# Discontinuity of Fermi liquid occupancy

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?

• Assuming you have a delta function at thre Fermi surface, then you get a discontinuity as one of the answers says. But it's worth mentioning that at finite temperature this will always be smeared out, so experimentally it has not been possible to directly measure $Z$ this way. But there are relations between $Z$ anas the effective mass which have been verified. – KF Gauss Apr 9 '20 at 5:36

When the quasiparticle is not exactly on the Fermi surface (FS), the spectral function will indeed contain a peak with finite width. However, once we have $$|\vec{k}|=k_F$$, we will have a $$\delta$$ peak, which corresponds to infinitely long lived quasiparticle. The existence of the $$\delta$$ peaks will give rise to the finite step in the distribution at $$k_F$$.
We can look at it the other way around. Assuming the step function is completely smeared out, i.e. no finite jumps like the finite $$T$$ case, then differentiation of the distribution will not lead to any $$\delta$$ function, meaning there will be no quasiparticles with infinite lifetime. This is a contradiction, because we do have this kind of quasiparticles living right on the FS.