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When trying to solve the Schrödinger equation for hydrogen, one usually splits up the wave function into two parts:

$$\psi(r,\phi,\theta)= R(r)Y_{l,m}(\phi,\theta).$$

I understand that the radial part usually has a singularity for the 1s state at $r=0$ and this is why you remove it by writing:

$$R(r) = \frac{u(r)}{r}$$

But what is the physical meaning of

$$R(r=0) = \infty~?$$

Wouldn't this mean that the electron cloud is only at the centre of the atomic nucleolus?

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The infinitesimal probability for the electron to be in the volume $dV$ around a point $(r,\theta,\phi)\leftrightarrow (x,y,z)$ is given by $$ dP = dV\cdot |\psi(x,y,z)|^2 = dV\cdot |R(r)|^2\cdot |Y_{lm}(\theta,\phi)|^2 =\dots$$ as you can see if you substitute your Ansatz for the wave function. However, the infinitesimal volume $dV=dx\cdot dy\cdot dz$ may be rewritten in terms of differentials of the spherical coordinates as $$ dV = dr\cdot r^2 \cdot d\Omega = dr\cdot r^2 \cdot \sin\theta\cdot d\theta\cdot d\phi $$ where the small solid angle $d\Omega$ was rewritten in terms of the spherical coordinates. You see that for dimensional reasons (or because the surface of a sphere scales like $r^2$), there is an extra factor of $r^2$ in $dV$ and therefore also in $dP$ which suppresses the probability. There is simply not enough volume for small values of $r$.

So $|R(r)|^2$ may still go like $1/r^2$ for small $r$ and in that case, $dV$ will be proportional to $dr$ times a function that is finite for $r\to 0$. Such $dP$ may be integrated and there's no divergence at all near $r=0$.

That's why one should allow the wave function to go like $1/r$ near $r=0$ which is the true counterpart of one-dimensional wave function's being finite near a point. However, Nature doesn't use this particular loophole because the wave function $\psi$ for small $r$ actually scales like $r^l$ where $l$ is the orbital quantum number and the wave function actually never diverges even though it could.

Update 2016: I should have and could have written it four years ago but I didn't. While the normalizability allows $1/r$ around $r=0$, such singular functions ultimately can't be in stationary or nearly stationary states, for the following reason which differs from various reasons above and those in the comments.

For example, someone mentioned that $1/r$ could lead to a continuous spectrum or some surprising degeneracies. But if the correct wave functions predicted a continuous or degenerate spectrum in a box, then it would be how Nature works. The actual reason why $1/r$ is not finally allowed as a stationary wave function near $r=0$ is that the Laplacian of this wave function (and Schrödinger's equation contains such a Laplacian) is proportional to a delta-function at the origin (or contains such a term) and no other term in Schrödinger's equation can cancel this delta-function, so the Schrödinger equation must be violated.

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  • $\begingroup$ Consider a spherical box with constant potential inside. The general solution of Schrödinger's equation for electron in such a box is a sum of Spherical Bessel and Neumann functions. Both functions (of zero order, as for S states) are square integrable with weight of $r^2$. If you do allow the wavefunction to go like $1/r$ near $r=0$ like you propose in the last paragraph, then you'll get 1) continuous spectrum (although the box is finite!), 2) degeneracy for S states. So one shouldn't allow such behavior of the wavefunction, and this isn't because Nature doesn't use some loophole. $\endgroup$ – Ruslan Nov 14 '16 at 9:40
  • $\begingroup$ Yes and no. These $1/r$ wave functions are ultimately not appearing in reality but the reason is different than you state. If the correct wave functions predicted a continuous or degenerate spectrum in a box, then it would be how Nature works. The actual reason why $1/r$ is not finally allowed as a stationary wave function near $r=0$ is that the Laplacian of this wave function is a delta-function at the origin and no term in Schrödinger's equation can cancel this delta-function, so the Schr. equation must be violated. $\endgroup$ – Luboš Motl Nov 17 '16 at 8:14
  • $\begingroup$ Hm, this is an interesting argument, never thought about it this way. I stand corrected. Too bad I can't remove the downvote now. If you added this argument into your answer, I'd be able to do it. $\endgroup$ – Ruslan Nov 17 '16 at 9:16
  • $\begingroup$ @Ruslan - I added an update to my answer with the same argument. You may change vote but you don't have to. $\endgroup$ – Luboš Motl Nov 19 '16 at 16:26
  • $\begingroup$ Your comment has actually helped me answer my own question. Thank you. $\endgroup$ – Ruslan Nov 19 '16 at 19:52
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The physical observable is not the wavefunction, but its integral over a finite area. In spherical coordinates, this is:

$$P({\vec x})=\int~\mathrm dr\, \mathrm d\theta\, \mathrm d\phi ~r^{2}\sin\theta~ \psi^{*}\psi$$

This integrand is manifestly finite at $r=0$, even if $R(r)$ has a $\frac{1}{r}$ divergence.

