57
1 Introduction to Quantum Theory of Angular Momentum
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1
Introduction to Quantum Theory of Angular Momentum
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.
3
Definition
An angular momentum, J, is a linear operator with 3 components (Jx, Jy, Jz) whose commutation properties are defined as
JiJJ
4
Or in component form
ˆ ˆ,
ˆ ˆ,
ˆ ˆ,
y z y z z y x
z x z x x z y
x y x y y x z
J J J J J J i J
J J J J J J i J
J J J J J J i J
5
Convention
Jz is diagonal
For example:
10
01
2
zJ
100
000
001
zJ
6
Therefore
2
1,2
1
2
1
2
1,2
1
2
1,2
1
2
1
2
1,2
1
z
z
J
J
Where |jm> is an eigenketh-bar m is an eigenvalue
For a electron with spin up
mjmmjJ z
Or spin down
7
Definition
2222zyx JJJJ
These Simple Definitions have some major consequences!
8
THM
2 2 2 2
2
2
2
, , , ,
Recall , , ,
, , , , ,
, ( ) ( ) ( )
, 0
x x x y x z x
x x y y y x y x z z z x z
x z y y z y z z y
z
J J J J J J J J
A BC A B C B A C
J J J J J J J J J J J J J J
J J i J J J i J i J J J i J
J J
Proof:
zyxiwhereJJ i ,,0,2
QED
9
Raising and Lowering Operators
x yJ J iJ
Lowering Operator
yx iJJ J
Raising Operator
10
Product of J and J+
zyx
zyx
yxyx
yxxyyx
JJJ
obviouslyand
JJJ
JJiJJ
JiJJiJJJ
22
22
22
22
,
],[
JJ
JJ
JJ
JJ
11
Fallout
z
z
z
yx
J,
differencetheand
JJ
JJ
JJ
2][
,2
1
)(2
)(2
22
22
22
JJ
J)J(JJ
JJJJ
JJJJ
12
Proof that J is the lowering operator
1)1(
)1(
))(1(
)]()[(
)]()[(
,1
)(
)(
mjmmjJ
mjmmjJ
mjiJJmmjJ
mjJimJiJimJmjJ
mjmmjJand
mjJiJJiJiJJmjJ
JiJJJJdefinitionfrom
jmJiJJJjmJ
jmiJJJjmJ
z
z
yxz
xyyxz
z
xzyyzxz
xzyyzst
yzxzz
yxzz
J
JJ
J
J
J
J
J
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
13
[J2,Jz]=0 indicates J2 and Jz are simultaneous observables
mjmmjJJ
mjJJmjJJ
mjmjJ
yx
zyx
)()(
)()(
22222
2222
22
Since Jx and Jy are Hermitian, they must have real eigenvalues so -m2 must be positive!
is both an upper and LOWER limit to m!
14
Let msmall=lower bound on m andlet mlarge=upper bound on m
smallsmallsmallsmall
smallzz
smallzyx
small
small
mjmmmjJ
mjJJJ
mjJJJ
mj
mj
)(
0)(
0)(
0
0
222
22
22
JJ
J
largelargelargemjmmmjJ
mjJJJ
mjJJJ
mj
mj
large
largezz
largezyx
large
large
)(
0)(
0)(
0
0
222
22
22
JJ
J
el
elsmall
elsmall
m
mm
mmmm elsmall
arg
arg
2arg
2
1
arg
mlarge cannot any larger
15
Final Relation
smalllargelargesmall
smalllargelargesmall
largesmall
smallsmallsmallsmall
small
large
large
mjmmmjJ
mjmmmjJ
mm
mjmmmjJ
mforand
mjmmmjJ
mjmmmjJ
largelargelarge
largelargelarge
)1(
)1)((
)1(
)1(
)(
22
22
22
22
222
So the eigenvalue is mlarge*(mlarge +1) for any value of m
mjjjmjJ )1(22
16
Four Properties
,2,23,1,2
1,0
,)12()4
)12()3
,1,,1,)2
)1()1 22
j
thenintegerjSince
possiblevaluesjExactly
jjjjm
mjjjmjJ
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 half-integer values of spin, or orbital angular momentum
18
END OF LECTURE 1
19
Matrix Elements of J
mmz
mm
mmjJmj
jjmjJmj
,
,22 )1(
)(
1
1
22
2*
*
zz JJJbut
cccmjmj
mjcmj
mjcmj
JJ
JJ
J
J
Indicates a diagonal matrix
22
222
22222
222
)1)((
))1((
))1((
)(
mjmjc
cmmjj
cmjmmjjmj
cmjJJJmjand zz
20
Theorems
1)1)((
1)1)((
mjmjmjmj
and
mjmjmjmj
J
J
1,
1,
)1)((1
)1)((1
mm
mm
mjmjmjmj
and
mjmjmjmj
J
J
And we can make matrices of the eigenvalues, but these matrices are NOT diagonal
21
Fun with the Raising and Lowering Operators
yx iJJ J
yx iJJ J
2
JJxJ
2
)( iJ y
J-J
22
A matrix approach to Eigenvalues
0
1
2
1
2
1
1
0
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
12
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
J
J
J
and
If j=0, then all elements are zero! B-O-R-I-N-G!
