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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
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2 Answers

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.

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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.

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