I am dealing with electron/phonon interaction in QM. In particular, given the Hamiltonian of a solid,


we have that the el-phonon Hamiltonian is treated perturbatively with respect to $H_\text{el}+H_\text{ion}$

and, neglecting Umklapp processes we have

$$H_\text{el-ion}=\Sigma_{\vec{q}} \;\nu(\vec{q})\cdot (a^\dagger(-\vec{q})+ a(\vec{q}))\cdot c^\dagger_{\vec{k}+\vec{q}}c_{\vec{k}}$$

According to this Hamiltonian we can see that only two first order processes are admitted (emission of a phonon of momentum $-\vec{q}$ and absorption of a phonon of momentum $\vec{q}$).

Then, supposing to know all the states of the unperturbed Hamiltonian $H_\text{el}+H_\text{ion}$, denoting them with $|\psi_n^{(0)}\rangle$ we compute the correction to the ground state of this Hamiltonian using perturbation theory, obtaining


meaning that 1st order processes (absorption/emission) do not change energy levels, while

$$E_{GS}^2=\Sigma_{n>0}\frac{|\langle\psi_{n}^{(0)}|H_\text{el-ion}|\psi_0^{(0)}\rangle|^2}{E_0^{(0)}-E_n^{(0)}}\neq 0$$

meaning that the 2nd order process (el/el effective attractive coupling due to an exchange of a phonon) changes energy.

It seems me that there is a relation between order of correction in perturbation theory and order of the process that caused that correction (and physically this is intuitive because in 1st order correction calculations involve only one wavefunction while in second order correction we have 2 different wavefunctions involved).

Is what I am saying correct? If yes, what is the formal way to say this? In other words, is there a relationship between the order of a process and order in perturbation theory correction?

  • 1
    $\begingroup$ Yes you are correct and it becomes more obvious in the path integral formulation. I'm not proficient enough to demonstrate this on my own but you should look up things on path integral and Feynman diagrams. $\endgroup$ Commented Jul 31, 2015 at 16:03
  • $\begingroup$ I'm a little puzzled by what you're asking. You say "It seems me that there is a relation between order of correction in perturbation theory and order of the process that caused that correction" but that's trivially true, in a sense by definition. The "order of a correction in perturbation theory" and the "order of a process" are literally the same thing. Are you trying to ask why, physically, energy levels are affected by 2nd order processes? $\endgroup$
    – DanielSank
    Commented Feb 23, 2016 at 3:09

1 Answer 1


1) Your perturbation operator does not conserve the particle number of the phonons, so only even powers of it will contribute to equilibrium expectation values. Since you are interested only in the ground state, which doesn't have any phonons excited, this means that you have to create a phonon first. After that either another phonon can be created or the former can be destroyed. Only processes that return to the ground state contribute to the expectation value and for this you obviously need to apply the interaction operator an even amount of times.

2) In a path integral the phonons can be integrated out, leaving an effective electron-electron interaction. That too is proportional to the square of the perturbations matrix element.

  • $\begingroup$ Item #1 is spot on, and it would be really nice to give a physical intuition for it, even if that "physical" intuition is still a bit mathematical. Item #2 is also spot on, and I wonder if you could indicate why the effective couplings are second order with respect to the coupling Hamiltonian. I mean, it's sort of obvious based on #1, but if you could show a little generic example or toy problem it would really help. $\endgroup$
    – DanielSank
    Commented Feb 23, 2016 at 3:03

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