Chem 526
NMR for Analytical Chemists
Lecture 10
(Chapter 2)
Home work6A (Due 10/6)
• Prove the following relationships
Q1. [A, A] = 0
Q2. [A+B, C]= [A, C] + [B, C]
Q3. [A, BC]= [A, B]C + B[A, C]
Q4. [IX, IX] = 0
Q5. [IX + IY, IX] = -iIz
Q6. [IY ,[ IY, IX] ]= IX
Q7. [IY, [IY ,[ IY, IX] ]]= -iIZ
1
Homework 6B (Due 10/6)
Q1. Confirm
exp(-iαI
e
p( iαIX)
= E + {-iα/2}2IX + {(-iα/2)2/2}E + {(-iα/2)3/3!} 2IX + …
= Ecos(α/2) –i2IXsin(α/2)
Q2. Prove that
exp(-iαIX) IZ exp(iαIX) = Izcosα - IYsinα
Ex. Free Particle
• The Schrödinger equation for a free particle is
given by
(-2/2m)(d2(x) /dx2)= E(x), (2)
where
h
E iis a constant.
t t
• A solution of this equation is given by
(x) = Acos(kx) + Bsin(kx)
(3)
or
=Cexp(ikx) +D exp(-ikx) (3’)
with
k = (2mE)1/2/
(4)
• From E =p2/2m, the wave length, 2/k is
 = 2/k = h/p
(5)
2
Wave function of particle in a box
e-
-C=C-C=C922/(2mL2)
E3
422/(2mL2)
E2
22/(2mL2)
E1
En= n222/(2mL2)
E (n  n+1) = En+1 - En
= {(n+1)2 – n2}22/(2mL2)
Q1. Assume that you put one electron or neutron in a box.
Is E (n  n+1) larger for an electron or a neutron?
2.1.2 Eigenvalue Equation
 Eigen values & eigen functions
For operator A, eigen function j(x) is given
[2.10]
Aj(x) = ajj(x)
aj is called [Q1]
[Q1], (j = 1
1, 2
2, .., N)
N).
 Orthonormality of eigen functions
[2.14]
*
*
 dx j i   dx i  j  [Q 2]
 Completeness of eigen set
Ψ(x) =  cjj(x) for arbitrary function
• German 101: In German, “ei” is pronounced as
“EYE”. (Ex. Albert Einstein.)
3
2.2.1 Dirac notation
• In Dirac notation, Schrödinger equation is given by
d |  (t)>
i
= H|  (t)>
dt
where |(t)> is denoted as a ket.
• The ket corresponds to a wave function (x, t). We define a
bra, which corresponds to *(x, t) as <(t)|.
• In Dirac notation, the bra |(t)> and ket <(t)| have a
relationship with wave functions *(x,t) and (x,t) as
follows:

<  (t)|
 (t)> =
dx
 *(x,t)
 (x,t)
[2.32]
• As you know well the right hand side of the above equation
is a complex number.
Hence,
<(t)|(t)> = c,
where c is a complex number.
-
Eigen vector & Eigen function
 2 0  i   i 

   2 
 0 3  0   0 
A pk
Operator
↓
A |k >
 i
i
[Q1]]
0    1[Q
0
pjt *·pk
= akpk
= [Q2]
jk
Eigen value
↓
= ak | k >
  j |  k   dx *j k [Q3]
 jk
4
Adjoint & Hermitian
When  = Aj(x),
* = j*(x)A†,
where A† is called an adjoint of A.
When A† = A (called self adjoint),
<A> is real number (<A>* = <A>)
In such a case, A is called an Hermitian operator
(pronounced ~”ermitian”)
Adjoint, Hermitian
i / 2  i   i / 2 
 0

   
 
  i / 2 0 1 1 / 2 
A |k>
1 i
 
2 1
= ak | k >
i / 2
 0

A  
i / 2 0 
 i, 1
0
i / 2 1
    i, 1
i / 2 0  2
<k |(A*)t = ak*<k |
t
i / 2
 0  i / 2  0
  

( A*)  
i
/
2
0

i
/
2
0

 

t
 A is an Hermitian
5
Quiz about Dirac notation
• In Dirac notation, Schrödinger equation is given by
[Q1]
id|(t)>/dt = H|(t)>
where |(t)> is denoted as a[Q2]
ket and H is Hamiltonian.
[Q2]
• The
ket corresponds to a wave function (x, t). We
define a[Q3]
bra, which corresponds to *(x, t) as <(t)|.
• In Dirac notation, |(t)> and <(t)| have a relationship
with wave functions *(x,t) and (x,t) as follows:

