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For signature $(+,-,-,-)$, translation invariance of the Maxwell Lagrangian gives this tensor. It is generally neither symmetric nor gauge invariant. The antisymmetric superpotential improvement $-{\partial }_{\rho }(F^{\rho \mu }{A}^{\nu })$ produces, on the source-free equations of motion, the symmetric electromagnetic stress-energy tensor ${\Theta }^{\mu \nu }=-F^{\mu \rho }{F}^{\nu }{}_{\rho }+{\eta }^{\mu \nu }{F}^2/4$, which is traceless in four spacetime dimensions.

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 signature $+---$, the Maxwell Lagrangian yieldsIt is symmetric after raising its second index, gauge-invariant, and conserved on the source-free Maxwell equations. Its energy density is $\frac{1}{2}({\mathbf{E}}^2+{\mathbf{B}}^2)$ and its energy flux is the Poynting vector. Its difference from the canonical stress-energy tensor is an antisymmetric superpotential divergence plus an equation-of-motion term, so integrated charges coincide when the constraints and vanishing boundary terms are imposed.

For a spatial potential $\mathbf{A}$ with $\mathbf{E}=-\dot{\mathbf{A}}-\nabla A_0$, the Maxwell Lagrangian gives $\mathbf{\Pi }=-\mathbf{E}$ and ${\Pi }^0=0$. Its Hamiltonian is $\int [({\mathbf{\Pi }}^2+{\mathbf{B}}^2)/2+A_0\nabla \cdot \mathbf{\Pi }],d^{3}x$. Preserving the temporal momentum first-class constraint gives the Gauss law constraint in gauge theory. In free space, Coulomb gauge and decay at spatial infinity set $A_0=0$ and leave only transverse modes. The reduced Dirac bracket uses the projector ${\delta }_{ij}-k_i{k}_j/|\mathbf{k}{|}^2$, so quantization gives the transverse equal-time commutator and the positive Hamiltonian displayed above.

The source-free Maxwell Lagrangian gives $S=-\frac{1}{4}\int F_{\mu \nu }{F}^{\mu \nu }{d}^{4}x$ in mostly-minus Lorentzian signature. Its Euclidean action is the displayed positive quadratic functional of the electromagnetic field tensor. Variation, with boundary terms vanishing, yields ${\partial }_{\mu }{F}_{\mu \nu }=0$.

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 Maxwell field has the gauge symmetry $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha (x)$ for an arbitrary smooth scalar function $\alpha$. Commuting the two derivatives shows that $F_{\mu \nu }$ is unchanged, so the Maxwell Lagrangian is invariant. Vary the action, use antisymmetry of $F$ and integrate by parts with variations vanishing on the boundary:The Euler-Lagrange field equations are therefore ${\partial }_{\mu }{F}^{\mu \nu }=0$. Expanding $F$ givesIn Lorenz gauge the second term vanishes, leavingThe PDF calls this “Lorentz gauge”; the standard name is Lorenz gauge. A residual gauge symmetry preserves this condition when $\square \alpha =0$.

Put $D={\partial }_{\rho }{A}^{\rho }$. The gauge-fixing variation isAfter integration by parts its contribution to the Euler-Lagrange field equations is $+{\partial }^{\nu }D$. It cancels the $-{\partial }^{\nu }D$ from the Maxwell Lagrangian, giving $\square A^{\nu }=0$ for all four components. This is the Feynman gauge equation; the physical Lorenz gauge condition must still be imposed on states rather than inferred as an additional classical equation from this gauge-fixed density.

Treating the covariant components $A_{\mu }$ as coordinates, the derivative of the actual displayed density with respect to ${\partial }_0{A}_{\mu }$ gives the canonical momentaIn terms of lower-index spatial coordinates,The gauge fixing has supplied the nonzero momentum of $A_0$, absent from the ungauge-fixed Maxwell Lagrangian.

