Volume charge density of hydrogen atom The potential of the hydrogenatom is given by:
$$\phi(r) = \frac{1}{4 \pi \epsilon_0} \frac{e}{r} ( 1 + \frac{r}{a_0}) e^{-\frac{2r}{a_0}}$$
I'm no supposed to find the volume charge density $\psi$ which produces such a potential, which conceptually is fairly simple, just slap on the Laplacian from the left side and you're basically done. But I'm having some trouble actually computing it, here's how far I've gotten so far:
$$\nabla ^2 \phi(r) = \frac{1}{4 \pi \epsilon_0} \nabla ( \nabla \frac{e}{r} ( 1 + \frac{r}{a_0}) e^{-\frac{2r}{a_0}} = \frac{1}{4 \pi \epsilon_0} \nabla ( \nabla (\frac{e}{r} + \frac{e}{a_0})e^{-\frac{2r}{a_0}} + (\frac{e}{r} + \frac{e}{a_0}) \nabla e^{-\frac{2r}{a_0}}) = \frac{1}{4 \pi \epsilon_0} ( \nabla (\frac{e}{r} + \frac{e}{a_0}) \nabla e^{-\frac{2r}{a_0}} + \nabla^2 (\frac{e}{r} + \frac{e}{a_0})e^{-\frac{2r}{a_0}} + \nabla(\frac{e}{r} + \frac{e}{a_0})\nabla e^{-\frac{2r}{a_0}} + (\frac{e}{r} + \frac{e}{a_0}) \nabla^2 e^{-\frac{2r}{a_0}})$$
We also have a hint which says that $\nabla^2 \frac{1}{r} = - 4\pi\delta^3(\vec r)$ where $\delta$ is the dirac delta function.
I have troubles calculating the Laplacian for the exponentials and get really weird results which I don't know how to use. Can someone finish my calculation from this point out and tell me how they did it? I'd appreciate it greatly
 A: For simplicity, let's define
$$
  f(r,\theta,\phi) = \frac{1}{r}, \qquad g(r,\theta,\phi) = re^{-2 r/\alpha_0}, \qquad \kappa = \frac{e}{4\pi\epsilon_0} %comment for the 6 character rule; because just "r" was missing in the formula of "g"; so you can delete this
$$
so that
$$
  \phi = \kappa\left( fg +\frac{1}{\alpha_0}g \right)
$$
Now recall that
$$
  \nabla^2(fg) = g\nabla^2f + f\nabla^2g + 2\nabla f\cdot\nabla g,
$$
so that
$$
  \nabla^2\phi = \kappa\left(g\nabla^2f + f\nabla^2g + 2\nabla f\cdot\nabla g + \frac{1}{\alpha_0}\nabla^2 g\right)
$$
Note now that the spherical coordinate gradient and Laplacian and divergence, when acting on a function with no angular dependence ($\partial_\theta\psi = 0, \partial_\phi\psi=0$), are simply
$$
  \nabla\psi = (\partial_r\psi)\mathbf e_r, \qquad \nabla^2\psi = \frac{1}{r^2}\partial_r(r^2\partial_r\psi)
$$
and finally use
$$
  \nabla^2 \frac{1}{r} = 4\pi\delta^{(3)}(\mathbf r)
$$
The rest is algebra.  Try it again with these facts and see if you can get it to work!
A: I noticed by dimensional analysis that the expansion of the Laplacian product is not correct. Actually, it should read
\begin{equation}
\nabla^{2} (fg) = g\nabla^{2}f + f\nabla^{2}g+2\nabla g \cdot \nabla f
\end{equation}
With this expression, it is now possible to get the right answer.
