this is a quantum lecturer

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