lagrangian formalism - Why does a four-divergence not contribute in perurbation theory?

The QCD Lagrangian is $$\mathcal{L}=-\frac{1}{4}G^{a}_{\mu\nu}G^{a\mu\nu}+\sum\limits_{j=1}^n \left[\bar{q}_j\gamma^{\mu}iD_{\mu}q_j - (m_jq^{\dagger}_{Lj}q_{Rj}+h.c.)\right]+\frac{\theta g^2}{32\pi^2}G^{a}_{\mu\nu}\widetilde{G}^{a\mu\nu}$$ And in the book Particle Dark Matter, Sikivie states that the last term is a 4-divergence hence does not contribute in perturbation theory. Why is it a 4-divergence, and that being the case why does it not contribute in perturbation theory?

Answer

Why a divergence does not contribute to the equation of motion is trivially due to the assumption that the fields should vanish at infinity. By Stokes' theorem, any volume integral over a region $\Sigma$ of a divergence $dF$ is equal to a surface integral over its boundary $\partial \Sigma$ of the function $F$.

$$\int_\Sigma d^4x\ \partial_\mu F^\mu = \int_{\partial \Sigma} d^3\sigma_\mu\ F^\mu =0\quad \text{if }\ F^\mu\Big|_{\partial \Sigma}\equiv 0.$$

The more interesting question is why the theta term is a total divergence. Turns out that the theta term is a topological charge, and like any charge it is sourced by a topological current

$$K^\mu = \frac1{16\pi^2} \epsilon_{\mu\alpha\beta\gamma} (A^a_\alpha\partial_\beta A^a_\gamma + \frac13 f^{abc}A^a_\alpha A^b_\beta A^c_\gamma)\,,$$

known as the Chern-Simons current.

You can verify that the divergence of this current gives the following.

$$Q = \int d^4x\ \partial_\mu K^\mu = \frac1{32\pi^2} \int d^4x\ G^a{}_{\mu\nu}\tilde G^{a\ \mu\nu}\,,$$

where $\tilde G^{\mu\nu} = \frac12 \epsilon_{\mu\nu\rho\sigma}G^{\rho\sigma}$. As expected, being a total divergence, the equation of motion will be blind to this term.