The Time Independent Quantum Mechanics of NMR
 As with all spectroscopies NMR is concerned with the Schrödinger equation. It has a
Hamiltonian, H, which contains terms describing the interaction between the spins and the
magnetic field and interactions between the spins themselves. This Hamiltonian has
corresponding set of eigenfunctions, n which describe the states of the spin system with energy
En.
Hn  E n n

The eigenfunctions form an orthonormal set of functions meaning that two functions describing
different states are orthogonal to each other, and each individual function is normalized.
1


d


 n m 0

for n  m
for n  m
In Dirac notation taking the expectation value of an operator is depicted in a short hand form.
The expectation value of an operator O is conventionally written as:
O   *nO m d ,
whereas in Dirac notation it is written as:
n O m
.
The action of taking this integral is decomposed into two parts,
n
and
m
which are
referred to as the bra and the ket.

For a single spin-1/2 system, the eigenfunctions represent the spin either aligned or against the
field, referred to as or , respectively. These functions describe the total spin angular
momentum and its z-component of a spin-1/2 system, and thus is an eigenfunction of the spinangular momentum operator, I, and its z-component, Iz as well as the Hamiltonian.

Each state is described in terms of 2 quantum numbers, I, the spin angular momentum quantum
number, and mI, the magnetic quantum number. A spin-1/2 system has I =1/2, and has mz = ½
and - ½, that correspond the and  respectively, which form a complete set of eigenfunctions
providing a description for all the possible states of this system.

Taking the expectation value of the Iz operator with the  state:

*
I z  d   I z   m  1 / 2

Repeating this for all possible combinations of and , we get:
 I z   1 / 2 ,  I z   0 , and  I z   0 .
Which is a complete description of the operator Iz, for a spin-1/2 system.

In the Dirac notation all the possible expectation values of an operator are summarized in the
form of a matrix, which is like a table with a column and rows corresponding to the
eigenfunctions, and with the expectation values as entries corresponding to the eigenfunctions
indicated by the row and column.

 

1 0 
I z   1 / 2
0   1 / 2



0  1
 
0  1 / 2
This is known as the matrix form of the operator.

In this matrix notation the bra is a row vector and the ket is a column vector. For instance for a
spin-1/2 system:
1 

In this notation taking expectation value can be expressed a vector-matrix-vector multiplication .
For example:


*
0  1 
1/ 2
I z  d   I z   1 0 
    1/ 2
 0 1/ 2  0
Recall that the interaction of a spin-1/2,  
E    B  

 I nucleus with a magnetic field is given by:
 IB  
The corresponding Hamiltonian operator is of the form:
H    Bo I z

Lets include the chemical shift at this stage:
H    Bo (1   )I z

0 
  1 0 ,   0 1 ,     and    
0 
1 
The above equation states that :
 BoIz

H  

 




     Bo (1   ) 1 0 
 Bo (1   ) / 2
0
0  1

2


0
Bo (1   ) / 2
Often it is preferable to work in frequency units, thus the Hamiltonian is divided by Planck’s
constant, h, giving

H  

 



   (ν o - ν) 1 0 
 (ν o - ν) / 2
0

2 0  1
0
(ν o - ν) / 2

In summary this Hamiltonian states that there are two energy levels characterized by the frequencies
- (ν o - ν ) / 2 and (ν o - ν ) / 2 . The matrix is diagonal and thus the functions and  are the
corresponding eigenfunction. The resonance frequency is given by the energy difference between
the two states, which is (ν o - ν ) .

The above Hamiltonian is a satisfactory description of a single spin, but how is one to consider
multiple spins that are interacting with each other. Now, recall that the operator Iz represents,
completely, the spin angular momentum of a single spin, and can be thought of classically as being
directly related to the z-magnetization (i.e. it can be used to represent the z-magnetization). The
indirect spin-spin coupling is an interaction between two nuclear magnetic moments, by an
interaction strength J,
E   A   B  JI A  I B  JI zA  I B
z
thus a new corresponding term in the Hamiltonian is JI zA  I B
z in frequency units.

What does a product between the two magnetizations mean when represented as matrices. Lets start
with how to represent the spin states for two simultaneous spins. Like in other quantum mechanical
methods a product basis is employed, in other words the state of a two-spin system is taken to be the
product between two one spin states. Thus a two-spin system has four possible states
, where the state of spin A is first and that of spin B is second.

