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?)

  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please use that tag on this type of question. $\endgroup$ – user4552 Sep 24 '19 at 22:38
  • $\begingroup$ Hint: Choose polar coordinates where the $\vartheta' = 0$ axis points along $\vec r$. Then you can determin $\left| \vec r - \vec r' \right|$ in terms of $\vartheta'$, $r$ and $r'$ by trigonometrics and the integrals become elementary (with the substitution $d\big(\cos(\theta')\big) = d\vartheta' \sin(\theta')$. $\endgroup$ – Sebastian Riese Sep 24 '19 at 23:00
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    $\begingroup$ @SebastianRiese Respectfully, that is very, very far from the easiest way to approach this problem. $\endgroup$ – tparker Sep 25 '19 at 2:51
  • $\begingroup$ Well, the solution to your problem is the integral. Unfortunately very few functions can be integrated, that's why the multipole expansion is employed for points far from the charge distribution: it's easier to calculate! $\endgroup$ – ErickShock Sep 25 '19 at 4:10

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


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.

  • $\begingroup$ Thank you for the answer, I did not consider splitting the integral in a way that you suggested. I might be able to solve this now. About the homework comment: it did not occur me that this could be a homework to someone, I've never studied physics so I assumed this is a reasonable question ... sorry for omitting the tag $\endgroup$ – Ezze Sep 25 '19 at 11:52
  • $\begingroup$ Yes, it’s a common homework problem. I was assigned it as homework in a course on electromagnetism, and later when I taught such a course I assigned it to my students. $\endgroup$ – G. Smith Sep 25 '19 at 16:24
  • $\begingroup$ Another method to solve it is to assume that $\Phi$ is a function of $r$ only, in which case the Laplacian becomes an ODE for $\Phi(r)$ that can easily be integrated twice. I believe this was the method that tparker's answer was driving at. $\endgroup$ – Michael Seifert Sep 26 '19 at 21:06

Hint: the Laplacian of a spherically symmetric function is $$\nabla^2 \phi = \frac{1}{r} \frac{d^2}{dr^2} (r \phi).$$

Don't forget about the minus sign.

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    $\begingroup$ The OP asked how to find the potential by doing the integral involving the charge density. $\endgroup$ – G. Smith Sep 25 '19 at 7:35
  • $\begingroup$ @G.Smith No they didn't. $\endgroup$ – tparker Sep 25 '19 at 11:28
  • $\begingroup$ @G.Smith Oh, I see, you're looking at the bounty memo rather than at the OP. I believe that when the bounty memo contains a question that's different from the OP, the guidance is that you should answer the question in the OP, but I'm willing to go to the refs about this one. $\endgroup$ – tparker Sep 25 '19 at 11:33
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    $\begingroup$ The OP says “I have not been able to solve the integral by hand and Mathematica can not tell me the result either.” The bounty memo merely repeats the request. It didn't ask something different. $\endgroup$ – G. Smith Sep 25 '19 at 16:21
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    $\begingroup$ In any case, I don’t understand your answer. The OP knows the formula for the Laplacian in spherical coordinates. That’s how they got a (slightly incorrect) $\rho$. $\endgroup$ – G. Smith Sep 25 '19 at 23:11

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