Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 304 / 4

Treat the BRST transformation $Q$ as an odd graded derivation. In matrix notation the first three transformations areThenafter substituting $Qc$ and using the Jacobi identity. Similarly, the two terms incancel because the ghost components anticommute and only the Lie-algebra commutator survives. Applying $Q$ once more to $Qc$ gives a sum proportional to $f^d{}_{e[a}{f}^e{}_{bc]}{c}^a{c}^b{c}^c$, which vanishes by Jacobi. Finally,Thus $Q^2=0$ on every field.

The gauge-invariant Yang-Mills and matter part ${\mathcal{L}}_0$ is BRST invariant because a BRST variation is a gauge transformation with ghost-valued parameter. For the gauge-fixing fermionthe graded Leibniz rule givesup to the common sign convention used to define the ghost term. This is precisely the gauge-fixing, auxiliary-field, and ghost sector of the displayed Lagrangian. Therefore

Under $\Psi \mapsto \Psi +\delta \Psi$, the transition amplitude changes to first order by an insertionIfwith the corresponding condition on the bra, the two terms vanish after moving $\hat{Q}$ to the external states. The amplitude is then independent of the gauge-fixing choice.

Conversely, invariance for arbitrary changes of the gauge-fixing fermion requires physical external states to be BRST closed. States differing by a BRST-exact state have identical matrix elements against closed states, so the physical state space is the cohomology