There is a boundary-term convention needed for the next part's printed momentum expansion. Direct expansion gives the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction:To verify it, the difference of the densities is $\frac{1}{2}[({\partial }_{\rho }{A}_{\sigma })({\partial }^{\sigma }{A}^{\rho })-D^2]$; differentiating $K^{\rho }$ gives these two terms, while the remaining second-derivative terms cancel after renaming indices. The subtracted density has ${\tilde{\pi }}^{\mu }=-{\dot{A}}^{\mu }$, which is exactly the momentum used by the mode expansion in part (c). The literal momenta asked for here are the boxed expression, not $-{\dot{A}}^{\mu }$. Both densities have the same field equations, and their canonical commutation relations agree through the boundary-induced canonical transformation in Feynman gauge.

Use the free Maxwell Lagrangian $\mathcal{L}=-F_{\mu \nu }{F}^{\mu \nu }/4$ in mostly-minus signature. In terms of the spatial electromagnetic four-potential, define $\mathbf{E}=-\dot{\mathbf{A}}-\nabla A_0$ and $\mathbf{B}=\nabla \times \mathbf{A}$. Thus $\mathcal{L}=({\mathbf{E}}^2-{\mathbf{B}}^2)/2$. Taking the Cartesian components of $\mathbf{A}$ as coordinates gives the canonical momentaThis sign convention treats $A_i$ here as Cartesian spatial-vector labels; it is not a claim that covariant and contravariant four-vector components coincide.

The temporal momentum vanishes because no ${\dot{A}}_0$ occurs. This is the primary momentum constraint of the electromagnetic potential, so one cannot treat all four components as independent unconstrained oscillators. The Legendre transformation for the spatial components gives ${\Pi }_i{\dot{A}}_i-\mathcal{L}=({\mathbf{\Pi }}^2+{\mathbf{B}}^2)/2-\mathbf{\Pi }\cdot \nabla A_0$. After integration by parts, the Hamiltonian isSurface terms vanish for decaying fields or the corresponding regulated boundary conditions. The total constrained Hamiltonian also contains an arbitrary multiplier of ${\Pi }^0$. Preserving ${\Pi }^0=0$ under time evolution requireswhich is the Gauss law constraint in gauge theory. Both constraints are first-class constraints; they express the gauge invariance $\mathbf{A}\mapsto \mathbf{A}+\nabla \alpha$, $A_0\mapsto A_0-\dot{\alpha }$. The electric and magnetic fields are unchanged, so different gauge potentials need not describe different physical states.

One consistent canonical quantization of the electromagnetic field first reduces this redundancy. Choose Coulomb gauge $\nabla \cdot \mathbf{A}=0$. In the absence of charges, the Gauss law constraint in gauge theory gives ${\nabla }^2{A}_0=0$; decay at infinity sets $A_0=0$. Residual spatially constant gauge parameters do not change $\mathbf{A}$. The remaining variables are the transverse pairs ${\mathbf{A}}^T,{\mathbf{\Pi }}^T={\dot{\mathbf{A}}}^T$. Their gauge reduction of the Maxwell Hamiltonian isThe two transverse coordinates for each nonzero momentum are the physical degrees of freedom. Treating the constraints as operator identities while retaining unconstrained componentwise commutators would be inconsistent: differentiating $[A_i^T,{\Pi }_j^T]=i{\delta }_{ij}{\delta }^3$ would give a nonzero right side even though ${\partial }_i{A}_i^T=0$.

The reduced Dirac bracket, promoted to a canonical commutation relation, instead gives the transverse equal-time commutator:whereIndeed the spatial transverse projector of a vector field $P_{ij}^T={\delta }_{ij}-k_i{k}_j/|\mathbf{k}{|}^2$ is symmetric, has $P^T\mathbf{k}=0$ and $(P^T{)}^2=P^T$. Projecting both entries of the unconstrained canonical bracket onto the transverse subspace therefore produces $P^T{\delta }^3$, equivalently the constrained Dirac bracket. The spatial zero mode is excluded by the decay condition; in a finite-volume regulator it must be treated separately.

