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The two electromagnetic tree-level Feynman diagrams for $q\gamma \to q\gamma$ exchange a quark in the $s$ and $u$ channels. Their sum obeys both photon Ward identities. Averaging over the initial quark spin, color charge and two physical incoming photon polarization vectors gives $-2e^4{Q}_q^4(s/u+u/s)$ after final-state sums. The color average is one because the interaction preserves the single open quark colour index. The interference trace is proportional to $(p_1+p_2)\cdot (p_1-p_4)=(s+t+u)/2=0$ for massless external states.

Put $Q_u=2/3$, $s=q_s^2$, $u=q_u^2$, and suppress the common colour factor ${\delta }_{ji}$. Define $i\mathcal{M}$ by the product of Feynman rules. With massless external Dirac spinors the two graphs giveThe massless Dirac equation gives $/p_{1}u(p_1)=0$ and $\bar{u}(p_3)/p_3=0$. ThusBy four-momentum conservation, $p_2=p_3-q_u$, soThe two terms cancel:This is the Ward identity for the incoming photon; replacing its photon polarization vector by its four-momentum produces zero. The outgoing photon obeys the analogous identity. Each exchange graph separately fails this test, so the cancellation requires their sum. Consequently the physical scattering amplitude is invariant under ${\epsilon }^{\mu }(p_2)\mapsto {\epsilon }^{\mu }(p_2)+c{p}_2^{\mu }$.

There are four tree-level Feynman diagrams: three photon exchange graphs with pairings $12|34$, $13|24$, $14|23$, and the four-photon contact interaction. They are drawn in the bottom row of the figure. With no external fermions or Faddeev-Popov ghost fields, an internal matter or ghost line would have to close into a loop; it cannot add a tree-level Feynman diagram here.

Use all-incoming null four-momenta $k_i$ with ${\sum }_i{k}_i=0$. Let ${\epsilon }_i$ be the external photon polarization vectors, including the complex conjugation for physical outgoing photons, and impose $k_i\cdot {\epsilon }_i=0$. For one pairing $ij|kl$, put $K=k_i+k_j=-(k_k+k_l)$. At the first cubic Feynman vertex, the terms proportional to $k_i\cdot {\epsilon }_i$ and $k_j\cdot {\epsilon }_j$ vanish, leavingAt the other end, the incoming internal four-momentum is $+K$, giving $+2\lambda ({\epsilon }_k\cdot {\epsilon }_l)K^{\sigma }$. The Feynman gauge photon propagator then givesThe last expression is evaluated at generic kinematics, with singular channel limits taken after cancellation. Summing the three pairings produces $+4i{\lambda }^{2}P$, whereThe quartic Feynman vertex contributes $i{\mathcal{M}}_{\mathrm{contact}}=-4i{\lambda }^{2}P$. ThusThis is the tree photon-scattering cancellation in a quadratic Abelian gauge: the individual gauge fixing interactions cancel in the physical scattering amplitude. The argument uses transverse external photon polarization vectors; it does not assert that an arbitrary off shell four-point correlation function vanishes. Physical light-by-light scattering first occurs through charged-particle loop Feynman diagrams.

There are exactly four tree-level Feynman diagrams: one four-photon contact graph and one exchange graph for each of the three partitions $12|34$, $13|24$ and $14|23$ of the external legs. The exchange graphs contain two cubic Feynman vertices. A fermion loop or a closed Faddeev-Popov ghost line has positive loop order, so neither enters here.

Write $p_1+p_2+p_3+p_4=0$ and take physical photon polarization vectors ${\epsilon }_i$ with $p_i\cdot {\epsilon }_i=0$ and $p_i^2=0$. At the cubic Feynman vertex attached to legs $i,j$, the two terms containing $p_i\cdot {\epsilon }_i$ or $p_j\cdot {\epsilon }_j$ vanish. If $q=p_i+p_j$ flows into the internal line, the remaining factor at that end isAt the other end the internal incoming four-momentum is $q$ rather than $-q$, giving $+2\lambda q^{\nu }({\epsilon }_k\cdot {\epsilon }_l)$. Multiplication by the photon propagator givesThe cancellation holds first at generic nonzero channel Mandelstam invariants and then by continuation at degenerate kinematics. Summing the three exchange graphs produces $4i{\lambda }^{2}P$, whereThe quartic Feynman vertex contributes $-4i{\lambda }^{2}P$. ThereforeThis is a cancellation for physical on shell photon polarization vectors. The off shell four-photon correlation function need not vanish. Physical light-by-light scattering first occurs through charged-particle loop Feynman diagrams.

