This document provides an introduction to the quantum theory of angular momentum. It begins by discussing how angular momentum arises in quantum mechanics when moving from one-dimensional to three-dimensional systems. It then defines the components of angular momentum as linear operators and establishes their commutation relations. Next, it introduces raising and lowering operators and shows how they relate to the components. It also proves several theorems, such as that the square of the total angular momentum commutes with its individual components. Finally, it discusses how angular momentum eigenstates can be rotated using rotation operators.

1. Introduction to Quantum
Theory of Angular Momentum
1

2. Angular Momentum
 AM
begins to permeate QM when you
move from 1-d to 3-d
 This discussion is based on postulating
rules for the components of AM
 Discussion is independent of whether
spin, orbital angular momenta, or total
momentum.
2

3. Definition
An angular momentum, J, is a linear
operator with 3 components (Jx, Jy, Jz)
whose commutation properties are defined
as
 

J × J = iJ
3

4. Or in component form
ˆ ˆ
 J y , J z  = J y J z − J z J y = iJ x


ˆ ˆ
 J z , J x  = J z J x − J x J z = iJ y


ˆ ˆ
 J x , J y  = J x J y − J y J x = iJ z


4

5. Convention
Jz is diagonal
For example:
 1 0 
Jz = 
 0 − 1

2

1 0 0 


J z =  0 0 0 
 0 0 − 1


5

6. Therefore
J z j m = m j m
Where |jm> is an eigenket
h-bar m is an eigenvalue
For a electron with spin up
1 1
1 1 1
J z ,+
= +  ,+
2 2
2 2 2
Or spin down
1 1
1 1 1
J z ,− = −  ,−
2 2
2 2 2
6

7. Definition
J =J +J +J
2
2
x
2
y
2
z
These Simple Definitions
have some major
consequences!
7

8. THM
[J , J ] = 0
2
i
where i = x, y, z
Proof:
 J 2 , J x  =  J x2 , J x  +  J y2 , J x  +  J z2 , J x 

 
 
 

Recall [ A, BC ] = [ A, B ] C + B [ A, C ]
Jx, J 2  = [ Jx, J y ] J y + J y [ Jx, J y ] + [ Jx, Jz ] Jz + Jz [ Jx, Jz ]


 J x , J 2  = iJ z J y + J y (iJ z ) + ( −iJ y ) J z + J z (−iJ y )


Jz , J 2  = 0


QED
8

9. Raising and Lowering Operators
Lowering Operator
J = J x − iJ y
Raising Operator
J = J x + iJ y
+
9

10. Product of J and J +
JJ+ = J x + J y − iJ y J x + iJ x J y
2
2
JJ = J x + J y + i[ J x , J y ]
+
2
2
JJ = J x + J y − J z
+
2
2
and obviously ,
J J = J x + J y + J z
+
2
2
10

11. Fallout
JJ + J J = 2( J x + J y )
+
+
2
2
JJ + J J = 2( J − J z )
+
+
2
2
1
2
+
+
(JJ + J J) + J z = J 2
2
and the difference,
[ J,J+ ] = −2J z
11

12. Proof that J is the lowering operator
J z J jm = J z ( J x − iJ y ) jm
J z J jm = ( J z J x − iJ z J y ) jm
from 1st definition, J z J y = J y J z + iJ x
J z J j m = [( J x J z + iJ y ) − i ( J y J z + iJ x )] j m
and J z j m = m j m
J z J j m = [(mJ x + iJ y ) − (imJ y + iJ x )] j m
J z J j m = (m − 1)( J x − iJ y ) j m
J z J j m = (m − 1) J j m
J z J j m = (m − 1) j m − 1
It is a lowering operator
since it works on a state
with an eigenvalue, m, and
produces a new state with
eigenvalue of m-1
12

13. [J2,Jz]=0 indicates J2 and Jz are
simultaneous observables
J 2 j m =  2λ j m
2
2
(J x + J y ) j m = (J − J ) j m
2
2
z
( J x + J y ) j m = ( λ −  m ) j m
2
2
2
2
2
Since Jx and Jy are Hermitian, they must have real
eigenvalues so λ-m2 must be positive!
λ is both an upper and LOWER limit to m!
13

