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Vary the antisymmetric auxiliary electromagnetic field tensor and the electromagnetic four-potential independently. The first variation sets $F_{ab}={\nabla }_a{A}_b-{\nabla }_b{A}_a$ algebraically, and variation of the potential gives ${\nabla }_b{F}^{ab}=0$. Substitution of the algebraic equation reduces the first-order action to the usual negative quadratic Maxwell action. Metric variation gives the same stress tensor after eliminating the auxiliary field, because the chain-rule term multiplying its variational equation then vanishes.

Use $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Vary the Maxwell action with respect to the electromagnetic four-potential. Antisymmetry of the electromagnetic field tensor gives $\delta \mathcal{L}=-F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. After integration by parts, the Euler-Lagrange field equation isA gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. It therefore leaves the Maxwell action and observable electromagnetic fields unchanged: different potentials represent the same physical field. This redundancy matters for canonical quantization: before gauge fixing, the momentum conjugate to $A_0$ vanishes and the four potential components cannot be treated as four independent physical oscillators.

Choose the Lorenz gauge $D={\partial }_{\mu }{A}^{\mu }=0$, attainable by solving $\square \chi =-D$ with suitable boundary data. The Euler-Lagrange field equation then reduces toResidual Lorenz gauge symmetry consists of transformations with $\square \chi =0$. The gauge condition and this residual redundancy ultimately leave two transverse polarizations of a free photon.

For the specified kinetic density ${\mathcal{L}}_C=-\frac{1}{2}{\partial }_{\mu }{A}_{\nu }{\partial }^{\mu }{A}^{\nu }$, one has $\partial {\mathcal{L}}_C/\partial ({\partial }_{\mu }{A}_{\nu })=-{\partial }^{\mu }{A}^{\nu }$. Its Euler-Lagrange field equation is therefore $-\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction, sinceThe divergence follows by expanding both sides and commuting the second derivatives. Thus the wave equation follows from a Feynman gauge action; the additional physical-state condition will implement the Lorenz gauge. One must not impose $D=0$ inside the original action before varying it.

Take the covariant components $A_{\nu }$ as canonical coordinates. The canonical momentum computed from ${\mathcal{L}}_C$ isThe equal-time canonical Poisson brackets are $\{A_{\mu }(t,\mathbf{x}),{\Pi }^{\nu }(t,\mathbf{y})\}={\delta }_{\mu }{}^{\nu }{\delta }^3(\mathbf{x}-\mathbf{y})$, with coordinate-coordinate and momentum-momentum brackets zero. Applying canonical quantization givesIn particular $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The negative temporal sign is the source of the indefinite state-space norm in covariant photon quantization.

The spatial Fourier transform of the wave equation gives an oscillator of frequency $\omega =|\mathbf{k}|$ for each component. The Hermitian adjoint condition on the electromagnetic four-potential then yields the mode expansion of a free fieldThese are massless on shell modes. Consistently with the preceding equal-time canonical commutation relation,The normalization is the same as for the scalar modes, with $-g_{\mu \nu }$ replacing the species identity matrix.

Gupta-Bleuler quantization starts with the Fock vacuum annihilated by every $a_{\mu }$ and imposes only the annihilation, or positive-frequency part of a quantum field, of the Lorenz gauge condition:Imposing the entire Hermitian field divergence as a strong operator identity would be incompatible with the four-component canonical commutation relation. The weaker condition suffices: between two physical states its positive-frequency part vanishes on the ket, and its negative-frequency part vanishes on the bra. Hence $⟨{\mathrm{phys}}^{\prime }|{\partial }_{\mu }{A}^{\mu }|\mathrm{phys}⟩=0$.

For a one-photon state $|\mathbf{k},ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(\mathbf{k})|0⟩$, commuting the constraint through the creator givesThus the allowed photon polarization vectors satisfy $\boxed{k\cdot ϵ=0}$. Their state inner product, with the common momentum-normalization factor suppressed, is $⟨{ϵ}^{\prime }|ϵ⟩=-{ϵ}^{\prime \ast }\cdot ϵ$. Momentum eigenstates are delta-normalized; the same statements hold with finite norms for normalizable wave packets.

Rotate the spatial axes so that $k=\omega (1,0,0,1)$. The constraint is ${ϵ}^0={ϵ}^3$, so the complete allowed family isThe first vector is the allowed longitudinal gauge polarization, proportional to the null four-momentum $k$. Its state has zero norm because $k^2=0$; it is also orthogonal to every constrained state because $k\cdot ϵ=0$. A purely spatial vector $(0,0,0,1)$ alone is not an allowed physical polarization: its time component must accompany it to satisfy the constraint. The other two vectors are transverse polarizations, perpendicular to $\mathbf{k}$, with positive unit norms. For the displayed family,The timelike negative-norm excitation has been excluded by the constraint, and the remaining form is positive semidefinite. The Gupta-Bleuler null-state quotient identifies $ϵ$ and $ϵ+\lambda k$, removing the null direction. It leaves exactly two transverse physical photon polarizations with a positive inner product; their bosonic Fock space is the physical photon state space.

