Download 4– Quantum Mechanical Description of NMR 4.1 Mathematical Tools∗

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4– Quantum Mechanical Description
of NMR
Up to this point, we have used a semi-classical
description of NMR (semi-classical, because we
obtained M0 using the fact that the magnetic dipole
moment is quantized). Though this approach is
useful, in particular for understanding NMR in terms
of vector diagrams, it cannot be used to describe all
NMR phenomena. For instance, it is not possible to
describe the evolution of the magnetization in
coupled spins (e.g. via scalar coupling) using Bloch
equations.
Therefore, we need to be familiar with quantum
mechanics in order to describe all NMR phenomena
properly.
4.1 Mathematical Tools∗
* Special thanks to Scotty for this section!
Much of the maths used in QM uses complex
numbers. Here are some useful definitions/relations
to remember:
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• A complex number is defined as
z = a + ib = r(cos φ + i sin φ), where i2 = −1.
The real and imaginary parts of z are:
Re(z) = a, Im(z) = b. The norm and argument
of z are: r and arg (z) = φ.
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• The Euler representation of a complex number z:
r(cos φ + i sin φ) = reiφ
• The complex conjugate of z = a + ib is
z ∗ = a − ib.
• The norm or modulus of
z = a + ib = r(cos φ + i sin φ):
p
√
∗
|z| = zz = a2 + b2 = r
2
Note that |z| = zz ∗ = a2 + b2 .
Functions, Operators and Functionals
• A function maps numbers 7→ numbers. For
example, if we define
f (x) = sin(x) ,
we can evaluate f (0) = sin(0).
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NOTE: For the purposes of this discussion,
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we do not differentiate between scalar numbers,
vectors, matrices or tensors. These are all
’numbers’.
• An operator maps functions 7→ functions. For
example, if we define
2
F̂ {g(x)} = (g(x)) ,
then F̂ {sin(x)} = sin2 (x) and F̂ {ln(x)} = ln2 (x)
This notation deviates from the standard
notation:
F̂ {g(x)} = F̂ g(x)
Example:
The ∇ (’nabla’) operator is defined as


N
∂g(~
x )
∂x1
∂g(~
xN)
∂x2



∇{g(~x N )} = 
 ...

∂g(~
xN)
∂xN
for a scalar function g(~x N ).






