Potential energy of electron density at the nucleus

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?

• 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. – Jon Custer Aug 19 at 13:37
• 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. – Ivan Neretin Aug 19 at 14:12
• You are taking into account that the nucleus, even a hydrogen atom, has a finite size? – Karl Aug 19 at 20:00

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.

• Actually, some integrals of unbounded functions converge all right. – Ivan Neretin Aug 19 at 15:07
• @IvanNeretin Argh. You're absolutely right. I will clean up the language here. – Zhe Aug 19 at 15:12
• 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). – Buck Thorn Aug 20 at 6:58
• $\frac{x^{2}}{x}$ is effectly just $x$ except at $x = 0$ where it is not defined, since it is $\frac{0}{0}$. – Zhe Aug 20 at 16:53
• Hmmm, while I see what you are doing, I wonder if you aren't allowed to use L'Hopitals rule. – Buck Thorn Aug 20 at 20:47