6
$\begingroup$

Analytically, I calculated a self-energy $\Sigma(\omega)$, for which I verified that

1) $\text{Im}\big[\Sigma(\omega)\big] \leq 0$ for all $\omega$ and specifically $\text{Im}\big[\Sigma(0)\big] = 0$,

2) $\text{Im}\big[\Sigma(\omega)\big] = 0$ for $|\omega|>\omega_C$,

3) $\text{Re}\big[\Sigma(\omega)\big] \sim \frac{1}{\omega}$ for $\omega\rightarrow\infty$,

4) $\text{Re}\big[\Sigma(\omega)\big]$ and $\text{Im}\big[\Sigma(\omega)\big]$ obey the Kramers-Kronig relations.

Besides the analytical calculation, I did a numerical calculation yielding the same results.

I was under the impression that obeying these requirements would be enough to expect the sum rule, i.e., $$ \int_{-\infty}^{\infty}d\omega \frac{-2\text{Im}\big[\Sigma(\omega)\big]}{\big(\omega - \text{Re}\big[\Sigma(\omega)\big]\big)^2 + \big(\text{Im}\big[\Sigma(\omega)\big]\big)^2} = 1, $$ to be satisfied. Unfortunately, it is not satisfied in my case.

Hence my question: I thought that an analytic self-energy that vanishes for large $\omega$ and has no poles in the uppper half plane always satisfies the sum rule. Apparently the above 4 results are not strong enough to conclude that the sum rule has to be obeyed. What am I missing here?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.