14. Let msmall=lower bound on m and
let mlarge=upper bound on m
J+ j mlarge = 0
J j msmall = 0
JJ+ j mlarge = 0
J J j msmall = 0
+
( J + J + J z ) j msmall = 0
2
x
2
y
( J 2 − J z2 + J z ) j msmall = 0
2
J 2 j msmall =  2 (msmall − msmall ) j msmall
2
2
( J x + J y − J z ) j mlarge = 0
( J 2 − J z2 − J z ) j mlarge = 0
2
J 2 j mlarge =  2 (m large + m large ) j m large
2
msmall − m small = ml2arg e + m l arg e
msmall
1 + ml arg e
=
 − ml arg e
mlarge cannot any larger
14

15. Final Relation
2
J 2 j mlarge =  2 (m large + m large ) j m large
J 2 j mlarge =  2 m large (m large + 1) j m large
and for msmall
J 2 j msmall =  2 msmall (msmall − 1) j msmall
msmall = −mlarge
J 2 j msmall =  2 (− mlarge )(− mlarge − 1) j msmall
J 2 j msmall =  2 mlarge (mlarge + 1) j msmall
So the eigenvalue is mlarge*(mlarge +1) for any value of m
J
2
j m =  j ( j + 1) j m
2
15

16. Four Properties
1) J
2
j m = j ( j + 1) j m
2
2) m = − j , − j + 1,  , j − 1, j
3) Exactly (2 j + 1) values possible
4) Since (2 j + 1) = integer , then
j = 0 , 1 , 1 , 3 , 2 ,
2
2
16

17. Conclusions
 As
a result of property 2), m is called the
projection of j on the z-axis
 m is called the magnetic quantum
number because of the its importance in
the study of atoms in a magnetic field
 Result 4) applies equally integer or halfinteger values of spin, or orbital angular
momentum
17

18. END OF LECTURE 1
18

19. Matrix Elements of J
j m′ J 2 j m =  2 j ( j + 1)δ m ,m′
j m′ J z j m = mδ m ,m′
J j m = c j m −1
Indicates a diagonal
matrix
and j m ( J 2 − J z2 + J z ) j m = c
j m J+ = c* j m − 1
2
j m ( j ( j + 1) 2 −  2 m 2 +  2 m) j m = c
2
( j ( j + 1) − m 2 + m) 2 = c
but J J = ( J − J + J z )
2
c = ( j + m)( j − m + 1) 2
jm J J jm =c c= c
+
+
*
2
2
z
2
2
19

20. Theorems
J j m = ( j + m)( j − m + 1)  j m − 1
and
J+ j m = ( j − m)( j + m + 1)  j m + 1
And we can make matrices of the eigenvalues, but these matrices are
NOT diagonal
j m − 1 J j m = ( j + m)( j − m + 1) δ m ,m −1
and
j m + 1 J+ j m = ( j − m)( j + m + 1)  δ m,m +1
20

21. Fun with the Raising and Lowering
Operators
J = J x − iJ y
J = J x + iJ y
+
J+ J
Jx =
2
+
⇒
( J - J )i
Jy =
2
+
21

22. A matrix approach to Eigenvalues
If j=0, then all elements are zero! B-O-R-I-N-G!
j= 1/2
1
11
= 
 0
22
 
final m
1 −1  0
= 
1
2 2
 
1 −1 1 1
 1 1  1 1 
J
=  +  − + 1
2 2
2 2
 2 2  2 2 
1 −1 1 1
J
=
2 2
2 2
and
1 1 + 1 −1
J
=
2 2
2 2
m1
m2
1
2
−1
2
Initial m
m1
m2
so
1
0
2
−1

2
0
0
Therefore,
 0 0
J = 
1 0



What does
J + look like?
22

23. Using our relations,
Answer:
0 1
J = 
0 0



+
J + J+   0 1 
Jx =
= 
1 0

2
2

and
(J − J )i   0 − i 
Jy =
= 
i 0 

2
2

and finally ,
+
Jz =
 1 0 

 0 − 1

2

3 2  1 0 

J2 =
0 1

4 

Pauli Spin Matrices
0 1
σx =
1 0



0 − i
σy =
i 0 



1 0 
σz =
 0 − 1



23

24. J=1, An Exercise for the Students
Hint:
1
 
11 =  0 
 0
 
0
 
1 0 = 1
0
 
 0
 
1 −1 =  0 
1
 
 0 0 0


J = 2  1 0 0 
 0 1 0


24

25. Rotation Matices
 We
want to show how to rotate
eigenstates of angular momentum
 First, let’s look at translation
 For a plane wave:
ψ ( x) = e ik x x for 1 − d
px
since k x =
then for 3 − d

