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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?

Thanks in advance!

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5 Answers 5

The physical observable is not the wavefunction, but its integral over a finite area. In spherical coordinates, this is:

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

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

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Dear @Jerry, you were a minute faster but shorter ;-). I think that $\sin^2\theta$ should be just $\sin\theta$. –  Luboš Motl Apr 26 '12 at 16:32
    
Indeed! I'm so used to writing the metric that I forgot the square root. –  Jerry Schirmer Apr 26 '12 at 17:16
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Good way to phrase the priorities. ;-) –  Luboš Motl Apr 26 '12 at 18:09
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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.

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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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Re:positronium: wouldn't sub-Compton-wavelength renormalization of the Coulomb law soften the singularity? –  Slaviks Apr 26 '12 at 19:28
    
@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. –  dmckee Apr 26 '12 at 19:53
    
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. –  Slaviks Apr 26 '12 at 19:56
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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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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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