
∂
∂z
+ QAz ˆγ3
Ψ
−c2
i
∂
∂x
+ QAx i
∂
∂y
+ QAy − i
∂
∂y
+ QAy i
∂
∂x
+ QAx ˆγ1
ˆγ2
Ψ
−c2
i
∂
∂y
+ QAy i
∂
∂z
+ QAz − i
∂
∂z
+ QAz i
∂
∂y
+ QAy ˆγ2
ˆγ3
Ψ
−c2
i
∂
∂z
+ QAz i
∂
∂x
+ QAx − i
∂
∂x
+ QAx i
∂
∂z
+ QAz ˆγ3
ˆγ1
Ψ = 0,
(6.180)
where the second line and the third line cancel each other. To proceed, we need to evaluate the products of operators
on the last three lines (we must be very careful with differentiation).
i
∂
∂x
+ QAx i
∂
∂y
+ QAy − i
∂
∂y
+ QAy i
∂
∂x
+ QAx ψ =
− 2 ∂
∂x
∂ψ
∂y
−
∂
∂y
∂ψ
∂x
+ Q2
(AxAy − AyAx)ψ + i Q
∂(Ayψ)
∂x
+ Ax
∂ψ
∂y
−
∂(Axψ)
∂y
− Ay
∂ψ
∂x
=
i Q
∂Ay
∂x
ψ + Ay
∂ψ
∂x
+ Ax
∂ψ
∂y
−
∂Ax
∂y
ψ − Ax
∂ψ
∂y
− Ay
∂ψ
∂x
= i Q
∂Ay
∂x
−
∂Ax
∂y
ψ = i QBzψ (6.181)
because ∂2ψ/∂x∂y = ∂2ψ/∂y∂x. The combinations on the last two lines of Eq. 6.180 are obtained in the same
manner. The products ˆγ1ˆγ2, ˆγ2ˆγ3, and ˆγ3ˆγ1 can be calculated from Eq. 6.147
6.9. HAMILTONIAN OF SPIN MAGNETIC MOMENT 93
ˆγ1
ˆγ2
=
ˆ0 ˆσ1
−ˆσ1 ˆ0
ˆ0 ˆσ2
−ˆσ2 ˆ0
= −
ˆσ1 ˆσ2 ˆ0
ˆ0 ˆσ1 ˆσ2 = −i
ˆσ3 ˆ0
ˆ0 ˆσ3 , (6.182)
ˆγ2
ˆγ3
=
ˆ0 ˆσ2
−ˆσ2 ˆ0
ˆ0 ˆσ3
−ˆσ3 ˆ0
= −
ˆσ2 ˆσ2 ˆ0
ˆ0 ˆσ2 ˆσ3 = −i
ˆσ1 ˆ0
ˆ0 ˆσ1 , (6.183)
ˆγ3
ˆγ1
=
ˆ0 ˆσ3
−ˆσ3 ˆ0
ˆ0 ˆσ1
−ˆσ1 ˆ0
= −
ˆσ3 ˆσ1 ˆ0
ˆ0 ˆσ3 ˆσ1 = −i
ˆσ2 ˆ0
ˆ0 ˆσ2 , (6.184)
where the following important properties of the ˆσj matrices were used in the lasts steps:
ˆσ1
ˆσ2
=
0 1
1 0
0 −i
i 0
=
i 0
0 −i
= iˆσ3
(6.185)
ˆσ2
ˆσ3
=
0 −i
i 0
1 0
0 −1
=
0 i
i 0
= iˆσ1
(6.186)
ˆσ3
ˆσ1
=
1 0
0 −1
0 1
1 0
=
0 1
−1 0
= iˆσ2
. (6.187)
After inserting everything into Eq. 6.180, we get
i
∂
∂t
− QV
2 ˆ1 ˆ0
ˆ0 ˆ1
Ψ = c2
i
∂
∂x
+ QAx
2
+ c2
i
∂
∂y
+ QAy
2
+ c2
i
∂
∂z
+ QAz
2
+ m2
0c4
ˆ1 ˆ0
ˆ0 ˆ1
Ψ
− c2
Q Bx
ˆσ1 ˆ0
ˆ0 ˆσ1 + By
ˆσ2 ˆ0
ˆ0 ˆσ2 + Bz
ˆσ3 ˆ0
ˆ0 ˆσ3 Ψ. (6.188)
Now we have a relativistic equation describing our particle in an electromagnetic field. Let us now separate the mass
contribution to the energy from the operator i ∂/∂t and let us call the difference ˆH (it becomes clear soon why we choose
the same symbol as the symbol used for the Hamiltonian in the Schr¨odinger equation):
ˆH = i
∂
∂t
− m0c2
, (6.189)
Eq. 6.188 can be rewritten as
ˆH + m0c2
− QV
2 ˆ1 ˆ0
ˆ0 ˆ1
Ψ = ( ˆH − QV )2
+ 2m0c2
( ˆH − QV ) + m2
0c4
ˆ1 ˆ0
ˆ0 ˆ1
Ψ =
c2
i
∂
∂x
+ QAx
2
+ c2
i
∂
∂y
+ QAy
2
+ c2
i
∂
∂z
+ QAz
2
+ m2
0c4
ˆ1 ˆ0
ˆ0 ˆ1
Ψ
− c2
Q Bx
ˆσ1 ˆ0
ˆ0 ˆσ1 + By
ˆσ2 ˆ0
ˆ0 ˆσ2 + Bz
ˆσ3 ˆ0
ˆ0 ˆσ3 Ψ. (6.190)
Dividing both sides of the equation by 2m0c2 gives
( ˆH − QV )2
2m0c2
+ ˆH − QV
ˆ1 ˆ0
ˆ0 ˆ1
Ψ =
1
2m0
i
∂
∂x
+ QAx
2
+ i
∂
∂y
+ QAy
2
+ i
∂
∂z
+ QAz
2 ˆ1 ˆ0
ˆ0 ˆ1
Ψ
−
Q
2m0
Bx
ˆσ1 ˆ0
ˆ0 ˆσ1 + By
ˆσ2 ˆ0
ˆ0 ˆσ2 + Bz
ˆσ3 ˆ0
ˆ0 ˆσ3 Ψ. (6.191)
Note that the rest energy of particles m0c2 is huge. Unless the eigenvalue of ˆH is very large (which is not expected
in a standard NMR experiment), the term with m0c2 in the denominator of the first term can be safely neglected. For
the same reason, the factors ±cpz/(Et + m0c2) and c(px ± ipy)/(Et + m0c2) in Eq. 6.170 are close to zero for v c.
94 CHAPTER 6. SPIN
The block-diagonal form of all matrices reveals that the first two equations and the last two equations can be solved
separately. Therefore, we can write
ˆH
u1ψ
u2ψ
≈
1
2m0
i
∂
∂x
+ QAx
2
+ i
∂
∂y
+ QAy
2
+ i
∂
∂z
+ QAz
2
+ QV ˆ1 −
Q
2m0
Bx ˆσ1
+ By ˆσ2
+ Bz ˆσ3 u1ψ
u2ψ
.
(6.192)
Now it is easy to identify the Hamiltonian of potential electric energy QV and, by comparison with the corresponding
equation for an orbital magnetic moment (Eq. 6.104), the Hamiltonian of interactions with the magnetic field
−
Q
2m0
Bx ˆσ1
+ By ˆσ2
+ Bz ˆσ3
. (6.193)
The mentioned comparison also helps us to identify the operator of the components of the spin magnetic moment:
ˆµx =
Q
2m0
ˆσ1
=
Q
2m0
0 1
1 0
, (6.194)
ˆµy =
Q
2m0
ˆσ2
=
Q
2m0
0 −i
i 0
, (6.195)
ˆµz =
Q
2m0
ˆσ3
=
Q
2m0
1 0
0 −1
. (6.196)
6.9.1 Operators of spin angular momentum
Our final task is to find the operators of the components of the spin angular momentum, which also gives us the value
of the magnetogyric ratio. Eq. 6.192 itself is not sufficient because it does not say which constants belong to the spin
angular momentum and which constitute the magnetogyric ratio. We cannot use the classical definition either because
our case does not have a classical counterpart. But we can use the commutation relations Eqs. 6.78–6.81, which define
the operators of any angular momentum components. In order to distinguish it from the orbital angular momentum L, we
label the spin angular momentum I. Operators of Ix, Iy, Iz must fulfill the same commutation relations as the operators
of Lx, Ly, Lz:
ˆIx
ˆIy − ˆIy
ˆIx = i ˆIz, ˆIy
ˆIz − ˆIz
ˆIy = i ˆIx, ˆIz
ˆIx − ˆIx
ˆIz = i ˆIy. (6.197)
Comparison with Eqs. 6.185–6.187 shows that the right choice is
ˆIx =
2
0 1
1 0
ˆIy =
2
0 −i
i 0
ˆIz =
2
1 0
0 −1
ˆI2
=
3 2
4
1 0
0 1
.
(6.198)
Comparison of Eq. 6.192 with Eq. 6.104 shows that the magnetogyric ratio differs by
a factor of 2 from the value for orbital magnetic moment:
γ = 2
Q
2m
. (6.199)
6.9.2 Eigenfunctions and eigenvalues of ˆIz
The fact that ˆIz is diagonal tells us that we have written the matrix representations of
the operators of the spin angular momentum in the basis formed by the eigenfunctions
of ˆIz. This basis is a good choice if the matrix representing Hamiltonian is also diagonal
6.9. HAMILTONIAN OF SPIN MAGNETIC MOMENT 95
in this basis and eigenfunctions of ˆIz are the same as eigenfunctions of the Hamiltonian,
representing stationary states. These eigenfunctions can be
1
L3
ψ
0
,
1
L3
0
ψ
, (6.200)
i.e., the two-dimensional variants of the free-particle wavefunctions from Eq. 6.170
in the low-energy approximation. However, the normalization coefficients and ψ (or ψ∗
)
can be canceled out in the eigenvalue equations and the eigenfunctions can be replaced
by the vectors
1
0
,
0
1
, (6.201)
corresponding to the first and second wavefunctions in Eq. 6.170.
Traditionally, eigenfunctions of ˆIz are labeled as |α or | ↑ and |β or | ↓ .
ˆIz|α = +
2
|α ˆIz| ↑ = +
2
| ↑
2
1 0
0 −1
1
0
= +
2
1
0
, (6.202)
ˆIz|β = −
2
|β ˆIz| ↓ = −
2
| ↓
2
1 0
0 −1
0
1
= −
2
0
1
. (6.203)
Note that the vectors used to represent |α and |β in Eqs. 6.202 and 6.203 are not
the only choice. Vectors in Eqs. 6.202 and 6.203 have a phase set to zero (they are made
of real numbers). Any other phase φ would work as well, e.g.
1
0
→
eiφ
0
. (6.204)
• If the particle is in state |α , the result of measuring Iz is always + /2. The
expected value is
Iz = α|Iz|α = 1 0
2
1 0
0 −1
1
0
= +
2
. (6.205)
• If the particle is in state |β , the result of measuring Iz is always − /2. The
expected value is
Iz = β|Iz|β = 0 1
2
1 0
0 −1
0
1
= −
2
. (6.206)
96 CHAPTER 6. SPIN
• Any state cα|α + cβ|β is possible, but the result of a single measurement of Iz is
always + /2 or − /2. However, the expected value of Iz is
Iz = α|Iz|β = c∗
α c∗
β
2
1 0
0 −1
cα
cβ
= (|cα|2
− |cβ|2
)
2
. (6.207)
Wave functions |α and |β are not eigenfunctions of ˆIx or ˆIy.
The eigenvalues ± /2 are closely related to the fact that spin is a relativistic effect. Special relativity requires that
the Dirac equation must not change if we rotate the coordinate frame or if it moves with a constant speed (Lorentz
transformation). This is true in general, but for the sake of simplicity, we just check rotation about the z axis.
We start by writing explicitly the Dirac equation as a set of four equations8
i
∂(u1ψ)
∂t
= −ic
∂(v1ψ∗)
∂z
− ic
∂(v2ψ∗)
∂x
− ic
∂(−iv2ψ∗)
∂y
+ m0c2
u1ψ, (6.208)
i
∂(u2ψ)
∂t
= +ic
∂(v2ψ∗)
∂z
− ic
∂(v1ψ∗)
∂x
+ ic
∂(−iv1ψ∗)
∂y
+ m0c2
u2ψ, (6.209)
i
∂(v1ψ∗)
∂t
= −ic
∂(u1ψ)
∂z
− ic
∂(u2ψ)
∂x
− ic
∂(iu2ψ)
∂y
− m0c2
v1ψ∗
, (6.210)
i
∂(v2ψ∗)
∂t
= +ic
∂(u2ψ)
∂z
− ic
∂(u1ψ)
∂x
+ ic
∂(iu1ψ)
∂y
− m0c2
v2ψ∗
. (6.211)
Let us assume that we have an original coordinate frame t, x, y, z and a rotated frame t , x , y , z . If we rotate about
z by an angle ϕ,
t = t (6.212)
z = z (6.213)
x = cos ϕx − sin ϕy (6.214)
y = sin ϕx + cos ϕy (6.215)
and
∂f
∂t
=
∂f
∂t
(6.216)
∂f
∂z
=
∂f
∂z
(6.217)
∂f
∂x
=
∂x
∂x
∂f
∂x
+
∂y
∂x
∂f
∂y
= cos ϕ
∂f
∂x
+ sin ϕ
∂f
∂y
(6.218)
∂f
∂y
=
∂x
∂y
∂f
∂x
+
∂y
∂y
∂f
∂y
= − sin ϕ
∂f
∂x
+ cos ϕ
∂f
∂y
(6.219)
and consequently
∂f
∂x
+ i
∂f
∂y
= e−iϕ ∂f
∂x
+ i
∂f
∂y
, (6.220)
∂f
∂x
− i
∂f
∂y
= eiϕ ∂f
∂x
− i
∂f
∂y
. (6.221)
8
Note that we use the form of the Dirac equation which directly defines the relativistic Hamiltonian
(Eq. 6.139).
6.9. HAMILTONIAN OF SPIN MAGNETIC MOMENT 97
We also need to transform the wavefunction Ψ to the rotated frame. We already know that rotation of a complex
function f by an angle φ can be written as f = feiφ. Let us assume that each of component of Ψ rotates by some angle
(ϕ1, ϕ2, ϕ3, ϕ4,) – the key step of our analysis will be to relate values of these angles the actual angle of rotating the
coordinate frames ϕ.
Now we have everything that we need to write the set of Eqs. 6.208–6.211 in the rotated coordinate frame:
i
∂(eiϕ1 u1ψ )
∂t
= −ic
∂(eiϕ3 v1ψ ∗
)
∂z
− ic
∂(ei(ϕ4+ϕ)v2ψ ∗
)
∂x
− ic
∂(−iei(ϕ4+ϕ)v2ψ ∗
)
∂y
+ m0c2
eiϕ1 u1ψ , (6.222)
i
∂(eiϕ2 u2ψ )
∂t
= +ic
∂(eiϕ4 v2ψ ∗
)
∂z
− ic
∂(ei(ϕ3−ϕ)v1ψ ∗
)
∂x
+ ic
∂(−iei(ϕ3−ϕ)v1ψ ∗
)
∂y
+ m0c2
eiϕ2 u2ψ , (6.223)
i
∂(eiϕ3 v1ψ ∗
)
∂t
= −ic
∂(eiϕ1 u1ψ )
∂z
− ic
∂(ei(ϕ2+ϕ)u2ψ )
∂x
− ic
∂(iei(ϕ2+ϕ)u2ψ )
∂y
− m0c2
eiϕ3 v1ψ
∗
, (6.224)
i
∂(eiϕ4 v2ψ ∗
)
∂t
= +ic
∂(eiϕ2 u2ψ )
∂z
− ic
∂(ei(ϕ1−ϕ)u1ψ )
∂x
+ ic
∂(iei(ϕ1−ϕ)u1ψ )
∂y
− m0c2
eiϕ4 v2ψ
∗
. (6.225)
According to the first postulate of the special theory of relativity, Eqs. 6.222–6.225 must have the same form as Eqs.
6.208–6.211. In other words, we must eliminate the complex exponential expressions from Eqs. 6.222–6.225. Let us first
multiply both sides of the first equation by e−iϕ1 , both sides of the second equation by e−iϕ2 , both sides of the third
equation by e−iϕ3 , and both sides of the last equation by e−iϕ4 :
i
∂(u1ψ )
∂t
= −ic
∂(ei(ϕ3−ϕ1)v1ψ ∗
)
∂z
− ic
∂(ei(ϕ4−ϕ1+ϕ)v2ψ ∗
)
∂x
− ic
∂(−iei(ϕ4−ϕ1+ϕ)v2ψ ∗
)
∂y
+ m0c2
u1ψ ,(6.226)
i
∂(u2ψ )
∂t
= +ic
∂(ei(ϕ4−ϕ2)v2ψ ∗
)
∂z
− ic
∂(ei(ϕ3−ϕ2−ϕ)v1ψ ∗
)
∂x
+ ic
∂(−iei(ϕ3−ϕ2−ϕ)v1ψ ∗
)
∂y
+ m0c2
u2ψ ,(6.227)
i
∂(v1ψ ∗
)
∂t
= −ic
∂(ei(ϕ1−ϕ3)u1ψ )
∂z
− ic
∂(ei(ϕ2−ϕ3+ϕ)u2ψ )
∂x
− ic
∂(iei(ϕ2−ϕ3+ϕ)u2ψ )
∂y
− m0c2
v1ψ
∗
, (6.228)
i
∂(v2ψ ∗
)
∂t
= +ic
∂(ei(ϕ2−ϕ4)u2ψ )
∂z
− ic
∂(ei(ϕ1−ϕ4−ϕ)u1ψ )
∂x
+ ic
∂(iei(ϕ1−ϕ4−ϕ)u1ψ )
∂y
− m0c2
v2ψ
∗
. (6.229)
This cleared the t and m0 terms. The exponential expressions disappear from the z term if ϕ1 = ϕ3 and ϕ2 = ϕ4
(i.e., if the rotation of u1ψ and v1ψ∗ is identical and the same applies to u2ψ and v2ψ∗). In order to fix the x and
y terms, we assume that ϕ1 = −ϕ2 and ϕ3 = −ϕ4, i.e., that the rotation of u1ψ and u2ψ is opposite and the same
applies to v1ψ∗ and v2ψ∗. This implies that u1ψ and u2ψ describe states with opposite spins (and v1ψ∗ and v2ψ∗ too).
Then, u1ψ and v1ψ ∗
in the x and y terms are multiplied by ei(2ϕ1−ϕ), and u2ψ and v2ψ ∗
in the x and y terms are
multiplied by e−i(2ϕ1−ϕ). In both cases, the exponential expersions disappear (are equal to one) if ϕ1 = ϕ/2. What does
it mean? If we rotate the coordinate system by a certain angle, the components of the wavefunction rotate only by half
of this angle! The function describing rotation of the wavefunction about z has the form
Rj = ei
Iz,j ϕ
2 . (6.230)
This looks very similar to Eq. 6.86, but with one important difference: rotation by 2π (360 ◦) does not give the same
eigenfunction Rj as no rotation (ϕ = 0), but changes its sign. Only rotation by 4π (720 ◦) reverts the system to the initial
state!
Eq. 6.86 tells us that the eigenvalues of the operator of the spin angular momentum are half-integer multiples of :
Iz,1 =
2
Iz,2 = −
2
. (6.231)
6.9.3 Eigenfunctions of ˆIx and ˆIy
Eigenfunctions of ˆIx are the following linear combinations of |α and |β :
1
√
2
|α +
1
√
2
|β =
1
√
2
1
1
≡ | → , (6.232)
−
i
√
2
|α +
i
√
2
|β =
1
√
2
−i
i
≡ | ← , (6.233)
98 CHAPTER 6. SPIN
or these linear combinations multiplied by a phase factor eiφ. E.g., | ← can be represented by
eiπ/2 1
√
2
−i
i
= i
1
√
2
−i
i
=
1
√
2
1
−1
. (6.234)
Eigenvalues are again /2 and − /2:
ˆIx| → = +
2
| →
2
0 1
1 0
1
√
2
1
1
= +
2
·
1
√
2
1
1
, (6.235)
ˆIx| ← = +
2
| ←
2
0 1
1 0
1
√
2
−i
i
= −
2
·
1
√
2
−i
i
. (6.236)
Eigenfunctions of ˆIy are the following linear combinations of |α and |β :
1 − i
2
|α +
1 + i
2
|β =
1
2
1 − i
1 + i
≡ |⊗ , (6.237)
−
1 + i
2
|α +
1 − i
2
|β =
1
2
1 + i
1 − i
, ≡ | (6.238)
or these linear combinations multiplied by a phase factor eiφ. E.g., |⊗ can be represented by
eiπ/4 1
2
1 − i
1 + i
=
1 + i
√
2
1
2
1 − i
1 + i
=
1
√
2
1
i
. (6.239)
Eigenvalues are again /2 and − /2:
ˆIy|⊗ = +
2
|⊗
2
0 −i
i 0
1
2
1 − i
1 + i
= +
2
·
1
2
1 − i
1 + i
, (6.240)
ˆIy| = −
2
|
2
0 −i
i 0
1
2
1 + i
1 − i
= −
2
·
1
2
1 + i
1 − i
. (6.241)
6.10 Real particles
Eq. 6.192, used to derive the value of γ, describes interaction of a particle with an
external electromagnetic field. However, charged particles are themselves sources of
electromagnetic fields. Therefore, γ is not exactly twice Q/2m. In general, the value of
γ is
γ = g
Q
2m
, (6.242)
where the constant g include corrections for interactions of the particle with its own
field (and other effects). For electron, the corrections are small and easy to calculate.
The current theoretical prediction is g = 2.0023318361(10), compared to a recent experimental
measured value of g = 2.0023318416(13). On the other hand, ”corrections” for
the constituents of atomic nuclei, quarks, are two orders of magnitude higher than the
basic value of 2! It is because quarks are not ”naked” as electrons, they are confined in
protons and nucleons, ”dressed” by interactions, not only electromagnetic, but mostly
strong nuclear with gluon. Therefore, the magnetogyric ratio of proton is difficult to
calculate and we rely on its experimental value. Everything is even more complicated
when we go to higher nuclei, consisting of multiple protons and neutrons. In such cases,
6.11. STATIONARY STATES AND ENERGY LEVEL DIAGRAM 99
Table 6.1: Values of the magnetogyric ratios of selected nuclei
Nucleus magnetogyric ratio
1
1H 267.513 × 106
rad.s−1
.T−1
13
6C 67.262 × 106
rad.s−1
.T−1
15
7N −27.116 × 106
rad.s−1
.T−1
19
9F 251.662 × 106
rad.s−1
.T−1
31
15P 108.291 × 106
rad.s−1
.T−1
adding spin angular momenta represents another level of complexity. Fortunately, all
equations derived for electron also apply to nuclei with the same eigenvalues of spin
magnetic moments (spin-1/2 nuclei), if the value of γ is replaced by the correct value
for the given nucleus.9
Magnetogyric ratios of the nuclei observed most frequently are
listed in Table 6.1
6.11 Stationary states and energy level diagram
In the presence of a homogeneous magnetic field B0 = (0, 0, B0), the evolution of the system is given by the Hamiltonian
ˆH = −γB0
ˆIz. The Schr¨odinger equation is then
i
∂
∂t
cα
cβ
= −γB0
2
1 0
0 −1
cα
cβ
, (6.243)
which is a set of two equations with separated variables
dcα
dt
= +i
γB0
2
cα, (6.244)
dcβ
dt
= −i
γB0
2
cβ, (6.245)
with the solution
cα = cα(t = 0)e+i
γB0
2
t
= cα(t = 0)e−i
ω0
2
t
, (6.246)
cβ = cβ(t = 0)e−i
γB0
2
t
= cβ(t = 0)e+i
ω0
2
t
. (6.247)
If the initial state is |α , cα(t = 0) = 1, cβ(t = 0) = 0, and
cα = e−i
ω0
2
t
, (6.248)
cβ = 0. (6.249)
Note that the evolution changes only the phase factor, but the system stays in state |α (all vectors described by
Eq. 6.204 correspond to state |α ). It can be shown by calculating the probability that the system is in the |α or |β
state.
9
NMR in organic chemistry and biochemistry is usually limited to spin-1/2 nuclei because signal
decays too fast if the spin number is grater than 1/2.
100 CHAPTER 6. SPIN
Pα = c∗
αcα = e+i
ω0
2
t
e−i
ω0
2
t
= 1, (6.250)
Pβ = c∗
βcβ = 0. (6.251)
If the initial state is |β , cα(t = 0) = 0, cβ(t = 0) = 1, and
cα = 0, (6.252)
cβ = e+i
ω0
2
t
. (6.253)
Again, the evolution changes only the phase factor, but the system stays in state |β . The probability that the system
is in the |α or |β state is
Pα = c∗
αcα = 0, (6.254)
Pβ = c∗
βcβ = e−i
ω0
2
t
e+i
ω0
2
t
= 1. (6.255)
The states described by basis functions which are eigenfunctions of the Hamiltonian
do not evolve (are stationary). It makes sense to draw energy level diagram for
such states, with energy of each state given by the corresponding eigenvalue of the
Hamiltonian. Energy of the |α state is − ω0/2 and energy of the |β state is
+ ω0/2. The measurable quantity is the energy difference ω0, corresponding to
the angular frequency ω0.
6.12 Oscillatory states
In the presence of a homogeneous magnetic field B1 = (B1, 0, 0), the evolution of the system is given by the Hamiltonian
ˆH = −γB0
ˆIx. The Schr¨odinger equation is then
i
∂
∂t
cα
cβ
= −γB1
2
0 1
1 0
cα
cβ
, (6.256)
which is a set of two equations
dcα
dt
= i
γB1
