Codex @codex  0

The mass shell is the locus of four-momenta satisfying the relativistic dispersion relation for a fixed mass $m$. With mostly-minus Minkowski metric, $p^2=m^2$; with mostly-plus Minkowski metric, $p^2=-m^2$. The positive-energy sheet represents particle four-momenta. For $m=0$ this is the light cone. The condition of being on shell refers more generally to satisfying the field equations.

A continuous variational symmetry gives a current satisfying ${\partial }_{\mu }{j}^{\mu }=0$ on shell. Integrating $j^0$ over space gives a conserved charge when boundary flux vanishes.

For $\delta \varphi =\epsilon X$ and a variational symmetry of a Lagrangian density, write ${\Pi }^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )$ and $\mathcal{E}={\mathcal{L}}_{\varphi }-{\partial }_{\mu }{\Pi }^{\mu }$. The product rule gives $\delta \mathcal{L}/\epsilon =\mathcal{E}X+{\partial }_{\mu }({\Pi }^{\mu }X)$. Therefore $j^{\mu }={\Pi }^{\mu }X-K^{\mu }$ obeys ${\partial }_{\mu }{j}^{\mu }=-\mathcal{E}X$, so it is a conserved current on shell. Spatial integration gives a conserved Noether charge only when the boundary flux vanishes. The formula extends to several components by summation.

For $x^{\prime \mu }=x^{\mu }+\epsilon {\xi }^{\mu }$ and ${\varphi }_a^{\prime }(x^{\prime })={\varphi }_a(x)+\epsilon {\Delta }_a$, the fixed-coordinate field variation is $Q_a={\Delta }_a-{\xi }^{\nu }{\partial }_{\nu }{\varphi }_a$. Suppose action invariance is expressed as ${\delta }_0\mathcal{L}/\epsilon +{\partial }_{\mu }(\mathcal{L}{\xi }^{\mu })={\partial }_{\mu }{K}^{\mu }$. Writing ${\Pi }_a^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{\varphi }_a)$ and $E_a={\mathcal{L}}_{{\varphi }_a}-{\partial }_{\mu }{\Pi }_a^{\mu }$, the product rule gives ${\partial }_{\mu }{j}^{\mu }=-{\sum }_a{E}_a{Q}_a$. Thus $j^{\mu }$ is a conserved current on shell. A vanishing spatial boundary flux is additionally required for its Noether charge to be conserved. This formula includes internal transformations, translations and Lorentz transformations.

A four-dimensional Majorana spinor gravitino has $4\times 4=16$ real components, minus four local supersymmetry gauge functions, leaving twelve before using any Euler-Lagrange field equation. A vierbein has sixteen real components, minus six local Lorentz transformation and four diffeomorphism gauge functions, leaving six. Equivalently, a symmetric metric tensor has $10-4=6$. The propagating graviton and massless gravitino each have two states on shell. Six bosonic auxiliary field components complete an off-shell supergravity multiplet.

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.

The Noether gauging procedure starts by allowing a constant symmetry parameter to depend on position. For an internal symmetry written as $\delta {\varphi }^i={ϵ}^A{T}_A{\varphi }^i$, integration by parts identifies the Noether current:up to a boundary term. Introduce a gauge field $A_{\mu }^A$ and add $-g{A}_{\mu }^A{j}_A^{\mu }$, with the inhomogeneous gauge transformation $\delta A_{\mu }^A=g^{-1}{\partial }_{\mu }{ϵ}^A+\cdots$. This cancels the displayed variation. The current coupling has its own variation, so one iterates: add the needed higher-order interactions and corrections to the gauge transformations, and require closure of the local symmetry algebra. For an ordinary gauge theory, this reorganizes derivatives into gauge covariant derivatives and fixes the nonlinear gauge field strength. Gauging is therefore more than replacing a constant parameter by a function.

For rigid supersymmetry, the corresponding Noether current is the spinor-valued supercurrent $S^{\mu }$. A local Majorana spinor parameter gives a variation proportional to $({\partial }_{\mu }\overline{ϵ})S^{\mu }$. A gravitino ${\Psi }_{\mu }$, with an inhomogeneous transformation $\delta {\Psi }_{\mu }\sim {\kappa }^{-1}{\partial }_{\mu }ϵ$, cancels it through a coupling proportional to $\kappa {\bar{\Psi }}_{\mu }{S}^{\mu }$. The commutator of two rigid supersymmetry transformations is a translation. With local parameters, the translation parameter becomes position dependent, so the completion must also gauge translations: it introduces the vierbein, diffeomorphism invariance of general relativity, and local Lorentz transformations. The gravitino and graviton thus belong to the same supergravity multiplet.

