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A relativistic theory must accommodate the energy-momentum relation $E^2={\mathbf{p}}^2+m^2$, causal interactions and the possibility of particle creation. Promoting the squared relation to a wave equation gives the Klein-Gordon field equationA plane wave $\varphi =N{e}^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ hasThe positive and negative branches both solve the local differential equation. Its being second order in time means that specifying only a wavefunction, as in the nonrelativistic Schrödinger equation, is not the complete initial datum.

For a complex scalar, the conserved Klein-Gordon inner product comes from the currentFor the plane wave, $j^0=2E|N{|}^2$, which is negative on the negative-frequency branch. Thus this conserved density cannot be a positive single-particle probability density for arbitrary Klein-Gordon solutions. Merely replacing it by $|\varphi {|}^2$ does not supply the missing conserved relativistic probability law. The indefinite Klein-Gordon charge density is an obstacle to that naive one-particle interpretation, not evidence that a consistent scalar field theory has negative probabilities.

The Dirac equation is first order,The gamma-matrix anticommutation relations imply $H_D^2=({\mathbf{p}}^2+m^2)I$, so its spectrum again contains $\pm E_{\mathbf{p}}$. But its conserved Dirac current has density $j^0={\psi }^{†}\psi \ge 0$. The Dirac equation solves the density-sign difficulty while leaving both energy branches. Treating every negative-energy solution as an ordinary accessible one-electron state would leave the energy unbounded below and permit radiative descent without a stable ground state.

The historical Dirac sea filled the negative-energy electron states and interpreted holes as positive-energy antiparticles. It motivates the Positron but is not the general modern explanation: bosons do not obey an exclusion principle and cannot be stabilized by filling a sea. Quantum field theory instead promotes the fields to operators and interprets negative-frequency modes using antiparticle creation operators.

For a complex scalar field, schematically,Bosonic commutation relations give a normal-ordered Hamiltonian and conserved charge of the formBoth particle species have positive excitation energies. The sign-changing scalar current now represents charge, rather than particle probability; opposite charges are physically allowed. The ordinary positive Fock space norm gives nonnegative probabilities. A real scalar field is its own antiparticle and has no corresponding complex-field U(1) charge.

For the Dirac field, the analogous expansion contains electron annihilation operators and positron creation operators. Fermionic anticommutation relations turn the negative-frequency contribution into a positive antiparticle excitation energy after the vacuum constant is removed. The Pauli exclusion principle and relativistic locality are thereby incorporated in a field description, rather than a postulated infinite population of physical negative-energy electrons. The normal-ordering step removes an additive vacuum constant; it does not by itself settle the separate gravitational vacuum-energy problem.

Interactions can change particle number while conserving total energy, momentum and charge. Trying to localize a particle on its Compton wavelength scale requires momenta of order its rest mass, where accessible interactions can involve pair creation. A fixed-particle wavefunction is consequently not a general description of such processes. Creation and annihilation operators act between different particle-number sectors, and field interactions describe scattering and production within one framework.

Finally, relativistic causality concerns observables: spacelike separated local bosonic observables commute, while fermionic fields satisfy the appropriate anticommutation conditions. This microcausality prevents a controllable influence through spacelike separated local operations. Free relativistic one-particle theories remain useful, and low-energy fixed-number approximations can be excellent. What fails is the naive insistence that unrestricted relativistic interactions, positive probabilities, stability and arbitrary particle production all fit an unchanged one-particle Schrödinger picture.^[ref]

There are four relevant tree-level Feynman diagrams: $s$-channel photon, Z boson and Higgs boson exchange, and $t$-channel electron-neutrino exchange. The electron and Positron are incoming, and $W^{-}$ and $W^{+}$ outgoing. The Higgs diagram is normally tiny because its electron coupling is proportional to $m_e$; it should nevertheless be included in the requested list.

Let $v=2m_W/g$. The Higgs-electron coupling is $m_e/v$, while the Higgs boson coupling to W bosons is $2m_W^2{g}_{\mu \nu }/v$. Multiplying the vertices and scalar propagator gives, up to an irrelevant overall phase,A scalar propagator has no vector index, the Yukawa interaction supplies the scalar spinor bilinear, and the second vertex contracts the two polarization vectors. We retain the tree propagator and neglect widths, consistently with the stated leading-order calculation.