• A functional maps functions 7→ numbers, e.g. if
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we define
F [g(x)] =
then F [sin(x)] =
R 2π
0
Z
$
2π
g(x) dx ,
0
sin(x) dx = 0.
We will use these definitions in this and subsequent
chapters to describe our spin systems, where:
• the state of the system is given by the wavefunction
~ se , sn , t)
ψ(~q, Q,
• every observable has an operator  associated with
it,
~ refers to
where ~q refers to the electron coordinates, Q
the nuclear coordinates, se refers to the electron spin,
and sn refers to the nuclear spin.
We can manipulate the state of the system using the
postulates of quantum mechanics.
4.2 Postulates of Quantum Mechanics
Wavefunction
If a string is stretched between two fixed points, the
displacement of the string from its equilibrium
horizontal position is given classically by u(x, t). If a
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perturbation moves along the string with velocity v,
then we can write a one-dimensional wave equation:
1 ∂2u
∂2u
= 2 2.
2
∂x
v ∂t
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(4.1)
Separating u(x, t) into a temporal and a spatial part,
we can rewrite it as
u(x, t) = ψ(x)cos(ωt)
(4.2)
which in turn allows us to rewrite the wave equation
as:
∂ 2 ψ(x) ω 2
+ 2 ψ(x) = 0.
(4.3)
2
∂x
v
which, we will see later, is the time-independent
Schrödinger equation.
Probability
Given the wavefunction above, let us now consider a
slightly more complex picture: a string stretched
between two fixed points as before, with additional
points (nodes) where it can be fixed.
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psiN(x)
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Three Wavefunctions of a 1-D Quantum Particle in a Box (L=10)
0.5
sqrt(2.0/10.0)*sin(1*3.14*x/10)
0.4
sqrt(2.0/10.0)*sin(2*3.14*x/10)
sqrt(2.0/10.0)*sin(3*3.14*x/10)
0.3
$
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
-0.5
0
2
4
6
8
10
x
In this case, we have many ψn (x) with n = 1, 2, . . ..
This is analogous to the particle in a one-dimensional
box problem, where the wavefunction ψn (x)
describes the microstate of a quantum particle of
mass m in a box of length L (x ∈ [0 . . . L]) for every
discrete energy level En , n = 1, 2, . . . (the more
nodes, the higher the energy - excited state)
• (Born Interpretation): the probability of finding
the particle at a certain position x is
2
ρn (x) = |ψn (x)| = ψn∗ (x)ψn (x)
NOTE: We have used an operator to map ψn (x)
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onto ρn (x).
The wavefunction is normalised so that
Z L
ρn (x) dx = 1
0
psiN^2(x)
The Probability of Finding the Quantum Particle at Position x
0.2
(2.0/10.0)*sin(1*3.14*x/10)**2
0.18
(2.0/10.0)*sin(2*3.14*x/10)**2
(2.0/10.0)*sin(3*3.14*x/10)**2
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
2
4
6
8
10
x
NOTE: This is where the image of an ’electron
cloud’ around a nucleus comes from: the cloud is
dense where the probability of finding an electron is
high.
Operator postulate
For any observable a there is an associated
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Hermitian operator Â{ψn (x)} such that
Â{ψn (x)} = an ψn (x)
where an is the observed value of a when the system
is at energy En .
NOTE: The operator of an observable must be
Hermitian because this guarantees that the
observable is a real, rather than an imaginary
number. An operator  is called Hermitian if
∗
Z
Z
ψn∗ (x)Â{ψn (x)}
ψn∗ (x)Â{ψn (x)} =
(4.4)
This makes sense when you consider the next
postulate:
Expectation value postulate
The expectation value i.e. the ’microscopic
experimentally measured value’ of any observable a is
Z L
ψn∗ (x)Â{ψn (x)} dx
Expa [ψn (x)] =
=
&
Z
0
L
0
ψn∗ (x)an ψn (x) dx
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=
=
Z
$
L
an ρn (x) dx
0
an
(4.5)
NOTE: This is a functional.
Example 1 - Let’s assume we have some given
potential V (x) that interacts with the particle at x.
In order to calculate the potential energy, we need a
potential energy operator, which is
Êpot {ψ(x)} = V (x)ψ(x) ;
then the expectation value of the potential energy is
Z L
ExpEpot [ψn (x)] =
ψn∗ (x)Êpot {ψ(x)} dx
=
=
Z
Z
0
L
ψn∗ (x)V (x)ψ(x) dx
0
L
V (x)ρ(x) dx
0
• A microscopic expectation value of any x
dependent function f (x) is a weighted average
Z L
f (x)ρ(x) dx
Exp[f (x)] =
&
(4.6)
0
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Example 2 - In order to calculate the kinetic energy
of the particle, we need a kinetic energy operator,
which is
$
h̄2 2
∇ {ψ(x)} ,
Êkin {ψ(x)} = −
2m
h
where h̄ = 2π
. Now we can calculate the expectation
value for the kinetic energy of the system by
evaluating the functional
Z L
ψn∗ (x)Êkin {ψ(x)} dx =
ExpEkin [ψn (x)] =
0
2
h̄
−
2m
h̄2
−
2m
Z
L
ψn∗ (x)∇2 {ψ(x)} dx
0
Z
L
0
2
d
ψn (x)
∗
ψn (x)
dx ,
d2 x
=
(4.7)
which we can solve if we know ψn (x).
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In general:
To calculate a system property of a quantum system:
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① Start with the wavefunction
② Choose the appropriate operator for the system
property
③ Evaluate the expectation value functional
A Quantum Particle in a Box
As we saw above, a quantum particle of mass m in a
one-dimensional box of length L (x ∈ [0 . . . L]) can
be likened to a string stretched between two fixed
points, with nodes. Starting from equation 4.3, we
