Physics Asked on December 3, 2020
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(pmtau) =mp int_{-infty}^{+infty} frac{domega}{2pi i} frac{e^{-iomega tau} }{omega pm ieta}, $
where $eta$ is assumed infinitesimal, one gets the Fourier Trasform of the propagator. Expressing it in Lehmann representation:
$g(r, r’, omega) = sum_{n} frac{<Psi_{0}^{N}|psi(r)|Psi_{n}^{N+1}> <Psi_{n}^{N+1}|psi^{dagger}(r’) |Psi_{0}^{N}>}{hbar omega – (E_{n}^{N+1}-E_{0}^{N})+ieta} + same ; but ; for ; the ; quasihole ; part.$
$eta$ 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
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