Kramers-Kronig relations for the electron Self-Energy Σ

Question

I'm currently studying an article by Maslov, in particular the first section about higher corrections to Fermi-liquid behavior of interacting electron systems. Unfortunately, I've hit a snag when trying to understand an argument concerning the (retarded) self-energy $\Sigma^R(ε,k)$.

Maslov states that in a Fermi liquid, the real part and the imaginary part of the self-energy $\Sigma^R(ε,k)$ are given by

$$ \mathop{\text{Re}}\Sigma^R(ε,k) = -Aε + B\xi_k + \dots $$ $$ -\mathop{\text{Im}}\Sigma^R(ε,k) = C(ε^2 + \pi^2T^2) + \dots $$

(equations 2.4a and 2.4b). These equations seem reasonable: when plugged into the fermion propagator,

$$ G^R(ε,k) = \frac1{ε + i\delta - \xi_k - \Sigma^R(ε,k)} $$

the real part slightly modifies the dispersion relation $ε = \xi_k$ slightly and the imaginary part slightly broadens the peak. That's what I'd call a Fermi liquid: the bare electron peaks are smeared out a bit, but everything else stays as usual.

Now, Maslov goes on to derive higher-order corrections to the imaginary part of the self-energy, for instance of the form

$$ \mathop{\text{Im}}\Sigma^R(ε) = Cε^2 + D|ε|^3 + \dots .$$

First, I do not quite understand how to interpret this expansion.

How am I to understand the expansions in orders of $ε$? I suppose that $ε$ is small, but in relation to what? The Fermi level seems to be given by $ε=0$.

Second, he states that this expansion is to be understood "on the mass-shell".

I take it that "on the mass shell" means to set $\xi_k=ε$? But what does the expansion mean, then? Maybe I am supposed to expand in orders of $(ε-\xi_k)$?

Now the question that is the most important to me. Maslov argues that the real part of the self-energy can be obtained via the Kramers-Kronig relation from the imaginary part of self-energy. My problem is that the corresponding integrals diverge.

How can $$ \mathop{\text{Re}}\Sigma^R(ε,k) = \mathcal{P}\frac1{\pi}\int_{-\infty}^{\infty} d\omega \frac{\mathop{\text{Im}}\Sigma^R(\omega,k)}{\omega-ε} $$ be understood for non-integrable functions like $\mathop{\text{Im}}\Sigma^R(ε,k) = ε^2$?

It probably has to do with $ε$ being small, but I don't really understand what is going on.

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I should probably mention my motivation for these questions: I have calculated the imaginary part of the self-energy for the one-dimensional Luttinger liquid $\xi_k=|k|$ as

$$ \mathop{\text{Im}}\Sigma^R(ε,k) = (|ε|-|k|)θ(|ε|-|k|)\mathop{\text{sgn}}(ε) $$

and would like to make the connection to Maslov's interpretation and results. In particular, I want to calculate the imaginary part of the self-energy with the Kramers-Kronig relations.

Answer 1

I can't speak knowledgeably about the specifics of your problem but I can offer some thoughts.

Regarding your first question, you will need to have dimensions of $\text{energy}^{-1}$ for $C$ and $\text{energy}^{-2}$ for $D$. In specific, this means that $C/D$ has units of energy. This gives meaning to the statement that $$ D|\epsilon|^3 \ll C\epsilon^2 \quad\Leftrightarrow\quad |\epsilon| \ll C/D\,. $$

As far as a divergent Kramers-Kronig relation goes, you should read about once or more subtracted dispersion relations. Then, instead of writing $$ \mathop{\text{Re}}\Sigma^R(\epsilon,k) = \mathcal{P}\frac1{\pi}\int_{-\infty}^{\infty} d\omega \frac{\mathop{\text{Im}}\Sigma^R(\omega,k)}{\omega-\epsilon}\,, $$ you can write $$ \mathop{\text{Re}}\Sigma^R(\epsilon,k) - \mathop{\text{Re}}\Sigma^R(\epsilon_0,k) = \mathcal{P}\frac1{\pi}\int_{-\infty}^{\infty} d\omega \frac{(\epsilon-\epsilon_0)\mathop{\text{Im}}\Sigma^R(\omega,k)}{(\omega-\epsilon)(\omega-\epsilon_0)}\,, $$ where $\epsilon_0$ is some convenient subtraction point, presumably one at which you know $\mathop{\text{Re}}\Sigma^R(\epsilon_0,k)$. You can extend to twice or more subtracted dispersion relations too. Weinberg Vol. 1 has a lot about dispersion relations where you can read more about this.