j= 1/2
01
00
,
02
1
002
12
1
2
1
2
1
21
J
Therefore
so
m
m
mm Initial m
final m
What doesJ+ look like?
23
Using our relations,
00
10J
10
01
2
,
0
0
22
)
01
10
22
JJ
JJ
z
y
x
J
finallyand
i
ii(J
and
J
10
01
4
3 22 J
Answer:
Pauli Spin Matrices
10
01
0
0
01
10
z
y
x
i
i
24
J=1, An Exercise for the Students
Hint:
010
001
000
2
1
0
0
11
0
1
0
01
0
0
1
11
J
25
Rotation Matices
We want to show how to rotate eigenstates of angular momentum
First, let’s look at translationFor a plane wave:
)(
)(
3
1)(
pri
xx
xik
er
dforthenp
ksince
dforex x
26
A translation by a distance, A, then looks like
pa
ipar
ierear
)()()(
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, , in direction of
27
So
Jn
i
xx
enU
likelooksoperatorrotationaso
Jthenxif
)ˆ(
),ˆ(
:
,ˆˆ
Where n-hat points along the axis of rotation
Suppose we rotated through an angle about the z-axis
zJi
ezU
),(
28
Using a Taylor (actually Maclaurin) series expansion
im
mi
n
nn
nnz
n
nz
n
Ji
x
ezU
so
mjemjzU
mjn
mimjzU
mjmmjJbut
mjn
JimjzU
mjemjzU
so
xxxe
z
),(
),(
!),(
!),(
),(
!3!21
0
0
32
29
What if = 2?
,2
7,2
5,2
3,2
1""
,3,2,1,0"")2,(
)2,( 2
jfor
jformjmjzU
mjemjzU im
The naïve expectation is that thru 2 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
30
Let j=1/2 (for convenience it could be any value of j)
11601
10
8
1410
01
401
10
201
10
2
01
10
2
44
33
222
xx
x
x
JJ
J
J
!52!32201
10
!42!221
10
01),(
10
01
!4201
10
!3210
01
!2201
10
21),(
)(!
