1. 8. Spin and Adding Angular Momentum 8A. Rotations Revisited
The Assumptions We Made
We assumed that |r formed a basis and R()|r = |r
From this we deduced R()(r) = (Tr)
Is this how other things work?
Consider electric field from a point particle
Can we rotate by R()E(r) = E(Tr)?
Let’s try it
This is not how electric fields rotate
It is a vector field, we must also rotate the field components
R()E(r) =  E(Tr)
Maybe we have to do something similar with ? +

2. Spin Matrices How do the D()’s behave?
We want to find all matrices satisfying this relationship
Easy to show: when  = 1, D() = 1
As before, Taylor expand D for small angles
In a manner similar to before, then show

3. We Already Know the Spin Matrices
We used to have identical expressions for the angular momentum L
From these we proved that L has the standard commutation relations
It follows that S has exactly the same commutation relations
The three S’s are generalized angular momentum
But in this case, they really are finite dimensional matrices
Logically, our wave functions would now be labeled
But s is a constant, so just label them
There are 2s + 1 of them total:

4. Restrictions on s?
Recall, for angular momentum, we had to restrict l to integers, not half-integers
Why? Because wave functions had to be continuous
Can we find a similar argument for spin? Consider s = ½
Consider a rotation by 2:
This would imply if you rotate by 2, the state vector changes by |   – | 
But these states are indistinguishable, so this is okay!
Any value of s, integer or half-integer, is fine
The basic building blocks of matter are all s = ½
Other particles have other spins
part. s
e- ½
p+ ½
n0 ½
part. s
 1
, 0 0
’s 3/2

5. Basis States for Particles With Spin
Basis states used to be labeled by |r
But now we must label them also by which component we are talking about |r,ms
Comment: for spin ½, it is common to abbreviate the ms label:
The spin operators affect only the spin label:
Operators that concern position, like R, P, and L, only affect the position label
All these position operators must commute with spin operators

6. Sample Problem
Define J = L + S. Find all commutators of J, J2, S2, and L2
That’s 6 operators, so 65/2 = 15 possible commutators
I’ll just do five of them to give you the idea
Recall, for any angular momentum-like set of operators, [J2,J] = 0

7. Hydrogen Revisited Recall our Hamiltonian:
Note that S commutes with the Hamiltonian
We can diagonalize simultaneously H, L2, Lz, S2, and Sz:
It is silly to label them by s, because s = ½
Degeneracy: ms takes on two values, doubling the degeneracy
Do all Hamiltonians commute with spin?
No! Magnetic interactions care about spin
Even hydrogen has small contributions (spin-orbit coupling) that depend on spin

8. 8B. Total Angular Momentum and Addition What Generates Rotations?
Recall that:
Rewrite this in ket notation
Define J:
J is what actually generates rotations
If a problem is rotationally invariant, we would expect J to commute with H
Not necessarily L or S

9. What are L, S and J?
Consider the rotation of the Earth around the Sun:
It has orbital angular momentum from its orbit around the Sun: L
It has spin angular momentum from its rotation around the axis: S
The total angular momentum is
It is another set of angular momentum-like operators
It will have eigenvectors |j,m with eigenvalues:
Because L and S typically don’t commute with the Hamiltonian, we might prefer to label our states by J eigenvalues, which do
To keep things as general as possible, imagine any two angular momentum operators adding up to yield a third:

10. Adding Angular Momentum
Commutation relations:
We could label states by their eigenvalues under the following four commuting operators:
Instead, we’d prefer to label them by the operators
These all commute with each other
These have the same j1 and j2 values, so we’ll abbreviate them:
Two things we want to know:
Given j1 and j2, what will the states |j,m be?
How do we convert from one basis to another, i.e., what is:
Clebsch-Gordan coefficients

11. The procedure
It is easy to figure out what the eigenvalues of Jz are, because
For each basis vector |j1,j2;m1,m2, there will be exactly one basis vector |j,m with m = m1 + m2
The ranges of m1 and m2 are known
From this we can deduce exactly how many basis vectors in the new basis have a given value of m
By looking at the distribution of m values, we can deduce what j values must be around
Easier illustrated by doing it than describing it

12. Sample Problem
Suppose j1 = 2 and j2 = 1, and we change basis from |j1,j2;m1,m2 to |j,m.
(a) What values of m will appear in |j,m, and how many times?
(b) What values of j will appear in |j,m, and how many times?
First, find a list of all the m1 and m2 values that occur
I will do it graphically
Now, use the formula m = m1 + m2 to find the m value for each of these points
From these, deduce the m values and how many there are
Note where the transitions are:
m=3
m=2
m=-3
m=-1
m=1
m=-2
m=0