Now lets explore the matrix representation of IzA, that is the z-component of the magnetization of
spin A.
 I zA    A  B I zA  A  B   A I zA  B  A  B
1
1
  A I zA  A  B  B  *1 
2
2
 I zA    A  B I zA  A  B   A I zA  B  A  B
1
  A I zA  A  B  B  * 0  0
2
A
A
 I z    A  B I z  A  B   A I zA  B  A  B
  A I zA  A  B  B  0 *1  0
 I zA    A  B I zA  A  B   A I zA  B  A 
  A I zA  A  B  B  0 * 0  0
1
2
1
1
 *1 
2
2
 0*0  0
 I zA    A I zA  A  B  B  * 0  0
 I zA    A I zA  A  B  B
 I zA    A I zA  A  B  B
 I zA    A I zA  A  B  B  0 *1  0
 I zA   0
 I zA   0
 I zA   
1
2
 I zA   0

 I zA   0
 I zA   0
 I zA   0
 I zA   
The matrix representation of the IzA operator is therefore:
 


1
0

2

1
 0
A
IZ  
2

0
  0

0
0
 


0
0

1
2
0
 

0 

0 


0 
1
 
2
1
2

Notice that the matrix form of IzA can be broken up into four 2-by-2 matrices, in the following
way:
 1
 2

 0
A
I Z  

 0
 0


 1 1
 0
2
 
 1
 0 0
 


0
1

2
0
0
0
1
0
1

0 0  
0 0   1 1 0
0 0  

   

0 0  

 
   2 0 1 
 1
   0 0 
1 1 0 

0

 2
   0 0 
2 0 1 




1 
 0  

2 
1 0   1

0

0

 1 0
0 1     2

 Iz  E
1 1 0   0  1  0 1


 




2
2 0 1  
This decomposition is expressed in terms of a matrix direct product. In performing a direct
product one copy of the second matrix multiplied the position of each entry of the first matrix,
and multiplies by the value of that entry. Notice that the two matrices in the direct product are
the Iz operator along with the 2-by-2 identity matrix E.
Question 1
Derive the matrix form of the IzB operator.
  


1

0
0

2

 0  1 0
B
IZ  
2
1


0
0


2

0
0
0
 

 
 1
 
0  2

  0

0  


0   0
 
1   0
  
2

0 
1
 
2
0
0

0 0  
0 0  

 

1

0
2


1 
 0  

2 
 1

1

0
0
 2

2

1
0
 

1
1 
1
 0  
 0    1 0 

2
2  
I ZB   
 2


 1
  0 1 
 1
0
0
 2 0  
 2

1


0 

1
1
 0   0   
2 
2 
 


0 
 E  Iz
1
 
2
Thus IzAIzB will be the matrix product of the IzA and IzB. Lets first derive the matrix form of this
operator, then verify that it is such a product.
 I zAI Bz    A  B I zAI Bz  A  B   A I zA  B I Bz  A  B ...
1 1 1
  A I zA  A  B I Bz  B  * 
2 2 4
Similarly
1
2
1
2
 I zAI Bz    A I zA  A  B I Bz  B  *   
1
2
1
4
1
2
 I zAI Bz    A I zA  A  B I Bz  B   *   
1
2
1
2
 I zAI Bz    A I zA  A  B I Bz  B   *  

1
4
1
4
All the other entries can be show to be zero. Therefore the matrix form of the IzAIzB operator is:
 


1

0
4


1
0 
A B 
IZ IZ 
4


0
  0

  0
0


0
0

1
4
0
 

0 


0 


0 

1 
4 

The matrix product of IzA and IzB is derived as:
1
2

0
A B
Iz Iz  
0


 0
1
0 
0
2

1
1
0
0 0 
2
2

1
0 
0 0 0

2
1 
0 0  0 0
2
0
0
 1
0  
0
0
4
 
1
0 0  0 
0
4

1
1
0  0 0 
 
2
4
1 
0   0 0
0
2 
0

0

0

0

1

4
I z  I z which can be derived from I zA  I z  E
I Bz  E  I z since I zAI Bz  I z  EE  I z   I z  E  E  I z  I z  I z .
The above matrix is of the form

The Hamiltonian of a two spin system is thus the sum the terms involving interaction between the
spin and the magnetic field and between the spins and thus can be summarized as:
A B
H LF  (ν A  ν0 )I zA  (ν B  ν0 )I B
z  JI z I z