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.

The original PDF contains three algebraic misprints in this question. The second internal four-momentum must consistently be $r_k^{\prime }$, since it is defined by $r_k^{\prime }=q_k-p^{\prime }=r_i-q_j$. Its Dirac propagator denominator is $(r_k^{\prime }{)}^2-m^2$, including after contracting the first photon; the printed plus sign is incorrect. Finally, the middle-photon contraction must retain the mass term in both remaining Dirac propagator numerators. The following calculation keeps the electron mass and proves the corrected identities.

Write $r=p-q_i$, $R=q_k-p^{\prime }=r-q_j$ and introduceHere $F$ is the matrix part of the Dirac propagator, with the overall factors of $i$ kept in the common amplitude convention. The Clifford algebra proves $D(l)F(l)=F(l)D(l)=I$. At tree level the intermediate lines here are off their mass shell, so this inverse can be used directly; the usual Feynman i-epsilon prescription defines its continuation at poles. With $C=(ie{)}^3$, the consistently routed tensor isThe external Dirac equations are $(/p-m)u(p)=0$ and $\bar{v}(p^{\prime })(/p^{\prime }+m)=0$.

For the photon at the electron end, $/q_i=/p-/r$ and hence $/q_{i}u=-D(r)u$. Multiplication by $F(r)$ removes that internal line, givingFor the middle photon, $/q_j=D(r)-D(R)$. Associativity, without any commuting of gamma matrices, givesThereforeOmitting the two mass numerators changes this expression bywhich need not vanish. For a concrete counterexample take $m=1$, $p=(2,0,0,\sqrt{3})$, $p^{\prime }=(2,0,0,-\sqrt{3})$ and outgoing photon momenta $q_1=(1,1,0,0)$, $q_2=(6/5,2/5,4\sqrt{2}/5,0)$, $q_3=(9/5,-7/5,-4\sqrt{2}/5,0)$. All external momenta are on shell and conserve four-momentum. Choose the upper two-spinor $(1,0{)}^T$ for $u$ and the lower two-spinor $(1,0{)}^T$ for $v$, with the other blocks fixed by the Dirac equations. For $(i,j,k)=(1,2,3)$ and $(\mu ,\sigma )=(0,3)$, the correct middle contraction divided by $C$ is $1/3$, whereas the expression with the mass numerators omitted is $5/27$. Thus the printed middle identity is not valid for a massive electron.

At the positron end, $q_k=R+p^{\prime }$. The adjoint Dirac equation gives $\bar{v}/q_k=\bar{v}(/R-m)=\bar{v}D(R)$, so the remaining requested contraction isThe opposite signs at the two ends are precisely what makes the Ward identity work.

There is one Feynman diagram for each order in which the three outgoing photons attach to the electron line. Their orders, encountered from the incoming electron to the incoming Positron, are $(1,2,3)$, $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$ and $(3,2,1)$.

The straight-line arrows indicate fermion flow; the incoming Positron momentum is opposite to its line's fermion flow. Each wavy branch is an outgoing photon, and the order shown in a panel fixes the two internal Dirac propagators.

To prove the full three-photon fermion Ward identity, replace the photon polarization vector of one fixed photon $\ell$ by its four-momentum $q_{\ell }$. Fix the order $(a,b)$ of the other two photons and write ${\Gamma }_a=/{ϵ}_a$, ${\Gamma }_b=/{ϵ}_b$. There are exactly three positions for photon $\ell$ relative to that order. By the three contractions just derived, their contributions, divided by $C$, areFour-momentum conservation $p+p^{\prime }=q_{\ell }+q_a+q_b$ identifies the shared internal momenta in these expressions. The three terms cancel exactly. Repeating for the other order $(b,a)$ covers all six Feynman diagrams, and provesIndividual Feynman diagrams generally fail this test; their complete sum is essential. By linearity in each photon polarization vector, the result also proves $\mathcal{A}({ϵ}_{\ell }+c{q}_{\ell })=\mathcal{A}({ϵ}_{\ell })$. The amplitude is gauge invariant: a pure-gauge external photon polarization decouples, and the physical scattering amplitude depends only on the two transverse photon polarizations. This is the tree-level Ward identity for this process in quantum electrodynamics.