<  (t)| [Q4]
(t)> =
dx
 *(x,t)
 (x,t)
-
• As you know well the right hand side of the above
equation is a complex number. Hence,
<(t)|(t)> = c,
where c is [Q5]. From this definition,
<(t)|(t)> = c*.
[Q5]
Eigen kets and basis kets
An arbitrary wave function |Ψ(t)> can be written as
a superposition of a set of orthonormal timeindependent eigenkets or basis kets |n>
|Ψ(t)> = cn(t)|n>=cn(t)|n>,
where <m|n> = δmn and cm = <m|Ψ(t)>
[2.33]
[2.34]
Q. Fill [[Q1]] when eq.
q [2.33]
[
] is true,,
<Ψ(t)| =[Q1] <n|,
[2.33B]
6
Expectation value in Dirac notation (p38-39)
d|Ψ(t)>/dt = iH(t)|Ψ(t)>
In general, state ket |Ψ(t)> is exanded as
|Ψ(t)> = = Σck(t)| φk > = Σck(t)|k >,
where
here |φk> is an eigen ket (k =1,
1 2
2..))
The expectation value <A> is obtained by use of operator
A as
<A>=  dx  ( x , t ) * A  ( x , t ) =< Ψ(t)| A|Ψ(t)>
<A> is given by A as
 <A> = <Ψ(t)|A|Ψ(t)>
= (Σck*(t)<k|)A(Σcj*(t)|j>)
= ΣΣcj(t)ck*(t)<k|A|j>
= ΣΣcj(t)ck*(t)Akj
 In Dirac notation, operator
can be generally denoted by a
matrix Akj
Spin Operator
• For a spin-½ system, we define two basis kets:
|> and |>.
> For these basis kets,
kets spin
operators IX, IY, IZ satisfies the following
relationships:
♦ Iz|α> = ½|α> & Iz|β > = -1/2 |β>
♦ IX|α>
| = 1/2|β>
1/2|β & IX|β > = 1/2 ||α >
♦ IY|α> = i/2|β> & IY|β > = -i/2 |α>
7
Spin Operator in Matrix Forms
|
 1 


 0 
 | 1 
| 
   | I x |
 
   | I x |
1 / 2 
 0
 

0 
1 / 2
I
I
I
| 
x
1 / 2
 
 0
Z
c1 | 

| 
| 
x
x
 

 
Q. In an matrix form,
A† = (At)*
P
Prove
A = A† for
f IX, IY, IZ


 1 / 2 
0
  c
2
| 

 | c 1 *    |c
  | 
  | I
  | I


 i / 2 

0

 0
 
 i / 2
y
 0 


 1 
 | 2 
2
 c1 


 c2 
*  c 1*
c
*
2

Practice
I
| 
Z

0

 
  1 / 2
I
X
|

 0 

 
1 / 2 
1 / 2

 0

 1
 
2

 0

1 / 2
1  0


2  1
 0 


 1 / 2   1 
0 
 1
 
| 
1 
2
0




1 / 2  1 


0   0 

1
 
| 
2


|>  ½|>
8
Spin Operator
• For a spin-½ system,
[Q1] |β>
Iz|α> = ½|α> & Iz|β > = -1/2
IX|α> = 1/2|β> & IX|β > = [Q2]
1/2 |α >
[Q3]
IY|α> = i/2|β> & IY|β > = -i/2 |α>
[Q1] What is an eigen ket for Iz ?
[Q2] What is an eigen value for Iz?
[Q3] Is |α> an eigen ket for IX?
Homework 5
2) < |Iz|> = [Q2]
4) < |Iz|> = [Q4]
6) < |IX|> = [Q6]
8) < |Iy|> = [Q8]
1) <|Iz|> = [Q1]
3) < |Iz|> = [Q3]
5) < |IX|> = [Q5]
7) < |Iy|> = [Q7]
9) When |> =c1|>+ c2| >
Obtain <IX>, <IY>, <IZ>
 

