Why $n-\ell-1$ nodes?

Question

Can someone tell, why the radial part of $H$-atom wavefunction has exactly $n-\ell-1$ nodes? I know this comes by solving but is there some physical reason attached to this also?

There is a related question on Physics SE, but that's somewhat different than what I'm asking.

Answer 1

1. The number of nodes has to be an integer (because the wavefunction must cross the axis an integer number of times before tending to zero as $r \rightarrow \infty$)
2. let's call that integer $k$
3. $l$ is also an integer.
4. Now define $n$ as $$ n = k + l + 1 $$ The number $n$ was introduced into the study of the hydrogen atom because the energy eigenvalues are simply related to this $n$. The fact that they have this simple pattern is a special property of the $1/r$ potential: they had to depend on $k$ and $l$, but it happens that for the Coulomb potential, states with different $k$ and $l$ have the same energy if they have the same $k+l$.

Answer 2

There's not a whole lot of physics here. For given $\ell$, the radial part of the Schrödinger equation is equivalent to that of a one-dimensional problem, and for such a problem it is known from the properties of the solutions to this type of differential equation that solutions with higher energies will have a higher number of nodes (irrespective of the potential, provided it is confining). The argument is qualitatively that, with increasing energy, you need to "fit" an increasing number of wavelength in the confining region so the solutions must have increasing curvatures and thus increasing number of nodes.

It is only for the hydrogen atom that the number of nodes is given by $n-\ell-1$. For the 3d harmonic oscillator, the number of nodes $n_r=\frac12(n-\ell)$, where $n=0,1,2\ldots$ is related to the energy by $E_n=(n+\frac32)\hbar\omega$ and $\ell=n,n-2,n-4,\ldots$ so that $n_r\ge 0$. For the infinite spherical well, the situation is even more obscure.