Most quantum mechanics books deal with perturbation theory and give the result that in the non-degenerate case, the first order corrections to the energy eigenvectors is given by $$\left | n \right \rangle = \left | n^0 \right \rangle + \sum_{m \neq n} \frac{\left \langle m^0 \Big | \hat H' \Big | n^0 \right \rangle}{E_n^0 - E_m^0} \left | m^0 \right \rangle.$$
Most books then move the degenerate case. My understanding is that they usually give an argument of the form:
- Write out the perturbation in the degenerate subspace.
- Solve for the eigenvectors and eigenvalues.
- The perturbation will break the degeneracy. This is the 'correct basis' to use for non-degenerate perturbation theory. Or these are the 'zeroth order correction to the states'.
I'm unsure of how we would then go on to do higher order perturbation theory with these states. For example if these are the 'correct basis states for non-degenerate perturbation theory', then how do we find the first order corrections? The expression above diverges if we have degenerate states. (I have sometimes seen it written that we can ignore contributions from degenerate states. Why is this so / how can one prove that this is the case.)
The second question is: what happens in the case when the degeneracy is not fully broken by the perturbation?
This Stack Exchange answer tries to answer my question, but I am not fully convinced by this derivation. For example they seem to take the operator $\left ( \hat H^0 - E_n^0 \right )$ and invert it, however the operator is non-invertible because due to the degeneracy it has zero eigenvalues. They seem to brush it under the rug by saying we effectively have
$$\left | n^1 \right \rangle = \sum_m \left | m^0 \right \rangle \overbrace{\left \langle m^0 \right | \frac{1}{\hat H^0 - E_n^0} \left | m^0 \right \rangle}^{\rightarrow \, \infty} \underbrace{ \left \langle m^0 \right | \left ( E^1_n - \hat H' \right ) \left | n^0 \right \rangle}_{\rightarrow \, 0}$$
and it seems like they claim that since the last term is zero we only need to sum over $m \not\in \mathcal D$ where $\mathcal D$ is the degenerate subspace. This doesn't seem very convincing to me since we seem to have an infinity and a zero 'fighting'.
Maybe I'm missing the point of the Stack Exchange answer and someone could answer my question by explaining the answer in more detail.
