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
R  R   r   D  R   R T r 
similar with ?
Spin Matrices
R  R   r   D  R   R T r 
• How do the D()’s behave?
R  R1  R  R 2   r   R  R1R 2   r 
R  R1  D  R 2   R r   D  R1R 2 

T
2

 R R  r 
T
1
2
D  R 1  D  R 2  R 2T  R 1T r   D  R 1R 2   R 2T R 1T r 
• We want to find all matrices satisfying this relationship
• Easy to show: when  = 1, D() = 1
• As before, Taylor expand D for small angles
D  R  nˆ ,    1  i  nˆ  S
 O  2 
• In a manner similar to before, then show
D  R  nˆ ,    exp  i  nˆ  S

D  R1  D  R 2   D  R1R 2 
We Already Know the Spin Matrices
D  R1  D  R 2   D  R1R 2 
D  R  nˆ ,    exp  i  nˆ  S

• 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
 Si , S j    i  ijk S k
k
•
•
•
•
•
 S x , S y   i S z , etc.
  s r  
The three S’s are generalized angular momentum


But in this case, they really are finite dimensional matrices
 s 1  r  

Logically, our wave functions would now be labeled  s ,ms   r, t   


But s is a constant, so just label them ms


  s r  
There are 2s + 1 of them total:
Restrictions on s?
D  R1  D  R 2   D  R1R 2 
D  R  nˆ ,    exp  i  nˆ  S

• Recall, for angular momentum, we had to restrict l to integers, not half-integers
• Why? Because wave functions had to be continuous
Yl m ~ eim
• Can we find a similar argument for spin? Consider s = ½
S  12 σ
D  R  nˆ ,    exp   12 i nˆ  σ   cos  12    inˆ  σ sin  12  
•
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•
•
•
•
Consider a rotation by 2: D  R  nˆ , 2    cos    inˆ  σ sin    1
This would imply if you rotate by 2, the state vector changes by |   – | 
But these states are indistinguishable, so this is okay!
part. s
part. s
Any value of s, integer or half-integer, is fine
e½ 
1
The basic building blocks of matter are all s = ½
p+
½ , 0 0
Other particles have other spins
n0
½ ’s
3/2
Basis States for Particles With Spin
• Basis states used to be labeled by |r
 r   r 
• But now we must label them also by which
 ms  r   r, ms 
component we are talking about |r,ms
• Comment: for spin ½, it is common to abbreviate the ms label: r,  12  r, 
• The spin operators affect only the spin label:
S 2 r, ms 
2
s
2
 s  r, ms , S z r, ms  ms r, ms ,
S r, ms 
s 2  s  ms2
ms r, ms  1
• Operators that concern position, like R, P, and L, only affect the position label
R r, ms  r r, ms
• All these position operators must commute with spin operators
 Si , R j    Si , Pj    Si , L j   0
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
 J x , J y    Lx  S x , Ly  S y    Lx , Ly    Lx , S y    S x , Ly    S x , S y 
 i Lz  0  0  i S z  i
 Lz  Sz   i
Jz
S 2 , J   S 2 , L  S   S 2 , S   0
• Recall, for any angular momentum-like set of operators, [J2,J] = 0
S 2 , J 2   0
Hydrogen Revisited
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•
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•
Recall our Hamiltonian:
1 2 ke e 2
H
P 
Note that S commutes with the Hamiltonian
2
R
2
2
We can diagonalize simultaneously H, L , Lz, S , and Sz:
n, l , m, s, ms
It is silly to label them by s, because s = ½
S2   34
2

n, l , m, ms
• Degeneracy: ms takes on two values,
doubling the degeneracy
D  En   2n2
H n, l , m, s, ms  En n, l , m, s, ms
L2 n, l , m, s, ms 
2
l
2
 l  n , l , m , s , ms
Lz n, l , m, s, ms  m n, l , m, s, ms
S 2 n, l , m, s, ms 
2
s
2
 s  n, l , m, s, ms
S z n, l , m, s, ms  ms n, l , m, s, ms
Do all Hamiltonians commute with spin?
• No! Magnetic interactions care about spin
• Even hydrogen has small contributions (spin-orbit coupling) that depend on spin
8B. Total Angular Momentum and Addition
What Generates Rotations?
T
R
R

