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At a nonzero Euclidean momentum $k$, the longitudinal projector of a vector field is $P_L^{ab}=k^a{k}^b/k^2$. Its Lorentzian mixed-index form is $(P_L{)}^a{}_b=p^a{p}_b/p^2$ for non-null four-momentum. Together with the transverse projector of a vector field, it obeys $P_T+P_L=I$, $P_T{P}_L=0$, $P_T^2=P_T$, and $P_L^2=P_L$. This formula is not a pointwise decomposition at null four-momentum.

For a constant unitary matrix $h$, ${\varphi }^{\prime }=h\varphi$ gives ${\varphi }^{\prime †}{\varphi }^{\prime }={\varphi }^{†}{h}^{†}h\varphi ={\varphi }^{†}\varphi$. Also ${\partial }_{\mu }{\varphi }^{\prime }=h{\partial }_{\mu }\varphi$, so the kinetic contraction is unchanged. The mass term and the squared norm in the potential are unchanged for the same reason. The original scalar action has the stated global symmetry. For a spacetime-dependent $h$, however, ${\partial }_{\mu }(h\varphi )=h{\partial }_{\mu }\varphi +({\partial }_{\mu }h)\varphi$, and the extra term prevents local invariance of the ordinary-derivative kinetic term.

Choose the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+ig{A}_{\mu }$, with $A_{\mu }=A_{\mu }^{\alpha }{T}_{\alpha }$ Hermitian. Demand $D_{\mu }^{\prime }(h\varphi )=h{D}_{\mu }\varphi$. Expanding both sides yieldsso the necessary Yang-Mills gauge transformation isUnitarity makes the transformed connection Hermitian. This formula includes the inhomogeneous derivative term; mere conjugation would be insufficient for a local transformation.

Define the Yang-Mills field strength by $[D_{\mu },D_{\nu }]=ig{F}_{\mu \nu }$. Direct expansion givesChoose an orthonormal Hermitian generator basis with $\mathrm{tr}(T_{\alpha }{T}_{\beta })=\kappa {\delta }_{\alpha \beta }$. The compact simple Lie algebra has an invariant positive trace metric, so this basis can be chosen; its structure constants are totally antisymmetric. With these conventions a locally invariant dynamical action isEach matter contraction is invariant because $D\varphi$ transforms like $\varphi$, and the gauge kinetic term is invariant by cyclicity of the trace. It is $-\frac{1}{4}{F}_{\mu \nu }^{\alpha }{F}_{\alpha }^{\mu \nu }$ in components. Choosing $\kappa =1/2$ gives the familiar $-\frac{1}{2}\mathrm{tr}{F}^2$ convention.

Formally introduce a gauge-field source $J_{\alpha }^{\mu }$ and a normalized generating functionalA gauge-field Green function is obtained by source differentiation:Here $\mathcal{N}$ is the unnormalized integral at zero source, so the normalized functional obeys $Z[0]=1$. This expression needs gauge fixing: the integral repeats every physical configuration over its whole gauge orbit, producing an irrelevant but divergent gauge-group volume, and the free gauge kinetic operator has longitudinal zero modes, so it has no inverse propagator on that unrestricted space. Ordinary gauge-field Green functions are gauge-dependent; physical gauge-invariant quantities must be obtained consistently from the gauge-fixed theory.

Use the Lorenz condition ${\chi }^{\alpha }[A]={\partial }^{\mu }{A}_{\mu }^{\alpha }$. To keep signs explicit, parametrize a small transformation by $h=1-ig{\omega }^{\alpha }{T}_{\alpha }+O({\omega }^2)$, equivalently the original parameter is ${\theta }^{\alpha }=g{\omega }^{\alpha }$. The finite transformation above gives the adjoint gauge variation for a positive-sign covariant derivative:Therefore $\delta {\chi }^{\alpha }={\partial }^{\mu }{\mathcal{D}}_{\mu }^{\alpha \beta }{\omega }^{\beta }$. Define the Faddeev-Popov operator as $\mathcal{M}=-{\partial }^{\mu }{\mathcal{D}}_{\mu }$, absorbing the field-independent minus sign of its determinant into normalization.

In a perturbative gauge patch, with residual zero modes removed by boundary conditions, the Faddeev-Popov gauge-orbit identity isIt follows by changing coordinates from the gauge parameter to the gauge condition: the functional Jacobian is precisely the determinant of its linearization. Insert the identity into the integral and change variables along the orbit. The action and measure are invariant, so the gauge-group volume factors out and is divided away. Averaging the gauge-slice value $f$ with Gaussian weight $\exp [-i\int f^2/(2\xi )]$ changes the delta-functional restriction to the covariant gauge-fixing termThe remaining determinant depends on $A$ through ${\mathcal{D}}_{\mu }$. Omitting it while adding only ${\mathcal{L}}_{\mathrm{gf}}$ would weight different orbits incorrectly.

