Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  Kramer's-Kronig relations for the electron Self-Energy Σ

+ 6 like - 0 dislike
1465 views

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.


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.

This post has been migrated from (A51.SE)
asked Feb 28, 2012 in Theoretical Physics by Greg Graviton (775 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Feb 29, 2012 by josh (205 points) [ no revision ]
Hm, I'm not entirely happy with your condition on $|ε|$ since it arises only a posteriori, once you have one particular expansion. Thanks a lot for the subtracted dispersion relations, that looks very useful. I'll check out what Weinberg writes.

This post has been migrated from (A51.SE)
I'm glad the subtracted dispersion relations were useful. Regarding the expansion condition, I would say rather that it emerges simultaneously with taking the expansion itself, and is part of the justification for truncating the expansion at a given level.

This post has been migrated from (A51.SE)
I have pondered the subtracted dispersion relations and it appears to me that we following situation: the subtracted dispersion relation gives better convergence, but we lose information about lower-order terms. For instance, if we know that the imaginary part vanishes faster than $|z|\to∞$, we can reconstruct the second derivative of the real part, but we cannot gain any information about the linear or constant part. This is a fundamental limitation. Unfortunately, this calls into the question the whole approach of trying to reconstruct a low-order expansion. Do you have any thoughts on this?

This post has been migrated from (A51.SE)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverflo$\varnothing$
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...