### J 2

```Angular Momentum
Classical
J
p
J r p
from origin momentum
iˆ
J x
px
ˆj
y
py
kˆ
z
pz
r
determinant form of cross product i  xˆ
j  yˆ
k  zˆ
J x  y pz  z p y
J y  z px  x pz
J z  x p y  y px
J  J  J 2  J x2  J y2  J z2
Copyright – Michael D. Fayer, 2007
Q.M. Angular Momentum
In the Schrödinger Representation, use Q.M. operators for x and p, etc.

x
P x=  i
x=x
Substituting
J  i
iˆ
ˆj
kˆ
x
y
z

x

 y

z
 
 
J x  i  y
z


z

y


 
 
J z  i  x
y


y

x


 
 
J y  i  z
x


x

z


J J  Jx  J y  Jz
2
2
2
Copyright – Michael D. Fayer, 2007
Commutators
Consider
 J x , J y   J x J y  J y J x
substituting operators in units of 
 
  
 
J x J y   y
z
z
x



z

y

x

z



 






 
  y
z
y
x
z
z
z
x


z

x

z

z

y

x

y

z


Similarly
 






 
J y J x   z
y
z
z
x
y
x
z
x  y
z z
 z  y 
  x z
Subtracting
   
 
  
 
 J x , J y     y
z

z

x
z

z 



 z
 y   z  z 
  x z
Copyright – Michael D. Fayer, 2007
   
 
  
 
 J x , J y     y
z

z

x
z

z 



 z
 y   z  z 
  x z
 
   
  y
x
, z


 y   z 
  x
 
   
x
y
, z


 x   z 
  y


 iJz 
, z
 z 
But

Therefore,


 z , z  1


1
1
P
,
z


 1  z, P z 


z
i
i

 J x , J y   i J z
 J x , J y   i J z
because
1
i
i
P

 z
 z i
Using  z , P z   i
 1
in conventional units
Copyright – Michael D. Fayer, 2007
The commutators in units of  are
 J x , J y   i J z
 J y , J z   i J x
J z , J x   i J y .
Using these it is found that
J 2 , J z   J 2 , J x   J 2 , J y   0

 
 

Components of angular momentum do not commute.
J2 commutes with all components.
Copyright – Michael D. Fayer, 2007
Therefore,
J2 and one component of angular momentum
can be measured simultaneously.
Call this component Jz.
Therefore,
J2 and Jz can be simultaneously diagonalized
by the same unitary transformation.
Furthermore,
H,J   0
(J looks like rotation)
Therefore,
H,J 2   0


H, J2, Jz are all simultaneous observables.
Copyright – Michael D. Fayer, 2007
Diagonalization of J2 and Jz
J2 and Jz commute.
Therefore, set of vectors
m
are eigenvectors of both operators.
J
2
and J z
are simultaneously diagonal in the basis  m
J m  m
2
Jz m  m m
(in units of )
Copyright – Michael D. Fayer, 2007
Form operators
J  J x i J y
J  J x  i J y
From the definitions of J  and J  and the angular momentum commutators,
the following commutators and identities can be derived.
Commutators
J  , J z   J 
J  , J z   J 
 J  , J    2J z
Identities
JJ  J  Jz  Jz
2
2
JJ  J  Jz  Jz
2
2
Copyright – Michael D. Fayer, 2007
Expectation value
m J m  m Jz m
2
2
Because
m J m  m Jz m  m Jx m  m J y m
2
2
2
2
Positive numbers because J’s are Hermitian –
give real numbers. Square of real numbers – positive.
Therefore,
the sum of three positive numbers is great than or equal to one of them.
Now
m J m 
2
 m J z  m  m2
2
Therefore,
  m2
Eigenvalues of J2 is greater than or equal to
square of eigenvalues of Jz.
Copyright – Michael D. Fayer, 2007
Using
J  , J z   J 
JzJ  JJz  J
Consider
J z  J   m   J  J z  m  J   m
 J m  m  J   m
  m  1  J   m 
eigenvalue
Furthermore,
J 2 , J    0


Then
eigenvector
J2 commutes with J+ because it commutes with Jx and Jy.
J  J   m   J  J  m
   J   m 
eigenvalue
2
2
eigenvector
Copyright – Michael D. Fayer, 2007
Thus,
J   m is eigenvector of Jz with eigenvalue m + 1
and of J2 with eigenvalue .
