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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.

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    $\begingroup$ Would you accept that it is due to the need to be orthogonal? $\endgroup$
    – naturallyInconsistent
    Commented May 19, 2023 at 11:45
  • $\begingroup$ @LalitHooda This is the orthogonal condition: $\int \psi * \psi^* dV = 0$. This applies when $P=0$, since that's the definition of a node; regions with zero probability. I haven't tried this out myself - but see if you can substitute $\psi = \psi_n * \psi_l * \psi_m $ $\endgroup$
    – Karthik
    Commented May 19, 2023 at 12:59
  • $\begingroup$ Related: physics.stackexchange.com/q/572215/2451 , physics.stackexchange.com/q/724784/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented May 20, 2023 at 13:31

2 Answers 2

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  1. The number of nodes has to be an integer (because the wavefunction crosses the axis an integer number of times)
  2. let's call that integer $k$
  3. $l$ is also an integer.
  4. Now define $n$ as $$ n = k + l + 1 $$
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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.

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