Gauge invariance gives $Q\int\mathcal L=0$. Nilpotence gives $Q^2\Psi=0$, and therefore
$$
QS=Q\int\mathcal L+Q^2\Psi=0.
$$
Thus writing the gauge-fixing contribution as a BRST-exact term makes the complete action <BRST symmetry>[BRST invariant].

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-304/4/c/ii/solution