Struggling to understand degenerate perturbation theory

This is a good question, on an issue that seem to confuse many students. I will try to clarify to the best of my ability the issue with degenerate perturbation theory.

When we have a degeneracy in the Hamiltonian, a problem is that there is no 'true' preferred basis to work with. Any rotation within a degenerate sub-space is allowed without changing the fact that the Hamiltonian is still diagonal.

Let's consider, for example, the angular-momentum Hamiltonian $\mathcal{H} = \frac{\omega}{\hbar}{\bf L}^2$, and consider a system with total angular momentum of $l=1$. We approach the problem and decide to diagonalize it in the usual basis of $|l=1, l_z = -1\rangle$, $|l=1, l_z = 0\rangle$ and $|l=1, l_z = +1\rangle$. However all this states have identical energy of $E = 2\hbar\omega$. Therefore, the basis $$|l=1, l_z = -1\rangle, \frac{1}{\sqrt{2}}(|l=1, l_z = +1\rangle \pm |l=1, l_z = 0\rangle)$$ is also legitimate and diagonalize the Hamiltonian.

When we apply perturbation theory, we always seek corrections to a specific state with specific energy. But in the degenerate case we face a problem - what is the correct state to start working from? What is the 'true' basis state to which we need to seek perturbative corrections? This question does not arise in the case of a nondegenerate subspace of the Hamiltonian.

So, here, the perturbation itself determines this for us. Perturbation theory is a tool to help us analyze the Hamiltonian with the perturbation, so it might be (and indeed, in many cases) that the full Hamiltonian is non degenerate. The thing that lifts the degeneracy is the perturbation. So we look in each degenerate subspace of the Hamiltonian, and diagonalie the perturbation within this subspace - this will tell us what is the correct unperturbed basis to work with. This is another important point - the state that we find at the first step are still eigenstates of the unperturbed Hamiltonian. They do not include the perturbation!

Now, once we have the correct basis, we can proceed in a similar manner to the nondegenerate perturbation theory: the first correction to the energy is just $\langle \psi^0 | V | \psi^0 \rangle$ with $|\psi^0\rangle$ the (correct) eigenstates of the unperturbed Hamiltonian, and the first order correction to the state itself comes from summing over matrix elements of the perturbation with states outside the degenerate subspace.

In the example above, let's say we added a perturbation $\omega_0 L_x$ with $\omega_0 \ll \omega$. Then it is clear that the correct basis to work with is the $L_x$ basis.

Lastly, it might be that the perturbation will not lift the degeneracy, or only partially lift it (for example, $\omega_0/{\hbar}L_z^2$ will leave $|l=1,l_z=\pm 1\rangle$ degenerate). In that case, we still have the freedom to choose the basis, and the perturbation theory will work for every choice.

From an operational point of view, the problem is that terms of the form

$$\frac{\langle n^{(0)} | \hat V | m^{(0)} \rangle}{E_n - E_m}$$

arise when summing over all $|m\rangle \neq |n\rangle$. If $|m^{(0)}\rangle$ and $|n^{(0)}\rangle$ are degenerate, then $E_n=E_m$ and we are dividing by zero, as you say.

If we choose the basis of the degenerate subspace such that $\hat V$ is diagonalized, however, then there are no such terms because $\langle n^{(0)} | \hat V | m^{(0)} \rangle=0$ for all $|n^{(0)}\rangle \neq |m^{(0)}\rangle$.