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 signature $(+,-,-,-)$ and electron charge $q=-e$, with $e>0$. Then ${\mathcal{L}}_{\mathrm{int}}=-q\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }=+e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. Write the process as $e^{-}(p,s)+\gamma (k,ϵ)\to e^{-}(p^{\prime },s^{\prime })+\gamma (k^{\prime },{ϵ}^{\prime })$. Physical external states satisfyand the photon polarization vectors obey $k\cdot ϵ=k^{\prime }\cdot {ϵ}^{\prime }=0$. Choose unit physical polarizations with ${ϵ}^{\ast }\cdot ϵ={ϵ}^{\prime \ast }\cdot {ϵ}^{\prime }=-1$; representatives differing by a multiple of the corresponding null momentum describe the same physical polarization. The external spinors obey $(/p-m)u(p)=0$, $\bar{u}(p^{\prime })(/p^{\prime }-m)=0$, with ${\bar{u}}_s{u}_r=2m{\delta }_{sr}$. Outgoing polarization appears complex conjugated, so this formula includes helicity states as well as real linear polarizations.

The needed QED Feynman rules are a vertex $-iq{\gamma }^{\mu }=+ie{\gamma }^{\mu }$, an internal Dirac propagator $i(/q+m)/(q^2-m^2+i0)$, the external spinors, and the incoming/outgoing photon polarization vectors. There is no internal photon at this order and no elementary two-photon–Dirac seagull vertex. The two photon orderings along the electron line give internal momenta $q_s=p+k$ and $q_u=p-k^{\prime }$. Defining the matrix element by $S_{fi}-{\delta }_{fi}=i(2\pi {)}^4{\delta }^4(p+k-p^{\prime }-k^{\prime })\mathcal{M}$, their sum isThe denominators are $2p\cdot k+i0$ and $-2p\cdot k^{\prime }+i0$. Both terms are essential; this is the tree-level Compton amplitude.

To prove the incoming Ward identity, replace $ϵ$ by $k$ and write $R(q)=(/q+m)/(q^2-m^2)$, with the Feynman boundary prescription understood. The physical nonzero-photon kinematics keeps these tree denominators off the electron pole. Sincethe external Dirac equations implyTherefore the two contracted terms are $\bar{u}(p^{\prime })/{ϵ}^{\prime \ast }u(p)$ and its negative, and cancel. By linearity,This is the two-photon fermion Ward identity. An analogous cancellation holds for the outgoing photon. Gauge-equivalent polarization representatives have the same scattering amplitude, so a longitudinal pure-gauge component is not an extra physical photon state. Individual diagrams do not generally satisfy this identity; their sum does.

For perturbative order, each elementary QED vertex has one photon leg. If $I_{\gamma }$ is the number of internal photon lines and $E_{\gamma }=2$, the vertex parity of a QED amplitude givesThus there is no third-order contribution. At fourth order, $V=4$ and, counting all three legs per vertex with four external legs, $I=(3V-4)/2=4$. A connected graph has loop number $I-V+1=1$.

The nonzero one-loop classes on the open electron line are an internal-electron self-energy insertion, a vertex correction at either external-photon vertex, and a virtual photon spanning both photon vertices, giving the four-point box class. The first figure sketches each class for one photon ordering; interchange the two external photon attachments for its crossed partner. This yields two internal-self-energy graphs, four vertex graphs and two spanning-box graphs. They use the usual photon propagator, for example $-i{g}_{\mu \nu }/({\ell }^2+i0)$ in Feynman gauge.

For completeness, one-loop diagram classes for Compton scattering also depend on whether one is drawing an unamputated correlation function or a physical amplitude. Electron self-energies can be attached to either external electron leg, and photon vacuum polarization can be attached to either external photon leg, for both photon orders. In an on-shell renormalized S-matrix, these are accounted for by LSZ reduction and external-field residues rather than counted again as independent amputated graphs. Photon-leg corrections are not vacuum-polarization insertions on an internal Born photon: there is no such Born line.

At the same fourth order, a renormalized calculation has one mass or electron-field counterterm on the internal electron line, or one vertex counterterm at either Born vertex; the crossed partners are included. Each counterterm is a relative order-$e^2$ correction to the order-$e^2$ tree amplitude. External-field normalization must follow the same convention.