 
(r ⋅ p)
i

ψ (r ) = e 
25

26. A translation by a distance, A, then
looks like
 
ψ (r − a ) = e
  
( r − a )⋅ p
i


= ψ ( r )e
 
a⋅ p
−i

translation operator
Rotations about a given axis commute, so a
finite rotation is a sequence of infinitesimal
rotations
Now we need to define an operator for rotation
that rotates by amount, θ, in direction of θ
26

27. So
ˆ ˆ
if θ = x, then θ x J x
so a rotation operator looks like :
ˆ
U ( n, θ ) = e

ˆ
(θ n )⋅ J
−i

Where n-hat points along the axis of rotation
Suppose we rotated through an angle φ about the z-axis
U ( z, φ ) = e
φJ z
−i

27

28. Using a Taylor (actually Maclaurin)
series expansion
x 2 x3
e = 1+ x + + +
2! 3!
so
x
U ( z, φ ) j m = e
− iφJ z

jm
n
n
 − iφ  J z
U ( z, φ ) j m = ∑ 
jm

  n!
n =0 
∞
but
J zn j m = ( m ) j m
n
 − iφ  ( m )
U ( z, φ ) j m = ∑ 

  n!
n =0 
∞
n
n
jm
U ( z , φ ) j m = e −iφm j m
so
U ( z , φ ) = e −imφ
28

29. What if φ = 2π?
U ( z ,2π ) j m = e −2πim j m
U ( z ,2π ) j m = ± j m
"+" for j = 0, 1, 2, 3, 
1 3 5 7
"−" for j = , , , , 
2 2 2 2
The naïve expectation is
that thru 2π and no
change.
This is true only if j=
integer. This is called
symmetric
BUT for ½ integer, this
is not true and is called
anti-symmetric
29

30. Let j=1/2 (for convenience it could
be any value of j)
 0 1


2 1 0


  0 1    0 1  2  1 0 2 
2
 
= 
= 1
Jx = 
2 1 0 2 1 0 4 0 1 4

 



3  0 1 
4 
3
4
 Jx = 1
Jx = 
8 1 0
16


Jx =
− iθ 1
( J x )n
n!
n =0 
∞
U ( x, θ ) = ∑
iθ
U ( x, θ ) = 1 −
2
 0 1  θ 2  1 0  iθ 3  0 1  θ 4  1 0 

 1 0  − 2 2 2!  0 1  + 233!  1 0  + 2 4 4!  0 1  + 
















 1 0 
θ2
θ4
1 − 2 + 4 + 
U ( x, θ ) = 
 0 1  2 2! 2 4!





 0 1  θ θ 3
θ5
 − 3 + 5 + 
− i
 1 0  2 2 3! 2 5!




30

31. Using the sine and cosine relation
so
x3 x5
sin x = x − + +  and
3! 5!
θ

 cos
2
U ( x, θ ) = 
 i sin θ

2

x2 x4
cos x = 1 − + + 
2! 4!
θ
− i sin 
2  and if θ = 2π , U ( x,2π ) = − 1 0 

0 1

θ 


cos 
2 
And it should be no surprise, that a rotation of β around the y-axis is
β

cos

2
U ( y, β ) = 
 sin β

2

β

2
β
cos 

2 
− sin
31

32. Consequences




If one rotates around y-axis, all real numbers
Whenever possible, try to rotate around zaxis since operator is a scalar
If not possible, try to arrange all non-diagonal
efforts on the y-axis
Matrix elements of a rotation about the y-axis
are referred to by
d m′ m ( β )
j
32

33. And
d mj′ m ( β ) ≡ j m′ U ( y, β ) j m
Example :
1
11
= 
 0
22
 
then
β

 cos
2
d 1 1 ( β ) = (1 0 ) 
β
 sin
22

2

1
2
1
2
d 1 1 ( β ) = cos
22
β

2  1 
 
β  0 
cos  
2 
− sin
β
2
Wigner’s Formula (without proof)
β
β
(− 1) k (cos ) 2 j + m− m '− 2 k (sin ) m '− m+ 2 k
2
2
d mj′ m ( β ) = ( j + m)!( j − m)!( j + m' )!( j − m)'!∑
k ( j − m'− k )!( j + m − k )!( k + m'− m)! k!
33