2
cβ, (6.257)
dcβ
dt
= i
γB1
2
cα. (6.258)
These equations have similar structure as Eqs. 6.91 and 6.92. Adding and subtracting them leads to the solution
cα + cβ = C+e+i
γB1
2
t
= C+e−i
ω1
2
t
, (6.259)
cα − cβ = C−e−i
γB1
2
t
= C−e+i
ω1
2
t
. (6.260)
If the initial state is |α , cα(t = 0) = 1, cβ(t = 0) = 0, C+ = C− = 1, and
cα = cos
ω1
2
t , (6.261)
cβ = −i sin
ω1
2
t . (6.262)
6.12. OSCILLATORY STATES 101
Probability that the system is in the |α or |β state is calculated as
Pα = c∗
αcα = cos2 ω1
2
t =
1
2
+
1
2
cos(ω1t), (6.263)
Pβ = c∗
βcβ = sin2 ω1
2
t =
1
2
−
1
2
cos(ω1t). (6.264)
If the initial state is |β , cα(t = 0) = 0, cβ(t = 0) = 1, C+ = 1, C− = −1, and
cα = −i sin
ω1
2
t , (6.265)
cβ = cos
ω1
2
t . (6.266)
Probability that the system is in the |α or |β state is calculated as
Pα = c∗
αcα = sin2 ω1
2
t =
1
2
−
1
2
cos(ω1t), (6.267)
Pβ = c∗
βcβ = cos2 ω1
2
t =
1
2
+
1
2
cos(ω1t). (6.268)
In both cases, the system oscillates between the |α and |β states.
The states described by basis functions different from eigenfunctions of the Hamiltonian
are not stationary but oscillate between |α and |β with the angular frequency
ω1, given by the difference of the eigenvalues of the Hamiltonian (− ω1/2
and ω1/2).
102 CHAPTER 6. SPIN
Chapter 7
Mixed state of non-interacting spins
Literature: A nice short introduction is given in K3.1. The topic is clearly described
in K6, L11, C2.2. The mixed state is introduced nicely in B17.2, K6.8, L11.1, and
C2.2.2. More specific references are given in the individual sections below.
7.1 Mixed state
So far, we worked with systems in so-called pure states, when we described the whole
studied system by its complete wave function. It is fine if the system consists of one
particle or a small number of particles. However, the complete wave function of whole
molecules (or ensembles of whole molecules) is very complicated, represented by multidimensional
state vectors and their properties are described by operators represented by
multidimensional matrices. In NMR spectroscopy, we are interested only with properties
of molecules associated with spins of the observed nuclei. If we assume that motions of
the whole molecule, of its atoms, and of electrons and nuclei in the atoms, do not depend
on the spin,1
we can divide the complete wave function into spin wave functions and wave
function describing all the other degrees of freedom. In general, the spin wave functions
for different molecules are not identical. Therefore, the spin wave function describing
the whole set of nuclei in different molecules is represented by multidimensional vectors
and with properties described by operators represented by multidimensional matrices.
Also, the magnetic fields may depend on the position of the molecule in the sample.
This can be simplified dramatically if
1. the measured quantity does not depend on other coordinates that spin coordinates
α or β – true for magnetization in homogeneous magnetic fields (contributions of
1
This is a very reasonable assumption in most cases. However, note that it is not true completely:
if motions of the magnetic moments and of the molecules were independent, it could not be explained
how the magnetic moments reach their equilibrium distribution.
103
104 CHAPTER 7. MIXED STATE OF NON-INTERACTING SPINS
individual nuclei to the magnetization then do not depend on their positions is
space)
2. the interactions of the observed magnetic moments change only eigenvalues, not
eigenfunctions – true for interactions with fields which can be described without
using spin eigenfunctions
Using the same basis for different nuclei allows us to use two-dimensional operator
matrices (for spin-1/2 nuclei) instead of multidimensional operator matrices.
Expected value A of a quantity A for a single nucleus can be calculated using
Eq. 6.17 as a trace of the following product of matrices:
A = Tr
cαc∗
α cαc∗
β
cβc∗
α cβc∗
β
A11 A12
A21 A22
. (7.1)
Expected value A of a quantity A for multiple nuclei with the same basis is
A = Tr
cα,1c∗
α,1 cα,1c∗
β,1
cβ,1c∗
α,1 cβ,1c∗
β,1
A11 A12
A21 A22
+
cα,2c∗
α,2 cα,2c∗
β,2
cβ,2c∗
α,2 cβ,2c∗
β,2
A11 A12
A21 A22
+ · · ·
= Tr
cα,1c∗
α,1 cα,1c∗
β,1
cβ,1c∗
α,1 cβ,1c∗
β,1
+
cα,2c∗
α,2 cα,2c∗
β,2
cβ,2c∗
α,2 cβ,2c∗
β,2
+ · · ·
A11 A12
A21 A22
= NTr
cαc∗
α cαc∗
β
cβc∗
α cβc∗
β
ˆρ
A11 A12
A21 A22
ˆA
= NTr ˆρ ˆA . (7.2)
The matrix ˆρ is the (probability) density matrix, the horizontal bar indicates average
over the whole ensemble of nuclei in the sample, and N is the number of non-interacting
nuclei described in the same operator basis.
Why probability density? Because the probability P = Ψ|Ψ , the operator of probability
can be written as the unit matrix ˆ1: Ψ|Ψ ≡ Ψ|ˆ1|Ψ . Therefore, the expectation
value of probability can be also calculated using Eq. 6.17 as Tr{ˆρˆ1} = Tr{ˆρ}.
The most important features of the mixed-state approach are listed below:
• Two-dimensional basis is sufficient for the whole set of N nuclei (if they do not
interact with each other).
• Statistical approach: the possibility to use a 2D basis is paid by loosing the information
about the microscopic state. The same density matrix can describe an
astronomic number of possible combinations of individual angular momenta which
give the same macroscopic result. What is described by the density matrix is called
the mixed state.
7.2. COHERENCE 105
• Choice of the basis is encoded in the definition of ˆρ (eigenfunctions of ˆIz).
• The state is described not by a vector, but by a matrix, ˆρ is a matrix like operators.
• Any 2 × 2 matrix can be written as a linear combination of four 2 × 2 matrices.
Such four matrices can be used as a basis of all 2×2 matrices, including operators
(in the same manner as two selected 2-component vectors serve as a basis for all
2-component vectors).
• Good choice of a basis is a set of orthonormal matrices.2
• Diagonal elements of ˆρ (or matrices with diagonal elements only) are known as
populations. They tell what populations of pure α and β states would give the
same polarization along z.
• Off-diagonal elements (or matrices with diagonal elements only) are known as
coherences. They tell what combinations of coefficients cα and cβ would give the
same coherence of phases of the rotation about z.
7.2 Coherence
Coherence is a very important issue in NMR spectroscopy. It is discussed in K6.9, L11.2,
C2.6.
• In a pure state, cαc∗
β is given by amplitudes and by the difference of phases of cα
and cβ: cαc∗
β = |cα||cβ|e−i(φα−φβ)
.
• In a mixed state, cα,j and cβ,j is different for the observed nucleus in each molecule
j. If cα,j and cβ,j describe stationary states, only phases of cα,j and cβ,j change
as the system evolves. Therefore, cαc∗
β = |cα||cβ| · e−i(φα−φβ). The phase of cαc∗
β is
given by e−i(φα−φβ). If the evolution of phases is coherent, φα,j and φβ,j vary but
φα,j − φβ,j is constant. In such a case, cαc∗
β = |cα||cβ|ei(φα−φβ)
. However, if the
phases φα,j and φβ,j evolve independently, e−i(φα−φβ) = e−iφα · eiφβ = 0 · 0 (because
φα,j and φβ,j can be anywhere between 0 and 2π and the average value of both real
component cos(φα,j) and imaginary component sin(φα,j) of eiφα,j
in the interval
(0, 2π) is zero). Obviously, cαc∗
β = 0 in such a case.
2
Orthonormality for a set of four matrices ˆA1, ˆA2, ˆA3, ˆA4 can be defined as Tr{ ˆA†
j
ˆAk} = δj,k, where
j and k ∈ {1, 2, 3, 4}, δj,k = 1 for j = k and δj,k = 0 for j = k, and ˆA†
j is an adjoint matrix of ˆAj,
i.e., matrix obtained from ˆAj by exchanging rows and columns and replacing all numbers with their
complex conjugates.
106 CHAPTER 7. MIXED STATE OF NON-INTERACTING SPINS
7.3 Basis sets
Usual choices of basis matrices are (C2.7.2):
• Cartesian operators, equal to the operators of spin angular momentum divided by
. In this text, these matrices are written as Ix, Iy, etc. In a similar fashion, we
write H = ˆH/ for Hamiltonians with eigenvalues expressed in units of (angular)
frequency, not energy. The normalization factor
√
2 is often omitted (then the
basis is still orthogonal, but not orthonormal):
√
2It =
1
√
2
1 0
0 1
√
2Iz =
1
√
2
1 0
0 −1
√
2Ix =
1
√
2
0 1
1 0
√
2Iy =
1
√
2
0 −i
i 0
. (7.3)
• Single-element population
Iα = It + Iz =
1 0
0 0
Iβ = It − Iz =
0 0
0 1
(7.4)
and transition operators
I+ = Ix + iIy =
0 1
0 0
I− = Ix − iIy =
0 0
1 0
. (7.5)
• A mixed basis
√
2It =
1
√
2
1 0
0 1
√
2Iz =
1
√
2
1 0
0 −1
I+ =
0 1
0 0
I− =
0 0
1 0
.
(7.6)
7.4 Liouville - von Neumann equation
In order to describe the evolution of mixed states in time, we must find an equation describing
how elements of the density matrix change in time. Derivation of such equation
is nicely described in C2.2.3.
We start with the Schr¨odinger equation for a single spin in matrix representation:
i
d
dt
cα
cβ
=
Hα,α Hα,β
Hβ,α Hβ,β
cα
cβ
=
Hα,αcα + Hα,βcβ
Hβ,αcα + Hβ,βcβ
. (7.7)
Note that the Hamiltonian matrix is written in a general form, the basis functions are not necessarily eigenfunctions
of the operator. However, the matrix must be Hermitian, i.e., Hj,k = H∗
k,j:
Hα,β = H∗
β,α Hβ,α = H∗
α,β. (7.8)
7.4. LIOUVILLE - VON NEUMANN EQUATION 107
If we multiply Eq. 7.7 by the basis functions from left, we obtained the differential equations for cα and cβ (because
the basis functions are orthonormal):
( 1 0 )i
d
dt
cα
cβ
= i
dcα
dt
= Hα,αcα + Hα,βcβ (7.9)
( 0 1 )i
d
dt
cα
cβ
= i
dcβ
dt
= Hβ,αcα + Hβ,βcβ. (7.10)
In general,
dck
dt
= −
i
l
Hk,lcl (7.11)
and its complex conjugate (using Eq. 7.8) is
dc∗
k
dt
= +
i
l
H∗
k,lc∗
l = +
i
l
Hl,kc∗
l . (7.12)
Elements of the density matrix consist of the products cjc∗
k. Therefore, we must calculate
d(cjc∗
k)
dt
= cj
dc∗
k
dt
+ c∗
k
dcj
dt
=
i
l
Hl,kcjc∗
l −
i
l
Hj,lclc∗
k. (7.13)
For multiple nuclei with the same basis,
d(cj,1c∗
k,1 + cj,2c∗
k,2 + · · · )
dt
= cj,1
dc∗
k,1
dt
+ c∗
k,1
dcj,1
dt
+ cj,2
dc∗
k,2
dt
+ c∗
k,2
dcj,2
dt
+ · · · (7.14)
=
i
l
Hl,k(cj,1c∗
l,1 + cj,2c∗
l,2 + · · · ) −
i
l
Hj,l(cl,1c∗
k,1 + cl,2c∗
k,2 + · · · ). (7.15)
Note that
l
(cj,1c∗
l,1 + cj,2c∗
l,2 + · · · )Hl,k = N
l
ρj,lHl,k (7.16)
is the j, k element of the product N ˆρ ˆH, and
l
Hj,l(cl,1c∗
k,1 + cl,2c∗
k,2 + · · · ) = N
l
Hj,lρl,k (7.17)
is the j, k element of the product N ˆH ˆρ. Therefore, we can write the equation of motion for the whole density matrix
as
dˆρ
dt
=
i
(ˆρ ˆH − ˆH ˆρ) =
i
[ˆρ, ˆH] = −
i
[ ˆH, ˆρ] (7.18)
or in the units of (angular) frequency
dˆρ
dt
= i(ˆρH − Hˆρ) = i[ˆρ, H] = −i[H, ˆρ]. (7.19)
Eqs. 7.18 and 7.19 are known as the Liouville - von Neumann equation.
108 CHAPTER 7. MIXED STATE OF NON-INTERACTING SPINS
7.5 Rotation in operator space
The Liouville - von Neumann equation can be solved using techniques of linear algebra.
However, a very simple geometric solution is possible (K7.3, C2.7.3, L11.8) if the Hamiltonian
does not change in time and consists solely of matrices which commute (e.g., It
and Iz, but not Ix and Iz).
Let us look at an example for H = εtIt + ω0Iz and ˆρ = cxIx + cyIy + czIz + ctIt.
Let us first evaluate the commutators from the Liouville - von Neumann equation:
It is proportional to a unit matrix ⇒ it must commute with all matrices:
[It, Ij] = 0 (j = x, y, z, t). (7.20)
Commutators of Iz are given by the definition of angular momentum operators:
[Iz, Iz] = [Iz, It] = 0 [Iz, Ix] = iIy [Iz, Iy] = −iIx. (7.21)
Let us write the Liouville - von Neumann equation with the evaluated commutators:
dcx
dt
Ix +
dcy
dt
Iy +
dcz
dt
Iz +
dct
dt
It = i (−iω0cxIy + iω0cyIx) . (7.22)
Written in a matrix representation (noticing that cz and ct do not evolve because the czIz and ctIt components of
the density matrix commute with both matrices constituting the Hamiltonian),
1
2
0 dcx
dt
dcx
dt
0
+
1
2
0 −i
dcy
dt
i
dcy
dt
0
+ 0 + 0 =
i
2
0 −ω0cx
ω0cx 0
+
i
2
0 iω0cy
iω0cy 0
, (7.23)
0
d(cx−icy)
dt
d(cx+icy)
dt
0
= iω0
0 −(cx − icy)
cx + icy 0
. (7.24)
This corresponds to a set of two differential equations
d(cx − icy)
dt
= −iω0(cx − icy) (7.25)
d(cx + icy)
dt
= +iω0(cx + icy) (7.26)
with the same structure as Eqs. 6.94 and 6.95. The solution is
cx − icy = (cx(0) − icy(0))e−iω0t
= c0e−i(ω0t+φ0)
(7.27)
cx + icy = (cx(0) + icy(0))e+iω0t
= c0e+i(ω0t+φ0)
(7.28)
with the amplitude c0 and phase φ0 given by the initial conditions. It corresponds to
cx = c0 cos(ω0t + φ0) (7.29)
cy = c0 sin(ω0t + φ0). (7.30)
We see that coefficients cx, cy, cz play the same roles as coordinates rx, ry, rz in Eqs. 6.91–6.93, respectively, and
operators Ix, Iy, Iz play the same role as unit vectors ı, , k, defining directions of the axes of the Cartesian coordinate
system.
The evolution of ˆρ can be described as a rotation in an abstract three-dimensional
operator space with the dimensions given by Ix, Iy, and Iz.
7.6. GENERAL STRATEGY OF ANALYZING NMR EXPERIMENTS 109
7.6 General strategy of analyzing NMR experiments
The Liouville - von Neumann equation is the most important tool in the analysis of
evolution of the spin system during the NMR experiment. The general strategy consists
of three steps:
1. Define ˆρ at t = 0
2. Describe evolution of ˆρ using the relevant Hamiltonians – this is usually done in
several steps
3. Calculate the expectation value M of the measured quantity according to Eq.
6.17
Obviously, the procedure requires knowledge of
1. relation(s) describing the initial state of the system (ˆρ(0))
2. all Hamiltonians
3. the operator representing the measurable quantity
In the next chapter, we start from the end and define first the operator of the
measurable quantity. Then we spend a lot of time defining all necessary Hamiltonians.
Finally, we use the knowledge of the Hamiltonians and basic thermodynamics to describe
the initial state.
110 CHAPTER 7. MIXED STATE OF NON-INTERACTING SPINS
Chapter 8
Chemical shift, NMR experiment
Literature: The general strategy is clearly outlined in C2.4, Hamiltonians discussed
in L8, thermal equilibrium in L11.3, C2.4.1, K6.8.6, relaxation due to the chemical shift
in C5.4.4, K9.10 (very briefly, the quantum approach to relaxation is usually introduced
using dipole-dipole interactions as an example). The one-pulse experiment is analyzed
in K7.2.1, L11.11 and L11.12.
8.1 Operator of the observed quantity
The quantity observed in the NMR experiment is the bulk magnetization M, i.e., the
sum of magnetic moments of all nuclei divided by volume of the sample, assuming
isotropic distribution of the nuclei in the sample. Technically, we observe oscillations
in the plane perpendicular to the homogeneous field of the magnet B0. The associated
oscillations of the magnetic fields of nuclei induce electromotive force in the detector
coil, as described by Eq. 1.50. Since a complex signal is usually recorded, the operator
of complex magnetization M+ = Mx + iMy is used (M− = Mx − iMy can be used as
well).
ˆM+ = Nγ(ˆIx + iˆIy) = Nγ ˆI+, (8.1)
where N is the number of nuclei in the sample (per unit volume).
8.2 Static field B0
We already defined the Hamiltonian of the static homogeneous magnetic field B0, following
the classical description of energy of a magnetic moment in a magnetic field
(Eq. 1.50). Since B0 defines direction of the z axis,
ˆH0,lab = −γB0
ˆIz. (8.2)
111
112 CHAPTER 8. CHEMICAL SHIFT, NMR EXPERIMENT
8.3 Radio-frequency field B1
The oscillating magnetic field of radio waves irradiating the sample is formally decomposed
into two rotating magnetic fields (with the same speed given by the frequency of
the radio waves ωradio, but with opposite sense of rotation, as described in Section 3.1.3).
The component resonating (approximately) with the precession frequency of the observed
nuclei usually defines the x axis of the rotating coordinate frame used most often
in NMR spectroscopy. In this system, frequency of the resonating component1
is subtracted
from the precession frequency and the difference Ω = ω0 − ωrot = −γB0 − ωrot is
the frequency offset defining the evolution in the rotating frame in the absence of other
fields:
In the absence of other fields than B0:
ˆH0,rot = (−γB0 − ωrot)ˆIz = ΩˆIz. (8.3)
During irradiation by waves with the phase defining x
ˆH1,rot = (−γB0 − ωrot)ˆIz − γB1
ˆIx = ΩˆIz + ω1
ˆIx. (8.4)
During irradiation by waves shifted by π/2 from the phase defining x
ˆH1,rot = (−γB0 − ωrot)ˆIz − γB1
ˆIy = ΩˆIz + ω1
ˆIy. (8.5)
If the radio frequency is close to resonance, −γB0 ≈ ωrot, Ω ω1, and the ˆIz
component of the Hamiltonian can be neglected.
8.4 Hamiltonian of chemical shift
Using the description of the magnetic fields of moving electrons discussed in Section 3.2,
definition of the chemical shift Hamiltonian is straightforward:
ˆHδ = −γ(ˆIxBe,x + ˆIyBe,y + ˆIzBe,z) = −γ( ˆIx
ˆIy
ˆIz )


