# Can I integrate out the fermion field that is not gapped?

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?