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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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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. –  Ryan Thorngren Aug 28 '12 at 9:55
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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. –  Ron Maimon Aug 28 '12 at 10:28
    
I think the poles are not masses but dispersion relations. (@user404153) –  ChenChao Aug 29 '12 at 2:54
    
Yes, it is the correct answer. But can you explain more about "nonlocal action?" @RonMaimon –  ChenChao Aug 29 '12 at 2:56
    
@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 –  Ron Maimon Aug 29 '12 at 3:13
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