Be,x
Be,y
Be,z

 =
= −γ( ˆIx
ˆIy
ˆIz )


δxx δxy δxz
δyx δyy δyz
δzx δzy δzz




B0,x
B0,y
B0,z

 = −γ
ˆ
I · δ · B0. (8.6)
The Hamiltonian of a homogeneous magnetic field aligned with the z-axis of the coordinate frame can be decomposed
into
1
Formally opposite to ωradio.
8.5. SECULAR APPROXIMATION AND AVERAGING 113
• isotropic contribution, independent of rotation in space:
ˆHδ,i = −γB0δi(ˆIz) (8.7)
• axial component, dependent on ϕ and ϑ:
ˆHδ,a = −γB0δa(3 sin ϑ cos ϑ cos ϕˆIx + 3 sin ϑ cos ϑ sin ϕˆIy + (3 cos2
ϑ − 1)ˆIz)
= −γB0δa(3axaz
ˆIx + 3ayaz
ˆIy + (3a2
z − 1)ˆIz) (8.8)
• rhombic component, dependent on ϕ, ϑ, and χ:
ˆHδ,r = −γB0δr( (−(2 cos2
χ − 1) sin ϑ cos ϑ cos ϕ + 2 sin χ cos χ sin ϑ cos ϑ sin ϕ)ˆIx +
(−(2 cos2
χ − 1) sin ϑ cos ϑ sin ϕ − 2 sin χ cos χ sin ϑ cos ϑ cos ϕ)ˆIy +
((2 cos2
χ − 1) sin2
ϑ)ˆIz)
= γB0δr((cos(2χ)ax − sin(2χ)ay)az
ˆIx + (cos(2χ)ay + sin(2χ)ax)az
ˆIy + cos(2χ)(a2
z − 1)ˆIz)
The complete Hamiltonian of a magnetic moment of a nucleus not interacting with magnetic moments of other nuclei
in the presence of the static field B0 but in the absence of the radio waves is given by
ˆH = ˆH0,lab + ˆHδ,i + ˆHδ,a + ˆHδ,r. (8.9)
8.5 Secular approximation and averaging
The Hamiltonian including the chemical shift is complicated, but can be simplified in
many cases.
• The components of the induced fields Be,x and Be,y are perpendicular to B0. The contributions of ˆHδ,i are constant
and the contributions of ˆHδ,a and ˆHδ,r fluctuate with the molecular motions changing values of ϕ, ϑ, and χ. Since
the molecular motions do not resonate (in general) with the precession frequency −γB0, the components ˆIxBe,x
and ˆIyBe,y of the Hamiltonian oscillate rapidly with a frequency close to −γB0. These oscillations are much faster
than the precession about Be,x and Be,y (because the field Be is much smaller than B0) and effectively average to
zero on the timescale given by 1/(γB0) (typically nanoseconds). Therefore, the ˆIxBe,x and ˆIyBe,y terms can be
neglected if the effects on longer timescales are studied. Such a simplification is known as secular approximation.2
The secular approximation simplifies the Hamiltonian to
H = −γB0(1 + δi + (3 cos2
ϑ − 1)δa + cos(2χ) sin2
ϑδr)ˆIz (8.10)
2
In terms of quantum mechanics, eigenfunctions of ˆIxBe,x and ˆIyBe,y differ from the eigenfunctions
of ˆH0,lab (|α and |β ). Therefore, the matrix representation of ˆIxBe,x and ˆIyBe,y contains off-diagonal
elements. Terms proportional to ˆIz represent so-called secular part of the Hamiltonian, which does not
change the |α and |β states (because they are eigenfunctions of ˆIz). Terms proportional to ˆIx and ˆIy are
non-secular because they change the |α and |β states (|α and |β are not eigenfunctions of ˆIx or ˆIy).
However, eigenvalues of ˆIxBe,x and ˆIyBe,y, defining the off-diagonal elements, are much smaller than the
eigenvalues of ˆH0,lab (because the field Be is much smaller than B0). Secular approximation represents
neglecting such small off-diagonal elements in the matrix representation of the total Hamiltonian and
keeping only the diagonal secular terms.
114 CHAPTER 8. CHEMICAL SHIFT, NMR EXPERIMENT
• If the sample is an isotropic liquid, averaging over all molecules of the sample further simplifies the Hamiltonian.
As no orientation of the molecule is preferred, all values of χ are equally probable and independent of ϑ. Therefore,
the last term in Eq. 8.10 is averaged to zero. Moreover, average values of a2
x = cos2 ϕ sin2 ϑ, of a2
y = sin2 ϕ sin2 ϑ,
and of a2
z = cos2 ϑ must be the same because none of the directions x, y, z is preferred:
a2
x = a2
y = a2
z. (8.11)
The consequence has been already discussed when we described relaxation classically (Eq. 8.12):
a2
x + a2
y + a2
z = 1 ⇒ a2
x + a2
y + a2
z = 3a2
z ⇒ 3a2
z − 1 = (3 cos2 ϑ − 1) = 0, (8.12)
and the anisotropic and rhombic contributions can be neglected.
The Hamiltonian describing the effects of the static external magnetic field and
coherent effects of the electrons in isotropic liquids reduces to
H = −γB0(1 + δi)ˆIz. (8.13)
Note that the described simplifications can be used only if they are applicable.
Eq. 8.13 is valid only in isotropic liquids, not in liquid crystals, stretched gels, polycrystalline
powders, monocrystals, etc.!
8.6 Thermal equilibrium as the initial state
Knowledge of the Hamiltonian allows us to derive the density matrix at the beginning
of the experiment. Usually, we start from the thermal equilibrium. If the equilibrium is
achieved, phases of individual magnetic moments are random and the magnetic moments
precess incoherently. Therefore, the off-diagonal elements of the equilibrium density
matrix (proportional to Ix and Iy) are equal to zero.
We use the mixed state approach to define the state of the sample in thermal equilibrium. Populations of the states
can be evaluated using statistical arguments similar to the Boltzmann law in the classical molecular statistics:
Peq
α =
e−Eα/kBT
e−Eα/kBT + e−Eβ /kBT
, (8.14)
Peq
β =
e−Eβ /kBT
e−Eα/kBT + e−Eβ /kBT
, (8.15)
where kB = 1.38064852 × 10−23 m2 kg s−2 K−1 is the Boltzmann constant.
The energies Eα and Eβ are the eigenvalues of the energy operator, the Hamiltonian. Since we use eigenfunctions of
ˆIz as the basis, eigenfunctions of H = −γB0(1 + δi)ˆIz are the diagonal elements of the matrix representation of ˆH:
ˆH = −γB0(1 + δi)ˆIz = −γB0(1 + δi)
2
1 0
0 −1
=
−γB0(1 + δi) 2
0
0 +γB0(1 + δi) 2
. (8.16)
The thermal energy at 0 ◦C is more than 12 000 times higher than γB0 /2 for the most sensitive nuclei (protons) at
spectrometers with the highest magnetic fields (1 GHz). The effect of chemical shift is four orders of magnitude lower.
Therefore, we can approximate
e
±
γB0(1+δi)
kBT
≈ 1 ±
γB0
2kBT
(8.17)
8.7. RELAXATION DUE TO CHEMICAL SHIFT ANISOTROPY 115
and calculate the populations as
Peq
α =
e−Eα/kBT
e−Eα/kBT + e−Eβ /kBT
=
1 + γB0
2kBT
1 + γB0
2kBT
+ 1 − γB0
2kBT
=
1 + γB0
2kBT
2
, (8.18)
Peq
β =
e−Eβ /kBT
e−Eα/kBT + e−Eβ /kBT
=
1 − γB0
2kBT
1 + γB0
2kBT
+ 1 − γB0
2kBT
=
1 − γB0
2kBT
2
. (8.19)
Writing the populations as the diagonal elements, the equilibrium density matrix is
ˆρeq
=
1
2
+ γB0
4kBT
0
0 1
2
− γB0
4kBT
=
1
2
1 0
0 1
+
γB0
4kBT
1 0
0 −1
= It + κIz, (8.20)
where
κ =
γB0
2kBT
. (8.21)
Note that we derived the quantum description of a mixed state. Two populations of
the density matrix provide correct results but do not tell us anything about microscopic
states of individual magnetic moments. Two-dimensional density matrix does not imply
that all magnetic moments are in one of two eigenstates.
8.7 Relaxation due to chemical shift anisotropy
The simplified Eq. 8.13 does not describe the effects of fast fluctuations, resulting in
relaxation. In order to derive quantum description of relaxation caused by the chemical
shift, the Liouville - von Neumann equation must be solved for the complete Hamiltonian
including the axial and rhombic contributions.
The Liouville - von Neumann equation describing the relaxing system of magnetic moments interacting with moving
electrons in a so-called interaction frame (corresponding to the rotating coordinate frame in the classical description) has
the form
d∆ˆρ
dt
= −
i
[ ˆHδ,a + ˆHδ,r, ∆ˆρ], (8.22)
where ˆHδ,a and ˆHδ,r are defined by Eqs. 8.8 and 8.9, respectively, and ∆ˆρ is a difference (expressed in the interaction
frame) between density matrix at the given time and density matrix in the thermodynamic equilibrium. Writing ∆ˆρ in
the same bases as used for the Hamiltoninan,
∆ˆρ = dt
ˆIt + dz
ˆIz + d+
ˆI+eiω0t
+ d−
ˆI−e−iω0t
. (8.23)
If the chemical shift is axially symmetric and its size or shape do not change,
d(dz
ˆIz + d+
ˆI+eiω0t + d−
ˆI−e−iω0t)
dt
= −
ib
cz ˆIz +
3
8
c+ ˆI+eiω0t
+
3
8
c− ˆI−e−iω0t
, dz
ˆIz + d+
ˆI+eiω0t
+ d−
ˆI−e−iω0t
,
(8.24)
116 CHAPTER 8. CHEMICAL SHIFT, NMR EXPERIMENT
where ˆI±e±iω0t are operators ˆI± = ˆIx ± ˆIy in the interaction frame, ω0 = −γB0(1 + δa), and
cz
=
1
2
(3 cos2
ϑ − 1) = Θ (8.25)
c+
=
3
2
sin ϑ cos ϑe−iϕ
=
2
3
Θ⊥
e−iϕ
(8.26)
c−
=
3
2
sin ϑ cos ϑe+iϕ
=
2
3
Θ⊥
e+iϕ
(8.27)
Analogically to the classical analysis, the evolution can be written as
d∆ˆρ
dt
= −
1
2
∞ˆ
0
[ ˆHδ,a(0), [ ˆHδ,a(t), ∆ˆρ]]dt. (8.28)
The right-hand side can be simplified dramatically by the secular approximation: all terms with e±iω0t are averaged to
zero. Only terms with (cz)2 and c+c− are non zero (both equal to 1/5 at tj = 0).3 These are the terms with [ˆIz, [ˆIz, ∆ˆρ]],
[ˆI+, [ˆI−, ∆ˆρ]], and [ˆI−, [ˆI+, ∆ˆρ]]. Moreover, averaging over all molecules makes all three correlation functions identical in
isotropic liquids: cz(0)cz(t) = c+(0)c−(t) = c−(0)c+(t) = c(0)c(t).
In order to proceed, the double commutators must be expressed. We start with
[ˆIz, ˆI±] = [ˆIz, ˆIx] ± i[ˆIz, ˆIy] = ± (ˆIx ± iˆIy) = ± ˆI± (8.29)
and
[ˆI+, ˆI−] = [ˆIx, ˆIx] − i[ˆIx, ˆIy] + i[ˆIy, ˆIx] + [ˆIy, ˆIy] = 2 ˆIz. (8.30)
Our goal is to calculate relaxation rates for the expectation values of components parallel (Mz) and perpendicular
(M+ or M−) to B0.
Let us start with Mz. According to Eq. 6.17,
∆Mz = Tr{ ˆMz∆ˆρ} (8.31)
where ∆ Mz is the difference from the expectation value of Mz in equilibrium. The operator of Mz for one magnetic
moment observed is (Eq. 9.32)
ˆMz = Nγ ˆIz, (8.32)
where N is the number of molecules per volume element detected by the spectrometer. Since the basis matrices are
orthogonal, products of ˆIz with the components of the density matrix different from ˆIz are equal to zero and the left-hand
side of Eq. 8.28 reduces to
ddz
dt
ˆIz (8.33)
when calculating relaxation rate of Mz . In the right-hand side, we need to calculate three double commutators:
[ˆIz, [ˆIz, ˆIz]] = 0 [ˆI+, [ˆI−, ˆIz]] = 2 2 ˆIz [ˆI−, [ˆI+, ˆIz]] = 2 2 ˆIz (8.34)
After substituting into Eq. 8.28,
ddz
dt
Tr{ˆIz
ˆIz} = −

3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt +
3
4
b2
∞ˆ
0
c−(0)c+(t)e−iω0t
dt

 dzTr{ˆIz
ˆIz} (8.35)
d∆ Mz
dt
= −

3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt +
3
4
b2
∞ˆ
0
c−(0)c+(t)e−iω0t
dt

 ∆ Mz (8.36)
The relaxation rate R1 for Mz, known as longitudinal relaxation rate in the literature, is the real part of the expression
in the parentheses
3
We have factored out 3/8 in order to make c+c− = (cz)2.
8.7. RELAXATION DUE TO CHEMICAL SHIFT ANISOTROPY 117
R1 =
3
4
b2



∞ˆ
0
c+(0)c−(t)eiω0t
dt +
∞ˆ
0
c−(0)c+(t)e−iω0t
dt



(8.37)
As already discussed in the classical description of relaxation, if the fluctuations are random, they are also stationary:
the current orientation of the molecule is correlated with the orientation in the past in the same manner as it is correlated
with the orientation in the future. Therefore,
∞ˆ
0
c+(0)c−(t)eiω0t
dt =
1
2


∞ˆ
0
c+(0)c−(t)eiω0t
dt +
0ˆ
−∞
c+(0)c−(t)eiω0t
dt

 =
1
2
∞ˆ
−∞
c+(0)c−(t)eiω0t
dt. (8.38)
∞ˆ
0
c−(0)c+(t)e−iω0t
dt =
1
2


∞ˆ
0
c−(0)c+(t)e−iω0t
dt +
0ˆ
−∞
c−(0)c+(t)e−iω0t
dt

 =
1
2
∞ˆ
−∞
c−(0)c+(t)e−iω0t
dt, (8.39)
The right-hand side integrals are identical with the mathematical definition of the Fourier transform of the correlation
functions and real parts of such Fourier transforms are the spectral density functions J(ω).
The relaxation rate R1 can be therefore written in the same form as derived classically.
R1 =
3
4
b2 1
2
J(ω0) +
1
2
J(−ω0) ≈
3
4
b2
J(ω0) (8.40)
What is the physical interpretation of the obtained equation? Relaxation of Mz is
given by the correlation functions c+(0)c−(t) and c−(0)c+(t), describing fluctuations of
the components of the chemical shift tensor perpendicular to B0 (ax and ay). Such
fluctuating fields resemble the radio waves with B1 ⊥ B0. If the frequency of such
fluctuations matches the precession frequency ω0, the resonance condition is fulfilled and
(random) transitions between the |α and |β states can take place. If the molecular
motions are assumed to be completely random and independent of the distribution of
magnetic moments, Mz is expected to decay to zero, which doe not happen in reality. If
the coupling between molecular motions and magnetic moment distribution is described
correctly by the quantum theory, the transition |β → |α is slightly more probable.4
This drives the system back to the equilibrium distribution of magnetic moments.
Let us continue with M+. According to Eq. 6.17,
∆ M+ ≡ M+ = Tr{ ˆM+∆ˆρ} (8.41)
The expectation value of M+ in equilibrium is zero, this is why we do not need to calculate the difference for M+
and why we did not calculate the difference in the classical analysis.
The operator of M+ for one magnetic moment observed is
ˆM+ = Nγ ˆI+ = Nγ(ˆIx + iˆIy). (8.42)
Due to the orthogonality of basis matrices, the left-hand side of Eq. 8.28 reduces to
dd+
dt
ˆI+eiω0t
(8.43)
when calculating relaxation rate of ∆ M+ ≡ M+ . In the right-hand side, we need to calculate three double
commutators:
4
It can be described as J(ω0) = e− ω0/kBT
J(−ω0) and taken into account by working with ∆ˆρ and
∆Mz instead of ˆρ and Mz .
118 CHAPTER 8. CHEMICAL SHIFT, NMR EXPERIMENT
[ˆIz, [ˆIz, ˆI+]] = 2 ˆI+ [ˆI+, [ˆI−, ˆI+]] = 2 2 ˆI+ [ˆI−, [ˆI+, ˆI+]] = 0. (8.44)
After substituting into Eq. 8.28,
dd+
dt
Tr{ˆI+
ˆI+} = −

b2
∞ˆ
0
cz(0)cz(t)dt +
3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt

 d+Tr{ˆI+
ˆI+} (8.45)
d M+
dt
= −

b2
∞ˆ
0
cz(0)cz(t)dt +
3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt

 M+ (8.46)
The relaxation rate R2 for M+, known as transverse relaxation rate in the literature, is the real part of the expression
in the parentheses.
R2 = b2
∞ˆ
0
cz(0)cz(t)dt +



3
4
b2
∞ˆ
0
c+(0)c−(t)eiω0t
dt



. (8.47)
Note that the first integral in 8.47 is a real number, equal to R0 derived by the classical analysis.
Using the same arguments as for Mz,
R2 = b2 1
2
J(0) +
3
8
J(ω0) ≈ R0 +
1
2
R1. (8.48)
What is the physical interpretation of the obtained equation? Two terms in Eq. 8.48
describe two processes contributing to the relaxation of M+. The first one is the loss
of coherence with the rate R0, given by the correlation function cz(0)cz(t) and describing
fluctuations of the components of the chemical shift tensor parallel with B0 (az).
This contribution was analyzed in Section4.2 using the classical approach. The second
contribution is transitions between the |α and |β states due to fluctuations of the
components of the chemical shift tensor perpendicular to B0 (ax and ay), returning the
magnetization vector M to its direction in the thermodynamic equilibrium. As M is
oriented along the z axis in the equilibrium, the transitions renew the equilibrium value
of Mz, as described above, but also make the Mx and My components to disappear. Note
however, that only one correlation function (c+(0)c−(t)) contributes to the relaxation of
M+, while both c+(0)c−(t) and c−(0)c+(t) contribute to the relaxation of Mz and only
R1/2 contributes to R2. If we defined R2 as a relaxation rate of M−, c−(0)c+(t) would
contribute5
:
R2 = b2 1
2
J(0) +
3
4
1
2
J(−ω0) ≈ R0 +
1
2
R1. (8.49)
5
Fluctuations with frequency +ω0 affect M+ and fluctuations with frequency −ω0 affect M−, but
both affect Mz. Alternatively, we could define R2 as a relaxation rate of Mx or My. Fluctuations of the
Be,y component affect Mx but not My, while fluctuations of the Be,x component affect My but not Mx.
On the other hand, both fluctuations of Be,x and Be,y affect Mz. Working with M+, M− or Mx, My,
the relaxation of Mz due to Be,x and Be,y is always twice faster.
8.8. THE ONE-PULSE EXPERIMENT 119
8.8 The one-pulse experiment
At this moment, we have all we need to describe a real NMR experiment for sample
consisting of isolated magnetic moments (not interacting with each other). The basic
NMR experiment consists of two parts. In the first part, the radio-wave transmitter is
switched on for a short time, needed to rotate the magnetization to the plane perpendicular
to the magnetic filed B0 (a radio-wave pulse). In the second time, the radio-wave
transmitter is switched off but the receiver is switched on in order to detect rotation of
the magnetization vector about the direction of B0. We will analyze evolution of the
density matrix during these two periods and calculate the magnetization contributing
to the detected signal.
8.8.1 Excitation by radio wave pulses
At the beginning of the experiment, the density matrix describes thermal equilibrium
(Eq. 8.20):
ˆρ(0) = It + κIz. (8.50)
The Hamiltonian governing evolution of the system during the first part of the experiments
consists of coherent and fluctuating terms. The fluctuating contributions result in
relaxation, described by relaxation rates R1 and R2. The coherent contributions include
H = εt · It − γB0(1 + δi)Iz − γB1(1 + δi) cos(ωrott)Ix − γB1(1 + δi) sin(ωrott)Iy, (8.51)
where εt is the total energy of the system outside the magnetic field, and the choice
of the directions x and y is given by the cos(ωrott) and sin(ωrott) terms.
The Hamiltonian simplifies in a coordinate system rotating with ωrot = −ωradio
H = εt · It + (−γB0(1 + δi) − ωrot)
Ω
Iz + (−γB1(1 + δi))
ω1
Ix, (8.52)
but it still contains non-commuting terms (Ix vs. Iz). Let us check what can be
neglected to keep only commuting terms, which allows us to solve the Liouville - von
Neumann equation using the simple geometric approach.
• The value of εt is unknown and huge, but It commutes with all matrices (it is
proportional to the unit matrix). As a consequence, this term can be ignored
because it does not have any effect on evolution of ˆρ.
• The value of ω1 defines how much magnetization is rotated to the x, y plane. The
maximum effect is obtained for ω1τp = π/2, where τp is the length of the radiowave
pulse. Typical values of τp for proton are approximately 10 µs, corresponding
120 CHAPTER 8. CHEMICAL SHIFT, NMR EXPERIMENT
to frequency of rotation of 25 kHz (90◦
rotation in 10 µs corresponds to 40 µs for a
full circle, 1/40 µs = 25 kHz).
• Typical values of R1 are 10−1
s−1
to 100
s−1
and typical values of R2 are 10−1
s−1
to
102
s−1
for protons in organic molecules and biomacromolecules. Therefore, effects
of relaxations can be safely neglected during τp.
• When observing a single type of proton (or other nucleus), Ω can be set to zero by
the choice of ωradio. However, variation of Ω is what we observe in real samples,
containing protons (or other nuclei) with various δi. The typical range of proton
δi is 10 ppm, corresponding to 5 kHz at a 500 MHz spectrometer.6
The carrier
frequency ωradio is often set to the precession frequency of the solvent. In the case
of water, it is roughly in the middle of the spectrum (4.7 ppm at pH 7). So, we
need to cover ±2.5 kHz. We see that |Ω| < |ω1|, but the ratio is only 10 % at the
edge of the spectrum.
In summary, we see that we can safely ignore It and fluctuating contributions, but
we must be careful when neglecting ΩIz. The latter approximation allows us to use the
geometric solution of the Liouville - von Neumann equation, but is definitely not perfect
for larger Ω resulting in offset effects.
Using the simplified Hamiltonian H = ω1Ix, evolution of ˆρ during τp can be described
as a rotation about the ”Ix axis”:
ˆρ(0) = It + κIz −→ ˆρ(τp) = It + κ(Iz cos(ω1τp) − Iy sin(ω1τp)). (8.53)
For a 90◦
pulse,
ˆρ(τp) = It − κIy. (8.54)
8.8.2 Evolution of chemical shift after excitation
After switching off the transmitter, ω1Ix disappears from the Hamiltonian, which now
contains only commuting terms. On the other hand, signal is typically acquired for a
relatively long time (0.1 s to 10 s) to achieve a good frequency resolution. Therefore, the
relaxation effects cannot be neglected.
The coherent evolution can be described as a rotation about the ”Iz axis” with the
angular frequency Ω
ˆρ(t) = It + κ(−Iy cos(Ωt) + Ix sin(Ωt)). (8.55)
The measured quantity M+ can be expressed as (Eq. 6.17)
6
Chosen as a compromise here: spectra of small molecules are usually recored at 300 MHz–500 MHz,
while spectra of biomacromolecules are recorded at 500 MHz–1 GHz.
8.8. THE ONE-PULSE EXPERIMENT 121
M+ = Tr{ ˆM+ ˆρ(t)} = Nγ Tr{I+(It + κ(−Iy cos(Ωt) + Ix sin(Ωt))}. (8.56)
The relevant traces are
Tr{I+It} = Tr
0 1
0 0
1
2
0
0 1
2
= Tr
0 1
2
0 0
= 0 (8.57)
Tr{I+Ix} = Tr
0 1
0 0
0 1
2
1
2
0
= Tr
1
2
0
0 0
=
1
2
(8.58)
Tr{I+Iy} = Tr
0 1
0 0
0 − i
2
i
2
0
= Tr
i
2
0
0 0
=
i
2
(8.59)
Including relaxation and expressing κ
M+ =
Nγ2 2
B0
4kBT
e−R2t
(sin(Ωt) − i cos(Ωt)). (8.60)
which can be rewritten as
M+ =
Nγ2 2
B0
4kBT
e−R2t
cos Ωt −
π
2
+ i sin Ωt −
π
2
) =
Nγ2 2
B0
4kBT
e−R2t
eiΩt
e−iπ
2 .
(8.61)
We know that in order to obtain purely Lorentzian (absorption) real component of
the spectrum by Fourier transformation, the signal should evolve as e−R2t
eiΩt
. We see
that magnetization described by Eq. 8.61 is shifted from the ideal signal by a phase of
−π/2. However, this is true only if the evolution starts exactly at t = 0. In practice,
this is impossible to achieve for various technical reasons (instrumental delays and phase
shifts, evolution starts already during τp, etc.). Therefore, the rotation has an unknown
phase shift φ (including the π/2 shift among other contributions), which is removed by
an empirical correction during signal processing (corresponding to multiplying Eq. 8.61
by eiπ/2
). It tells us that we can ignore the phase shift and write the phase-corrected
signal as
M+ =
Nγ2 2
B0
4kBT
e−R2t
(cos(Ωt) + i sin(Ωt)) =
Nγ2 2
B0
4kBT
e−R2t
eiΩt
. (8.62)
8.8.3 Spectrum and signal-to-noise ratio
Knowing the expected magnetization, we can try to describe the one-dimensional NMR spectrum quantitatively. To do
it, we need to know
1. how is the detected signal related to the magnetization. Here, Eq. 1.50 helps us: µ in Eq. 1.50 is simply magnetization
multiplied by the volume sensed by the detector coil.
122 CHAPTER 8. CHEMICAL SHIFT, NMR EXPERIMENT
2. how is the time-dependent signal converted to a frequency spectrum. Here, the answer is described in Chapter 5,
the most important step is the Fourier transform.7
3. how is the noise defined. Here, we use the result of statistical mechanics showing that the square of the thermal
noise of electrons is n2 = 4RkBT∆f, where R is resistance, T is temperature, and ∆f is the frequency bandwidth
of the detector.
If we do not include relaxation, neglect effect of the preamplifier, resistance of the sample, and assume that the
receiver coil and sample have the same temperature, the result of the analysis is
Signal/noise = K
Nγ5/2 2B
3/2
0
k
3/2
B T
3/2
sample(∆f)1/2
, (8.63)
where K includes geometry and construction factors, and ∆f is the receiver bandwidth.
In the most sensitive NMR probes, the motions of the electrons are suppressed by cooling the receiver coil to a very
low temperature, approximately 20 K. Therefore, we have to include the sample and coil temperature separately. If the
effect of preamplifier is included, we get a bit more complex relation
Signal/noise = K
Nγ5/2 2B
3/2
0
k
3/2
B Tsample (Tcoil + TsampleR /R + (1 + R /R) T )∆f
, (8.64)
where R is the resistance of the coil, R is the resistance induced by the sample in the coil (proportional to the
conductivity and therefore to the ionic strength of the sample), and T is so called noise temperature of the amplifier.8
The actual sensitivity also depends on relaxation,9 apodization (or other tricks of processing), and is also proportional to
square root of the ratio of the time of signal acquisition to the overall time of the experiment.10
The numerical values given by Eqs. 8.63–8.64 are of little practical use. However, it is useful to notice how sensitivity
depends on individual factors (temperature, field, magnetogyric ratio of the observed nucleus).
8.8.4 Conclusions
In general, the analysis of an ideal one-pulse experiment leads to the following conclu-
sions:
• The analysis of a one-pulse NMR experiment shows that the density matrix evolves
as
ˆρ(t) ∝ (Ix cos(Ωt + φ) + Iy sin(Ωt) + φ) + terms orthogonal to I+, (8.65)
the magnetization rotates during signal acquisition as
M+ = |M+|e−R2t
eiΩt
(8.66)
(with some unimportant phase shift which is empirically corrected).
7
We already assumed that the phase correction was applied. Another factor determining the shape
of the spectrum in practice is apodization, but we can ignore it now for the sake of simplicity.
8
The input noise is amplified by the factor (1 + T /T)G, where G is the gain of the preamplifier.
9
At low temperatures, Boltzmann distribution is more favorable but line broadening more severe
(mostly due to higher viscosity of the solvent at low temperature). Therefore, the temperature dependence
of sensitivity on the temperature has a maximum (interestingly close to room temperature for
medium-size proteins in aqueous solutions).
10
In many experiments (but not necessarily in the one-dimensional experiment), recycle delay (waiting
for the sample to return close to the equilibrium before the next measurement) is much longer than the
actual signal acquisition.
8.8. THE ONE-PULSE EXPERIMENT 123
• Fourier transform gives a complex signal proportional to
Nγ2 2
B0
4kBT
R2
R2
2 + (ω − Ω)2
− i
ω − Ω
R2
2 + (ω − Ω)2
. (8.67)
• The cosine modulation of Ix can be taken as the real component of the signal and
the sine modulation of Iy can be taken as the imaginary component of the signal:
Signalinchannel1:{y(t)}
t
Signalinchannel2:{y(t)}
t
After Fourier transformation:
{Y(ω)}
ω
Ω
{Y(ω)}
ω
Ω
• The signal-to-noise ratio (without relaxation) is proportional to γ5/2
B
3/2
0 , with the
optimal temperature given by relaxation properties (close to room temperatures
for proteins in aqueous solutions).
124 CHAPTER 8. CHEMICAL SHIFT, NMR EXPERIMENT
Chapter 9
Product operators, dipolar coupling
Literature: The product operator formalism for multi-spin systems is described in
B17.4, B18, C2.5.1, C2.7, L15. The dipole-dipole Hamiltonian is discussed in L9.3.
Relaxtion is described K9, L19–L20, C5 in different manners. All texts are excellent. It
is very helpful to read them all if you really want to get an insight. However, the topic
is difficult and absorbing the information requires a lot of time.
9.1 Dipolar coupling
So far, we analyzed effects of various fields on nuclei, but we assumed that all nuclei
are independent and their properties can be described by operators composed of twodimensional
matrices. Now we take into account also mutual interactions – interactions
with fields generated by other nuclei.
If spin magnetic moments of two spin-1/2 nuclei interact with each other, the magnetic
moment of nucleus 1 is influenced by the magnetic field B2 of the magnetic moment
of nucleus 2.
B2 is given by the classical electrodynamics as
B2 = × A2, (9.1)
where
≡
∂
∂x
,
∂
∂y
,
∂
∂z
. (9.2)
Let us assume (classically) that the source of the magnetic moment of nucleus 2 is a current loop. It can be derived
from Maxwell equations that the vector potential A2 in a distance much larger than radius of the loop is
A2 =
µ0
4π
µ2 × r
r3
, (9.3)
where r is a vector defining the mutual position of nuclei 1 and 2 (inter-nuclear vector).
Calculation of B2 thus includes two vector products
B2 =
µ0
4π
×
µ2 × r
r3
. (9.4)
As a consequence, each component of B2 depends on all components of µ2:
125
126 CHAPTER 9. PRODUCT OPERATORS, DIPOLAR COUPLING
B2,x =
µ0
4πr5
((3r2
x − r2
)µ2,x + 3rxryµ2,y + 3rxrzµ2,z) (9.5)
B2,y =
µ0
4πr5
(3rxryµ2,x + (3r2
y − r2
)µ2,y + 3ryrzµ2,z) (9.6)
B2,z =
µ0
4πr5
(3rxrzµ2,x + 3ryrzµ2,y + (3r2
z − r2
)µ2,z), (9.7)
which can by described by a matrix equation


