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

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.

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. $$c_{m_1}=c_{-m_1}$$.