r

D
R

R






 r
• Recall that:
• Rewrite this in ket notation
R  R  nˆ ,     D  R  nˆ ,   exp  i  nˆ  L

 exp  i  nˆ  S
 exp  i  nˆ  L  
 exp  i  nˆ   S  L   
• Define J:
J  LS
R  R  nˆ ,    exp  i  nˆ  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
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
J  LS
• The total angular momentum is
 J i , J j    i  ijk J k
k
• It is another set of angular momentum-like operators
• It will have eigenvectors |j,m with eigenvalues:
J 2 j, m  2  j 2  j  j, m
• Because L and S typically don’t commute with the J z j , m  m j , m
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:
J  J1  J 2
Adding Angular Momentum
J  J1  J 2
 J ai , J bj    ab  i  ijk J ak
k
• Commutation relations:
• We could label states by their eigenvalues under the following four commuting
operators: J12 , J 22 , J1z , J 2 z 
j1 , j2 ; m1 , m2
J 2a j1 , j2 ; m1 , m2 
2
2
j
 a  ja  j1 , j2 ; m1 , m2 , J az j1 , j2 ; m1, m2  ma j1, j2 ; m1, m2 .
• Instead, we’d prefer to label them by the operators J 2 , J 2 , J 2 , J

1
2
z
– These all commute with each other
• These have the same j1 and j2 values, so we’ll abbreviate them: j, m
J 2 j, m 
2
j
2
 j  j, m , J z j, m  m j, m .
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
j1 , j2 ; m1 , m2 j, m
The procedure
J  J1  J 2
• It is easy to figure out what the eigenvalues of Jz are, because
J z j1 , j2 ; m1 , m2   J1z  J 2 z  j1 , j2 ; m1, m2 
 m1  m2 
J z  J1z  J 2 z
j1 , j2 ; m1 , m2
• For each basis vector |j1,j2;m1,m2, there will be
 m1, m2   m  m1  m2
exactly one basis vector |j,m with m = m1 + m2
m1   j1 , , j1 m2   j2 , , j2
• 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 N  m  N   m1 , m2  : m1  m2  m
• By looking at the distribution of m values, we
m   j , , j for each value of j .
can deduce what j values must be around
• Easier illustrated by doing it than describing it
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?
ma   ja , , ja
• First, find a list of all the m1 and m2 values that occur
– I will do it graphically
m2
• 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
m1
m=3
there are
– I will do it graphically
m=2
• Note where the transitions are:
m=-3
m=-1
m=1
m=-2
m=0
m
0
j1  j2
 j1  j2 j2  j1
j1  j2
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
m   j,
,j
m
0
j
0
General Addition of Angular Momentum
•
•
•
•
•
J  J1  J 2
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 |
m2
So, j runs from | j1 – j2 | to
j2
j1 + j2 in steps of size 1
 j1  j2
 j1  j2 j1  j2
j1  j2
j  j1  j2 , j1  j2  1,
 j1    j2   
m  j1
j1
 j2
, j1  j2
j1  j2    j1  j2  1 
  j1  j2 
m1
Check Dimensions
j  j1  j2 , j1  j2  1,
, j1  j2
• 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
D
j1  j2
  2 j  1
j  j1  j2

j1  j2

j  j1  j2
 2 j1  1 2 j2  1
 j  12  j 2 


2
2
2
2



  j1  j2  1   j1  j2    j1  j2    j1  j2  1  

 

