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We've been learning in class that when approaching a perturbed Hamiltonian with degeneracies you should try to choose a basis made of eigenstates of both the unperturbed Hamiltonian and the perturbation. But at that point, why do you need perturbation theory? Don't you now have the exact solutions to the Schrodinger equation?

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    $\begingroup$ Note that you only diagonalize the perturbation in the degenerate subspace. So if $P$ is the projection onto the degenerate eigenspace (of the unperturbed hamiltonian) and $V$ is the perturbation, then you try to find a basis of the degenerate eigenspace, that diagonalizes $PVP$. $\endgroup$
    – jd27
    Commented Oct 11, 2023 at 6:07
  • $\begingroup$ related to physics.stackexchange.com/q/225727/36194 $\endgroup$ Commented Oct 11, 2023 at 15:57

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You should use only the basis of eigenstates of unperturbed Hamiltonian $H_0$. However in case of degeneracies that will be broken by perturbation $V$ you need to choose the appropriate starting point. I.e. you need to chose those states that the exact eigenstates of $H$ would tend in the limit of the turned off perturbation. Obviously for that you should use the information about this perturbation, from $H_0$ point of view all degenerate states are equally good.

So among all the eigenstates of $H_0$ we select the particular sector of degenerate states, say $\Psi_m$ (all having the same energies). In this sector you construct the matrix $V_{mn}=\langle\Psi_m|V|\Psi_n\rangle$. Diagonalizing it you get the superpositions of $\Psi_m$ that are the appropriate starting points that will become eigenstates of $H$ after perturbation.

However note that these superpositions diagonalizing $V_{mn}$ are still eigenstates of $H_0$. They are not yet eigenstates of $H$, just the appropriate starting points to find them. They are not even the eigenstates of $V$, just its eigenstates when you restrict it to the degenerate sector under consideration.

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  • $\begingroup$ I think I'm starting to get it. What about when you have a perturbation like spin orbit coupling, where $V$ commutes with $H_0$ , so your eigenstates of $H_0$ are exactly eigenstates of $V$. Does first order perturbation just give the complete picture in that case? $\endgroup$
    – Aggle
    Commented Oct 11, 2023 at 9:37

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