In the rotating frame the term can be removed, giving:
A B
H RF  ν AI zA  ν BI B
z  JI z I z

and
The matrix form can be derived from the following direct product:
H RF  ν AI z  E  ν BE  I z  J I z  I z
1
2

0
H  νA 
0


 0

1
0 
2 0


1
1
0
0 
0 
2
2
  νB 
1
0 0
0 
0 


2

1
0 0  
 0 0
2
0
0
 1
0  
0
0
4
 
1
0 0  0 
0
4
  J
1
1
0  0 0 
 
2
4
1 
0   0 0
0
2 
0

0

0

0

1

4
(νA  νB ) J


2
4

0

H

0


0





1
J
(νA  νB ) 
0
0

2
4

J

0
 (νA  νB ) 
0

4
(ν  νB ) J 
0
0
 A
 
2
4
0
0
0
Notice that this Hamiltonian is diagonal as well, which implies that the product basis is the
eigenbasis of this system. Eigenvalue, i.e. energies, are the diagonal entries of the matrix. The
corresponding energy level diagram is shown below.
Energy Level Diagram for an AX Spin System

A + /2 + J/4
B
B - J/2
A
A - J/2

A - /2 - J/4
A
A + J/2

A - /2 - J/4
B
B + J/2


A + /2 + J/4
In the above diagram the energy levels are shown with the corresponding eigenfunction.
Transitions between these energy levels corresponds to flipping either spin A or B (is indicated in
figure), where the energy of the transition is the difference between the two levels involved. In
the above case there are four distinct transitions, two centered at A and two centered at B each
set split by J, giving rise to the doublet-of-doublet pattern typical of AX spin systems.
Question 2
Derive the Hamiltonian for a three-spin, AMX, system. Show the energy level diagram and predict
the spectrum. Assume that JAM = JMX = JAX.

The x and y components of the spin angular momentum don’t have a simple intuitive picture from
which they can be derived. Recall that there are only two quantum numbers, the total angular
moment and the z-component, thus there is no information about the remaining components. This
arises from the fact that the x, y and z component operators do not commute. At this stage it will
have to suffice to introduce them without further explanation.

The matrix form of the x and y component of spin angular momentum, Ix and Iy are given by:

0
Ix  
1

2
1
2

0


0
and I y  
i

2
i
 
2

0 

Each represent the magnetization along the x and y axis respectively. (add in cyclic permutation,
and more background on properties of angular momentum)

Thus the total angular momentum is given by the vector:

0
I
1

2

1 
i  1

0

0
ˆ
2  ˆi  
2  ˆj   2
k
 i
 
1
0 
0  0  
 2
 
2
Another way to represent components in the transverse plane is to take combination of the x and y
components, Ix + iIy and Ix - iIy these can be thought of a positive and negative rotating
component, respsectively. These have the matrix form:

0
Ix  i Iy  
1

2
1 
0
2  i
 i
0 
 2
 i
2   0 1   I
 0 0  

0 


0
Ix  i I y  
1

2
1 
0
2  i
 i
0 
 2
 i
2   0 0   I
 
 
0  1 0

which are referred to as the raising and lowering operator, i.e. they represent transitions between 
and Ultimately it is the expectation value of these operators that give rise to the transition, but
more of that later when time-dependent quantum-mechanics is discussed.



For example:
0 1 0 1
0 1 1
I   



I


and

 1 0
0 0   0   0 
0
0

   

 
0 0 1 0
0 0  0 
I   



I


and

 0 1
1 0 1  0 
1
0

   

 
This illustrates that the raising operator, I+, serves to raise the spin state from to, flipping the
spin from against to with the field, and the lowering operator, I-, serves to lower the state from  to
flipping the spin from with to against the field


Strictly speaking the coupling interaction does not occur between just the z-magnetization but the
total magnetization, thus the term is of the from J IAIB. Which can be rewritten as:



I A  I B  I xA , I Ay , I zA  I Bx , I By , I Bz  I xAI Bx  I AyI By  I zAI Bz 

Which can be shown to be equivalent to
I

A B
 I

 I AI B
 I zAI Bz where the first two terms
2
are referred to as the flip-flop terms. They represent the simultaneous raising and lowering and
lowering and raising of spin A and B, respectively. It is terms like these that are responsible for
transition between spins, giving rise to relaxation.

The flip-flop terms only become important when the energy-levels are closely spaced to each
other. In other words when the interaction strength is of the same order as the differences in the
frequencies of the nuclei. In fact it is these terms that give rise to the second-order effects.

