For <axial gauge>, $G^a[A]=n^\mu A_\mu^a$. Its Faddeev-Popov operator is
$$
\frac{\delta G^a[A^\alpha]}{\delta\alpha^b}
=\frac1g n^\mu D_\mu^{ab}
=\frac1g n\mathbin\cdot\partial\,\delta^{ab}
+n^\mu f^{acb}A_\mu^c.
$$
On the gauge slice $n\mathbin\cdot A^c=0$, the second term vanishes. Hence
$$
\boxed{\Delta_{\rm FP}[A]\big|_{n\cdot A=0}
=\det\left(\frac1g n\mathbin\cdot\partial\right),}
$$
This <functional determinant> is independent of the gauge field and may be absorbed into the normalization of the <path integral>. Any introduced ghosts are free and decouple, so \b[no ghost fields are needed in axial gauge].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 4 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-304/4/e/solution