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For a hydrogen-like atom in 3 spatial dimensions, the rewriting of the radial part

$$R(r)~=~\frac{u(r)}{r}$$

is not performed to keep the $u(r)$ part regular, as OP suggests, but usually because the 3D radial equation in terms of the $u$ function has the same form as a 1D Schrödinger equation.

Imagine that the radial wave function goes as a power

$$R(r) ~\sim ~ r^{p} \qquad {\rm for} \qquad r~\to~ 0, \qquad p~\in~\mathbb{R}.$$

On general grounds, one can impose the following list of consistency conditions, listed with the weakest condition first and the strongest condition last.

  1. Normalizability of the wave function $$\infty~>~\langle\psi|\psi\rangle~=~\int d^3r~|\psi(\vec{r})|^2 ~\propto~ \int_0^{\infty} r^{2}dr~|R(r)|^2 .$$ Integrability at $r=0$ yields that the power $p>-\frac{3}{2}$. In other words, this normalizability condition does not by itself imply that $R(r)$ or $u(r)$ should be regular at $r=0$, which is also the conclusion of many of the other answers.

  2. The expectation value of the potential energy $V$ should be bounded from below, $$-\infty~<~\langle\psi| V|\psi\rangle~=~\int d^3r~V(r)|\psi(\vec{r})|^2~\propto~-\int_0^{\infty} rdr~|R(r)|^2. $$ Integrability at $r=0$ yields that the power $p>-1$. In other words, $u(r)$ should be regular for $r\to 0$.

  3. The kinetic energy operator (or equivalently, the Laplacian $\Delta$) should behave self-adjointly for two wave functions $\psi_1(\vec{r})$ and $\psi_2(\vec{r})$, $$\langle\psi_1| \Delta\psi_2\rangle~=~-\langle\vec{\nabla}\psi_1| \cdot\vec{\nabla}\psi_2\rangle,$$ without picking up pathological contributions at $r=0$. A detailed analysis shows that the powers of the radial parts of $\psi_1(\vec{r})$ and $\psi_2(\vec{r})$ should satisfy $p>-\frac{1}{2}$.

In comparison, the actual bound state solutions have non-negative $p=\ell\in \mathbb{N}_0$, and therefore satisfy these three conditions.

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In addition to the simply geometric constraints that Jerry and Lubos talk about, the derivation used to illustrate the problem almost always assumes that the proton is a point particle which is a pretty good approximation but not strictly true. Working the problem again with a realistic proton charge density function (roughly constant inside a radius of about 1 fm) would be another way to remove the singularity.

Mind you, you this argument does not hold true for the positronium so you still need the geometric constraint.

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  • $\begingroup$ Re:positronium: wouldn't sub-Compton-wavelength renormalization of the Coulomb law soften the singularity? $\endgroup$ – Slaviks Apr 26 '12 at 19:28
  • $\begingroup$ @Slaviks: I'm a little on thin ice here, but I think that renormalization does solve the problem, but that's in the context of QFT, while this question seem to be phrased in the language of introductory QM. $\endgroup$ – dmckee --- ex-moderator kitten Apr 26 '12 at 19:53
  • $\begingroup$ Sure, I was just entertaining the concept :) There is no singularity in the w.f., worrying in about the radial part is just staring at a singularity if the coordinate system, imho. $\endgroup$ – Slaviks Apr 26 '12 at 19:56
  • $\begingroup$ @Slaviks there is a singularity, it's just that it's not as severe as unboundedness — merely a cusp. $\endgroup$ – Ruslan Apr 24 '17 at 12:18
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For Hydrogen, $R(r)$ does not diverge, as $U(r)$ vanishes as fast as (or faster than) $r$ as $r\rightarrow 0$. In fact, it's only for the $s$ orbitals that the wavefunction is non zero at $r=0$. But as pointed out before, a non-zero radial wavefunction does not mean a non-zero probability of finding the electron at the center.

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