Covariant gauge

Adding $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\alpha )$ gives a family of Lorentz-covariant gauges; $\alpha =1$ is Feynman gauge.

The Fourier-space quadratic Maxwell kernel after adding $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$. For nonzero $\xi$ and non-null four-momentum, its transverse and longitudinal momentum projectors give $K=-k^2(P_T+{\xi }^{-1}{P}_L)$ and its inverse is $-(P_T+\xi P_L)/k^2$. The inversion of the gauge-fixed Maxwell kinetic operator supplies the photon propagator with its vacuum boundary prescription.

The $\xi \to 0$ limit of covariant gauge fixing by $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$. It imposes the divergence condition sharply and gives a transverse photon propagator. This is distinct from the vector-potential choice called Landau gauge for a uniform magnetic field when computing Landau levels.

The covariant Landau gauge propagator is the transverse limit of the inversion of the gauge-fixed Maxwell kinetic operator. Contracting with $k^{\mu }$ gives zero away from the pole. The Lorenz gauge condition still has residual transformations satisfying $\square \chi =0$, so boundary conditions and the pole prescription complete the choice.

Use $F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$ and $(D_{\mu }c{)}^a={\partial }_{\mu }{c}^a+g{f}^{abc}{A}_{\mu }^b{c}^c$. The Faddeev-Popov gauge-orbit identity with Lorenz gauge function $G^a={\partial }^{\mu }{A}_{\mu }^a$, followed by Gaussian averaging of $G^a=f^a$, supplies the quadratic gauge-fixing term. Represent the determinant of $-{\partial }^{\mu }{D}_{\mu }$ using a Grassmann Gaussian integral over a Faddeev-Popov ghost field and antighost field to obtain the ghost term. The minus sign in this operator differs by a field-independent determinant factor from the direct gauge-condition Jacobian; it is a consistent antighost convention.

The ghost term equals $({\partial }^{\mu }{\bar{c}}^a)(D_{\mu }c{)}^a$ up to a boundary term. Thus it contains a ghost-gluon vertex with the sign matching this kinetic convention. An equivalent auxiliary-field form replaces the gauge-fixing term by $B^a{G}^a+\xi B^a{B}^a/2$. The Nakanishi-Lautrup field equation gives $B^a=-G^a/\xi$ and recovers the displayed Lagrangian. With left-acting BRST transformations $sA=Dc$, $sc=-g[c,c]/2$, $s\bar{c}=B$, and $sB=0$, the gauge-fixing and ghost terms together are $s[{\bar{c}}^a(G^a+\xi B^a/2)]$.

Gauge fixing removes the noninvertible longitudinal direction of the Maxwell quadratic action. In a gauge with parameter $\xi$, the Fourier kernel is $-k^2{\eta }^{\mu \nu }+(1-{\xi }^{-1})k^{\mu }{k}^{\nu }$. Multiplying its inverse by $i$ gives the photon propagator, with the Feynman i-epsilon prescription. At $\xi =1$ the answer is $-i{\eta }_{\mu \nu }/(k^2+i0)$.

For fixed $\xi \ne 0$, the Lorentzian quadratic density $-F^2/4+(\partial \cdot A{)}^2/(2\xi )$ has momentum kernel $K^{ab}=-p^2{\eta }^{ab}+(1+1/\xi )p^a{p}^b$. At $p^2\ne 0$ its inverse, defined by $K^{ac}{G}_{cb}={\delta }_b^a$, is $G_{ab}=-{\eta }_{ab}/p^2+(1+\xi )p_a{p}_b/(p^2{)}^2$. A vacuum photon propagator is $iG$ with the appropriate Feynman i-epsilon prescription. The usual parameter in the negative-sign covariant gauge term is $\alpha =-\xi$, so Feynman gauge is $\xi =-1$.

For the positive-sign covariant gauge-fixing inverse, $p^a{G}_{ab}=\xi p_b/p^2$. It is not identically zero for fixed $\xi \ne 0$ and is a nonzero vector at a nonzero non-null four-momentum. Null momenta require the chosen pole prescription, and individual zero components are possible. The covariant Landau gauge limit suppresses this longitudinal part. The Ward identity removes longitudinal contributions from physical conserved-current amplitudes.

With $D={\partial }_{\mu }{A}^{\mu }$, expand the Maxwell Lagrangian and its Feynman gauge term to findCommuting partial derivatives verifies the divergence identity. Both densities give $\square A^{\mu }=0$ under the usual variational boundary conditions. Their canonical momentum components differ: the original density gives $-F^{0\nu }-{\eta }^{0\nu }D$, whereas the subtracted density gives $-{\dot{A}}^{\nu }$. Thus a free-photon momentum expansion using the latter requires a stated boundary-term convention.

In mostly-minus signature, adding the boundary term ${\partial }_{\mu }{K}^{\mu }$ with $K^{\mu }=\frac{1}{2}(A_{\nu }{\partial }^{\nu }{A}^{\mu }-A^{\mu }{\partial }_{\nu }{A}^{\nu })$ to the subtracted Feynman gauge density changes ${\tilde{\pi }}^{\mu }=-{\dot{A}}^{\mu }$ to the displayed canonical momenta. For fields with vanishing spatial boundary terms, $\int K^0{d}^{3}x=\int A_0{\partial }_i{A}_i{d}^{3}x$, whose functional derivatives are these momentum shifts. They depend only on coordinates, so $[A_{\mu },{\pi }^{\nu }]=i{\delta }_{\mu }{}^{\nu }{\delta }^3$ is unchanged. The only potentially nonzero mixed momentum bracket is $[{\pi }^0(x),{\pi }^i(y)]=i({\partial }_{y^i}+{\partial }_{x^i}){\delta }^3(x-y)=0$. Thus both density representatives define canonical commutation relations, although their momenta must not be equated componentwise.

