# Momentum distribution Fermi liquid and spectral representation

In a Fermi liquid the momentum distribution shows a jump at the Fermi surface, i.e. $$\begin{equation}\langle n_{k_F-\delta k} - n_{k_F+\delta k}\rangle = Z_{k_F}\end{equation}$$ with $$Z_k$$ the strength of the pole in the Green's function $$G\left(k,\omega\right) = \frac{Z_{k}}{\omega - \epsilon_k + i \delta \mathrm{sgn}\left(k-k_F\right)} + g\left(k,\omega\right)$$ and where $$g$$ is supposed to be regular near $$k_F$$. By relating $$\langle n_k \rangle$$ to the Green's function and closing the contour in the complex plane this is straightforward to show - the integral over $$g$$ cancels and the only remaining contribution is $$Z_k$$ for $$\langle n_{k_F-\delta k}\rangle$$.

However I don't see how this is reflected in writing the spectral representation (I'm using Coleman's book) $$\begin{equation} G\left(k,\omega\right) = \sum_\lambda \frac{|M_{\lambda} \left(k\right)|^2}{w- \epsilon_{\lambda} + i \delta \mathrm{sgn} \left(\epsilon_\lambda\right)} \end{equation}$$ which would contribute $$\sum_{\lambda'} |M_{\lambda'} \left(k\right)|^2$$ to $$\langle n_k \rangle$$, where $$\lambda'$$ is such that $$\epsilon_{\lambda'} < 0$$. This suggests

$$\begin{equation} \langle n_{k_F-\delta k} - n_{k_F+ \delta k} \rangle = \sum_{\lambda'} |M_{\lambda'}\left(k_F-\delta k\right)|^2 - |M_{\lambda'} \left(k_F+ \delta k\right)|^2. \end{equation}$$

I wonder if we can actually use the spectral representation, because it seems to suggest a series of sharp poles, but I would expect them to broaden into a continuum especially away from the Fermi surface.

So: can we actually use the spectral representation written this way? If so, how does the last equation simplify to $$Z_{k_F}$$?