Calculating Clebsch Gordon Coefficients

[quasar_4]

Homework Statement

Find the Clebsch Gordon coefficients of 1/2 \otimes 1 = 3/2 \oplus 1/2

Homework Equations

We have some properties of the CG coefficients which might be useful:

1) they are nonzero only if j is between j1-j2 and j1+j2
2) m = m1+m2 for nonzero coefficients
3) they are real

The Attempt at a Solution

I am horribly confused. I know that the CG coefficients are given as the coefficients in the expansion

|j m, j_1 j_2 \rangle = \sum_{m_1} \sum_{m_2} | j_1 m_1, j_2 m_2 \rangle \langle j_1 m_1, j_2 m_2| jm, j_1 j_2\rangle

or

\langle j_1 m_1, j_2 m_2 | j m \rangle

and I know that the possible |j m j1 j2> states are for the product space with j1 = 1, j2 = 1/2:

| \frac{3}{2} \frac{3}{2} ,1 \frac{1}{2} \rangle, | \frac{3}{2} \frac{1}{2},1 \frac{1}{2} \rangle, | \frac{3}{2} \frac{-1}{2} ,1 \frac{1}{2} \rangle, | \frac{3}{2}\frac{-3}{2} ,1 \frac{1}{2} \rangle, | \frac{1}{2} \frac{1}{2} ,1 \frac{1}{2} \rangle, | \frac{1}{2} \frac{-1}{2} ,1 \frac{1}{2} \rangle

but I don't understand what on Earth to do or where to even start. Any help would be great.

 

[nickjer]

You usually just look them up in a table. I hope your assignment doesn't want you to actually derive them. Also, I am a bit confused on your notation in the first sentence.

 

[quasar_4]

Yes, I think the notation is what's confusing me. I computed some of the coefficients a couple of years ago in an undergraduate class, but I've never seen this notation. I guess what the author means is to find the coefficients for all the basis kets in the product space where j1 = 1/2, j2 =1. There are (2*j1+1)*(2j2+1)=6 such kets.

I actually have figured it out - but boy, I think Shankar has weird notation...

All one needs do is to start with the maximal possible z component, and apply lowering operators for each given j value, along with applying orthogonality and normalization conditions. A bit tricky but doable...

 

[vela]

Start with the | 3/2\mbox{ }3/2 \rangle state. You know that it has to correspond to |1/2\mbox{ }1/2; 1\mbox{ }1\rangle. Then apply the lowering operator to get | 3/2\mbox{ }1/2 \rangle, and so on.Edit: Oh sure, figure it out right before I post! ;)

 

[quasar_4]

Thanks anyway! It's good to know that my solution wasn't just pure nonsense :-)