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What's the time evolution process between two different degenerate states? Is it also described by Schrodinger equation?

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They evolve exactly the same way and their evolution is determined by the Schroedinger equation. A Eigenstate of an Hamiltonian $\left|n\right\rangle$ evolves according to \begin{align} \left|n\right\rangle(t) = \exp(-i E_n t) \left|n\right\rangle \end{align} For two eigenstates with the same eigenvalue $E_n$ the evolution is therefore the same. Another argument is that the superposition of two degnerate eigenstate is also an eigenstate and is therefore stationary, from which we can conclude that their time evolution must be the same.

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According to basic postulate of Statistical mechanics, all degenerate states are equally probably filled up. That means all these states have same probability of occupation.

As for Schrodinger equation, it describes the evolution of energy states, so all states of same energy evolves the same way. There is no distinction.

A simple way to understand this would be to think that since degenerate states are identical in every way, you can't number them or specify them. In order to do so, you have to lift the degeneracy by applying some other observable.

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  • $\begingroup$ The question doesn't ask about statistical mechanics. Also, "As for Schrodinger equation, it describes the evolution of energy states, so all states of same energy evolves the same way. There is no distinction" is incorrect. $\endgroup$ – DanielSank Feb 6 '17 at 6:46
  • $\begingroup$ It is also not true that "degenerate states are identical in every way, you can't number them or specify them..." Actually the complete opposite is true. The reason we talk about degenerate states is that in systems with symmetry there usually exist distinguishable preparations that nevertheless have the same energy. $\endgroup$ – Mark Mitchison Feb 6 '17 at 9:54

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