Doubt regarding Degenerate perturbation theory

Question

McIntyre, quantum mechanics,pg360

Suppose states $\left|2^{(0)}\right\rangle$ and $\left|3^{(0)}\right\rangle$ are degenerate eigenstates of unperturbed Hamiltonian $H$.

The first-order perturbation equation we want to solve is $$ \left.\left(H_{0}-E_{n}^{(0)}\right)\left|n^{(1)}\right\rangle=\left(E_{n}^{(1)}-H^{\prime}\right) \mid n^{(0)} \right >) $$

But the energy degeneracy of these two states creates an ambiguity. Both $\left|2^{(0)}\right\rangle$ and $\left|3^{(0)}\right\rangle$ satisfy the zeroth-order energy eigenvalue equation for the energy $E_{2}^{(0)}$, but so does any linear combination of the two states. If we are trying to find the energy correction to the state with zeroth-order energy $E_{2}^{(0)}$, how do we know whether to use the state $\left|2^{(0)}\right\rangle$ or the state $\left|3^{(0)}\right\rangle$ in the perturbation equation?

Why can't I use $\left|2^{(0)}\right\rangle$ and $\left|3^{(0)}\right\rangle$individually in the perturbation equation and carry on?

Answer 1

There are an infinite number of linear combinations of $|2^{(0)}\rangle$ and $|3^{(0)}\rangle$ which are eigenstates of $H_0$. However, only two specific linear combinations are eigenstates of $H'$. The main difference between degenerate perturbation theory and nondegenerate perturbation theory is that you need to find those special linear combinations that are eigenstates of $H'$. Once you have them, then the rest of the calculation proceeds in the same way as nondegenerate perturbation theory.