Consider the hydrogen atom for simplicity.

The electronic density at the nucleus is not null.

The attractive potential between a small volume of electronic density $\mathrm dV$ (at position $\vec r$) and the nucleus (at position $\vec R$) is:

$$\frac{-Z\rho(\vec r)}{|\vec R - \vec r|}\mathrm dV$$

How is the energy not going to (negative) infinity near the nucleus then? How is this accounted for to give the traditional $\pu{-0.5 Ha}$ energy value of the hydrogen atom?

  • 1
    $\begingroup$ It is accounted for by solving the Schroedinger equation. In a very small region the attraction is quite high. In other regions not so much. And it all volume averages out. $\endgroup$
    – Jon Custer
    Aug 19, 2019 at 13:37
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    $\begingroup$ I'd say it is accounted for by integration. Yes, an integral of an unbounded function can pretty well have a meaningful, finite value, and those familiar with calculus can even find it. $\endgroup$ Aug 19, 2019 at 14:12
  • $\begingroup$ You are taking into account that the nucleus, even a hydrogen atom, has a finite size? $\endgroup$
    – Karl
    Aug 19, 2019 at 20:00

1 Answer 1


I think it's useful to expand slightly on the comments because I think the language here should be as clear as possible.

You're concerned about the convergence of the integral of a function when some piece of the function is unbounded, i.e., for $f(x)$ on domain $[a, b]$, $\exists c \in [a,b]$ such that $f(c) = \infty$. Frequently, integrals of this type do not converge.

Your function certainly looks like that but the resolution of this problem is actually hidden in the $dV$. Typically, we treat the integration of the wave function as a separable problem, i.e, we integrate the radial and spherical components separately. When you look at the integral over the distance from the nucleus, you get a factor $4\pi r^{2}$ because you are integrating shells and $dV = 4\pi r^{2} dr$. So even though $\frac{1}{r}$ diverges at zero, $\frac{r^{2}}{r}$, while not defined at zero, does converge.

$$\lim_{r\rightarrow 0}\frac{r^{2}}{r} = 0$$

From this perspective, no piece of the integral is infinite, and the integral converges.

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    $\begingroup$ Actually, some integrals of unbounded functions converge all right. $\endgroup$ Aug 19, 2019 at 15:07
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    $\begingroup$ @IvanNeretin Argh. You're absolutely right. I will clean up the language here. $\endgroup$
    – Zhe
    Aug 19, 2019 at 15:12
  • $\begingroup$ Good explanation, the volume shell element dV goes to zero at the origin. But I don't get why you should write that $r^2/r$ is not defined at the origin? It doesn't diverge, it is zero (because the volume of the shell is zero). $\endgroup$
    – Buck Thorn
    Aug 20, 2019 at 6:58
  • $\begingroup$ $\frac{x^{2}}{x}$ is effectly just $x$ except at $x = 0$ where it is not defined, since it is $\frac{0}{0}$. $\endgroup$
    – Zhe
    Aug 20, 2019 at 16:53
  • $\begingroup$ Hmmm, while I see what you are doing, I wonder if you aren't allowed to use L'Hopitals rule. $\endgroup$
    – Buck Thorn
    Aug 20, 2019 at 20:47

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