homework and exercises - How to find the electrostatic potential of a hydrogen-like charge density?

I've been trying to find the scalar potential that would correspond to the charge density of a ground state hydrogen atom. The result is known, and the inverse of my problem can be found e.g. in Jackson's electrodynamics problem 1.5 or many questions here on this site.

The problem asks you to find the charge density that corresponds to the following potential: $$ \Phi(r) = \frac{q \exp{(-\alpha r)}}{4 \pi \epsilon _0 r}\left(1+\frac{\alpha r}{2}\right) .$$

As far as I can tell, according to Poisson's equation you basically just need to get the Laplacian of this potential. This is not hard to do, and the result is what you would expect:

$$ \rho(r) = \frac{-q\alpha^3}{8\pi}\exp{(-\alpha r)}. $$

• My question is the inverse: given $\rho(r)$, how would you find $\Phi(r)$?

The most obvious approach that I had is to use the integral you get from Coulomb's law:

$$ \Phi(r) = \int \frac{\rho(r')}{|r-r'|}d^3r' $$

However, I have not been able to solve the integral by hand and Mathematica can not tell me the result either. My guess is that this potential never goes to zero, so the direct integration is not feasible? If so, then how else would you go about solving this problem?

(My next task would be to solve the same thing for a Gaussian density, for which I can again find the result on Wikipedia. Is that problem easier or harder than this one?)

Answer

We don't provide complete answers to homework-like questions, even for a bounty.

You've got the wrong charge density because, when you took the Laplacian of the potential, you didn't take into account the fact that

$$\nabla^2\frac{1}{r}=-4\pi\delta^3(\vec r).$$

To understand this, think of the potential of a point charge.

If you simply use the spherical-coordinates expression for the Laplacian,

$$\nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}+...$$

you would calculate that

$$\nabla^2\frac{1}{r}=0$$

and this is incorrect. Basically, the spherical-coordinates expression for the Laplacian isn't valid at $r=0$.

The correct charge density is

$$\rho(\vec r)=q\delta^3(\vec r)-\frac{q\alpha^3}{8\pi}e^{-\alpha r},$$

where the Dirac delta function represents the positive charge density of the proton and the second term the negative charge density of the electron cloud.

Note that if you integrate this over all space, you get zero; a hydrogen atom has no net charge.

The integral you want to do is

$$\Phi(\vec r)=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec r')}{|\vec{r}-\vec{r}'|}d^3\vec r'.$$

Note the vector signs which you left out. They're important; $|\vec{r}-\vec{r}'|$ and $|r-r'|$ are two different things. Note also the $1/4\pi\epsilon_0$ that you omitted.

Using spherical polar coordinates for $\vec r'$, with the polar axis through $\vec r$, this is

$$\Phi(\vec r)=\frac{q}{4\pi\epsilon_0}\left\{\frac{1}{r}-\frac{\alpha^3}{8\pi}\int_0^\infty r'^2 dr' \int_0^\pi \sin{\theta'}d\theta' \int_0^{2\pi} d\phi' \frac{e^{-\alpha r'}}{(r^2+r'^2-2r r'\cos\theta')^{1/2}}\right\}.$$

The $\phi'$ integration is trivial.

The $\theta'$ integration can be performed by letting $u=\cos\theta'$. The result will involve the absolute value $|r-r'|$.

The $r'$ integration can be performed by splitting the integral into two parts,

$$\int_0^r dr'...+\int_r^\infty dr'...$$

so that you can take $|r-r'|$ to be either $r-r'$ or $r'-r$.

With all these hints, you can fill in the details.

Since the charge density is spherically symmetric, another approach would be to use Gauss’ Law to compute the field, and then integrate a radial path integral in from infinity to compute the potential.