# 2nd order Perturbation theory for Electron-Phonon interactions?

My understanding of perturbation theory has always been that up to second order, we can calculate the energy of a perturbed system with the following formula $$E=E_n^0+\langle n|H'|n\rangle+\sum_{m\neq n}\frac{\langle n|H'|m\rangle \langle m|H'|n\rangle}{E^0_n-E^0_m}\tag{1}$$

However, when dealing with the Frohlich Hamiltonian for electron phonon interactions, My textbook (a quantum approach to Condensed matter physics by Philip Taylor) says we can use perturbation to calculate the energy correction to second order via the following $$E=E_n^0+\langle \phi|H_{e-p}|\phi\rangle +\langle \phi|H_{e-p}(E^0_n-H^0)^{-1}H_{e-p}|\phi\rangle\tag{2}$$ where $$H_{e-p}=i\sum_{k,k'}M_{k,k'}(a^{\dagger}_{-q}+a_{q})c^\dagger_kc_{k'}$$ and $$q=k-k'$$ and $$\phi$$ is the unperturbed state. How do we get from eq 1 to eq 2? Why has the summation been dropped? Does this have something to do with how perturbation theory works in second quantization?

Using the resolution of identity: $$1=\sum_m|m\rangle\langle m|,$$ we can write $$\langle n|H'(E_n^0-H^0)^{-1}H'|n\rangle = \sum_{m,m'}\langle n|H'|m\rangle\langle m|(E_n^0-H^0)^{-1}|m'\rangle\langle m'|H'|n\rangle = \sum_{m,m'}\frac{\langle n|H'|m\rangle\langle m'|H'|n\rangle}{\langle m|E_n^0-H^0|m'\rangle}=\\ \sum_{m,m'}\frac{\langle n|H'|m\rangle\langle m'|H'|n\rangle}{(E_n^0-E_m^0)}\delta_{m,m'}= \sum_{m}\frac{\langle n|H'|m\rangle\langle m|H'|n\rangle}{(E_n^0-E_m^0)}$$ Thus, the expression in the book is essentially correct: the one thing missing is the projection operator that assures that $$m\neq n$$, but it might be that this is mentioned somewhere in the text or superfluous due to the nature of the problem.
• Thanks for the great response! Okay I think I understand now. As for the assurance that $m \neq n$, the author actually defines the operator $(E^0_n-H^0)^{-1}$ as that operator whose eigenfunctions are $|m\rangle$ with corresponding eigenvalues $(E^0_n-E^0_m)^{-1}$ "provided $E^0_n \neq E^0_m$" so I think that clears up the dividing by zero issue? Sep 22, 2021 at 12:23