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  Discontinuity of Fermi liquid occupancy

+ 2 like - 0 dislike

In Fermi liquid theory, the electron spectral function is often represented by $$A(k,\omega) = Z\delta(\omega-\epsilon_k)\ + \text{incoherent background} $$ where $Z$ is the weight in the quasiparticle peak. Consequently, the zero-temperature occupancy $$ n(k)=\int_{-\infty}^0 d\omega A(k,\omega) $$ has a discontinuity of $Z$ at the Fermi level.

However, since the spectral function is actually more accurately described by a Lorentzian with a non-zero width (except right at the Fermi level), is the occupancy, in fact, continuous, and the purported discontinuity only approximate?

This post imported from StackExchange Physics at 2017-02-18 11:09 (UTC), posted by SE-user leongz

asked Feb 17, 2017 in Theoretical Physics by leongz (70 points) [ revision history ]
edited Feb 18, 2017 by Dilaton

The discontinuity is exact in the thermodynamic limit, which is usually presupposed.

Seconding Arnold, it means it depends on the number of electrons in question.

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