Choose two orthonormal transverse photon polarization vectors ${\mathbf{ϵ}}_r(\mathbf{k})$, so thatThe canonical transverse photon field has mode expansionDifferentiate in time to obtain ${\Pi }_i^T$: the annihilation coefficient acquires $-i{\omega }_{\mathbf{k}}$ and the creation coefficient $+i{\omega }_{\mathbf{k}}$. The oscillator canonical commutation relations areSubstitution into $[A_i^T,{\Pi }_j^T]$ gives $i/2$ times the sum of the two Fourier kernels with phases $\pm i\mathbf{k}\cdot (\mathbf{x}-\mathbf{y})$. The polarization completeness relation and the evenness of $P^T(\mathbf{k})$ make that sum exactly $i{\delta }_{ij}^T$. This derives, rather than presumes, the correspondence between the oscillator and field commutators.

Substitute the mode expansions into $H_T$. Spatial integration sets paired momenta equal or opposite. The $aa$ and $a^{†}{a}^{†}$ terms from the electric and magnetic energies cancel because ${\omega }_{\mathbf{k}}^2=|\mathbf{k}{|}^2$ and $\mathbf{k}\cdot {\mathbf{ϵ}}_r=0$. The remaining terms are harmonic-oscillator energies. Removing the state-independent zero-point energy by normal ordering givesThe corresponding momentum operator is $\mathbf{P}={\sum }_r\int d^{3}k\mathbf{k}{a}_r^{†}{a}_r/(2\pi {)}^3$. These expressions imply $[H_T,a_r^{†}(\mathbf{k})]={\omega }_{\mathbf{k}}{a}_r^{†}(\mathbf{k})$ and $[\mathbf{P},a_r^{†}(\mathbf{k})]=\mathbf{k}{a}_r^{†}(\mathbf{k})$.

Define the Fock vacuum by $a_r(\mathbf{k})|0⟩=0$. Applying $a_r^{†}(\mathbf{k})$ creates a photon of energy $|\mathbf{k}|$, momentum $\mathbf{k}$, and one of two transverse polarizations. Its mass is zero because $k^2=0$. The relativistically normalized state is $|\mathbf{k},r⟩=\sqrt{2{\omega }_{\mathbf{k}}}{a}_r^{†}(\mathbf{k})|0⟩$, with inner product $2{\omega }_{\mathbf{k}}(2\pi {)}^3{\delta }_{rs}{\delta }^3(\mathbf{k}-\mathbf{q})$. Smearing these momentum eigenstates gives normalizable wave packets. Repeated commuting bosonic creation operators construct the symmetric multiphoton bosonic Fock space. Circular combinations of the two transverse photon polarization vectors transform with opposite phases under rotations about $\mathbf{k}$, giving the two physical helicities $\boxed{h=\pm 1}$.

The same physical states can be obtained while retaining manifest Lorentz covariance through Gupta-Bleuler quantization. Add the Feynman gauge term $-({\partial }_{\mu }{A}^{\mu }{)}^2/2$ and quantize four auxiliary oscillator polarizations. Their algebra has an indefinite metric, including a negative-norm temporal excitation. Impose $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. Temporal and longitudinal excitations then do not survive as independent physical states. The result agrees with the positive, two-polarization space obtained by Coulomb gauge reduction; covariant gauge fixing is a different implementation of the same physical constraint, not four physical photon states.

For fields ${\Phi }_a$ with a first-derivative Lagrangian density, suppose an infinitesimal global transformation at fixed coordinates is $\delta {\Phi }_a=\alpha {\Delta }_a$ and $\delta \mathcal{L}=\alpha {\partial }_{\mu }{K}^{\mu }$, with constant infinitesimal $\alpha$. Noether theorem gives the on shell conserved currentThis follows by varying the Lagrangian, integrating the terms with ${\partial }_{\mu }{\Delta }_a$ by parts, and setting the Euler-Lagrange expressions to zero. If spatial boundary flux vanishes, $\int d^{3}x{J}^0$ is a conserved Noether charge. Spacetime symmetries are covered by including the induced field variation at fixed coordinates and the corresponding total derivative $K^{\mu }$.

