The final result of many-body perturbation theory based on Green's function method can be organized into the famous Dyson equation: $$G = G_0 + G_0 \Sigma G=G_0 + G \Sigma G_0 \tag{1}$$ where $G/G_0$ represents the Green's function and $\Sigma$ the self-energy, which have a one-to-one correspondence with your Hamiltonian $H$.

  1. Hubbard model Hamiltonian: $$H=-t\sum_{\langle ij\rangle,\sigma} \left[ c^\dagger_{i,\sigma}c_{j,\sigma} +h.c. \right] + \boxed{u\sum_i n_{i\uparrow}n_{i\downarrow}}=H_0+H_1 \tag{2}$$ where we are treating the interaction term $H_1$ as perturbation and we can organized the infinite perturbations into the compact Dyson equation $(1)$.
  2. Two-probe transport (or resonant level) model Hamiltonian (see Ref. [1]): \begin{align} H & = \sum_{n\alpha} \epsilon_{n\alpha}(t) c^\dagger_{n\alpha} c_{n\alpha}+\sum_m \epsilon_m(t) d^\dagger_m d_m+ \boxed{\sum_{n\alpha,m}\left[V_{n\alpha,m}c^\dagger_{n\alpha}d_m + h.c. \right]} \\ & = H_L+H_C+H_T \tag{3} \end{align} For this Hamiltonian $H$, we are treating $H_T$ as the perturbation term and can also arrive at the same Dyson equation $(1)$.

One can see the big difference is self-energy, which are closely related to the perturbation term in your Hamiltonian. So my first question is how do we partition our Hamiltonian and how to decide which part should be treated as perturbation? And my second question is: What's the difference for the self-energy due to the first boxed perturbation term (Coulomb interaction term) and the second boxed perturbation term (Coupling term between the central region and the left/right electrodes)?

[1] Jauho, A.-P., N. S. Wingreen, and Y. Meir. “Time-dependent transport in interacting and noninteracting resonant-tunneling systems”. Phys. Rev. B 50.8 (1994): 5528 [arXiv:cond-mat/9404027].

  • $\begingroup$ Hint: if $u\gg t$ then you definitely don't want to treat $H_1$ as a perturbation. See, for example, Cleveland & Medina, Am. J. Phys. 44, 44 (1976) (pdf). $\endgroup$ Commented Feb 22, 2018 at 10:08
  • $\begingroup$ Does my post answer all your questions, or should I add/elaborate on something? $\endgroup$
    – AlQuemist
    Commented Feb 26, 2018 at 11:49
  • $\begingroup$ @AlQuemist: Can you draw the Feynman diagrams for self-energy $\Sigma$ to demonstrate the difference corresponding to $(2)$ and $(3)$? If it is tedious, you can ignore this requirement. $\endgroup$
    – Jack
    Commented Feb 26, 2018 at 11:56
  • $\begingroup$ For a detailed discussion of the diagrammatics of the two cases, I'd refer you to chps. 11 & 12 of Bruus & Flensberg. “Many-body quantum theory in condensed matter physics” (2002) [ here ]. There, the details are very well explained -- I cannot do better. $\endgroup$
    – AlQuemist
    Commented Feb 26, 2018 at 12:07
  • $\begingroup$ Ok, thanks, I will do some homework to completely understand your answers. $\endgroup$
    – Jack
    Commented Feb 26, 2018 at 12:11

2 Answers 2


Perturbation theory assumes that there is a small dimensionless parameter in the theory. To perform perturbative calculations, one would make an asymptotic expansion in terms of this dimensionless parameter. The first term in the expansion represents the unperturbed result; the second term is the first perturbation; and so forth.

The small parameter (expansion parameter) is usually (or at least often) given by the coupling constant in the interaction term of your Hamiltonian (or Lagrangian). Perturbation theory therefore only works well in the weak coupling limit. In the strong coupling limit, one can try to use so-called non-perturbative methods, of which the Schwinger-Dyson equation approach is an example (not to be confused with the Dyson expansion or Dyson series in the context of perturbation theory).

