Can the non-degenerate perturbation theory formula for higher-order energy corrections be used in case of degenerate perturbation theory?

Question

Consider the system given by, $$ H|n\rangle = E|n\rangle$$ where:

1. $H$ is the hamiltonian.
2. $|n\rangle$ is the eigenstate.
3. $E$ is the energy of the eigenstate.

Now from $\underline{\textbf{non-degenerate perturbation theory}}$, we get the following formula for the higher-order corrections to the energy, $$ E^{(s)} = \langle n^{(0)}|\delta H|n^{(s-1)}\rangle\tag{1.1.26}$$

where:

1. $E^{(s)}$ is the $s^{th}$ order correction to the energy.
2. $|n^{(s-1)}\rangle$ is the $(s-1)^{th}$ order correction to the eigenstate.
3. $\delta H$ is the perturbation applied to the hamiltonian $H$.

A reference to this formula can be found here: https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/a0889c5ca8a479c3e56c544d646fb770_MIT8_06S18ch1.pdf, equation number (1.1.26).

$\textbf{Can we use the same equation (with same bra-eigenstate (non-degenerate)}$ $\\$ $\textbf{but the ket-eigenstate replaced by the degenerate eigenstate) to calculate}$ $\\$ $\textbf{the higher order corrections}$ $\textbf{to the energies in}$ $\underline{\textbf{degenerate perturbation theory}}$ $\textbf{with the } $'$\textbf{correct}$' $\textbf{eigenstates?}$

I can find no reason that this equation will not hold for the degenerate case if the correct eigenbasis is used, but my experience in quantum mechanics or linear algebra is only limited.

$\\$ $\\$

$\textbf{To be specific, is the following equation (with the mentioned definitions) correct? :}$

$$ E_{k}^{(s)} = \langle p^{(0)}|\delta H|(n;k)^{(s-1)}\rangle\tag{1}$$ where:

1. $\delta H$ is the perturbation applied to the hamiltonian $H$.
2. $|p^{(0)}\rangle$ is an eigenstate in the non-degenerate subspace $\hat{V}$.
3. $|(n;k)\rangle$ is the $k^{th}$ eigenstate among the $N$ degenerate eigenstates.
4. The degenerate state $|(n;k)^{(s-1)}\rangle$, will have a component in non-degenerate subspace $\hat{V}$ as well as the degenerate subspace $\mathbb{V}_N$.
5. $E_{k}^{(s)}$ is the $s^{th}$ correction to the energy of the $k^{th}$ degenerate state among the $N$ degenerate states.

Note: For the case of s=1, the bra-eigenstate must also be the degenerate eigenstate $(n;k)^{0}$ (instead of the non-degenerate state $p^{0}$).

For the non-degenerate correction to the higher order energy $E_{p}^{(s)}$, we can use the non-degenerate equation to calculate the $s^{th}$ correction to the $p^{th}$ (one of the non-degenerate) state.

Answer 1

1. OP's linked reference (Ref. 1) shows in eq. (1.2.5) that OP's eq. (1.1.26) with $s=1$ may be violated for degenerate (time-independent) perturbation theory. For $s=1$ one needs to impose an additional condition (1.2.24) on the free basis $|n^{(0)};k\rangle$ in the degenerate $\mathbb{V}_N$ subspace (1.2.14).

2. The situation for $s>1$ is more involved and depends on whether the degeneracy is totally lifted, partially lifted, or not lifted to 1st order, 2nd order, 3rd order, etc, cf. Ref. 1.

References:

1. B. Zwiebach, MIT OCW 8-06 QM3 spring 2018; Chap. 1.