Adding angular momenta

1. Presentation

Two angular momenta J1 and J2. The 
set of basis vectors of J1 is {j1,m1}, and he 
set of basis vectors of J2 is {j2,m2}.

J1 = Σ(i) ci |j1 mi> [mi from - j1 to + j1] = 
Σ(m1) cm1 |j1 m1> [m1 from - j1 to + j1]

Similarly,
J2 = Σ(m2) cm2 |j2 m2> [m2 from - j2 to + j2]

The eigenstates of J12 and J1z satisfy:

J12 = j1(j1 + 1)ℏ2|j1 m1>
J1z = m1ℏ|j1 m1>
and
J1+ |j1,m1>  = ℏ[j1(j1 + 1) - m1(m 1+ 1)]1/2|j1,m1+1> 
J1- |j1,m1>  = ℏ[j1(j1 + 1) - m1(m1 - 1)]1/2|j1,m1-1>

The eigenstates of J22 and J2z satisfy:

J22 = j2(j2 + 1)ℏ2|j2 m2>
J2z = m2ℏ|j2 m2>

and
J2+ |j2,m2>  = ℏ[j2(j2 + 1) - m2(m2 + 1)]1/2|j2,m2+1> 
J2- |j2,m2>  = ℏ[j2(j2 + 1) - m2(m2 - 1)]1/2|j2,m2-1>


Now, we are interested to add J1 to J2. 
J = J1 + J2
With J = Σ cm |j m> [m from - j to + j]
Where |j m> is the eigenstates of J.

J must satisfy:
J2 |J M> = J(J + 1)ℏ2 |J M>
Jz |J M> = M ℏm |J M>
M: from - J to + J

and
J+ |J,M>  = ℏ[J(J + 1) - M(M + 1)]1/2|J,M+1> 
J- |J,M>  = ℏ[J(J + 1) - M(M- 1)]1/2|J,M-1>


What is then the transformation coefficients from 
{|j1,m1>} and {|j2,m2>} to {|J,M>}?

2. Clebsch-Gordan coefficients

The transformation needed is just the expansion the new basis 
vectors {J,M} in terms of the old  {j1,m1} and {j2,m2}.

We have:
|J,M> = Σ(m1) Σ(m2) |j1 m1; j2m2><j1m1;j2m2|J M>

m1 from - j1 to + j1 and m2 from - j2 to + j2 

The coefficients <j1m1;j2m2|J M> are called the 
Clebsch-Gordan coefficients (CGC). They are null unless M = m1 + m1.


General expression of the Clebsch-Gordan Coefficients:

<j1m1;j2m2|J M>  = δ(m1+m1,M) x 
[(2J+1) x (s-2J)!(s-2j2)!(s-2j1)!/(s+1)!) x 
(j1+m1)!(j1-m1)!(j2+m2)!(j2-m2)!(J+M)!(J-M)!]1/2  x
Σ(-1)k[1/k!(j1+j2-J-k)!(j1-m1-k)!(j2+m2-k)!(J-j2+m1+k)!(J-j1-m2+k)!]

Where s = j1 + j2 + J, and k runs from 0 to an integer 
while the factorial factors are grater or equal to 0.

The minimum value of J is |j1 - j2|, while its maximum is j1 = j2.

3. Special cases for the CGC

Case1:
<j1 j2;j1 j2|J J> = 1

Case2:
If m1 = +j1 or m2 = +j2  and M = +J, or
m1 = -j1 or m2 = -j2  and M = -J, then:
<j1 m1;j2 m2|J J> = <j1 -m1;j2 -m2|J -J> = 
(-1)j1 - m1  x [(2J+1)!(j1+j2-J)!/(j1+j2+J+1)!(J +j1 - j2)!(J-j1+j2)!]1/2 x 
[(j1+m1)!(j2+m2)!/(j1-m1)!(j2 - m2)!]1/2

Case3:
If j1 + j2 = J, then:
<j1 m1;j2 m2|J M> = [(2j1)!(2j2)!/(2J)!]1/2 x
 [(J+M)!(J-M)!/(j1+m1)!(j1-m1)!(j2+m2)!(j2 - m2)!]1/2

4. Example: Two spin 1/2

j1 = 1/2 and j2 = 1/2
J = j1 + j2 
The max value of J is j1 + j2 = 1/2 + 1/2 = 1. 
The min value is |1/2 - 1/2| = 0

For J, we have:
|0,0>, |1,-1> |1,0>, and |1,1>

j1 = 1/2, then m1 = - j1, 0, + j1, that is -1/2, 0, + 1/2. 
Similarly for j2. Then: 

1.
|0,0> = ΣΣ|1/2 m1;1/2 m2><1/2 m1; 1/2 m2|0 0>
m1 and m2 from -1/2 to + 1/2
Since m1 + m2 = M = 0, then only m1 = -1/2 and m2 = +1/2 , 
and m1 = +1/2 and n2 = -1/2 contribute.