can rewrite the term ω 2 /v 2 as
ω2
(2πν)2
4π 2
=
= 2 .
2
2
v
v
λ
Using de Broglie’s formula, which states that
h
λ=
p
(4.8)
(4.9)
and using the definition of the total energy:
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p2
+ V (x)
E=
2m
(4.10)
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where p is momentum, equation 4.3 can be written
as the time-independent Schrödinger equation:
h̄2 2
∇ {ψ(x)} + V (x)ψ(x) = Eψ(x) ,
−
2m
$
(4.11)
h
, and h is Planck’s constant, ψ(x) is the
where h̄ = 2π
complex wavefunction, (V (x) is the potential and E
is the energy (a scalar, real value).
d2 ψ(x)
2
As V (x) = 0 inside the box, and ∇ {ψ(x)} = d2 x
we have
h̄2 d2 ψ(x)
−
= Eψ(x) ,
(4.12)
2
2m d x
for which the general solution is known to be:
ψ(x)
=
A sin (kx) + B cos (kx)
k 2 h̄2
E =
,
(4.13)
2m
where A and B are complex numbers and k is real.
• We use boundary conditions to determine A, B
and k. We choose ψ(0) = ψ(L) = 0.
• ψ(0) = 0:
0 = ψ(0)
&
=
A sin (0) + B cos (0)
=
A·0+B·1
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(4.14)
$
⇒ B = 0.
• Now set ψ(L) = 0
0 = ψ(L) = A sin (kL)
(4.15)
Either trivial, uninteresting solution A = 0, or for
A 6= 0
sin (kL) = 0
which has many solutions:
nπ
k=
L
(4.16)
for n = 1, 2, . . ..
NOTE: This means only ’complete waves’ are
allowed in the box, which explains how only discrete
energy levels are allowed for the quantum system.
Analogy with sound waves: resonance.
• The probability of finding the particle at a point x
is |ψ(x)|2 , so we know that
Z L
|ψ(x)|2 dx = 1
&
0
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Lets use this to determine A:
Z L
nπ 2
2
1 =
A [sin ( x)] dx
L
0
(4.17)
• We use the fact that
Z
1
1
sin (2ax)
[sin (ax)]2 dx = x −
2
4a
(4.18)
So then
A
−2
=
−
and with a =
nπ
L
1
1
L−
sin (2aL)
2
4a
1
1
0−
sin (2a0)
2
4a
(4.19)
we determine that
r
2
|A| =
L
So:qA is any complex number with an absolute value
of L2 ;
&
En
=
ψn
=
n2 h̄2 π 2
n2 h2
=
n = 1, 2, . . .
2mL2
8mL2
r
nπx 2
sin
(4.20)
L
L
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Hamiltonians
Given the definition of an operator, we can see that
equation 4.12 can be written in general form as
Â{ψn (x)} = an ψn (x)
(4.21)
Ĥψn (x) = Eψn (x)
(4.22)
or
In other words, the classical energy function is
related to the Hamilton operator or energy operator.
Classical
energy
function, E
^
H
For example, recall from Chapter 1, the definition of
the energy of a magnetic moment in an external
magnetic field B~0 :
~ 0 = −γ L
~ ·B
~0
E = −~
µ·B
(4.23)
The corresponding quantum mechanical definition is
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given by
Ĥ
~ ~
= −γh̄L̂ · B
0
= −γh̄[L̂x Bx + L̂y By + L̂z Bz ]
(4.24)
In general, one can write
~ ~
Ĥ = −γ Jˆ · B
0
(4.25)
~
where Jˆ can be the angular momentum associated
~
with the electron spin, which we denote as h̄Ŝ or the
angular momentum associated with the nuclear spin,
~ˆ
which we denote as h̄I.
Quantum - Classical LINK:
Recall from chapter 1, that we said that NMR involves measuring the total magnetic dipole moments
in our samples. Translating this into QM lingo - since
~
µ
~ “translates” into Iˆ for nuclear spins - NMR involves
~
measuring the expectation value of Iˆ (usually only one
component of it actually, see below).
4.3 Angular Momentum in MR
~ ~
~ˆ
The angular momentum operators (L̂, Ŝ, and I)
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have the special property that their components do
not commute with each other. The commutation
relation for two operators  and B̂ is given by
[Â, B̂] = ÂB̂ − B̂ Â.
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(4.26)
If  and B̂ commute with each other, then the
equation above is zero. The physical implication of
this is that the two observables can be measured at
the same time - measuring one does not affect the
outcome of the other, and vice versa.
~
Applying this to the components of Iˆ (the same
applies for the other two), which are given by
∂
∂
−z
Iˆx = −i y
∂z
∂y
∂
∂
ˆ
Iy = −i z
−x
∂x
∂z
∂
∂
Iˆz = −i x
−y
∂y
∂x
(4.27)
we can obtain the following cyclic commutation
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relations:
h
i
Iˆx , Iˆy
h
i
Iˆy , Iˆz
h
i
Iˆz , Iˆx
$
= iIˆz
= iIˆx
= iIˆy
(4.28)
A good way to remember this is with the picture:
^I
x
I^z
^
Iy
Another important property of momentum is the
definition
Iˆ2 = Iˆx2 + Iˆy2 + Iˆz2
(4.29)
It can be shown that
i
h
2
Iˆx , Iˆ
h
i
Iˆy , Iˆ2
h
i
2
Iˆz , Iˆ
&
=
0
=
0
=
0
(4.30)
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i.e. Iˆ2 commutes with the components Iˆj where
j = x, y, orz.
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4.4 Matrix representation of operators
It is possible to express the wavefunction ψ as a
weighted sum of known, appropriately chosen
functions ϕk , such that
ψ=
D
X
c k ϕk ,
k=1
where D is the dimension of the space. If our chosen
functions are orthonormal, then
Z
ϕk ϕ∗l = δk,l ,
where δk,l is the Kroenecker delta, i.e. is equal to 1 if
k = l, and 0 otherwise.
NOTE: This is like the equation for probability
we defined above.
Operating on these “new” functions yields:
&
Âϕk =
D
X
k=1
alk ϕl .
(4.31)
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Consequently, equation 4.21 can be rewritten as
XX
X
X
Â{ψ} = Â
c k ϕk =
alk ck ϕl =
d l ϕl
k
k
l
$
l
(4.32)
where
dl
X
=
alk ck
k
d~ = (A)~c
(4.33)
where the latter equation is in matrix form.
Examples of (A) are