B2,x
B2,y
B2,z

 =
µ0
4πr5


3r2
x − r2
3rxry 3rxrz
3rxry 3r2
y − r2
3ryrz
3rxrz 3ryrz 3r2
z − r2

 ·


µ2,x
µ2,y
µ2,z

 . (9.8)
The matrix in Eq. 9.8 represents a tensor describing the geometric relations of the dipolar coupling and has the same
form as the matrix in Eq. 3.41, describing the anisotropic contribution to the chemical shift tensor: the vector defining
the symmetry axis of the chemical shift tensor a is just replaced with the inter-nuclear vector r in Eq. 9.8. Like the
anisotropic part of the chemical shift tensor, the matrix in Eq. 9.8 simplifies to
µ0
4πr3


−1 0 0
0 −1 0
0 0 2

 (9.9)
in a coordinate system with axis z r. Rotation to the laboratory frame is described by angles ϕ and ϑ defining
orientation of r in the laboratory frame


−1 0 0
0 −1 0
0 0 2

 −→
1
r2


3r2
x − r2 3rxry 3rxrz
3rxry 3r2
y − r2 3ryrz
3rxrz 3ryrz 3r2
z − r2

 , (9.10)
where rx = r sin ϑ cos ϕ, ry = r sin ϑ sin ϕ, and rz = r cos ϑ.
Classical electrodynamics tells us that the energy of the interaction with the field of the second dipolar magnetic
moments (E = −µ1 · B2) is
E = −
µ0γ1γ2
4πr3
(3 sin2
ϑ cos2
ϕ − 1)µ1xµ2x + (3 sin2
ϑ sin2
ϕ − 1)µ1yµ2y + (3 cos2
ϑ − 1)µ1zµ2z+
+ 3 sin2
ϑ sin ϕ cos ϕµ1xµ2y + 3 sin ϑ cos ϑ cos ϕµ1xµ2z + 3 sin ϑ cos ϑ sin ϕµ1yµ2z
+ 3 sin2
ϑ sin ϕ cos ϕµ1yµ2x + 3 sin ϑ cos ϑ cos ϕµ1zµ2x + 3 sin ϑ cos ϑ sin ϕµ1zµ2y . (9.11)
Having the classical description of the interaction of two magnetic dipolar moments,
we can ask how the quantum description should be modified if the magnetic moments
interact. If two spin magnetic moments interact mutually, they cannot be described
using the same basis. Eigenfunctions are influenced by the interactions. State of the
first spin depends on the state of the second spin. For two spin-1/2 nuclei, there are
2 × 2 = 4 states. Therefore, the probability density matrix should describe the pair of
magnetic moments and its four states as one entity. Furthermore, the classical expression
of the interaction energy (Eq. 9.11) suggests that the Hamiltonian should be built from
operators representing products of individual components of the interacting magnetic
moments. Let us know look for a basis that fulfils these requirements.
9.2. PRODUCT OPERATORS 127
9.2 Product operators
The density matrix for four states is a 4×4 matrix. Basis used for such density matrices
must consist of 42
= 16 matrices. The density matrix for N states is a N × N matrix.
Basis used for such density matrices must consist of 4N
matrices.
The basis can be derived by the direct product1
of basis matrices of spins without
mutual interactions. For example, Cartesian single-spin operators can be used to create
a basis for two spins (see Table 9.1) using the following direct products:
2 · It(1) ⊗ It(2) = It(12) (9.12)
2 · Ix(1) ⊗ It(2) = I1x(12) (9.13)
2 · Iy(1) ⊗ It(2) = I1y(12) (9.14)
2 · Iz(1) ⊗ It(2) = I1z(12) (9.15)
2 · It(1) ⊗ Ix(2) = I2x(12) (9.16)
2 · It(1) ⊗ Iy(2) = I2y(12) (9.17)
2 · It(1) ⊗ Iz(2) = I2z(12) (9.18)
2 · Ix(1) ⊗ Ix(2) = 2I1xI2x(12) (9.19)
2 · Ix(1) ⊗ Iy(2) = 2I1xI2y(12) (9.20)
2 · Ix(1) ⊗ Iz(2) = 2I1xI2z(12) (9.21)
2 · Iy(1) ⊗ Ix(2) = 2I1yI2x(12) (9.22)
2 · Iy(1) ⊗ Iy(2) = 2I1yI2y(12) (9.23)
2 · Iy(1) ⊗ Iz(2) = 2I1yI2z(12) (9.24)
2 · Iz(1) ⊗ Ix(2) = 2I1zI2x(12) (9.25)
2 · Iz(1) ⊗ Iy(2) = 2I1zI2y(12) (9.26)
2 · Iz(1) ⊗ Iz(2) = 2I1zI2z(12), (9.27)
where the numbers in parentheses specify which nuclei constitute the spin system
described by the given matrix (these numbers are not written in practice). The matrices
on the right-hand side are known as product operators. Note that It, equal to2 1
2
ˆ1,
is not written in the product operators for the sake of simplicity. Note also that e.g.
Ix(1) and Ix(2) are the same 2 × 2 matrices, but I1x(12) and I2x(12) are different
4 × 4 matrices. Basis matrices for more nuclei are derived in the same manner, e.g.
2I1zI2x(12) ⊗ Iy(3) = 4I1zI2xI3y(123).
The basis presented in Table 9.1 represents one of many possible choices. Another
1
Direct product ˆA ⊗ ˆB is a mathematical operation when each element of the matrix ˆA is multiplied
by the whole matrix ˆB.
2ˆ1 is the unit matrix.
128 CHAPTER 9. PRODUCT OPERATORS, DIPOLAR COUPLING
Table 9.1: Cartesian basis of product operators for a pair of spin-1
2 nuclei
It = 1
2




+1 0 0 0
0 +1 0 0
0 0 +1 0
0 0 0 +1



 I1z = 1
2




+1 0 0 0
0 +1 0 0
0 0 −1 0
0 0 0 −1



 I2z = 1
2




+1 0 0 0
0 −1 0 0
0 0 +1 0
0 0 0 −1



 2I1zI2z = 1
2




+1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 +1




I1x = 1
2




0 0 +1 0
0 0 0 +1
+1 0 0 0
0 +1 0 0



 2I1xI2z = 1
2




0 0 +1 0
0 0 0 −1
+1 0 0 0
0 −1 0 0



 I1y = 1
2




0 0 −i 0
0 0 0 −i
+i 0 0 0
0 +i 0 0



 2I1yI2z = 1
2




0 0 −i 0
0 0 0 +i
+i 0 0 0
0 −i 0 0




I2x = 1
2




0 +1 0 0
+1 0 0 0
0 0 0 +1
0 0 +1 0



 2I1zI2x = 1
2




0 +1 0 0
+1 0 0 0
0 0 0 −1
0 0 −1 0



 I2y = 1
2




0 −i 0 0
+i 0 0 0
0 0 0 −i
0 0 +i 0



 2I1zI2y = 1
2




0 −i 0 0
+i 0 0 0
0 0 0 +i
0 0 −i 0




2I1xI1x = 1
2




0 0 0 +1
0 0 +1 0
0 +1 0 0
+1 0 0 0



 2I1yI1y = 1
2




0 0 0 −1
0 0 +1 0
0 +1 0 0
−1 0 0 0



 2I1xI1y = 1
2




0 0 0 −i
0 0 +i 0
0 −i 0 0
+i 0 0 0



 2I1yI1x = 1
2




0 0 0 −i
0 0 −i 0
0 +i 0 0
+i 0 0 0




possible basis is shown in Table 9.2. Eqs. 7.4 and 7.5 can be used to convert product
operators of the basis sets in Tables 9.1 and 9.2.
9.3 Liouville - von Neumann equation
The Liouville - von Neumann equation can be written in the same form as for spins
without mutual interactions (Eq. 7.19):
dˆρ
dt
= i(ˆρH − Hˆρ) = i[ˆρ, H] = −i[H, ˆρ], (9.28)
but the density matrix and Hamiltonian are now N × N matrices described in the
appropriate basis. The same simple geometric solution as for spins without mutual interactions
is possible if the Hamiltonian does not vary in time and consists of commuting
matrices only. However, the operator space is now N2
dimensional (16-dimensional for
two spin-1/2 nuclei). Therefore, the appropriate three-dimensional subspace must be selected
for each rotation. The subspaces are defined by the commutator relations, which
can be defined for spin systems consisting of any number of spin-1/2 nuclei using the
following equations.
[In,x, In,y] = iIn,z [In,y, In,z] = iIn,x [In,z, In,x] = iIn,y (9.29)
[In,j, 2In,kIn ,l] = 2[In,j, In,k]In ,l (9.30)
[2In,jIn ,l, 2In,kIn ,m] = [In,j, In,k]δlm, (9.31)
9.4. OPERATOR OF THE OBSERVED QUANTITY 129
Table 9.2: Single-element basis of product operators for a pair of spin-1
2 nuclei
I1αI2α =




1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0



 I1αI2β =




0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 0



 I1βI2α =




0 0 0 0
0 0 0 0
0 0 1 0
0 0 0 0



 I1βI2β =




0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1




I1αI2+ =




0 1 0 0
0 0 0 0
0 0 0 0
0 0 0 0



 I1αI2− =




0 0 0 0
1 0 0 0
0 0 0 0
0 0 0 0



 I1βI2+ =




0 0 0 0
0 0 0 0
0 0 0 1
0 0 0 0



 I1βI2− =




0 0 0 0
0 0 0 0
0 0 0 0
0 0 1 0




I1+I2α =




0 0 1 0
0 0 0 0
0 0 0 0
0 0 0 0



 I1−I2α =




0 0 0 0
0 0 0 0
1 0 0 0
0 0 0 0



 I1+I2β =




0 0 0 0
0 0 0 1
0 0 0 0
0 0 0 0



 I1−I2β =




0 0 0 0
0 0 0 0
0 0 0 0
0 1 0 0




I1+I2+ =




0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0



 I1+I2− = −




0 0 0 0
0 0 1 0
0 0 0 0
0 0 0 0



 I1−I2+ = −




0 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0



 I1−I2− =




0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0




where n and n specify the nucleus, j, k, l ∈ {x, y, z}, and δlm = 1 for l = m and
δlm = 0 for l = m.
9.4 Operator of the observed quantity
In order to describe the observed signal for a system of different nuclei, Eq. 9.32, defining
the operator of complex magnetization, must be slightly modified
ˆM+ =
n
Nnγn(ˆIn,x + iˆIn,y) =
n
Nnγn
ˆIn,+, (9.32)
where the index n distinguishes different types of nuclei and Nn is the number of
nuclei of each type in the sample (per unit volume).
9.5 Hamiltonian of dipolar coupling
As usually, Hamiltonian of the dipolar coupling can be obtained using the classical description of the energy (Eq. 9.11).
Describing the magnetic moments by the operators ˆµ1,jγ1
ˆI1,j and ˆµ2,jγ1
ˆI2,j, where j is x, y, and z, the Hamiltonian of
dipolar coupling ˆHD can be written as
ˆHD = −γ1(ˆI1,xB2,x + ˆI1,yB2,y + ˆI1,zB2,z) = −γ1( ˆI1,x
ˆI1,y
ˆI1,z )


B2,x
B2,y
B2,z

 =
= −
µ0γ1γ2
4πr5
( ˆI1,x
ˆI1,y
ˆI1,z )


