# Computation of relation between angular power spectrum and matter power spectrum

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.

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 $$\delta(z-z')$$.

• 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
– user87745
Apr 14, 2021 at 11:26
• Yes. $k$ or $q$ is a dummy index. You may replace $q$ by $k$ in the end.
– ytlu
Apr 14, 2021 at 11:28
• ok, it is kind from your part.
– user87745
Apr 14, 2021 at 11:29
• 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 ?
– user87745
Apr 14, 2021 at 12:07
• 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.
– ytlu
Apr 14, 2021 at 12:33