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 $P$?
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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:
if $\psi$ is zero, so $P$ is zero
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.
John Rennie
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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.
– Jonathan Clark Mar 20 '24 at 12:38