The complete Hamiltonian for a two spin-1/2 system will have to be amended to include these
terms as follows:
H  ν A I zA  ν BI Bz  JI zAI Bz 

J A B
I  I   I AI B
2

The corresponding Hamiltonian can be derived from the following direct products:
H  ν A I z  E  ν B E  I z  JI z  I z 

J
I  I  I  I
2

ν 1 0  1 0  ν B 1
H A


2 0 1 0 1  2 0
J  0 1  0



2  0 0  1
 ν A  ν B  J


2
4

0

H

0


0

0  1

1  0
0 0

0  1
0  J 1 0  1 0 


1 4 0 1 0 1
0 0 1  


0  0 0  



ν A  ν B   J
J
0

2
4
2


J
νA  νB  J



0

2
2
4
ν  ν B   J 
0
0
 A

2
4
0
0
0

Notice that this Hamiltonian is not diagonal. This means that the product basis is not suited as an
eigenbasis for this system. In order to find the eigenvalues and hence the eigenbasis this matrix
will have to be diagonalized. Notice that just the center two row and columns make the system
non-diagonal, so just diagonalizing this part would make the whole matrix diagonal. This is matrix
is referred to a being block diagonal.

To find the eigenvalue the following two-by-two system has to be solved:
lets refer to
νA  νB
as
J
 νA  νB J


 2

4
2


J
νA  νB  J 

 


2
2
4
ν . Thus the eigenvalue problem becomes:
ν J
 
2
4
det
J
2
J
2
ν J

 
2 4
0
ν
J
 
2
det 2
0
J
ν

 
2
2
 ν
 ν
 J
    
    0 

 2
 2
 4
2
Which obeys the characteristic equation:

ν 2
J2
ν 2 J 2
2
2

     0 which can be rearranged to   
 . Therefore
4
4
4
4
ν 2 J 2
ν 2 J 2 J
  
 which implies that   

  This means that the energy
4
4 4
4
4
ν 2 J 2 J
  The energy level diagram is shown
levels correspond to frequencies ν  
4
4 4
below.
The Energy Level Diagram of an AB Spin System
A + /2 + J/4

J/4 - J2/4)1/2
C3 C4
J/4 + J2/4)1/2
C1 C2
A + /2 + J/4


Now the corresponding eigen-vectors must be found. In other word the extent to which  and 
mix to form 2 and 3 has be determined, which involved solving for C1, C2, C3, and C3. This is
done be placing the values of each eigen value back into the eigen-equation as:
J
 ν




1
2
  C1 
ν 2 J 2 D
2


 J
 C   0 where 1 
ν
4
4
2

 1  2 

 2

2
 ν D
2 2
 J

 2

Now we have to solve for C1 and C2. Lest reduce the matrix to a upper-triangular form:


1




0



J

  C1 
2
0
ν D  C2 

 
2 2



ν D
 C 

2 2
 1   0.
ν D
J  C2 


2 2 
2 
J
ν D 
 
2
2 2
J
2
The first row states that:
 D ν 
  
D  ν  C
2 2 
C1 
C2  0  C2  
C1  
1
ν D
J
J

2 2
2
J
2

Now lets assume that C1 and C2 are of trigonometric form, such as cos and sin, respectively. Then
the following process can be pursued:
sin  

Recall that:
tan 2 
tan 2 
 D  ν  cos
J
2 tan 
1  tan 2 
. Therefore
 tan  
 D  ν 
J
 D  Δν  
2 

J

 .
tan 2 
2
 D  Δν  
1 

J


 2JD  Δν 
 2JD  Δν 
 2JD  Δν 


2
J 2  - D 2 - 2Dν  ν 2 J 2   ν 2  J 2 - 2Dν  ν 2
J 2  D  Δν 




tan 2 

 2JD  Δν 
 2JD  Δν 
J
.