2
2

  j1  j2  1  j1  j2 


  j1  j2  1   j1  j2    j1  j2  1  j1  j2  j1  j2  1  j1  j2    2 j1  1 2 j2  1
2
2
• So dimensions work out
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?
s1  s2  s3  12
• Electrons have spin s = ½, so
• Combine the first two electrons: s12  s1  s2 , , s1  s2  0, ,1  0,1
• Now add in the third: s  s12  s3 , , s12  s3
• If s1+2 = 0, this says: s  0  12 ,
,0  12  12 ,
, 12  12
s  1  12 ,
,1  12  12 ,
, 32  12 , 32
• If s1+2 = 1, this says
• Final answer for s: s  12 , 12 , 32
– The repetition means there are two ways to combine to make s = ½
• For S2: S 2 s, ms  2  s 2  s  s, ms
S 2  34 2 , 34 2 , 154 2
Hydrogen Re-Revisited
• Recall hydrogen states labeled by n, l , m, ms
• Because of relativistic corrections, these aren’t eigenstates
• Closer to eigenstates are basis states n, l , j , m j
– 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
d  n, l   2  2l  1
l
0
1
2
3
4
5
6
7
8
9
10
11
12
let
s
p
d
f
g
h
i
k
l
m
n
o
q
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
j, m   j1 , j2 ; m1 , m2
j1 , j2 ; m1 , m2 j, m
j1 , j2 ; m1 , m2   j , m
j , m j1 , j2 ; m1 , m2
m1 m2
j
m
• The coefficients are called Clebsch-Gordan
j1 , j2 ; m1 , m2 j, m
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
Convention Confusion
• If you ever have to look them up, be warned, different sources use different
notations
j1 , j2 ; m1 , m2  j1 , m1; j2 , m2  j1 , m1  j2 , m2
• Recall that the other states are also
j, m  j1 , j2 ; j, m
eigenstates of J12 and J22
• People also get lazy and drop some commas
j1 j2 m1m2 jm
• 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 j, m j1 , j2 ; m1 , m2  j1 , j2 ; m1 , m2 j, m
• Because of this ambiguity, people get lazy and often use what is logically the
wrong one
Nonzero Clebsch-Gordan (C-G) Coefficients
j1 , j2 ; m1 , m2 j, m
J  J1  J 2
j  j1  j2 , j1  j2  1,
, j1  j2
When are the coefficients meaningful and (probably) non-zero?
(1) j range: j1  j2  j  j1  j2
j1 + j2 – j is an integer
(2) m range:  j1  m1  j1 ,  j2  m2  j2 ,  j  m  j
j – m is an integer, etc.
(3) conservation of Jz: m1  m2  m
• Let’s prove the last one using
J1z  J 2 z  J z
j, m J z j1 , j2 ; m1 , m2  j, m  J1z  J 2 z  j1 , j2 ; m1 , m2
• Act on the left with Jz and on the right with J1z and J2z:
m j, m j1 , j2 ; m1 , m2  j, m j1, j2 ; m1, m2

0   m1  m2  m j, m j1 , j2 ; m1 , m2
• Must be zero unless
m1  m2  m
m1  m2 
Finding C-G Coefficients for m = j
j1 , j2 ; m1 , m2 j, m
J  J1  J 2
• Largest value for m is j, therefore
• Recall in general J  j, m 
j  j1  j2 , j1  j2  1,
J  j, j  0
, j1  j2
 0  j, j J 
j 2  j  m2  m j , m  1
• We therefore have 0  j, j J  j1 , j2 ; m1 , m2  j, j  J1  J 2  j1 , j2 ; m1 , m2
0
j12  j1  m12  m1 j, j j1 , j2 ; m1 1, m2 
j22  j2  m22  m2 j, j j1 , j2 ; m1 , m2 1
• 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
1  j, j j, j   j, j j1 , j2 ; m1 , m2 j1 , j2 ; m1 , m2 j , j   j, j j1 , j2 ; m1 , j  m1
m1 m2
• This determines everything up to a phase
– We arbitrarily pick
m1
j, j j1 , j2 ; j1 , j  j1  0
2
Finding C-G Coefficients for m – 1 from m
j1 , j2 ; m1 , m2 j, m
J  J1  J 2
• 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
J j, m 
j 2  j  m2  m j , m  1

j 2  j  m2  m j , m  1  j , m J 
• Dagger this
• So
j 2  j  m2  m j, m  1 j1 , j2 ; m1 , m2  j, m J  j1 , j2 ; m1 , m2
 j, m  J1  J 2  j1 , j2 ; m1 , m2

