Electron - phonon interaction and perturbation theory

It might be not so clear what the question actually is, so I'll start with that: why do we often use perturbation theory for non-degenerate spectrum in solid state physics, while the spectrum of $H_0$ is actually degenerate?

Typical example: electron - phonon interactions:

$$H = H_0 + H^\prime$$

Where $H_0$ is the Hamiltonian of the free electrons and phonons:

$$H_0 = \sum_{\vec{k}, \sigma} \varepsilon_{\vec{k}} \, c^{\sigma \, \dagger}_{\vec{k}} \, c^{\, \sigma}_{\vec{k}} + \sum_{\vec{k}, s} \hbar \omega_{\vec{k}} \left( a^{\, \dagger}_{\vec{k}} \, a_{\vec{k}} + \frac{1}{2} \right)$$

And $H^\prime$ describes the interactions between the electrons ($c$) and phonons ($a$):

$$H^\prime = \frac{1}{\sqrt{V}} \sum_{\vec{k}, \sigma} \sum_{\vec{q}} g_{\vec{k} + \vec{q}, \vec{k}} \: c^{\sigma \, \dagger}_{\vec{k} + \vec{q}} \; c^\sigma_{\vec{k}} \left( a_{\vec{q}} + a^{\, \dagger}_{- \vec{q}} \right)$$

Where $g$ is now the not so important function.

People usually write:

$$E^{(0)} = \left\langle \psi \right| H_0 \left| \psi \right\rangle$$

$$E^{(1)} = \left\langle \psi \right| H^\prime \left| \psi \right\rangle = 0$$ (this is zero because $H^\prime$ does not conserve the number of particles)

$$E^{(2)} = - \sum_{\left| \chi \right\rangle \neq \left| \psi \right\rangle} \frac{\left| \left\langle \chi \right| H^\prime \left| \psi \right\rangle \right|^2}{E_\chi - E_\psi}$$

These formulas are used when the spectrum for $H_0$ is non-degenerate. But we can easily contradict this: imagine state $\left| \psi \right\rangle$ which consists of Fermi sea and one phonon in state $\left| \vec{k} \right\rangle$. If we rotate $\vec{k}$ in momentum space, we get the same one-particle phonon energy, so the energy of $\left| \psi \right\rangle$ doesn't change, but this state is orthogonal to the previous (because different phonon states are orthogonal to each other). That means the spectrum of $H_0$ is degenerate. The only state that is non-degenerate is the ground state, $\left| FS \right\rangle \left| 0_{ph} \right\rangle$, but people sometimes calculate corrections to higher states without thinking about some degeneracy. Why is that so?

• What people usually do in this situation is to use Feynman diagrams. – Hosein Nov 26 '16 at 16:48
• Well and I adress the approach that doesn't use Feynman diagrams. Perturbation theory is somewhat more basic and we don't need full Eliashberg theory to work out just the first or second contribution of the interactions to the electron spectrum. – user16320 Nov 26 '16 at 16:57