13. Sample Problem (2)
Suppose j1 = 2 and j2 = 1, and we change basis from |j1,j2;m1,m2 to |j,m.
(a) What values of m will appear in |j,m, and how many times?
(b) What values of j will appear in |j,m, and how many times?
For any value of j, m will run from –j to j
Clearly, there is no j bigger than 3
But since m = 3 appears, there must be j= 3
This must correspond to m’s from –3 to 3
Now, there are still states with m up to 2
It follows there must also be j = 2
This covers another set of m’s from –2 to 2
What remains has m up to 1
It follows there must be j = 1
And that’s it.
Why did it run from j = 3 to j = 1?
Because it went from j1 + j2 down to j1 – j2

14. General Addition of Angular Momentum
The set of all (m1,m2) pairs forms a rectangle
The largest value of m is m = j1 + j2, which can only happen one way
As m decreases from the max value, there is one more way of making each m value for each decrease in m until you get to | j1 – j2 |
This implies that you get maximum jmax = j1 + j2 and minimum jmin = | j1 – j2 |
So, j runs from | j1 – j2 | to j1 + j2 in steps of size 1

15. Check Dimensions
For fixed j1 and j2, the number of basis vectors |j1,j2;m1,m2 is
How many basis vectors |j,m are there?
For each value of j, there are 2j + 1.
Therefore the total is
So dimensions work out

16. Sample Problem
Suppose we have three electrons. Define the total spin as S = S1 + S2 + S3. What are the possible values of the total spin s, the corresponding eigenvalues of S2, and how many ways can each of them be made?
Electrons have spin s = ½, so
Combine the first two electrons:
Now add in the third:
If s1+2 = 0, this says:
If s1+2 = 1, this says
Final answer for s:
The repetition means there are two ways to combine to make s = ½
For S2:

17. Hydrogen Re-Revisited
Recall hydrogen states labeled by
Because of relativistic corrections, these aren’t eigenstates
Closer to eigenstates are basis states
j = l  ½
States with different mj are related by rotation
Indeed, the value of mj will depend on choice of x, y, z axis
And they are guaranteed to have the same energy
Therefore, when labelling a state we need to specify n, l, j
We label l values by letters, in a not obvious way
Good to know the first four: s, p, d, f
We then denote j by a subscript, so example state could be 4d3/2
Remember restrictions: l < n and j = l  ½
Often, we don’t care about j, so just label it 4d
Remember, number of states for given n,l is
l let
0 s
1 p
2 d
3 f
4 g
5 h
6 i
7 k
8 l
9 m
10 n
11 o
12 q

18. 8C. Clebsch-Gordan Coefficients
How do we change bases?
We wish to interchange bases |j1,j2;m1,m2  |j,m
These are complete orthonormal basis states in the same vector space
We can therefore use completeness either way
The coefficients are called Clebsch-Gordan coefficients, or CG coefficients for short
Our goal: Show that we can find them (almost) uniquely
Note that the states |j1,j2;m1,m2 are all related by J1 and J2
There are no arbitrary phases concerning how they are related
The |j,m states with the same j’s and different m’s are related by J
But there is no simple relation between |j,m’s different j’s – convention choice

19. Convention Confusion
If you ever have to look them up, be warned, different sources use different notations
Recall that the other states are also eigenstates of J12 and J22
People also get lazy and drop some commas
In addition, the Clebsch-Gordan coefficients are defined only up to a phase
Everyone agrees on phase up to sign
As long as you use them consistently, it doesn’t matter which convention you use.
They will turn out to be real, and therefore
Because of this ambiguity, people get lazy and often use what is logically the wrong one

20. Nonzero Clebsch-Gordan (C-G) Coefficients
When are the coefficients meaningful and (probably) non-zero?
(1) j range:
(2) m range:
(3) conservation of Jz:
Let’s prove the last one using
Act on the left with Jz and on the right with J1z and J2z:
Must be zero unless
j1 + j2 – j is an integer
j – m is an integer, etc.

21. Finding C-G Coefficients for m = j
Largest value for m is j, therefore
Recall in general
We therefore have
Recall: only if m1 + m2 = m (= j) are non-zero
This relates all the non-zero terms for m = j, all relative sizes determined
To get overall scale, use normalization
This determines everything up to a phase
We arbitrarily pick

22. Finding C-G Coefficients for m – 1 from m
We now have CG coefficients when m = j
I will now demonstrate that if we have them for m, we can get them for m – 1
First note
Dagger this So
So if we know them for m, we know them for m – 1
Since we know them for m = j, we know them for m = j – 1, j – 2, etc.
Hence we have a (painful) procedure for finding all CG coefficients
Sane people don’t do it this way, they look them up or use computers