3r2
x − r2 3rxry 3rxrz
3rxry 3r2
y − r2 3ryrz
3rxrz 3ryrz 3r2
z − r2




ˆI2,x
ˆI2,y
ˆI2,z

 =
ˆ
1I · D ·
ˆ
2I, (9.33)
where D is the tensor of direct dipole-dipole interactions (dipolar coupling).
130 CHAPTER 9. PRODUCT OPERATORS, DIPOLAR COUPLING
The Hamiltonian can be written in spherical coordinates as
ˆHD = −
µ0γ1γ2
4πr3
(3 sin2
ϑ cos2
ϕ − 1)ˆI1x
ˆI2x + (3 sin2
ϑ sin2
ϕ − 1)ˆI1y
ˆI2y + (3 cos2
ϑ − 1)ˆI1z
ˆI2z+
+ 3 sin2
ϑ sin ϕ cos ϕˆI1x
ˆI2y + 3 sin ϑ cos ϑ cos ϕˆI1x
ˆI2z + 3 sin ϑ cos ϑ sin ϕˆI1y
ˆI2z
+ 3 sin2
ϑ sin ϕ cos ϕˆI1y
ˆI2x + 3 sin ϑ cos ϑ cos ϕˆI1z
ˆI2x + 3 sin ϑ cos ϑ sin ϕˆI1z
ˆI2y . (9.34)
9.6 Secular approximation and averaging
Similarly to the chemical-shift Hamiltonian, the Hamiltonian of dipolar coupling can be
simplified in many cases.
• Magnetic moments with the same γ and chemical shift precess about the z axis with the same precession frequency.
In addition to the precession, the magnetic moments moves with random molecular motions, described by reorientation
of r. In a coordinate system rotating with the common precession frequency, r quickly rotates about
the z axis in addition to the random molecular motions. On a time scale slower than nanoseconds, the rapid
oscillations of rx, ry, and rz are neglected (secular approximation). The values of r2
x and r2
y do not oscillate
about zero, but about a value r2
x = r2
y , which is equal to3 (r2 − r2
z )/2 because r2
x + r2
y + r2
z = r2 = r2.
Therefore, the secular approximations (i.e., neglecting the oscillations and keeping the average values) simplifies
the Hamiltonian to
ˆHD = −
µ0γ1γ2
4πr5
3 r2
z − r2 ˆI1,z
ˆI2,z −
1
2
ˆI1,x
ˆI2,x −
1
2
ˆI1,y
ˆI2,y (9.35)
= −
µ0γ1γ2
4πr3
3 cos2 ϑ − 1
2
2ˆI1,z
ˆI2,z − ˆI1,x
ˆI2,x − ˆI1,y
ˆI2,y . (9.36)
• Magnetic moments with different γ and/or chemical shift precess with different precession frequencies. Therefore,
the x and y components of µ2 rapidly oscillate in a frame rotating with the precession frequency of µ1 and vice
versa. When neglecting the oscillating terms (secular approximation), the Hamiltonian reduces to
ˆHD = −
µ0γ1γ2
4πr5
3 r2
z − r2 ˆI1,z
ˆI2,z = −
µ0γ1γ2
4πr3
3 cos2 ϑ − 1
2
2ˆI1,z
ˆI2,z. (9.37)
• Averaging over all molecules in isotropic liquids has the same effect as described for the anisotropic part of the
chemical shielding tensor because both tensors have the same form:
r2
x = r2
y = r2
z. (9.38)
Finally,
r2
x + r2
y + r2
z = r2
⇒ r2
x + r2
y + r2
z = 3r2
z = r2
⇒ 3r2
z − r2 = r(3 cos2 ϑ − 1) = 0. (9.39)
Unlike the chemical shift Hamiltonian, the Hamiltonian of the dipolar coupling does
not have any isotropic part. As a consequence, the dipole-dipole interactions are not
observable in isotropic liquids. On the other hand, their effect is huge in solid state
NMR and they can be also be measured e.g. in liquid crystals or mechanically stretched
gels. Last but not least, the dipole-dipole interactions represent a very important source
of relaxation.
3
Note that r2
x = r2
y = r2
z in general.
9.7. DIPOLE-DIPOLE RELAXATION 131
9.7 Dipole-dipole relaxation
Rotation of the molecule (and internal motions) change the orientation of the internuclear
vector and cause fluctuations of the field of magnetic moment µ2 sensed by
the magnetic moment µ1. It leads to the loss of coherence in the same manner as
described for the anisotropic part of the chemical shift (cf. Eqs 3.41 and 9.8. However,
the relaxation effects of the dipole-dipole interactions are more complex, reflecting the
higher complexity of the Hamiltonian of the dipolar coupling.
In order to describe the dipole-dipole relaxation on the quantum level, it is useful to work in spherical coordinates
and to convert the product operators to a different basis. Single quantum operators are transformed using the relation
ˆI± = ˆIx ± iˆIy):
ˆI1x
ˆI2z =
1
2
(+ˆI1+
ˆI2z + ˆI1−
ˆI2z), (9.40)
ˆI1y
ˆI2z =
i
2
(−ˆI1+
ˆI2z + ˆI1−
ˆI2z), (9.41)
ˆI1z
ˆI2x =
1
2
(+ˆI1z
ˆI2+ + ˆI1z
ˆI2−), (9.42)
ˆI1z
ˆI2y =
i
2
(−ˆI1z
ˆI2+ + ˆI1z
ˆI2−). (9.43)
Since
cos ϕ + i sin ϕ = eiϕ
, (9.44)
cos ϕ − i sin ϕ = e−iϕ
, (9.45)
3 sin ϑ cos ϑ(ˆI1x
ˆI2z cos ϕ + ˆI1y
ˆI2z sin ϕ + ˆI1z
ˆI2x cos ϕ + ˆI1z
ˆI2y sin ϕ).
=
3
2
sin ϑ cos ϑ(ˆI1+
ˆI2ze−iϕ
+ ˆI1−
ˆI2zeiϕ
+ ˆI1z
ˆI2+e−iϕ
+ ˆI1z
ˆI2−eiϕ
) (9.46)
The double-quantum/zero-quantum operators are transformed in a similar fashion
ˆI1x
ˆI2y =
i
4
(+ˆI1+
ˆI2− − ˆI1−
ˆI2+ − ˆI1+
ˆI2+ + ˆI1−
ˆI2−),
ˆI1y
ˆI2x =
i
4
(−ˆI1+
ˆI2− + ˆI1−
ˆI2+ − ˆI1+
ˆI2+ + ˆI1−
ˆI2−),
ˆI1x
ˆI2x =
1
4
(+ˆI1+
ˆI2− + ˆI1−
ˆI2+ + ˆI1+
ˆI2+ + ˆI1−
ˆI2−),
ˆI1y
ˆI2y =
1
4
(+ˆI1+
ˆI2− + ˆI1−
ˆI2+ − ˆI1+
ˆI2+ − ˆI1−
ˆI2−),
and
3 sin2
ϑ(ˆI1x
ˆI2x cos2
ϕ + ˆI1y
ˆI2y sin2
ϕ + ˆI1x
ˆI2y sin ϕ cos ϕ + ˆI1y
ˆI2x sin ϕ cos ϕ) − (ˆI1x
ˆI2x + ˆI1y
ˆI2y)
132 CHAPTER 9. PRODUCT OPERATORS, DIPOLAR COUPLING
=
3
4
sin2
ϑ ( ˆI1+
ˆI2−(cos2
ϕ + sin2
ϕ + i sin ϕ cos ϕ − i sin ϕ cos ϕ)
+ˆI1−
ˆI2+(cos2
ϕ + sin2
ϕ − i sin ϕ cos ϕ + i sin ϕ cos ϕ)
+ˆI1+
ˆI2+(cos2
ϕ − sin2
ϕ − i sin ϕ cos ϕ − i sin ϕ cos ϕ)
+ˆI1−
ˆI2−(cos2
ϕ − sin2
ϕ + i sin ϕ cos ϕ + i sin ϕ cos ϕ) )
−
1
4
(2ˆI1+
ˆI2− + 2ˆI1−
ˆI2+)
=
1
4
ˆI1+
ˆI2−(3 sin2
ϑ − 2) +
1
4
ˆI1−
ˆI2+(3 sin2
ϑ − 2)
+
3
4
ˆI1+
ˆI2+ sin2
ϑe−i2ϕ
+
3
4
ˆI1−
ˆI2− sin2
ϑei2ϕ
= −
1
4
ˆI1+
ˆI2−(3 cos2
ϑ − 1) −
1
4
ˆI1−
ˆI2+(3 cos2
ϑ − 1)
+
3
4
ˆI1+
ˆI2+ sin2
ϑe−i2ϕ
+
3
4
ˆI1−
ˆI2− sin2
ϑei2ϕ
. (9.47)
Using Eqs. 9.46 and 9.47 and moving to the interaction frame (ˆIn± → ˆIn±e±iω0,nt), Eq. 9.34 is converted to
ˆHI
D = −
µ0γ1γ2
4πr3
ˆI1z
ˆI2z(3 cos2
ϑ − 1)
−
1
4
ˆI1+
ˆI2−(3 cos2
ϑ − 1)ei(ω0,1−ω0,2)t
−
1
4
ˆI1−
ˆI2+(3 cos2
ϑ − 1)e−i(ω0,1−ω0,2)t
+
3
2
ˆI1+
ˆI2z sin ϑ cos ϑe−iϕ
ei(ω0,1)t
+
3
2
ˆI1−
ˆI2z sin ϑ cos ϑeiϕ
e−i(ω0,1)t
+
3
2
ˆI1z
ˆI2+ sin ϑ cos ϑe−iϕ
ei(ω0,2)t
+
3
2
ˆI1z
ˆI2− sin ϑ cos ϑeiϕ
e−i(ω0,2)t
+
3
4
ˆI1+
ˆI2+ sin2
ϑe−i2ϕ
ei(ω0,1+ω0,2)t
+
3
4
ˆI1−
ˆI2− sin2
ϑei2ϕ
e−i(ω0,1+ω0,2)t
= −
µ0γ1γ2
4πr3
2ˆI1z
ˆI2zczz
−
1
2
c+− ˆI1+
ˆI2− −
1
2
c−+ ˆI1−
ˆI2+
+
3
2
c+z ˆI1+
ˆI2z + c−z ˆI1−
ˆI2z + cz+ ˆI1z
ˆI2+ + cz− ˆI1z
ˆI2− + c++ ˆI1+
ˆI2+ + c−− ˆI1−
ˆI2− . (9.48)
Similarly to Eq. 8.28, the dipole-dipole relaxation is described by
d∆ˆρ
dt
= −
1
2
∞ˆ
0
[ ˆHD(0), [ ˆHD(t), ∆ˆρ]]dt. (9.49)
The right-hand side can be simplified dramatically by the secular approximation as in Eq. 8.28: all terms with
e±iω0,nt are averaged to zero. Only terms with (czz)2, cz+cz−, c+zc−z, c+−c−+, and c++c−− are non zero (all equal
to 1/5 at tj = 0).4 This reduces the number of double commutators to be expressed from 81 to 9 for each density matrix
component. The double commutators needed to describe relaxation rates of the contributions of the first nucleus to the
magnetization M1z and M1+ are, respectively,
4
Averaging over all molecules makes all correlation functions identical in isotropic liquids.
9.7. DIPOLE-DIPOLE RELAXATION 133
ˆI1z
ˆI2z, [ˆI1z
ˆI2z, ˆI1z] = 0, (9.50)
ˆI1−
ˆI2+, [ˆI1+
ˆI2−, ˆI1z] = 2
(ˆI1z − ˆI2z), (9.51)
ˆI1+
ˆI2−, [ˆI1−
ˆI2+, ˆI1z] = 2
(ˆI1z − ˆI2z), (9.52)
ˆI1+
ˆI2z, [ˆI1−
ˆI2z, ˆI1z] =
1
2
2 ˆI1z, (9.53)
ˆI1−
ˆI2z, [ˆI1+
ˆI2z, ˆI1z] =
1
2
2 ˆI1z, (9.54)
ˆI1z
ˆI2+, [ˆI1z
ˆI2−, ˆI1z] = 0, (9.55)
ˆI1z
ˆI2−, [ˆI1z
ˆI2+, ˆI1z] = 0, (9.56)
ˆI1+
ˆI2+, [ˆI1−
ˆI2−, ˆI1z] = 2
(ˆI1z + ˆI2z), (9.57)
ˆI1−
ˆI2−, [ˆI1+
ˆI2+, ˆI1z] = 2
(ˆI1z + ˆI2z), (9.58)
ˆI1z
ˆI2z, [ˆI1z
ˆI2z, ˆI1+] =
1
4
2 ˆI1+, (9.59)
ˆI1+
ˆI2−, [ˆI1−
ˆI2+, ˆI1+] = 2 ˆI1+, (9.60)
ˆI1−
ˆI2+, [ˆI1+
ˆI2−, ˆI1+] = 0, (9.61)
ˆI1+
ˆI2z, [ˆI1−
ˆI2z, ˆI1+] =
1
2
2 ˆI1+, (9.62)
ˆI1−
ˆI2z, [ˆI1+
ˆI2z, ˆI1+] = 0, (9.63)
ˆI1z
ˆI2+, [ˆI1z
ˆI2−, ˆI1+] =
1
2
2 ˆI1+, (9.64)
ˆI1z
ˆI2−, [ˆI1z
ˆI2+, ˆI1+] =
1
2
2 ˆI1+, (9.65)
ˆI1+
ˆI2+, [ˆI1−
ˆI2−, ˆI1+] = 0, (9.66)
ˆI1−
ˆI2−, [ˆI1+
ˆI2+, ˆI1+] =
1
2
2 ˆI1+. (9.67)
The relaxation rates can be then derived as described for the relaxation due to the chemical shift.
134 CHAPTER 9. PRODUCT OPERATORS, DIPOLAR COUPLING
The following equations describe relaxation due to the dipole-dipole interactions in
a pair of nuclei in a rigid spherical molecule:
d∆ M1z
dt
= −
1
8
b2
(2J(ω0,1 − ω0,2) + 6J(ω0,1) + 12J(ω0,1 + ω0,2))∆ M1z
+
1
8
b2
(2J(ω0,1 − ω0,2) − 12J(ω0,1 + ω0,2))∆ M2z
= −Ra1∆ M1z − Rx∆ M2z , (9.68)
d∆ M2z
dt
= −
1
8
b2
(2J(ω0,1 − ω0,2) + 6J(ω0,2) + 12J(ω0,1 + ω0,2))∆ M2z
+
1
8
b2
(2J(ω0,1 − ω0,2) − 12J(ω0,1 + ω0,2))∆ M1z
= −Ra2∆ M2z − Rx∆ M1z , (9.69)
d M1+
dt
= −
1
8
b2
(4J(0) + 6J(ω0,2) + J(ω0,1 − ω0,2) + 3J(ω0,1) + 6J(ω0,1 + ω0,2)) M1+
= −R2,1 M1+ = − R0,1 +
1
2
Ra1 M1+ , (9.70)
where
b = −
µ0γ1γ2
4πr3
. (9.71)
The relaxation rate R1 of the dipole-dipole relaxation is the rate of relaxation of the
z-component of the total magnetization Mz = M1z + M2z . R1 is derived by solving
the set of Eqs. 9.68 and 9.69. The solution is simple if J(ω0,1) = J(ω0,2) = J(ω0) ⇒
Ra1 = Ra2 = Ra (this is correct e.g. if both nuclei have the same γ, if the molecule
rotates as a sphere, and if internal motions are negligible or identical for both nuclei).5
d∆ Mz
dt
= −
1
8
b2
(6J(ω) + 24J(2ω))∆ Mz = − (Ra + Rx)
R1
∆ Mz . (9.72)
There are several remarkable differences between relaxation due to the chemical shift
anisotropy and dipole-dipole interactions:
• The rate constants describing the return to the equilibrium polarization is more
complex than for the chemical shift anisotropy relaxation. In addition to the
3b2
J(ω0,1)/4 term, describing the |α ↔ |β transition6
of nucleus 1, the autorelaxation
rate Ra1 contains terms depending on the sum and difference of the
5
The general solution gives R1 = 1
2 Ra1 + Ra2 + (Ra1 − Ra2)2 + 4R2
x .
6
The |αα ↔ |βα and |αβ ↔ |ββ transitions in a two-spin system
9.7. DIPOLE-DIPOLE RELAXATION 135
precession frequency of µ1 and µ2. These terms correspond to the zero-quantum
(|αβ ↔ |βα ) and double-quantum (|αα ↔ |ββ ) transitions, respectively.
• Return to the equilibrium polarization of nucleus 1 depends also on the actual
polarization of nucleus 2. This effect, resembling chemical kinetics of a reversible
reaction, is known as cross-relaxation, or nuclear Overhauser effect (NOE), and
described by the cross-relaxation constant Rx. The value of Rx is proportional to
r−6
and thus provides information about inter-atomic distances. NOE is a useful
tool in analysis of small molecules and the most important source of structural
information for large biological molecules.
• The relaxation constant R0, describing the loss of coherence, contains an additional
term, depending on the frequency of the other nucleus, 3b2
J(ω0,2)/4. This
term has the following physical significance. The field generated by the second
magnetic moment depends on its state. The state is changing due to |α ↔ |β
transitions the with the rate given by 3b2
J(ω0,2)/4 (single quantum contribution
to R1, analogous to R1 due to the chemical schift anisotropy). Such changes have
the similar effect as the chemical or conformational exchange, modifying the size
of the chemical shift tensor. Therefore, 3b2
J(ω0,2)/4 adds to R0 like the exchange
contribution.
136 CHAPTER 9. PRODUCT OPERATORS, DIPOLAR COUPLING
Chapter 10
2D spectroscopy, NOESY
Literature: A very nice explanation of the principles of two-dimensional spectroscopy
can be found in K8.1–K8.2. The idea of 2D spectroscopy, but for a different type of
experiment (COSY) is also presented in C4.1, L5.6 and L5.9.
10.1 Two-dimensional spectroscopy
In order to describe principles of 2D spectroscopy, we first analyze an experiment consisting
of three 90◦
pulses and two delays before data acquisition:
a(π/2)xb − t1 −c (π/2)xd − τm −e (π/2)xf − t2(acquire).
We describe the density matrix just before and after pulses, as labeled by letters ”a”
to ”f”.
10.1.1 Thermal equilibrium
Before we analyze evolution of the density matrix in a 2D experiment, we must define
its initial form.
Again, we start from the thermal equilibrium and use the Hamiltonian. The difference from the case of isolated nuclei
is that we need to define a 4 × 4 density matrix in order to describe a pair of mutually interacting nuclei. As explained
above, the off-diagonal elements of the equilibrium density matrix (proportional to Ix and Iy) are equal to zero. The
four diagonal elements describe average populations of four stationary states of a system composed of (isolated) nuclear
pairs: αα, αβ, βα, and ββ. These populations are:
137
138 CHAPTER 10. 2D SPECTROSCOPY, NOESY
Peq
αα =
e−Eαα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 − Eαα
kBT
4
, (10.1)
Peq
αβ =
e−Eαβ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eαβ
kBT
4
, (10.2)
Peq
βα =
e−Eβα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eβα
kBT
4
, (10.3)
Peq
ββ =
e−Eββ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eββ
kBT
4
. (10.4)
The energies Eαα, Eαβ, Eβα, and Eββ are the eigenvalues of the total Hamiltonian, which should include effects of
the external field B0, of chemical shifts of both nuclei, and of their dipolar coupling. However, the dipolar coupling in
isotropic liquids is averaged to zero. It is therefore sufficient to write the total Hamiltonian as
ˆH = −γ1B0(1 + δ1,i)ˆI1,z − γ2B0(1 + δ2,i)ˆI2,z = −γ1B0(1 + δ1,i)
2




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 − γ2B0(1 + δ2,i)
2




1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 −1



 =
B0
2




−γ1(1 + δ1,i) − γ2(1 + δ2,i) 0 0 0
0 −γ1(1 + δ1,i) + γ2(1 + δ2,i) 0 0
0 0 +γ1(1 + δ1,i) − γ2(1 + δ2,i) 0
0 0 0 +γ1(1 + δ1,i) + γ2(1 + δ2,i)



 ,
(10.5)
where the diagonal elements (eigenavalues) are the energies of the individual states. Therefore,
Peq
αα ≈
1 − Eαα
kBT
4
=
1
4
+ γ1(1 + δ1,i)
B0
8kBT
+ γ2(1 + δ2,i)
B0
8kBT
, (10.6)
Peq
αβ ≈
1 −
Eαβ
kBT
4
=
1
4
+ γ1(1 + δ1,i)
B0
8kBT
− γ2(1 + δ2,i)
B0
8kBT
, (10.7)
Peq
βα ≈
1 −
Eβα
kBT
4
=
1
4
− γ1(1 + δ1,i)
B0
8kBT
+ γ2(1 + δ2,i)
B0
8kBT
, (10.8)
Peq
ββ ≈
1 −
Eββ
kBT
4
=
1
4
− γ1(1 + δ1,i)
B0
8kBT
− γ2(1 + δ2,i)
B0
8kBT
. (10.9)
(10.10)
Neglecting the chemical shifts (δ1,i 1, δ2,i 1)
ˆρeq
=





1
4
+ γ1B0
8kBT
+ γ2B0
8kBT
0 0 0
0 1
4
+ γ1B0
8kBT
− γ2B0
8kBT
0 0
0 0 1
4
− γ1B0
8kBT
+ γ2B0
8kBT
0
0 0 0 1
4
− γ1B0
8kBT
− γ2B0
8kBT





(10.11)
=
1
4




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1



 +
γ1B0
8kBT




+1 0 0 0
0 +1 0 0
0 0 −1 0
0 0 0 −1



 +
γ2B0
8kBT




+1 0 0 0
0 −1 0 0
0 0 +1 0
0 0 0 −1



 (10.12)
10.1. TWO-DIMENSIONAL SPECTROSCOPY 139
=
1
2
(It + κ1I1,z + κ2I2,z) , (10.13)
where
κj =
γjB0
4kBT
. (10.14)
10.1.2 Evolution in the absence of dipolar coupling
We start with an analysis for two non-interacting magnetic moments, e.g. of two protons
with different chemical shift δi too far from each other. The pair of protons is an example
of a homonuclear system, where all nuclei have the same γ. Since we have neglected the
very small effect of different chemical shifts in Eq. 10.13, the values of κ are also the
same for both protons. As in the one-pulse experiment, we follow the coherent evolution
of ˆρ step-by-step, and add the effect of relaxation ad hoc.
• ˆρ(a) = 1
2
It + 1
2
κ(I1z + I2z)
We start from the thermal equilibrium described by Eq. 10.13. Note that the
matrices are different than for the single-spin mixed state, but the constant is the
same.
• ˆρ(b) = 1
2
It + 1
2
κ(−I1y − I2y)
Here we describe the effect of the 90◦
pulse. For detailed analysis, see the one-pulse
experiment.
• ˆρ(c) = 1
2
It + 1
2
κ (− cos(Ω1t1)I1y + sin(Ω1t1)I1x − cos(Ω2t1)I2x + sin(Ω2t1)I2y)
Here we describe evolution during t1 exactly as in the one-pulse experiment.To
keep the equations short, we replace the goniometric terms describing the evolution
by (time-dependent) coefficients c11, c21, s11, and s21:
ˆρ(c) = 1
2
It + 1
2
κ (−c11I1y + s11I1x − c21I2y + s21I2x)
The coefficients c11, c21, s11, and s21 deserve some attention. First, note that the
first subscript specifies the nucleus and the second subscript specifies the time
period (so-far, it is always 1 because we have analyzed only evolution during t1).
Second, we include the effect of relaxation into the coefficients:
c11 → e−R2,1t1
cos(Ω1t1) s11 → e−R2,1t1
sin(Ω1t1)
c21 → e−R2,2t1
cos(Ω2t1) s21 → e−R2,2t1
sin(Ω2t1)
• ˆρ(d) = 1
2
It + 1
2
κ (−c11I1z + s11I1x − c21I2z + s21I2x)
Here we analyze the effect of the second 90◦
pulse, similarly to the step a → b.
The x-pulse does not affect x magnetization, rotates −y magnetization further to
−z. The final magnetization is parallel with B0, but the equilibrium polarization
is inverted.
140 CHAPTER 10. 2D SPECTROSCOPY, NOESY
• ˆρ(e) =?
This is a new case, it should be analyzed carefully. Here perform the analysis for a
large molecule such as a small protein: In proteins, Mx, My relax with R2 > 10 s−1
and Mz with R1 ≈ 1 s−1
. The delay τm is usually longer than 0.1 s. Let us assume
τm = 0.2 s and R2 = 20 s−1
. After 0.2 s, e−R2τm
= e−20×0.2
= e−4
≈ 0.02. We see
that Mx, My relaxes almost completely ⇒ I1x, I1y, I2x, I2y can be neglected. On
the other hand, e−R1τm
= e−1×0.2
= e−0.2
≈ 0.82. We see that Mz does not relax
too much. Therefore, we continue analysis with I1z, I2z. The I1z, I2z terms do not
evolve because they commute with H = Ω1I1z + Ω2I2z. Consequently,
ˆρ(e) = 1
2
It + 1
2
κ −e−R1τm
c11I1z − e−R1τm
c21I2z = 1
2
It − A1I1z − A2I2z
We further simplified the notation by introducing the factors A1 and A2. Again,
we include the relaxation effects into A1 and A2 when we express the measurable
signal:
A1 → κ
2
e−R1,1τm
c11 = κ
2
e−R1,1τm
e−R2,1t1
cos(Ω1t1)
A2 → κ
2
e−R1,2τm
c21 = κ
2
e−R1,2τm
e−R2,2t1
cos(Ω2t1)
• ˆρ(f) = 1
2
It + A1I1y + A2I2y
Here we analyze the effect of the third pulse, in the same manner as we analyzed
the first pulse.
• ˆρ(t2) = 1
2
It + A1(cos(Ω1t2)I1y − sin(Ω1t2)I1x) + A2(cos(Ω2t2)I2y − sin(Ω2t2)I2x)
In the last step, we analyze evolution during data acquisition.
Having ˆρ(t2), we can calculate M+ . As the size of the matrices increased, it is more
convenient to use the orthonormality of the basis than to calculate all matrix products.
It follows from the definition of orthonormal matrices that for the two-spin matrices
Tr {(In,x + iIn,y)In,x} = 1, (10.15)
Tr {(In,x + iIn,y)In,y} = i, (10.16)
and traces of products with other matrices are zero. Applying the orthonormality
relations to the obtained ˆρ(t2) and introducing relaxation, we get
M+ = Nγ A1(e−R2,1t2
cos(Ω1t2)I1y − e−R2,1t2
sin(Ω1t2)I1x)
+A2(e−R2,2t2
cos(Ω2t2)I2y − e−R2,2t2
sin(Ω2t2)I2x) . (10.17)
Note that the resulting phase is shifted by π/2 similarly to Eq. 8.65, but in the
opposite direction. After applying the phase correction, Fourier transform of the signal
provides spectrum in the form (cf. Eq. 8.67)
10.2. NOESY 141
Nγ
A1R2,1
R2
2,1 + (ω − Ω1)2
+
A2R2,2
R2
2,2 + (ω − Ω2)2
− i
A1(ω − Ω1)
R2
2,1 + (ω − Ω1)2
+
A2(ω − Ω2)
R2
2,2 + (ω − Ω2)2
.
(10.18)
In the one-dimensional experiment, A1 and A2 just scale the peak height. However,
they depend on the length of the delay t1. If the measurement is repeated many times
and t1 is increased by an increment ∆t each time, the obtained series of 1D spectra
is amplitude modulated by c11 = e−R2,1t2
cos(Ω1t1) and c21 = e−R2,2t2
cos(Ω2t1). Since
the data are stored in a computer in a digital form, they can be treated as a twodimensional
array (table), depending on the real time t2 in one direction and on the
length of the incremented delay t1 in the other directions. These directions are referred
to as direct dimension and indirect dimension. Fourier transform can be performed in
each dimension.
Since we acquire signal as a series of complex numbers, it is useful to introduce the
complex numbers in the indirect dimension as well. It is possible e.g. by repeating the
measurement twice for each value of t1, once with the x-phase (the same phase as the
first pulse) of the second pulse, as described above, and then with the y-phase (phaseshifted
from the first pulse by 90◦
). In the latter case, the I1y and I2y components are
not affected and relax during τm, while the I1x and I2x are rotated to −I1z and −I2z,
respectively, and converted to the measurable signal by the third pulse. Because the I1x
and I2x coherences are modulated by s11 and s21, A1 and A2 oscillate as a sine function,
not cosine function, in the even spectra. So, we obtain cosine modulation in odd spectra
and sine modulation in even spectra. The cosine- and sine- signals are then treated
as the real and imaginary component of the complex signal in the indirect dimension.
Complex Fourier transform in both dimensions provides a two-dimensional spectrum.
10.2 NOESY
The two-dimensional spectra described in the preceding section are not very useful because
they do not bring any new information. The same frequencies are measured in
the direct and indirect dimension and all peaks are found along the diagonal of the
spectrum. What makes the experiment really useful is the interaction between magnetic
moments during τm. Such approach is known as Nuclear Overhauser effect spectroscopy
(NOESY) and is used frequently to measure distances between protons in molecules.
As described by Eq. 9.68, relaxation of nucleus 1 is influenced by the state of nucleus
2 (and vice versa):
142 CHAPTER 10. 2D SPECTROSCOPY, NOESY
Transmitter on
Transmitter off
Receiver on
Receiver off
t1 τm t2
t1 f1 f1
t1
f1
Figure 10.1: Principle of two-dimensional spectroscopy (experiment NOESY). The acquired signal is
shown in red, the signal after Fourier transform in the direct dimension is shown in magenta, and the
signal after Fourier transform in both dimensions is shown in blue.
10.2. NOESY 143
−
d∆ M1z
dt
= Ra1∆ M1z + Rx∆ M2z (10.19)
−
d∆ M2z
dt
= Ra2∆ M2z + Rx∆ M1z . (10.20)
The analysis greatly simplifies if the auto-relaxation rates are identical for both
magnetic moments. Then,
−
d∆ M1z
dt
= Ra∆ M1z + Rx∆ M2z , (10.21)
−
d∆ M2z
dt
= Ra∆ M2z + Rx∆ M1z . (10.22)
Such set of differential equations can be solved easily e.g. by the substitutions ∆+ =
∆ M1z + ∆ M2z and ∆− = ∆ M2z − ∆ M1z .
The result is
∆+ = ∆+(0)e−(Ra+Rx)t
, (10.23)
∆− = ∆−(0)e−(Ra−Rx)t
. (10.24)
Returning back to ∆ M1z and ∆ M2z ,
∆ M1z = ((1 − ζ)∆ M1z (0) + ζ∆ M2z (0)) e−(Ra+Rx)t
, (10.25)
∆ M2z = ((1 − ζ)∆ M2z (0) + ζ∆ M1z (0)) e−(Ra+Rx)t
, (10.26)
where ζ = (1−e2Rxt
)/2. Therefore, the amplitudes A1 and A2 in our two-dimensional
experiment are
A1 =
κ
2
((1 − ζ)c11 + ζc21)e−(Ra+Rx)τm
, (10.27)
A2 =
κ
2
((1 − ζ)c21 + ζc11)e−(Ra+Rx)τm
. (10.28)
Now, the amplitudes A1 and A2 depend on both frequencies Ω1 and Ω2 (contain both
c11 and c21. Therefore, the spectrum contains both diagonal peaks (with the frequencies
of the given magnetic moment in both dimensions) and off-diagonal cross-peaks (with
the frequencies of the given magnetic moment in the direct dimension and the frequency
of its interaction partner in the indirect dimension). The overall loss of signal (”leakage”)
due to the R1 relaxation is given by e−(Ra−Rx)τm
and intensities of the cross-peaks are
given by the factor
144 CHAPTER 10. 2D SPECTROSCOPY, NOESY
ζe−(Ra+Rx)τm
= −
1
2
eRxτm
− e−Rxτm
e−Raτm
. (10.29)
For short τm, eRxτm
− e−Rxτm
≈ 1 + Rxτm − 1 + Rxτm and e−Raτm
is close to one.
Therefore, the cross-peak intensities are approximately proportional to
−
1
2
eRxτm
− e−Rxτm
e−Raτm
≈ −Rxτm =
µ0
8π
2 γ4 2
r6
(J(0) − 6J(2ω0))τm, (10.30)
where the difference of the precession frequencies due to different chemical shifts was
neglected (ω0,1 = ω0,2 because γ1 = γ2). Hence, the cross-peak intensity is proportional
to r−6
in the linear approximation. If the molecular motions are slow, 2ω0τC 1 ⇒
J(0) > 6J(2ω0), and cross-peaks have the same sign as diagonal peaks. However, if the
molecular motions are fast (e.g., if the molecule is small), the sign is opposite.
Chapter 11
J-coupling, spin echoes
Literature: The through-bond coupling (J-coupling) is described in L14 and L15, the
Hamiltonian is presented in L9.4 and J-coupled spins are described in L14.2, L14.3, and
L14.5. Spin echoes are nicely described in K7.8 and also presented in LA.10.
11.1 Through-bond coupling
Magnetic moments of nuclei connected by covalent bonds interact also indirectly, via
interactions with magnetic moments of the electrons of the bonds. This type of interaction
is known as J-coupling, through-bond coupling, hyperfine coupling, or scalar coupling
(see below).
In principle, both orbital and spin magnetic moments of electrons can contribute to the J-coupling, but the contribution
of the orbital magnetic moments is usually negligible (coupling between hydrogen nuclei in water is an interesting
exception). The contribution of the electron spin can be described in the following manner. Energy of the interaction
between the (spin) magnetic moment of nucleus µn and the magnetic field generated by the spin magnetic moment of
electron Be is given by
E = −µn · Be. (11.1)
Distribution of the electron density in the molecule is described by orbitals. The interaction between the nucleus and
electron outside the nucleus is nothing else but the through-space interaction between two dipolar magnetic moments.
This interaction is averaged to zero. Contributions of p-, d- and higher orbitals to the density inside the nucleus are
negligible because these wave functions are close to zero in the center. Therefore, only s-orbitals are important. The
interaction between the nucleus and electron inside the nucleus can be simulated by a hypothetical current loop giving
the correct magnetic moment when treated classically. If B of such a loop is calculated, its integral over the volume inside
the nucleus is proportional to the magnetic moment:
Be,in =
2
3
µ0µe,in. (11.2)
The expected value of the electron magnetic moment in the volume of the nucleus Vn is given by
µe inside =
ˆ
Vn
ψ(r)∗
ˆµeψ(r)dV, (11.3)
where ψ is the wave function (s orbital in our case) and r is the distance from the center of the atom. Since µe is
constant in the nucleus,
ψ(r)∗
ˆµeψ(r) = ˆµeψ(r)∗
ψ(r) = ˆµe|ψ(r)|2
. (11.4)
145
146 CHAPTER 11. J-COUPLING, SPIN ECHOES
Combining Eqs. 11.1–11.4 gives the energy
E = −
2µ0
3
µn · µe |ψ(inside)|2
(11.5)
and the corresponding Hamiltonian
ˆHF = −
2µ0γnγe
3
ˆ
In ·
ˆ
Ie finside, (11.6)
where
ˆ
In and
ˆ
Ie are operators of the spin of the nucleus and the electron, respectively, γn and γe are magnetogyric
ratios of the spin of the nucleus and the electron, respectively, and finside is equal to one inside the nucleus and to zero
outside the nucleus. This type of interaction is known as the Fermi contact interaction and does not depend on orientation
of the molecule in the magnetic field, as documented by the scalar vecor in Eq. 11.6.
The simplest example of the J-coupling in chemical compounds is J-interaction in a
pair of nuclei (e.g., 1
H and 13
C) connected by a σ bond. In such system, the states |αβ
and |βα allow all interacting particles to be in the opposite1
state (H↑
-e↓
-e↑
-C↓
and H↓
-
e↑
-e↓
-C↑
, respectively) and their interactions are are energetically more favorable than
those of the |αα and |ββ states, which require too interacting particles to be in the
same state (H↑
-e↓
-e↑
-C↑
or H↑
-e↑
-e↓
-C↑
and H↓
-e↓
-e↑
-C↓
or H↓
-e↑
-e↓
-C↓
, respectively).
The relations are more complex in the case of interactions through multiple bonds.
However, the sign of the coupling constant can be predicted easily if only the Fermi
contact interaction is considered: J < 0 for a two-bond coupling (H↑
-e↓
-e↑
-C↓
-e↑
-e↓
-H↑
and H↓
-e↑
-e↓
-C↑
-e↓
-e↑
-H↓
) and J > 0 for a three-bond coupling (H↑
-e↓
-e↑
-C↓
-e↑
-e↓
-C↑
-e↓
-
e↑
-H↓
and H↓
-e↑
-e↓
-C↑
-e↓
-e↑
-C↓
-e↑
-e↓
-H↑
).
In general, each component of the field felt by magnetic moment 1 (e.g. of 1
H)
depends on all components of the magnetic moment 2 (e.g. of 13
C), similarly to the
chemical shift and through-space dipole-dipole coupling. Therefore, the interaction is
described by tensors (like chemical shift or dipolar coupling):
ˆHJ = −γ(ˆIx1B2,x + ˆIy1B2,y + ˆIzB2,z1) = −γ( ˆIx1
ˆIy1
ˆIz1 )