•
•
•
•
j12  j1  m12  m1 j, m j1 , j2 ; m1  1, m2 
j22  j2  m22  m2 j , m j1 , j2 ; m1 , m2  1
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
Properties of CG-coefficients
j1 , j2 ; m1 , m2 j, m
• Adding j1 and j2 is the
j j j
j1 , j2 ; m1 , m2 j , m   1 1 2
j2 , j1 ; m2 , m1 j , m
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
j1  j2  j
down from m = j:
j1 , j2 ; m1 , m2 j , m   1
j1 , j2 ; m1 , m2 j , m
• Adding j1 = 0 or j2 = 0 is pretty trivial,
j,0; m,0 j, m  0, j;0, m j, m   m,m
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);
1, 12 ;1,  12
3
2
, 12 
1
3
CG coefficients when j2 = ½
• For j2 small, we can find simple formulas for the CG coefficients
• If j2 = ½, then j = j1  ½
j1  12  m
1
1
1
1
1
1 1
1
j1 , 2 ; m  2 ,  2 j1 2 , m   j1 , 2 ; m  2 , 2 j1  2 , m 
2 j1  1
• Example:
j1  12 , m 
j1  12 m
j1 , 12 ; m  12 ,  12 
2 j1  1
j1  12  m
j1 , 12 ; m  12 , 12
2 j1  1
• For one electron, J = L + S. Let j1  l, m  mj, drop j2 = s = ½, m2 = ½  
l  12 , m j 
l  12 m j
l , m j  12 ,  
l  12  m j
l , m j  12 , 
2l  1
2l  1
• For adding two electron spins, drop s1 and s2, abbreviate mi = ½  
1, 1    ,
1,0 
1
2
0,0 
1
2
    ,
    
1, 1   
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
 0
1
for the first two:
1
0
 2,1,1,  R21  r  Y1  ,     ,  2,1,0,   R21  r  Y1  ,    
1
 0
1
1
• For the |j,mj
l

m
l

j
2
2  mj
1
1
1
l

,
m

l
,
m

,


l
,
m

states we have:
j
j
j
2
2
2 ,
2l  1
2l  1
2,1, 32 , 12
1
1
1  12  12
1


2
2

2,1, 12  12 ,  
2,1, 12  12 ,  
2 1  1
2 1  1
1
3
2,1,1,  
2
3
2,1, 0, 
2,1, 12 , 12
1
1
1  12  12
1


2
2

2,1, 12  12 ,  
2,1, 12  12 ,  
2 1  1
2 1  1
2
3
2,1,1,  
1
3
2,1, 0, 
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
 2,1,1,
2,1, 32 , 12 
1
3
 0
1
0
 R21  r  Y  ,     ,  2,1,0,   R21  r  Y1  ,    
1
 0
2,1,1,  
• You can also get the CG
coefficients from Maple:
 2,1,
3,1
2 2
1
1
2
3
2,1, 0, 
>
>
>
>
 2Y10  ,   
1

R21  r   1


3
 Y1  ,   
2,1, 12 , 12 
2
3
2,1,1,  
1
3
2,1, 0, 
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);
 2,1,
1,1
2 2
0



Y
1
1  ,  

R21  r  

1
 2Y  ,   
3
1


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
2
2
2
2
2
2
2
15 2 J  m  1
13.6 eV
L

l

l

2
J

j

j

z
j
2




4
E
 3.40 eV
2
n
S 2   s 2  s  2  34 2
• The eigenstate will be unchanged by this measurement
• For the last two, we write it in terms of 2,1, 32 , 12  13 2,1,1,   32 2,1, 0, 
eigenstates of Lz or Sz
2
3 1
1
P  Lz    P  S z   2   2,1,1,  2,1, 2 , 2  13
• Then we have
• State afterwards is
2,1,1, 
2
3 1
1
P  Lz  0   P  S z  2   2,1, 0,  2,1, 2 , 2  23
• Or we have
• State afterwards is 2,1,0, 
8D. Scalar, Vector, Tensor
Definition and Commutation with J
•
•
•
•
•
•
•
•
A scalar operator S is anything that is unchanged under rotation R†  R  SR  R   S
Examples: R 2 , P 2 ,V  R  , L2 , S 2 , J 2 , L  S
Scalar operators commute with the generator of rotations J:  J, S   0
Vector operators V are operators that rotate like a vector: R†  R  VR  R   R V
Examples: R , P, L, S, J
They have commutation relations with J given by  J i , V j   i  k  ijkVk
A rank 2 tensor Tij under rotation rotates as: R†  R  Tij R  R    R ik R jlTkl
k
l
Can show that  J , T   i

