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Suppose $V$ is a Killing vector field, $F$ is an electromagnetic field tensor satisfying $dF=0$ and ${\mathcal{L}}_{V}F=0$, and a particle of nonzero mass $m$ obeys the Lorentz force equation ${\nabla }_{u}u=(q/m)F^a{}_b{u}^b$. Cartan's magic formula gives $d({\iota }_{V}F)=0$, so the local potential of a closed differential one-form proof gives $d\Phi ={\iota }_{V}F$ locally. The Killing equation eliminates $u^a{u}^b{\nabla }_a{V}_b$, and the force term gives $d(V\cdot u)/d\tau =(q/m)d\Phi /d\tau$. Therefore $mV\cdot u-q\Phi$ is locally conserved. The sign assumes contraction in the first argument of $F$. A global potential additionally requires the closed one-form to have zero periods; invariance alone is insufficient without the homogeneous Maxwell equations.

For a constant electromagnetic field tensor, the relativistic Lorentz force is a constant-coefficient linear system on four-velocity, with solutionThe electric-type blocks of $F$ generate hyperbolic rotations, while magnetic-type blocks generate ordinary rotations.

The dual electromagnetic field tensor is obtained by applying the Hodge star operator to the electromagnetic field tensor. For metric signature $(+---)$ and ${ϵ}^{0123}=+1$, define ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. With $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, it has components ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. The homogeneous Maxwell equations become ${\partial }_{\mu }{\tilde{F}}^{\mu \nu }=0$.

Because the electromagnetic field tensor is the exterior derivative $F=dA$ of a potential, $dF=d^{2}A=0$. For a torsion-free connection this iswhich is the covariant form of the two homogeneous Maxwell equations.

The fermion electric dipole moment operator couples a fermion pseudotensor bilinear to the electromagnetic field tensor. Its nonrelativistic interaction is proportional to $\mathbf{S}\cdot \mathbf{E}$, so it is odd under parity and time reversal and signals CP violation.

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 ${\partial }_{\mu }{F}_{\mu \nu }=0$.

The source-free electromagnetic Lagrangian density is quadratic in the electromagnetic field tensor. Its Euler-Lagrange field equation is ${\partial }_{\mu }{F}^{\mu \nu }=0$. Antisymmetry of $F$ makes the density invariant under gauge transformations, and contraction of its indices makes it a Lorentz scalar.

Perpendicular fields with $\mathbf{E}\cdot \mathbf{B}=0$ and $E^2-c^2{B}^2=0$ form a null electromagnetic field. The constant-field Lorentz-force generator is nilpotent, so its exponential terminates as a polynomial.

Vary the four-potential with compactly supported variation. Antisymmetry of the electromagnetic field tensor gives $\delta (F_{\mu \nu }{F}^{\mu \nu })=4F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. Integration by parts therefore givesThus the Euler-Lagrange field equation is $\boxed{{\partial }_{\mu }{F}^{\mu \nu }-{\lambda }^2{A}^{\nu }=-{\mu }_0{J}^{\nu }}$.

A gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. Commuting derivatives leave $F_{\mu \nu }$ unchanged. At $\lambda =0$, the source term changes by $c^{-1}\int {\partial }_{\mu }\chi J^{\mu }$, a boundary term because the current is conserved. With suitable boundary conditions the action is gauge invariant. For $\lambda >0$, the mass term changes by terms involving $2A^{\mu }{\partial }_{\mu }\chi +(\partial \chi {)}^2$, so the same arbitrary gauge transformations are not a symmetry.

Take the divergence of the field equation. Antisymmetry gives ${\partial }_{\nu }{\partial }_{\mu }{F}^{\mu \nu }=0$, and conservation gives ${\partial }_{\nu }{J}^{\nu }=0$. Thus ${\lambda }^2{\partial }_{\nu }{A}^{\nu }=0$. For $\lambda >0$ this is the Lorenz constraint in Proca theory, and expanding $F$ givesFor $\lambda =0$ the same reduced equation holds when the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$ is imposed. More generally it is enough that the divergence be spacetime constant, but the conventional vanishing boundary conditions set that constant to zero.

Use SI units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. Write $x^{\mu }=(ct,x,y,z)$, ${\partial }_{\mu }=(c^{-1}{\partial }_t,\nabla )$ and the four-current $J^{\mu }=(c\rho ,\mathbf{J})$, with Greek indices from $0$ to $3$. The electromagnetic field tensor is the antisymmetric tensor whose components areIndices are raised and lowered with $\eta$. Define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$, with ${ϵ}^{0123}=+1$. Then the covariant Maxwell equations areHere $\rho$ and $\mathbf{J}$ are charge and current density, ${\mu }_0$ is the vacuum magnetic permeability, and ${ϵ}_0{\mu }_0{c}^2=1$. For example the $\nu =0$ equation gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, and the spatial equations give $\nabla \times \mathbf{B}-c^{-2}{\partial }_t\mathbf{E}={\mu }_0\mathbf{J}$. With the stated dual convention, ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$; the second tensor equation gives the two homogeneous Maxwell equations.

