Title: Introduction to Quantum Theory of Angular Momentum
1Introduction to Quantum Theory of Angular Momentum
2Angular 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.
3Definition
- An angular momentum, J, is a linear operator with
3 components (Jx, Jy, Jz) whose commutation
properties are defined as -
-
4Or in component form
5Convention
- Jz is diagonal
- For example
6Therefore
Where jmgt is an eigenket h-bar m is an
eigenvalue For a electron with spin up
Or spin down
7Definition
These Simple Definitions have some major
consequences!
8THM
Proof
QED
9Raising and Lowering Operators
Lowering Operator
Raising Operator
10Product of J and J
11Fallout
12Proof that J is the lowering operator
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
13J2,Jz0 indicates J2 and Jz are simultaneous
observables
Since Jx and Jy are Hermitian, they must have
real eigenvalues so l-m2 must be positive! l is
both an upper and LOWER limit to m!
14Let msmalllower bound on m andlet mlargeupper
bound on m
mlarge cannot any larger
15Final Relation
So the eigenvalue is mlarge(mlarge 1) for any
value of m
16Four Properties
17Conclusions
- 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
18END OF LECTURE 1
19Matrix Elements of J
Indicates a diagonal matrix
20Theorems
And we can make matrices of the eigenvalues, but
these matrices are NOT diagonal
21Fun with the Raising and Lowering Operators
22A matrix approach to Eigenvalues
If j0, then all elements are zero! B-O-R-I-N-G!
Initial m
j 1/2
final m
What does J look like?
23Using our relations,
Answer
Pauli Spin Matrices
24J1, An Exercise for the Students
Hint
25Rotation Matices
- We want to show how to rotate eigenstates of
angular momentum - First, lets look at translation
- For a plane wave
26A translation by a distance, A, then looks like
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, q, in direction
of q
27So
Where n-hat points along the axis of rotation
Suppose we rotated through an angle f about the
z-axis
28Using a Taylor (actually Maclaurin) series
expansion
29What if f 2p?
The naïve expectation is that thru 2p 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
30Let j1/2 (for convenience it could be any value
of j)
31Using the sine and cosine relation
And it should be no surprise, that a rotation of
b around the y-axis is
32Consequences
- 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
33And
Wigners Formula (without proof)
34Certain symmetry properties of d functions are
useful in reducing labor and calculating rotation
matrix
35Coupling of Angular Momenta
- We wish to couple J1 and J2
- From Physics 320 and 321, we know
- But since Jz is diagonal, m3m1m2
36Coupling contd
- 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
37Coupling contd
- 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
38Proof of commutative property
39The expansion is written as
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
40A simple formula for C-G coefficients
- Proceeds over all integer values of k
- Begin sum with k0 or (j1-j2-m3) (which ever is
larger) - Ends with k(j3-j1-j2) or kj3m3 (which ever is
smaller) - Always use Stirlings formula log (n!) nlog(n)
Best approach use a table!!!
41- What if I dont have a table?
- And Im afraid of the simple formula?
- Well, there is another path a 9-step path!
429 Steps to Success
- Get your values of j1 and j2
- Identify possible values of j3
- Begin with the stretched cases where j1j2j3
and m1j1, m2j2 , and m3j3, thus j3 m3gtj1
m1gtj2 m2gt - From J3J1J2,, it follows that the lowering
operator can be written as J3J1J2
439 Steps to Success, contd
- Operate J3j3 m3gt(J1J2 )j1 m1gtj2 m2gt
- Use
- Continue to lower until j3j1-j2, where m1-j1
, m2 -j2, and m3 -j3 - Construct j3 m3 gt j1j2 -1 j1j2-1gt so that
it is orthogonal to j1j2 j1j2-1gt - Adopt convention of Condon and Shortley,
- if j1 gt j2 and m1 gt m2 then
- Cm1 m2j1 j2 j3 gt 0
- (or if m1 j1 then coefficient positive!)
449 Steps to Success, contd
- Continue lowering and orthogonalizin until
complete! - Now isnt that easier?
- And much simpler
- You dont believe me Im hurt.
- I know! How about an example?
45A CG Example j1 1/2 and j2 1/2
Step 1
Step 2
Step 3
46Steps 4 and 5 and 6-gt
47Step 7Keep lowering
As low as we go
48An aside to simplify notation
Now we have derived 3 symmetric states
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!
49The infamous step 8
- Construct j3 m3 gt j1j2 -1 j1j2-1gt so that
it is orthogonal to j1j2 j1j2-1gt - j1j21 and j1j2-10 for this case so we want to
construct a vector orthogonal to 1 0gt - The new vector will be 0 0gt
50Performing Step 8
An orthogonal vector to this could be
or
Must obey Condon and Shortley if m1j1,, then
positive value j11/2 and gt represents m ½ ,
so only choice is
51Step 9 The End
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
52A 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?
53CG Helper Diagram
j3 m3 C
m1 m2
It is understood that a C means square root of
C (i.e. all radicals omitted)
54Solution to Part 1
- Look at 1 x 1 table
- Find j3 1 and m3 0
- There 3 values under these
m1 m2
1 -1 1/2
0 0 0
-1 1 -1/2
55So the final part
m2 C Prob
-1 1/2 ½
0 0 0
1 -1/2 ½
56Part 2
- An electron is spin up in a state, y5 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?
57Solution
- Look at the 2 x ½ table since electron is spin ½
and orbital angular momentum is 2 - Now find the values for m11 and m21/2
- There are two values across from these
- 4/5 which has j3 5/2
- -1/5 which has j3 3/2
- So j35/2 has probability of 4/5
- So j3 3/2 has probability of 1/5