# Lifetime of quasiparticles in propagators

I have a rather technical question about lifetimes and propagators. The definition of the single particle propagator is:

$$g(r, r', t, t') = -i <\Psi_{0}^{N}|T[\psi(t, r)\psi^{\dagger}(t', r')]|\Psi_{0}^{N}> = - i \theta(t-t') <\Psi_{0}^{N}|\psi(t, r)\psi^{\dagger}(t', r')|\Psi_{0}^{N}> \mp i \theta(t'-t) <\Psi_{0}^{N}|\psi(t', r')^{\dagger} \psi(t, r)|\Psi_{0}^{N}>,$$

where - is for bosons and + for fermions.

Using the identity $$\theta(\pm\tau) =\mp \int_{-\infty}^{+\infty} \frac{d\omega}{2\pi i} \frac{e^{-i\omega \tau} }{\omega \pm i\eta},$$

where $$\eta$$ is assumed infinitesimal, one gets the Fourier Trasform of the propagator. Expressing it in Lehmann representation:

$$g(r, r', \omega) = \sum_{n} \frac{<\Psi_{0}^{N}|\psi(r)|\Psi_{n}^{N+1}> <\Psi_{n}^{N+1}|\psi^{\dagger}(r') |\Psi_{0}^{N}>}{\hbar \omega - (E_{n}^{N+1}-E_{0}^{N})+i\eta} + same \; but \; for \; the \; quasihole \; part.$$

$$\eta$$ was always considered infinitesimal, so I would conclude that the poles of the propagator are all real (apart from an infinitesimal imaginary part). But then it is proven that the poles are not real and have a finite imaginary part. This imaginary part is strictly related to the lifetime of the quasiparticle. How is it possible? The Lehmann representation says the propagator must have real poles.

Thanks

• Note that $\eta$ being non-zero is absolutely essential. You should not take it to zero and conclude "that the poles of the propagator are all real". For this reason also your statement "The Lehmann representation says the propagator must have real poles" is false. Regarding the lifetime, keeping $\eta$ finite corresponds to having a finite width of a delta function. This can be interpreted as a lifetime. Commented Aug 30, 2020 at 13:52
• But in order to have the integral representation of $\theta$ shouldn't I have $lim_{\eta \to 0^{+}}$?This would mean I have to consider $\eta$ as being infinitesimal, which would contradict a finite lifetime Commented Aug 30, 2020 at 13:58
• I'm so glad to see someone has the same issue with me. Do you have a good solution now? It seems in Lehmann representation the Green's function will not decay at all since all of its poles are near the real axis. I think the problem may hide in the infinite summation but I have no exact proof...... If the free state is coupled to finite number of other states by the full hamiltonian, then the situation will resemble Rabi oscilation to some extent. In that case, the Green's function should oscilate and do not decay. The infinite coupling indeed play a role somehow. Commented May 27, 2022 at 9:14