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Varying the Dirac action with respect to the spinor, rather than its Dirac adjoint, gives the left-acting field equation after integration by parts. Together with the Dirac equation it proves conservation of the vector current.

The Dirac scalar bilinear is invariant under the simultaneous spinor and adjoint transformations because ${\bar{\Psi }}^{\prime }{\Psi }^{\prime }=\bar{\Psi }{S}^{-1}S\Psi$. It is the mass term of the Dirac action. At fixed coordinates it transforms as a scalar field, with an orbital transport term and no spin term.

Replace the derivative in the Dirac action by the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$. The interaction is then $-q{A}_{\mu }\bar{\Psi }{\gamma }^{\mu }\Psi$. With $\Psi \mapsto e^{-iq\alpha }\Psi$ and $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha$, the covariant derivative transforms in the same way as the spinor, proving local gauge invariance.

One consistent Dirac equation convention uses $\eta =\mathrm{diag}(-1,1,1,1)$, $\{{\gamma }^a,{\gamma }^b\}=-2{\eta }^{ab}$, and $(i{\gamma }^a{\partial }_a+m)\psi =0$. Taking the negatives of the standard mostly-minus gamma matrices, and defining the Dirac adjoint with the new ${\gamma }^0$, gives the same physical Dirac action. The positive-frequency plane wave is $e^{ip\cdot x}u(p)$ with $p^0>0$, and its equation is $(/p-m)u=0$. The mass shell is $p^2=-m^2$. These signs must be translated together when using a formula stated in another convention.

A symmetry of a classical field theory maps solutions to solutions; for the Noether theorem we require a continuous variational symmetry of the action, allowing its density to change by a total divergence. This is stronger than simply observing an accidental symmetry of one solution. Consider first-derivative fields ${\varphi }_a$ and defineThe Euler-Lagrange field equations are $E_a=0$. For a fixed-coordinate variation ${\delta }_0{\varphi }_a=\epsilon Q_a$, the product rule gives the off-shell identityThis identity is the proof's essential integration-by-parts step.

For a general spacetime symmetry write $x^{\prime \mu }=x^{\mu }+\epsilon {\xi }^{\mu }$ and ${\varphi }_a^{\prime }(x^{\prime })={\varphi }_a(x)+\epsilon {\Delta }_a$. At a fixed point $Q_a={\Delta }_a-{\xi }^{\nu }{\partial }_{\nu }{\varphi }_a$, while the volume-element variation contributes ${\partial }_{\mu }(\mathcal{L}{\xi }^{\mu })$. Suppose action invariance takes the formCombining with the preceding identity provesThis is the Noether current for a spacetime symmetry. It also covers internal symmetries, for which $\xi =0$. A constant parameter corresponds to one Noether current; several independent parameters give several currents. Integrating the continuity equation givesThus the Noether charge is conserved when the fields have suitable falloff or no flux through the spatial boundary. These are classical identities; quantum symmetries may additionally require absence of an quantum anomaly.

Translations conserve energy and momentum. If the density has no explicit coordinate dependence, take the active fixed-coordinate variation $Q_a={\partial }_{\nu }{\varphi }_a$ for translation in direction $\nu$. Then ${\delta }_0\mathcal{L}/\epsilon ={\partial }_{\nu }\mathcal{L}$, and the internal-form proof with $K^{\mu }={\delta }^{\mu }{}_{\nu }\mathcal{L}$ gives the canonical energy-momentum tensorIts four spatial integrals are the conserved energy and momentum. For a real scalar field with density $\frac{1}{2}(\partial \varphi {)}^2-V(\varphi )$ this becomes $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Direct differentiation gives ${\partial }_{\mu }{T}^{\mu \nu }=(\square \varphi +V^{\prime }(\varphi )){\partial }^{\nu }\varphi$, which vanishes by the field equation.

Lorentz symmetry conserves angular momentum and boost charges. For that scalar example, $T^{\mu \nu }$ is symmetric, so the Lorentz transformation currents can be writtenTheir divergence is $T^{\rho \sigma }-T^{\sigma \rho }=0$. Spatial pairs give angular momentum, and time-space pairs give the conserved boost charges. For a spinor field or a vector field, the intrinsic field variation supplies an additional spin current; equivalently the Belinfante-Rosenfeld stress-energy tensor combines spin and orbital pieces into a symmetric stress tensor. The nonunitarity of the finite component Spinor representation of the Lorentz group does not obstruct conservation of these physical spacetime charges.

Global phase symmetry conserves particle charge. For a complex scalar field with density ${\partial }_{\mu }{\Phi }^{\ast }{\partial }^{\mu }\Phi -V({\Phi }^{\ast }\Phi )$, the constant transformation $\Phi \mapsto e^{-i\alpha }\Phi$ leaves the density invariant. Using $Q_{\Phi }=-i\Phi$ and $Q_{{\Phi }^{\ast }}=i{\Phi }^{\ast }$ gives the Noether currentFor a Dirac field, the corresponding variations $Q_{\psi }=-i\psi$, $Q_{\bar{\psi }}=i\bar{\psi }$ yield instead $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, the Dirac current. Its divergence vanishes by the Dirac equation and adjoint Dirac equation. The Yukawa interaction of part 2 also preserves this phase symmetry, because each bilinear has one $\psi$ and one $\bar{\psi }$. Thus it preserves fermion number and forbids annihilation of two fermions into a single scalar. These currents express charge or particle-minus-antiparticle-number conservation, rather than conservation of total particle count in arbitrary interacting processes.