for D= 2 and
(A) = 

for D= 3.
a11
a12
a21
a22
a11

(A) = 
 a21
a31


a12
a13
a22
a23
a32
a33
(4.34)




(4.35)
In general, the elements of a matrix are obtained
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using the definition
X
X
< ϕj |Â|ϕk >=
alk < ϕj |ϕl >=
alk δjl = ajk
l
$
l
(4.36)
where we use the Dirac notation (see box 1).
NOTE: This is just another way of writing the
expectation value of the operator Â, operating on
the functions ϕk .
Box 1. Dirac Notation
Dirac notation is a short form to save having to write
out the integral:
Z
∗
(x)ψn (x).
< m|n >= dxψm
Since changing the representation of an operator into
matrix form should not change its properties, the
commutation relations given in equation 4.28 should
still be valid, i.e.
&
[(Ix ), (Iy )] =
i(Iz )
[(Iy ), (Iz )] =
i(Ix )
[(Iz ), (Ix )] =
i(Iy )
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(4.37)
$
where the () are there to emphasize that we are
writing the equations in terms of matrices.
We can use these relations and the convention that
the (Iz ) matrix is diagonal to derive the three
operators in matrix representation. The dimension D
of our matrices is determined by the spin quantum
number, which in chapter 1 we defined as
mJ = −J, −J + 1, ...J,
(4.38)
such that D = 2J + 1. This corresponds to the
number of energy levels we drew for a given value of
J.
NOTE: In many textbooks, J is replaced by I in
order to emphasize that we are talking about nuclear
spins. Recall, that J is the property of a nucleus.
Let us assume, that (Iz ) and (Ix ) are given by


