Some Clebsch-Gordan coefficients for $j_{1}=1$ and $j_{2}=1$

Question

I've successfully derived every coefficient, but not the one that has $j=0$. Starting from $|J=2,M=2⟩$ and applying $J_{-}$ we derive $|2,1⟩$ and $|2,0⟩$ and using orthonormality (and the Condon-Shortley convention) we can obtain $|1,1⟩$ and $|1,0⟩$.

The "difficulty" lies in obtaining $|0,0⟩$, because now I have to impose 2 conditions of orthonormality, one for $|2,0⟩$ and the other for $|1,0⟩$.

My professor proposed to use $$J_{-}|0,0⟩$$ because it makes everything more compact, the two conditions of orthonormality in one expression. (Applying $J_{-}$ to $|0,0⟩$ gives you zero)

How is it possible to get the coefficients for $|0,0⟩$ with this method? I have tried but failed.

Answer 1

Write $\vert 0,0\rangle$ as a general combination of $\vert j_1m_1\rangle \vert j_2m_2\rangle$ states so that $m_1+m_2=0$: $$ \vert 0,0\rangle=\sum_{m_1} c_{m_1} \vert 1,m_1\rangle \vert 1,-m_1\rangle $$ This guarantees $J_z$ has eigenvalue $0$. Then use not only $J_-\vert 0,0\rangle=0$ but also $J_+\vert 0,0\rangle=0$ to obtain conditions on the $c_{m_1}$ coefficients. You now have enough to solve, up to a normalization and a phase. Indeed the $c_{m_1}$ will be the CGs when normalized if you choose the CS phase convention.

Additionally, because you are tensoring two $j=1$ states, the resulting tensor product states will be either symmetric or antisymmetric. It turns out there are $3$ antisymmetric combinations and $6$ symmetric combinations. The $6$ symmetric states are the $J=2$ and the $J=0$ states, i.e. your $\vert 0,0\rangle$ state will be symmetric, i.e. $c_{m_1}=c_{-m_1}$.