For a finite matrix $M$, expansion of a Grassmann integral selects the product containing every $c$ and $\bar{c}$. Anticommuting the variables into a fixed order attaches the sign of each permutation, giving $\int d\bar{c}dc{e}^{i\bar{c}Mc}$ proportional to $\det M$, with only a matrix-independent phase. The regulated functional version therefore represents the Faddeev-Popov determinant asThe independent Faddeev-Popov ghost fields and antighost fields obey $c_{\alpha }{c}_{\beta }=-c_{\beta }{c}_{\alpha }$, with the corresponding mixed and antighost anticommutation. A commuting complex Gaussian integral instead gives an inverse determinant, so it cannot supply this Jacobian. The required local ghost term isAfter integration by parts,This exhibits the free ghost kinetic term and the ghost-gluon vertex. Ghosts are Grassmann-valued scalar auxiliary fields in the Adjoint representation, not physical fermionic particles or external asymptotic states. Their loop signs arise from anticommutation. In an Abelian theory the adjoint derivative loses the field-dependent part, so the determinant is field-independent and the ghosts decouple; in the non-Abelian theory they must be retained.

The correctly gauge-fixed source functional thus uses $S_{\mathrm{inv}}+S_{\mathrm{gf}}+S_{\mathrm{gh}}$ and integrates over $A,\varphi ,{\varphi }^{†},c,\bar{c}$. Expanding its interaction terms gives the gauge, matter and ghost vertices, and differentiating its source functional computes the gauge-field Green functions. For example, at non-null momentum the quadratic gauge kernel iswhere $P_T$ and $P_L$ are the transverse projector of a vector field and longitudinal projector of a vector field. With the Feynman prescription this gives the gauge-boson propagatorFor the chosen ghost normalization the quadratic kernel is $-{\partial }^2$, so the ghost propagator is $i{\delta }^{\alpha \beta }/(k^2+i0)$. Redefining the antighost sign changes ghost propagator and vertex signs together, without changing the determinant or physical conclusions. Gauge fixing requires the accompanying determinant, and its local representation requires anticommuting ghost and antighost fields. The gauge-orbit argument is local/perturbative: global gauge copies, the Gribov ambiguity, prevent interpreting it as a globally unique gauge slice without further treatment.

A Non-Abelian gauge theory describes fields with a local redundancy, not an independent physical degree of freedom for each component of its gauge field. Take a compact gauge group with Hermitian generators $T^a$ and Lie algebra structure constants defined by $[T^a,T^b]=i{f}^{abc}{T}^c$. In a mostly-plus metric signature, the pure Yang-Mills theory hasWith the matter gauge covariant derivative $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$, a finite gauge transformation acts as $A_{\mu }\mapsto U{A}_{\mu }{U}^{-1}-(i/g)({\partial }_{\mu }U)U^{-1}$. Expanding $U=e^{ig{\omega }^a{T}^a}$ givesThe Yang-Mills action is constant along each gauge orbit. Consequently a naive functional integral $\int \mathcal{D}A{e}^{iS[A]}$ repeatedly integrates the same physical configuration and contains the infinite formal volume of the local gauge group. At quadratic order the same problem appears as noninvertibility: the free kinetic operator annihilates $A_{\mu }={\partial }_{\mu }\omega$, so no unique gauge-boson propagator exists. In canonical language the Gauss law constraint in gauge theory and the absence of an independent momentum for the time component show that the field components cannot all be independent oscillators. In covariant quantization, time-like and longitudinal polarizations must not be counted as additional physical states.

The Faddeev-Popov determinant implements a local quotient by the gauge orbit. Choose $G^a[A]={\partial }^{\mu }{A}_{\mu }^a$ and first impose $G^a[A]=f^a(x)$ for a specified function $f$. This is the Lorenz gauge condition when $f=0$. Its infinitesimal change isThe direct Faddeev-Popov operator is therefore $M_0^{ab}={\partial }^{\mu }(D_{\mu }{)}^{ab}$. Locally, if the condition cuts each orbit once and residual zero modes have been removed by boundary conditions, changing variables from $\omega$ to $G[A^{\omega }]-f$ proves the Faddeev-Popov gauge-orbit identityThe determinant is the functional Jacobian of this change of variables. A finite-dimensional version is $\int d\omega \delta (G(\omega )-f)=1/|\det (\partial G/\partial \omega )|$ near a unique root; the perturbative determinant has a fixed sign or phase, absorbed into normalization. This is the same change-of-variables mechanism used in Bastianelli's derivation of the Faddeev–Popov construction.

Insert the identity into the normalized functional integral for a gauge-invariant observable $\mathcal{O}[A]$:For each fixed $\omega$, change the integration variable from $A$ to $A^{\omega }$. The action, observable and measure are invariant, and the determinant is the orbit Jacobian at the resulting gauge slice. The integrand is now independent of the separately integrated orbit parameter. Its integral is $\mathrm{Vol}\mathcal{G}$, canceling the denominator. Thus, up to normalization,This assumes an anomaly-free measure and a local perturbative gauge slice; it is not an assertion of a globally unique representative on every gauge orbit.