J+ is a raising operator.
It increases m by 1
and leaves  unchanged.
Copyright – Michael D. Fayer, 2007
Repeated applications of
J  to
m
gives new eigenvectors of Jz (and J2) with larger and larger
values of m.
But,
this must stop at a largest value of m, mmax
because
  m2 .
(m increases,  doesn’t change)
Call largest value of m (mmax) j.
mmax = j
For this value of m, that is, m = j
J  j  0
with
j 0
Can’t raise past max value.
Copyright – Michael D. Fayer, 2007
In similar manner can prove
J m
is an eigenvector of J z with eigenvalues m – 1
and of J2 with eigenvalues .
Therefore,
J  is a lowering operator.
It reduces the value of m by 1 and leaves  unchanged.
Operating J  repeatedly on  j
J  j
largest value of m
gives eigenvectors with sequence of m eigenvalues
m  j , j  1, j  2, 
Copyright – Michael D. Fayer, 2007
But,
  m2
Therefore, can’t lower indefinitely.
Must be some
 j
such that
J   j  0
with
 j  0
Smallest value of m.
Can’t lower below smallest value.
Thus,
j = j' + an integer.
largest value
of m
Went from largest value to smallest
value in unit steps.
smallest value
of m
Copyright – Michael D. Fayer, 2007
We have
largest value of m
J  j  0
J   j  0
smallest value of m
Left multiplying top equation by J  and bottom equation by J 
JJ  j  0
2
identities
J  J   j  0
Then
JJ  J  Jz  Jz
2
JJ  J  Jz  Jz
2

 0  J
2

 J   j
JJ  j  0  J  Jz  Jz  j
J  J   j
and operating

2
2
2
Jz
2
z

J  J   j  0    j2  j  j


J  J   j   0    j 2  j   j 
Copyright – Michael D. Fayer, 2007


J  J   j  0    j2  j  j


J  J   j   0    j 2  j   j 
Because  j  0 and  j   0
the coefficients of the kets must equal 0.
Therefore,
  j ( j  1)
and
  (  j)(  j  1)
Because j > j'
j   j
and
2j = an integer
because we go from j to j' in unit steps with lowering operator J  .
Thus, the eigenvalues of J2 are
  j ( j  1)
and
The eigenvalues of Jz are
largest m
j  0,
1
3
, 1, ,
2
2
m  j , j  1,
(largest m for a )
,  j  1,  j
change by unit steps
smallest value of m
Copyright – Michael D. Fayer, 2007
Final results
J
2
jm  j ( j  1) j m
Jz jm  m jm
There are (2j + 1) m-states for a given j.
Can derive
J jm 
 j  m  jm1
j m1
J jm 
 jm  j  m1
jm 1
Copyright – Michael D. Fayer, 2007
Angular momentum states can be grouped by the value of j.
Eigenvalues of J2,  = j(j + 1).
j  0, 1/2, 1, 3/2, 2, 
j0
m0
00
j  1/2
m  1/2,  1/2
1 1
2 2
j 1
m  1, 0,  1
1 1 1 0 1 1
j  3/2 m  3/2, 1/2,  1/2,  3/2
3 3
2 2
j2
22
m  2, 1, 0,  1,  2
1 1

2 2
31
2 2
21
3 1

2 2
20
3 3

2 2
2 1
2 2
etc.
Copyright – Michael D. Fayer, 2007
Eigenvalues of J 2 are the square of the total angular momentum.
The length of the angular momentum vector is
j ( j  1)
j ( j  1)
or in conventional units
Example
j=1
Eigenvalues of Jz are the
projections of the angular
momentum on the z axis.
z
m=1
2
m=0
m = -1
Copyright – Michael D. Fayer, 2007
The matrix elements of J
jm J
2
2
J z J  J  are
jm  j ( j  1)  jj m ,m
j m  J z jm  m  jj m ,m
jm J  jm 
 j  m  j  m  1
  m,m 1
jm J  jm 
 j  m  j  m  1
  m,m 1
jj
jj
The matrices for the first few values of j are (in units of  )
j = 1/2
j = 0
J   (0)
J   (0)
J z  (0)
J  (0)
 0 1
J  

 0 0
 0 0
J  

 1 0
2
3
0 
 1/2
2
 4
J

Jz  

 0
 0 1/2 

0 

3 
4
Copyright – Michael D. Fayer, 2007
j = 1
0

J  0
0


2
0
0
0 
 0


2 J   2
 0
0 


1 0 0 


Jz  0 0 0 
 0 0 1 


0
0
2
0

0
0 
 2 0 0
2


J   0 2 0
 0 0 2


The j m are eigenkets of the J 2 and J z operators – diagonal matrices.
