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For a massive particle of rest mass $m$, its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$, where the four-velocity is $U^{\mu }=\gamma (c,\mathbf{v})$, $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. With metric signature $(-,+,+,+)$, $P\cdot P=-m^2{c}^2$. A photon has four-momentum $(E/c,(E/c)\hat{\mathbf{n}})$, a future-pointing null vector. Four-momentum conservation states that the total incoming and outgoing four-momenta agree for an isolated interaction; this includes both energy and momentum conservation in every inertial frame.

Take the incoming photon direction as $x$. Its energy is $\hslash \omega =\mu m{c}^2$, so the total initial four-momentum isThe invariant mass available after photon absorption is therefore $M=m\sqrt{1+2\mu }$. In the centre-of-momentum frame, the two equal daughters have opposite momenta. Their total energy is at least their combined rest energy $2\alpha m{c}^2$, with equality precisely when both daughters have zero momentum in that frame. Thus the mass threshold after photon absorption isThe bound is attainable kinematically by taking each daughter four-momentum equal to $P/2$; each then has the required rest mass. It immediately gives ${\alpha }_{\max }\to 1/2$ as $\mu \to 0$.

At this threshold both daughters move with the centre-of-momentum frame. The common laboratory velocity follows from $\mathbf{v}=c\mathbf{P}/P^0$:For $\mu =0$, its continuous limiting value is zero. At smaller allowed daughter rest masses, some of the centre-of-momentum energy appears as their relative kinetic energy, so their four-momenta need not be $P/2$.

Choose the Fourier transform conventionThen ${\partial }_a$ becomes $i{q}_a$. Applying this to the Lorenz constraint in Proca theory yieldsThus every physical Fourier transform mode is orthogonal to its four-momentum in the Minkowski metric.

For non-null four-momentum, the transverse projector of a vector field and the complementary longitudinal projector of a vector field areIndeed, $q_a{P}^a{}_b=0$, $P^2=P$, $L^2=L$, and $P+L=I$. Acting on an arbitrary four-vector,So $P$ removes the unwanted component, whereas $L$ extracts it. On the massive mass shell, $q^2=-m^2$ and $P^a{}_b={\delta }_b^a+q^a{q}_b/m^2$. This on-shell form should not be used as an off-shell linear projection: away from $q^2=-m^2$ it is not idempotent. The non-null hypothesis matters; this decomposition is undefined at $q^2=0$.

There are two closely related objects to distinguish. Inverting the quadratic Proca action gives the usual covariant Proca propagator. After integration by parts, its kernel isThe matrix inverse with the Feynman i-epsilon prescription givesMultiplication by $K^{ab}$ gives $i{\delta }_b^a$ in the distributional limit. The numerator agrees with the polarization sum for a massive vector boson at the poles, but is not a transverse linear projection at arbitrary four-momentum.

For the literal canonical time-ordered product of $A_a$ and $A_b$, the nondynamical component $A_0$ produces the Proca time-ordering contact term. In the chosen time coordinate the full answer isTo see the local term directly, the three physical polarization vectors give ${\sum }_{\lambda }{ϵ}_{\lambda 0}^2=|\mathbf{q}{|}^2/m^2$. The canonical $00$ two-point correlation function therefore has Fourier transformBy contrast, ${\eta }_{00}+q_0^2/m^2=|\mathbf{q}{|}^2/m^2-(q^2+m^2)/m^2$, so the covariant expression contains an additional $i/m^2$. The mixed and spatial components have no additional contact term. ThusIf $T$ is used to mean covariant time ordering, commonly denoted $T^{\ast }$, the conventional answer is instead just ${\Delta }^{\mathrm{cov}}$. Both conventions have the same propagating poles and agree away from coincidence; explicitly separating them respects the printed definition as an ordinary time-ordered product.

Put $q^a=(E,0,0,k)$ with $E=\sqrt{k^2+m^2}$. An orthonormal real basis of polarization vectors isEach polarization vector obeys $q_a{ϵ}_{\lambda }^a=0$. The Minkowski metric gives ${ϵ}_{\lambda }\cdot {ϵ}_{{\lambda }^{\prime }}={\delta }_{\lambda {\lambda }^{\prime }}$, since $E^2-k^2=m^2$. The first two are transverse to the spatial momentum; the longitudinal polarization of a massive vector boson is spatially longitudinal but still orthogonal to the full four-momentum. The polarization sum for a massive vector boson isIn the rest frame all three polarization vectors are spatial unit vectors. Their three independent positive-norm states are the physical spin states of a massive spin-one particle.

