# Asymptotic relation of Green's function for diverging self energy

I am considering the derivation on pages 64 to 66 of Zagoskin's Quantum Theory of Many-Body Systems. They consider a Green's function in the Lehmann representation:

$$G(p,\,\omega)=(2\pi)^3 \sum_s \left( \frac{A_s \delta(p-P_s)}{\omega-\epsilon_s^++\mu+i0} \pm \frac{B_s \delta(p+P_s)}{\omega-\epsilon_s^-+\mu-i0} \right)$$

where $\epsilon_s^+=E_s(N+1)-E_0(N)>\mu$ and $\epsilon_s^-=E_0(N)-E_s(N-1)<\mu$.

They they derive $G(\omega)\sim 1/\omega$ by stating that

in this limit we can neglect all other terms in the denominator...so that

$$G(p,\,\omega)\sim \frac{1}{\omega}(2\pi)^3 \sum_s \left(A_s\delta(p-P_s)\pm B_s \delta(p+P_s)\right)$$

The problem here is that, for some general interacting Green's function, I feel as if they are making an assumption about the self energy--i.e., that $\omega>\Sigma(k,\,\omega)\,\forall \omega$. If $\Sigma(\omega)\sim k,\,\omega^2$, for example, then the above analysis fails, although I can't find any explanation of such divergent behavior in my books.

What makes me concerned is the behavior of the real and imaginary parts of $G(k,\,\omega)$ as $|\omega|\rightarrow \infty$. I have seen authors use the above argument to state that the imaginary part of the Green's function disappears faster than the real part at diverging $\omega$. However, for some self energy that grows faster than $\omega$, does this argument break down?. If feel as if it would, but I can't find any references that talk about it. Any explanation or references (preferably original papers that can be easily accessed online) would be greatly appreciated.

EDIT 1: Am I correct in stating that the self energy is completely contained in $A_s$ and $B_s$, and therefore $G(p,\,\omega)\sim 1/\omega$ as $|\omega|\rightarrow \infty$ regardless of the self energy behavior?

EDIT 2: Is it simply because the limit of $\Sigma(\omega,\,k)\rightarrow \infty$ is unphysical for $\omega\rightarrow \pm \infty$ (even though $\frac{\partial \Sigma}{\partial \omega}\rightarrow \infty$ is legal in a non-Fermi liquid)?

## 1 Answer

A good solution to this question can be found in Andre-Marie Tremblay's A refresher in many-body theory. On page 93, Section 3.6, he says that the high-frequency asymptotic behavior of the Green's function is determined by sum rules. Namely, using Tremblay's notation,

$$\lim_{ik_n\rightarrow \infty}G(k,\,ik_n)=\lim_{ik_n\rightarrow \infty}\frac{1}{ik_n}$$ To get this correct behavior, the self energy cannot diverge at high frequency; i.e.,

$$\lim_{ik_n\rightarrow \infty}\Sigma(k,\,ik_n)=\textrm{constant}$$

Therefore, the fermionic Green's function will always go as one over the frequency in the large frequency limit.

• That's a really good reference, but in this case it is slightly off: a self-energy that goes, e.g., as the square root of frequency for large frequencies is also allowed. The above requirement is too strict, it suffices for the self-energy to have a frequency behavior that is slower than linear. – Funzies Dec 18 '18 at 8:48
• @Funzies Do you happen to have any references that talk about the sub-linear behavior of the self energy? – Joshuah Heath Dec 19 '18 at 0:56
• Have a look at this paper: arxiv.org/pdf/1112.5074.pdf. The discussion around Eqs. (1.1), (1.3), (1.4) and (3.9) might be useful. In the end it boils down to the fact that the Green's function should go exactly like 1/\omega for large \omega. The reason is that if you close the frequency integral (that is, the sum rule) by a contour in the upper half plane and there are no poles there, then the integral is given by the contribution of the semi circle at infinity, which gives exactly 2\pi i, but only if the Green's function goes as 1/\omega. – Funzies Dec 19 '18 at 9:05