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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-304/4/c/solution