For unpolarized incoming beams the spin average is $1/4$, not a sum without normalization. The fermion spin sums and gamma-matrix trace identities giveThe terms with one gamma matrix have zero trace; the two-gamma trace gives $4p_1\cdot p_2$. Both final polarization states are specified as longitudinal, so no final polarization average is appropriate.

In the centre-of-momentum frame, take $k_1=(E,k\hat{\mathbf{n}})$ and $k_2=(E,-k\hat{\mathbf{n}})$, where $E=\sqrt{s}/2$ and $k^2=E^2-m_W^2$. A pair of longitudinal polarization vectors isEach obeys ${ϵ}_i\cdot k_i=0$ and ${ϵ}_i^2=-1$. Their contraction is angle independent:Consequently

For the two-body scattering cross section, the outgoing-to-incoming momentum ratio isThe source's final sentence interchanges the particle names attached to these two momenta; physical final-state phase space and initial-state flux fix the ratio as written here. Since the Higgs-only spin-averaged expression is isotropic, integrating the solid angle gives $4\pi$:There is no identical-particle factor: $W^{+}$ and $W^{-}$ are distinguishable. The threshold factor ${\beta }_W$ makes this contribution vanish when the outgoing momentum vanishes. This is the isolated squared Higgs amplitude for a specified longitudinal pair, not the complete physical cross section: the full result squares the coherent sum of all four amplitudes and includes their interference.

For a photon gas in thermal and chemical equilibrium, photon number is not a conserved thermodynamic charge. Number-changing processes can create and absorb photons without a corresponding conserved-number constraint, so the equilibrium photon chemical potential is zero.

At $T=5\times 10^9\mathrm{K}$, $k_{\mathrm{B}}T\simeq 0.431\mathrm{MeV}$. The other light species are electrons, Positrons, three neutrino species and their antiparticles. Neutrinos are ultrarelativistic; electrons and Positrons are only moderately relativistic at this particular temperature because $m_e{c}^2=0.511\mathrm{MeV}$. The later simple counting formulas use the intended ultrarelativistic approximation for them. Baryons, muons and heavier particles are not an abundant relativistic bath at this temperature.

Pair equilibrium implies ${\mu }_{e^{-}}+{\mu }_{e^{+}}=2{\mu }_{\gamma }=0$. Approximate charge symmetry, with the net electron excess tied to the much smaller baryon asymmetry, then makes both electron and Positron chemical potentials small compared with $k_{\mathrm{B}}T$. For neutrinos one additionally assumes negligible lepton asymmetry. Their near-zero chemical potentials are not implied merely by their statistics. After neutrino decoupling, a distribution initially having negligible chemical potential retains a redshifted Fermi-Dirac distribution, with its own temperature.

Let $V$ be proper volume, let $g_i$ count the populated internal states, and take ${\mu }_i=0$ and $ϵ=pc$. Spherical integration in momentum space givesUsing the fermion-to-boson thermal integral ratio, the minus-denominator boson integral is $2\zeta (3)$ and the plus-denominator fermion integral is $3\zeta (3)/2$. HenceThe energy integral has one extra factor $pc=k_{\mathrm{B}}Tx$:The supplied distribution identity gives the fermion integral as $(1-2^{-3}){\pi }^4/15=7{\pi }^4/120$. ThereforeFor an isotropic ultrarelativistic gas, momentum flux gives $P_i=\frac{1}{3}\int pv{f}_i{d}^{3}p$. Since $v=c$ and $pc=ϵ$,Photons have $g_{\gamma }=2$, electrons and Positrons each have $g=2$, and each neutrino or antineutrino species has one populated helicity in the standard thermal bath.

For zero chemical potential, each species contributes $S_i=(U_i+P_{i}V)/T=4U_i/(3T)$. ThusEntropy density scales as $T^3$ at fixed relativistic particle content; total entropy also includes the changing proper volume. An isolated adiabatically expanding comoving gas has conserved total entropy and $aT$ constant, rather than its total entropy falling as $T^3$.

