Induced "ungauged" Chern-Simons terms from a massive Dirac fermion

Question

It is a well-known fact that a massive Dirac fermion minimally coupled to a gauge field $A_\mu$ induces a Chern-Simons term when integrating out the fermion: \begin{align} i\bar{\psi}\gamma^\mu(\partial_\mu + A_\mu)\psi + m\bar{\psi} \psi \rightarrow \frac{i\operatorname{sign}(m)}{4\pi}\epsilon_{\mu\nu\gamma}A_\mu\partial_\nu A_\gamma \end{align}

What happens in the case that $A_\mu$ is no longer gauged, but is just a vector field? Would it simply generate a non-quantized Chern-Simons term, or is gauging necessary for a Chern-Simons term at all? As far as I can tell, it seems that perturbative calculations at leading order are insensitive to whether or not $A_\mu$ is gauged.

Answer 1

The (2+1)-dimensional QED, after integrating out the fermions, is still a theory with a gauge symmetry. This is not obvious from the expressions that are given in your question, however, because what appears in the question are the Lagrangian densities for the respective theories. The gauge symmetry is not a symmetry of the Lagrangian density, but only of the integrated action, $$ S=\int d^{3}\,{\cal L}.$$ The Chern-Simons term proportional to $\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}A^{\rho}$ is not gauge invariant, but it changes by a total derivative under a gauge transformation. Under $A^{\mu}\rightarrow A^{\mu}+\partial^{\mu}\Lambda$, the Chern-Simons term undergoes $$\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}A^{\rho}\rightarrow\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}A^{\rho}+\epsilon_{\mu\nu\rho}\partial^{\mu}\Lambda\partial^{\nu}A^{\rho}+\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}\partial^{\rho}\Lambda=\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}A^{\rho}+\partial^{\mu}\left(\epsilon_{\mu\nu\rho}\Lambda\partial^{\nu}A^{\nu}\right).$$

That means that the integrate action does not change under the gauge transformation, $S\rightarrow S$. Consequently, the equations of motion of the theory still possess gauge invariance. In spite of the explicit appearance of the potential $A^{\mu}$ in ${\cal L}$, the theory is still a gauge theory.