Average over $f$ with weight $\exp [-i\int d^{4}x{f}^a{f}^a/(2\xi )]$. The delta functional sets $f=G[A]$, so the average produces the covariant gauge termNext represent the Faddeev-Popov determinant by a Grassmann Gaussian integral. Choose $M=-M_0=-{\partial }^{\mu }{D}_{\mu }$; changing $M_0$ to $-M_0$ changes the regulated determinant by a field-independent factor that is absorbed in normalization. For a Faddeev-Popov ghost field $c$ and independent antighost field $\bar{c}$,To see why it is a determinant rather than its reciprocal, diagonalize a finite-dimensional quadratic form formally: for each Grassmann pair only the term proportional to ${\bar{c}}_j{c}_j$ survives integration, supplying one eigenvalue. Multiplying the surviving eigenvalues gives $\det M$, with an overall factor of $i$ per pair. The functional version is defined with the same regulator as the rest of the theory.

Combining these steps gives the gauge-fixed Yang–Mills LagrangianThis is the gauge-fixed Yang-Mills Lagrangian in a covariant gauge. Integrating its ghost term by parts givesThe second term is the ghost-gluon vertex in the antighost convention chosen here. The Faddeev-Popov ghosts are anticommuting Lorentz scalars introduced to represent a determinant; they are not extra asymptotic particles. A closed ghost loop has a minus sign from their Grassmann parity. In a Non-Abelian gauge theory the determinant depends on $A$, so these loops are essential. In an Abelian gauge theory $D_{\mu }$ in the adjoint sector reduces to ${\partial }_{\mu }$, the determinant is field independent, and its ghost sector decouples.

The quadratic gauge field kernel at non-null momentum isDefine the mixed-index longitudinal projector of a vector field $P_L{}^{\mu }{}_{\nu }=k^{\mu }{k}_{\nu }/k^2$ and transverse projector of a vector field $P_T=1-P_L$. The kernel is $-k^2(P_T+{\xi }^{-1}{P}_L)$ and hence has inverse $-k^{-2}(P_T+\xi P_L)$. Multiplying by $i$ gives the gauge-boson propagator, off its poles,The vacuum Feynman i-epsilon prescription completes this off-pole formula; with the mostly-plus convention its physical transverse denominator is $k^2-i0$. The choice $\xi =1$ is Feynman gauge; $\xi \to 0$ is covariant Landau gauge. The invertible quadratic kernel supplies perturbation theory, while the ghost determinant ensures the gauge-orbit measure has been accounted for.

The consistency of the unphysical sector is organized by BRST symmetry. Introduce a bosonic Nakanishi-Lautrup field $B^a$ and a left-acting BRST differentialThe graded Leibniz rule and the Jacobi identity give $s^2=0$: for example $s^{2}A=D(sc)+g[Dc,c]=0$ since $D[c,c]=2[Dc,c]$, and $s^{2}c$ vanishes by the graded Jacobi identity. Nilpotence on $\bar{c}$ and $B$ is immediate. Using the odd antighost field, the same graded Leibniz rule givesEliminating $B^a$ by its algebraic field equation $B^a=-G^a/\xi$ recovers the displayed gauge-fixing and ghost terms. Thus they are BRST-exact together, and the total action preserves BRST symmetry. At the level of states, the BRST charge $Q$ implements physical states as the ghost-number-zero BRST cohomology $\ker Q/\mathrm{im}Q$. This prescription removes longitudinal, time-like and ghost excitations from physical amplitudes while allowing them on internal lines. For a BRST-closed observable, changing $\xi$ inserts $s({\bar{c}}^a{B}^a/2)$; its expectation vanishes by the BRST change-of-variables identity when the measure preserves the symmetry. This explains the independence of physical quantities from the arbitrary gauge parameter.

Finally, a local condition such as Lorenz gauge can have several solutions on the same gauge orbit, the Gribov ambiguity. The preceding Faddeev-Popov gauge-orbit identity and perturbative gauge fixing construction apply in a neighborhood where the linearized operator is invertible after residual modes are excluded. They resolve perturbative gauge overcounting and supply consistent covariant Feynman rules; they do not by themselves prove a globally unique nonperturbative gauge fixing.

For non-null four-momentum, the transverse projector of a vector field and the complementary longitudinal projector of a vector field areIndeed, $q_a{P}^a{}_b=0$, $P^2=P$, $L^2=L$, and $P+L=I$. Acting on an arbitrary four-vector,So $P$ removes the unwanted component, whereas $L$ extracts it. On the massive mass shell, $q^2=-m^2$ and $P^a{}_b={\delta }_b^a+q^a{q}_b/m^2$. This on-shell form should not be used as an off-shell linear projection: away from $q^2=-m^2$ it is not idempotent. The non-null hypothesis matters; this decomposition is undefined at $q^2=0$.

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.

The quadratic gauge-field action in momentum space isIn terms of the transverse projector of a vector field and longitudinal projector of a vector field,the operator is $k^2(P_T+{\xi }^{-1}{P}_L)$. Its inverse, the gauge-boson propagator, isTherefore $X=1$ and $Y=\xi -1$.