Finally use Feynman slash notation, $/a={\gamma }^{\mu }{a}_{\mu }$, and impose four-momentum conservation $p+q=k+k^{\prime }$. The external Dirac spinors obeySet $r=p-k$ and $r^{\prime }=p-k^{\prime }$. Replacing $ϵ$ by $k$ in the first term of the electron-positron annihilation into two photons amplitude, the rightmost contracted vertex obeysSince the Clifford algebra implies $(/r+m)(/r-m)=(r^2-m^2)I_4$, that term becomesFor the second term, four-momentum conservation gives $k=q+r^{\prime }$. The contracted vertex is now adjacent to the outgoing positron spinor on the left, soConsequentlyThe two terms cancel, proving the two-photon fermion Ward identityExchanging the two photon labels proves the same result for ${ϵ}^{\prime }=k^{\prime }$. By linearity the amplitude is invariant under $ϵ\mapsto ϵ+\lambda k$ on either external photon leg. Therefore the null longitudinal gauge polarization has zero production amplitude: physical scattering produces only the two transverse photon polarizations. This cancellation also shows why both photon orderings in the tree-level Feynman diagrams are essential; either term alone fails the Ward identity.

Use the metric Levi-Civita connection and hold the covariant potential $A_a$ fixed when varying the metric. Symmetry of the connection cancels its terms in the electromagnetic field tensor:The Electromagnetic Bianchi identity follows from commuting partial derivatives, or $F=dA$ and $d^2=0$. Written covariantly it is ${\nabla }_c{F}_{ab}+{\nabla }_a{F}_{bc}+{\nabla }_b{F}_{ca}=0$, since total antisymmetrization also cancels the symmetric connection terms. Hence $\boxed{F_{[ab;c]}=0}$ without using any field equation.

For the potential variation, $\delta F_{ab}={\nabla }_a\delta A_b-{\nabla }_b\delta A_a$. Antisymmetry in the Maxwell action giveswhere the second equality is integration by parts followed by relabeling; compactly supported variations, or vanishing boundary variations, remove the surface term. Arbitrary $\delta A_a$ therefore give the source-free Maxwell equationsNo gauge fixing of $A$ is needed for this variational calculation.

In signature $(+---)$ define the Hilbert stress-energy tensor by ${\delta }_{g}S=\frac{1}{2}\int \sqrt{-g}{T}_{ab}\delta g^{ab}{d}^{4}x$. Since the independent covariant potential makes $F_{ab}$ metric independent,Combining these two terms yields the electromagnetic stress-energy tensorEquivalently $\delta S=-\frac{1}{2}\int \sqrt{-g}{T}^{ab}\delta g_{ab}$. The overall stress-variation sign has been matched to the signature and action. As a check, in an orthonormal frame $F_{cd}{F}^{cd}=2(B^2-E^2)$ and $T^{00}=(E^2+B^2)/(8\pi )>0$ for a nonzero field; copying a mostly-plus formula without changing conventions would fail this check.

Let $J^a={\nabla }_b{F}^{ab}$. Contract the Bianchi identity with $F^{bc}$ to obtain $2F^{bc}{\nabla }_b{F}_{ac}=\frac{1}{2}{\nabla }_a(F_{cd}{F}^{cd})$, hence $F^{bc}{\nabla }_b{F}^a{}_c=\frac{1}{4}{\nabla }^a(F_{cd}{F}^{cd})$. Differentiating the stress tensor cancels these derivative terms and gives the electromagnetic stress-energy divergence identityThus Maxwell's equation implies stress conservation for any field. Conversely, where the matrix $F^a{}_c$ is nonsingular, conservation implies $J^c=0$ by matrix inversion. When it is singular, conservation only says that $J$ lies in its kernel, so the converse cannot be inferred without the printed nonsingularity assumption.

For the first-order formulation, temporarily call the independently constructed potential curvature $f_{ab}(A)={\nabla }_a{A}_b-{\nabla }_b{A}_a$. In the first-order Maxwell action, vary the independent antisymmetric $F^{ab}$, with $F_{ab}=g_{ac}{g}_{bd}{F}^{cd}$. Its variation isso $F_{ab}=f_{ab}(A)$. Varying $A$ independently gives exactly ${\nabla }_b{F}^{ab}=0$, by the same integration-by-parts calculation as before. Eliminating the auxiliary field algebraically makes the integrand $f^{ab}{f}_{ab}-2f^{ab}{f}_{ab}=-f^{ab}{f}_{ab}$, recovering the original Maxwell action. The Bianchi identity then follows from $f=dA$. Metric variation also agrees after elimination: the chain-rule term from the eliminated field multiplies its already vanishing field equation. Thus the two actions give the same source-free electromagnetic dynamics and stress tensor. The Palatini procedure here means independent first-order field variables, not an additional variation of the spacetime connection.