At equal temperature, single electrons and photons each have two spin/polarization states. In the ultrarelativistic approximation,Positrons are a separate species with the same contribution, so $(S_{e^{-}}+S_{e^{+}})/S_{\gamma }=7/4$. This distinction is needed for the heating count in part d. At the exact supplied temperature the electron rest mass gives corrections to the ideal $7/8$ ratio; it is the limiting relativistic value requested by the problem.

There is also a sign convention issue in the introductory entropy formula. With the chemical-potential convention used in the supplied distribution, the Euler relation is $U=TS-PV+{\sum }_i{\mu }_i{N}_i$, so $S=(U+PV-{\sum }_i{\mu }_i{N}_i)/T$. Its chemical-potential term has the opposite sign to the printed one. For a concrete counterexample, a dilute monatomic ideal gas with one internal state and $n{\lambda }_T^3=e^{-1}$, where ${\lambda }_T$ is the thermal de Broglie wavelength, has $\mu =-k_{\mathrm{B}}T$, $U=3N{k}_{\mathrm{B}}T/2$ and $PV=N{k}_{\mathrm{B}}T$. Its entropy is $7N{k}_{\mathrm{B}}/2$; the printed plus sign would instead give $3N{k}_{\mathrm{B}}/2$. All chemical potentials have been neglected here, so this source error does not change any of the above results.

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 the printed hypercharge convention; reversing every hypercharge relative to another common convention does not change the gauge invariants. Write $U,D,E,N$ for the singlet chiral superfields labelled with a superscript $R$ in the paper, and define the weak-doublet contraction $A\cdot B={ϵ}_{ab}{A}^a{B}^b$. The most general homogeneous cubic superpotential with right-handed-neutrino superfields separates as $W_3=W_{B,L}+W_{\mathrm{viol}}$, whereandRepeated generation and color indices are summed. Because the chiral superfields commute while the gauge contractions are antisymmetric,The factors $1/2$ and $1/6$ remove the corresponding overcounting.

Here is a completeness check, not merely a list of possible terms. A color-singlet cubic can have no colored fields, a fundamental–antifundamental pair, or three fundamentals or three antifundamentals. A colored pair must be $QU$ or $QD$: the remaining weak doublet and hypercharge select respectively $H_2$ or $H_1,L$. Three antifundamentals allow only $UDD$ at zero hypercharge; three $Q$ doublets have no weak singlet. With no colored fields, weak invariance requires zero or two doublets. Two positive-hypercharge doublets require $E$, giving $L{H}_{1}E$ and $LLE$, while $H_1\cdot H_1=0$. Opposite-hypercharge doublets require $N$, giving $L{H}_{2}N$ and $H_1{H}_{2}N$. With no doublets, only $NNN$ has zero hypercharge. These are exactly the nine structures above.

For baryon number and lepton number, assignand zero charges to the Higgs fields. This is the conjugate-singlet convention, so the neutrino Yukawa coupling conserves lepton number. The first four terms conserve both charges. The $LLE$ and $LQD$ terms carry $L=1$, the $UDD$ term carries $B=-1$, the $N{H}_1{H}_2$ term carries $L=-1$, and $NNN$ carries $L=-3$. The gauge-singlet neutrino interactions must therefore be included in the violating sector.

If “cubic” instead means a polynomial of degree at most three, there are also the lower-degree gauge invariantsThe Higgs mass term conserves both charges; $t_i{N}_i$, $M_{ij}{N}_i{N}_j$ and ${\mu }_i{L}_i\cdot H_2$ violate lepton number. These additional terms distinguish the full renormalizable superpotential from its homogeneous cubic part.

