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Consider the fermionic Matsubara sum for $$\frac{1}{\beta}\sum_n G^2(k,\omega_n) = \frac{1}{\beta}\sum_n \frac{1}{(i\omega_n - \epsilon_k)^2}$$ where $G(k,\omega_n)$ is the free fermion Green's function. To evaluate this sum, one can do the standard procedure by introducing the fermionic distribution function $n_F$, providing poles at $(2n+1)\pi/\beta$ with residue $-1/\beta$. This gives the result $$\frac{1}{\beta}\sum_n G^2(k,\omega_n) = n'_F(\epsilon_k)$$ In the $T\rightarrow 0$ limit, this evaluates to $\delta(\epsilon_k)$.

However, now let us first take the $T\rightarrow 0$ limit first. We can then change the sum into an integral, giving us $$\int_{-\infty}^\infty \frac{d\omega}{2\pi} \frac{1}{(i\omega - \epsilon_k)^2}$$ Evaluating this integral by contour integration, however, gives $0$ due to the fact that it is a double pole.

Is there something wrong with my manipulations here, or is there something inherently tricky about taking the $T\rightarrow 0$ limit?

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  • $\begingroup$ In the absence of answers, I'll say this: I believe I've heard that there might be something tricky about the limit, but I don't remember the specifics or why that'd be. $\endgroup$ – Anyon Feb 10 at 23:01

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