quantum mechanics - "Reversed" Clebsch Gordan coefficients

Consider the sum of angular momenta $J=J_1+J_2$.

When one has a state $|J,M\rangle$ of an eigenbasis of a common eigenspace of $J^2$ and $J_z$, one can write it in terms of the elements $|j_1,m_1;j_2,m_2\rangle$ of an eigenbasis of a common eigenspace of $J_1^2,\,J_2^2,\,J_{1z},\,J_{2z}$: $$|J,M\rangle=\underset{m_1,m_2}{\sum} C^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}\ \ |j_1,m_1;j_2,m_2\rangle$$ Where $C^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}$ are the Clebsch Gordan coefficients.

I can do this easily using a Clebsch Gordan Coefficient's table. What if I want to express the kets $|j_1,m_1;j_2,m_2\rangle$ in terms of the kets $|J,M\rangle$? In this case, I can write all the $|J,M\rangle$ in terms of the states $|j_1,m_1;j_2,m_2\rangle$ and combine them in such a way that I get one of the $|j_1,m_1;j_2,m_2\rangle$ states. This can lead to long calculations.

I wonder if there is a table like the one for Clebsch Gordan coefficients but for a kind of "reversed" Clebsch Gordan coefficients $B^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}$ such that $$|j_1,m_1;j_2,m_2\rangle=\underset{J,M}{\sum} B^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}\ \ |J,M\rangle$$

Answer

If I understand well, the coefficients $B$ are just in fact CG's: to be explicit $$ \vert j_1m_1; j_2m_2\rangle = \sum_{J(M)} C^{j_1j_2J}_{m_1m_2M} \vert JM\rangle $$ Note the sum of $M$ is not really a sum as $M$ must satisfy $M=m_1+m_2$.

The best way to see this is to start with $\vert j_1m_1; j_2m_2\rangle$ and just insert the unit $I=\sum_{JM}\vert JM\rangle\langle JM\vert$. One then has $$ \vert j_1m_1; j_2m_2\rangle=\sum_{JM}\vert JM\rangle\langle JM\vert j_1m_1;j_2m_2\rangle $$ with $$ \langle JM\vert j_1m_1;j_2m_2\rangle = \langle j_1m_1;j_2m_2\vert JM\rangle = C^{j_1j_2J}_{m_1m_2M} $$ since the CGs are real.