Nodes for a hydrogen atom probability density?

It is said that the wave function $$\psi_{n,\ell,m}$$ has $$n-1$$ nodes; $$n-\ell-1$$ from the radial part of the wavefunction and $$\ell$$ from the angular part. However, the probability density of finding a particle at a given location is: $$P=|\psi_{n,\ell,m}|^2 r^2 \sin(\theta) dr d\theta d\phi$$ This goes to $$0$$ at $$r=0$$ and at $$\theta=0,\pi$$ which is not always true for $$\psi_{n,\ell,m}$$ so what in general can be said about the nodes of the probability density $$P$$?

The quantity $P = |\psi^*\psi|$ is the probability density i.e. the probability of finding the object in a volume element $dV$ is $PdV$.

There are two ways this quantity can go to zero:

1. if $\psi$ is zero, so $P$ is zero

2. if the volume of the element $dV$ is zero

In polar coordinates the volume of the element bounded by $r$ and $r+dr$, $\theta$ and $\theta+d\theta$ and $\phi$ and $\phi+d\phi$ is:

$$dV = dr\,rd\theta\,r\sin(\theta)d\phi = r^2 \sin(\theta) dr d\theta d\phi$$

as in your question. What happens at $r=0$, $\theta=0$ and $\theta=\pi$ is that the volume element $dV$ goes to zero. There is no physical significance to this, it's just a side effect of the coordinates that we've chosen.

• "There is no physical significance to this." It might be useful to comment that this does actually have physical implications. We often just write the probability density using the wave function, but graphing the n=1 Hydrogen ground state suggests that it it likely to be found at the proton. This is not true; multiplying by the radius squared causes the probability distribution to go to zero at the origin, which we expect from the uncertainty principle. So, using spherical coordinates does play a role in interpreting the solutions physically. Commented Mar 20 at 12:38