Transform the electromagnetic field tensor by $F^{\prime \mu \nu }={\Lambda }^{\mu }{}_{\alpha }{\Lambda }^{\nu }{}_{\beta }{F}^{\alpha \beta }$. For example,Reading off the other components in the same way yields the transverse field transformationThe longitudinal components remain $E_x^{\prime }=B_x^{\prime }=0$. These signs correspond to the stated boost in which $S^{\prime }$ moves with velocity $v$ relative to $S$.

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.

Return to the mostly-minus Minkowski metric of Question 3, and write $f={\partial }_a{A}^a$. The field $\xi (x)$ is assumed real and nonzero wherever its inverse occurs. Under the usual Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$, with $\xi$ inert, the electromagnetic field tensor is invariant but $f\mapsto f+\square \lambda$. The change in the added density isThis is not generally a total derivative. The action is not invariant under arbitrary gauge transformations. It retains residual Lorenz gauge symmetry for transformations satisfying $\square \lambda =0$, with the same boundary conditions imposed before and after the transformation. The action itself is undefined at $\xi =0$; that value can only be considered as a limiting gauge.

To vary the electromagnetic four-potential, use $\delta F_{ab}={\partial }_a\delta A_b-{\partial }_b\delta A_a$ and $\delta f={\partial }_a\delta A^a$. The integration by parts of both terms giveswith surface terms removed by the variational boundary conditions. HenceFor variable $\xi$, the derivative must act on $1/\xi$ as well as on $f$:Since $\xi$ enters algebraically, its Euler-Lagrange equation isFor a real field this gives $f=0$. Thus a dynamical gauge-fixing parameter imposes a constraint equation in field theory; it is not an ordinary propagating scalar. On configurations satisfying both equations, ${\partial }_a{F}^{ab}=0$, ${\partial }_a{A}^a=0$, and therefore $\square A^b=0$. There is no independent kinetic equation that determines $\xi$.

For the momentum-space equation at a prescribed constant $\xi \ne 0$, use $A_a(x)=\int d^{4}p{e}^{-ip\cdot x}{A}_a(p)/(2\pi {)}^4$. Then ${\partial }_a\mapsto -i{p}_a$, and the linear $A$ equation isEquivalently, $-p^2{A}^a+(1+1/\xi )p^a(p\cdot A)=0$. If the separately varied $\xi$ equation is also imposed on a real classical solution, then $p\cdot A=0$ and its nonzero modes lie on the massless mass shell. For constructing a full-field Green function, however, invert the fixed-background quadratic operator before imposing that on-shell constraint. Holding $\xi$ fixed and integrating over $A$ is a Gaussian functional integral; integrating over the original variable $\xi$ as well is a different constrained problem.

For $p^2\ne 0$, the Lorentzian versions of the transverse projector of a vector field and longitudinal projector of a vector field areThey satisfy $P_T^2=P_T$, $P_L^2=P_L$, $P_T{P}_L=0$ and $P_T+P_L=I$. The mixed-index operator isIts inverse follows by inverting these two scalar eigenvalues. Define $G_{ab}$ by $K^{ac}{G}_{cb}={\delta }_b^a$. The positive-sign covariant gauge-fixing inverse isThis is the algebraic inverse of the kinetic operator. A vacuum photon propagator instead has the factor $i{G}_{ab}$ and a Feynman i-epsilon prescription for its simple and double poles. If the name $G$ is used for the vacuum two-point function, include that factor $i$ consistently in its defining source equation. A retarded Green function would use different boundary conditions at the same poles. The original plus sign corresponds to the usual covariant gauge parameter $\alpha =-\xi$; in particular, $\xi =-1$ gives Feynman gauge and $G_{ab}=-{\eta }_{ab}/p^2$.

Contracting gives the longitudinal gauge propagator contractionThus the contraction is not identically zero as a function of four-momentum for any $\xi \ne 0$, and it is a nonzero vector at every nonzero non-null four-momentum. A particular component may vanish when $p_b=0$. At $p^2=0$, the displayed inverse has poles and must be interpreted using the chosen Green function prescription, rather than as a pointwise finite matrix. In the limiting Landau gauge, $\xi \to 0$, the longitudinal part vanishes. A covariant photon Green function may have a longitudinal component even though physical photon polarizations are transverse.