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I am wondering about the origin of the $\ell \leq n-1$ orbital filling rule. For the hydrogen atom, I believe the reason is because in the spatial wave function there is the term

$$\psi \propto \sqrt{(n-\ell-1)!}$$

so if $\ell > n-1$, by the definition of the factorial, $\psi$ goes to $0$. However, what about atoms with more protons and electrons? Will we always have this sort of term in our wavefunction or is there are more general argument for why this inequality has to be true?

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  • $\begingroup$ you should Check molecular Quantum Mechanics, Oxford, I have seen a derivation there $\endgroup$
    – Dorothea
    Commented Aug 9, 2020 at 16:51

1 Answer 1

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TL;DR: The quantization condition $$n_r ~:=~ n-\ell -1 ~\in\mathbb{N}_0 \tag{A}$$ follows by looking for normalizable wavefunction solutions $R(r)$ to the radial TISE with a fixed $\ell$ value. Here the radial quantum number $n_r$ is the number of nodes, cf. the node theorem in 1D QM.

In more detail: Schematically, one first solves $R(r)$ in the regions for small & large radial coordinate $r$. After factoring out the newly found asymptotic behaviours, one obtains a function $v(r)$, where (due to the TISE) the coefficients of its power series satisfy a recursion relation. It turns out that the series $v(r)$ must truncate in order for the solution $R(r)$ not to alter its asymptotic behaviour. This leads to the quantization condition (A), cf. e.g. Ref. 1. $\Box$

Classically, $n_r=0$ corresponds to circular orbits, i.e. no radial motion.

References:

  1. D. Griffiths, Intro to QM, 1995; subsection 4.2.1.
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    $\begingroup$ moreover, the relation between $\ell$ and $n$ is specific to this potential: for the $3D$ h.o. $\ell=n,n-2,\ldots$ $0$ or $1$ depending if $n$ is even or odd. $\endgroup$ Commented Aug 9, 2020 at 22:52
  • $\begingroup$ @ZeroTheHero, I think that's the key point here $\endgroup$
    – KF Gauss
    Commented Aug 10, 2020 at 2:16

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