2
2
 2ν - 2Dν  2ν - 2Dν ν
The eigen-function corresponding to energy level 2 is of the form:
2  cos   sin  
which has frequency :
ν 2 J 2 J
ν
  
4
4 4


The other eigen-function is found in a similar way by solving the following system of equations:
 ν D
2 2
 J

 2
Thus
C3 
J

 C3 
2
0
ν D  C4 

 
2 2
D  ν  C
J
C4  0  C4 
3
ν  D
J
now assume that C3 and C4 are trigonometric function of the form -sin and cos, then it can be seen
that:
cos  
 D  ν  sin
J
tan 2 

 tan 
J
 ν-D 
2J/ Δν - D
J

1 - (J/ Δν - D) 2 ν
The eigen function corresponding to this eigen value is therefore:
3   sin    cos 

The remainder of the Hamiltonian is diagonal so therefore:
1  

and
4  
The transition frequencies are determined from differences between the energy levels, there are
four such transitions possible giving rise to four spectral lines they are:
ν ν
D J
1  2 ν12  A B  
2
2
2
ν ν
D J
1  3 ν13  A B  
2
2  4 ν 24 
2
2
νA  νB D J
 
2
2 2
ν ν
D J
3  4 ν34  A B  
2
2 2

The intensities of each transition is directly related to the transition probability. The probability of
a transition can be derived using time-dependent perturbation theory, which is beyond the scope of
this course. It will suffice to state that this transition probability is proportional to the transition
moment given by the square of the expectation value of the I+ operator between the two state in
question,
M ij  i I   j

The intensity of the transition from 1
M12  1 I  2
2
 2
2
can be evaluated as follows:
  F cos   sin 

 cos    I
2
 cos   I1  I 2   sin   I1  I 2 
1


2
   I 2    sin    I1    I 2 
 cos  0      sin      0
2
 cos   sin    1  2 cos  sin   1  sin 2
2

Similarly it can be shown that:
M13  1  sin 2 , M 24  1  sin 2 , and M 34  1  sin 2 .

2
The Frequencies of transition in an AB Spectrum
2 to 4
-J/2
1 to 2
+J/2
1 to 3
+J/2
3 to 4
-J/2

D/2
-D/2

As the coupling constant increases with respect to the frequency difference in the spectrum, the
central two line come closer together an become more intense while the outer lines diverge and
diminish in size.

In the limit of very large couplings, or vanishingly small chemical shift differences, the two central
lines coalesce into one line at the average chemical shift, while the outer line disappear, i.e. 2 =
90o. When 2 = 90o degrees 2 and  3 are degenerate, thus they will form symmetric and
antisymmetric combinations of as follows:
2  0.71   0.71 
and 3   0.71   0.71 
.
The energy level diagram in the extreme limit of strong coupling.
A + /2 + J/4

J/4
 0.71
J/4
 0.71
A + /2 + J/4

Symmetric
Anitisymmetric
The AB spectrum as the ratio of J to  is varied from 10 to 0.
/J = 10
/J = 3
/J = 2
600
500
400
300
Frequency (Hz)
200
100
0
/J = 0.5
/J = 0.1
/J = 0
600
500
400
300
Frequency (Hz)
200
100
0
Three spin systems

There are three main possibilities for a three spin system, that is non equivalent, it can be either a
AMX, ABX, ABC the Hamiltonians are given below:
X
A M
A X
M X
ˆ  ν AI zA  ν MI M
AMX: H
z  ν X I z  J AMI z I z  J AXI z I z  J MX I z I z
A 0 0 0 0 0 0 0 
0 B 0 0 0 0 0 0


0 0 C 0 0 0 0 0


0
0
0
D
0
0
0
0

H
0 0 0 0 E 0 0 0


0
0
0
0
0
F
0
0


0 0 0 0 0 0 G 0


0
0
0
0
0
0
0
H


ν A  ν M  ν X J AM  J MX  J AX

2
4
ν  ν M  ν X  J AM  J MX  J AX
C A

2
4
 ν A  ν M  ν X  J AM  J MX  J AX
E

2
4
 ν A  ν M  ν X  J AM  J MX  J AX
G

2
4
A
ν A  ν M  ν X J AM  J MX  J AX

2
4
ν  ν M  ν X  J AM  J MX  J AX
D A

2
4
 ν A  ν M  ν X  J AM  J MX  J AX
F

2
4
 ν A  ν M  ν X  J AM  J MX  J AX
H

2
4
B

This is a diagonal matrix thus the product basis is the eigen basis and the diagonal element are
the eigen value, the intensity of each transition is the same. In other word this is a first-order
spectrum composed of the familiar three sets of doublets of doublets centered at A, andX,
shown below, as suggested by the rules for determining coupling pattern discussed previously.

The corresponding energy level diagram is composed of the pure product basis states with their
corresponding energies. There is no mixing between the energy levels. This does occur for the
ABX and ABC case.

