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

Substitution of the gauge transformations and $U{U}^{†}=I$ givesA direct commutator calculation givesCovariance of the left side then implies $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{†}$. The cyclic property of the trace makes the traces of $F_{\mu \nu }{F}^{\mu \nu }$, $D_{\mu }\varphi D^{\mu }\varphi$, and ${\varphi }^3$ invariant, so the Lagrangian is gauge invariant.

The product of generators satisfiesand thereforeBecause ${\varphi }^a{\varphi }^b{\varphi }^c$ is symmetric, its contraction with the antisymmetric structure constant of a Lie algebra $f_{abc}$ vanishes. Hence

For the Grassmann-odd ghost $c$, the BRST transformations areup to a simultaneous convention-dependent sign for $Q$ and $c$. The ghost transformation makes the BRST charge nilpotent: $Q^2{A}_{\mu }=Q^2\varphi =0$.

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.

An infinitesimal change of gauge-fixing fermion, $\Psi \mapsto \Psi +\delta \Psi$, changes the action by the BRST-exact term $Q\delta \Psi$. If $\mathcal{O}$ is gauge invariant, then $Q\mathcal{O}=0$. BRST invariance of the measure gives the BRST Ward identity $⟨QX⟩=0$, so the normalized variation isTherefore correlation functions of gauge-invariant operators do not depend on the choice of gauge, provided there is no BRST anomaly or boundary contribution.