34. Certain symmetry properties of d functions are useful
in reducing labor and calculating rotation matrix
j
j
d m m ' ( β ) = (−1) m − m ' d m ' m ( β )
d
j
−m' −m
d
j
m'm
−1
(β )
(π − β ) = (−1)
1
∫d
(β ) = d
j
m'm
j
m m'
(β ) d
j'
m m'
j −m'
d
j
m'm
(β )
2
( β ) d (cos β ) =
δ j j'
2 j +1
34

35. Coupling of Angular Momenta
 We
wish to couple J1 and J2
 

J1 + J 2 = J 3
 From
Physics 320 and 321, we know
j1 − j2 ≤ j3 ≤ j1 + j2
      
2 + 3 = 1, 2, 3, 4, 5
 But
since Jz is diagonal, m3=m1+m2
35

36. Coupling cont’d



The resulting eigenstate
is called
And is assumed to be
capable of expansion of
series of terms each of
with is the product of 2
angular momentum
eigenstates conceived of
riding in 2 different
vector spaces
Such products are called
“direct products”
j3 m3
j1 m1 j2 m2
36

37. Coupling cont’d



The separateness of
spaces is most apparent
when 1 term is orbital
angular momentum and
the other is spin
Because of the
separateness of spaces,
the direct product is
commutative
The product is
sometimes written as
j1 m1 ⊗ j2 m2
37

38. Proof of commutative property
Let
c = a ⊗b = a b
cc = a b a b = a a b b
cc = b b a a = b a b a
c = b a = b ⊗a
38

39. The expansion is written as
j j
j3 m3 = ∑ Cm11 m22 j3 j1 m1 j2 m2
m1
C
j1 j2 j3
m1 m2
Is called the Clebsch-Gordan coefficient
Or Wigner coefficient
Or vector coupling coefficient
Some make the C-G coefficient look like an inner product, thus
j j
Take j3 m3 = ∑ Cm11 m22 j3 j1 m1 j2 m2
and multiply by j1 m1 j2 m2
m1
and thus
j j
j1 j2 j3 m1 m2 j3 m3 = Cm11 m22 j3
39

40. A simple formula for C-G
coefficients
j j
Cm11 m22 j3 =
( j3 + j1 − j2 )!( j3 − j1 + j2 )!( j1 + j2 − j3 )!( j3 − m3 )!( j3 + m3 )!( 2 j3 − 1) ×
( −1) k + j2 + m2 ( j3 + j2 − m1 − k )! ( j1 − m1 + k )!
∑ ( j − j + j − k )! ( j + m − k )! k! (k + j − j − m )!
k
3
1
2
3
3
1
2
3
•Proceeds over all integer values of k
•Begin sum with k=0 or (j1-j2-m3) (which ever is larger)
•Ends with k=(j3-j1-j2) or k=j3+m3 (which ever is smaller)
•Always use Stirling’s formula log (n!)= n*log(n)
Best approach: use a table!!!
40

41. What if I don’t have a table?
And I’m afraid of the “simple” formula?
Well, there is another path… a 9-step
path!
41

42. 9 Steps to Success
1.
Get your values of j1 and j2
2.
Identify possible values of j3
3.
Begin with the “stretched cases” where
j1+j2=j3 and m1=j1, m2=j2 , and m3=j3, thus |j3
m3>=|j1 m1>|j2 m2>
4.
From J3=J1+J2,, it follows that the
lowering operator can be written as
J 3=J 1+J 2
42

43. 9 Steps to Success, cont’d
5.
6.
7.
8.
Operate J 3|j3 m3>=(J 1+J 2 )|j1 m1>|j2 m2>
Use
j m − 1 J j m = ( j + m)( j − m + 1)
Continue to lower until j3=|j1-j2|, where m1=-j1 , m2=
-j2, and m3= -j3
Construct |j3 m3 > = |j1+j2 -1 j1+j2-1> so that it is
orthogonal to |j1+j2 j1+j2-1>
Adopt convention of Condon and Shortley,
if j1 > j2 and m1 > m2 then
Cm1 m2j1 j2 j3 > 0
(or if m1 =j1 then coefficient positive!)
43

44. 9 Steps to Success, cont’d
9.
Continue lowering and orthogonalizin’ until
complete!
Now isn’t that easier?
And much simpler…
You don’t believe me… I’m hurt.
I know! How about an example?
44

45. A CG Example: j1 =1/2 and j2 =1/2
Step 1
Step 2
Step 3
1 1
In CG speak :
⊗
2 2
so
1 1
1 1
j3 = +
j3 = −
2 2
2 2
The stretched case is
11
11 3 =
22
1
11
22
2
45

