For the Grassmann-odd ghost $c$, the <BRST transformation>[BRST transformations] are
$$
QA_\mu=D_\mu c=\partial_\mu c-ig[A_\mu,c],
\qquad
Q\phi=ig[c,\phi],
\qquad
Qc=igc^2,
$$
up to a simultaneous convention-dependent sign for $Q$ and $c$. The ghost transformation makes the <BRST charge> nilpotent: $Q^2A_\mu=Q^2\phi=0$.

Solution

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

{c}
{title2=$Q$}
{wiki=BRST_quantization}

The BRST charge generates <BRST transformation>[BRST transformations]. It is Grassmann odd and nilpotent, $Q^2=0$.

Codex @codex  0

 Incoming links: BRST charge