23. Properties of CG-coefficients
Adding j1 and j2 is the same as adding j2 and j1
Corollary: if j1 = j2, then the combinations of spins is symmetric if j1 + j2 – j is even, anti-symmetric if it is odd
You can work your way up from m = –j in the same way we worked our way down from m = j:
Adding j1 = 0 or j2 = 0 is pretty trivial, because these imply J1 = 0 or J2 = 0
If you ever look things up in tables, they will assume j1  j2 > 0, and assume you will use the first or third rule to get other CG coefficients
Or you can use computer programs to get them
> clebsch(1,1/2,1,-1/2,3/2,1/2);

24. CG coefficients when j2 = ½
For j2 small, we can find simple formulas for the CG coefficients
If j2 = ½, then j = j1  ½
Example:
For one electron, J = L + S. Let j1  l, m  mj, drop j2 = s = ½, m2 = ½  
For adding two electron spins, drop s1 and s2, abbreviate mi = ½  

25. Sample Problem
Hydrogen has a single electron in one of the states |n,l,m,ms = |2,1,1,– or |2,1,0,+ , or in one of the states |n,l,j,mj = |2,1,3/2,1/2 or |2,1,1/2,1/2 . In all four cases, write explicitly the wave function
For s = ½, wave function looks like
Spin state ms tells us which component exists
This lets us immediately write the wave function for the first two:
For the |j,mj states we have:

26. Sample Problem (2)
… or in one of the states |n,l,j,mj = |2,1,3/2,1/2 or |2,1,3/2,1/2 . In all four cases, write explicitly the wave function
You can also get the CG coefficients from Maple:
> clebsch(1,1/2,1,-1/2,3/2,1/2);
> clebsch(1,1/2,0,1/2,3/2,1/2);
> clebsch(1,1/2,1,-1/2,1/2,1/2);
> clebsch(1,1/2,0,1/2,1/2,1/2);

27. Sample Problem
Hydrogen has a single electron in the state |n,l,j,mj = |2,1,3/2,1/2. If one of the following is measured, what would the outcomes and corresponding probabilities be, and what would the state afterwards look like: E, J2, Jz, L2, S2, Lz,Sz
For the first five choices, our state is an eigenstate of the operator
The eigenstate will be unchanged by this measurement
For the last two, we write it in terms of eigenstates of Lz or Sz
Then we have
State afterwards is
Or we have

28. Definition and Commutation with J
8D. Scalar, Vector, Tensor
Definition and Commutation with J
A scalar operator S is anything that is unchanged under rotation
Examples:
Scalar operators commute with the generator of rotations J:
Vector operators V are operators that rotate like a vector:
They have commutation relations with J given by
A rank 2 tensor Tij under rotation rotates as:
Can show that
Rank k tensor has k indices and commutation relations:
Scalar = rank 0 tensor, Vector = rank 1 tensor
Rank 2 tensor is sometimes just called a “tensor”

29. How to Make a Tensor From Vectors
If V and W are any two vector operators, then we can define a rank-2 tensor operator:
One can similarly define higher rank tensor operators
This tensor has nine independent components
But it has pieces that aren’t very rank-2 tensor-like:
Dot product VW is a scalar operator
Cross product VW is a vector operator
The remaining five pieces are the truly rank-2 part
We want figure out how to extract the various pieces

30. Spherical Tensors We start with a vector operator V
Define the three operators Vq by:
You can then show the following:
Proof by homework problem
Compare this with:
Another way to write it:
Generalize this formula
Define a spherical tensor of rank k as 2k + 1 operators:
It must have commutation relations:
Trivial example: A scalar is a spherical tensor of rank 0

31. Combining Spherical Tensors (1)
Theorem: Let V and W be spherical tensors of rank k1 and k2 respectively. Then we can build a new spherical tensor T of rank k defined by:
Those matrix elements are CG coefficients
Proof:

32. Combining Spherical Tensors (2)
We have complete set of states |k1,k2;q1,q2
Now insert complete set of states |k’,q’:
J doesn’t change the k value, so k’ = k
So we have proven it

33. How it Comes Out
This sum only makes sense if CG coefficients are non-zero
Only non-zero terms are when q1 + q2 = q
So it’s really just a single sum
By combining two vectors, we can get k = 0, 1, 2
k = 0: Scalar (dot product)
k = 1: Vector (cross product)
k = 2: Truly rank 2 tensor part
We can then combine rank 2 tensors with more vectors to make rank 3 spherical tensors

34. Sample Problem
If we combine two copies of the position operator R, what are the resulting components of the rank-2 spherical tensor Tq(2)?