Keep the mostly-plus Minkowski metric and the Fourier transform $e^{ip\cdot x}$. Write $/p={\gamma }^a{p}_a$, with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$. To fix the otherwise unspecified phase of ${\gamma }^5$ and the Dirac adjoint, takewhere ${\Gamma }^a$ are ordinary mostly-minus gamma matrices. Then $({\gamma }^5{)}^2=-1$, $\{{\gamma }^5,{\gamma }^a\}=0$, and the fermionic time-derivative term is $i{\psi }^{†}{\partial }_t\psi$. In these conventions the interaction with real $g$ is Hermitian: in mostly-minus notation it is $ig{\bar{\psi }}_{\mathrm{std}}{\Gamma }^5\psi \varphi$. If one instead calls $i{\gamma }^5$ the square-one chirality matrix, its coefficient must be $-ig$ to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding $e^{i{S}_{\mathrm{int}}}$ yields the following Feynman rules with relativistically normalized external states:

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.

Use the physical rapidity strip $0<\mathrm{Im}\vartheta <\pi$ for poles of the two-body S-matrix, and write $\vartheta$ for scattering rapidity to distinguish it from the theta angle in question 2. A denominator in the kink-antikink product vanishes atNo numerator cancels these poles. For two equal-mass constituents with rapidities $\chi \pm i{u}_k/2$, their four-momentum vectors sum toThis gives the relativistic bound-state mass from a rapidity pole. The ordered breather spectrum isIt increases strictly with $k$. The hypothetical $k=n$ state would lie at the two-kink threshold $2M$ and is not a bound state; it is absent from the pole product. At these couplings $M=mn/\pi$ and ${\beta }^2=8\pi /(n+1)$. In particular $m_1\to m$ at weak coupling $n\to \infty$. This is the Sine-Gordon breather spectrum at reflectionless couplings. At $n=1$ there are no breathers; the subsequent processes involving a physical ${\mathcal{B}}_2$ require $n\ge 3$.

For two identical neutral ${\mathcal{B}}_1$ particles, exchanging the two outgoing labels does not produce a distinguishable channel. In one spatial dimension the elastic final momenta are the incoming pair, up to interchange. Thus there is one scalar identical-particle amplitude, rather than separately observable transmission and reflection amplitudes.

Put $a=\pi /(2n)$, so the basic amplitude uses $\sin (2a)$. Its poles in the physical strip occur at $\vartheta =2ia$ and $\vartheta =i(\pi -2a)$. The first is the direct bound-state pole. Choosing constituent rapidities $\chi \pm ia$ gives real total energy-momentumSince $m_1=2M\sin a$,Both energy and momentum therefore match an on shell ${\mathcal{B}}_2$ with rapidity $\chi$. The complementary pole is its crossed-channel partner. At $n=2$ the would-be ${\mathcal{B}}_2$ is a threshold state, so this physical fusion interpretation must not be imposed there.

The bound-state fusion of factorized S-matrices treats a bound particle as its on shell constituents with analytically continued rapidities. If equal-mass particles $A,A$ fuse to $C$ at relative rapidity $iu$, use constituent rapidities $\chi \pm iu/2$. To scatter a third particle $D$ off $C$, multiply its scattering amplitudes with each constituent and take the bound-state residue or projection in the constituent channel. For a scalar amplitude this givesFor particles with internal indices, the product is projected using the bound-state coupling tensors. The heuristic reason is factorized scattering: conserved higher charges prevent particle production and fix the rapidity data, so the third particle scatters through the constituents by successive two-body processes. Consistency of different orders is the Yang-Baxter equation. Without integrability, an independent three-body interaction would invalidate this simple bootstrap product.

For ${\mathcal{B}}_2=({\mathcal{B}}_1{\mathcal{B}}_1)$, the constituent shifts are $\pm ia$, givingIt is useful to write this Sine-Gordon breather fusion amplitude in explicitly factorized form:To check the reduction, put $z=\sinh \vartheta$. Multiplying the shifted factors gives numerator and denominator $z^2\pm 2iz\sin (2a)\cos a+{\sin }^{2}a-{\sin }^2(2a)$. Use $\sin a+\sin 3a=2\sin 2a\cos a$ and $\sin a\sin 3a={\sin }^{2}2a-{\sin }^{2}a$ to factor them as $(z\pm i\sin a)(z\pm i\sin 3a)$. The product tends to one at large positive real rapidity, fixing its overall phase in this bootstrap convention.