The raising and lowering operators J  and J  have matrix elements
one step above and one step below the principal diagonal, respectively.
Copyright – Michael D. Fayer, 2007
Particles such as atoms
  R( r )Y m ( , )
spherical harmonics from solution of H atom
The Y m ( , ) are the eigenvectors of the operators
L2 and Lz.
The
Y m ( , )  j m 
m
L Y m ( , )  (  1)Y m ( ,  )
2
L z Y m ( , )  mY m ( ,  )
Copyright – Michael D. Fayer, 2007
Examples
Orbital and spin angular momentum -  and s.
These are really coupled – spin-orbit coupling.
NMR – one proton spin coupled to another. Not independent.
ESR – electron spins coupled to nuclear spins
Inorganic spectroscopy – unpaired d electrons
Molecular excited triplet states – two unpaired electrons
Could consider separate angular momentum vectors
j1 and j2.
These are distinct.
But will see, that when they are coupled, want to combine
the angular momentum vectors into one resultant vector.
Copyright – Michael D. Fayer, 2007
Specific Case
j1 
1
2
m1  
1
2
j2 
1
2
m2  
1
2
Four product states
j1 m1 j2 m2
1 1 1 1
2 2 2 2
m1m2

1 1 1 1

2 2 2 2
1 1 1 1

2 2 2 2
1 1 1 1


2 2 2 2


1 1
2 2
j1 and j2 omitted because
they are always the same.
1 1

2 2
Called the m1m2 representation

1 1
2 2
The two angular momenta are
considered separately.
1 1
  
2 2
Copyright – Michael D. Fayer, 2007
m1 m 2
j1 j2 m1 m2
Want different representation
m1m2 representation
Unitary Transformation to
coupled rep.
New States labeled jm
j1 j2 jm  jm
jm representation
Copyright – Michael D. Fayer, 2007
jm
Eigenkets of operators
J
2
and
Jz
where
J  J1  J 2
J z  J 1z  J 2 z
J
2
jm  j  j  1 jm
vector sum of j1 and j2
J z jm  m jm
Want unitary transformation from the m1m2 representation
to the jm representation.
Copyright – Michael D. Fayer, 2007
Want
jm 
C
m1m2
m1m2
m1m2
C m1m2  m1 m2 jm
Cm1m2 are the Clebsch-Gordan coefficients; Wigner coefficients;
vector coupling coefficients
m1 m 2
are the basis vectors
N states in the m1m2 representation
N states in the jm representation.
N  (2 j1  1)(2 j2  1)
J2 and Jz obey the normal commutator relations.
Prove by using J  J 1  J 2
and cranking through commutator relations
using the fact that J1 and J2 and their components commute.
Operators operating on different state spaces commute.
Copyright – Michael D. Fayer, 2007
Finding the transformation
J z  J 1z  J 2 z
m  m1 m2 or coupling coefficient vanishes.
To see this consider
jm 
C
m1m2
m1m2
m1m2
Operate with Jz
equal
J z jm  m jm   J 1 z  J 2 z   Cm1m2 m1m2
m1m2

These must be equal.
Other terms
C m1m2  0
 m
1
 m2  Cm1m2 m1m2
m1m2
if
m1  m2  m
Copyright – Michael D. Fayer, 2007
Largest value of m
m  j1 + j2  m1max  m2max
since largest
m1  j1
and
m2  j2
Then the largest value of j is
j  j1  j2
because the largest value of j equals the largest value of m.
There is only one state with the largest
j and m.
There are a total of (2j + 1) m states associated with the largest j  j1  j2 .
Copyright – Michael D. Fayer, 2007
Next largest m (m – 1)
m  j1  j2  1
But
m  m1 m2
Two ways to get m - 1
m1  j1 and m2  j2  1
m1  j1  1 and m2  j2
Can form two orthogonal and normalized combinations.
One of the combinations belongs to
j  j1  j2
Because this value of j has m values
m  ( j1  j2 ), ( j1  j2  1),
,(  j1  j2 )
Other combination with m  j1  j2  1
j  j1  j2  1
with
m  ( j1  j2  1), ( j1  j2  2),
largest
,( j1  j2  1)
smallest
Copyright – Michael D. Fayer, 2007
Doing this repeatedly
j values from
j  j1  j2 to j1  j2
in unit steps
Each j has associated with it,
its 2j + 1 m values.