T


T


 l ijl lk ikl jl
 i jk 

• Rank k tensor has k indices 
J i , T j1 j2 jk   i  l  ij1lTlj2 j3

and commutation relations:
• Scalar = rank 0 tensor, Vector = rank 1 tensor
• Rank 2 tensor is sometimes just called a “tensor”
jk
  ij2lT j1l
jk

 ij lTj j
k
1 2
l

How to Make a Tensor From Vectors
 J i , T jk   i
 
l
T   iklT jl 
ijl lk
• If V and W are any two vector operators,
i
j
then we can define a rank-2 tensor operator: Tij  VW
– 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
Spherical Tensors
• We start with a vector operator V
V0  Vz , V1  12  Vx  iVy 
• Define the three operators Vq by:
• You can then show the following:
2




J
,
V

qV
,
J
,
V

2

q
qVq 1 .
q
– Proof by homework problem  z q 
  q
• Compare this with:
J z 1, q  q 1, q , J  1, q  2  q 2 q 1, q  1
• Another way to write it:
1
J,Vq    Vq 1, q J 1, q
q1
Generalize this formula
k
• Define a spherical tensor of rank k as 2k + 1 operators: Tq  q  k , k  1, , k
• It must have commutation relations:
k  
k 

J
,
T

T
k , q J k , q

q 
q
k
k



 J z , Tq   qTq
q


 J  , Tq k    k 2  k  q 2 qTqk1


• Trivial example: A scalar is a spherical tensor of rank 0
 J, S0 0   0


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:
 k1   k2 
V
 q1 Wq2 k1 , k2 ; q1 , q2 k , q
Tq k  
 J, Tq k     Tqk  k , q J k , q

 q
q1 , q2
• Those matrix elements are CG coefficients
• Proof:
 J, Tq k      J, Vq k1 Wq k2   k1 , k2 ; q1 , q2 k , q
1
2




q1 , q2


 k1    k2 
 k1  
 k2  

J
,
V
W

V
J
,
W

q1  q2
q1
q2  k1 , k2 ; q1 , q2 k , q


q1 , q2


 k1   k2 
 k1   k2 
    Vq1 Wq2 k1 , q1 J k1 , q1  Vq1 Wq2 k 2 , q2 J k 2 , q2  k1 , k 2 ; q1 , q2 k , q
q1 , q2  q1
q2




 J , Tq k      Vqk1 Wqk2  k1 , q1 J k1 , q1  q ,q  k2 , q2 J k2 , q2  q ,q k1 , k 2 ; q1 , q2 k , q
2
2 2
1 1

 q ,q q ,q 1
1
2
1
2
Combining Spherical Tensors (2)
Tq k  
 k1   k2 
V
 q1 Wq2 k1 , k2 ; q1 , q2 k , q
q1 , q2
 J, Tq k     Tqk  k , q J k , q

 q

 J , Tq k      Vqk1 Wqk2  k1 , q1 J k1 , q1  q ,q  k2 , q2 J k2 , q2  q ,q
2
2 2
1 1

 q ,q q ,q 1
1
2
1
 k ,k ;q ,q
1
2
1
2
k, q
2
 k1 , k2 ; q1, q2 J1 k1 , k2 ; q1 , q2
   Vq1 Wq2 
  k , k ; q , q  J k , k ; q , q
q1 , q2 q1 , q2
1
2
1
2
2
1
2
1
2

• We have complete set of states |k1,k2;q1,q2
 k1 
 k2 

 k1 , k2 ; q1 , q2 k , q

 J, Tq k     Vqk1 Wqk2  k1 , k2 ; q1, q2  J1  J 2  k , q   Vqk1 Wqk2  k1 , k2 ; q1, q2 J k , q
1
2
2

 q , q 1


q
,
q
1 2
1 2
• Now insert complete set of states |k’,q’:
 J, Tq k      Vqk1 Wqk2  k1 , k2 ; q1, q2 k , q k , q J k , q   Tqk  k , q J k , q
2

 k , q q , q 1
k  , q
1
2
• J doesn’t change the k value, so k’ = k
• So we have proven it
  Tqk  k , q J k , q
q
How it Comes Out
Tq k  
 k1   k2 
V
 q1 Wq2 k1 , k2 ; q1 , q2 k , q
q1 , q2
• This sum only makes sense if CG coefficients are non-zero
k1  k2  k  k1  k2
• 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
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)?
k 
Tq 
 Vq1 Wq2
 k1 
 k2 
V0  Vz , V1 
k1 , k2 ; q1 , q2 k , q
R0(1)  Z , R 1 
q1 , q2
 2
1 1
T2  R1 R1 1,1;1,1 2, 2 
1
1
2
  X  iY 
2