a 0


(Iz ) =
(4.39)
0 b
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and

(Ix ) = 
c
d
∗
d
e

.
$
(4.40)
We can get (Iy ) from the relation


 
ca db
ac ad




−
− i[(Iz ), (Ix )] = −i
d∗ a eb
bd∗ be


0 −d

 = (Iy )
= i(a − b)
d∗ 0
(4.41)
Using this definition of (Iy ), we can write


0
(a − b)d


− i[(Iy ), (Iz )] = (a − b)
(a − b)d∗
0


c d
 = (Ix )

(4.42)
=
∗
d e
Using this last relation, we see therefore that
&
c
= e=0
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(a − b)2
= 1
(4.43)
$
Using the final commutation relation,


2dd∗
0


− i[(Ix ), (Iy )] =
0
−2dd∗


a 0

 = (Iz ), (4.44)
=
0 b
we can solve for a, b, and d:
a =
a−b
yields a =
1
2
and b =
=
−b
1
(4.45)
1
2dd = ,
2
(4.46)
−1
2 .
Finally, given that
∗
we obtain d =
1
2
as well.
Therefore, we can write the matrix form of (Iz ), (Ix ),
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and (Iy ), as
(Iz )
(Ix )
(Iy )
=
=
=






1
2
0
0
−1
2
0
1
2
1
2
0
0
−i
2
i
2
0






(4.47)
These three matrices represent the Pauli matrices for
spin 1/2.
In general, we should be able to find functions
ϕk = φm such that
Iˆz φmJ = mJ φmJ .
(4.48)
Given the equation above, we say that φmJ are
eigenfunctions and mJ are eigenvalues, because the
result of operating on φmJ is a constant times the
same function φmJ .
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Therefore in matrix representation, (Iz ) =


$
< φJ |Îz |φJ >
< φJ−1 |Îz |φJ >
< φJ |Îz |φJ−1 >
< φJ−1 |Îz |φJ−1 >
...
...
< φJ |Îz |φ−J >
< φJ−1 |Îz |φ−J >
...
...
...
...
< φ−J |Îz |φJ >
< φ−J |Îz |φJ−1 >
...


< φ−J |Îz |φ−J >
(4.49)
or after evaluating the expectation values,


J
0
... 0


 0 J − 1 ... 0 


(Iz ) = 
.
 ...

...
...
...


0
0
... −J
(4.50)
Putting in J = 21 , we see that we get the same (Iz )
defined in 4.47.
At this point, we can introduce two new operators
Iˆ+ and Iˆ− , which are defined as
Iˆ+
Iˆ−
=
=
Iˆx + iIˆy
Iˆx − iIˆy
(4.51)
These operators have the special property that
p
+
J(J + 1) − mJ (mJ + 1)φmJ +1
Iˆ φmJ =
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Iˆ− φmJ
=
p
J(J + 1) − mJ (mJ − 1)φmJ −1 ,
$
(4.52)
i.e. the Iˆ+ operator raises the magnetic quantum
number by 1 and Iˆ− lowers the magnetic quantum
number by one. Let’s see what this means by way of
the spin 1/2 example:
•Eigenfunctions: φ1/2 , φ−1/2 . (see box 2)
Therefore,
r
1 1
+
ˆ
J(J + 1) − ( + 1)φ1/2+1 = 0
I φ1/2 =
2 2
r
−1 −1
+
ˆ
(
+ 1)φ−1/2+1
J(J + 1) −
I φ−1/2 =
2 2
= φ1/2
r
1 1
Iˆ− φ1/2 =
J(J + 1) − ( − 1)φ1/2−1
2 2
= φ−1/2
r
−1 −1
J(J + 1) −
(
− 1)φ−1/2−1 = 0
Iˆ− φ−1/2 =
2 2
(4.53)
where J = 1/2.
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Box 2. Special Case: Spin 1/2
For spin 1/2, the eigenfunctions φ1/2 and φ−1/2 are
often written in terms of two new symbols:
$
φ1/2 = α
φ−1/2 = β.
In matrix form, for spin 1/2, the raising operator,
Iˆ+ , and lowering operator, Iˆ− , are