Copyright – Michael D. Fayer, 2007
Example
j1 
1
,
2
j2 
1
2
j  j1  j2 to j1  j2
j values
j
1 1
 1
2 2
j
1 1
 0
2 2
j 1
m  1, 0,  1
j0 m0
jm rep. kets
11 , 1 0 , 0 0 , 1 1
m1m2 rep. kets
11 1 1
1 1
1 1
,  , 
,  
22 2 2
2 2
2 2
Know jm kets
still need correct combo’s of m1m2 rep. kets
Copyright – Michael D. Fayer, 2007
Generating procedure
Start with the jm ket with the largest value of j and the largest value of m.
11
J z 11  1 11
But
m = 1
m  m1 m2
Therefore,
1
1
m2 
2
2
because this is the only way to get
m1 
m1 m2  1
Then
11 
jm
1 1
2 2
m 1 m2
Clebsch-Gordan coefficient = 1
Copyright – Michael D. Fayer, 2007
Use lowering operators
J   J 1  J 2
jm
m 1 m2
11 
1 1
2 2
jm
m 1 m2
J  11  2 10
from
lowering
op. expression
  J 1  J 2 
11
11
11
 J 1
 J 2
22
22
22
1 
1 1
1 1
1 
2 2
2 2
Then
10 
1 1 1
1
1 1
 

2 2 2
2 2 2
Clebsch-Gordan Coefficients
from
lowering
op. expression
(Use correct ji and mi values.)
Copyright – Michael D. Fayer, 2007
Plug into raising and lowering op. formulas correctly.
J jm 
 j  m  jm1
j m1
J jm 
 jm  j  m1
jm 1
For jm rep.
plug in j and m.
For m1m2 rep.
m1 m2
jm
m1 m 2
means
j1 j2 m1m2
For J 1 and J 2 must put in
j1 and m1 when operating with J 1
and
j2 and m2 when operating with J 2
Copyright – Michael D. Fayer, 2007
Lowering again
J  1 0  2 1 1
m1 m2
  J 1  J 2  
m1 m2
1  1 1
1 1 





2 2 
2 2 2
 1
1 1
1
1 1 



00
  
2 2 2 
 2 2 2
Therefore,
1 1  
jm
1 1

2 2
m 1 m2
Have found the three m states for j = 1 in terms of the m1m2 states.
Still need 00
m  0  m1  m2
Copyright – Michael D. Fayer, 2007
Need jm
00
m0
 m1 m2  0
Two m1m2 kets with m1 m2  0
1 1
11
 , 
2 2
22
The 0 0 is a superposition of these.
Have already used one superposition of these to form 10
10 
1 1 1
1
1 1
 

2 2 2
2 2 2
0 0 orthogonal to 10 and normalized. Find combination of 1  1 ,  1 1
normalized and orthogonal to 10 .
2 2
22
1 1 1
1
1 1

00 


2
2
2
2 2 2
Clebsch-Gordan Coefficients
Copyright – Michael D. Fayer, 2007
Table of Clebsch-Gordan Coefficients
1
1
j1=1/2
j2=1/2


1
1
2
1
2
1

1
0
0
0
1
1
1
2
1
2
1
2
1
2
1
2
1
2
1
2
2

2
2
m1
m2
1 j
1 m
1
Copyright – Michael D. Fayer, 2007
Next largest system
1
2
1 1
m2  , 
2 2
j1  1
j2 
m1  1,0,  1
m1m2 kets
1
1
2
1
1
2
0
1
2
0
1
2
1
1
2
1 
1
2
jm states
j  j1  j2 
3
2
m
j  j1  j2 
1
2
m
jm kets
33
22
3 1
1
3
,
,  , 
2 2
2
2
31
22
1
1
, 
2
2
3 1

2 2
3 3

2 2
11
22
1 1

2 2
Copyright – Michael D. Fayer, 2007
Table of Clebsch-Gordan Coefficients
3
2
3
2
1
2
3
2
1
2
1
2
1
2
1
3
2
3
0
1
2
2
3
0
2
1
2
3
1
1
2
1
3
j1 = 1
j 2 = 1/2
1
2
1
1

3
2

1
2
1
2

1
2

3
2
j
3
2
m
1

1
3
1
3

2
3
1
1  2
1
m1 m2
jm
Example
11

22
m1 m2
m1 m2
2
1
1 1
1

0
3
2
3 2
Copyright – Michael D. Fayer, 2007
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