The are 12 possible transitions in the AMX spectrum. We will see that there are more transitions
possible in the case of ABX and AMX. There are four transitions corresponding to each nucleus.
The Energy Level Diagram for an AMX Spin System
H

8
G

7
F
6

5
E
11

12
D
9

10
C
4

3
2
B

1
A
M
JAM
JAM
JAX
9
JAX
10
11
12
A

JAX
JMX
2
JMX
3
X
6 7
JMX
1
JMX
4 5 8
ˆ   ν A I zA  ν BI Bz  ν X I zX  J ABI zAI Bz  J AXI zAI zX  J BX I Bz I zX
H
ABX:



J AB A B
I  I   I AI B
2

This matrix is mostly diagonal, apart from two two-by-two blocks, which have to be
diagonalized. This can be done analytically just as was done with the AB case.
0
0
0
0
0 0
A 0
0 B
0
0
0
0
0 0


0 0
C
0
J AB / 2
0
0 0


0
0
0
D
0
J
/
2
0
0
AB

H
 0 0 J AB / 2
0
E
0
0 0


0
0
0
J
/
2
0
F
0
0
AB


0 0
0
0
0
0
G 0


0
0
0
0
0
0
0
H



This matrix can be rearranged to show the block-diagonal structure as follows:
0
0
0
0
0 0
A 0
0 B
0
0
0
0
0 0


0 0
C
J AB / 2
0
0
0 0


0
0
J
/
2
E
0
0
0
0
AB

H
0 0
0
0
D
J AB / 2 0 0 


0
0
0
0
J
/
2
F
0
0
AB


0 0
0
0
0
0
G 0


0
0
0
0
0
0
0
H



The eigen functions of this Hamiltonian are the product basis function when the matrix is
diagonal and a mixture of two when for the 2-by-2 blocks, i.e.
C1 C2C3C4D1D2D3D4

 where the coefficients Ci and Di are trigonometric function of an angle determined by in the
inverse tangent of a ration between the coupling constants and the chemical shift differences
between the nuclei in question as seen with the AB spin system.
Energy Level Diagram For an ABX Spin System

Fz= -3/2


mixing
Fz= -1/2
Fz= 1/2


Fz= -1/2


mixing

Fz= 1/2
Fz= 3/2

There are two groups of energy levels corresponding to the two possible states of the X nucleus.

Each group of energy levels resemble a two-spin-1/2 energy level diagram, where the middle
two energy levels mix.

Mixing occurs only between level with the same total z-component of angular momentum, Fz,
and that have comparable energies. This occurs between the level corresponding to  and the
since they both have Fz = -½ , however  does not mix since it is too far away in energy
despite having Fz = -½. An analogues situation exists where mixing occurs between  and
 but not for  where all have Fz= ½ .
Energy Level Diagram For an ABX Spin System
8

8
7
7
6
cossin
sincos
6
5
12
5

14
11
13
10
4
9

3
4
3
2
cossin
sincos
1
2
1


There are 14 possible transitions, 2 more than with AMX, due to the mixing between two sets of
two energy levels. Whether a transition occurs or not can determined from the transition
moment between the states in question. In the AMX spin system the transition between
andis forbidden whilst the corresponding transition in the ABX between level 2 and 7
is allowed.

The transitions an be groups into two sets one set of 6, corresponding to transition of the X,
nucleus and the remaining set of 8 corresponds to transition in A and B simultaneously, since the
energy levels are intermixed. Two of the six X have small transition moments leading to rather
small line intensity and consequently are frequently over looked.

The AB transitions are divided into two groups of four lines which appear as AB quartets (an AB
quartet centered at the average of the A and B chemical shift, where each side of the quartet is
split by the scalar coupling of X to either A or B depending on the side.).
In many cases
estimates of the chemical shifts and coupling constants for this system can be determined by
inspection of the AB and X subsystems.
AB Region of an ABX Spectrum
325
300
275
250
225
200
175
150
125
100
75
50
25
X Region of an ABX Spectrum

1675
1675
1650
1650
1625
1625
1600
1600
1575
1575
1550
1550
1525
1525
1500
1500
1475
1475
1450
1450
1425
1425
1400
1400
1375
1375
1350
1350
1325
1325
1675
1650
1625
1600
1575
1550
1525
1500
1475
1450
1425
1400
1375
1350
1325
Spectral parameters A = 100, B= 200, X = 1500, JAX= 20, JBX = 30 and JAB = 100.
ˆ   ν A I zA  ν BI Bz  ν C I Cz  J ABI zAI Bz  J ACI zAI Cz  J BC I Bz I Cz
H
ABC:







