Product of two Clebsh-Gordan coefficients

Notation: let us have a particle of spin(more generally angular momentum, but for simplicity just consider spin) $j_1$ with projection to the third axis $m_1$ and a particle with spin $j_2$ with projection of $m_2$. We want to combine them into a doublet with total spin $J$ and its projection $M$. We write clebsch-gordan coefficient $( j_1 m_1 j_2 m_2 | J M )$.

Now consider we have two clebsh-gordan coefficients that we want to multiply: $$\sum_{M\in\{-J,..,J\}}( j_1 m_1 j_2 m_2 | J M )(j'_1 m'_1 j'_2 m'_2 | J M )$$ my question is: Is there a simple formula this can be reduced into?

$$\sum^{j_1+j_2}_{J=|j_1-j_2|}\sum^{J}_{M=-J}<j_1,m_1,j_2,m_2|J,M><J,M|j_1,m'_1,j_2,m'_2>=\\=<j_1,m_1,j_2,m_2|j_1,m'_1,j_2,m'_2>=\delta_{m_1,m'_1}\delta_{m_2,m'_2}$$
• Yeah, thanks, that really helped. I actually had to compute a special case when J=2 and $j_1=j_2=j'_1=j'_2=1$ and sum over the rest of $m$(in other words a product of two spherical tensors made of vector operators), so I know, that $( 1 1 1 -1 | 1 0 )=-( 1 -1 1 1 | 1 0 )$ and simmilarly for J=M=1, this will sum to zero and I am left with only the case of J=M=0, which is much easier to compute than 5 cases of M for J=2 – user74200 May 25 '18 at 0:00
The simplest way to look at such sums is to start with $$\langle j_1m_1; j_2m_2\vert j_1’m_1’; j_2’m_2’\rangle = \delta_{j_1j_1’}\delta_{j_2j_2’}\delta_{m_1m_1’}\delta_{m_2m_2’}\, ,$$ insert the unit operator in the form of $$1= \sum_{JM}\vert JM\rangle\langle JM\vert$$ where $$J$$ runs over all the allowed value in the coupling $$j_1\otimes j_2$$, $$M$$ runs from $$-J$$ to $$J$$, so as to get $$\sum_{JM}\langle j_1m_1; j_2m_2\vert JM\rangle \langle JM\vert j_1’m_1’; j_2’m_2’\rangle = \delta_{j_1j_1’}\delta_{j_2j_2’}\delta_{m_1m_1’}\delta_{m_2m_2’} \tag{1}$$ and recall that the CGs are real, i.e. $$\langle JM\vert j_1’m_1’;j_2’m_2’\rangle = \langle j_1’m_1’;j_2’m_2’\vert JM\rangle \tag{2}\, .$$ Inserting (2) in (1) gives the desired orthogonality relation.