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I am trying to manipulate an expression involving summations in order to take advantage of a particular identity, but my result is off by some factors of -1. How can I convert:

$$f=\sum_{\kappa\psi\rho}(-1)^{2p+\psi+3r}\left(\begin{matrix}p&a&q\\\psi&\alpha&-\kappa\end{matrix}\right)\left(\begin{matrix}q&b&r\\\kappa&\beta&-\rho\end{matrix}\right)\left(\begin{matrix}r&c&p\\\rho&\gamma&-\psi\end{matrix}\right)$$

into something akin to:

$$f=A\sum_{\kappa\psi\rho}(-1)^{p-\psi+q-\kappa+r-\rho}\left(\begin{matrix}p&a&q\\\psi&\alpha&-\kappa\end{matrix}\right)\left(\begin{matrix}q&b&r\\\kappa&\beta&-\rho\end{matrix}\right)\left(\begin{matrix}r&c&p\\\rho&\gamma&-\psi\end{matrix}\right)?$$

Where $A$ is a multiplicative factor or some expression that does not involve summations (the whole point of this exercise is to eliminate the summations, and there is an identity that simplifies the sum in the second equation to a compact closed form expression).

If I multiply by $$(-1)^{-p+2\psi+q-\kappa-2r-\rho}$$ then do I have to multiply the other side of the equation by $$\sum_{\kappa\psi\rho}(-1)^{-p+2\psi+q-\kappa-2r-\rho}\;\;?$$

Note: $\left(\begin{matrix}a&b&c\\\alpha&\beta&\gamma\end{matrix}\right)$ is a Wigner 3j symbol. As such, the following relationships between the arguments to the 3j symbol hold:

  1. $0 \leq a, 0 \leq b, 0 \leq c$
  2. Any/all permutations of the triangular inequalities $\left| a-c \right| \leq b \leq a+b$
  3. $\left|\alpha\right| \leq a, \left|\beta\right| \leq b, \left|\gamma\right| \leq c$
  4. $\alpha+\beta+\gamma=0$

Additionally, in my case all arguments to the 3j symbol are integers.

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Your question is unclear. By "convert", do you mean, how do you show the two are equal, or do you mean, how are the two related, that is, what operations do you have to perform on the one to make it equal the other? Also, you refer to "the other side of the equation," but you don't actually write any equation, and it is not clear what equation you have in mind. Please think a little more about what you want to ask, and express it in such a way that others might understand you. –  Gerry Myerson Jun 29 '11 at 0:55
    
If the sum is over $\kappa \psi \rho$ then you could factor out the part of the power of $-1$ that doesn't depend on those variables. Knowing any relationships that hold between the indexes would also help. –  trutheality Jun 29 '11 at 2:36
    
@Gerry Myerson: I've update the question, hopefully it is more clear now, I want to manipulate the first expression so that I can take advantage of an identity for the summation in the form of the second expression, naturally other terms will pop out. –  okj Jun 29 '11 at 13:54
    
@trutheality: Once I factor out $(-1)^{p+2r}$ I still need to get $(-1)^{-2\psi+q-\kappa-\rho}$ into the summation some how. Any ideas? I've updated the question to include the relationships between the arguments now. –  okj Jun 29 '11 at 13:58
    
By "the following relationships between the arguments to the 3j symbol hold", you mean that the value is $0$ unless these relationships hold? –  joriki Jun 29 '11 at 14:13
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1 Answer 1

up vote 1 down vote accepted

I'll use different variables for your condition $4$ so as not to get them mixed up with the $\alpha$, $\beta$, $\gamma$ in your summation. So $\left(\begin{matrix}x&y&z\\ \lambda&\mu&\nu\end{matrix}\right)$ vanishes unless $\lambda+\mu+\nu=0$, and hence the parity of $\lambda+\nu$ is that of $\mu$.

Thus, in the exponent of $-1$ you can replace $\psi-\kappa$, $\kappa-\rho$ and $\rho-\psi$ by $\alpha$, $\beta$ and $\gamma$, respectively. If these three expressions were linearly independent, you could eliminate all occurrences of $\psi$, $\kappa$ and $\rho$ in this way, and then take the parity factors outside of the summation. They aren't, however, since their sum is zero (which leads to a constraint $\alpha+\beta+\gamma=0$; the sum vanishes unless this is fulfilled).

But you can still use these replacements to rewrite the sum. With $\psi=\psi+\psi-\psi=(\kappa-\alpha)+(\rho+\gamma)-\psi$, you get

$$ \begin{eqnarray} \sum_{\kappa\psi\rho}(-1)^{2p+\psi+3r}\,\Gamma &=& (-1)^{2p+3r}\sum_{\kappa\psi\rho}(-1)^{\psi}\,\Gamma\\ &=& (-1)^{2p+3r}\sum_{\kappa\psi\rho}(-1)^{(\kappa-\alpha)+(\rho+\gamma)-\psi}\,\Gamma\\ &=& (-1)^{2p+3r+\gamma-\alpha}\sum_{\kappa\psi\rho}(-1)^{\kappa+\rho-\psi}\,\Gamma\\ &=& (-1)^{2p+3r+\gamma-\alpha}\sum_{\kappa\psi\rho}(-1)^{-\kappa-\rho-\psi}\,\Gamma \end{eqnarray} $$

Here $\Gamma$ is the product of the three $3j$ symbols. (I'm not sure why you have $2p$ and $3r$ in the exponent, since these are equivalent to $0$ and $r$, respectively, if $p$ and $r$ are integers, but I've left them as you had them.)

So the factor $A$ that you're looking for is $(-1)^{p+q+\alpha+\gamma}$.

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