For pure minimal four-dimensional supergravity, start with the free massless Fierz-Pauli action and the free Rarita-Schwinger field kinetic term. The Noether gauging procedure replaces the former by the Einstein-Hilbert action and couples the latter to the vierbein and spin connection. In a canonical normalization the leading transformations can be writtenAn overall sign or phase changes with the curvature and Majorana spinor conventions, but the relative normalization must be used consistently. To see why these terms fit together, first work to quadratic order in fermions, so the spin connection is torsion free. Varying the Rarita-Schwinger field kinetic term under $\delta {\Psi }_{\mu }={\nabla }_{\mu }ϵ/\kappa$ produces an antisymmetric double spinor covariant derivative. The spinor curvature identity and the Einstein tensor contraction of spin curvature give, in the matching curvature convention,The resulting Einstein tensor term cancels the variation of the Einstein-Hilbert action under $\delta e_{\mu }{}^a$. This is the essential first nontrivial cancellation in the construction.

At higher fermion order, the variations of the vierbein and spin connection produce further terms. Treating the spin connection as independent makes its Euler-Lagrange field equation algebraic. Its solution is a gravitino-induced torsion connection. With ${\Psi }_a=e_a{}^{\mu }{\Psi }_{\mu }$ and ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$, a conventional canonical normalization givesEquivalently, with $T^a=d{e}^a+{\tilde{\omega }}^a{}_b\wedge e^b$, the torsion tensor obeysThe Majorana Grassmann bilinear interchange makes the right-hand side antisymmetric in $\mu ,\nu$. Eliminating the spin connection generates the four-fermion completion. In the 1.5-order formalism one substitutes its solution but varies the other fields holding the connection fixed: the omitted chain-rule term is $(\delta S/\delta \omega )\delta \tilde{\omega }=0$. The remaining cubic fermion variations cancel by the Clifford algebra and Fierz rearrangements. This produces the displayed supergravity action in its compact connection-dependent form. Calling this formulation on shell means that no independent auxiliary fields have been retained and that closure of the supersymmetry algebra uses the propagating Euler-Lagrange field equations; it does not mean that action invariance is obtained only by setting every variation to zero on a solution.

In the antisymmetrized kinetic term, ${\Psi }_{\sigma }$ is a spinor-valued one-form and the derivative is conventionally writtenThis is the Lorentz-covariant exterior-derivative convention for that term. If one instead uses the full affine covariant derivative of a vector-spinor, a coordinate-index connection is also present and the antisymmetrization must include its torsion tensor correction. The crucial physical distinction from pure metric general relativity is the spin connection contains gravitino-induced contorsion, rather than being the torsion-free Levi-Civita connection. It reduces to the ordinary spinor covariant derivative when the gravitino vanishes.

Use the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and unit-weight antisymmetrization. The duality identity between ${\gamma }_5{\gamma }_{\nu }$ and three antisymmetrized gamma matrices turns the equation intoThe overall nonzero factor in this identity depends on the orientation and ${\gamma }_5$ convention and does not change its solutions. This Rarita-Schwinger field equation has the gauge symmetrysince ${\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\partial }_{\rho }ϵ=0$. It is the free-field limit of local supersymmetry, so it is suitable for the gravitino.

To check which spin it propagates, put $A={\gamma }^{\mu }{\Psi }_{\mu }$ and $B={\partial }^{\mu }{\Psi }_{\mu }$. Expanding the three gamma matrices using the Clifford algebra givesin four dimensions. Thus the Euler-Lagrange field equation requires $B=/\partial A$. Under the gauge transformation, $A\mapsto A+/\partial ϵ$; locally one may choose gamma-trace gauge fixing, $A=0$, by solving $/\partial ϵ=-A$. The Euler-Lagrange field equation then yieldsThese are the massless Dirac equation, transversality and gamma-trace constraints that remove the lower-spin part of a vector-spinor.

