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I know that up to constant factors ($\pi$'s and such), the Fourier transform of the Coulomb potential $$\frac{1}{4\pi}\frac{1}{r}$$ in 3-dimensions is proportional to $$\int d^3\vec r\frac{e^{i\vec k \cdot \vec r}}{4\pi r}\propto\frac{1}{\vec k^2}.$$

I am trying to reproduce this result using the plane-wave expansion formula which states $$e^{i\vec k \cdot \vec r} = 4\pi \sum_{\ell=0}^\infty \sum_{m=-\ell}^{\ell} j_\ell(k r)Y^m_\ell (\hat k) Y^{m*}_\ell (\hat r).$$

Using this, $$ \int d^3\vec r\frac{e^{i\vec k \cdot \vec r}}{4\pi r} = \int d\Omega dr r^2\left(\sum_{\ell=0}^\infty \sum_{m=-\ell}^{\ell} j_\ell(k r)Y^m_\ell (\hat k) Y^{m*}_\ell (\hat r)\right)\frac{1}{r} $$ $$= \int_0^\infty dr r j_0(kr) = \frac{1}{k^2}\int_0^\infty d\rho \rho j_0(\rho) = \frac{1}{k^2}\int_0^\infty d\rho \sin(\rho).$$

Here I used the fact that $\int d\Omega Y^{m}_\ell(\hat r)$ vanishes for all but $\ell = m = 0$ and the explicit form of the sphericla Bessel function $j_0(\rho) = \sin(\rho)/\rho$. The problem is then that the last integral does not converge and certainly does not give 1, so I cannot reproduce the result.

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  • $\begingroup$ These Fourier transforms normally need to be regulated. Have you tried just putting a Tauberian regulator $e^{-\alpha\rho}$ in the integrand and taking $\rho\rightarrow0$ at the end? $\endgroup$
    – Buzz
    Aug 23, 2023 at 2:47
  • $\begingroup$ Yes, looks like that's it. I'm used to using the regulator when I do the standard contour integration way to take this transform, but it looks like you need the regulator when using this plane wave expansion too. $\endgroup$ Aug 23, 2023 at 3:47

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