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Vary the Maxwell Lagrangian with respect to $A_{\nu }$. Since $\delta F_{\mu \nu }={\partial }_{\mu }\delta A_{\nu }-{\partial }_{\nu }\delta A_{\mu }$ and $F$ is antisymmetric,After integration by parts, the Euler-Lagrange field equation isThe gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. This gauge invariance means that potentials related in this way represent the same electromagnetic field; all four potential components are not independent physical photon degrees of freedom. In particular, the ungauged kinetic operator is singular and its temporal canonical momentum vanishes.

Choose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. The field equation then becomes $\boxed{\square A_{\mu }=0}$. Residual gauge transformations with $\square \chi =0$ preserve this condition. The correctly named Lorenz gauge condition concerns the divergence of the potential; Lorentz covariance is the separate transformation property of the equations.

For the alternative quadratic Lagrangian density, direct differentiation givesso its Euler-Lagrange field equation is also $\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction. Indeed, with $D={\partial }_{\mu }{A}^{\mu }$,The divergence identity follows by expanding ${\partial }_{\mu }{K}^{\mu }$ and commuting its second derivatives. The quadratic density alone describes four wave equations; the gauge constraint must still be incorporated in the physical states. It is not an unconstrained four-positive-polarization replacement for Maxwell equations.

Using this specified boundary-term convention, the canonical momentum conjugate to $A_{\nu }$ isThe canonical commutation relations are thereforewith field-field and momentum-momentum brackets zero. Equivalently $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The temporal component has the opposite sign to the spatial components; replacing the metric by a positive Euclidean one would destroy these brackets.

The wave equation gives massless frequencies ${\omega }_{\mathbf{k}}=|\mathbf{k}|$. Hermiticity pairs its positive- and negative-frequency solutions, so the Heisenberg picture field has the expansionThis is a real field expansion for each component, but its oscillator metric is indefinite. The assumed brackets, with the dagger's vector index restored, areThey verify the photon oscillator completeness and canonical brackets. Inserting ${\Pi }^{\nu }=-{\dot{A}}^{\nu }$ gives at equal timesThe other brackets cancel between the two frequency contributions. Thus both the $2\omega$ normalization and the sign $-g_{\mu \nu }$ are needed.

The Gupta-Bleuler formalism imposes the Lorenz gauge condition weakly on states:Here $(+)$ is the annihilation, positive-frequency part. Setting the full divergence to zero as an operator would conflict with the four-component canonical commutation relations. The subsidiary condition instead selects a physical pre-space inside the covariant photon Fock space, followed by a quotient of its null states.

Apply it to $|k,ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(k)|0⟩$. Commuting $k^{\mu }{a}_{\mu }$ through the creator gives a multiple of $-k_{\nu }{ϵ}^{\nu }|0⟩$. Hence the allowed photon polarization vectors obeyTake $k=(\omega ,0,0,\omega )$. The complete solution iswith complex coefficients. The last two vectors are transverse linear photon polarizations, and $(0,1,\pm i,0)/\sqrt{2}$ gives the two circular photon polarizations. The first vector is $k/\omega$, the allowed temporal-longitudinal combination. A purely spatial longitudinal vector $(0,0,0,1)$ by itself does not satisfy the subsidiary condition.

The oscillator brackets give, apart from the positive continuum normalization factor,Momentum eigenstates are distributionally normalized; a box or a wave packet makes the norm statement literal. The allowed longitudinal combination has zero norm because $k^2=0$, and it is orthogonal to every constrained state since $k\cdot ϵ=0$. It also gives a pure-gauge field wave: the field strength of $A_{\mu }\propto k_{\mu }{e}^{-ik\cdot x}$ vanishes. Thus $ϵ$ and $ϵ+ck$ represent the same physical photon polarization. The Gupta-Bleuler null-state quotient removes this null direction and leaves two positive-norm transverse photon photon polarizations. The negative-norm temporal oscillator of the unrestricted space is not an extra physical state. This is the transverse one-photon physical quotient.

