The variation found above uses the <adjoint covariant derivative>:
$$
\delta A_\mu^a=\frac1g(D_\mu\alpha)^a,
\qquad
(D_\mu c)^a=\partial_\mu c^a+g f^{abc}A_\mu^b c^c.
$$
For the <Lorenz gauge> functional $G^a[A]=\partial^\mu A_\mu^a$,
$$
\frac{\delta G^a[A^\alpha](x)}{\delta\alpha^b(y)}
=\frac1g\partial^\mu D_\mu^{ab}\delta(x-y).
$$
The field-independent factor $1/g$ may be absorbed into normalization. The <Grassmann Gaussian integral> exponentiates the <Faddeev-Popov determinant> with anticommuting <Faddeev-Popov ghost fields>:
$$
\det(\partial^\mu D_\mu)
=\int\mathcal D\bar c\,\mathcal Dc\,
e^{-\int d^4x\,\bar c^a\partial^\mu(D_\mu c)^a}.
$$
A Gaussian average over the gauge condition supplies the covariant gauge-fixing term, so
$$
\boxed{S=S_g+\int d^4x\left[
\frac1{2\xi}(\partial^\mu A_\mu^a)^2
+\bar c^a\partial^\mu(D_\mu c)^a
\right],
\qquad
D_\mu^{ac}=\delta^{ac}\partial_\mu+g f^{abc}A_\mu^b.}
$$

Solution

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

 Solution

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