How to calculate $|1,1\rangle \otimes |1,-1\rangle=\alpha|2,0\rangle+\beta|1,0\rangle+\gamma|0,0\rangle$ for $\mathfrak{su}(2)$?

Question

At $\mathfrak{su}(2)$, tensor product $3 \otimes 3$ can be decomposed to $3\otimes3=5\oplus3\oplus1$.

In this space, $|1,1\rangle \otimes |1,-1\rangle$ has weight $0$.

So we should be able to write down $$|1,1\rangle \otimes |1,-1\rangle=\alpha|2,0\rangle+\beta|1,0\rangle+\gamma|0,0\rangle.$$

How to calculate $\alpha,\beta,\gamma$?

Answer 1

Apply $\langle 2,0\mid$ on both sides so you get $\alpha$ on the right hand side (assuming standard normalization) and the inner product

$$ \langle 2,0 \mid 1,1 ; 1,-1 \rangle $$

in the

$$ \langle J M \mid j_1 j_2 ; m_1 , m_2 \rangle $$

notation. Now look at the Clebsch-Gordan recurrence relations (the recursion @JyrkiLahtonen refers to).

This is the special case of $j_1=j_2=m_1=-m_2$ in the article. But check that you are using the same normalization convention as the article before using that to check your answer.

Answer 2

You are given, 4th relation, $$ \langle 1 \, 1 \, 1 \, (- 1) | J \, 0 \rangle = 2 \sqrt{\frac{2 J + 1}{(J + 2 + 1)! (2 - J)!}}~~, $$ so read off, as instructed, $$ \alpha=1/\sqrt{6}, \qquad \beta=1/\sqrt{2},\qquad \gamma=1/\sqrt{3}. $$