To obtain squark-mediated proton decay, combine a lepton number-violating $LQD$ vertex with a baryon number-violating $UDD$ vertex. A nonzero example is ${\lambda }_{112}^{\prime }$ with ${\lambda }_{112}^{\prime \prime }$, exchanging the singlet strange quark squark of mass $M_{\tilde{s}}$. The antisymmetry forbids ${\lambda }_{111}^{\prime \prime }$, so using three first-generation singlets would not be a valid example. The relevant component vertices contain $e_L{u}_L{\tilde{s}}^c$ and $u^c{d}^c{\tilde{s}}^c$, with the required color contraction. Connecting one vertex to the complex conjugate of the other through the massive scalar propagator produces schematicallyParentheses denote Lorentz scalar two-spinor contractions, dotted for the right-handed pair and undotted for the left-handed pair. This dimension-six proton-decay operator contains the three quarks $uud$ and an electron field that can create a Positron. Its hadronic matrix element $⟨{\pi }^0|{ϵ}_{abc}(u_R^a{d}_R^b)u_L^c|p⟩$ gives the channel $p\to e^{+}{\pi }^0$. The internal strange flavor need not appear in the final state. The two interactions together violate both continuous charges, as required for this channel.

The coefficient of the dimension-six effective operator has mass dimension $-2$. Thus dimensional analysis giveswhere $c_h$ is a dimensionless hadronic and phase-space factor. Its detailed value cannot be fixed by dimensional reasoning. Applying the historical lower bound supplied in the paper gives the proton-lifetime bound on a product of R-parity-violating couplingsThe numerical coefficient is only the accuracy of the requested dimensional estimate; the squark mass was not fixed by the question.

Finally, R-parity is $R_p=(-1{)}^{3(B-L)+2s}$. At the superfield level use matter parity $P_M=(-1{)}^{3(B-L)}$: $Q,U,D,L,E,N$ are odd and $H_1,H_2$ are even. Each conserving Yukawa coupling contains two matter chiral superfields and one Higgs field, so it is even. Every violating homogeneous cubic above is odd: $LLE,LQD,UDD,NNN$ contain three matter fields, while $N{H}_1{H}_2$ contains one. R-parity forbids all the listed cubic violations and preserves the Yukawa terms that generate fermion masses. The Higgs bilinear is also even, whereas the lepton bilinear $L{H}_2$ and the singlet linear term are odd.

The final assertion must not be read as forbidding every possible violation of continuous baryon number or lepton number. In particular, matter parity allows Majorana neutrino masses: $N_i{N}_j$ is even but carries $L=-2$, a concrete counterexample once quadratic terms are included. Higher-dimensional operators such as $QQQL$ and $(L{H}_2{)}^2$ are likewise even and violate the continuous charges. The verified statement here concerns the homogeneous cubic sector, including the neutrino singlet terms.

A Positron is the electron's antiparticle: it has the same mass and spin one-half, opposite electric charge $+e$, and fermionic statistics. A state $b_{\mathbf{p}}^{s†}|0⟩$ has positive energy $E_{\mathbf{p}}$ and physical momentum $\mathbf{p}$, as follows from the Hamiltonian and momentum operatorsThe charge commutator gives $[Q_{\mathrm{em}},b^{†}]=+e{b}^{†}$.

The term $b^{†}v(p)e^{ip\cdot x}$ in the field has negative frequency, but it creates this positive-energy Positron. Treating its coefficient as an annihilator of a physical negative-energy electron would give an unbounded energy interpretation. Fermionic quantization instead reorders that sector, giving positive antiparticle occupation energies and the opposite signed charge. The historical Dirac sea describes a Positron as a hole in filled negative-energy electron levels; the field-theory interpretation uses an antiparticle creation operator and normal ordering, without requiring a literal infinite material sea.

Under charge conjugation of a Dirac field, ${\psi }^C=C{\bar{\psi }}^T$, where $C^{-1}{\gamma }^{\mu }C=-({\gamma }^{\mu }{)}^T$, particle and antiparticle modes interchange. Electron–Positron pair creation and annihilation change total particle number while conserving electric charge. Positrons are positive-energy antiparticles, not physical negative-energy electrons.

Use natural units, assume negligible neutrino chemical potentials, and approximate neutrino decoupling as occurring before electron-positron annihilation in cosmology. At high temperature, weak interactions maintain a common temperature for neutrinos, antineutrinos, electrons, Positrons and photons. Since ${\Gamma }_{\nu }/H\propto T^3$, the reaction rate eventually loses against expansion. Decoupled relativistic neutrinos then free stream; their momenta redshift as $a^{-1}$, so their distribution retains a temperature parameter satisfying $a{T}_{\nu }=\mathrm{constant}$.

