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