| I
c 1 *,
X
|
c

2
c 1 *,
c
2
 0
* 
1 / 2
1 / 2  c1

0   c 2



 c / 2 

*  2
 c1 / 2 
9
Spin Operator in Matrix Forms
|
 1 


 0 
 | 1 
| 
   | I x |
 
   | I x |
1 / 2 
 0
 

0 
1 / 2
I
I
I
| 
x
y
Z

  | 
c1 | 
  | I
  | I
 0 


 1 
| 
| 
x
x
 

 
Q. In an matrix form,
A† = (At)*
Prove A = A† for IX, IY, IZ
 i / 2 

0

 0
 
 i / 2
1 / 2
 
 0


 | 2 


 1 / 2 
  c
0
2
| 

 | c 1 *    |c
2
 c1 


 c2 
*  c 1*
c
*
2

Home work (due 10
• Prove the following relationships
Q1. [A, A] = 0
Q2. [A+B, C]= [A, C] + [B, C]
Q3. [A, BC]= [A, B]C + B[A, C]\
Q4. [IX, IX] = 0
Q5. [IX + IY, IX] = -iIz
Q6. [IY ,[ IY, IX] ]= IX
Q7. [IY, [IY ,[ IY, IX] ]]= -iIZ
10
Commutation Rule of Spin Matrix
• [A, B] = AB -BA
• [Iz, IZ] = 0
IZIZ =
 1

 2
 0

 1

 2
1 

 0
2 
IXIX =

 0

 1
 2
1 
 0
2 
1
0  
 2
0


 Q 1
  
1 
 Q 3


2 
0
1 

2    Q 5
 Q 7
0 

Q 2 

Q 4 
Q 6 

Q 8 
• IXIX = IZIZ = [Q1]
Commutation Rule of Spin Matrix
• [Iz, IX] = iIY
 1
IzIX =  2
 0


 0

1  1


2  2
0
1 

2  
0 


 0

  1

4
1 

4 
0 

IXIZ = [Q1]
•
•
•
•
[IY, IZ] = iIX
[IZ, IX] = [Q2]
[IY, IX] = [Q3]
[IZ, IZ] = [Q4]
11
Spin Operator in Matrix Forms
1
|   | 1   
0
|   | 2 
   | Ix | 
Ix  
   | Ix | 
1Q
/ 2 2 
 0
  Q 1

