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The Dirac electromagnetic current is the electric charge times the Dirac current. Its coupling $-e{j}^{\mu }{A}_{\mu }$ to the electromagnetic four-potential is the matter interaction of quantum electrodynamics.

At $\xi =1$, the Euclidean photon propagator is $D_{\mu \nu }(k)={\delta }_{\mu \nu }/k^2$. Its Lorentz-index contraction simplifies loop numerators in quantum electrodynamics.

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.

Choose the phase convention in the question and use the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. For a charged Dirac field, the quantum electrodynamics Lagrangian isThe simultaneous local gauge transformation isDirect differentiation gives $D_{\mu }^{\prime }{\psi }^{\prime }=e^{ie\chi }{D}_{\mu }\psi$: the derivative of the phase cancels the shift of $A_{\mu }$. The mass term and covariant kinetic term are therefore invariant. The electromagnetic field tensor is unchanged because mixed partial derivatives of $\chi$ commute. This proves classical gauge invariance of the coupled field action. A charged scalar similarly uses $(D_{\mu }\varphi {)}^{\ast }{D}^{\mu }\varphi$ and a potential depending only on $|\varphi {|}^2$.

The electromagnetic coupling is $-e{A}_{\mu }\bar{\psi }{\gamma }^{\mu }\psi$. The matter equations give a conserved current $j^{\mu }=e\bar{\psi }{\gamma }^{\mu }\psi$, with ${\partial }_{\mu }{j}^{\mu }=0$. Constant $\chi$ gives the global symmetry and its conserved charge. When applying Noether's theorem, the sign of the parameterized Noether generator can be reversed without changing this conservation law; the transformation and coupling signs above remain fixed.

In quantum electrodynamics, gauge invariance connects local phase freedom, electric-charge conservation and the restrictions on photon interactions. For definiteness takeThe gauge transformationhas $D_{\mu }\psi \mapsto e^{-ie\alpha }{D}_{\mu }\psi$: the derivative of the local phase cancels the added gauge potential. Also $F_{\mu \nu }$ is unchanged because mixed partial derivatives commute. Consequently every term in the Lagrangian density is invariant. This derives the minimal electromagnetic coupling of a Dirac field, including the interaction $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. A fermion mass is compatible with the symmetry, whereas a local Proca photon mass $m_{\gamma }^2{A}_{\mu }{A}^{\mu }/2$ is not.

The classical Maxwell equations are ${\partial }_{\mu }{F}^{\mu \nu }=j^{\nu }$, with $j^{\nu }=e\bar{\psi }{\gamma }^{\nu }\psi$. Taking a divergence gives ${\partial }_{\nu }{j}^{\nu }=0$ because $F^{\mu \nu }$ is antisymmetric. The same conservation law follows directly from the charged Dirac equation and its adjoint. For vanishing flux at spatial infinity, $Q=\int d^{3}x{j}^0$ is conserved. Gauss law identifies this charge with the electric flux through a surrounding surface. Particle-antiparticle creation is allowed, but its net electric charge is zero.

Local gauge invariance is also a redundancy in the potentials, rather than an extra propagating degree of freedom. The canonical momentum of $A_0$ vanishes, and its field equation enforces the Gauss law constraint in gauge theory, $\nabla \cdot \mathbf{E}=j^0$. After imposing the constraint and quotienting gauge freedom, a massless photon has two transverse physical polarizations. Gauge transformations vanishing at the boundary identify equivalent descriptions; constant phase transformations at infinity can act on charged states and generate their conserved charge. This distinction explains why gauge redundancy and electric charge are compatible.

In perturbative QED, the quadratic photon operator cannot be inverted before gauge fixing. Adding $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$ gives the covariant gauge propagatorThis propagator contains longitudinal and timelike components, but these are not external physical photon states. A Gupta-Bleuler null-state quotient, or the equivalent BRST construction, removes them. In linear Lorenz gauge, $\delta ({\partial }_{\mu }{A}^{\mu })=\square \alpha$, so the Faddeev-Popov determinant is independent of $A_{\mu }$ and the Faddeev-Popov ghost fields decouple. This last statement concerns the linear gauge; Abelian gauge groups alone do not guarantee decoupling in a nonlinear gauge condition.