An example of a theory with a Lorentz four-vector is the source-free Maxwell field with electromagnetic four-potential $A_{\mu }$. Under an Abelian gauge transformation,The change in $F_{\mu \nu }$ is ${\partial }_{\mu }{\partial }_{\nu }\chi -{\partial }_{\nu }{\partial }_{\mu }\chi =0$, so the Maxwell Lagrangian is gauge-invariant. For $A_{\mu }=(\Phi ,-\mathbf{A})$ define $E_i=F_{0i}$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$. With signature $+---$,The minus sign in $-F^2/4$ gives the positive electric kinetic term and positive physical field energy; reversing it would reverse the Hamiltonian's sign for the physical modes. The normalization $1/4$ accounts for antisymmetry and yields the conventional equations. Varying $A_{\nu }$ gives ${\partial }_{\mu }{F}^{\mu \nu }=0$.

For time translation at fixed coordinates take ${\Delta }_{\nu }={\partial }_0{A}_{\nu }$. Since $\alpha$ is constant, $\delta F_{\mu \nu }=\alpha {\partial }_0{F}_{\mu \nu }$, and hence $\delta \mathcal{L}=\alpha {\partial }_0\mathcal{L}$. Thus $K^{\mu }={\delta }^{\mu }{}_0\mathcal{L}$. The momentum derivative isNoether theorem therefore gives the canonical stress-energy tensor's time-translation currentHere ${\dot{A}}_i=E_i+{\partial }_i{A}_0$ was used. The canonical momentum of $A_0$ vanishes, while that of the lower spatial component $A_i$ is $E_i$: $A_0$ acts as a multiplier for Gauss law, not a propagating degree of freedom. The density above is not manifestly gauge-invariant, but its charge agrees with the physical energy on the constraint surface and with appropriate boundary conditions.

The gauge-covariant time translation instead has ${\Delta }_{\nu }={\partial }_0{A}_{\nu }-{\partial }_{\nu }{A}_0=F_{0\nu }$. It is time translation combined with a gauge transformation of parameter $\chi =-\alpha A_0$. Directly,using the definition of $F$, equivalently its Bianchi identity. The total derivative in $\delta \mathcal{L}$ is unchanged. The improved current isIt is the gauge-invariant Maxwell stress-energy tensor. In particular,The spatial current is the Poynting vector $\mathbf{E}\times \mathbf{B}$, so conservation gives ${\partial }_{t}u+\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. This is the familiar positive local electromagnetic field energy density, expressed entirely in measurable fields.

To compare the currents,The last term vanishes on the source-free Maxwell equations, leaving a stress-energy tensor improvement generated by a stress-energy superpotential. At the charge level,when the surface term vanishes and Gauss law holds. Thus the symmetries yield the same conserved energy under these conditions while the second supplies a gauge-invariant local density. Boundary terms would have to be retained if those boundary conditions were not imposed.

Use units $\hslash =c=1$, the Minkowski metric ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, and the Fourier transform convention $f(x)=\int d^{4}p{e}^{-ip\cdot x}\tilde{f}(p)/(2\pi {)}^4$. Here gamma matrices satisfy $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}{I}_4$, and $\bar{\psi }={\psi }^{†}{\gamma }^0$ is the Dirac adjoint. A single Dirac field describes both electrons and Positrons; these are the particle and antiparticle sectors of that field.

The free Maxwell Lagrangian and Dirac action, with a linear covariant gauge condition for the photon, giveBefore gauge fixing, the Maxwell Lagrangian has zero modes in field theory along $A_a\mapsto A_a+{\partial }_a\lambda$, so its quadratic operator cannot be inverted on all potentials. In the linear Lorenz gauge, the Faddeev-Popov determinant is $\det \square$. It is independent of $A$ and may be absorbed into the normalization; the corresponding Faddeev-Popov ghost fields have no interacting vertices in this Abelian linear gauge. A residual gauge symmetry is removed by the specified boundary conditions.