1
2
1
2
1  12 X 2  iXY  12 Y 2
T12  R11 R01 1,1;1, 0 2,1  R01 R11 1,1;0,1 2,1 
1
2
  X  iY  Z  12 
1
2
Vx  iVy 

X  iY 
   XZ  iYZ
T0 2  R11 R11 1,1;1, 1 2, 0  R01 R01 1,1;0, 0 2, 0  R11 R11 1,1; 1,1 2, 0
 12   X  iY  X  iY 

1
6

1
6

 Z2
2
3
T12  R01 R11 1,1;0, 1 2, 1  R11 R01 1,1; 1,0 2, 1 
 2
1 1
T2  R1 R1 1,1; 1, 1 2, 2 
1
2
 X  iY 
2

1
2
1
6
 2Z
2
 X 2 Y 2 
 X  iY  Z
1  12 X 2  iXY  12 Y 2
2
2
 XZ  iYZ
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  , j, m
• It is common to need matrix elements
 , j, m O  , j, m
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,
 , j, m Tq k   , j, m
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
Similarities With CG coefficients (1)
 , j, m Tq k   , j, m
j, m j, k ; m, q
• I want to compare the matrix element above to the CG coefficient above
• Recall relations for Jz:
 J z , Tq k    qTq k 



J z  , j, m  m  , j, m

• Use commutation
 , j, m J zTq k   Tq k  J z  , j, m   , j, m qTq k   , j , m
relation:
• Let Jz act on the bra
m  m   , j, m Tq k   , j , m  q  , j , m Tq k   , j , m

or the ket on the left:
m  q  m
• Hence matrix elements are zero unless
• Compare to the CG coefficient above:
– This vanishes unless
m  q  m
Similarities With CG coefficients (2)
 , j, m Tq k   , j, m
j, m j, k ; m, q
 J  , Tq k    k 2  k  q 2 qTqk1


J   , j, m 
j 2  j  m2 m  , j , m  1
• Recall relations for J:
• For m = j, J   , j, j  0
• Implies:  , j, j J   0
• Our commutation relations tell us:


 , j, j J Tq k   Tq k  J   , j , m 
0
k 2  k  q 2  q  , j , j Tq k   , j , m
k
k
j2  j  m2  m  , j, j Tq   , j, m  1  k 2  k  q 2  q  , j, j Tq1  , j, m
• Compare to the CG coefficients:
0
j2  j  m2  m j, j j, k ; m  1, q  k 2  k  q 2  q j, j j , k ; m, q 1
Similarities With CG coefficients (3)
 , j, m Tq k   , j, m
 J  , Tq k   


qTqk1
k 2  k  q2
j, m j, k ; m, q
J   , j, m 
j 2  j  m2 m  , j , m  1
• We have:
J   , j, m 
j 2  j  m2  m  , j , m  1
• Equivalent to
 , j , m J   j 2  j  m2  m  , j , m  1
• Our commutation relations tell us:
1
 , j, m J Tq k   Tq k  J   , j , m  k 2  k  q 2  q  , j , m Tqk1  , j , m


j 2  j  m 2  m  , j , m  1 Tq k   , j , m 
j 2  j   m2  m  , j , m Tq k   , j , m  1  k 2  k  q 2  q  , j , m Tqk1  , j , m
j 2  j  m 2  m j , m  1 j , k ; m, q

