# Degenerate Perturbation Theory Wavefunction Correction

1. Aug 30, 2011

### McLaren Rulez

Hi,

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

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

where we sum over all $m$ such that $m\neq n$.

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

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

only this time the summation exculdes m if $\psi_{m}$ 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?

2. Aug 30, 2011

### DrDu

The matrix elements $$\langle \psi_m^{(0)}|V|\psi_n^{0)}\rangle$$ will be different for different "good states" n from the degenerate subspace, no?

3. Aug 30, 2011

### McLaren Rulez

Oh dear, I don't know what I was thinking! Sorry about that. It is obviously different for different n. Thank you DrDu.