The free generating functional isThe Dirac field variables and their sources are independent Grassmann fields in this path integral. They anticommute; replacing them by ordinary commuting fields would give the wrong statistics and the wrong functional determinant. The bosonic Gaussian functional integral contributes an inverse square root of a determinant, and the Grassmann Gaussian integral contributes a determinant. If $K_A$ is the photon quadratic operator and $K_D=i{\gamma }^a{\partial }_a-m$, completing the square givesIn the last expression $D_F=i{K}_A^{-1}$ and $S_F=i{K}_D^{-1}$ are the photon propagator and Dirac propagator, with Feynman i-epsilon prescription. The products include the appropriate spacetime integrals and index contractions. Functional derivatives with respect to $J$, and consistently ordered left or right Grassmann derivatives with respect to the fermionic sources, generate the time ordering of the corresponding fields. In Feynman gauge, $\alpha =1$, the momentum-space two-point functions areThe Feynman i-epsilon prescription specifies vacuum boundary conditions rather than an arbitrary inverse of the differential operator.

The covariant photon propagator uses four potential components. Its operator formalism counterpart is Gupta-Bleuler quantization: impose $({\partial }_a{A}^a{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. This leaves a positive physical state space with two transverse photon polarization vectors. The temporal and longitudinal oscillator components occur in intermediate covariant expressions; the Ward identity removes their dependence from physical amplitudes. The free Dirac field uses the canonical anticommutation relations, producing the same fermionic signs as its Grassmann fields in the path integral.

The operator-path-integral equivalence can be seen directly with a regulator. Divide time into small intervals and insert complete sets of field-coordinate states for bosons, and resolutions in fermionic coherent states for fermions. The bosonic matrix elements produce the phase-space factor $\exp i\sum [\pi \Delta \varphi -H\Delta t]$; integrating out the quadratic canonical momentum produces the bosonic action. The fermionic coherent state overlaps produce the first-order term $i{\psi }^{†}\dot{\psi }$ and the Berezin integral measure. Multiplying the short-time kernels recovers the path integral. Projecting the remote endpoints onto the Fock vacuum with an infinitesimal damping selects the same Feynman propagators as the operator formalism. Field insertions become time-ordered products under this construction. Conversely, their quadratic generating functional obeys the canonical free-field equations and has precisely the oscillator two-point functions, so its higher free correlators agree by the Wick theorem. This establishes the equivalence for the regulated free theory and order by order in the perturbation series.

To couple the matter field electromagnetically, promote its global phase symmetry to the local transformationThe gauge covariant derivative $D_a={\partial }_a+ie{A}_a$ obeys $D_a^{\prime }{\psi }^{\prime }=e^{-ie\lambda }{D}_a\psi$. Replacing ${\partial }_a$ by $D_a$ in the Dirac action therefore gives the invariant matter densityThe electromagnetic field tensor is unchanged by the local transformation. Thus the unfixed quantum electrodynamics action is gauge-invariant. The Dirac current $j^a=\bar{\psi }{\gamma }^a\psi$ is conserved by the matter equations, and the electron and Positron excitations carry opposite charges. The added gauge fixing density selects a representative and is not itself invariant under arbitrary local transformations; it does not change gauge-invariant observables. At a free fermion vertex,For external on-shell Dirac spinors the corresponding current contraction vanishes. The quantum extension is the Ward identity, which makes physical amplitudes insensitive to adding a multiple of the photon momentum to its polarization vector. A gauge-compatible regularization preserves this vector-current identity.

With ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^a\psi A_a$, expand $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$ in powers of $e$. A term of order $n$ contains $n$ spacetime integrations and $1/n!$. Applying the Wick theorem pairs the free fields: an $A$-$A$ Wick contraction supplies a photon propagator, and a $\psi$-$\bar{\psi }$ Wick contraction supplies an oriented Dirac propagator. Each insertion supplies an interaction vertex. The permutations of contractions cancel the expansion factorials except for the Feynman-diagram symmetry factor. Interchanging Grassmann fields produces the fermionic sign, including a minus sign for every closed fermion loop. This is how Feynman diagrams arise from expectation values, rather than an extra dynamical assumption.

