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  Connection between Matsubara frequencies and Landau Quasiparticle Interpretation

+ 1 like - 0 dislike

In a zero-temperature Fermi liquid, I understand that Landau quasiparticles correspond to poles in the interacting retarded Green's function, with the quasiparticle weight given by the residue of said function:

$$ Z=\left(1-\frac{\partial \Re \Sigma}{\partial \omega}\bigg|_{\omega=0}\right)^{-1} $$

where $\Sigma$ is the self-energy. However, this only is for zero temperature. At finite temperature, we get smearing, and Landau quasiparticles gain a lifetime $\tau$, where $1/\tau \sim T^2$. To incorporate finite temperature behavior into the Green's function approach, we have to perform a sum over all Matsubara frequencies when we take the Fourier transform of the finite-temperature Green's function $\mathcal{G}$:

$$ \mathcal{G}(\tau)=\frac{1}{\beta}\sum_{\omega_n} e^{-i\omega_n \tau}\mathcal{G}(i\omega_n) $$ where $\omega_n=(2n+1)\pi/\beta$ for fermions and $\tau$ is imaginary time. From the question Matsubara frequencies as poles of distribution fucntion and other sources, I know that the Matsubara frequencies are the poles of the distribution function. Therefore, is there some connection between the quasiparticle concept and the Matsubara frequencies? That is, if quasiparticles are the poles of the interacting Green's function and $\omega_n$ are the poles of the distribution function, is there an explicit connection between the two quantities? Can I use the Matsubara frequencies themselves to describe the requirements for a finite-temperature Fermi liquid? Any explanation or references at the level of Abrikosov, Gorkov, & Dzyaloshinskii would be greatly appreciated.

This post imported from StackExchange Physics at 2019-08-21 22:46 (UTC), posted by SE-user Joshuah Heath

asked Aug 13, 2019 in Theoretical Physics by Joshuah Heath (70 points) [ revision history ]
edited Aug 21, 2019 by Dilaton

1 Answer

+ 1 like - 0 dislike

Correction: Finite temperature is not the only source of adding smearing/lifetime. One can also get smearing/lifetime from interaction terms, by doing perturbation theory. This is most easily seen if expanding the Dyson-Schwinger equation $G = (G_0^{-1} - \Sigma)^{-1}$, and perturbatively finding $\Sigma$ to some order. It will in general have both real and imaginary parts. It turns out $\Sigma(\omega)$ disappears for $\omega \rightarrow 0$ for 0 temperature, however, which means at 0 temperature there is still a well-defined Fermi surface (quasiparticles with infinite lifetime).

Onto the question. In finite temperature calculations, we often use the Matsubara formalism, which is to say we evaluate $e^{-\beta H}$ or observables w.r.t. it by a path integral method, but since the argument of the exponent is real, the time parameter seen in Feynman path integrals is replaced by $\tau$, which people sometimes call "imaginary time" in this context. A Fourier transform yields a description in terms of Matsubara frequencies, as you write.

The existence of the Matsubara frequencies has little/nothing to do with the quasiparticle nature of the particles, but the (anti)-symmetry of exchange of fermions/bosons (quantum statistics) and can be traced back to how the variables (anti)-commute or not, when performing the Fourier transform.

In practice, if one wants to query quasiparticle poles within the Matsubara formalism, one must compute $G(\tau)$ or $G(i\omega_n)$ and analytically continue this object to one defined on the real-frequency axis, $G(\omega)$, and then make the same prescription as OP notes; the poles correspond to quasiparticles. One can instead compute $\Sigma(i\omega_n)$ and analytically continue this, plugging it into the Dyson-Schwinger equation above. This analytic continuation is a trick, wherein you say "if I know the correlation functions on the imaginary axis (a discrete set of Matsubara points for ex.), then simply by treating the temperature as imaginary, I can get out what they would be for a zero-temperature theory, and regain the quasiparticle poles."

answered Mar 24, 2022 by Slender (20 points) [ no revision ]

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