j 2  j   m2  m j , m j , k ; m  1, q  k 2  k  q 2  q j , m j , k ; m, q  1
These Matrix Elements are CG Coefficients
 , j, m Tq k   , j, m
j, m j, k ; m, q
• We have three relations that are identical for these two expressions: m  q  m
0
j2  j  m2  m  , j, j Tq   , j, m  1  k 2  k  q 2  q  , j, j Tq1  , j, m
k
k
j 2  j  m 2  m  , j , m  1 Tq k   , j , m 
j 2  j   m2  m  , j , m Tq k   , j , m  1  k 2  k  q 2  q  , j , m Tqk1  , j , m
• 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
 , j, m Tq k   , j, m  j, m j , k ; m, q
The Wigner-Eckart Theorem
 , j, m Tq k   , j, m  j, m j , k ; m, q
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:
1
 , j, m Tq  , j, m 
 , j T  k   , j j, k ; m, q j, m
2 j 1
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
k 
•
•
•
•
•
Why Is the Wigner-Eckart Theorem Useful?
k 
 , j, m Tq
1
 , j, m 
 , j T  k   , j
2 j 1
j, k ; m, q j, m
• The number of matrix elements is  2 j  1 2k  1 2 j  1
– 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
, j T
k 
 , j   , j T
k 
, j
*
Sample Problem
The magnetic dipole transition of hydrogen causing the 21 cm line is governed by
the matrix element 1s, F  0, mF  0  S  12 L  1s, F  1, mF , 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
1
1
• S and L are vector operators V  S  12 L
V0   Vz , V1   12  Vx  iVy 
1s,0,0 Vq1 1s,1, mF 
1
201
1s,0 V 1 1s,1 1,1; mF , q 0,0  A 1,1; mF , q 0,0
• Call reduced matrix element A:
• Non-vanishing only if q + mF = 0
• Get the CG coefficients from program
• All other matrix elements vanish
1s, 0, 0 V11 1s,1, 1 
1s, 0, 0 V0  1s,1, 0  
1
1s, 0, 0 V1  1s,1, 1 
1
1
3
1
3
1
3
A
A
A
Sample Problem
Deduce as much as you can about these matrix elements for mF = +1, 0, or –1.
1s, F  0, mF  0  S  12 L  1s, F  1, mF
• For mF = + 1, we also have
1s, 0, 0
1
2
 V
x
 iVy  1s,1, 1  0
1s, 0, 0 Vz 1s,1, 1  0
• Vx and Vy: two equations, two unknowns
• We therefore have:
1s, 0, 0  S  12 L  1s,1, 1 
1
6
A  xˆ  iyˆ 
• Similarly:
1s,0,0  S  12 L  1s,1, 1 
1s, 0, 0  S  12 L  1s,1, 0 
1
3
1
6
A  xˆ  iyˆ 
Azˆ
1s, 0, 0
1
2
V
x  iV y  1s,1, 1 
1s, 0, 0 Vz 1s,1, 0  
1s, 0, 0
1
2
 V
x
1
3
1
3
A
A
 iVy  1s,1, 1 
1s, 0, 0 Vx 1s,1, 1 
1
6
A
1s, 0, 0 Vy 1s,1, 1 
1
6
iA
1
3
A
8F. Integrals of Spherical Harmonics
Products of Spherical Harmonics
m1
m2
Y

,

Y


• Consider the product of any two spherical harmonics:
l1
l2  ,  
• By completeness, this can be written as a sum of spherical harmonics:
Yl1m1  ,   Yl2m2  ,     clmYl m  ,  
l ,m
clm   d  Yl
m
 , 
*
Yl1m1  ,   Yl2m2  ,  
• The coefficients clm can be found usingorthogonality:
• Think of the expression Yl1m1  ,   Yl2m2  ,  
m2
m1
Y

,

Y


as an operator acting on a wave function:
l2
l1
• It is not hard to see that this operator is a spherical tensor operator
 Lz , Yl2m2  ,    m2Yl2m2  ,   ,  L , Yl2m2  ,   
• Think of clm then as a matrix element
• By the Wigner-Eckart theorem:
• All that remains is to find
the reduced matrix elements
l22  l2  m22
m2 Yl2m2 1  ,  
clm  l , m Yl2m2 l1 , m1
clm 
1
2l 1
l Yl2 l1 l , m l1 , l2 ; m1 , m2
Working on the Reduced Matrix Element
clm   d  Yl
m
 , 
*
Yl1m1  ,   Yl2m2  ,   
1
2l 1
l Yl2 l1 l , m l1 , l2 ; m1 , m2
Yl1m1  ,   Yl2m2  ,     clmYl m  ,  
• Substitute the top equation in the bottom
Yl1m1  ,   Yl2m2  ,     21l 1 l Yl2 l1 l , m l1 , l2 ; m1 , m2 Yl m  ,  
l ,m
l ,m
• Multiply this expression by l1 , l2 ; m1 , m2 l ,0 and sum over m1, m2:
m1
m2