B2,x
B2,y
B2,z

 =
= 2π( ˆIx
ˆIy
ˆIz )


Jxx Jxy Jxz
Jyx Jyy Jyz
Jzx Jzy Jzz




ˆIx1
ˆIy1
ˆIz1

 = 2π
ˆ
1I · J ·
ˆ
2I. (11.7)
However, the anisotropic part of the J-tensor is usually small (and difficult to distinguish
from the dipolar coupling) and is neglected in practice. Therefore, only the
isotropic (scalar2
) part of the tensor is considered and the interaction is called scalar
coupling:
1
Note, however, that the magnetogyric constants of electrons are negative. Therefore, the orientations
of angular momenta and magnetic moments of electrons are opposite. The energetically more
favorable orientation of the magnetic moments of electrons is parallel with the magnetic moments of
the nuclei.
2
Note that the Fermi contact interaction depends on scalar product of magnetic moment vectors,
Eq. 11.6 does not include tensors.
11.2. SECULAR APPROXIMATION AND AVERAGING 147
2π


Jxx 0 0
0 Jyy 0
0 0 Jzz

 = 2π
Jxx + Jyy + Jzz
3


1 0 0
0 1 0
0 0 1

 = 2πJ


1 0 0
0 1 0
0 0 1

 . (11.8)
The scalar coupling is observed as splitting of peaks by 2πJ in NMR spectra. Protonproton
coupling is significant (exceeding 10 Hz) up to three bonds and observable for 4
or 5 bonds in special cases (planar geometry like in aromatic systems). Interactions of
other nuclei are weaker, but the one-bond couplings are always significant (as strong
as 700 Hz for 31
P-1
H, 140 Hz to 200 Hz for 13
C-1
H, 90 Hz for 15
N-1
H in amides, 30 Hz to
60 Hz for 13
C-13
C, 10 Hz to 15 Hz for 13
C-15
N). The value of J is given by the distribution
of electrons in bonds and thus reflect the local geometry of the molecule. Three-bond
scalar couplings can be used to measure torsion angles in molecules.
11.2 Secular approximation and averaging
If the anisotropic part of the J-tensor is neglected, the J-coupling does not depend
on orientation (scalar coupling) and no ensemble averaging is needed. The secular
approximation is applied like in the case of the dipolar coupling.
• In the case of magnetic moments with the same γ and chemical shift, precessing
about the z axis with the same precession frequency,
ˆHJ = πJ 2ˆI1,z
ˆI2,z + 2ˆI1,x
ˆI2,x + 2ˆI1,y
ˆI2,y . (11.9)
• In the case of magnetic moments with different γ and/or chemical shift, precessing
about the z axis with different precession frequencies,
ˆHJ = 2πJ ˆI1,z
ˆI2,z = πJ 2ˆI1,z
ˆI2,z . (11.10)
11.3 Relaxation due to the J-coupling
In principle, the anisotropic part of the J-tensor would contribute to relaxation like
the anisotropic part of the chemical shift tensor, but it is small and usually neglected.
Scalar coupling (isotropic part of the J-tensor) does not depend on the orientation.
Therefore, it can contribute to the relaxation only through a conformational or chemical
exchange. Conformational effects are usually small: one-bond and two-bond couplings do
not depend on torsion angles and three-bond coupling constants are small. In summary,
relaxation due to the J-coupling is rarely observed.
148 CHAPTER 11. J-COUPLING, SPIN ECHOES
11.4 Spectroscopy in the presence of J-coupling
11.4.1 Thermal equilibrium
Derivation of the initial form of a density matrix for two nuclei interacting through
bonds (J-coupling) is very similar to that described for two nuclei interacting through
space (dipolar coupling) in Section 10.1.1.
Before we analyze evolution of the density matrix in a 2D experiment, we must define its initial form. Again, we
start from the thermal equilibrium and use the Hamiltonian. The difference from the case of isolated nuclei is that we
need to define a 4 × 4 density matrix in order to describe a pair of mutually interacting nuclei. As explained above, the
off-diagonal elements of the equilibrium density matrix (proportional to Ix and Iy) are equal to zero. The four diagonal
elements describe average populations of four stationary states of a system composed of (isolated) nuclear pairs: αα, αβ,
βα, and ββ. These populations are:
Peq
αα =
e−Eαα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 − Eαα
kBT
4
, (11.11)
Peq
αβ =
e−Eαβ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eαβ
kBT
4
, (11.12)
Peq
βα =
e−Eβα/kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eβα
kBT
4
, (11.13)
Peq
ββ =
e−Eββ /kBT
e−Eαα/kBT + e−Eαβ /kBT
+ e−Eβα/kBT
+ e−Eββ /kBT
≈
1 −
Eββ
kBT
4
. (11.14)
In principle, the total Hamiltonian also includes the term ˆHJ , which describes the J coupling and which is not
averaged to zero.
ˆH = −γ1B0(1 + δ1,i)ˆI1,z − γ2B0(1 + δ2,i)ˆI2,z + 2πJ ˆI1,z
ˆI2,z = (11.15)
−γ1B0(1 + δ1,i)
2




1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 −1



 − γ2B0(1 + δ2,i)
2




1 0 0 0
0 −1 0 0
0 0 1 0
0 0 0 −1



 +
πJ
2 2




1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 1



 .
(11.16)
where the diagonal elements (eigenavalues) are the energies of the individual states. Therefore, the populations
(diagonal elements of the density matrix) should be given by
Peq
αα ≈
1 − Eαα
kBT
4
=
1
4
+ γ1(1 + δ1,i)
B0
8kBT
+ γ2(1 + δ2,i)
B0
8kBT
−
πJ
16kBT
, (11.17)
Peq
αβ ≈
1 −
Eαβ
kBT
4
=
1
4
+ γ1(1 + δ1,i)
B0
8kBT
− γ2(1 + δ2,i)
B0
8kBT
+
πJ
16kBT
, (11.18)
Peq
βα ≈
1 −
Eβα
kBT
4
=
1
4
− γ1(1 + δ1,i)
B0
8kBT
+ γ2(1 + δ2,i)
B0
8kBT
+
πJ
16kBT
, (11.19)
Peq
ββ ≈
1 −
Eββ
kBT
4
=
1
4
− γ1(1 + δ1,i)
B0
8kBT
− γ2(1 + δ2,i)
B0
8kBT
−
πJ
16kBT
. (11.20)
(11.21)
However, the values of J in typical organic compounds are at least six orders of magnitude lower than the frequencies
measured even at low-field magnets (J ≤ 200 Hz vs. 200 MHz). As a consequence, the contribution of J-coupling can be
11.4. SPECTROSCOPY IN THE PRESENCE OF J-COUPLING 149
safely neglected, and the initial density matrix is identical to that derived for a pair of nuclei interacting through space
(Eq. 10.13).
Neglecting the chemical shifts (δ1,i 1, δ2,i 1)
ˆρeq
=





1
4
+ γ1B0
8kBT
+ γ2B0
8kBT
0 0 0
0 1
4
+ γ1B0
8kBT
− γ2B0
8kBT
0 0
0 0 1
4
+ γ1B0
8kBT
− γ2B0
8kBT
0
0 0 0 1
4
+ γ1B0
8kBT
− γ2B0
8kBT





(11.22)
=
1
4




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1



 +
γ1B0
8kBT




+1 0 0 0
0 +1 0 0
0 0 −1 0
0 0 0 −1



 +
γ2B0
8kBT




+1 0 0 0
0 −1 0 0
0 0 +1 0
0 0 0 −1



 (11.23)
=
1
2
(It + κ1I1,z + κ2I2,z) , (11.24)
where
κj =
γjB0
4kBT
. (11.25)
11.4.2 Evolution in the presence of J-coupling
In the presence of the scalar coupling, the Hamiltonian describing evolution after a 90◦
pulse is complicated even in a coordinate system rotating with ωrot = −ωradio
H = −γ1B0(1 + δi1)
Ω1
I1z −γ1B0(1 + δi2)
Ω2
I2z + πJ (2I1zI2z + 2I1xI2x + 2I1yI2y) . (11.26)
However, if the precession frequencies differ, the Hamiltonian simplifies to a form
where all components commute. Therefore, the Liouville - von Neumann equation can be
solved geometrically as rotations in three-dimensional subspaces of the 16-dimensional
operator space. Rotations described by different components of the Hamiltonian are
independent and can be performed consecutively, in any order.
For a density matrix ˆρ(b) = 1
2
It + 1
2
κ(−I1y − I2y) after a 90◦
pulse, the evolution
due to the chemical shift (described by Ω1 and Ω2) and scalar coupling (described by
πJ) can be analyzed as follows
150 CHAPTER 11. J-COUPLING, SPIN ECHOES
−2 ˆI1zI2z
ω1
ω1 ω1
ω1
ˆI1z
−ˆI1x
−ˆI1z
−ˆI1x
ˆI1z
−ˆI1x
−ˆI1z
ˆI1z
−ˆI1z
ˆI1z
−ˆI1x
−ˆI1z
Ω1t Ω1t
πJ πJ
−2 ˆI1x
ˆI2z
2 ˆI1z
ˆI2z
−2 ˆI1x
ˆI2z
−2 ˆI1z
ˆI2z
2 ˆI1z
ˆI2z
−2 ˆI1z
ˆI2z
2 ˆI1z
ˆI2z
−2 ˆI1x
ˆI2z
−ˆI1y
ˆI1x
ˆI1y
2 ˆI1y
ˆI2z
ˆI1x
−ˆI1y
ˆI1y
ˆI1x
ˆI1y−ˆI1y
ˆI1x
2 ˆI1x
ˆI2z
−ˆI1y
ˆI1y
2 ˆI1x
ˆI2z
2 ˆI1z
ˆI2z
2 ˆI1y
ˆI2z
2 ˆI1x
ˆI2z
2 ˆI1y
ˆI2z−2 ˆI1y
ˆI2z
−2 ˆI1y
ˆI2z
−2 ˆI1y
ˆI2z
−ˆI1y
ˆI1y
ˆI1x
−ˆI1x
−2 ˆI1zI2z
Figure 11.1: Rotations in product operator space
11.4. SPECTROSCOPY IN THE PRESENCE OF J-COUPLING 151
I1t −→ I1t −→ I1t (11.27)
−I1y −→



−c1I1y −→
−c1cJ I1y
+c1sJ 2I1xI2z
+s1I1x −→
+s1cJ I1x
+s1sJ 2I1yI2z
(11.28)
−I2y −→



−c2I2y −→
−c2cJ I2y
+c2sJ 2I2xI1z
+s2I2x −→
+s2cJ I2x
+s2sJ 2I2yI1z
(11.29)
where the first arrows represent rotation ”about” I1z or I2z by the angle Ω1t or Ω2t,
the second arrows represent rotation ”about” 2I1zI2z by the angle πJt, and
c1 = cos(Ω1t) s1 = sin(Ω1t) (11.30)
c2 = cos(Ω2t) s2 = sin(Ω2t) (11.31)
cJ = cos(πJt) sJ = sin(πJt) (11.32)
Only I1x, I1y, I2x, I2y contribute to the expected value of M+, giving non-zero trace
when multiplied by ˆI+ (orthogonality).
Including relaxation and applying a phase shift by 90 ◦
, the expected value of M+
evolves as
κ
4
e−R2,1t
ei(Ω1−πJ)t
+ ei(Ω1+πJ)t
+ e−R2,2t
ei(Ω2−πJ)t
+ ei(Ω2+πJ)t
(11.33)
which gives two doublets in the spectrum after Fourier transform:
Nγ2 2
B0
16kBT
R2,1
R2
2,1 + (ω − Ω1 + πJ)2
+
R2,1
R2
2,1 + (ω − Ω1 − πJ)2
+
R2,2
R2
2,2 + (ω − Ω2 + πJ)2
+
R2,2
R2
2,2 + (ω − Ω2 − πJ)2
,
−i
Nγ2 2
B0
16kBT
(ω − Ω1 + πJ)
R2
2,1 + (ω − Ω1 + πJ)2
+
(ω − Ω1 − πJ)
R2
2,1 + (ω − Ω1 − πJ)2
+
(ω − Ω2 + πJ)
R2
2,2 + (ω − Ω2 + πJ)2
+
(ω − Ω2 − πJ)
R2
2,2 + (ω − Ω2 − πJ)2
.
(11.34)
152 CHAPTER 11. J-COUPLING, SPIN ECHOES
11.5 Spin echoes
Experiments utilizing scalar coupling are based on ”spin alchemy” - artificial manipulations
of quantum states of the studied system.
Spin echoes are basic tools of spin alchemy, providing the possibility to control evolution
of the chemical shift and scalar coupling separately.
Here we analyze three types of spin echoes for a heteronucler system (two nuclei
with different γ, 1
H and 13
C in our example). In order to distinguish the heteronuclear
systems from homonuclear ones, we will use symbols Ij and Sj for operators of nucleus
1 and 2, respectively, if γ1 = γ2. For the sake of simplicity, relaxation is not included.
The vector analysis is shown in Figure 11.2, where the solid arrows represent components
of µ1 ⊥ B0 for spin 2 in |α , dashed arrows represent components of µ1 ⊥ B0
for spin 2 in |β , and colors indicate different δi.
11.5.1 Free evolution
Evolution of the system of two nuclei in the presence of scalar coupling was already
described in Section 11.4.2. The vector analysis is shown in Figure 11.2A).
• ˆρ(a) = 1
2
It + 1
2
κ1Iz + 1
2
κ2Sz
thermal equilibrium, the constants κ1 and κ2 are different because the nuclei have
different γ.
• ˆρ(b) = 1
2
It − 1
2
κ1Iy + 1
2
κ2Sz
90◦
pulse applied to nucleus 1 only
• ˆρ(e) = 1
2
It + 1
2
κ1 (−c1cJ Iy + s1cJ Ix + c1sJ 2IxSz + s1sJ 2IySz) + 1
2
κ2Sz
free evolution during 2τ (t → 2τ in c1 etc.)
For nuclei with γ > 0, magnetizations of nucleus 1 (proton) evolve faster if nucleus
2 (13
C) is in |β (the energy difference between |αβ and |ββ is larger than the energy
difference between |αα and |βα ) - solid arrows rotated by a larger angle than dashed
arrows in Fig. 11.2A.
The 2IxSz, 2IySz coherences do not give non-zero trace when multiplied by I+
(they are not measurable per se), but cannot be ignored if the pulse sequence continues
because they can evolve into measurable coherences later (note that the scalar coupling
Hamiltonian 2πJIzSz converts them to Iy, Ix, respectively).
11.5.2 Refocusing echo
The refocusing echo consists of a 90◦
pulse exciting magnetic moment 1 and a 180◦
pulse
applied to the excited nucleus in the middle of the echo:
11.5. SPIN ECHOES 153
A
B
C
D
Figure 11.2: Vector analysis of spin echoes for 1
H (nucleus 1) and 13
C (nucleus 2) in an isolated
–CH– group. In individual rows, evolution of magnetization vectors in the plane ⊥ B0 is shown for
three protons (distinguished by colors) with slightly different precession frequency due to the different
chemical shifts δi. The protons are bonded to 13
C. Solid arrows are components of proton magnetization
for 13
C in |β , dashed arrow are components of proton magnetization for 13
C in |α . The first column
shows magnetization vectors at the beginning of the echo (after the initial 90◦
pulse at the proton
frequency), the second column shows magnetization vectors in the middle of the first delay τ, the third
and fourth columns show magnetization immediately before and after the 180◦
pulse(s) in the middle
of the echo, respectively, the fifths column shows magnetization vectors in the middle of the second
delay τ, the sixth column shows magnetization vectors at the end of the echo. Row A corresponds to an
experiment when no 180◦
pulse is applied, row B corresponds to the echo with the 180◦
pulse applied at
the proton frequency, row C corresponds to the echo with the 180◦
pulse applied at the 13
C frequency,
and row D corresponds to the echo with the 180◦
pulses applied at both frequencies. The x-axis points
down, the y-axis points to the right.
154 CHAPTER 11. J-COUPLING, SPIN ECHOES
a(π/2)1xb − τ −c (π)1xd − τ−e
The vector analysis is shown in Figure 11.2B). The middle 180◦
pulse flips all vectors
from left to right (rotation about the vertical axis x by 180 ◦
). The faster vectors start
to evolve with a handicap at the beginning of the second delay τ and they reach the
slower vectors at the end of the echo regardless of the actual speed of rotation.
Even without a detailed analysis of product operators, we see that the final state of
the system does not depend on chemical shift or scalar coupling: the evolution of both
chemical shift and scalar coupling is refocused during this echo.
The initial state of protons was described (after the 90◦
pulse) by −Iy in terms of
product operators and by an arrow with the −y orientation. As the vector only changed
its sign at the end of the experiment (arrow with the +y orientation), we can deduce
that the final state of protons is +Iy:
ˆρ(e) = 1
2
It + 1
2
κ1Iy + 1
2
κ2Sz
11.5.3 Decoupling echo
The refocusing echo consists of a 90◦
pulse exciting magnetic moment 1 and a 180◦
pulse
applied to the other nucleus in the middle of the echo:
a(π/2)1xb − τ −c (π)2xd − τ−e
The vector analysis is shown in Figure 11.2C). The middle 180◦
is applied at the 13
C
frequency. It does not affect vectors of proton magnetization but inverts polarization
(populations) of 13
C (solid arrows change to dashed ones and vice versa). The faster
vectors become slower, the slower vectors become faster, and they meet at the end of
the echo.
Without a detailed analysis of product operators, we see that the final state of the
system does not depend on scalar coupling (the difference between solid and dashed
arrows disappeared) but the evolution due to the chemical shift took place (arrows of
different colors rotated by different angles 2Ω1τ). As the effects of scalar coupling are
masked, this echo is known as the decoupling echo.
As the vectors at the end of the echo have the same orientations as if the nuclei
were not coupled at all, we can deduce that the final state of protons is identical to the
density matrix evolving due to the chemical shift only:
ˆρ(e) = 1
2
It + 1
2
κ1 (c1Iy − s1Ix) − 1
2
κ2Sz
11.5.4 Coupling echo
The coupling echo consists of a 90◦
pulse exciting magnetic moment 1 and 180◦
pulses
applied to both nuclei in the middle of the echo:
11.5. SPIN ECHOES 155
a(π/2)1xb − τ −c (π)1x(π)2xd − τ−e
The vector analysis is shown in Figure 11.2D). 180◦
pulses are applied at 1
H and 13
C
frequencies in the middle of the echo, resulting in combination of both effects described
in Figs. 11.2B and C. The proton pulse flips vectors of proton magnetization and the
13
C flips polarization (populations) of 13
C (solid arrows change to dashed ones and vice
versa). As a result, the average direction of dashed and solid arrows is refocused at the
end of the echo but the difference due to the coupling is preserved (the handicapped
vectors were made slower by the inversion of polarization of 13
C).
Without a detailed analysis of product operators, we see that the effect of the chemical
shift is removed (the hypothetical arrows showing average direction of vectors of the
same color just change the sign), but the final state of the system depends on scalar
coupling (the solid and dashed arrows disappeared).
We can deduce that the final state of the system is obtained by rotation ”about”
2IzSz, but not ”about” Iz in the product operator space, and by changing the sign of
the resulting coherences as indicated by the vector analysis:
ˆρ(e) = 1
2
It + 1
2
κ1 (cJ Iy − sJ 2IxSz) − 1
2
κ2Sz
156 CHAPTER 11. J-COUPLING, SPIN ECHOES
Chapter 12
INEPT, HSQC, APT
Literature: INEPT, HSQC, and APT experiments are nicely described in K7.10,
K8.7, and K12.4.4., respectively. INEPT is discussed in detail in L16.3., HSQC in
C7.1.1. Decoupling trains are reviewed in C3.5.
12.1 INEPT
INEPT is an NMR experiment based on the recoupling echo. It differs from the simple
echo in two issues:
• The length of the delay τ is set to 1/4|J|
• The echo is followed by two 90◦
pulses, one at the frequency of the excited nucleus
– this one must be phase-shifted by 90 ◦
from the excitation pulse, and one at the
frequency of the other nucleus (15
N in Fig. 12.1).
y
H
N
τ τ
Figure 12.1: INEPT pulse sequence applied to 1
H and 15
N.
157
158 CHAPTER 12. INEPT, HSQC, APT
With τ = 1/4J, 2πτ = π/2, cJ = 0,sJ = 1 if J > 0, and sJ = −1 if J < 0. Therefore,
the density matrix at the end of the echo is1
ˆρ(e) = 1
2
It − 1
2
κ1 (2IxSz) − 1
2
κ2Sz
−→ 1
2
It + 1
2
κ1 (2IzSz) − 1
2
κ2Sz after the first pulse and
−→ 1
2
It − 1
2
κ1 (2IzSy) + 1
2
κ2Sy after the second pulse.
If the experiment continues by acquisition, the density matrix evolves as
It −→ It −→ It (12.1)
−2IzSy −→