J
J
J AB A B
I  I   I AI B  AC I AI C  Iˆ AIˆ C  BC I B I C  I B I C
2
2
2

This matrix contains two three-by-three block along the main diagonal which cannot be solved
analytically, and thus have to be simulated numerically.
0
0
0
0
0
0
0
A
0
B
J BC / 2
0
J AC / 2
0
0
0


 0 J BC / 2
C
0
J AB / 2
0
0
0


0
0
0
D
0
J
/
2
J
/
2
0
AB
AC

H
 0 J AC / 2 J AB / 2
0
E
0
0
0


0
0
0
J
/
2
0
F
J
/
2
0
AB
BC


0
0
0
J AC / 2
0
J BC / 2
G
0


0
0
0
0
0
0
0
H



This matrix can be rearranged to show the 3-by-3 block structure as follows:
0
0
0
0
0
0
0
A
0
B
J BC / 2 J AC / 2
0
0
0
0


 0 J BC / 2
C
J AB / 2
0
0
0
0


0 J AC / 2 J AB / 2
E
0
0
0
0

H
0
0
0
0
D
J AB / 2 J AC / 2 0 


0
0
0
0
J
/
2
F
J
/
2
0
AB
BC


0
0
0
0
J AC / 2 J BC / 2
G
0


0
0
0
0
0
0
0
H



Thus the energy level diagram is composed of four level corresponding to total spin of –3/2, -1/2,
½ and 3/2, where the middle three levels are composed of three levels that are close in energy,
which will mix with one another. There are 15 transitions possible as shown in the diagram
below.
Energy Level Diagram of an ABC Spin System


mixing

mixing


Fz= -3/2
mixing
mixing


Fz= -1/2

Fz= 1/2
Fz= 3/2

There is only one group of energy levels divided according to their Fz value.

Mixing occurs only between level with the same total z-component of angular momentum, Fz,
and that have comparable energies. This occurs between the level corresponding to 
and  since they all have Fz = -½. An analogues situation exists where mixing occurs
between ,  and  where all have Fz= ½.

The resulting eigen functions will be of the form:




C1C2C3
C4C5 C6 C7C8 C9
D1D2D3
D4D5D6






D7D8D9
Energy Level Diagram of an ABC Spin System
8
8
7
6
12
7
4
14
10
5
15
13
6
5
3
11
3
2
1
9
4
2
1

There are 15 possible transitions, 3 more than with AMX, and 1 more than ABX due to complete
mixing of all levels with Fz= -½ and ½. In this case an additional transition is possible between
levels 4 and 5.
The Spectrum of a Sample ABC System
700
700
600
600
500
500
400
400
300
300
200
200
100
100
700
600
500
400
300
200
100

The spectral parameters are:
A = 290, B= 325, C = 375, JAC= 2=50, JBC = 150 and JAB = 100 Hz.
Equivalence: The AB2 spin system
Hamiltonian for an AB2 spin system:

 
H  ν AI z  E  E  ν B E  I z  E  E  E  I z  J I z  I z  E  I z  E  I z



J
I  I  E  I  I  E  I  E  I  I  E  I
2
which can be shown to be of the form:
J
 νA
 2  νB  2


0


0


0

H

0


0



0


0




0
0
0
0
0
0
νA
2
0
0
J/2
0
0
0
νA
2
0
J/2
0
0
0
0
νA
J
 νB 
2
2
0
J/2
J/2
0
0
νA
2
0
J/2 J/2
0

νA
J
 νB 
2
2

0
0
J/2
0
0
0
J/2
0
0
0
0
0
0
0

νA
2
0




0


0


0



0


0



0

νA
J

 νB  
2
2
0
This matrix can be rearranged to show the diagonal blocking pattern as:
J
 νA
 2  νB  2


0


0


0


H

0


0



0


0


0
0
0
0
0
0
νA
2
0
J/2
0
0
0
νA
2
J/2
0
0
0
0
0
0
J/2
J/2
0
J/2 J/2 
νA
J
 νB 
2
2
0
0
0
νA
J
 νB 
2
2
0
0
0
J/2
0
0
0
J/2
0
0
0
0
0
0

