Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 4 / c / ii

Gauge invariance gives $Q\int \mathcal{L}=0$. Nilpotence gives $Q^2\Psi =0$, and thereforeThus writing the gauge-fixing contribution as a BRST-exact term makes the complete action BRST invariant.