−c2 2IzSy −→
−c2cJ 2IzSy
+c2sJ Sx
+s2 2IxSz −→
+s2cJ 2IzSx
+s2sJ Sy
(12.2)
Sy −→



+c2Sy −→
+c2cJ Sy
−c2sJ 2SxIz
−s2Sx −→
−s2cJ Sx
−s2sJ 2SyIz
(12.3)
Both the ”blue” coherence 2IzSy and the ”green” coherence Sy evolve into measurable
product operators, giving non-zero trace when multiplied by S+.
After calculating the traces, including relaxation, and applying a phase shift by 90 ◦
,
the expected value of M2+ evolves as
κ2
4
e−R2t
e−i(Ω2−πJ)t
− e−i(Ω2+πJ)t
+
κ1
4
e−R2t
e−i(Ω2−πJ)t
+ e−i(Ω2+πJ)t
(12.4)
The real part of the spectrum obtained by Fourier transform is
Nγ2
2 2
B0
16kBT
R2
R2
2 + (ω − Ω2 + πJ)2
−
R2
R2
2 + (ω − Ω2 − πJ)2
+
Nγ1
2 2
B0
16kBT
+
R2
R2
2 + (ω − Ω2 + πJ)2
+
R2
R2
2 + (ω − Ω2 − πJ)2
(12.5)
• The ”blue” coherence 2IzSy gives a signal with opposite phase of the peaks at
Ω2 − πJ and Ω2 + πJ. Accordingly, it is called the anti-phase coherence.
• The ”green” coherence Sy gives a signal with the same phase of the peaks at
Ω2 − πJ and Ω2 + πJ. Accordingly, it is called the in-phase coherence.
1
The analysis is done for J > 0. If J < 0 (e.g. for one-bond 1
H-15
N coupling), all blue terms have
the opposite sign.
12.2. HSQC 159
• More importantly, the ”blue” coherence 2IzSy gives a signal proportional to γ2
1
while the ”green” coherence Sy gives a signal proportional to γ2
2. The amplitude of
the ”green” signal corresponds to the amplitude of a regular 1D 15
N spectrum. The
”blue” signal ”inherited” the amplitude with γ2
1 from the excited nucleus, proton.
In case of 1
H and 15
N, γ1 is approximately ten times higher than γ2. Therefore,
the blue signal is two orders of magnitude stronger. This is why this experiment
is called Insensitive Nuclei Enhanced by Polarization Transfer (INEPT).
• As described, the ”blue” and ”green” signals are combined, which results in different
heights of the Ω2 − πJ and Ω2 + πJ peaks. The ”blue” and ”green” signals
can be separated if we repeat the measurement twice with the phase of the proton
y pulse shifted by 180 ◦
(i.e., with −y). It does not affect the ”green” signal, but
changes the sign of the ”blue” signal. If we subtract the spectra, we obtained
a pure ”blue” signal. This trick - repeating acquisition with different phases is
known as phase cycling and is used routinely in NMR spectroscopy to remove
unwanted signals.
12.2 HSQC
Heteronuclear Single-Quantum Correlation (HSQC) is a 2D pulse sequence using scalar
coupling to correlate frequencies of two magnetic moments with different γ (Fig. 12.2).
• After a 90◦
pulse at the proton frequency, polarization is transfered to the other
nucleus (usually 15
N or 13
C). The density matrix at the end of the INEPT is
ˆρ(e) = 1
2
It − 1
2
κ1 (2IzSy) + 1
2
κ2Sy
• During an echo with a decoupling 180◦
pulse at the proton frequency (red pulse in
Fig. 12.2), anti-phase single quantum coherences evolve according to the chemical
shift
ˆρ(e) −→ 1
2
It + 1
2
κ1 (c212IzSy − s212IzSx) + 1
2
κ2 (c11Sy − s11Sx).
• Two 90◦
pulses convert 2IzSy to 2IySz and 2IzSx to 2IySx. The magenta operator
is a multiple quantum coherence (a combination of zero-quantum and doublequantum
coherence), which can be converted to a single quantum coherence only
by a 90◦
pulse. Since the pulse sequence does not contain any more 90◦
pulses
and since no multiple-quantum coherence is measurable, we ignore 2IySx. The 90◦
pulse applied to S converts Sy to Sz. Since the pulse sequence does not contain any
more 90◦
pulses, this does not contribute to the signal and we can ignore it. Also,
we ignore the red term which never evolves to a measurable coherence because it
commutes with all Hamiltonians. The Sx is not affected by the 90◦
pulses and we
should follow its fate.
160 CHAPTER 12. INEPT, HSQC, APT
y
2
τ τ
t1
2
t1
2
τ τ
H
N
t
Figure 12.2: 1
H,15
N HSQC pulse sequence.
τ
1
t2
t
H
N
τ τ
y
τ
Figure 12.3: Idea of the decoupling in the direct dimension.
• The last echo allows the scalar coupling to evolve but refocuses evolution of the
scalar coupling. If the delays τ = 1/4J, the measurable components of the density
matrix evolve to 1
2
κ1 cos(Ω2t1)Ix (rotation ”about” 2IzSz by 90 ◦
and change of
the sign by the last 180◦
pulse at the proton frequency). The green coherence
Sx evolves to −2IzSy. As no 90◦
pulse is applied to S after the last echo, this
coherence does not contribute to the signal and can be ignored.
• During acquisition, both chemical shift and scalar coupling evolve in the experiment
described in Fig. 12.2. Therefore, we obtain a doublet in the proton dimension
of the spectrum. The second dimension is introduced by repeating the
measurement with t1 being incremented. Each increment is measured twice with
a different phase of one of the 90◦
pulses applied to nucleus 2, which provides real
(modulated by cos(Ω2t1)) and imaginary (modulated by sin(Ω2t1)) component of
a complex signal, like in the NOESY experiment. After calculating the trace, including
relaxation (with different rates R2 in the direct and indirect dimensions),
phase shift by 90 ◦
and Fourier transforms in both t1 and t2 dimensions, we obtain
a 1D spectrum with peaks at Ω2 chemical shift in the indirect dimension and a
doublet at Ω1 ± πJ in the direct (proton) dimension. Note that the splitting by
±πJ was removed by the red decoupling pulse in the indirect dimension.
12.3. APT 161
12.2.1 Decoupling trains
Splitting of peaks in the direct dimension in spectra recorded by the pulse sequence in
Fig. 12.2 is undesirable. On the other hand, we acquire signal in real time and cannot
remove the splitting by a decoupling echo. In principle, we can divide the acquisition
time into short fragments and apply a 180◦
pulse at the frequency of nucleus 2 (13
C or
15
N) in the middle of each such echo. In practice, imperfections of such a long series of
echoes, affecting especially magnetic moments with large Ω2, are significant. However,
more sophisticated series of pulses have much better performance. Typical examples of
decoupling pulse sequences are
• WALTZ - a series of 90◦
, 180◦
, and 270◦
pulses with phase of 0 ◦
(x), or 180 ◦
(−x),
repeating in complex patterns
• DIPSI - a similar series of pulses with non-integer rotation angles
• GARP - computer-optimized sequence of pulses with non-integer rotation angles
and phases.
12.2.2 Benefits of HSQC
• 13
C or 15
N frequency measured with high sensitivity (higher by (γ1/γ2)5/2
than
provided by the direct detection, cf. Section 8.8.3)
• expansion to the second dimension and reducing the number of peaks in spectrum
(only 13
C or 15
N-bonded protons and only protonated 13
C or 15
N nuclei are visible)
provides high resolution
• 1
H-13
C and 1
H-15
N correlation is important structural information (which proton
is attached to which 13
C or 15
N)
12.3 APT
The attached proton test (APT) is useful for analysis of systems with multiple protons,
most often CHn (C, CH, CH2, CH3). The experiment consists of 13
C excitation, recoupling
echo (discussed in Section 11.5.4), and 13
C acquisition with proton decoupling. In
the following analysis, the 13
C operators are labeled Sx, Sy, Sz, and relaxation is ignored
for the sake of simplicity.
• ˆρ(a) = 1
2n It + κ1
2n
n
j=1
(Ijz) + κ2
2n Sz
The probability density matrix at equilibrium is described in a similar manner
as for one or two magnetic moments, the extension to the multinuclear system is
162 CHAPTER 12. INEPT, HSQC, APT
reflected by the scaling constant 1/2n
, where n is the number of protons attached
to 13
C.
• ˆρ(b) = 1
2n It + κ1
2n
n
j=1
(Ijz) − κ2
2n Sy Excitation of 13
C is an analogy of cases discussed
above.
• Understanding of the next step is critical for the analysis. The general conclusions
of Section 11.5.4 apply, but the actual form of the density matrix must be derived
for each system. The general conclusions are: evolution of Ω2 (13
C frequency
offset) due tho the 13
C chemical shift is refocused, scalar coupling evolves for 2τ
as cos(2πJτ) and sin(2πJτ), nucleus 1 (proton) is never excited (no proton 90◦
pulse), therefore only Ijz contributions are present for protons.
• The actual analysis for 13
CH2 and 13
CH3 groups requires extension of the density
matrix to 2n+1
× 2n+1
dimensions. Construction of the basis matrices for such
4n+1
-dimensional operator space involves additional direct products with the matrices
It, Ix, Iy, Iz. Evolution of the 2n+1
× 2n+1
matrices is governed by their
commutation rules, three-dimensional subspaces where ”rotations” of operators
take place are define by these commutation rules.
• When the rules are applied, the analysis gives
ˆρ(e) = 1
2n It+κ1
2n
n
j=1
(Ijz)+κ2
2n



n = 0 : Sy
n = 1 : cSy − s2I1zSx
n = 2 : c2
Sy − sc(2I1zSx + 2I2zSx) − s2
4I1zI2zSy
n = 3 : c3
Sy − sc2
(2I1zSx + 2I2zSx + 2I3zSx)
−s2
c(4I1zI2zSy + 4I1zI3zSy + 4I2zI3zSy)
+s3
8I1zI2zI3zSx
where s = sin(2πJτ) and c = cos(2πJτ).
• Since decoupling is applied during acquisition, only the Sy coherences give a measurable
signal. Note that the fact that the proton decoupling is used tells us
in advance that the terms containing Ijz need not be analyzed. Therefore the
knowledge of exact commutation rules is not necessary, the only important conclusion
is that the observable contributions to the density matrix are modulated
by cosn
(2πJτ) for CHn. During acquisition, these terms evolve under the influence
of chemical shift, exactly like in a one-pulse experiment. If τ is set to τ = 2J, then
c = cos π = −1. Therefore, signals of C and CH2 are positive and signals of CH
and CH3 are negative ⇒ useful chemical information.
Chapter 13
COSY
Literature: COSY is described in detail in L16.1, C6.2.1., and K8.3 (with a detailed
discussion of DQF-COSY in K8.4).
13.1 Homonucler correlation: COSY
We started the discussion of experiments based on scalar couplings with heteronuclear
correlations because they are easier to analyze. The basic (and very popular) homonuclear
experiment is COSY (COrrelated SpectroscopY). Its pulse sequence is very simple,
consisting of only two 90◦
pulses separated by an incremented delay t1 (which provides
the second dimension), but the evolution of the density matrix is relatively complex.
Here, we analyze evolution for a pair of interacting nuclei (protons).
• ˆρ(a) = 1
2
It + 1
2
κ(I1z + I2z)
thermal equilibrium, the matrices are different than for the noninteracting spin,
but the constant is the same.
• ˆρ(b) = 1
2
It + 1
2
κ(−I1y − I2y)
90◦
pulse, see the one-pulse experiment
• ˆρ(c) = 1
2
It
+1
2
κ(−c11cJ1I1y + s11cJ1I1x + c11sJ12I1xI2z + s11sJ12I1yI2z)
+1
2
κ(−c21cJ1I2y + s21cJ1I2x + c21sJ12I1zI2x + s21sJ12I1zI2y),
where ci1 = cos(Ωit1), si1 = sin(Ωit1), cJ1 = cos(πJt1), and sJ1 = sin(πJt1) –
evolution of the chemical shift and coupling.
• The second 90◦
pulse creates the following coherences
ˆρ(d) = 1
2
It
+1
2
κ(−c11cJ1I1z+ s11cJ1I1x −c11sJ12I1xI2y− s11sJ12I1zI2y )
163
164 CHAPTER 13. COSY
+1
2
κ(−c21cJ1I2z+ s21cJ1I2x −c21sJ12I1yI2x− s21sJ12I1yI2z ).
The red terms contain polarization operators, not coherences, they do not contribute
to the signal. The green terms contain in-phase single-quantum coherences,
only they give non-zero trace when multiplied with ˆM+ ∝ (I1x + iI1y + I2x + iI2y).
The blue terms contain anti-phase single-quantum coherences, they do not contribute
to the signal directly, but they evolve into in-phase coherences during acquisition
due to the scalar coupling. The magenta terms contain multiple-quantum
coherences. They do not contribute to the signal, but can be converted to singlequantum
coherences by 90◦
pulses. Such pulses are not applied in the discussed
pulse sequence, but are used in some versions of the experiment.
• The terms in black frames evolve with the chemical shift of the first nucleus during
acquisition:
s11cJ1I1x → s11cJ1c12cJ2I1x + s11cJ1s12cJ2I1y+ unmeasurable anti-phase coherences
−s21sJ12I1yI2z → s21sJ1c12sJ2I1x + s21sJ1s12sJ2I1y+ unmeasurable anti-phase coherences ,
where ci2 = cos(Ωit2), si2 = sin(Ωit2), cJ2 = cos(πJt2), and sJ2 = sin(πJt2). Using
the following trigonometric relations
cikciJ =
c−
ik + c+
ik
2
siksiJ =
c−
ik − c+
ik
2
ciksiJ =
−s−
ik + s+
ik
2
sikciJ =
s−
ik + s+
ik
2
,
(13.1)
where c±
ik = cos((Ωi ± πJ)tk) and s±
ik = sin((Ωi ± πJ)tk), the terms contributing to
the signal can be written as

(s−
11 + s+
11)(c−
12 + c+
12)
[Ω1,Ω1]
+ (c−
21 − c+
21)(−s−
12 + s+
12)
[Ω2,Ω1]


 I1x
+


(s−
11 + s+
11)(s−
12 + s+
12)
[Ω1,Ω1]
+ (c−
21 − c+
21)(c−
12 − c+
12)
[Ω2,Ω1]


 I1y
The first and second line show coherences providing the real and imaginary component
of the complex signal acquired in the direct dimension (t2).
• Evaluation of the traces of ˆM + ˆρ(t2) gives the follwong modulation of the signal:
(s−
11 + s+
11) ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
[Ω1,Ω1]
+ i(c−
21 − c+
21) ei(Ω1−πJ)t2
− ei(Ω1+πJ)t2
[Ω2,Ω1]
The imaginary signal in the indirect dimension is obtained by repeating acquisition
for each increment of t1 with a different phase of the second 90 ◦
pulse (shifted by 90 ◦
,
which correspons to the direction y in the rotating coordinate system).
13.1. HOMONUCLER CORRELATION: COSY 165
• The second 90◦
pulse with the y phase creates the following coherences
ˆρ(d) = 1
2
It
+1
2
κ(− c11cJ1I1y −s11cJ1I1z− c11sJ12I1zI2x +s11sJ12I1yI2x)
+1
2
κ(− c21cJ1I2y −s21cJ1I2z− c21sJ12I1xI2z +s21sJ12I1xI2y).
• The terms in black frames evolve with the chemical shift of the first nucleus during
acquisition:
c11cJ1I1y → c11cJ1s12cJ2I1x − c11cJ1c12cJ2I1y+ unmeasurable anti-phase coherences
−s21sJ12I1xI2z → c21sJ1s12sJ2I1x − c21sJ1c12sJ2I1y+ unmeasurable anti-phase coherences .
The terms contributing to the signal can be written as

(c−
11 + c+
11)(s−
12 + s+
12)
[Ω1,Ω1]
+ (−s−
21 − s+
21)(c−
12 − c+
12)
[Ω2,Ω1]


 I1x
−


(c−
11 + c+
11)(c−
12 + c+
12)
[Ω1,Ω1]
+ (−s−
21 + s+
21)(−s−
12 + s+
12)
[Ω2,Ω1]


 I1y.
• Evaluation of the traces of ˆM + ˆρ(t2) gives the follwong modulation of the signal:
−i(c−
11 + c+
11) ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
[Ω1,Ω1]
+ (−s−
21 + s+
21) ei(Ω1−πJ)t2
− ei(Ω1+πJ)t2
[Ω2,Ω1]
Now we combine signals obtained with the different phases of the second pulse.
• The hypercomplex signal is modulated as
e−i π
2 ei(Ω1−πJ)t1
+ ei(Ω1+πJ)t1
ei(Ω1−πJ)t2
+ ei(Ω1+πJ)t2
[Ω1,Ω1]
+ ei(Ω2−πJ)t1
− ei(Ω2+πJ)t1
ei(Ω1−πJ)t2
− ei(Ω1+πJ)t2
[Ω2,Ω1]
,
where we replaced −i by e−iπ/2
.
• The green component of the signal evolves with the same chemical shift in both
dimensions, providing diagonal signal (at frequencies [Ω1, Ω1] in the 2D spectrum).
The blue (originally anti-phase) component of the signal also evolves with Ω1 in
the direct dimension, but with Ω2. It provides off-diagonal signal, a cross-peak at
frequencies [Ω1, Ω1] in the 2D spectrum. Note that the blue and green components
have the phase different by 90 ◦
. Therefore, either diagonal peaks or cross-peaks
have the undesirable dispersion shape (it is not possible to phase both diagonal
peaks or cross-peaks, they always have phases differing by 90 ◦
). Typically, the
spectrum is phased so that the cross-peaks have a nice absorptive shape because
166 CHAPTER 13. COSY
they carry a useful chemical information - they show which protons are connected
by 2 or 3 covalent bonds.
• The diagonal peaks are not interesting, but their dispersive shape may obscure
cross-peaks close to the diagonal. The problem with the phase can be solved
if one more 90 ◦
pulse is introduced. Such a pulse converts the magenta multiquantum
coherences to anti-phase single-quantum coherences, which evolve into
the measurable signal. The point is that other coherences can be removed by phase
cycling. The obtained spectrum contains diagonal peaks and cross-peaks, but (in
contrast to the simple two-pulse variant of the COSY experiment) both diagonal
peaks and cross-peaks have the same phase.1
This version of the experiment is
known as double-quantum filtered COSY (DQF-COSY). Its disadvantage is a lower
sensitivity – we lose a half of the signal.
• Also, note that each peak is split into doublets in both dimensions. More complex
multiplets are obtained if more than two nuclei are coupled. The distance of peaks
in the multiplets is given by the interaction constant J. In the case of nuclei
connected by three bonds, J depends on the torsion angle defined by these three
bonds. So, COSY spectra can be used to determine torsion angles in the molecule.
• The terms in gray frames evolve with the chemical shift of the second nucleus
during acquisition as
s21cJ1I1x → s21cJ1c12cJ2I1x + s21cJ1s12cJ2I1y+ unmeasurable anti-phase coherences
−s11sJ12I1yI2z → s11sJ1c12sJ2I1x + s11sJ1s12sJ2I1y+ unmeasurable anti-phase coherences
and give a similar type of signal for the other nucleus:
e−i π
2 ei(Ω2−πJ)t1
+ ei(Ω2+πJ)t1
ei(Ω2−πJ)t2
+ ei(Ω2+πJ)t2
[Ω2,Ω2]
+ ei(Ω1−πJ)t1
− ei(Ω1+πJ)t1
ei(Ω2−πJ)t2
− ei(Ω2+πJ)t2
[Ω1,Ω2]
.
This signal represents the other dianonal and off-diagonal peak in the spectrum.
1
We cannot use phase cycling to remove the green terms resulting in the unwanted diagonal peaks
because phase cycling can distinguish multi-quantum coherences from single-quantum ones, but it
cannot distinguish anti-phase single quantum coherences from in-phase single quantum coherences.
Chapter 14
Strong coupling
We have seen that secular approximation substantially simplifies Hamiltonian of the
J-coupling if γ and/or chemical shifts differ. However, the description of the system of
interacting nuclei changes dramatically if γ and chemical shifts are identical.
For a pair of nuclei, the Hamiltonian is given by Eq. 11.9. In the presence of (very similar) chemical shifts
H = +ω0,1I1z + ω0,2I2z + πJ (2I1zI2z + 2I1xI2x + 2I1yI2y) . (14.1)
Let us write the matrix representation of this Hamiltonian. If we use matrices listed in Table 9.1 to represent the
operators I1z, I2z, 2I1xI2x, 2I1yI2y, and 2I1zI2z, i.e., the matrices written in the basis constructed from the α and β
states of the interacting nuclei (i.e., |αα , |βα , |αβ , |ββ ), the Hamiltonian matrix is
H =
π
2




Σ + J 0 0 0
0 ∆ − J 2J 0
0 2J −∆ − J 0
0 0 0 −Σ + J



 , (14.2)
where Σ = (ω0,1 + ω0,2)/π and ∆ = (ω0,1 − ω0,2)/π. Obviously, the matrix is not diagonal. In order to draw the
energy level diagram, we need eigenvalues of the Hamiltonian. They can be obtained by changing the basis so that the
basis functions are eigenfunctions of the Hamiltonian. We are looking for a new, diagonal matrix representation of our
Hamiltonian H . From the mathematical point of view, diagonalization of our Hamiltonian can be described using a
transformation matrix ˆT:
H = ˆT−1
H ˆT. (14.3)
Multiplying by ˆT from left gives
ˆTH = H ˆT. (14.4)
The desired eigenvalues ωk and eigenvectors |ψk can be obtained by comparing the eigenvalue equation
H |ψk = ωk|ψk (14.5)
with the left-hand side of Eq. 14.4