46. Steps 4 and 5 and
j m −1 J j m =
6->
J3 11 3 = ( J + J2 )
1
11
22
1
11
22
=J
1
2
11
22
11
22
1
( j + m)( j − m + 1)
+ J2
2
11
22
1
11
22
2
One step at a time 
J3 11 3 = (1 + 1)(1 − 1 + 1) 1 0 3 = 2 1 0
3
Now J only operates on space 1 stuff and J2 only operates on space 2 stuff so
1
J
1
11
22
1
11
22
11
11
J2
22 1 22
= 1
2
= 1
2
1 −1
2 2
11
22
11
22
2
1 −1
2 2
2
1
1
so
2 1 0 3 = 1
10 3 =
1 −1
2 2
1  1 −1

2 2 2

1
1
11
22
11
22
+ 1
2
+
2
11
22
11
22
1
1
1 −1
2 2
1 −1
2 2
2



2
46

47. Step 7—Keep lowering
J3 1 0 3 = (1 + 0)(1 − 0 + 1) 1 − 1 3 = 2 1 − 1 3
J
1
1 −1
2 2
1
11
22
1 −1
11
J2
2 2 1 22
=0
2
= 1
2
11
22
1
1 −1
2 2
2
so
2 1 − 1 3 =
  1 −1

2 2 2

2 1 − 1
2 1 − 1 3 =
2 2 2
1
1 −1
2 2
1
1 −1
2 2
+
2
1 −1
2 2
1
As low as
we go
2
1 −1 3 =
1
1 −1
2 2
1 −1
1 −1
= 2
2 2
2 2



2
1 −1
2 2 2
1 −1
2 2
1
47

48. An aside to simplify notation
11
Let
= +
22
and
1 −1
= −
2 2
Now we have derived 3 symmetric states
11 = + +
1
(− + + + −
10 =
2
1−1 = − −
)
Note these are also
symmetric from the
standpoint that we can
permute space 1 and space
2
Which is 1? Which is 2?
“I am not a number; I am a
free man!”
48

49. The infamous step 8

“Construct |j3 m3 > = |j1+j2 -1 j1+j2-1> so
that it is orthogonal to |j1+j2 j1+j2-1>”
j1+j2=1 and j1+j2-1=0 for this case so we
want to construct a vector orthogonal
to |1 0>
 The new vector will be |0 0>

49

50. Performing Step 8
10 =
(
1
−1+ 2+ +1−
2
2
)
An orthogonal vector to this could be
−1+ 2− +1−
2
or
+1− 2− −1+
2
Must obey Condon and Shortley: if m1=j1,, then positive value
j1=1/2 and |+> represents m= ½ , so only choice is
(
1
00 =
+1− 2− −1+
2
2
)
50

51. Step 9– The End
11 = + +
1
(− + + + −
10 =
2
1−1 = − −
)
These three symmetric states
are called the “triplet” states.
They are symmetric to any
permutation of the spaces
00 =
(
1
+1− 2− −1+
2
2
)
This state is anti-symmetric
and is called the “singlet”
state. If we permute space 1
and space 2, we get a wave
function that is the negative of
the original state.
51

52. A CG Table look up Problem
Part 1—
Two particles of spin 1 are at rest in a
configuration where the total spin is 1
and the m-component is 0. If you
measure the z-component of the second
particle, what values of might you get
and what is the probability of each zcomponent?
52

53. CG Helper Diagram
j1 ⊗ j2
m1
j3
m3
m2
C
It is understood that a “C” means
square root of “C” (i.e. all radicals
omitted)
53

54. Solution to Part 1
 Look
at 1 x 1 table
 Find j3 = 1 and m3 = 0
 There
3 values under these
m1
m2
1
0
-1
0
1/2
0
-1
1
-1/2
54

55. So the final part
m2
C
Prob
-1
1/2
½
0
0
0
1
-1/2
½
55

56. Part 2
An electron is spin up in a state, ψ5 2 1 ,
where 5 is the principle quantum
number, 2 is orbital angular momentum,
and 1 is the z-component.
If you could measure the angular
momentum of the electron alone, what
values of j could you get and their
probabilities?
56

57. Solution
 Look
at the 2 x ½ table since electron is spin
½ and orbital angular momentum is 2
 Now find the values for m1=1 and m2=1/2
 There
are two values across from these:
 4/5 which has j3 = 5/2
 -1/5
which has j3 = 3/2
 So
j3=5/2 has probability of 4/5
 So
j3 = 3/2 has probability of 1/5
57