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How to write the 3D power spectrum, $\mathrm{P}(\mathrm{k})$, as an integral of the angular power spectrum, $\mathrm{C_\ell}$ ?

I have the following equation, $$ C_{\ell}\left(z, z^{\prime}\right)=\int_{0}^{\infty} d k k^{2} j_{\ell}(k z) j_{\ell}\left(k z^{\prime}\right) P(k) $$ where $j_{\ell}$ are the spherical Bessel functions.

I would like to invert this relation and write $\mathrm{P}(\mathrm{k})$ as a function of $\mathrm{C_\ell}$.

I don't know if this is a well known result, but I couldn't find anything.

Any ideas on how to tackle this problem

PS : maybe this post deserves to be placed in astrophysics exchange forum. Don't hesitate to move it on the appropriate forum.

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Given $$ \tag{1} C_{\ell}\left(z, z^{\prime}\right)=\int_{0}^{\infty} d k k^{2} j_{\ell}(k z) j_{\ell}\left(k z^{\prime}\right) P(k) $$ Question: how to invert the integral to find the function $P(k)$?


The closure relation for spherical Bessel function: $$ \tag{2} \int_0^\infty x^2 j_n(xu) j_n(xv) dx = \frac{\pi}{2u^2} \delta(u-v). $$

Multipy Eq.(1) with $z^2 j_\ell(qz)$ and integral over $z$: \begin{align} \int_0^\infty z^2 j_\ell(qz) C_{\ell}\left(z, z^{\prime}\right) dz =&\int_{0}^{\infty} d k k^{2} \left\{ \int^0_\infty z^2 dz j_\ell(qz) j_{\ell}(k z)\right\} j_{\ell}\left(k z^{\prime}\right) P(k) \\ =&\int_{0}^{\infty} d k k^{2} \left\{\frac{\pi}{2q^2} \delta(q-k)\right\} j_{\ell}\left(k z^{\prime}\right) P(k) \\ =& q^{2} \frac{\pi}{2q^2} j_{\ell}\left(q z^{\prime}\right) P(q) \tag{3}. \end{align}

Once again multiply Eq.(3) with $z'^2 j_\ell(q'z')$ and integral over $z'$ \begin{align} \int_0^\infty z'^2 dz' j_\ell(q'z') \int_0^\infty z^2 j_\ell(qz) C_{\ell}\left(z, z^{\prime}\right) dz =& \frac{\pi}{2} \left\{\int_0^\infty z'^2 dz' j_\ell(q'z') j_{\ell}(q z') \right\} P(q).\\ =& \frac{\pi}{2} \left\{ \frac{\pi}{2q'^2} \delta(q-q') \right\} P(q) \tag{4}.\\ \end{align}

To move the $\delta$ function in the right-hand-side, we multiply Eq. (4) (note that only $q=q'$ has contribution) with $q'^2$ and integral over $q'$: \begin{align} \int_0^\infty dq' q'^2\int_0^\infty z'^2 dz' j_\ell(q'z') \int_0^\infty z^2 j_\ell(q'z) C_{\ell}\left(z, z'\right) dz =& \frac{\pi^2}{4} \int_0^\infty dq' \delta(q-q') P(q).\\ =& \frac{\pi^2}{4} P(q) \tag{5}. \end{align}

The left-hand-side of Eq.(5); \begin{align} \int_0^\infty dq' & q'^2\int_0^\infty z'^2 dz' j_\ell(q'z') \int_0^\infty z^2 j_\ell(q'z) C_{\ell}\left(z, z'\right) dz \\ = & \int_0^\infty z'^2 dz' \int_0^\infty z^2 dz \left\{ \int_0^\infty dq' q'^2 j_\ell(q'z') j_\ell(q'z) \right\} C_{\ell}(z, z') \\ = & \int_0^\infty z'^2 dz' \int_0^\infty z^2 dz \left\{ \frac{\pi}{2z^2} \delta(z-z') \right\} C_{\ell}(z, z') \\ = & \frac{\pi}{2} \int_0^\infty z^2 dz C_{\ell}(z, z). \tag{6} \end{align}

Combine Eq.(5) and Eq.(6) $$ P(q) = \frac{2}{\pi} \int_0^\infty z^2 dz C_{\ell}(z, z) = P_0. $$

$P(k)$ is a constant independent of $k$. This renders Eq.(1) to be: \begin{align*} C_{\ell}\left(z, z^{\prime}\right) &=\int_{0}^{\infty} d k k^{2} j_{\ell}(k z) j_{\ell}\left(k z^{\prime}\right) P(k);\\ &= P_0 \int_{0}^{\infty} d k k^{2} j_{\ell}(k z) j_{\ell}\left(k z^{\prime}\right)\\ &= \frac{\pi P_0}{2 z^2} \delta \left( z- z'\right). \end{align*}

It means that if a two-variable function $C_\ell(z. z')$ can be bessel-fourier transformed with a single $k$, then $C_\ell(z, z')$ is a delta-function $\delta(z-z')$.

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    $\begingroup$ ytlu. what a beautiful demonstration (I am only a student) ! That would mean that we can pass from 3D to 2D and inversely ? and what does represent the variable $q$ : is it related to $k$ wave number ? Thanks a lot for your help, Regards $\endgroup$
    – user87745
    Apr 14, 2021 at 11:26
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    $\begingroup$ Yes. $k$ or $q$ is a dummy index. You may replace $q$ by $k$ in the end. $\endgroup$
    – ytlu
    Apr 14, 2021 at 11:28
  • $\begingroup$ ok, it is kind from your part. $\endgroup$
    – user87745
    Apr 14, 2021 at 11:29
  • $\begingroup$ just a question : $C_{\ell}$ has no dependence in $k$ scale ? only angular dependent and redshift dependent ? since only redshift $z$ appears in your expression ? $\endgroup$
    – user87745
    Apr 14, 2021 at 12:07
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    $\begingroup$ That bothers me either. But since I don't understand the physics of your equation, I have no clues at all. I just did it as a math problem. $\endgroup$
    – ytlu
    Apr 14, 2021 at 12:33

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