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Without gauge fixing, gauge-equivalent configurations are integrated repeatedly, and the quadratic gauge-field operator has zero modes along infinitesimal gauge transformations. It therefore has no inverse with which to define perturbative propagators. A gauge condition removes these directions locally in the perturbative expansion. The associated Faddeev-Popov determinant can be represented by anticommuting Faddeev-Popov ghost fields and Faddeev-Popov antighost fields.

First make the normalization explicit. For the fields literally occurring in the action with overall $1/g^2$, putThe resulting kinetic terms are canonical, and the interaction terms carry powers of $g$. We use signature $(-,+,+,+)$ and incoming Fourier modes $e^{ipx}$, consistent with the displayed $p^2-i0$ denominators. The quadratic action in the canonical fields isColor indices are diagonal. Introduce the longitudinal and transverse projectors $P_{L,\mu \nu }=p_{\mu }{p}_{\nu }/p^2$ and $P_T=\eta -P_L$. The gauge kinetic matrix is $-p^2(P_T+{\xi }^{-1}{P}_L)$, with inverse $-(P_T+\xi P_L)/p^2$. The ghost kinetic operator is $-p^2$. The Gaussian functional integral and Grassmann integral therefore giveThe longitudinal formula is understood as the Feynman-prescribed inverse of the kinetic operator; away from its poles the projector inversion is ordinary algebra. The propagators for the original unrescaled fields are $g^2$ times these expressions. Using $P_{T}p=0$ and $P_{L}p=p$ giveswith either normalization. This is the longitudinal gauge propagator contraction.

The canonical ghost interaction comes directly from the covariant derivative:For incoming antighost momentum $p$, the derivative supplies $i{p}^{\mu }$; the expansion of $e^{iS}$ supplies another $i$. Thus the ghost-gluon vertex is $g{p}^{\mu }{f}_{abc}$, as requested. For unrescaled fields its coefficient is instead $p^{\mu }{f}_{abc}/g^2$. The printed order-$g$ rule presupposes canonical normalization; keeping the overall $1/g^2$ and an order-$g$ vertex for the same unrescaled variables would mix conventions. Choosing Fourier modes $e^{-ipx}$ instead reverses the vertex sign and must be done consistently for all momentum assignments.

In Feynman gauge, $\xi =1$, the one-loop ghost two-point graph has one internal ghost line and one internal gauge line joined by two such vertices. The external ghost line is open, so it does not carry a closed ghost-loop minus sign. With an orthonormal adjoint basis, the structure constants are totally antisymmetric, and the color product along this line is $f_{adc}{f}_{cdb}=-C{\delta }_{ab}$. The two internal propagators supply $(-i{)}^2=-1$, leaving a positive color-and-vertex factor for the amputated insertion. DefineThe amputated insertion is $i{\delta }_{ab}{g}^2{I}_d(p)C$. Restoring the two external canonical ghost propagators gives the one-loop contributionThe literal unrescaled correlator in the question is $g^2{G}^{(1)}$, hence of order $g^4$, whereas its tree term is $-i{\delta }_{ab}{g}^2/(p^2-i0)$. In $d\ne 4$ the dimensionless renormalized coupling convention adds ${\mu }^{4-d}$ multiplying $g^2$ in the canonical loop term.

To evaluate the Feynman integral, combine the denominators with a Feynman parameter:Shift $\ell =k-(1-\alpha )p$. The denominator becomes $[{\ell }^2+\alpha (1-\alpha )p^2-i0{]}^2$, and the numerator is $p\cdot \ell +(1-\alpha )p^2$. The odd term vanishes in translation-invariant dimensional regularization. For spacelike $p$, a Wick rotation converts the remaining integral into a Euclidean one and cancels the prefactor $1/i$. The needed radial integral isIndeed, spherical coordinates and $u={\ell }_E^2/M^2$ leave a beta integral ${\int }_0^{\infty }{u}^{d/2-1}(1+u{)}^{-2}du=\Gamma (d/2)\Gamma (2-d/2)$; the angular factor cancels $\Gamma (d/2)$. Establish this first in a convergent range and then analytically continue in $d$. Substituting $M^2=\alpha (1-\alpha )p^2$ yieldsThis is the printed result, including the factor $1/i$ in its Minkowski-space definition. Equivalently the parameter integral is $B(d/2-1,d/2)$. Symmetry under $k\leftrightarrow p-k$ also gives $I_d(p)$ as $p^2/2$ times the massless scalar bubble, independently checking the numerator factor.