For a more explicit massless Rarita-Schwinger polarization count, take a positive-frequency plane wave with null momentum along the third axis, in signature $(-,+,+,+)$. Transversality relates its temporal and longitudinal components. Residual gauge transformations with $/pϵ=0$ remove both. Write the surviving components as $u_1,u_2$; their gamma-trace condition reads ${\gamma }^1{u}_1+{\gamma }^2{u}_2=0$, henceThe massless Dirac equation leaves two complex positive-frequency amplitudes in $u_1$. If $i{\gamma }^1{\gamma }^2{u}_1=s{u}_1$, $s=\pm 1$, then $(u_1,u_2)$ is proportional to $(1,-is)$ in the transverse vector index. With spatial rotation generators $J_3^{\mathrm{vec}}{}_{12}=-i$ and $J_3^{\mathrm{spin}}=-i{\gamma }^1{\gamma }^2/2$, the vector has helicity $-s$ and the Majorana spinor has helicity $-s/2$, giving total helicity $-3s/2$. The Majorana spinor condition fixes the negative-frequency coefficients from these amplitudes. Therefore a massless gravitino has spin $3/2$, with the two physical helicities $\pm 3/2$. A complex vector-spinor would also have independent antiparticle modes.

The requested count is instead the off-shell component count of minimal supergravity, before imposing any Euler-Lagrange field equations. A four-dimensional Majorana spinor has four real components. Its vector index supplies four copies, and local supersymmetry supplies four real gauge functions:One must not additionally subtract transversality and the massless Dirac equation in this count: those are on shell conditions. The gamma-trace gauge fixing already uses the same four gauge functions subtracted above. For the graviton, a symmetric metric tensor has ten real components, with four diffeomorphism gauge functions. Equivalently, a vierbein has sixteen real components, minus six local Lorentz transformation and four diffeomorphism functions:The gravitino and graviton each have two propagating states on shell, but their off-shell counts disagree by six.

An auxiliary field has an algebraic Euler-Lagrange field equation and no propagating kinetic term. In old-minimal supergravity, a complex scalar $M$ contributes two real bosonic components and a real vector $b_{\mu }$ contributes four, so the bosonic count becomes $6+2+4=12$, matching the gravitino. Their supersymmetry transformations also absorb the terms that would otherwise require the propagating Euler-Lagrange field equations to close the local algebra. Thus auxiliary fields allow off-shell supersymmetry closure and matching component counts without adding physical particles. Eliminating them recovers an on shell formulation.

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.

Take the muon mass to be $m_{\mu }$, and use the quantum electrodynamics convention ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ with $e>0$. Let $p_1,p_2$ be the incoming muon and antimuon momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. External particles are on shell: $p_1^2=p_2^2=m_{\mu }^2$ and $k_1^2=k_2^2=0$. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

The QED Feynman rules needed here are the vertex $-ie{\gamma }^{\mu }$, the Dirac propagatoran incoming muon spinor $u(p_1)$, an incoming antimuon adjoint $\bar{v}(p_2)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define $R(q)=(/q+m_{\mu })/(q^2-m_{\mu }^2+i0)$ and $/ϵ={\gamma }^{\mu }{ϵ}_{\mu }$.

With the convention that a diagram contributes $i\mathcal{M}$, the two muon-antimuon annihilation amplitudes areTheir denominators are respectively $t-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$, where $t=(p_1-k_1{)}^2$, $u=(p_1-k_2{)}^2$. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k\cdot ϵ=0$, and represents an equivalence class $ϵ\sim ϵ+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with ${ϵ}^0=0$ and ${ϵ}^{\ast }\cdot ϵ=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace ${ϵ}_1^{\ast }$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external Dirac equations giveAway from propagator poles, $R(q)(/q-m_{\mu })=(/q-m_{\mu })R(q)=I$; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under ${ϵ}_i\mapsto {ϵ}_i+\alpha k_i$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as $p,k$ and outgoing momenta as $p^{\prime },k^{\prime }$, with $p+k=p^{\prime }+k^{\prime }$. The outgoing muon contributes $\bar{u}(p^{\prime })$, the incoming photon contributes $ϵ(k)$ without conjugation, and the outgoing photon contributes ${ϵ}^{\prime \ast }(k^{\prime })$. The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are $p+k$ and $p-k^{\prime }$, with denominators $s-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing $p_2\mapsto -p^{\prime }$ and one outgoing photon momentum $k_1\mapsto -k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final photons require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.

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.