0 1
+


(I ) =
(4.54)
0 0
and

(I ) = 
−
0
0
1
0

.
(4.55)
We can write Iˆx and Iˆy in terms of the raising and
lowering operators:
&
Iˆx
=
Iˆy
=
1 ˆ+ ˆ−
[I + I ]
2
1 ˆ+ ˆ−
[I − I ]
2i
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(4.56)
$
We can also see from the spin 1/2 case, that in
general, the elements of (I + ) and (I − ) are given by
p
+
ˆ
< φi |I |φj >= J(J + 1) − j(j + 1)δj+1,i
p
−
ˆ
< φi |I |φj >= J(J + 1) − j(j − 1)δj−1,i
(4.57)
So if we recall that for a general operator Â, written
in matrix form as (A) =



< φJ |Â|φJ >
< φJ−1 |Â|φJ >
< φJ |Â|φJ−1 >
< φJ−1 |Â|φJ−1 >
...
...
< φJ |Â|φ−J >
< φJ−1 |Â|φ−J >
...
...
...
...
< φ−J |Â|φJ >
< φ−J |Â|φJ−1 >
...
< φ−J |Â|φ−J >
(4.58)
then the equations in 4.57 tell us that
(I + ) =

+
0


< φJ |Î
|φJ−1 >
0
|Î + |φ
...
0
...
0
0
0
0
0
0
...
0
...
...
...
...
...
0
0
0
...
0
< φJ−1
J−2 >
with the non-zero terms in blue and similarly
&




(4.59)
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(I − ) =



$
0
< φJ−1 |Î − |φJ >
0
0
< φJ−2
0
|Î − |φ
J−1 >
0
...
0
0
...
0
0
...
0
...
...
...
...
...
0
0
0
...
0
with the non-zero terms in again blue.