T11 T12 T13 T14
T21 T22 T23 T24
T31 T32 T33 T34
T41 T42 T43 T44








ω1 0 0 0
0 ω2 0 0
0 0 ω3 0
0 0 0 ω4



 =




ω1T11 ω2T12 ω3T13 ω4T14
ω1T21 ω2T22 ω3T23 ω4T24
ω1T31 ω2T32 ω3T33 ω4T34
ω1T41 ω2T42 ω3T43 ω4T44



 . (14.6)
The eigenvalue equation can be written as a set of four equations for k = 1, 2, 3, 4
H |ψk =
π
2




Σ + J 0 0 0
0 ∆ − J 2J 0
0 2J −∆ − J 0
0 0 0 −Σ + J








T1k
T2k
T3k
T4k



 =
π
2




(Σ + J)T1k
(∆ − J)T2k + 2JT3k
2JT2k − (∆ + J)T3k
(−Σ + J)T4k



 = ωk




T1k
T2k
T3k
T4k



 = ωk|ψk . (14.7)
167
168 CHAPTER 14. STRONG COUPLING
The first row of the middle equality allows us to identify
ω1 =
π
2
(Σ + J) =
ω0,1 + ω0,2
2
+
π
2
J (14.8)
if we set T21 = T31 = T41 = 0, i.e.,
|ψ1 =




T11
0
0
0



 . (14.9)
Similarly,
ω4 =
π
2
(−Σ + J) = −
ω0,1 + ω0,2
2
+
π
2
J (14.10)
for
|ψ4 =




0
0
0
T44



 . (14.11)
The ω2 and ω3 values can be calculated from the equations
2ωkT2k = π(∆ − J)T2k + 2πJT3k (14.12)
2ωkT3k = 2πJT2k − π(∆ + J)T3k, (14.13)
(setting T12 = T42 = T13 = T43 = 0).
T3k can be expressed from the first equation
T3k =
2ωk + π(J − ∆)
2πJ
T2k (14.14)
and inserted into the second equation
(2ωk + π(J + ∆))(2ωk + π(J − ∆))T2k = (2πJ)2
T2k, (14.15)
directly giving
ωk = −
π
2
J ± 4J2 + ∆2 . (14.16)
Choosing
ω2 = −
π
2
J − 4J2 + ∆2 =
(ω0,1 − ω0,2)2 + 4π2J2
2
−
π
2
J (14.17)
and
ω3 = −
π
2
J + 4J2 + ∆2 = −
(ω0,1 − ω0,2)2 + 4π2J2
2
−
π
2
J. (14.18)
completely defines the diagonalized Hamiltonian
H =
π
2




Σ + J 0 0 0
0
√
∆2 + 4J2 − J 0 0
0 0 −
√
∆ + 4J2 − J 0
0 0 0 −Σ + J



 . (14.19)
The new basis is given by Eqs. 14.12, 14.13, and the normalization condition
ψk|ψk = 1 ⇒
4
j=1
T2
jk = 1. (14.20)
14.1. EVOLUTION WITH THE STRONG COUPLING 169
The normalization conditions immediately defines T11 = T44 = 1. Substituting ω2 into Eqs. 14.12 and 14.13,
respectively, gives
T32
T22
=
√
4J2 + ∆2 − ∆
2J
(14.21)
T22
T32
=
√
4J2 + ∆2 + ∆
2J
. (14.22)
Consequently,
T2
32
T2
22
=
√
4J2 + ∆2 − ∆
√
4J2 + ∆2 + ∆
(14.23)
and applying the normalization condition T2
32 = 1 − T2
22
1 + T2
22
T2
22
=
√
4J2 + ∆2 − ∆
√
4J2 + ∆2 + ∆
(14.24)
defines
T2
22 =
1
1 −
√
4J2+∆2−∆
√
4J2+∆2+∆
=
√
4J2 + ∆2 + ∆
2
√
4J2 + ∆2
(14.25)
and
T2
32 = 1 − T2
22 =
√
4J2 + ∆2 − ∆
2
√
4J2 + ∆2
. (14.26)
Similarly, T2
23 and T2
33 can be calculated by substituting ω3 into Eqs. 14.12 and 14.13:
T2
23 =
√
4J2 + ∆2 − ∆
2
√
4J2 + ∆2
(14.27)
T2
33 =
√
4J2 + ∆2 + ∆
2
√
4J2 + ∆2
. (14.28)
Finally, the new basis consists of the following eigenvectors
|ψ1 =




1
0
0
0



 , |ψ2 =








0
1
2
+ ∆
2
√
4J2+∆2
1
2
− ∆
2
√
4J2+∆2
0








≡




0
cξ
sξ
0



 , |ψ3 =








0
− 1
2
− ∆
2
√
4J2+∆2
1
2
+ ∆
2
√
4J2+∆2
0








≡




0
−sξ
cξ
0



 , |ψ4 =




0
0
0
1



 .
(14.29)
14.1 Evolution with the strong coupling
As usually, the density matrix at the beginning of the experiment is given by the thermal
equilibrium. As mentioned in Section 11.4.1, the effect of the J-coupling on populations
is negligible. Therefore, the initial form of the density matrix and its form after the 90◦
excitation pulse are the same as in the case of a weak coupling:
170 CHAPTER 14. STRONG COUPLING
ˆρ(b) =
1
2
It −
κ
2
I1y −
κ
2
I2y. (14.30)
In order to describe evolution, we need to know the Hamiltonian. In the previous
section, we managed to find transformation to a basis where the Hamiltonian including
the strong coupling is diagonal (Eq. 14.19). The matrix describing the Hamiltonian can
be written as a linear combination of matrices listed in Table 9.1 (the Cartesian basis of
product operators):
H = ω+I1z + ω−I2z + πJ · 2I1zI2z, (14.31)
where
ω+ =
1
2
ω0,1 + ω0,2 + (ω0,1 − ω0,2)2 + 4π2J2 (14.32)
ω− =
1
2
ω0,1 + ω0,2 − (ω0,1 − ω0,2)2 + 4π2J2 . (14.33)
In order to proceed, we need to transform the density matrix to the same basis (i.e., to the basis where the Hamiltonian
is diagonal):
ˆρ (b) = ˆT−1
ˆρ(b) ˆT = −
κ
2
I1y + I2y =
κ
2
−cξ(I1y + I2y) + sξ(2I1yI2z + 2I1zI2y) . (14.34)
As all operators constituting H commute, we can use the simple geometric approach to describe the evolution of
ˆρ (b), only the number of rotations in the appropriate three-dimensional subspaces of the operator space is higher than
in the case of the weak coupling. The result of the rotations is
ˆρ (t) =
κ
2
( − cξc+cJ I1y + cξs+cJ I1x + cξc+sJ 2I1xI2z + cξs+sJ 2I1yI2z
− cξc−cJ I2y + cξs−cJ I2x + cξc−sJ 2I1zI2x + cξs−sJ 2I1zI2y
− sξs+sJ I1y − sξc+sJ I1x − sξs+cJ 2I1xI2z + sξc+cJ 2I1yI2z
+ sξs−sJ I2y + sξc−sJ I2x + sξc−cJ 2I1zI2x − sξc−cJ 2I1zI2y ) , (14.35)
where
c+ = cos(ω+t) s+ = sin(ω+t) (14.36)
c− = cos(ω−t) s− = sin(ω−t) (14.37)
cJ = cos(πJt) sJ = sin(πJt). (14.38)
Finally, we also need to transform the operator of the measured quantity ˆM+ = γN(I1x + I2x + i(I1y + I2y)):
ˆM+ = ˆT−1 ˆM+
ˆT = γN(cξ(I1x + I2x) + sξ(2I1zI2x − 2I1xI2z) + i(cξ(I1y + I2y) + sξ(2I1zI2y − 2I1yI2z)) (14.39)
The expected magnetization (neglecting relaxation) is then given by
M+ = Tr{ ˆM+ · ˆρ (t)}. (14.40)
14.1. EVOLUTION WITH THE STRONG COUPLING 171
Both M+ and ρ (t) consist of eight operator matrices suggesting that the product ˆM+ · ˆρ (t) should include 64 terms.
However, orthogonality1 simplifies the product to a sum of eight combinations of c’s and s’s:
s+cJ − 2sξcξc+sJ + s−cJ + 2sξcξc−sJ − i(c+cJ − 2sξcξs+sJ + c−cJ + 2sξcξs−sJ ) (14.41)
Including relaxation (assumed to be identical for all coherences), using the standard trigonometric relations and the
definition of the exponential form of a complex number, and applying a phase shift by 90 ◦, the expected value of M+
evolves as
γN
κ
8
e−R2t
(1 + 2sξcξ)ei(ω+−πJ)t
+ (1 − 2sξcξ)ei(ω++πJ)t
+ (1 − 2sξcξ)ei(ω−−πJ)t
+ (1 + 2sξcξ)ei(ω−+πJ)t
,
(14.42)
which gives four peaks of different intensities in the spectrum after Fourier transform:
1 −
J
√
∆2 + 4J2
Nγ2 2
B0
16kBT
R2
R2
2 + (ω − ω+ − πJ)2
+ 1 +
J
√
∆2 + 4J2
Nγ2 2
B0
16kBT
R2
R2
2 + (ω − ω− − πJ)2
+ 1 +
J
√
∆2 + 4J2
Nγ2 2
B0
16kBT
R2
R2
2 + (ω − ω+ + πJ)2
+ 1 −
J
√
∆2 + 4J2
Nγ2 2
B0
16kBT
R2
R2
2 + (ω − ω− + πJ)2
−i 1 −
J
√
∆2 + 4J2
Nγ2 2
B0
16kBT
(ω − ω+ − πJ)
R2
2 + (ω − ω+ − πJ)2
−i 1 +
J
√
∆2 + 4J2
Nγ2 2
B0
16kBT
(ω − ω− − πJ)
R2
2 + (ω − ω− − πJ)2
−i 1 +
J
√
∆2 + 4J2
Nγ2 2
B0
16kBT
(ω − ω+ + πJ)
R2
2 + (ω − ω+ + πJ)2
−i 1 −
J
√
∆2 + 4J2
Nγ2 2
B0
16kBT
(ω − ω− + πJ)
R2
2 + (ω − ω− + πJ)2
(14.43)
1
As we use normalized product operators, trace of the product of the same operator matrices is equal
to one, and product of two different operator matrices is equal to zero.
172 CHAPTER 14. STRONG COUPLING
Table 14.1: Cartesian basis for a three-spin system: polarization operators. Symbols
”+” and ”−” stand for +1 and −1, respectively.
It =
1√
8











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +











I1z =
1√
8











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −











I2z =
1√
8











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −











I3z =
1√
8











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −











2I1zI2z =
1√
8











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +











2I1zI3z =
1√
8











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +











2I2zI3z =
1√
8











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +











4I1zI2zI3z =
1√
8











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −











Table 14.2: Cartesian basis for a three-spin system: single/triple-quantum operators.
Symbols ”+” and ”−” stand for +1 and −1, respectively.
4I1xI2xI3x =
1√
8











0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0











4I1xI2yI3y =
1√
8











0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 − 0 0 0 0 0 0
− 0 0 0 0 0 0 0











4I1yI2xI3y =
1√
8











0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0











4I1yI2yI3x =
1√
8











0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0











4I1xI2xI3y =
1√
8











0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0











4I1xI2yI3x =
1√
8











0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0











4I1yI2xI3x =
1√
8











0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0











4I1yI2yI3y =
1√
8











0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0











14.1. EVOLUTION WITH THE STRONG COUPLING 173
Table 14.3: Cartesian basis for a three-spin system: single-quantum operators. Symbols
”+” and ”−” stand for +1 and −1, respectively.
I1x =
1√
8











0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0











2I1xI2z =
1√
8











0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0











2I1xI3z =
1√
8











0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0











4I1xI2zI3z =
1√
8











0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0











I1y =
i√
8











0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0











2I1yI2z =
i√
8











0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0











2I1yI3z =
i√
8











0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0











4I1yI2zI3z =
i√
8











0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0











I2x =
1√
8











0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0











2I1zI2x =
1√
8











0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0











2I2xI3z =
1√
8











0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0











4I1zI2xI3z =
1√
8











0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0











I2y =
i√
8











0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0











2I1zI2y =
i√
8











0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0











2I2yI3z =
i√
8











0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0











4I1zI2yI3z =
i√
8











0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0











I3x =
1√
8











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0











2I1zI3x =
1√
8











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0











2I2zI3x =
1√
8











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0











4I1zI2zI3x =
1√
8











0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0











I3y =
i√
8











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0











2I1zI3y =
i√
8











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0











2I2zI3y =
i√
8











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0











4I1zI2zI3y =
i√
8











0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0











174 CHAPTER 14. STRONG COUPLING
Table 14.4: Cartesian basis for a three-spin system: zero/double-quantum operators.
Symbols ”+” and ”−” stand for +1 and −1, respectively.
2I1xI2x =
1√
8











0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0











2I1yI2y =
1√
8











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
− 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0











4I1xI2xI3z =
1√
8











0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0











4I1yI2yI3z =
1√
8











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
− 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0











2I1xI3x =
1√
8











0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0











2I1yI3y =
1√
8











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0











4I1xI2zI3x =
1√
8











0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0











4I1yI2zI3y =
1√
8











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0











2I2xI3x =
1√
8











0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0











2I2yI3y =
1√
8











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0











4I1zI2xI3x =
1√
8











0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0











4I1zI2yI3y =
1√
8











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0











2I1xI2y =
i√
8











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0











2I1yI2x =
i√
8











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0











4I1xI2yI3z =
i√
8











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 − 0 0
0 0 − 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0











4I1yI2xI3z =
i√
8











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0











2I1xI3y =
i√
8











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0











2I1yI3x =
i√
8











0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0











4I1xI2zI3y =
i√
8











0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 − 0
0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 + 0 0 0 0
0 0 − 0 0 0 0 0











4I1yI2zI3x =
i√
8











0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0











2I2xI3y =
i√
8











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 − 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 + 0 0 0











2I2yI3x =
i√
8











0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 − 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0











4I1zI2xI3y =
i√
8











0 0 0 − 0 0 0 0
0 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 0
0 0 0 0 − 0 0 0











4I1zI2yI3x =
i√
8











0 0 0 − 0 0 0 0
0 0 − 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 0 0 0 0 − 0 0
0 0 0 0 − 0 0 0











Chapter 15
TOCSY
At the first glance, the strong coupling seems to represent a special case, limited to
certain molecules whose nuclei have very similar chemical shifts (by accident or as a
result of molecular symmetry). However, tricks discussed in the previous chapters allow
us to exploit advantages of strong coupling even if the chemical shifts are very different.
We have learnt that spin echoes allow us to suppress the effect of the chemical shift
evolution, which is exactly what we need: no chemical shift evolution corresponds to
zero difference in frequency offset. If we apply the echo that keeps the J coupling
evolution but refocuses evolution of the chemical shift, the state of the system of nuclei
at the end of the echo is the same as a state of a system of nuclei with identical chemical
shifts. Note, however, that a single application of a spin echo is not sufficient. Our
goal is to make the strong coupling to act continuously for a certain period of time,
not just in one moment. Therefore, we have to apply a series of radio-frequency pulses
to keep the strong coupling active for a whole mixing period. In principle, a series
of very short coupling echoes with very short 180 ◦
should work. However, specially
designed sequences of pulses with much weaker offset effects are used in practice.1
Twodimensional
experiment utilizing a mixing mimicking the strong coupling is known as
Totally Correlated Spectroscopy (TOCSY).
In order to describe the principle of the TOCSY experiment, we analyze a simple system of three nuclei (e.g. three
protons) where nuclei 1 and 2 are coupled, nuclei 2 and 3 are coupled, there is no coupling between nuclei 1 and 3. Let
us assume that both coupling constants are identical (J12 = J23 = J). The Hamiltonian describing the evolution of the
system during the TOCSY mixing period is
H = πJ (2I1zI2z + 2I1xI2x + 2I1yI2y + 2I2zI3z + 2I2xI3x + 2I2yI3y) . (15.1)
Let us now write the matrix representation of this Hamiltonian in the Cartesian basis, as we did in Eq. 14.2. For
three nuclei , we have eight states |ααα , |ααβ , |αβα , |αββ , |βαα , |βαβ , |ββα , |βββ ) and 64 basis matrices. The
matrices representing operators contributing to the Hamiltonian are
1
Technically, our task is very similar to decoupling during acquisition, shown in Section 12.2.1.
175
176 CHAPTER 15. TOCSY
I1xI2x =











0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0











I1yI2y =











0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 −
0 0 0 0 + 0 0 0
0 0 0 0 0 + 0 0
0 0 + 0 0 0 0 0
0 0 0 + 0 0 0 0
− 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0











I1zI2z =











+ 0 0 0 0 0 0 0
0 + 0 0 0 0 0 0
0 0 − 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 − 0 0
0 0 0 0 0 0 + 0
0 0 0 0 0 0 0 +











I2xI3x =











0 0 0 + 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
+ 0 0 0 0 0 0 0
0 0 0 0 0 0 0 +
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 + 0 0 0











I2yI3y =











0 0 0 − 0 0 0 0
0 0 + 0 0 0 0 0
0 + 0 0 0 0 0 0
− 0 0 0 0 0 0 0
0 0 0 0 0 0 0 −
0 0 0 0 0 0 + 0
0 0 0 0 0 + 0 0
0 0 0 0 − 0 0 0











I2zI3z =











+ 0 0 0 0 0 0 0
0 − 0 0 0 0 0 0
0 0 + 0 0 0 0 0
0 0 0 − 0 0 0 0
0 0 0 0 − 0 0 0
0 0 0 0 0 + 0 0
0 0 0 0 0 0 − 0
0 0 0 0 0 0 0 +











,
(15.2)
where + and − stand for +1/4 and −1/4, respectively (note that the matrices are orthogonal, but not orthonormal).
Consequently, the Hamiltonian matrix is
H = πJ











+1 0 0 0 0 0 0 0
0 0 +1 0 0 0 0 0
0 +1 −1 0 +1 0 0 0
0 0 0 0 0 +1 0 0
0 0 +1 0 0 0 0 0
0 0 0 +1 0 −1 +1 0
0 0 0 0 0 +1 0 0
0 0 0 0 0 0 0 +1











. (15.3)
We look for a matrix ˆT that transforms the Hamiltonian H to a diagonal form H :
H = ˆT−1
H ˆT ⇒ ˆTH = H ˆT. (15.4)
Elements of H are its eigenvalues and columns of ˆT are eigenvectors |ψk of H . The eigenvalues and eigenvectors
can be obtained from the normalization
ψk|ψk =
8
j=1
Tjk (15.5)
and from four separate sets of equations (note that H is block-diagonal if the fourth and fifth columns and rows are
replaced):
ω1T11 = πJT11 (15.6)
ωkT2k = πJT3k (15.7)
ωkT3k = πJ(T2k − T3k + T5k) (15.8)
ωkT5k = πJT3k (15.9)
ωlT4l = πJT6l (15.10)
ωlT6l = πJ(T4l − T6l + T7l) (15.11)
ωlT7l = πJT6l (15.12)
ω8T88 = πJT88. (15.13)
Equations 15.6 and 15.13 immediately provide ω1 = ω8 = πJ and
177
|ψ1 =











1
0
0
0
0
0
0
0











, |ψ8 =











0
0
0
0
0
0
0
1











. (15.14)
Eigenvalues ω2, ω3, ω5 can be obtained by standard methods of linear algebra, e.g. by writing Equations 15.7–15.9
as a matrix


−ωk πJ 0
πJ −πJ − ωk πJ
0 πJ −ωk




T2k
T3k
T5k




0
0
0

 (15.15)
and calculating ω2, ω3, ω5 as roots of the determinant
det


−ωk πJ 0
πJ −πJ − ωk πJ
0 πJ −ωk

 = −(πJ + ωk)(ωk)2
+ 2π2
J2
ωk = −ωk(ωk + 2πJ)(ωk − πJ). (15.16)
Choosing ω2 = πJ results in T22 = T32 = T52 and the normalization condition T2
22 + T2
32 + T2
52 gives
|ψ2 =












0
1/
√
3
1/
√
3
0
1/
√
3
0
0
0












. (15.17)
Choosing ω3 = 0 results in T33 = 0, T23 = −T53 and the normalization condition T2
23 + T2
53 gives
|ψ3 =












0
1/
√
2
0
0
−1/
√
2
0
0
0












. (15.18)
Choosing ω5 = −2πJ results in T35 = −2T25 = −2T55 and the normalization condition T2
25 + T2
35 + T2
55 = 6T25 gives
|ψ5 =












0
1/
√
6
−2/
√
6
0
1/
√
6
0
0
0












. (15.19)
Similarly, ω4 = πJ, ω6 = 0, ω7 = −2πJ and
178 CHAPTER 15. TOCSY
|ψ4 =














0
0
0
1/
√
3
0
1/
√
3
0
1/
√
3
0














, |ψ6 =














0
0
0
−1/
√
2
0
0
0
1/
√
2
0














, |ψ7 =














0
0
0
1/
√
6
0
−2/
√
6
0
1/
√
6
0














. (15.20)
Knowing ˆT, we can move to the ”primed” basis defined by the eigenvectors |ψk where the Hamiltonian is diagonal.
H = ˆT−1
H ˆT = πJ











+1 0 0 0 0 0 0 0
0 −2 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 +1 0 0 0 0
0 0 0 0 +1 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 −2 0
0 0 0 0 0 0 0 +1











= 2πJ(−I1zI2z + I1zI3z + 2I2zI3z). (15.21)
Here, we expressed H in its matrix form and as a linear combination of the Cartesian operator matrices.
Let us assume that the density matrix at the beginning of the TOCSY mixing is given by Inx for the n-th nucleus (we
ignore the proportionality constants for the sake of simplicity), and let us transform the Inx operators to the ”primed”
basis as well.2
I1x = ˆT−1
I1x
ˆT =
1
3
(2
√
3 4I1xI2zI3z +
√
6 I3x − 2
√
2 I2x −
√
2 4I1zI2xI3z − 8I1xI2yI3y + 8I1yI2yI3x), (15.22)
I3x = ˆT−1
I3x
ˆT =
1
3
(+2
√
3 I1x +
√
6 4I1zI2zI3x + 2
√
2 4I1zI2xI3z +
√
2 I2x − 8I1xI2yI3y + 8I1yI2yI3x). (15.23)
Finally, we can analyze evolution of Inx under the influence of H . Since we have expressed Inx and H as linear
combinations of Cartesian matrices, we can apply all familiar rules of solving Liouville - von Neumann equation by rotations
in the (now 64-dimensional) operator space. To simplify the analysis, let us assume that we apply the TOCSY mixing for
τ = 1/(3J). Then, evolution due to 2I1zI2z corresponds to the rotation by −π/3 rad (sin(−π/3) = −
√
3/2, cos(−π/3) =
1/2), evolution due to 2I1zI3z corresponds to the rotation by π/3 rad (sin(π/3) =
√
3/2, cos(π/3) = 1/2), and evolution
due to 2I2zI3z corresponds to the rotation by 2π/3 rad (sin(2π/3) =
√
3/2, cos(2π/3) = −1/2). Analysis for I1x shows
that it evolves into
+
1
12
(2
√
3 4I1xI2zI3z −
√
6 I3x + 2
√
2 I2x +
√
2 4I1zI2xI3z − 8I1xI2yI3y + 8I1yI2yI3x)
−
1
4
(+2
√
3 I1x +
√
6 4I1zI2zI3x + 2
√
2 4I1zI2xI3z +
√
2 I2x − 8I1xI2yI3y + 8I1yI2yI3x)
+ single quantum terms containing Iny (15.24)
Note that the second line contains a combination of terms corresponding to I3x. If I1x at the beginning of the
TOCSY mixing is modulated by the frequency of the first nucleus (Ω1), the modulation is transfered to I3x. When we
transform the density matrix at the end of the TOCSY mixing back to the original basis, it contains the I3x coherence
modulated by the frequency Ω1, despite the fact that there is no J coupling between nuclei 1 and 3. This greatly simplified
analysis shows that TOCSY allows us to observe correlation of all nuclei in the spin system even if they are not directly
coupled.
2
Note that ˆT is a real matrix. Therefore, real operators like Inx are converted only to real matrices,
while purely imaginary operators like Iny are transformed only to imaginary matrices.