 Solution

Start with the formally gauge-redundant path integral $Z=\mathrm{vol}(\mathcal{G}{)}^{-1}\int \mathcal{D}A{e}^{iS[A]}$. Use $A_{\mu }^{\theta }=A_{\mu }+{\partial }_{\mu }\theta$, and the generalized Lorenz gauge condition $F_{\omega }[A]={\partial }_{\mu }{A}^{\mu }-\omega$. In a perturbative neighborhood where residual zero modes are removed by boundary conditions, insert the Faddeev-Popov gauge-orbit identityThus ${\Delta }_{\mathrm{FP}}=\det \square$. Gauge invariance lets the $\theta$ integration factor out and cancel $\mathrm{vol}(\mathcal{G})$. Average over $\omega$ with the Gaussian integral weight $\exp [-i\int {\omega }^2/(2\xi )]$; the delta functional fixes $\omega =\partial \cdot A$ and givesRepresent the functional determinant by the independent Grassmann fields $\bar{c},c$. A field-independent determinant sign can be absorbed in normalization, so a convenient convention isThereforeThe Faddeev-Popov ghosts are free and decouple because $\square$ is independent of $A$. For a Non-Abelian gauge theory, an infinitesimal gauge transformation gives $\delta A_{\mu }^a=D_{\mu }^{ab}{\theta }^b$, so the Faddeev-Popov operator is ${\partial }^{\mu }{D}_{\mu }^{ab}[A]$ andIt contains a ghost-gluon vertex: these Faddeev-Popov ghosts do not decouple in covariant gauge fixing of Lorenz gauge type.