quantum mechanics - Why are the zeroth order terms in degenerate perturbation theory the eigenstates of the perturbing Hamiltonian?

I have for quite some time now tried to find a satisfactory answer to this, but I haven't yet. In perturbation theory, with small parameter $\lambda$, we expand the eigenstate as

$$| E \rangle=| E^{(0)} \rangle + \lambda | b \rangle + ...$$

Where $| E^{(0)} \rangle$ is an eigenstate of the unperturbed Hamiltonian. The problem is that when there is degeneracy, there is a choice of these eigenstates. I know the answer is that they are chosen to be eigenvectors of the perturbing Hamiltonian, but my question is why.

Answer

This is really a great question.

Take a look at the figure attached here. The circle indicates a 2-dimensional degenerate subspace at $\lambda=0$.

In red we indicate two possible basis states for the subspace. In blue we show another possible choice of basis states.

Now look at the curvy green lines, these are the states as they evolve for $\lambda > 0$. The green curves do not connect with the red states at $\lambda=0$. The perturbation series is a Taylor series, which is a continuous function (it's a polynomial), so there is no way to make the series go from the red states at $\lambda=0$ to the green states for $\lambda > 0$.

Clearly, we have to start the perturbation series at $\lambda=0$ with the blue states, because those connect up with the green ones when $\lambda > 0$. The green states are, by definition, eigenstates of the perturbed Hamiltonian, so the blue states at $\lambda=0$ must also be eigenstates of the perturbed Hamiltonian.

Degenerate perturbation theory states. We show the states as functions of the perturbation parameter $\lambda$ (green), and various choices of the unperturbed states (blue, red).