Product of two Clebsh-Gordan coefficients

Question

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?

Answer 1

Quoting Wikipedia on Clebsch-Gordan coefficients we have the orthogonality relation:

$$\sum^{j_1+j_2}_{J=|j_1-j_2|}\sum^{J}_{M=-J}<j_1,m_1,j_2,m_2|J,M><J,M|j_1,m'_1,j_2,m'_2>=\\=<j_1,m_1,j_2,m_2|j_1,m'_1,j_2,m'_2>=\delta_{m_1,m'_1}\delta_{m_2,m'_2}$$

This is the closest I could find resembling what you presented above. I hope this helps.

Answer 2

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.