At the quantum level, the vector symmetry gives a Ward identity. One way to obtain it is to change variables by a localized phase in a time-ordered correlation function containing $\psi (y)$ and $\bar{\psi }(z)$. The action variation becomes a current divergence; the two field insertions contribute opposite contact terms, since their charges are opposite. Fourier transformation turns the divergence into the photon momentum, and amputating the external fermion propagators turns the two contact terms into the difference of their inverse propagators. With the charge removed from the proper vertex and the overall propagator factor $i$ removed from its inverse, the result is the proper-vertex Ward-Takahashi identity in QEDAt tree level this is simply $k_{\mu }{\gamma }^{\mu }=(/p+/k-m)-(/p-m)$. Between on-shell external spinors, the inverse propagators vanish, so the full physical scattering amplitude satisfies $k_{\mu }{\mathcal{M}}^{\mu }=0$. Therefore replacing a photon polarization vector by ${\epsilon }_{\mu }+c{k}_{\mu }$ leaves the amplitude unchanged. The identity applies to the complete required sum of diagrams, not necessarily each diagram separately. It also ensures cancellation of the gauge parameter from physical observables when all terms at the chosen perturbative order are included.

For the photon two-point function, the Ward identity implies transverse photon vacuum polarization:Thus ultraviolet subtraction renormalizes the gauge-invariant kinetic term, rather than requiring a local photon-mass counterterm. Perturbative unbroken QED retains a massless photon. Transversality by itself would not exclude a nonlocal massless pole in a different theory; the assertion here includes the perturbative QED setting.

Taking $k\to 0$ in the proper-vertex identity gives ${\Gamma }^{\mu }(p,p)=\partial S^{-1}(p)/\partial p_{\mu }$. Matching the divergent coefficients of the derivative and vertex terms yields the Ward identity for QED renormalization constants, $Z_1=Z_2$, in a gauge-preserving prescription. If ${\psi }_0=\sqrt{Z_2}\psi$ and $A_0=\sqrt{Z_3}A$, comparison of the interaction coefficients givesHence charge renormalization is fixed by photon wave-function renormalization. Gauge invariance constrains counterterms and relates renormalizations; it does not make the electric charge independent of scale.

The Soft photon theorem gives another physical manifestation. Attaching a photon of small momentum $k$ to the external charged legs gives the leading factorwhere $Q_i$ is charge in units of $e$ and ${\eta }_i=+1$ for outgoing and $-1$ for incoming legs. Replacing $\epsilon$ by $k$ makes this factor proportional to ${\sum }_i{\eta }_i{Q}_i$, which vanishes by electric charge conservation. Massless photons also cause infrared divergences; physical inclusive measurements combine unresolved real emission with virtual corrections, rather than treating either contribution alone as an observable.

Finally, the vector gauge symmetry of a Dirac fermion in QED is anomaly-free, so it can be maintained by regularization and renormalization. This must be distinguished from the axial current: its divergence has explicit mass terms, as in Question 3, and can additionally have a quantum chiral anomaly. Conservation of the gauge current does not imply axial-current conservation. The resulting picture is that gauge invariance simultaneously organizes the physical photon states, enforces electric-charge conservation and ties together amplitudes and ultraviolet subtractions.

Use $\bar{\psi }={\psi }^{†}{\gamma }^0$, the Minkowski metric $(+---)$ and ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Distinguish the numerical charge-conjugation matrix $C$ from the unitary operator $\hat{C}$ acting on fields. The adjoint Dirac equation is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Transpose it and multiply by $C$; the identity $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$ givesThus the charge-conjugate field satisfies the same free Dirac equation.

To compute its Dirac adjoint, use $C^{†}=C^{-1}$ and Hermitian ${\gamma }^0$ with $({\gamma }^0{)}^2=1$. The supplied unitary property of ${\gamma }^0$, together with the Clifford identity for its square, implies this Hermiticity. Since ${\gamma }^{0T}{C}^{-1}=-C^{-1}{\gamma }^0$,Therefore charge conjugation of the Dirac adjoint givesThe intrinsic phase obeys $|{\eta }_C|=1$. Antisymmetry of $C$ also gives $({\psi }^c{)}^c=-C(C^{-1}{)}^T\psi =\psi$.

Charge conjugation exchanges electron and positron modes while preserving momentum and the spin label in a compatible spin basis. Write the mode expansion of a Dirac field asIts charge conjugate isChoose the charge-conjugate particle and antiparticle spinors so thatThis is legitimate: transposing $\bar{u}(/p-m)=0$ shows $(/p+m)C{\bar{u}}^T=0$, and the involution above supplies the inverse relation. Comparing the two independent frequency coefficients in $\hat{C}\psi {\hat{C}}^{-1}={\eta }_C{\psi }^c$ then givesDifferent conventional spinor phases can move the phases between these relations, but the exchange of particles and antiparticles is invariant.