Y

,

Y


 l1
l2  ,   l1 , l2 ; m1 , m2 l , 0 
m1 , m2

1
2 l 1
sum over complete states
l Yl2 l1 l , m l1 , l2 ; m1 , m2 l1 , l2 ; m1 , m2 l , 0 Yl m  ,  
m1 , m2 l , m
m1
m2

Y

,

Y


 l1
l2  ,   l1 , l2 ; m1 , m2 l , 0  
m1 , m2
• Rename l’ as l
1
2 l 1
l Yl2 l1 l , m l , 0 Yl m  ,  
l ,m
1

l  Yl2 l1 Yl 0  ,  
2l   1
Finishing the Computation
 d  Yl
m
 , 
*
Yl1m1  ,   Yl2m2  ,   
1
2l 1
l Yl2 l1 l , m l1 , l2 ; m1 , m2
1
0
Y

,

Y

,

l
,
l
;
m
,
m
l
,0

l
Y
l
Y
 l2   1 2 1 2

l1` 
l2 1
l  ,  
2l  1
m1 , m2
• Must be true at all angles
2l  1
• Evaluate at  = 0
m
Yl   0,   
 m,0
• Formula for the Y’s at  = 0 is simple
4
m1
m2
2l1  1 2l2  1
1
2l  1
l1 , l2 ;0, 0 l , 0 
l Yl2 l1
4
4
2l  1 4
• Now we solve for the
reduced matrix element
• We therefore have
1
l Yl2 l1 
 2l1  1 2l2  1 l1 , l2 ;0,0 l ,0
4
 d  Y  ,  Y  ,  Y  ,  
m
l
*
m1
l1
m2
l2
 2l1  1 2l2  1
4  2l  1
l1 , l2 ;0,0 l ,0 l1 , l2 ; m1 , m2 l , m
When doesn’t it vanish?
 d  Y  ,  Y  ,  Y  ,  
*
m
l
m1
m2
l1
l2
 2l1  1 2l2  1
4  2l  1
l1 , l2 ;0,0 l ,0 l1 , l2 ; m1 , m2 l , m
We want to know when this is non-zero, or likely to be non-zero:
• We need: m  m  m
1
• We need:
2
l1  l2  l  l1  l2
• Under parity, each of the spherical harmonics transforms to
• So the whole integral satisfies
 d  Yl
m
 , 
• We need
*
m1
Yl1
 ,  Yl  ,    1
m2
2
l1  l2  l even
l1 l2 l
Yl   1 Yl m
m
l
m1
m2
d

Y

,

Y

,

Y




l1
l2  ,  
 l
m
*
l  l1  l2 , l1  l2  2, l1  l2  4,
, l1  l2
Sample Problem
Atoms usually decay spontaneously by the electric dipole process, in which case
the rate is determined by the matrix element  F R  I , 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?
 4,l,m  R4l   r  Yl m  , 
• The initial state has n’ = 4 and l’ = 0, 1, 2, or 3
• Final state has unknown n and l, but
 n,l ,m  Rnl  r  Yl m  , 
– Must have n < 4 because energy goes down
l n4
– Must have l < n
• The position operators can be written in terms of l = 1 spherical harmonics

* q
2
m
m
• So we have  R 
R
r
rR
r
r
dr
d

Y

,

Y

,

Y
F
I
 n,l   4,l    l   1   l  , 
0
l  1  l  l   1 and l   1  l even
• To not vanish, we need
• For 4s, l = 0: 1  l  1 and 0  1  l even
l 1
• Must have l < n < 4
4s  2 p,3 p
Sample Problem (2)
Atoms usually decay spontaneously by the electric dipole process, in which case
the rate is determined by the matrix element  F R  I , 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?
l  1  l  l   1 and l   1  l even
l n4
• For 4p: l’ = 1, so 0  l  2 and 1  1  l even
• Must have l < n < 4
4 p  1s, 2s,3s,3d
l  0, 2
• For 4d: l’ = 2, so 1  l  3 and 2  1  l even
4d  2 p,3 p
• Must have l < n < 4
l  1,3
• For 4f:
2  l  4 and 3  1  l even
• Must have l < n < 4
4 f  3d
l  2, 4