1),(
5
5
3
3
4
4
2
2
4
4
3
3
2
2
0
i
xU
iixU
Jn
ixU n
xn
31
Using the sine and cosine relation
!4!2
1cos!5!3
sin4253 xx
xandxx
xx
10
01)2,(,2
2cos
2sin
2sin
2cos
),(
xUifandi
ixU
so
And it should be no surprise, that a rotation of around the y-axis is
2cos
2sin
2sin
2cos
),(
yU
32
Consequences
If one rotates around y-axis, all real numbers Whenever possible, try to rotate around z-
axis 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
)(j
mmd
33
And
2cos)(
0
1
2cos
2sin
2sin
2cos
01)(
0
1
2
1
2
1
:
2
1
2
1
2
1
2
1
2
1
2
1
d
d
then
Example
mjyUmjd j
mm),()(
Wigner’s Formula (without proof)
k
kmmkmmjk
j
kmmkkmjkmjmjmjmjmjd
mm !)!'()!()!'(
)2
(sin)2
(cos)1(!)'()!'()!()!()(
2'2'2
34
Certain symmetry properties of d functions are useful in reducing labor and calculating rotation matrix
'
1
1
'''
''
'
''
''
'
12
2)(cos)()(
)()1()(
)()(
)()1()(
jjjmm
jmm
jmm
mjjmm
jmm
jmm
jmm
mmjmm
jddd
dd
dd
dd
35
Coupling of Angular Momenta
We wish to couple J1 and J2
321 JJJ
From Physics 320 and 321, we know
5,4,3,2,132
21321
jjjjj
But since Jz is diagonal, m3=m1+m2
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”
2211
33
mjmj
mj
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 2211 mjmj
38
Proof of commutative property
ababc
ababaabbcc
bbaababacc
babacLet
39
The expansion is written as
1
321
21 221133m
jjjmm mjmjCmj
321
21
jjjmmC
Is called the Clebsch-Gordan coefficientOr Wigner coefficient Or vector coupling coefficient
Some make the C-G coefficient look like an inner product, thus
321
21
1
321
21
3321321
2211221133
jjjmm
m
jjjmm
Cmjmmjjjthusand
mjmjbymultiplyandmjmjCmjTake
40
A simple formula for C-G coefficients
k
mjk
jjjmm
mjjkkkmjkjjj
kmjkmjj
jmjmjjjjjjjjjjC
)!(!!)!(
)!()!()1(
12!!!!!
32133213
11123
33333321213213
22
321
21
•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!!!
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!
42
9 Steps to Success
1. Get your values of j1 and j22. Identify possible values of j33. 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 J3=J1+J2
43
9 Steps to Success, cont’d
5. Operate J3|j3 m3>=(J1+J2 )|j1 m1>|j2 m2> 6. Use
7. Continue to lower until j3=|j1-j2|, where m1=-j1 , m2= -j2, and m3= -j3
8. 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!)
)1)((1 mjmjmjmj J
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?
45
A CG Example: j1 =1/2 and j2 =1/2
213
33
2
1
2
1
2
1
2
111
2
1
2
1
2
1
2
1
2
1
2
1:
iscasestretchedThe
jj
so
speakCGInStep 1
Step 2
Step 3
46
Steps 4 and 5 and 6->
21213
21213
2122
1
21211
21
3333
212
211
212133
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
101
2
1
2
1
2
1
2
11
2
1
2
1
2
1
2
11012
2
1
2
1
2
1
2
11
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
11
2
1
2
1
2
1
2
1
21
01201)111)(11(11
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1)(11
so
sostuffspaceonoperatesonlyandstuffspaceonoperatesonlyNow
timeaatstepOne
J
J
JJ
J
JJJJJ
)1)((1 mjmjmjmj J
47
Step 7—Keep lowering
213
2113
21213
2122
1
211
3333
2
1
2
1
2
1
2
111
2
1
2
1
2
1
2
12
2
1
2
1
2
1
2
1
2
2112
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2112
2
1
2
1
2
1
2
11
2
1
2
1
2
1
2
1
02
1
2
1
2
1
2
1
11211)101)(01(01
so
J
J
J
As low as we go
48
An aside to simplify notation
2
1
2
1
2
1
2
1andLet
11
2
101
11
Now we have derived 3 symmetric states
Note these are also symmetric from the standpoint that we can permute space 1 and space 2Which is 1? Which is 2?“I am not a number; I am a free man!”
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>
50
Performing Step 8
21212
101
2121
2121
An orthogonal vector to this could be
or
Must obey Condon and Shortley: if m1=j1,, then positive value
j1=1/2 and |+> represents m= ½ , so only choice is
21212
100
51
Step 9– The End
11
2
101
11
21212
100
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.
These three symmetric states are called the “triplet” states. They are symmetric to any permutation of the spaces
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 z-component?
53
CG Helper Diagram
m1 m2
j3
m3
C
21 jj
It is understood that a “C” means square root of “C” (i.e. all radicals omitted)
54
Solution to Part 1
Look at 1 x 1 tableFind j3 = 1 and m3 = 0
There 3 values under these
m1 m2
1 -1 1/2
0 0 0
-1 1 -1/2
55
So the final part
m2 C Prob
-1 1/2 ½
0 0 0
1 -1/2 ½
56
Part 2
An electron is spin up in a state, , 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?
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