35. 8E. The Wigner-Eckart Theorem
Why it should work
Suppose we have an atom or other rotationally invariant system
Eigenstates should be eigenstates of J2, Jz, probably other stuff
It is common to need matrix elements of operators between these states:
We know how the ket and bra rotate
If we also know how the operator in the middle rotates, we should be able to find relations between these various quantities
Suppose the operator is a spherical tensor, or combinations thereof
Then we know how T rotates, and we should be able to find relations
This helps us because:
If the calculation is hard, we do it a few times and deduce the rest
If the calculation is impossible, we measure it a few times and deduce the rest

36. Similarities With CG coefficients (1)
I want to compare the matrix element above to the CG coefficient above
Recall relations for Jz:
Use commutation relation:
Let Jz act on the bra or the ket on the left:
Hence matrix elements are zero unless
Compare to the CG coefficient above:
This vanishes unless

37. Similarities With CG coefficients (2)
Recall relations for J:
For m = j,
Implies:
Our commutation relations tell us:
Compare to the CG coefficients:

38. Similarities With CG coefficients (3)
We have:
Equivalent to
Our commutation relations tell us:

39. These Matrix Elements are CG Coefficients
We have three relations that are identical for these two expressions:
These expressions were all that were used to find the CG coefficients
Plus, we had a normalization condition
Hence, these two expressions are identical
Up to normalization

40. The Wigner-Eckart Theorem
What can the proportionality constant depend on?
Not m, m’, nor q
It can depend on , ’, j, j’, and of course T
The Wigner-Eckart Theorem:
The square root in the denominator is a choice, neither right nor wrong
That other thing is called a “reduced matrix element”
You don’t calculate it (directly)
You may be able to calculate left side for one value of m, m’, q
Or you may be able to measure left side for one value of m, m’, q
Then you deduce the reduced matrix element from this equation
Then you can use it for all the other values of m, m’, q

41. Why Is the Wigner-Eckart Theorem Useful?
The number of matrix elements is
For example, if j = 3, j’ =2, k = 1, this is 105 different matrix elements
Calculating them computationally may be difficult or impossible
Measuring them may be a great deal of work
By doing one (difficult) computation or one (difficult) measurement you can deduce a lot of others
Comment: why is the factor of 2j + 1 there?
If T0(k) is Hermitian, then you can show

42. Sample Problem
The magnetic dipole transition of hydrogen causing the 21 cm line is governed by the matrix element , where F is the total angular momentum quantum number and mF is the corresponding z-component, and S and L are spin and orbital angular quantum operators for the electron. Deduce as much as you can about these matrix elements for mF = +1, 0, or –1.
We have no idea what most of this means, but it’s clear:
F and mF are angular quantum number, effectively, j  F and m  mF
S and L are vector operators
Call reduced matrix element A:
Non-vanishing only if q + mF = 0
Get the CG coefficients from program
All other matrix elements vanish

43. Sample Problem
Deduce as much as you can about these matrix elements for mF = +1, 0, or –1.
For mF = + 1, we also have
Vx and Vy: two equations, two unknowns
We therefore have:
Similarly:

44. 8F. Integrals of Spherical Harmonics Products of Spherical Harmonics
Consider the product of any two spherical harmonics:
By completeness, this can be written as a sum of spherical harmonics:
The coefficients clm can be found usingorthogonality:
Think of the expression
as an operator acting on a wave function:
It is not hard to see that this operator is a spherical tensor operator
Think of clm then as a matrix element
By the Wigner-Eckart theorem:
All that remains is to find the reduced matrix elements

45. Working on the Reduced Matrix Element
Substitute the top equation in the bottom
Multiply this expression by and sum over m1, m2:
Rename l’ as l
sum over complete states

46. Finishing the Computation
Must be true at all angles
Evaluate at  = 0
Formula for the Y’s at  = 0 is simple
Now we solve for the reduced matrix element
We therefore have

47. When doesn’t it vanish?
We want to know when this is non-zero, or likely to be non-zero:
We need:
Under parity, each of the spherical harmonics transforms to
So the whole integral satisfies
We need

48. Sample Problem
Atoms usually decay spontaneously by the electric dipole process, in which case the rate is determined by the matrix element , whereI and F are the initial and final states. For hydrogen in each of the following states, which states might be the final n and l quantum states if the initial state is: 4s, 4p, 4d, 4f?
The initial state has n’ = 4 and l’ = 0, 1, 2, or 3
Final state has unknown n and l, but
Must have n < 4 because energy goes down
Must have l < n
The position operators can be written in terms of l = 1 spherical harmonics
So we have
To not vanish, we need
For 4s, l = 0:
Must have l < n < 4

49. Sample Problem (2)
Atoms usually decay spontaneously by the electric dipole process, in which case the rate is determined by the matrix element , whereI and F are the initial and final states. For hydrogen in each of the following states, which states might be the final n and l quantum states if the initial state is: 4s, 4p, 4d, 4f?
For 4p: l’ = 1, so
Must have l < n < 4
For 4d: l’ = 2, so
For 4f: