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I would like to convince myself of the following relationship in an astrophysical context:

\begin{aligned} & \sum_{m}\sum_{m^{\prime}}\left\langle a_{\ell m} a_{\ell m}^* a_{\ell m^{\prime}} a_{\ell m^{\prime}}^*\right\rangle \\ & =\left[\sum_m\left\langle a_{\ell m} a_{\ell m}^* a_{\ell m} a_{\ell m}^*\right\rangle+\sum_{m^{\prime}\neq m} \left\langle a_{\ell m} a_{\ell m}^* a_{\ell m^{\prime}} a_{\ell m^{\prime}}^*\right\rangle\right] \\ & =\left(3 C_{\ell}^2(2 \ell+1)+2 \ell C_{\ell}^2(2 \ell+1)\right). \end{aligned}

I apply Wick's theorem for computing the expectation of $a_{\ell m} $:

Wick's theorem

EDIT 1: I think that my issue of understanding is about the second sum when $m\neq m'$. I can't represent myself the number of terms in this sum ( equal to $2\ell\ \times(2\ell +1)$)

EDIT 2: I think the keypoint is the number of different pairs ($m,m^{\prime}$).

$A_{n}^{p}= \dfrac{(2\ell+1)!}{(2\ell-1)!}= (2\ell+1)\times 2\ell$

Is this interpretation correct?

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    $\begingroup$ Are those supposed to be steps in an equality that just happens to be missing the all-important equals sign? Or is it a product of terms? What are $a_{\ell m}$? And $C_\ell$? $\endgroup$
    – Kyle Kanos
    Commented Jun 7, 2023 at 15:27
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    $\begingroup$ You seem to be summing over m s twice? $\endgroup$
    – Cosmas Zachos
    Commented Jun 7, 2023 at 15:47
  • $\begingroup$ @CosmasZachos .yes since I compute the case $m=m^{\prime}$n is it right ? $\endgroup$
    – guizmo133
    Commented Jun 7, 2023 at 16:43
  • $\begingroup$ Illustrate this to yourself by m ranging over 1, 2. What do you get? $\endgroup$
    – Cosmas Zachos
    Commented Jun 7, 2023 at 16:49
  • $\begingroup$ @KyleKanos . Sorry, the definition of a $$C_{\ell}=\dfrac{1}{2 \ell+1} \sum_{m=-\ell}^{+\ell}\left|a_{\ell m}\right|^{2}$$ $\endgroup$
    – guizmo133
    Commented Jun 7, 2023 at 17:00

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