A possible closed electron loop with three photon vertices, linked to the external electron line by a virtual photon, cancels between its two orientations by Furry's theorem. Transposing the loop and using $C^{-1}{\gamma }^{\mu }C=-({\gamma }^{\mu }{)}^T$ gives an extra sign $(-1{)}^3$, so its orientation-reversed partner is its negative. A one-photon tadpole likewise vanishes. Disconnected vacuum bubbles are removed by vacuum normalization. The first genuine radiative corrections are fourth order, with the internal self-energy, vertex and spanning-box classes and their consistently included renormalization terms.

In quantum electrodynamics, gauge invariance connects local phase freedom, electric-charge conservation and the restrictions on photon interactions. For definiteness takeThe gauge transformationhas $D_{\mu }\psi \mapsto e^{-ie\alpha }{D}_{\mu }\psi$: the derivative of the local phase cancels the added gauge potential. Also $F_{\mu \nu }$ is unchanged because mixed partial derivatives commute. Consequently every term in the Lagrangian density is invariant. This derives the minimal electromagnetic coupling of a Dirac field, including the interaction $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. A fermion mass is compatible with the symmetry, whereas a local Proca photon mass $m_{\gamma }^2{A}_{\mu }{A}^{\mu }/2$ is not.

The classical Maxwell equations are ${\partial }_{\mu }{F}^{\mu \nu }=j^{\nu }$, with $j^{\nu }=e\bar{\psi }{\gamma }^{\nu }\psi$. Taking a divergence gives ${\partial }_{\nu }{j}^{\nu }=0$ because $F^{\mu \nu }$ is antisymmetric. The same conservation law follows directly from the charged Dirac equation and its adjoint. For vanishing flux at spatial infinity, $Q=\int d^{3}x{j}^0$ is conserved. Gauss law identifies this charge with the electric flux through a surrounding surface. Particle-antiparticle creation is allowed, but its net electric charge is zero.

Local gauge invariance is also a redundancy in the potentials, rather than an extra propagating degree of freedom. The canonical momentum of $A_0$ vanishes, and its field equation enforces the Gauss law constraint in gauge theory, $\nabla \cdot \mathbf{E}=j^0$. After imposing the constraint and quotienting gauge freedom, a massless photon has two transverse physical polarizations. Gauge transformations vanishing at the boundary identify equivalent descriptions; constant phase transformations at infinity can act on charged states and generate their conserved charge. This distinction explains why gauge redundancy and electric charge are compatible.

In perturbative QED, the quadratic photon operator cannot be inverted before gauge fixing. Adding $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$ gives the covariant gauge propagatorThis propagator contains longitudinal and timelike components, but these are not external physical photon states. A Gupta-Bleuler null-state quotient, or the equivalent BRST construction, removes them. In linear Lorenz gauge, $\delta ({\partial }_{\mu }{A}^{\mu })=\square \alpha$, so the Faddeev-Popov determinant is independent of $A_{\mu }$ and the Faddeev-Popov ghost fields decouple. This last statement concerns the linear gauge; Abelian gauge groups alone do not guarantee decoupling in a nonlinear gauge condition.

At the quantum level, the vector symmetry gives a Ward identity. One way to obtain it is to change variables by a localized phase in a time-ordered correlation function containing $\psi (y)$ and $\bar{\psi }(z)$. The action variation becomes a current divergence; the two field insertions contribute opposite contact terms, since their charges are opposite. Fourier transformation turns the divergence into the photon momentum, and amputating the external fermion propagators turns the two contact terms into the difference of their inverse propagators. With the charge removed from the proper vertex and the overall propagator factor $i$ removed from its inverse, the result is the proper-vertex Ward-Takahashi identity in QEDAt tree level this is simply $k_{\mu }{\gamma }^{\mu }=(/p+/k-m)-(/p-m)$. Between on-shell external spinors, the inverse propagators vanish, so the full physical scattering amplitude satisfies $k_{\mu }{\mathcal{M}}^{\mu }=0$. Therefore replacing a photon polarization vector by ${\epsilon }_{\mu }+c{k}_{\mu }$ leaves the amplitude unchanged. The identity applies to the complete required sum of diagrams, not necessarily each diagram separately. It also ensures cancellation of the gauge parameter from physical observables when all terms at the chosen perturbative order are included.

For the photon two-point function, the Ward identity implies transverse photon vacuum polarization:Thus ultraviolet subtraction renormalizes the gauge-invariant kinetic term, rather than requiring a local photon-mass counterterm. Perturbative unbroken QED retains a massless photon. Transversality by itself would not exclude a nonlocal massless pole in a different theory; the assertion here includes the perturbative QED setting.