For $n\ge 3$, the nearest pole to the real axis is $\vartheta =ia$. It is simple and comes from the first factor. In the crossed, or t-channel, the momentum carried between the external particles is their difference. With the $+,-$ metric its invariant isAt the pole, substitute the breather masses:The trigonometric identity is applied with angles $2a$ and $a$. Thus the exchanged one-particle state is the lightest breather ${\mathcal{B}}_1$, on its mass shell. This is crossed-channel lightest-breather exchange. The external momenta at a bound-state pole are analytically continued; on shell here means the invariant mass relation and conservation of the continued energy-momentum, not a pole at real physical rapidity. At $n=3$ the more distant central factor has a double pole, but the nearest $ia$ pole and its ${\mathcal{B}}_1$ interpretation remain unchanged.

For a Lorentz transformation to an inertial frame moving at speed $v_B$ in the positive $x$ direction, write ${\beta }_B=v_B/c$ and ${\Gamma }_B=(1-{\beta }_B^2{)}^{-1/2}$. Any four-vector transforms asThis convention applies to $X=(ct,x,0,0)$ and uses the same $+---$ Minkowski metric as the four-momentum calculation.

The definition of proper time gives $d\tau =dt\sqrt{1-u^2/c^2}=dt/\gamma$ for $|u|<c$. Hence the four-velocity isSince $\dot{\gamma }={\gamma }^{3}u\dot{u}/c^2$, differentiating with respect to proper time givesHere $1+{\gamma }^2{u}^2/c^2={\gamma }^2$ simplifies the spatial component of the four-acceleration.

At each event, boost with $v_B=u$ to the instantaneous rest inertial frame. Then $A^{\prime 0}=0$ and $A^{\prime 1}=\gamma {\gamma }^4\dot{u}(1-u^2/c^2)={\gamma }^3\dot{u}$. Thus the signed proper acceleration for one-dimensional motion isThe last equality follows by differentiating $\gamma u$; its magnitude is the usual scalar proper acceleration.

If this signed proper acceleration is constant and the particle starts from rest, integration gives $\gamma u=at$, soIntegrating once more and imposing $x(0)=0$ yields the hyperbolic motion worldlineFor $a=0$ the continuous limiting solution is $x=0$. Finally, the Taylor series of the square root givesThis recovers constant-acceleration Newtonian mechanics at low speed while the exact relativistic speed remains less than $c$.

Use the Minkowski metric convention $+---$. A particle's four-momentum is $P=(E/c,\mathbf{p})$, with $P\cdot P=m^2{c}^2$ for rest mass $m$. For a photon, its energy and wavelength obey $E=hc/\lambda$, and its four-momentum is null. Here $h$ is the Planck constant.

The initially stationary electron has $P=(mc,0,0,0)$. Choose the scattering plane as the $xy$ plane. The incoming and outgoing photon four-momenta areBy four-momentum conservation, the final electron has $P^{\prime }=P+K_1-K_2$. Its rest mass is unchanged, so the relativistic energy-momentum relation gives $P^{\prime 2}=P^2=m^2{c}^2$. Since $K_1^2=K_2^2=0$, expansion yieldsorMultiplying through gives the Compton scattering shift:It is nonnegative and vanishes for forward scattering, consistent with the recoil energy being supplied by the photon.

Use flavor hypercharge $Y=B+S$, where $B$ is baryon number and $S$ is strangeness. The isospin coordinate is $I_3$. The baryon octet has coordinatesHere the nucleons have $S=0$, the Sigma baryons and Lambda baryon have $S=-1$, and the Xi baryons have $S=-2$. The central Sigma baryon belongs to an isospin triplet while the central Lambda baryon is an isospin singlet; equal coordinates do not identify the states.

The pseudoscalar meson octet isThe pions have $S=0$, the upper kaons have $S=+1$, and their lower antiparticles have $S=-1$. The two central states are the neutral pion and the Eta octet state. This is the octet basis of flavor symmetry; the physical eta can also mix with the flavor-singlet state.