Write $ϵ=4-d$. Since $\Gamma (ϵ/2)=2/ϵ+O(1)$ and the parameter integral tends to ${\int }_0^1(1-\alpha )d\alpha =1/2$, the pole isHere $p$ is off shell and nonzero, separating the ultraviolet pole from massless infrared issues. The unrescaled amplitude has an extra factor $g^2$. The divergence is proportional to the existing ghost kinetic operator, so the one-loop ghost kinetic counterterm in Feynman gauge is local:It inserts $-i\delta Z_c{p}^2$ into the inverse propagator, cancelling the divergent insertion $+i{g}^{2}C{I}_d(p)$. With the two external propagators attached, its contribution is $+i{\delta }_{ab}\delta Z_c/p^2$, cancelling the displayed connected amplitude pole. In the original variables the same action term is $-\delta Z_c\partial \bar{c}\cdot \partial c/g^2$. If instead one defines dimensional regularization by $d=4-2ϵ$, the identical counterterm is written $\delta Z_c=g^{2}C/(32{\pi }^2ϵ)$; the factor of two is purely the regulator convention.

Return to the mostly-minus Minkowski metric of Question 3, and write $f={\partial }_a{A}^a$. The field $\xi (x)$ is assumed real and nonzero wherever its inverse occurs. Under the usual Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$, with $\xi$ inert, the electromagnetic field tensor is invariant but $f\mapsto f+\square \lambda$. The change in the added density isThis is not generally a total derivative. The action is not invariant under arbitrary gauge transformations. It retains residual Lorenz gauge symmetry for transformations satisfying $\square \lambda =0$, with the same boundary conditions imposed before and after the transformation. The action itself is undefined at $\xi =0$; that value can only be considered as a limiting gauge.

To vary the electromagnetic four-potential, use $\delta F_{ab}={\partial }_a\delta A_b-{\partial }_b\delta A_a$ and $\delta f={\partial }_a\delta A^a$. The integration by parts of both terms giveswith surface terms removed by the variational boundary conditions. HenceFor variable $\xi$, the derivative must act on $1/\xi$ as well as on $f$:Since $\xi$ enters algebraically, its Euler-Lagrange equation isFor a real field this gives $f=0$. Thus a dynamical gauge-fixing parameter imposes a constraint equation in field theory; it is not an ordinary propagating scalar. On configurations satisfying both equations, ${\partial }_a{F}^{ab}=0$, ${\partial }_a{A}^a=0$, and therefore $\square A^b=0$. There is no independent kinetic equation that determines $\xi$.

For the momentum-space equation at a prescribed constant $\xi \ne 0$, use $A_a(x)=\int d^{4}p{e}^{-ip\cdot x}{A}_a(p)/(2\pi {)}^4$. Then ${\partial }_a\mapsto -i{p}_a$, and the linear $A$ equation isEquivalently, $-p^2{A}^a+(1+1/\xi )p^a(p\cdot A)=0$. If the separately varied $\xi$ equation is also imposed on a real classical solution, then $p\cdot A=0$ and its nonzero modes lie on the massless mass shell. For constructing a full-field Green function, however, invert the fixed-background quadratic operator before imposing that on-shell constraint. Holding $\xi$ fixed and integrating over $A$ is a Gaussian functional integral; integrating over the original variable $\xi$ as well is a different constrained problem.

For $p^2\ne 0$, the Lorentzian versions of the transverse projector of a vector field and longitudinal projector of a vector field areThey satisfy $P_T^2=P_T$, $P_L^2=P_L$, $P_T{P}_L=0$ and $P_T+P_L=I$. The mixed-index operator isIts inverse follows by inverting these two scalar eigenvalues. Define $G_{ab}$ by $K^{ac}{G}_{cb}={\delta }_b^a$. The positive-sign covariant gauge-fixing inverse isThis is the algebraic inverse of the kinetic operator. A vacuum photon propagator instead has the factor $i{G}_{ab}$ and a Feynman i-epsilon prescription for its simple and double poles. If the name $G$ is used for the vacuum two-point function, include that factor $i$ consistently in its defining source equation. A retarded Green function would use different boundary conditions at the same poles. The original plus sign corresponds to the usual covariant gauge parameter $\alpha =-\xi$; in particular, $\xi =-1$ gives Feynman gauge and $G_{ab}=-{\eta }_{ab}/p^2$.

Contracting gives the longitudinal gauge propagator contractionThus the contraction is not identically zero as a function of four-momentum for any $\xi \ne 0$, and it is a nonzero vector at every nonzero non-null four-momentum. A particular component may vanish when $p_b=0$. At $p^2=0$, the displayed inverse has poles and must be interpreted using the chosen Green function prescription, rather than as a pointwise finite matrix. In the limiting covariant Landau gauge, $\xi \to 0$, the longitudinal part vanishes. A covariant photon Green function may have a longitudinal component even though physical photon polarizations are transverse.