Crudely, one can derive the perturbations as follows.$^\star$ Assume that one is given a Hamiltonian of the form $$ H[\phi] = H_0[\phi] + \alpha H_i[\phi] , \ \ \ \ - (1) $$ where the dimensionless coupling constant $\alpha$ is made explicit. Then one can assume that the field can be expanded in terms of the coupling constant $$ \phi=\phi_0 + \alpha \phi_1 + \alpha^2 \phi_2 + ...$$ One can now substitute the expansion into the Hamiltonian. Then, to consider the unperturbed case, one sets $\alpha=0$, so that $$ H = H_0[\phi_0] . $$ From this one can obtain the unperturbed result $\phi_0$. For the first perturbation, one can take the first derivative of $(1)$ with respect to $\alpha$ and then set $\alpha=0$, $$ \partial_{\alpha}H|_{\alpha=0} = dH_0[\phi_0,\phi_1] + H_i[\phi_0] . $$ Thus one gets an expression that contains both $\phi_0$ and $\phi_1$. Since we now know $\phi_0$, one can substitute it in and proceed to obtain $\phi_1$. Next one proceeds to the second derivative, substitute in what is know on obtain the next term in the expansion. In this way, one can step-by-step obtain the different terms in the expansion for $\phi$.

$\star$ I'm not going to be working this out for your two examples.

  • $\begingroup$ “one can try to use so-called non-perturbative methods, of which the Dyson equation approach is an example”: Dyson equation is the basis of perturbative expansions. It cannot be called "non-perturbative". $\endgroup$
    – AlQuemist
    Commented Feb 26, 2018 at 9:02
  • $\begingroup$ @AlQuemist: Oh dear, sorry, I think these was some confusion. I clarified what I meant in my answer. $\endgroup$ Commented Feb 26, 2018 at 9:49

A perturbative approximation — as any other approximation — is based on some presuppositions about the defining parametres and the qualitative behaviour of a system. Without such assumptions you cannot take any step other than just tackling the whole Hamiltonian with all its complexities — usually an impossible task. Furthermore, notice that even the Hamiltonians you have provided as example (Hubbard and $c-d$ hybridization) are based on some deep assumptions about the physical behaviour of the systems under consideration. So, when analysing a physical system, one must always know well about such underlying presuppositions. For instance, the Hubbard model is an utter simplification of the original (intractable) model for Coulomb-interacting electrons on a lattice:

Full Hamiltonian for electrons and ions (source: chp. 2 of Ref. [1])

The simplifying assumptions in the Hubbard model are eg., that Coulomb interaction between delocalised conduction ($c$-) electrons is local, and the heavy ion cores form a static lattice. For $c-d$ hybridization model, the presumptions are that the states of delocalised $c$-electrons is almost orthogonal to that of the localised $d$-electrons, and that the local Coulomb interaction is only present for localised $d$-states.

How do we partition our Hamiltonian and how to decide which part should be treated as perturbation?

Such partitionings are often based upon deep physical insights gained from experimental observations of materials. For example, we know that the order of magnitude of Coulomb interaction for localised electrons is $\sim$ 5–10 eV, while their kinetic energy is $\sim$ 1 eV, and the hybridization strength, $V$, is $\sim$ 0.1–1 eV. Upon such a knowledge, one can decide which part of the Hamiltonian is the “main” part $H_0$ and which part is the perturbative part $H_1$. So, in general, one should properly compare the energy scales of the terms in the Hamiltonian (for instance, see Wigner–Seitz radius used to asses the importance of Coulomb interaction in the case of electron liquids).

What's the difference for the self-energy due to the ... Coulomb interaction term and coupling term between the central region and the left/right electrodes [= $c-d$ hybridization term]?

From a formal aspect, all self-energies are “born equal” — their origin/cause does not matter, although the technical nitty-gritties of a perturbative calculation depends on the particular form of the interaction [3]. Nonetheless, from a physical perspective, the self-energies are indeed very different in how they modify the behaviour of a system. For instance, the hybridization term $V$ (Eq. 3) merely “mixes” the states $c$ and $d$. The properties of the original electronic system, however, remain qualitatively the same as the unperturbed one (with $V \rightarrow 0$).

The rôle of the Coulomb interaction in the Hubbard model (Eq. 2) is manifestly different. When the strength of the Coulomb interaction overcomes the kinetic energy, $ |u / t| \gg 1 $, the original conducting system (with $ |u / t| \ll 1 $) becomes a Mott insulator — a qualitatively different phase.

[1] Bruus, H., and K. Flensberg. “Many-body quantum theory in condensed matter physics” (2002) [PDF].
[2] For details of the diagrammatic techniques, consult chps. 11 & 12 of Ref. [1].


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