The still-coupled electromagnetic bath receives the entropy of annihilating electrons and Positrons. Its effective number of relativistic entropy degrees of freedom is initiallywhere the two photon polarizations count as bosonic states and the four electron-positron spin states have the fermion entropy weight $7/8$. After the pairs disappear, $g_{\ast s}^{\mathrm{EM}}=2$. Cosmological entropy conservation in this coupled bath gives $g_{\ast s}^{\mathrm{EM}}(a{T}_{\gamma }{)}^3=\mathrm{constant}$. The three already decoupled neutrino species are excluded from this bath's entropy count. At decoupling $T_{\nu }=T_{\gamma }$, henceSubsequent expansion preserves this ratio in the idealized history. Residual interactions during annihilation give small corrections to the instantaneous-decoupling approximation. If a relic species later becomes massive, $T_{\nu }$ here remains the redshifting momentum-distribution parameter; it need not be its nonrelativistic kinetic temperature.

A single ordinary neutrino species includes one populated neutrino helicity and one antineutrino helicity. For negligible chemical potential, integration of the Fermi-Dirac distribution givesTherefore the number density of a relativistic neutrino relative to photons, with the antiparticle included, isThe factor $3/4$ is the fermionic number-density weight, distinct from the energy and entropy weight $7/8$. This relation continues to hold after this species becomes nonrelativistic if its comoving particle number is conserved.

Under the proposed matter-dominated, spatially flat model, the Friedmann equation requires the present relic-neutrino energy density to be approximately the critical density. Using the unreduced Planck mass $M_P=G^{-1/2}$ givesThus the critical-density mass of a thermal relic neutrino isTaking the paper's rounded values, $T_{\gamma 0}=2\times 10^{-13}\mathrm{GeV}$, $H_0=h/(4.5\times 10^{41}{\mathrm{GeV}}^{-1})$ and $M_P\sim 10^{19}\mathrm{GeV}$, givesFor $0.5<h<1$, this is of order $30$--$100\mathrm{eV}$ with these approximate constants. More precise constants change the coefficient, not the requested mass scale. This mass is much larger than the present momentum scale $T_{\nu 0}\simeq 1.4\times 10^{-4}\mathrm{eV}$, validating the nonrelativistic energy density approximation. The result assumes that one thermal relic species dominates the matter density; it is conditional on that model rather than a measurement of a neutrino mass.

A collisionless relativistic species redshifts with $T_{\nu }\propto a^{-1}$. The coupled plasma instead obeys cosmological entropy conservation, $g_{\ast s}{T}^3{a}^3=\mathrm{constant}$, and is heated relative to a decoupled species as massive particles annihilate. The effective number of relativistic energy degrees of freedom $g_{\ast }$ and effective number of relativistic entropy degrees of freedom $g_{\ast s}$ coincide while all species share a common relativistic temperature. Use $g_d:=g_{\ast s}(T_{\mathrm{dec}})\simeq 100$ for the Standard Model bath, excluding the already decoupled extra species.

Just before ordinary neutrino decoupling, the coupled bath contains two photon states, four electron/Positron states and six ordinary neutrino/antineutrino states. ThereforeThe extra-neutrino temperature relative to that bath is then $(43/(4g_d){)}^{1/3}$. After the ordinary neutrinos decouple, electron/Positron annihilation heats only the photon bath: its entropy count changes from $2+(7/8)4=11/2$ to $2$. Both neutrino populations miss that last heating, soThis is the two-stage entropy dilution of an early-decoupled neutrino. Numerically $T_{\nu ,0}\simeq 6.8\times 10^{-14}\mathrm{GeV}$. Relative to ordinary relic neutrinos it is cooler by $(43/(4g_d){)}^{1/3}\simeq 0.48$.

Let $g_{\nu }$ count the extra neutrino's populated helicity and antiparticle states. For one conventional neutrino plus antineutrino species $g_{\nu }=2$, the same state count as photons. The Fermi-Dirac distribution supplies the $3/4$ number-density factor, soThus a two-state species has about $2.9\%$ of the photon number density. Stating $g_{\nu }$ explicitly also covers a different choice of populated states.

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 $W=-i\log [Z[J,\bar{\eta },\eta ]/Z[0]]$.