# Sum relation for Clebsch-Gordan-Coefficients?

In the context of (numerically) calculating reduced density matrices in the Lipkin-Meshkov-Glick model (a model introduced to describe atomic nuclei, which has however found many other applications as well), I need to carry out a sum over Clebsch-Gordan coefficients.

If it is possible to find a simplification that replaces the sum below by a (simple) explicit expression, the numerical effort would be greatly reduced (the angular momenta in question are large, therefore the sum over $m_1$ has many terms). The sum in question is:

$\sum_{m_1} \langle J M|j_1 m_1 j_2 m_2\rangle \langle j_1 m_1 j_2 m_2'|J M'\rangle$

I have found sum formulas for Clebsch-Gordan coefficients, but none applicable to this one. Any insight would therefore be greatly appreciated!

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I doubt this is helpful, but on page 26 of "The Classical and Quantum 6j-symbols" we give a closed formula for the $sl_2(C)$ coefficients in a particular normalizations that is compatible with the quantum case.
If you sum on both $m_1$ and $m_2=m_2'$ then it does simplify into $\delta_{M,M'}$. Without $m_2$ summation, I don't think one can get much better than that. It all depends also on what larger tensor contraction your identity is inserted into. Is this only a small part of a larger formula with a product of a lot of CG coefficients and many sums over $m$'s?