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.

Solution

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

 Solution

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