= Σ(|1/2 m1;1/2 -1/2><1/2 m1; 1/2 -1/2|0 0> + 
1/2 m1;1/2 +/2><1/2 m1; 1/2 +1/2|0 0>)
 m1: -1/2 and + 1/2
= 
|1/2 -1/2 ;1/2 -1/2><1/2 -1/2 ; 1/2 -1/2|0 0> + 
|1/2 -1/2 ;1/2 +/2><1/2 -1/2 ; 1/2 +1/2|0 0> +
|1/2 1/2;1/2 -1/2><1/2 1/2; 1/2 -1/2|0 0> + 
|1/2 1/2;1/2 +/2><1/2 1/2; 1/2 +1/2|0 0> 

Only the following two terms contribute:
<1/2 -1/2 ; 1/2 +1/2|0 0> 
<1/2 1/2; 1/2 -1/2|0 0> 

Using the second case, we have:

<1/2 1/2; 1/2 -1/2|0 0>  = 
(-1)1/2 -1/2[1!1!/2!0!0!]1/2 [1!0!/0!1!]1/2 = 1/21/2
<1/2 -1/2 ; 1/2 +1/2|0 0>  = 
(-1)1/2 +1/2[1!1!/2!0!0!]1/2 [1!0!/0!1!]1/2 = - 1/21/2 

Hence:
!0,0> = 1/21/2(|1/2 1/2;1/2 -1/2> - |1/2 -1/2 ;1/2 +/2>)

2.
|1 -1> = |1/2 -1/2;1/2 -1/2>

3.
|1 0> =  ΣΣ|1/2 m1;1/2 m2><1/2 m1; 1/2 m2|1 0>
Here two terms contribute: m1 = 1/2, m2 = -1/2 and 
m1 = -1/2 , m2 = 1/2. Hence,
|1 0> =  |1/2 1/2;1/2 -1/2><1/2 1/2; 1/2 -1/2|1 0> + 
|1/2 -1/2;1/2 1/2><1/2 -1/2; 1/2 1/2|1 0>

Since j1 + j2 =1, we use the case 3, we find:
<1/2 1/2; 1/2 -1/2|1 0> = (1!1!/2!)1/2 (1!1!/1!0!0!1!)1/2 = 1/21/2
<1/2 -1/2; 1/2 1/2|1 0> = (1!1!/2!)1/2 (1!1!/0!1!1!0!)1/2 = 1/21/2

Therefore
|1 0>  = 1/21/2(|1/2 1/2; 1/2 -1/2> + |1/2 1/2; -1/2 1/2>)

4.
|1 1> =  ΣΣ|1/2 m1;1/2 m2><1/2 m1; 1/2 m2|1 1>
m1 + m2 = M = 1 is satified only by one couple: 
m1 = 1/2 and m2 = 1/2, so

|1 1> =  |1/2 1/2;1/2 1/2><1/2 1/2; 1/2 1/2|1 1>
The case 1 gives <1/2 1/2; 1/2 1/2|1 1> = 1, therefore:

|1 1> =  |1/2 1/2;1/2 1/2>

To recap:

Two spin 1/2: j1 = j2 = 1/2 → m1= -1/2, +1/2, 
and m2 = -1/2, + 1/2
Jmax = j1 + j2 = 1, and Jmin = |j2 - j1| = 0
J = 0 , 1 → M = 0 for J = 0, and J = -1, 0, +1 for J = 1, 
We have 4 kets:|0 0 >, |1  -1>|1 0>|1 1>
|0,0> = 1/21/2(|1/2 1/2;1/2 -1/2> - |1/2 -1/2 ;1/2 +1/2>)
|1 -1> = |1/2 -1/2;1/2 -1/2>
|1 0>  = 1/21/2(|1/2 1/2; 1/2 -1/2> + |1/2 1/2; -1/2 1/2>)
|1 1> =  |1/2 1/2;1/2 1/2>