Calculate the Laplacian of $\frac{e^{ikr}}r$ in spherical coordinates

Question

The expression has a divergence at the origin. The result shall contain a $\delta$-function. I tried 2 different ways to get 2 different answer. I guess neither of them is correct.

1. Use the identity: \begin{eqnarray} && \nabla^2 = \frac{1}{r^2}\bigg[\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{\sin\theta} \frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \varphi^2}\bigg] \\ && \nabla^2 f(r) =\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2\frac{\partial }{\partial r}\right)f(r) =\left(\frac{d^2}{d r^2} +\frac{2}{r}\frac{d}{dr}\right)f(r) . \end{eqnarray} So: \begin{equation} \nabla^2\left(\frac{e^{ikr}}{r}\right)=\left(\frac{d^2}{d r^2} +\frac{2}{r}\frac{d}{dr}\right)\left(\frac{e^{ikr}}{r}\right) =-k^2\frac{e^{ikr}}{r} . \end{equation}

2. Use: \begin{eqnarray} && \vec{\nabla} = \vec{e}_{\vec{r}} \frac{\partial}{\partial r} + \vec{e}_{\theta} \frac{1}{r}\frac{\partial}{\partial \theta} + \vec{e}_{\varphi} \frac{1}{r\sin\theta}\frac{\partial}{\partial \varphi}\\ && \vec{\nabla}f(r) = \left( \vec{e}_{\vec{r}} \frac{d}{dr} \right)f(r) = \frac{d f(r) }{dr} \vec{e}_{\vec{r}} \\ && \nabla^2 f(r) =\left(\frac{d^2}{d r^2} +\frac{2}{r}\frac{d}{dr}\right)f(r) \\ && \nabla^2 (f(r)g(r)) = (\nabla^2 f(r)) g(r) + 2(\vec{\nabla}f(r))\cdot(\vec{\nabla} g(r))+f(r)\nabla^2 g(r) . \end{eqnarray} So: \begin{eqnarray} && \nabla^2 \left(\frac{e^{ikr}}{r}\right) =\frac{1}{r}\nabla^2 e^{ikr} + e^{ikr}\nabla^2\left(\frac{1}{r}\right) + 2(\vec{\nabla}e^{ikr}) \cdot\vec{\nabla}\left(\frac{1}{r}\right) \end{eqnarray} in which I used: \begin{eqnarray} && \vec{\nabla}e^{ikr} = \left( \vec{e}_{\vec{r}} \frac{d}{dr} \right) e^{ikr}=ik e^{ikr}\vec{e}_{\vec{r}}\\ && \nabla^2 e^{ikr}=\left(\frac{d^2}{d r^2} +\frac{2}{r}\frac{d}{dr}\right)e^{ikr} = -k^2 e^{ikr}+\frac{2ik}{r}e^{ikr}\\ && \vec{\nabla}\left(\frac{1}{r}\right) = \left( \vec{e}_{\vec{r}} \frac{d}{dr} \right) \left(\frac{1}{r}\right) = -\frac{1}{r^2}\vec{e}_{\vec{r}} \\ && \nabla^2 \left(\frac{1}{r}\right) = -4\pi\delta(\vec{r}) . \end{eqnarray} The final result I get in this way is: \begin{eqnarray} && \nabla^2 \left(\frac{e^{ikr}}{r}\right) =\frac{1}{r}\left( -k^2 e^{ikr} +\frac{2ik}{r}e^{ikr}\right) + e^{ikr}\left[-4\pi\delta(\vec{r})\right] + 2 \left(ik e^{ikr}\vec{e}_{\vec{r}}\right)\cdot\left[-\frac{1}{r^2}\vec{e}_{\vec{r}}\right] \\ && = -k^2\frac{e^{ikr}}{r}-4\pi\delta(\vec{r}) e^{ikr} . \end{eqnarray} The result still does not look right to me because I would expect to get: \begin{eqnarray} && \nabla^2 \left(\frac{e^{ikr}}{r}\right) = -k^2\frac{e^{ikr}}{r}-4\pi\delta(\vec{r}) \end{eqnarray} Why the above 2 results are different and what did I do wrong?

L. H. Y.

Answer 1

With the suggestion of Brian Borchers I realize that my 2nd solution is correct. This is to say, $\delta(\vec{r})e^{ikr}=\delta(\vec{r})$ for the follow reasons:

1. $\delta(\vec{r})e^{ikr}$ is non-zero only at $r=0$, and

2. $\int \delta(\vec{r})e^{ikr} d^3 \vec{r}=1$.

Therefore $\delta(\vec{r})e^{ikr}$ fits the definition for the $\delta$-function.

But I still don't know why my 1st solution is wrong.

L. H. Y.