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This piece of argument has been repeated again and again by experts, that is

Since the fermions are gapped, then I can integrate it out.

but I have no idea of what will happen if the fermions are not gapped.

Can I still integrate it out safely?

For an ordinary insulator which is gapped, then after integrating out the fermions, we have an low energy (frequency) effective action

$$S=\int d^4x\, \frac{1}{16\pi} (F^{\mu\nu})^2 + \frac{1}{2} F^{\mu\nu}\mathcal{P}^{\mu\nu}-j^\mu A_\mu $$ where $\mathcal{P}_{\mu\nu}$ is the response to $\bf B$ and $\bf E$, i.e. $\bf M$ and $\bf P$. If we vary vector potential we have the good old Maxwell equations inside a medium.

It can be understood easily: if the external gauge field has frequency lower than the band gap, the medium can be viewed as a collections of dipoles; otherwise high energy photons will tear off electrons from atoms.

In the case of metal, it is not gapped (I think it is improper to call it gapless), photon of any frequency will excite a electron. We know that there is an additional Ohm's law added to describe a metal.

Can we obtain effective theory (Ohm+Maxwell) of metal by integrating out "ungapped" electrons?

In addition,

for a gapless conductor, say graphene, what happens if I integrate out Dirac electrons?

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  • $\begingroup$ Integrating out a Gaussian field amounts to inverting a Laplacian (the inverse is the propagator), whose zeroes (ie. poles of the propagator) correspond to the masses of excitations, so I think that this problem can be nicely understood from that perspective, but I am not sure. Mass gap subtleties are a mystery to me as well. $\endgroup$ Commented Aug 28, 2012 at 9:55
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    $\begingroup$ If you "integrate out" a field that isn't gapped, the action stays nonlocal at all scales, it doesn't reproduce an effective local Lagrangian theory. But your question is probably to derive the effective dissipative theory of light travelling through a weakly conducting medium, to derive the dissipative decay, and this requires a formalism that integrates statistically random fields that are not quantum-gapped. This is an interesting independent question. $\endgroup$
    – Ron Maimon
    Commented Aug 28, 2012 at 10:28
  • $\begingroup$ I think the poles are not masses but dispersion relations. (@user404153) $\endgroup$
    – Machine
    Commented Aug 29, 2012 at 2:54
  • $\begingroup$ Yes, it is the correct answer. But can you explain more about "nonlocal action?" @RonMaimon $\endgroup$
    – Machine
    Commented Aug 29, 2012 at 2:56
  • $\begingroup$ @ChenChao: I made a comment because I was too lazy to sit down and work out a precise answer, I'll try to fill it at some point. To integrate out a gapless fermion, you can just do it for a quadratic action, and you get a nonlocal determinant contribution to the action. This is reproduced in unquenched lattice QCD, in Feynman's papers from the early 1950s (where the photon field is integrated out, see especially the Feynman-Vernon collaboration), and other places. But I think the question of integrating out ungapped but thermally random fields, to leave an effective theory, is more interesting $\endgroup$
    – Ron Maimon
    Commented Aug 29, 2012 at 3:13

1 Answer 1

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The question of integrating out gapless fermions in an interacting theory of bosons and fermions usually arises when one studies

  1. the critical point of a 2nd order quantum phase transition from a metal to some ordered state, such as an anti-ferromagnet.
  2. a critical phase, such a gauge field interacting with fermions.

Gapless fermionic systems can be of two types based on the dispersion relation: energy and momentum of the low energy fermion excitations

A) vanish at the same point in the momentum space, as in the case of electrons in QED, electrons in graphene with suitably tuned chemical potential, "band touching" situations, etc. These systems are said to possess Fermi points in the momentum space.

B) vanish at different points in the momentum space, as is the case of fermions at finite density such as traditional metals that host extended Fermi surfaces and QCD at finite density.

In general, integrating out gapless modes is dangerous because the procedure may generate non-local relevant or marginal (in RG sense) terms in the effective action which cannot be renormalized within local RG schemes. However, it appears to be the case that if there is a single zero energy mode, as is the case for gapless bosons and cases 1A and 2A above, the integrating out procedure is "safe" and no problematic non-local terms are generated.

However, cases 1B and 2B are more subtle. In the field of quantum critical phenomena, works by Hertz, Moriya and Millis that resulted in the Hertz-Millis-Moriya (HMM) theory of quantum phase transition, relies on integrating out fermion thereby constructing an effective Landau-Ginzburg theory for the bosonic order parameter. This was considered "safe enough" for decades, until it was realized that the "safety" related to integrating out fermions around the Fermi surface crucially depends on the dimension and type of Fermi surface (extended vs. single point). For example, in (2+1)D integrating out fermions with momenta around an extended Fermi surface generates non-local marginal vertices in the effective bosonic theory, leading to the failure of the HMM paradigm. (Further reading: look here and here).

With our current understanding the moral of the story appears to be that integrating out fermions

  • with momenta around an extended Fermi surface is unsafe, especially in (2+1)D, although it seems to be safe in (3+1)D. The physical reason is that the extended Fermi surface provides the system with an infinite number of zero modes which contributes substantially to the low energy properties of the system. Also in lower dimensions increase in quantum fluctuations makes the Fermi surface contribution even more significant, such that integrating out fermions generates stronger IR singular vertices for the effective bosonic theory.
  • with momenta around one or several Fermi points is generally safe in any dimensions.

EDIT: It may not be safe (i.e. not lose out on important physics) even in the case of semi-metals (metallic states with Fermi points) to simply integrate out the fermions: see here for example (esp section V A).

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