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How does the application of the Schrödinger equation to model a system, such as a particle in a box, help us understand the origin of the degeneracy of atomic orbitals?

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    $\begingroup$ It doesn't. Unless we apply it to a model system which is a hydrogen atom, that is. $\endgroup$ – Ivan Neretin Apr 10 at 10:20
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    $\begingroup$ I would argue it does if you consdier a 2D (or 3D etc.) box and consider under what conditions you can get degeneracies. But I also would argue this isn't the world's greatest ever exam question. $\endgroup$ – Ian Bush Apr 10 at 10:37
  • $\begingroup$ I'd say the simplest model possessing degeneracy is the particle on a ring, which is quasi 1D for fixed radius. $\endgroup$ – Paul Apr 10 at 12:19
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Degeneracy refers to states having the same energy.

If you have two or more coordinates that can be related by a symmetry operation then the system will contain degenerate states because swapping the coordinates associated with those degrees will result in a Hamiltonian with the same solutions.

This is the case in both the hydrogen atom and in the particle in a cube.

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  • $\begingroup$ Non accidental Degeneracy implies symmetry but symmetry does not always imply degeneracy. The 1D particle in a box system does have inversion symmetry if you construct it with the center at the origin but it has no degenerate states. A good discussion is found here, physics.stackexchange.com/questions/319096/… $\endgroup$ – Hans Wurst Apr 10 at 11:38
  • $\begingroup$ I know what you mean and your answer is ok. My comment is intended for questionener's/readers that may want are more in depth view at symmetry and degeneracy. It's easy to jump to the wrong conclusion that symmetry always implies degeneracy so i thought it wouldn't hurt to mention it in a comment. $\endgroup$ – Hans Wurst Apr 10 at 11:56
  • $\begingroup$ @HansWurst You're right, and I repost my comment because I was unclear. I meant symmetric coordinates ie rotate x to y. If you have the same Hamiltonian after a symmetry operation that swaps coordinates then degeneracy follows. That's all I meant. $\endgroup$ – Buck Thorn Apr 10 at 11:59
  • $\begingroup$ This is not right. Consider the Hamiltoninan $H=-\partial_x^2-\partial_y^2+(x-y)^2$ with Dirichlet boundaries at $x,y=\pm1$. It's invariant with respect to reflection $x\leftrightarrow y$, but the symmetry group is abelian, so you don't get degeneracy (except maybe accidental). $\endgroup$ – Ruslan Apr 10 at 18:55
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    $\begingroup$ @Ruslan undoubtedly you are correct. However I don't think referring to Abelian groups will enlighten the OP (or me for that matter, who cannot explain it other than saying abelian group elements commute). Maybe I should add something hand-wavy about particular symmetry conditions that must be satisfied. I am looking for an explanation that is accessible to someone who does not know of group theory. $\endgroup$ – Buck Thorn Apr 11 at 6:12

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