For the flavor SU(3) Cartan generators, choose the Hermitian physics convention and the inner product $⟨A,B⟩=\mathrm{tr}(AB)$. Then an orthonormal diagonal basis isStrictly, $i{h}_1,i{h}_2$ are elements of the anti-Hermitian SU(3) Lie algebra; $h_1,h_2$ are the corresponding Hermitian observables. In the quark basis $(u,d,s)$, the up and down quarks form an isospin doublet and the strange quark is a singlet. Their $I_3$ values are $(1/2,-1/2,0)$. Each quark has baryon number $1/3$, and their strangeness values are $(0,0,-1)$. It follows thatThese flavor hypercharge conventions differ from the electroweak hypercharge convention. If instead the inner product is $2\mathrm{tr}(AB)$, the orthonormal basis is $t_3=h_1/\sqrt{2}$, $t_8=h_2/\sqrt{2}$, and the same operators are $I_3=t_3$, $Y=2t_8/\sqrt{3}$.

In the ordinary quark model, the proton has valence content $uud$ and charge $+1$, while the neutron has content $udd$ and charge zero. Additivity of electric charge gives $2q_u+q_d=1$ and $q_u+2q_d=0$, hence $q_u=2/3$, $q_d=-1/3$. The Sigma baryon ${\Sigma }^{-}$ has content $dds$ and charge $-1$, giving $2q_d+q_s=-1$. Thus the quark triplet's electric charges, in units of the positive elementary charge, areThe Gell-Mann--Nishijima formula is consequentlyIt also reproduces every baryon octet charge from the first diagram. On antiquarks the additive quantum numbers reverse sign, and combining a quark with an antiquark reproduces the meson octet charges.

Because the down and strange quarks have identical electric charge, $Q$ commutes with the U-spin generatorsThey satisfy $[U_a,U_b]=i{ϵ}_{abc}{U}_c$; equivalently the anti-Hermitian matrices $i{U}_a$ span an $\mathfrak{su}(2)$ subalgebra. The entries on the $d,s$ block of $Q$ are equal, so $[Q,U_a]=0$ for all three U-spin generators. The electric charge is therefore constant within each irreducible U-spin multiplet. For example, U-spin relates ${\pi }^{+}$ and $K^{+}$, and relates $p$ and ${\Sigma }^{+}$, without changing their charge. It does not imply exact mass degeneracy: unequal down- and strange-quark masses break U-spin.

For pion-nucleon octet channels, assume the collision is governed by the strong interaction. The initial baryon number is one and strangeness is zero, so an outgoing meson-baryon pair must preserve $B=1$, $S=0$, and electric charge. Thus its total flavor hypercharge is $Y=1$. The allowed types areA kaon of $S=+1$ can accompany a Lambda baryon or Sigma baryon of $S=-1$. An antikaon cannot balance the nonpositive strangeness of an octet baryon. A Xi baryon would require a meson of $S=+2$, which the meson octet does not contain.

Resolving these types by electric charge gives all possible pairs:In the isospin-symmetric approximation, total isospin is conserved as well: the incoming $1\otimes \frac{1}{2}$ contains $I=\frac{1}{2},\frac{3}{2}$. The $\pi N$ and $K\Sigma$ channels contain both values, while ${\eta }_{8}N$ and $K\Lambda$ contain only $I=\frac{1}{2}$. The extreme-charge initial states are pure $I=\frac{3}{2}$, consistently excluding ${\eta }_{8}N$ and $K\Lambda$. Clebsch-Gordan coefficients relate amplitudes in different charge channels; the table establishes permission, not equal probabilities. Electromagnetism and unequal up- and down-quark masses introduce small violations of isospin symmetry.

Finally, energy and momentum conservation require $\sqrt{s}\ge m_M+m_B$ for a particular pair, where $s$ is the squared total four-momentum. Only channels above their own threshold can occur. Total angular momentum and parity symmetry in quantum field theory constrain the partial waves of the meson-baryon scattering: a pseudoscalar meson and a positive-parity spin-one-half baryon have pair parity $-(-1{)}^L$ and total angular momentum $J=L\pm \frac{1}{2}$ (only $J=\frac{1}{2}$ for $L=0$). Initial and final partial waves must have matching $J$ and parity. Sufficient energy can open a channel, but cannot remove the electric charge conservation, baryon number, or strangeness constraints.

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 $G_{ab}=-{\eta }_{ab}/p^2$.