(4.60)
Therefore equations 4.50, 4.56, 4.59 and 4.60 can be
used to derive all five operators in matrix form for
any mJ .
As practice for your midterm, try this for spin 1!!!
4.5 Spin Hamiltonian
In the previous sections, we saw that the Hamilton
operator can be used to describe the total energy of
a system (recall: the particle in a box section). In
order to describe a molecule on which we want to
perform an NMR experiment, i.e. a molecule in a
~ 0 (and possibly an electric field E),
~
magnetic field B
we therefore need to describe the energy of the
system by writing the Hamiltonian:
n
e
n
e
en
+ Êkin
+ Êpot
+ Êpot
+ Êpot
+ ĤQ + ĤLL + ...
Ĥ = Êkin
(4.61)
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where the terms can be visualized by decomposing
the molecule in terms of its properties, i.e.
momentum, charge, position, angular momentum,
electron spin angular momentum,...
$
or graphically:
T^e
^
V
e
^pe
−e
^
H
LL
−β^Le
^
H
SS
^
−gβSe
Electrons
^
H
Q
E
B0
^
H
Q
p^n
ze
T^n
^
V
n
Qn
γnhI^n
Nuclei
^
H
II
Although this Hamiltonian is very descriptive, it is
also too cumbersome. Therefore, we must try to
simplify it:
① apply the Born-Oppenheimer approximation,
which assumes that nuclear motion can be
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neglected.
② for the magnetic properties of the molecule, we
need only to consider unpaired electron and
nuclear spins,
such that the spin Hamiltonian ĤS is
ĤS
ĤS
ĤS
= < ψ(~q, se , sn )|Ĥ|ψ(~q, se , sn ) >
~
~
= ĤS Ŝi , Iˆl
X ~
X ~
i
~
~ l Iˆl
=
A Ŝi +
B
i
+
XX ~
~
Ŝi (C ij )Ŝj
i
+
j
XX~
~
Iˆl (Dlk )Iˆk
l
+
l
k
XX ~
~
Ŝi (E il )Iˆl .
i
(4.62)
l
We will see in the “NMR Interactions” section what
some of these coefficients (vectors and matrices) are.
For example, (D lk ) includes terms such as scalar
spin-spin coupling, dipolar coupling, and
quadrupolar coupling.
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In any case, this equation clearly shows that it is
possible, with a spin Hamiltonian, to describe both
interactions of a spin with its “environment” (terms
~
in Iˆl ) and interactions between spins (terms in
~ ~
Iˆl X Iˆk ). Recall, that the whole reason for switching
to a quantum mechanical picture is that Bloch
equations cannot be used to describe spins
interacting with other spins (i.e. coupled spins).
$
SUMMARY
Up this point, in terms of NMR experiments, we have
described:
① the spin angular momentum, in both operator and
matrix form;
② the spin Hamiltonian.
What we want is a quantum mechanical equivalent to
~
dM
dt . So we still need to describe the total magnetization in our sample quantum mechanically, as well as
an equation of motion.
4.6 Density Operator/Matrix
As we saw in Chapter 1, if you place a single spin
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~ 0 , then
(let’s take J = 1/2) in a magnetic field B
there are 2J + 1 states (corresponding to two energy
levels for spin 1/2), which are described by the
eigenfunctions φ1/2 and φ−1/2 (or α and β). In a
NMR sample, we typically have on the order of 1022
spins (i.e. an ensemble), some being close to state
|α >, some being close to state |β >, while the rest
are in intermediate states. Therefore, we say that we
are not dealing with pure states, but with a
superposition of states. In this case, the
wavefunction for our ensemble is given by
X
c k ϕk .
ψ=
$
k
and the expectation value for a general operator  is
< Â > =
< ψ|Â|ψ >
< ψ|ψ >
=
< ψ|Â|ψ >
(4.63)
since the probability < ψ|ψ > is 1.
NOTE: This is different from equation 4.36,
which gives us the elements of the matrix (A). Here
we are dealing with the “total” wavefunction.
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Putting in the definition of ψ into the equation for
< Â >, we get
Z
< Â > =
ψ ∗ Â{ψ}
!
!∗
Z X
X
c k ϕk }
=
cl ϕl Â{
l
=
XX
l
=
=
k
XX
l
k
XX
l
c∗l ck
$
k
Z
ϕl ∗ Â{ϕk }
c∗l ck < ϕl |Â|ϕk >
c∗l ck alk
k
(4.64)
We can define a new parameter ρkl as the product
ρkl = c∗l ck
(4.65)
such that