Gupta-Bleuler quantization keeps Lorentz covariance by first quantizing all four photon polarizations in an indefinite inner-product space. The condition $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$, followed by quotienting null states, leaves the two transverse physical polarizations.

Choose $C_{\mathbf{p}}=a_{\mathbf{p}}^0-a_{\mathbf{p}}^3$ with the standard temporal and longitudinal polarization vectors. The physical pre-space is the common kernel of all $C_{\mathbf{p}}$. In each regulated mode, $a^0$ acts on creator polynomials as $-{\partial }_{a^{0†}}$ and $a^3$ as ${\partial }_{a^{3†}}$, so the constraint kernel consists of transverse creator polynomials and polynomials in $C^{†}=a^{0†}-a^{3†}$. Since $[C,C^{†}]=0$, these latter excitations remain constrained. They are orthogonal to every constrained state because $⟨\Phi |C^{†}=⟨C\Phi |=0$. Quotienting this radical of a Hermitian form leaves only the transverse bosonic Fock space with a positive inner product. The constraint alone gives a positive semidefinite Hermitian form; quotienting removes its null directions. If starting from finite-particle creator polynomials, take the Hilbert space completion of this positive quotient to obtain the physical Hilbert space.

Choose a temporal polarization vector, two transverse polarization vectors and a spatial longitudinal polarization vector. The Gupta-Bleuler quantization condition is $(a^0-a^3)|\chi ⟩=0$. Applied to ${\sum }_{\lambda }{c}_{\lambda }{a}^{\lambda †}|0⟩$, it gives $c_3=-c_0$. The remaining temporal-longitudinal excitation $a^{0†}-a^{3†}$ has zero norm and is orthogonal to every constrained state. Its field wavefunction is proportional to the null four-momentum, hence is pure gauge. Quotienting this direction leaves the two positive transverse photon states. Use wave packets or a box normalization to avoid treating continuum momentum eigenstates as normalizable vectors.

The covariant free-photon oscillator construction uses four polarization vectors and $[a^{\lambda },a^{{\lambda }^{\prime }†}]=-{\eta }^{\lambda {\lambda }^{\prime }}$. Starting from a positive Fock vacuum, it induces an indefinite Hermitian form on the multiparticle state space: the temporal oscillator has negative norm while the three spatial oscillators have positive norm. It is therefore not the physical positive Hilbert space. The Gupta-Bleuler null-state quotient selects a positive physical space with two transverse photon polarizations. Continuum momentum oscillators and their states are understood after smearing or finite-volume regularization.

With real complete polarization vectors, the covariant photon Fock space oscillator commutators and ${\sum }_{\lambda ,{\lambda }^{\prime }}{ϵ}_{\mu }^{\lambda }{ϵ}^{\nu {\lambda }^{\prime }}{\eta }_{\lambda {\lambda }^{\prime }}={\delta }_{\mu }{}^{\nu }$ imply the equal-time canonical commutation relations $[A_{\mu }(\mathbf{x}),-{\dot{A}}^{\nu }(\mathbf{y})]=i{\delta }_{\mu }{}^{\nu }{\delta }^3(\mathbf{x}-\mathbf{y})$. The two mixed creation-annihilation terms add; the two coordinate-coordinate or momentum-momentum terms cancel after their polarization sums. The sign of the time-component oscillator is essential to the identity. Boundary-induced canonical transformation in Feynman gauge explains the momenta for the unsubtracted density.

For a temporal photon wave packet $|f,0⟩=\int d^{3}pf(\mathbf{p})a_{\mathbf{p}}^{0†}|0⟩/(2\pi {)}^3$, the oscillator commutator gives $⟨f,0|f,0⟩=-\int d^{3}p|f(\mathbf{p}){|}^2/(2\pi {)}^3<0$ when $f\ne 0$. This is the negative direction of the indefinite Hermitian form, not the divergent normalization of an unsmeared momentum eigenstate. It is excluded by the physical condition in Gupta-Bleuler quantization.

For the Euclidean covariant gauge term $({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$, the free gauge-boson propagator is

With a mostly-plus metric signature, $e^{iS}$, and gauge-fixing term $-(\partial \cdot A{)}^2/2$, the free adjoint gauge-field kernel is $-p^2{\eta }_{\mu \nu }{\delta }_{ab}$. Inverting it gives the displayed Feynman propagator. This convention uses the same denominator and sign as a massless real scalar propagator.

At $\xi =1$, the Euclidean photon propagator is $D_{\mu \nu }(k)={\delta }_{\mu \nu }/k^2$. Its Lorentz-index contraction simplifies loop numerators in quantum electrodynamics.

At a nonzero Euclidean momentum $k$, the transverse projector of a vector field is $P_T^{ab}={\delta }^{ab}-k^a{k}^b/k^2$. In a Lorentzian Minkowski metric, its mixed-index form is $(P_T{)}^a{}_b={\delta }_b^a-p^a{p}_b/p^2$, defined for $p^2\ne 0$. It obeys $P_{T}p=0$ and $P_T^2=P_T$. A nonzero null four-momentum is insufficient for this particular decomposition because the denominator vanishes.

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.