Write the gauge-fixing and ghost action as the <BRST transformation> of the <gauge-fixing fermion>:
$$
S_{\rm gf}+S_{\rm gh}=s\int d^4x\,
\bar c^a\left(n^\mu A_\mu^a-\frac\xi2B^a\right),
$$
with harmless rescalings of $B$ accommodating the convention $s\bar c^a=\xi B^a$ used in the question. The gauge-invariant action obeys $sS[A]=0$, while nilpotence gives
$$
s(S_{\rm gf}+S_{\rm gh})=s^2\Psi_n=0.
$$
Thus the entire gauge-fixed action is <BRST symmetry>[BRST invariant]. Eliminating the auxiliary field $B^a$ by its algebraic field equation reproduces the axial gauge-fixing term and the ghost action found in part (b).

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 4 / c / Solution

Changing $n^\mu$ to $u^\mu$ changes the <gauge-fixing fermion> by
$$
\Psi_u-\Psi_n=\int d^4x\,\bar c^a(u-n)^\mu A_\mu^a.
$$
The corresponding change of the action is the BRST-exact term $s(\Psi_u-\Psi_n)$. For a BRST-closed observable $\mathcal O$, invariance of the functional measure implies that the variation of its expectation value is the expectation of a total BRST variation and vanishes:
$$
\delta\langle\mathcal O\rangle
\mathrel\propto\langle s[\mathcal O(\Psi_u-\Psi_n)]\rangle=0.
$$
Therefore physical states and observables, which are classes in <BRST cohomology>, do not depend on the fixed axial-gauge vector, provided there is no BRST anomaly or boundary contribution.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 4 / d / Solution

{title2=$\Psi$}
{wiki}

A gauge-fixing fermion is a Grassmann-odd functional whose <BRST transformation> $s\Psi$ supplies the gauge-fixing and ghost terms. Nilpotence gives $s(s\Psi)=0$ immediately.

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