< Â > =
XX
k
=
&
X
k
ρkl alk
l
((ρ)(A))kk
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=
T r{(ρ)(A)}
=
T r{ρ̂Â}
$
(4.66)
with the trace (Tr) being the sum of the diagonal
elements, and (ρ) is the density matrix and ρ̂ is the
density operator.
NOTE: This ρ is different from the ρn used in
the definition of the probability.
The implication of the last line in the above equation
(or equivalently the line above that) in NMR is that
any macroscopic observable (e.g. magnetization) can
be deduced from two operators:
① one which represents the observable being
measured;
② another representing the state of the entire spin
ensemble, independent of the number of spins in
it.
In other words, rather than having to define
microscopic states for each of our 1022 spins, we can
used a single operator ρ̂.
For spin 1/2, we can have a look at a few specific
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expectation values:
< φ1/2 |ρ̂|φ1/2 >
< φ−1/2 |ρ̂|φ−1/2 >
= cα c∗α
= cβ c∗β
(4.67)
which represent the population of the lower and
upper energy levels for a two spin system,
respectively. The line over the terms on the right
hand side of the equation represents an average over
all spins in the ensemble.
The other two possible terms,
< φ1/2 |ρ̂|φ−1/2 > =
< φ−1/2 |ρ̂|φ1/2 > =
cα c∗β
cβ c∗α
(4.68)
represent coherences, i.e. that a net spin polarization
exists in the direction perpendicular to the external
magnetic field. This requires that in our sample, we
have some spins which are in superposition states. In
other words, these spins do not point along z but in
the xy-plane, as shown below.
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M.H. Levitt, Spin Dynamics, p. 280.
We will see in the solution NMR section how
coherences are used.
4.7 Equations of motion in MR
The equation of motion in quantum systems is given
by the time-dependent Schrödinger equation,
∂ψ
−i
=
Ĥψ.
∂t
h̄
&
(4.69)
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Special case: time-dependent Schrödinger equation
When we can separate the wave function into a timeindependent component and a time-dependent term,
i.e.
fk (~q, Q,
~ se , sn , t) = ψ
~ s)e−iEk t/h̄ .
ψk (~q, Q,
$
(4.70)
then the time-dependent Schrödinger equation simplifies to
fk = Ek ψ
fk ,
Ĥψ
(4.71)
which is the time-independent Schrödinger equation.
Given the definition for the density matrix in
equation 4.65, we can write the differential with
respect to time as:
∂ρkl
∂ck ∗
∂c∗l
=
c + ck
∂t
∂t l
∂t
(4.72)
Now, given the time-dependent Schrödinger equation
above and the definition of the “NMR” wavefunction
ψ, we can determine that
∂ck
−i X
=
< ϕk |Ĥ|ϕj > cj
∂t
h̄ j
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=
∂c∗k
∂t
=
=
$
−i X
Hkj cj
h̄ j
i X
< ϕj |Ĥ|ϕk > c∗j
h̄ j
i X
Hjk c∗j
h̄ j
(4.73)
Therefore equation 4.72 can be rewritten as
∂ρkl
∂t
=
=
=
∂ck ∗
∂c∗l
c + ck
∂t l
∂t
−i X
i X ∗
∗
Hkj cj cl +
ck cj Hjl
h̄ j
h̄ j
X
−i X
[
Hkj ρjl −
ρkj Hjl ]
h̄ j
j
(4.74)
Rewritting in operator form, we can see that the
term in [ ] is a commutation relation, therefore
i
dρ̂(t)
−i h
=
Ĥ, ρ̂(t) .
(4.75)
dt
h̄
This equation is equivalent to the time-dependent
Schrödinger equation and is known as the
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Liouville-von Neumann equation.
The solution of this equation is
ρ̂(t) = e−iĤt/h̄ ρ̂(0)eiĤt/h̄
(4.76)
With this, we can finally describe a time-dependent
expectation value for a general operator  as
< Â > (t) = T r{e−iĤt/h̄ ρ̂(0)eiĤt/h̄ Â}
(4.77)
With these two last equations, we can describe how
our spin system evolves over time, given the initial
state of our spin system ρ̂(0), under a Hamiltonian
Ĥ.
This is exactly what NMR simulation packages like
GAMMA (http://gamma.magnet.fsu.edu) and
SIMPSON (http://www.bionmr.chem.au.dk/
bionmr/software/index.php) use. Here is an example
of a GAMMA program to simulate the outcome of a
one-pulse experiment in solution NMR:
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