The normalized vacuum generating functional removes components with no external insertions. In the operator formalism, for an interacting-vacuum expectation value of an inserted product $\mathcal{O}$, the same cancellation appears asVacuum projection and the Feynman i-epsilon prescription are implicit. The linked-cluster theorem exponentiates all connected vacuum bubbles into the same factor in numerator and denominator, so it cancels. Normalize by $Z[0]$ to remove every vacuum component. This cancellation of vacuum bubbles still leaves products of disconnected diagrams that each contain external insertions. If only connected correlators are wanted, differentiate the connected generating functional $W=-i\log [Z[J,\bar{\eta },\eta ]/Z[0]]$.

The resulting momentum-space QED Feynman rules for the bare theory can be stated in Feynman gauge as follows:

The illustrated interaction vertex has incoming fermion momentum $p$, incoming photon momentum $k$, and outgoing fermion momentum $p^{\prime }=p+k$. Its solid-line arrows indicate fermion flow. The photon wavy line has no fermion-flow arrow. The propagators, the vertex $-ie{\gamma }^a$, momentum conservation, and the fermionic signs determine the perturbative amplitudes.

Use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ throughout. An Abelian gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. For a sufficiently regular $\chi$, commuting the partial derivatives gives $F_{\mu \nu }\mapsto F_{\mu \nu }$, so the Maxwell Lagrangian is invariant pointwise.

Vary the action with respect to the electromagnetic four-potential, using variations of compact support or vanishing on the boundary. The antisymmetry of the electromagnetic field tensor givesAfter integration by parts, $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. The principle of stationary action therefore gives the source-free Maxwell equations,In the Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, this becomes $\square A_{\nu }=0$ with $\square ={\partial }_{\mu }{\partial }^{\mu }$. The PDF's name “Lorentz gauge” denotes the usual Lorenz gauge, named after Lorenz. Residual gauge transformations preserve it when $\square \chi =0$; this condition is not itself a complete removal of the gauge freedom.

Under a gauge transformation, the electromagnetic field tensor changes byThe partial derivatives commute for a smooth gauge parameter, or in the distributional formulation. Thus the Maxwell Lagrangian depends only on a gauge-invariant tensor:Strictly, the pointwise calculation requires the second derivatives to exist; smooth gauge parameters are the usual convention. Mere first differentiability in the wording can instead be understood through commuting distributional derivatives.

For the active Lorentz transformation of a vector field, coordinates are held fixed and $A^{\prime \mu }(x)={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }({\Lambda }^{-1}x)$. Its field strength transforms as a two-index Lorentz tensor. The index contractions in the Maxwell Lagrangian are invariant by ${\Lambda }^T\eta \Lambda =\eta$, so the density transforms as a Lorentz scalar:For $\Lambda =I+\alpha \omega$, its inverse is $I-\alpha \omega +O({\alpha }^2)$. Expanding the coordinate argument therefore givesSet $v^{\mu }(x)=\alpha {\omega }^{\mu }{}_{\nu }{x}^{\nu }$. Part (iii) implies ${\partial }_{\mu }{v}^{\mu }=\alpha {\omega }^{\mu }{}_{\mu }=0$. The product rule then converts the variation into a total derivative:Its integral changes the action only by a boundary term, which is the condition needed for Noether's theorem.

For a translation with the sign used in the paper, the field variation at fixed coordinates is $\delta A_{\rho }=\beta {ϵ}^{\nu }{\partial }_{\nu }{A}_{\rho }$, and $\delta \mathcal{L}={\partial }_{\mu }(\beta {ϵ}^{\mu }\mathcal{L})$. Differentiating the Maxwell Lagrangian with respect to the field gradient givesThus Noether's theorem yields $j^{\mu }={ϵ}_{\nu }{T}^{\mu \nu }$, where the canonical stress-energy tensor of the Maxwell field isIt is conserved on the source-free Maxwell equations, but it is not generally symmetric. Indeed,need not vanish. For example, take $A_0=a{x}^1$, $A_2=b{x}^0$, and the other components zero, with nonzero constants $a,b$. This produces constant field strength and satisfies the source-free equations, but $T^{12}=0$ and $T^{21}=ab$.