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If we have a non degenerate solution to a Hamiltonian and we perturb it with a perturbation V, we get the new solution by

[tex] |\psi_{n}^{(1)}> = \sum \frac{<\psi_{m}^{(0)}|V|\psi_{n}^{(0)}>}{E_n^{(0)} - E_m^{(0)}}\psi_m^{(0)}[/tex]

where we sum over all [itex]m[/itex] such that [itex]m\neq n[/itex].

When we do the same for a degenerate perturbation theory, we simply exclude all the [itex]\psi_m[/itex] that belongs to the degenerate subspace. That is

[tex] |\psi_{n}^{(1)}> = \sum \frac{<\psi_{m}^{(0)}|V|\psi_{n}^{(0)}>}{E_n^{(0)} - E_m^{(0)}}\psi_m^{(0)}[/tex]

only this time the summation exculdes m if [itex]\psi_{m}[/itex] is in the degenerate subspace. Am I right so far?

If so, my question is this. Assuming we are taking the "good" states that Griffiths mentions (those which are the eigenstates of V in the degenerate subspace), then each one has a different energy correction. But the wavefunction correction for all of them is exactly the same according to the expression given above. That seems a little fishy. The first order corrections to each of the "good" states is exactly the same but the first order energy correction is different?

Thank you for your help.

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# Degenerate Perturbation Theory Wavefunction Correction

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