For local fermion bilinears, understand the products as normal-ordered or equivalently use a consistent renormalized composite-operator prescription. The transformed product is initially $-{\psi }^T{C}^{-1}\Gamma C{\bar{\psi }}^T$. Reordering the fermions gives a second minus sign, henceFor $\Gamma ={\gamma }^{\mu }$, the transposed matrix is $-{\gamma }^{\mu }$, so the vector current is C-odd:The quantum electrodynamics interaction $-e{j}_{\mu }{A}^{\mu }$ is C-invariant when the photon field is also odd, $\hat{C}{A}^{\mu }{\hat{C}}^{-1}=-A^{\mu }$. Its field strength is then odd and its quadratic kinetic term even.

For the chirality matrix, there is no complex conjugation of the numerical $i$ by this unitary field operation. ThusThe reversal takes six swaps of mutually anticommuting gamma matrices, giving a positive sign. ThereforeThe axial current is C-even. Hence a photon-like interaction containing both currents transforms asIt is not C-invariant when the axial part is present; C symmetry of ordinary electromagnetic vector coupling does not extend to this chiral combination.

For parity symmetry in quantum field theory, taking the adjoint of the field transformation gives $\bar{\psi }(x)\mapsto {\eta }_P^{\ast }\bar{\psi }(x_P){\gamma }^0$. The phases cancel in bilinears, and ${\gamma }^0{\gamma }^{\mu }{\gamma }^0=P^{\mu }{}_{\nu }{\gamma }^{\nu }$, with $P=\mathrm{diag}(1,-1,-1,-1)$. Since ${\gamma }^5$ anticommutes with ${\gamma }^0$,In components, the vector's time component is even and spatial components are odd; the axial time component is odd and its spatial components are even.

Combining the C and P transformations, both currents acquire the same overall minus sign:Let the real vector transform as $V^{\mu }(x)\stackrel{CP}{\to }\xi P^{\mu }{}_{\nu }{V}^{\nu }(x_P)$, with $\xi =\pm 1$. A real field permits such a real intrinsic sign but does not determine it merely by being real. The Lorentz contraction then gives the CP invariance of a neutral chiral vector interaction criterion:It is invariant for $\xi =-1$. With ordinary vector parity and the C-odd neutral-gauge-field assignment, this is precisely the usual transformation, $V^0\mapsto -V^0(x_P)$ and $\mathbf{V}\mapsto +\mathbf{V}(x_P)$.

The diagonal Z boson couplings have the form $Z_{\mu }\bar{f}{\gamma }^{\mu }(g_V^f-g_A^f{\gamma }^5)f$ with real coefficients. The same calculation applies to any real vector and axial coefficients. Thus the tree-level flavour-diagonal neutral-current Z interaction preserves CP although it violates C and P separately when both current structures occur. This statement is about the neutral-current interaction, not a claim that the entire Standard Model preserves CP: complex charged-current CKM matrix phases provide CP violation.

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 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.

Take the quantum electrodynamics interaction density to be ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. The momentum-space Feynman-gauge photon propagator, with Lorentz indices $\mu ,\nu$ and momentum $k$, isThe electron-photon vertex from the QED Feynman rules iswith its spinor indices connecting the incident and outgoing fermion lines. The muon-photon vertex has the same coupling. The Feynman i-epsilon prescription specifies the pole boundary condition; using a different overall charge-sign convention changes both vertices consistently and leaves the squared amplitude unchanged.

For the electron-positron annihilation into a muon pair, the exchanged photon has $k=p+q$ and $k^2=s$. Up to the conventional overall phase, the scattering amplitude is

For a real vector relativistic quantum field, $V^{\mu }=V^{\mu †}$. Its transformed field must also be a Hermitian operator, so the intrinsic charge-conjugation phase satisfies ${\eta }_C={\eta }_C^{\ast }$; combined with $|{\eta }_C|=1$, this givesThe quantum electrodynamics interaction is $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. Since the Dirac electromagnetic current is odd under charge conjugation, charge conjugation invariance of this interaction requires the photon field to be odd too:

With $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$ and $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$, the quantum electrodynamics Lagrangian is