# Clebsch-Gordan Identity

I'm trying to take advantage of a particular identity for the sum of the product of three Clebsch-Gordan coefficients, however, the present form of my equation is slightly different. Is there a symmetry relation that will allow me to change:

$\sum_{\alpha\beta\delta}C_{a\alpha b\beta}^{c\gamma}C_{d\delta b\beta}^{e\epsilon}C_{d\delta f\phi}^{a\alpha}$

Into:

$\sum_{\alpha\beta\delta}C_{a\alpha b\beta}^{c\gamma}C_{d\delta b\beta}^{e\epsilon}C_{a\alpha f\phi}^{d\delta}$

Notice I need to swap $j_2m_2$ with $jm$ in the last Clebsh-Gordan coefficient. Does anyone know a way to do this?

Note: My notation follows that of Varshalovich, $C_{j_1 m_1 j_2 m_2}^{jm}$

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What are those sums supposed to add up to? –  Dan Jun 29 '11 at 0:09
What range are those sums over? –  Dan Jun 29 '11 at 0:18
@Dan: The sums are over all valid values of the arguments, specifically $-a\leq\alpha\leq a, -b\leq\beta\leq b, -d\leq\delta\leq d$ –  okj Jun 29 '11 at 13:27
In that case this equivalence is true only when $a=d$. –  Dan Jun 30 '11 at 0:41

Notice that $C^{22}_{1111}=1$ but $C^{11}_{2211}=0$. I don't think that this is true unless $a=d$ and the sums over $\alpha$ and $\delta$ have the same range.
In general you cannot make the change you suggest because of the condition on projections. In your first equation, the projections in your last CG must satisfy $\delta +\phi=\alpha$ whereas in your second equation, the projections in your last CG must satisfy $\alpha+\delta=\phi$. Thus, unless there is further symmetry that you have not mentioned in your problem, for instance $\alpha=\delta$, there is no way to transform the first into the second.