Derivation of 1st Order Degenerate Perturbation Equations

To deal with the problem of degenerate states, we will allow an arbitrary linear combination of those states at zeroth order. In the following equation, the sum over $i$ is the sum over all the states degenerate with $\phi_n$ and the sum over k runs over all the other states.

$$\bgroup\color{black}\psi_n=N(\lambda)\left(\sum\limits_{i\in{... ..._{k\not\in{\cal N}}\lambda c_{nk}^{(1)}\phi_k+...\right)\egroup$$

where ${\cal N}$ is the set of zeroth order states which are (nearly) degenerate with $\phi_n$. We will only go to first order in this derivation and we will use $\lambda$ as in the previous derivation to keep track of the order in the perturbation.

The full Schrödinger equation is.

$$\bgroup\color{black}(H_0+\lambda H_1)\left(\sum\limits_{i\in{... ...um\limits_{k\not\in{\cal N}}c_{nk}(\lambda)\phi_k\right)\egroup$$

If we keep the zeroth and first order terms, we have

$$\bgroup\color{black}(H_0+\lambda H_1)\sum\limits_{i\in{\cal N... ...sum\limits_{k\not\in{\cal N}}\lambda c_{nk}^{(1)}\phi_k.\egroup$$

Projecting this onto one of the degenerate states $\phi^{(j)}$, we get

$$\bgroup\color{black}\sum\limits_{i\in{\cal N}}\langle\phi^{(j... ...i)}\rangle\alpha_i= (E_n^{(0)}+\lambda E^{(1)})\alpha_j.\egroup$$

By putting both terms together, our calculation gives us the full energy to first order, not just the correction. It is useful both for degenerate states and for nearly degenerate states. The result may be simplified to

$$\bgroup\color{black}\sum\limits_{i\in{\cal N}}\langle\phi^{(j)}\vert H\vert\phi^{(i)}\rangle\alpha_i=E\alpha_j.\egroup$$

This is just the standard eigenvalue problem for the full Hamiltonian in the subspace of (nearly) degenerate states.