0Q 4 
 1Q/ 23

I
I
y
Z
  | Ix |   

  | Ix |   
IX = 1/2 |>< | + 1/2|>< |

 Q0 5
 
 iQ/ 27
i /6
2 
Q

0
Q 8 
1 / 2
 
 0
0


 1 / 2 

 0 
 
1 

 Q9 

|   c 1 |    c 2 |   
 Q10 
  |   | Q 3    |c 2 *  Q 11 , Q 12

Spin Operators in Ket-Bra Notation
| 
|
 0
 1
 | 


 Q 1
 Q 3
 | 
 1
Q 5  
 0
1 / 2
Q 6  
 0
0 

0 
Q 2 

Q 4 
0 

0 


 1 / 2 
0
12
Spin Operator in Ket-Bra Notation II
|
 1 
|   | 2 


 0 
0


 1 / 2 
 | 1 
1 / 2
 
 0
I
Z
I
Z


1 / 2 
I
Z


 1 / 2 

I
 1 / 2 
Z

[Q 1]  1 / 2 




[Q 2 ]   i / 2 
[Q 3 ] 
 0 


 1 
 1 / 2 
 1 / 2 




 i / 2 




2.1.3 Commutation Rule & Simultaneous Eigen Functions
Eigen ket of operator A is not necessarily for eigen ket for
operator B. (Q. List any example?)
When [[A, B]] =0, operators
p
A and B can be
diagonalized by the same basis set |k> if
eigen values of A are all different (not
degenerate).
Assume A|n>
A|n = an|n
|n> and <k|A
k|A = <k|a
k|ak
0 = <k|(AB – BA)|n> = <k|akB|n> - <k|Ban|n>
= (ak – an)<k|B|n>
 <k|B|n> = 0 when k  n
13
Expectation value in Dirac notation
d|Ψ(t)>/dt = iH(t)|Ψ(t)>
In general, state ket |Ψ(t)> is exanded as
|Ψ(t)> = = Σck(t)| φk > = Σck(t)|k >,
where
here |φk> is an eigen ket (k =1,
1 2
2..))
The expectation value <A> is obtained by use of operator
A as
<A>=  dx  ( x , t ) * A  ( x , t ) =< Ψ(t)| A|Ψ(t)>
<A> is given by A as
 <A> = <Ψ(t)|A|Ψ(t)>
= (Σck*(t)<k|)A(Σcj*(t)|j>)
= ΣΣcj(t)ck*(t)<k|A|j>
= ΣΣcj(t)ck*(t)Akj
 In Dirac notation, operator
can be generally denoted by a
matrix Akj
Density function & Expectation values
id|Ψ(t)>/dt = H(t)|Ψ(t)>
|Ψ(t)> = Σcj(t)|φj>
<A> = <Ψ(t)|A|Ψ(t)> = ΣΣcj(t)ck*(t)<k|A|j>
Density function approach:
Density matrix for a pure state
(t)  |Ψ(t)><Ψ(t)|
= Σck*(t)ci(t)|i><k|
<A> = Tr(A(t)) = Σ<m|A(t)|m>
= Σc*k(t)cm(t)<k|A|m>
Note: Tr(B) = <n|B|n>
14
Ensemble and Quantum Statistics
If more than one systems coexist, the
ensemble
bl average yields
i ld
<A> = Σ wm<Ψm(t)|A|Ψm(t)>
In a density function approach, we just need
to redefine (t)
(t) for a mixed state as
(t)  Σ wm m
= Σ wm|Ψm(t)><Ψm(t)|
<A> = ΣwmTr(Am(t)) =Tr(A(t))
Density Matrix for Spin-1/2
α = |α><α|
β = |β><β|
= wα|α><α| + wβ |β><β|
= (1/2 + /2) |α><α|+ (1/2 - /2) |β><β|
= (1/2)E +  Iz
β
α
15
Bulk Magnetic Moment (Spin Polarizatio
• Nα /Nβ = exp(-γħB0/kT)
~ 1 - γħB0/kT (γħB0/kT<<1)
(Nα - Nβ) = N(Nα - Nβ)/ (Nα + Nβ)
= N(Nα/Nβ - 1)/(Nα/Nβ+1)
= NγħB0/2kT
 Population Difference
 B0, N, 1/T, and γ
M0 = μZNm = - γħB0/2 (Nα - Nβ)
= Nγ2ħ2B0/4kT
Bulk Magnetic Moment
  B0, N, 1/T, and γ2
Density Matrix for Spin Operator
• Let’s think about the density matrix in an
equilibrium
q
state,, (0)
( ) for a single
g spin.
p
• (0) = P|><| + P|><|
~ {(1 - E/kT)/2}|><| +
{(1 - E/kT)/2}|>< |
= E/2 - γB0/(kT)
{1/2|><| - 1/2|>< |}
= E/2 - {γB0/kT} Iz
16
Comparison: density matrix (t) and state
ket |(t)>
Time dependence of a state ket
• d|(t)>/dt = -iH|(t)>
 |(t)> = exp(-iHt)|
exp( iHt)|(0)>
In contrast
• (t) = |(t)> <(t)| = exp(-iHt)|(0)><(0)|exp(iHt)
= exp(-iHt)[Q1] exp(iHt)
• U = exp(-iHt) is called time-evolution operator
Time dependence of density operator is governed by
• d(t)/dt = -i[H, (t)] (Liouville-von Neuman eq. )
Time-Evolution of Density Matrix
(t) = Pm|Ψm(t)><Ψm(t)|
• For each |Ψm(t)>, Schrödinger equation is given by
d|Ψm(t)>/dt = -iH|Ψ
iH|Ψm(t)>.
(t)>
This yields
|Ψm(t)> = exp(-iEmt)|Ψm(0)> = exp(-iHt)|Ψm(0)>.
The corresponding ket is
<Ψm(t)| = <Ψm(0)|exp(iHt)
• Hence,
(t)
=Pm exp(-iHt)|Ψm(0)><Ψm(0)|exp(iHt) =
exp(iHt){Pm|Ψm(0)><Ψm(0)|}exp(iHt)
= exp(-iHt) (0) exp(iHt)
[2.55]
17
Motions of σ(t)
d(t)/dt = -i[H, (t)]
In NMR, Zeeman Interaction yields
Emz = -γB0mz 
From Schrödinger Equation
E|> = H|>
H = -γB0Iz
• Assume σ(0) =Iz, what happens?
• If σ(0) =IX, what happens?
1.2 Bloch Equation
• The time dependence of the bulk magnetic
moment M(t) in a magnetic field is given by
Bloch equation:
dM(t)/dt = M(t)  B(t), [1.11]
When the magnetic field is a static magnetic field
applied along the z-axis, it is described as
dM(t)/dt = M(t)  B0.
• In the equilibrium state, M(0) is parallel to B0. In
this case,
dM(t)/dt = M(0)  B0 = 0.
Thus, no thing will happen
18
What is exp(-iHt)?
(t) = exp(-iHt) (0) exp(iHt)
exp(A)
p( ) =1  A  A

2
/ 2  A
3
[2.55]
/ 3 !  ..


