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Is there a good physical picture of why the energy levels in a hydrogen atom are independent of the angular momentum quantum number $\ell$ and $m$?

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Other than the fact that the Hamiltonian only depends on the radius and there are no velocity-dependent interaction terms? –  Jerry Schirmer May 31 at 22:15
@JerrySchirmer: That is not true. The kinetic term does depend on $l$ (though not on $m$). One could expect that the energy levels to depend on $l$. (And the solution of the radial equation does depend on $n$ and $l$.) I'm not sure if there is a physical argument to show that it should not be the case. –  Adam May 31 at 23:06
Well sort of, it comes into play because the Hamiltonian is fully separable for the hydrogen atom and the l dependence comes from these constants that must cancel out between the separated differential equations... I guess it depends on your perspective of dependence. –  Elvex May 31 at 23:12
@Adam: I said "interaction" terms. There's nothing to generate a spin-orbit or spin-spin coupling. –  Jerry Schirmer May 31 at 23:59
@JerrySchirmer: you said "Other than the fact that the Hamiltonian only depends on the radius" which seems independent from "and there are no velocity-dependent interaction terms". –  Adam Jun 1 at 0:02

3 Answers 3

The degeneracy of energy-levels can be traced to the fact that the hydrogen atom possesses an enhanced $SO(4)$ symmetry caused by (among other things) the conservation of the Laplace-Runge-Lenz vector operator, see e.g. this Phys.SE post and Ref. 1.


  1. G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapter 9. The pdf file is available here.
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I've seen that before, but I was wondering if perhaps there is a simple, descriptive explanation? –  malina May 31 at 23:31

The shortest and correct answer: this degeneracy is determined by the symmetry of the system.

The case of degeneracy in hydrogen atom is so-called "accidental degeneracy", when eigenfunctions belonging to different irreducible representations of the symmetry group of a Hamiltonian correspond to the same energy. This type of degeneracy can also occurs in larger systems, for instance, in molecules. This degeneracy can not be predicted only from the standart consideration of Hamiltonian symmetry. The reason for this degeneracy is the existance of hidden symmetry in the system.

Mathematically it means that one can construct for the systems with hidden symmetry some conserved quantities, so-called "integral invariant", which should be included in consideration of symmetry propetries in addition to symmetry of Hamiltonian. And in principle, it is possible to solve Schroedinger equation in more tricky manner with inclusion of these "integral invariants" and obtain solution, for which these "accidental degeneracies" will be strictly included.

In the case of hydrogen atom the reason is the invarience of the system not only to the three-dimensional rotation group 0(3), but also to the four-dimensional rotation group 0(4) - the system have unexpected at first glance hidden symmetry.

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Let me note that, strictly speaking, there is no degeneracy in $l$ in a hydrogen atom, if you use the more precise Dirac equation, rather than the Schroedinger equation. So any "explanation" of the degeneracy had better use some properties of the Schroedinger equation.

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