νA
2
0

νA
2
0




0


0


0



0


0



0

ν
J
 A  νB  
2
2
0


If we choose our basis carefully to exploit the symmetry of the basis representing an
equivalent pair of nuclei, we should be able to simplify the above Hamiltonian further.
Recall that an A2 spin system can be represented by a basis set composed of three symmetric
functions and one asymmetric function as follows:
and :symmetric with exchange between position 1 and 2.


antisymmetric with exchange between position 1 and 2.
(i.e it changes sign)

Using the following basis to represent the above Hamiltonian a new representation of the
Hamiltonian can be found:















The resulting Hamiltonian is:
J
 νA
 2  νB  2


0


0


0

H

0


0



0


0







0
0
0
0
0
0
νA
2
0
J
0
0
0
0
νA
2
0
0
0
0
J
0
0
0
0
0
0
0
νA
J
 νB 
2
2
0
J
0
0
0
0
νA
2
0
0
0
0
J
0
0
0
0
0
0

νA
J
 νB 
2
2


νA
2
0




0


0


0



0


0



0

νA
J

 νB  
2
2
0
This matrix can be rearranged to be block diagonal, having only two 2-by-2 diagonal block
the remainder is entirely diagonal which implied that in the energy level diagram there is
only mixing between the functions, andand , and
No
mixing
occurs
with
or
since they no longer have off-diagonal element in common with any
other basis function.
J
 νA
 2  νB  2


0


0


0


H

0


0



0


0



0
0
0
0
0
0
νA
2
0
0
0
0
0
0
νA
2
J
0
0
0
0
J
0
0
0
0
0
0
νA
J
 νB 
2
2
J
0
0
0
0
J
νA
2
0
0
0
0
0
0
0
0
0
0
0

νA
J
 νB 
2
2


νA
2
0




0


0


0



0


0



0

ν
J
 A  νB  
2
2
0
The energy level diagram is now split up into two groups which are either symmetric or
antisymmetric with respect to exchange between the two B spins. Transition between these
groups are symmetry forbidden.
The energy Level Diagram for an AB2 Spin System

Fz= -3/2
 
 
mixing

mixing
Fz= -1/2

 
 

Symmetric
Fz= 1/2
Fz= 3/2
Antisymmetric
The energy Level Diagram for an AB2 Spin System
8
7
12
6
10
5
15
5
4
7
3
11
2
3
1
9
1

Due to constraints imposed by symmetry there are only 9 possible transitions in the A2B
spectrum. Only one transition can be ascribed to a particular nucleus and that is the
transition between 3 and 7, which has frequency A.
Sample AB2 Spectrum
A
300
275

275
250
225
200
175
150
125
100
75
50
25
275
250
225
200
175
150
125
100
75
50
25
Spectral parameters are: A = 100, B = 180, and JAB= 50 Hz.
General Approaches to Simulating Static NMR Spectra.
 For an N spin system 2N product basis functions are constructed.
 These are divided into groups containing the same total z-component of angular momentum, Fz,
ranging between –N/2 to N/2. Mixing of basis function occurs within these groups. In each case
the top and bottom level contain 1 function, the number of functions in additional levels follow
the pattern of the binomial distribution. For example:
N=1 1:1
N = 2 1: 2: 1
N = 3 1: 3: 3: 1
N = 4 1: 4: 6: 4: 1
The number of transition possible between sets of levels is just the product of the number of
basis functions in each set. Consider N = 4, where there 5 sets of energy levels containing 1: 4:
6: 4: 1, members. The number of transition between these sets have the following pattern:
2 to 1 1 to 0 0 to –1 -1 to –2
Fz
Number of transitions
4
24
24
4
total = 56
In general there are: 2N!/(N+1)!(N-1)! Transitions.
N=1
1
N=2
N=3
2
3
9
2
Total 1
4
3
15
N=4
4
24
24
4
56
 Each level of total z-component of angular momentum can be further divided into groups, if
magnetic equivalence is present.
combination of the basis function.
This is achieved by using symmetry-adapted linear
 Transition are computed between all levels that differ by Fz = ± 1 and have the same symmetry
properties.
 In all for a N spin system, there are 2N basis function which gives rise to a Hamiltonian that is an
2N by 2N matrix, having 22N entries. This matrix has block structure as a result of Fz factorization
and symmetry. Each of these blocks are independently diagonalized to achieve a Hamiltonian
that is diagonal. The resulting eigen-basis and eigen-values are used to construct the spectrum.
The intensities of each line is determined from the aforementioned transition moment.