2
$\begingroup$

In the paper I am reading, they said

"If we want to evaluate a transport coefficient, in a typical situation the single-electron self-energy is not sufficient: one needs to know vertex corrections as well. "

I didn't understand what is the self-energy and the vertex corrections, can you help me?

the paper is:'Electronic properties of graphene-based bilayer systems'

$\endgroup$
4
  • $\begingroup$ can you give some more detail about your background, so an answer can be formulated in a beneficial way? are you familiar with quantum field theory? quantum mechanics at post-graduate level? $\endgroup$
    – user275556
    Commented Oct 20, 2022 at 10:53
  • 1
    $\begingroup$ I am a physics graduate student, I am familiar with quantum field theory but I am don't have a full understanding of it. $\endgroup$
    – nouha
    Commented Oct 20, 2022 at 10:56
  • 2
    $\begingroup$ Have you looked up self-energy on Wikipedia? If you are somewhat familiar with QFT the explanation there should suffice. $\endgroup$
    – ACuriousMind
    Commented Oct 20, 2022 at 12:00
  • $\begingroup$ The diagramatic formulation in the field of condensed matter theory distinguishes between interactions on the one- and two-particle level. The self-energy is the interaction on the one-particle level (one G in one G out), while vertex function usually refers to the two-particle diagrams (two G in two G out). $\endgroup$
    – DrCommando
    Commented Jul 4 at 9:10

3 Answers 3

5
$\begingroup$

in quantum field theory, the self-energy of a particle is the sum of all the connected irreducible diagrams that start and end with that particle. that is, it sums all the different ways in which a particle of certain type scatter back into itself in the system. Why it matters for transport can be elucidated by looking at the Dyson equation for the fully dressed Green function for a particle $\psi^{\dagger}_k$, starting from the time domain$^*$

$$ G_{\psi_k \psi^{\dagger}_k}(t) = g^0_{\psi_k \psi^{\dagger}_k}(t) + \int\! dt_1 G_{\psi_k \psi^{\dagger}_k}(t-t_1) V(t_1) g^0_{\psi_k \psi^{\dagger}_k}(t_1) \\= g^0_{\psi_k \psi^{\dagger}_k}(t) + \int\! dt' g^0_{\psi_k \psi^{\dagger}_k}(t-t') V(t') g^0_{\psi_k \psi^{\dagger}_k}(t') + \\ \int \! dt_1 dt_2 G_{\psi_k \psi^{\dagger}_k}(t-t_2) V(t_2) g^0_{\psi_k \psi^{\dagger}_k}(t_2-t_1)V(t_1)g^0_{\psi_k \psi^{\dagger}_k}(t_1) \\ = \ldots = g^0_{\psi_k \psi^{\dagger}_k}(t) + \int\! dt' g^0_{\psi_k \psi^{\dagger}_k}(t-t') V(t') g^0_{\psi_k \psi^{\dagger}_k}(t') + \\ \int \! dt_1 dt_2 g_{\psi_k \psi^{\dagger}_k}(t-t_2) V(t_2) g^0_{\psi_k \psi^{\dagger}_k}(t_2-t_1)V(t_1)g^0_{\psi_k \psi^{\dagger}_k}(t_1) + \cdots$$

where $g^0$ is the bare (i.e. free-theory) single particle Green function, $V$ is the many-body part of the theory that connects $\psi$ and $\psi^{\dagger}$ and this is the self-energy, which is labeled by $\Sigma$. We get an infinite series consisting of ever growing number of $\Sigma$ and $g^0$ in the integral. This equation basically tells us: a particle enters the system, it can propagate freely and thus scatter back into itself (by way of $g^0$) or it can interact via $V$ once at time $t_1$, and then again propagate freely, or multiple times at times $t_1, t_2$ and so forth. In frequency domain we get $G(\omega) = g^0(\omega) + G(\omega)\Sigma(\omega)g^0(\omega)$ which allows us to solve $G(\omega) = [(g^0(\omega))^{-1} - \Sigma(\omega)]^{-1}$.

To sum up to here: the self-energy of a particle represents that parts of the interaction that relate the particle back to itself, i.e. can "eat up" a particle and then "spit out" the same particle later. It is important because knowing it allows us to have the fully dressed single particle Green function, and usually transport involves terms related to this Green function (as physically in transport we scatter particles from one end and see how they come out at the other end of the sample).

It turns out that this is not enough, however, for some transport problems, as the interaction term itself has corrections. These are the vertex corrections mentioned. In diagrams, a vertex is an interaction term, so not a single-particle Green function, but a term that connects Green functions. It too is being affected by interactions and goes renormalization. In transport we (sometimes) need also the vertex corrections as well since they directly go into the expression for the transport.

Both the self-energy and the vertex corrections are results of interactions in the theory, and usually we calculate them numerically or perturbatively.

$^*$ I am making here some simplifying assumption for the sake of clarity, like that the interactions are only localized in time, or that the relevant fully dressed GF is the one particle one etc.

$\endgroup$
1
  • $\begingroup$ A bit silly while your answer is technically correct I doubt few could follow the example.We need concise answers. If the post isn't considered concise then I'd understand this complex answer. $\endgroup$ Commented Oct 22, 2022 at 15:49
2
$\begingroup$

In many-body physics there is a miracle that even in the presence of strong inter-particle interactions, the elementary excitations still resemble the "bare" non-interacting particles. For instance, the charge carriers in a conductor like graphene look $\textit{almost}$ like free electrons, and interact with electromagnetic field $\textit{almost}$ like free ones do. Almost, but not entirely so. So the difference in "self" properties of the electron in a material - coming from interactions with its ambience - as compared to the bare free particle (typically its effective kinetic energy $\epsilon(\vec{k})$, or "dispersion") is called self-energy (terminology from QED). Whereas the difference in the way the electron in the medium responds to the applied EM field (effective charge, most of the time) is called the "vertex correction" (the term becomes transparent if you look at the corresponding Feynman diagrams).

$\endgroup$
2
  • $\begingroup$ that's a much better answer than mine :) I like the conciseness! $\endgroup$
    – user275556
    Commented Oct 20, 2022 at 15:17
  • $\begingroup$ Thank you, yyy:). $\endgroup$
    – John
    Commented Oct 20, 2022 at 15:34
0
$\begingroup$

Indeed, put more simply, the self energy is the energy of an isolated particle by approximation. Take the electron self energy for example. If we reduce the radius to zero, then it obtains an infinite amount of energy. A shielding effect has been suggested by the background fluctuations of vacuum energy. Whether this is really case, I must disagree. We must give up on renormalization techniques because nature does not allow infinities in nature by first principles.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.