# Schrieffer-Wolff Transformation for conventional superconductors

I was trying to follow the discussion in Radi A. Jishi's book (Feynman Diagrams in Condensed Matter Physics), Chapter 12 on superconductors. They basically have a Hamiltonian that comprises of a quadratic electron part and a quadratic phonon part, which is coupled through the usual electron-phonon coupling. They then perform a Schrieffer-Wolff transformation to conclude that there is an effective interaction between electrons that can become attractive for appropriate momenta.

It seems that in this procedure one does a canonical transformation on the Hamiltonian to eliminate the term linear in the electron-phonon coupling strength (and replace it by more complicated higher order terms). However, the phonons are not completely integrated out. Moreover, this is not the original Hamiltonian - but a canonically transformed Hamiltonian. How does then one conclude that the original Hamiltonian also results in an effective interaction between electrons that can become attractive for appropriate momenta?

(Note: I do appreciate the effective e-e interaction once one completely integrates out the phonons, eg. as done in Bruus-Flensberg or Altland-Simons.)

## 1 Answer

As you describe, the Schrieffer-Wolff transformation does not "integrate out" the phonons in the sense of a path integral or partition function. Instead, the transformation returns an effective Hamiltonian where electrons and phonons are decoupled, up to 2nd order in perturbation theory. The phonons are still there, but they interact with the electrons only at 3rd order in perturbation theory, so the coupling can be ignored. This is especially true if your focus is on the electron dynamics, and the electrons are within the energy range where the effective interaction is attractive. In this regime, their attraction gives rise to the Cooper instability, which dominates their low-energy behavior (i.e., induces a superconducting transition).

• This is not the original Hamiltonian, but one obtained after a unitary transformation, right? How do I connect it to the original problem - by saying that the spectrum is the same? Commented May 7, 2020 at 1:03
• Precisely. Since the transformation is unitary, the spectra of both Hamiltonians are identical. Commented May 8, 2020 at 2:19