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Anonymous
Degeneracy in the one dimensional Schrodinger equation 2017-06-13 12:13:48 Post No. 8973296
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Degeneracy in the one dimensional Schrodinger equation 2017-06-13 12:13:48 Post No. 8973296
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Degeneracy in the one dimensional Schrodinger equation
Anonymous
2017-06-13 12:13:48
Post No. 8973296
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So I'm reading Zettili's book on quantum mechanics and in chapter 4 he is discussing the one dimensional Schrodinger equation. He then proceeds to discuss its solutions using a generalized potential (shown in the picture).
He concludes that for bound states, there is no degeneracy. Even though I find his method pretty meh, I agree and it's not particularly difficult to formally prove.
He also says that for unbound states where E>V2 (in the picture), the energy levels are doubly degenerate and I understand that due to solutions being two waves travelling in opposite directions.
However, how the hell does one conclude that for unbound states, V1<E<V2, there is no degeneracy? The picture shows his line of reasoning but I just don't seem to get it.
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>>8973296
There's only 1 solution to the d.e in that region. When you apply the boundary conditions you see there's no transmission or reflection going the other ways.
All of this is theoretical though because there's fine and hyper fine splitting which can cause degeneracy depending on the states
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>>8973401
Could you elaborate a bit more?
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