A
n
/ n!
n  0
exp(-iγB0Izt) =
1 -iγB0Izt
+(-iγB0Izt)2/2
+ (-iγB0Izt)3/3!
+ (-iγB0Izt)4/4!
+ ….
Matrix calculation of Izn
0 
1 / 2

I z  
 0  1/ 2
0 1 / 2
0  Q1 Q 2
1 / 2
2

  
I z  

0
1
/
2
0
1
/
2




 Q3 Q 4
Q5 Q 6
3
Iz  
  [Q9]I z
Q 7 Q8
Q10 Q11

  [Q14]
Q12 Q13
Q15 Q16
2 m 1
Iz

  [Q19]
Q17 Q18
Iz
2m
19
exp(-iγB0Izt) =
1 -iγB0Izt +(-iγB0Izt)2/2 + (-iγB0Izt)3/3! + ….
Using (2Iz)2 = E
exp(-iαI
exp(
iαIz)
= E +{-iα/2}2Iz + {(-iα/2)2/2}E + {(-iα/2)3/3!} 2Iz +
 aismatrix ?]0
= Ecos(α/2) –i2Izsin(α/2) = [Q1
 c in



0

c  is 
exp(-iαIz) Ix exp(iαIz)
= IXcosα+ IYsinα
exp(-iαIz) IY exp(iαIz)
= [Q1]
Iycosα - IXsinα
Homework 6B
Q1. Confirm
exp(-iαI
( i IX)
= E +{-iα/2}2IX + {(-iα/2)2/2}E + {(-iα/2)3/3!}
2IX + …
= Ecos(α/2) –i2IXsin(α/2)
Q2. Prove
exp(-iαIX) IZ exp(iαIX)
= Izcosα - IYsinα
20
Simple Description of Transition
exp(-i Iz)|> = [Q1]
exp(-iIX)|> = {Ecos(/2)–i2IXsin(/2)}|>
|>
= cos(/2)[Q2]
|> -isin(/2)[Q3]
IX = ½|><| + ½|><|
exp(-iαIz) Ix exp(iαIz)
= IXcosα+ IYsinα ??
0 
 c  is


c  is 
 0
 0 1/ 2


1 / 2 0 
1/
0 
 c  is


c  is 
 0


0
0  
c  is  c  is
1 0

 
 
2  c  is
0  0
c  is   c  is 2

2

cos   i sin  

0


2


 cos   i sin 

0


2


c  is 2 
2
0




21
Ensemble and Quantum Statistics
If more than one systems coexist, the
ensemble
bl average yields
i ld
<A> = Σ wm<Ψm(t)|A|Ψm(t)>
In a density function approach, we just need
to redefine (t)
(t) as
(t)  Σ wm m
= Σ wm|Ψm(t)><Ψm(t)|
<A> = ΣwmTr(Am(t)) =Tr(A(t))
Density Matrix for Spin-1/2
α = |α><α|
β = |β><β|
 = wα|α><α| + wβ |β><β|
22
Density Matrix for Spin Operator
• Let’s think about the density matrix in an
equilibrium
q
state,, (0)
( ) for a single
g spin.
p
• (0) = P|><| + P|><|
~ {(1 - E/kT)/2}|><| +
{(1 - E/kT)/2}|>< |
= E - γB0/(kT)
{1/2|><| - 1/2|>< |}
= E - {γB0/kT} Iz
General Formula
Baker Campbell-Hausdorff lemma
B’ = exp(-iαA) B exp(iαA)
= B(0) + ((-iα)[A,
iα)[A B]+ (-iα)
( iα)2/2[A,[A,
/2[A [A B]]+
3
+(-iα) /3![A,[A,[A, B]]]+
[2.70]
(t)= exp(-iαIZ) IX exp(iαIZ)
= IX + (-iα)[IZ, IX]+ (-iα)2/2[IZ, [IZ, IX]]+ …
= IX + (-iα)iIY + (-iα)2/2 [IZ, iIY] +
= IX + (-iα)iIY + (-iα)2/2 IX + (-iα)3/3! iIY+
= IX (-iα)2m/(2m)! + iIY (-iα)2m+1/(2m+1)!
= IX [Q1]
+ iIY [Q2]
= [Q3]
23