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A two-particle retarded correlation function is (its derivation is not related to my question here) $$ C^R(\omega) = \sum_{kq}\bigg(f(\epsilon_k )-f(\epsilon_{k+q} )\bigg)\frac{1}{\omega+\epsilon_k-\epsilon_{k+q}+i\eta} $$ here $f(\epsilon_k)$ is Fermi distribution function. This kind of function is usually seen when we derive electron current density. The imaginary part of this can be calculated by using identity $$\frac{1}{x+i\eta}=\mathcal{P}\frac{1}{x}-i\pi\delta(x).$$ $$ Im[C^R(\omega)] = -\pi\sum_{kq}\bigg(f(\epsilon_k )-f(\epsilon_{k+q} )\bigg)\delta(\omega+\epsilon_k-\epsilon_{k+q}) $$ I have seen that in many books (without any explanation), the above expression in the limit $\omega\to 0$ is written as $$ \lim_{\omega\to 0}Im[C^R(\omega)] = -\pi\sum_{kq}\bigg(-\frac{\partial f(\epsilon_k )}{\partial\epsilon_k}\hbar\omega\bigg) \delta(\epsilon_k-\epsilon_{k+q}) $$ My question is how is $$\big(f(\epsilon_k )-f(\epsilon_{k+q} )\big)=\big(-\frac{\partial f(\epsilon_k )}{\partial\epsilon_k}\hbar\omega\big)~?$$ I can see that $$\big(f(\epsilon_k )-f(\epsilon_{k+q} )\big)=-\frac{f(\epsilon_{k+q} )-f(\epsilon_k )}{\epsilon_{k+q}-\epsilon_k}\big(\epsilon_{k+q}-\epsilon_k\big)$$ which looks like a derivative if $\epsilon_{k+q}-\epsilon_k=\hbar\omega$. But I am looking for a more concrete explanation for this

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  • $\begingroup$ Consider to add a reference. $\endgroup$ Nov 13 '21 at 12:58
  • $\begingroup$ I think that your last comment tells it all: Due to the delta function one fix the that $\epsilon_k-\epsilon_{k+q}=-\omega$ therefore you can rewrite this as a derivative since $f(a)\delta(a-b)=f(b)\delta(a-b)$ $\endgroup$ Nov 24 '21 at 7:53

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