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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?

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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.

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    $\begingroup$ 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? $\endgroup$ Commented Sep 22, 2021 at 12:23
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    $\begingroup$ @SalahTheGoat Yes, as it is explicitly stated, there is no ambiguity. $\endgroup$
    – Roger V.
    Commented Sep 22, 2021 at 12:23
  • $\begingroup$ Just one last thing. If I understand correctly, this derivation has absolutely nothing to do with second quantization. So the formula in eq 2 should apply even in first quantization? I guess its just a little weird seeing the infinite sum in eq 1 reduced to such a simple single-term form like that in eq 2 $\endgroup$ Commented Sep 22, 2021 at 12:27
  • $\begingroup$ Indeed, the formula woudl also apply in the first quantization. However phonons are nearly always treated in the second quantization, but not necessarily the electron part of the coupling. $\endgroup$
    – Roger V.
    Commented Sep 22, 2021 at 12:30

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