Use a flat five-dimensional metric of signature $(+----)$ and a circle coordinate $y\sim y+2\pi r$. Start with the canonically normalized Maxwell actionExpand the real Maxwell field in a Fourier series:The circle integral makes different modes orthogonal. Writing $a^{(n)}=A_5^{(n)}$ and $k_n=n/r$, the resulting four-dimensional action isThe scalar sign is positive because the fifth direction is spacelike in the chosen mostly-minus metric. This expression follows directly by separating the $\mu \nu$ and the two identical $\mu 5$ contributions in $F_{MN}{F}^{MN}$.

A five-dimensional gauge transformation $A_M\mapsto A_M+{\partial }_M\Lambda$ acts on its modes byFor $n\ne 0$, define the gauge-invariant vectorThen $F_{\mu 5}^{(n)}=-i{k}_n{B}_{\mu }^{(n)}$, so its term is a Proca field mass term $+\frac{1}{2}{k}_n^2{B}_{\mu }^{(n)}{B}^{\mu (-n)}$. Equivalently, a periodic mode-dependent gauge transformation sets $a^{(n)}=0$. The nonzero modes are massive vectors whose longitudinal polarization is supplied by $a^{(n)}$ through the Stueckelberg mechanism:For each positive $n$ there is one complex vector, equivalently the two real sine/cosine vectors, each with three physical polarizations. The negative mode is its conjugate, not another independent complex field. This is the infinite Kaluza-Klein tower of the Maxwell reduction on a circle.

At $n=0$, $\delta a^{(0)}=0$ under periodic infinitesimal gauge transformations. The zero modes are a massless vector with two polarizations and a real massless scalar with one. The scalar is the circle component, or Wilson line degree of freedom; a generic constant value cannot be removed by a periodic gauge parameter. Both have $m_0=0$. At energies $E\ll 1/r$, the free theory's low-energy effective action is thereforeThere is no scalar potential in this free Maxwell theory. If the original action instead has coefficient $1/g_5^2$ and unrescaled constant zero-mode fields are used, integrating the circle gives $g_4^2=g_5^2/(2\pi r)$; canonical field rescaling recovers the normalization displayed above.

For the gravitational part, allow a circle radius field $R(x)$ with vacuum value $r$, and use a dimensionless angle $\chi \sim \chi +2\pi$ in the Kaluza-Klein theory ansatzThe vector comes from off-diagonal metric components; $R$ is the radion. Perform the coordinate transformation ${\chi }^{\prime }=\chi -\kappa \lambda (x)$, leaving $x$ unchanged. Since $d\chi =d{\chi }^{\prime }+\kappa {\partial }_{\mu }\lambda d{x}^{\mu }$, the metric has the same form withThus a higher-dimensional diffeomorphism becomes an Abelian gauge transformation in the reduced theory. Changing the sign convention for the internal coordinate shift reverses the displayed sign without changing its content.

More generally, continuous internal isometries give lower-dimensional gauge symmetries, with vector fields from metric components along the corresponding Killing vector fields. The circle translation group gives $U(1)$. A scalar mode ${\psi }_n(x)e^{in\chi }$ transforms as ${\psi }_n^{\prime }=e^{in\kappa \lambda }{\psi }_n$, so $D_{\mu }{\psi }_n=({\partial }_{\mu }-in\kappa A_{\mu }){\psi }_n$ transforms covariantly. This derives Kaluza-Klein charge quantization in the chosen vector normalization. The metric vector is a graviphoton; if a five-dimensional Maxwell field is present as well, its zero-mode photon is a separate field originating from that Maxwell gauge symmetry.

Choose one free, massless complex scalar field with unit charge magnitude and bosonic statistics, minimally coupled to $A$. To fix the sign, use the conventional gauge covariant derivative $D_{\mu }={\partial }_{\mu }-iq{A}_{\mu }$ and take $q=-1$, so $D_{\mu }={\partial }_{\mu }+i{A}_{\mu }$ and $\phi \mapsto e^{-i\alpha }\phi$. Its antiparticle has charge $+1$. The matter data areThere is no scalar self-interaction. For real fields before formal continuation, its action is $S_{\phi }=\int (D_{\mu }\phi {)}^{\ast }(D_{\mu }\phi )d^{4}x=\int \bar{\phi }(-D^2)\phi d^{4}x$, by integration by parts. The Gaussian functional integral gives the inverse functional determinant and thus ${\Gamma }_{\phi }=+{\mathrm{Tr}}^{\prime }\log (-D^2)$, up to normalization.

The key operator identity isOn the formal complex gauge slice, $-D^2=-M_A={\mathcal{O}}_A$. With matching regulators and omitted zero modes, this gives ghost-scalar determinant cancellation in a complex quadratic gauge, leaving the Maxwell action as the background functional:where gauge-fixing terms and field-independent constants are understood separately. For a fixed background the complex scalar field and Faddeev-Popov ghost field integrals are Gaussian, so this determinant cancellation is exact within the formal construction. It is a cancellation of closed bosonic matter loops against ghost loops in that background functional. It does not prove that physical scalar quantum electrodynamics has no radiative corrections: the printed real Euclidean gauge slice contains only $A=0$ and misses every nonzero-curvature orbit. On that literal slice the equality is trivial; a nontrivial version requires the additional complex-contour interpretation.