Finally verify the two-photon fermion Ward identity directly for electron-positron annihilation into two photons. Define $D(r)=/r-m$ and the reduced Dirac propagatorThe equality follows from the Clifford algebra, since $(/r-m)(/r+m)=(r^2-m^2)I$. The incoming Dirac spinors obey $D(p)u(p)=0$ and $\overline{v}(q)(/q+m)=0$. On replacing one inserted photon polarization vector by $k$, the spinor factor in the two diagrams isAt the electron end, $/k=D(p)-D(p-k)$, soFor the other diagram, four-momentum conservation gives $p-k^{\prime }=k-q$. HenceThe two contributions therefore cancel exactly:Individual diagrams need not vanish; their sum implements the Ward identity. Linearity then gives invariance under $ϵ\mapsto ϵ+ck$, as required by gauge invariance. In particular, the null longitudinal photon polarization proportional to $k$ has zero physical production amplitude. Physical scattering produces the transverse photon classes, not an independent longitudinal massless photon.

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.

Take incoming electron momentum $p$ and positron momentum $q$, with outgoing photon momenta $k_1,k_2$, so $p+q=k_1+k_2$. The leading electron-positron annihilation into two photons has two tree-level Feynman diagrams, differing by the order in which the two outgoing photons attach to the charged-fermion line.

The arrows indicate fermion flow. An incoming positron has the opposite flow direction to its physical momentum; the lower row uses an outgoing electron instead.

The required QED Feynman rules, with interaction $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$, are a vertex $-ie{\gamma }^{\mu }$, a Dirac propagator $i(/r+m)/(r^2-m^2+i0)$, an incoming electron spinor $u(p)$, an incoming positron adjoint $\bar{v}(q)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. Here Feynman slash notation means $/a={\gamma }^{\mu }{a}_{\mu }$, and the external Dirac equations are $(/p-m)u(p)=0$ and $\bar{v}(q)(/q+m)=0$.

Put $\mathcal{S}(r)=(/r+m)/(r^2-m^2+i0)$, omitting only the propagator's overall factor $i$. Multiplying the two vertices and the propagator givesThe diagrammatic expression is $i\mathcal{M}$. The relative sign is plus because the two photons are bosons; exchanging their attachments does not exchange external fermions. The amplitude is of order $e^2$.

A physical photon polarization vector obeys $k^2=0$ and $k\cdot ϵ=0$, with the gauge equivalence $ϵ\sim ϵ+\alpha k$. One can choose a transverse representative with ${ϵ}^0=0$ and normalize ${ϵ}^{\ast }\cdot ϵ=-1$. There are two physical transverse polarizations.

To prove the requested two-photon fermion Ward identity, replace ${ϵ}_1^{\ast }$ by $k_1$. Away from an internal pole, $\mathcal{S}(r)$ is the inverse of $/r-m$, with the boundary prescription understood. The first term simplifies becauseFor the second term, four-momentum conservation gives $k_1=(p-k_2)+q$, soThe two contributions are therefore $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Interchanging the photon labels proves the other contraction vanishes. By linearity, adding a multiple of either photon momentum to its polarization leaves the summed amplitude unchanged. This is gauge invariance: longitudinal gauge representatives do not describe an additional physical photon state. An individual diagram does not generally have this property.

For Compton scattering, let $p,k$ be incoming electron and photon momenta and $p^{\prime },k^{\prime }$ the outgoing ones. The diagrams in the lower row have intermediate electron momenta $p+k$ and $p-k^{\prime }$ respectively. The same QED Feynman rules giveTheir denominators are $(p+k{)}^2-m^2+i0$ and $(p-k^{\prime }{)}^2-m^2+i0$, the $s$- and $u$-channel denominators. In annihilation the two exchanged-electron channels instead carry $p-k_1$ and $p-k_2$. Crossing symmetry relates the two expressions by moving the positron to an outgoing electron and one outgoing photon to an incoming photon, with the corresponding momentum and polarization replacements. There is no two-photon contact vertex for the minimally coupled Dirac electron. Both processes require the sum of the two photon orders to satisfy the Ward identity.