A global symmetry in field theory uses a spacetime-independent parameter and ordinarily relates physically distinct configurations. A gauge symmetry instead has arbitrary local parameters and represents redundancy in the description. For example, the Maxwell field is unchanged physically by $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha (x)$: its electromagnetic field tensor is unchanged. For charged matter use $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$ together with $\psi \mapsto e^{-iq\alpha (x)}\psi$. Differentiating shows $D_{\mu }^{\prime }{\psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\psi$, so the coupled Dirac action is locally gauge invariant. A local phase applied to free matter alone produces extra derivatives of $\alpha$ and is not an invariance; the compensating gauge field is necessary.

Local arbitrariness gives identities among field equations, rather than an independent physical charge for every function $\alpha$. In the pure Maxwell field, action invariance under a compactly supported $\alpha$ impliesThus ${\partial }_{\nu }{E}^{\nu }=0$ holds off shell, directly because $F^{\mu \nu }$ is antisymmetric. This exemplifies the Noether second theorem. Gauge fixing and the Gupta-Bleuler null-state quotient remove redundant photon components; they do not remove a physical global charge. Gauge transformations that are nontrivial at a boundary can carry boundary charges, so the redundancy statement concerns transformations satisfying the chosen boundary conditions. Finally, discrete symmetries such as the sign reversal of a real scalar or parity symmetry in quantum field theory can be important without furnishing a continuous-parameter Noether current.

The Dirac field has the global symmetry $\psi \mapsto e^{i\alpha }\psi$, $\bar{\psi }\mapsto e^{-i\alpha }\bar{\psi }$, while the real scalar is unchanged. Both the free Dirac action and the pseudoscalar Yukawa interaction preserve this symmetry. Therefore Dirac fermion number conservation holds: its charge counts particles minus antiparticles.

The initial state has charge $+1$, whereas a final antiparticle and a neutral scalar have charge $-1$. Since the S-matrix commutes with that charge,This holds at every order, not just tree level. In the Feynman rules, a continuous fermion arrow cannot connect these specified external states. Replacing only the outgoing $\bar{u}$ by a $v$ in the previous expression would not give a physical amplitude. Changing both external fermions to antiparticles would instead produce an allowed process with reversed fermion flow and the appropriate $v$ spinors; moving a leg between initial and final states is a different operation governed by crossing symmetry.

Use gamma matrices satisfying the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with ${\gamma }^{0†}={\gamma }^0$ and ${\gamma }^{i†}=-{\gamma }^i$. The Dirac adjoint is $\bar{\Psi }={\Psi }^{†}{\gamma }^0$. Take the Dirac actionIt differs from the manifestly Hermitian form with $(i/2)\bar{\Psi }{\gamma }^{\mu }{\overleftrightarrow{\partial }}_{\mu }\Psi$ only by a boundary term. Treat $\Psi$ and $\bar{\Psi }$ as independent variables when applying the principle of stationary action. Varying $\bar{\Psi }$ givesVarying $\Psi$ and integrating by parts gives the adjoint Dirac equation, $i({\partial }_{\mu }\bar{\Psi }){\gamma }^{\mu }+m\bar{\Psi }=0$. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ also gives $(\square +m^2)\Psi =0$, so its dispersion relation is Lorentz invariant.

For a Lorentz transformation $x^{\prime }=\Lambda x$, the field transforms in the Spinor representation of the Lorentz group:Since ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$, this identity gives the Lorentz covariance of the Dirac operator:Thus every solution is carried to another solution. The field is a spinor rather than a four-vector; the transformation of the gamma matrices supplies the necessary covariance. The Dirac action is Lorentz invariant because its integrand is a scalar and $d^{4}x$ is invariant.

For electric charge $q$, use the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$. The minimal electromagnetic coupling of a Dirac field isThe local gauge transformations in this convention areDirect substitution gives $D_{\mu }^{\prime }{\Psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\Psi$, which proves gauge covariance of the charged Dirac equation. The mass and kinetic terms are invariant, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$, so the full action has gauge invariance. The conserved current is $j^{\mu }=q\bar{\Psi }{\gamma }^{\mu }\Psi$. It transforms as a four-vector under Lorentz transformations, and varying $A_{\mu }$ gives ${\partial }_{\nu }{F}^{\nu \mu }=j^{\mu }$.

To make the infinitesimal spinor transformation explicit, write ${\Lambda }^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+{\omega }^{\mu }{}_{\nu }$ with ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. DefineThe Lorentz generators from gamma-matrix commutators have precisely the needed algebra: $[{\sigma }^{\mu \nu },{\gamma }^{\rho }]=2i({\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu })$, which verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }$ to first order. The infinitesimal transformation of a Dirac field is thereforeAt the same coordinate argument, the orbital change must also be included:Both formulas describe the same transformation; their arguments differ.

Finally, the gamma adjoint identities imply ${\sigma }^{\mu \nu †}={\gamma }^0{\sigma }^{\mu \nu }{\gamma }^0$ and hence the pseudo-unitarity of the spinor Lorentz representation, $S^{†}{\gamma }^{0}S={\gamma }^0$. It follows thatThus the Dirac scalar bilinear $\bar{\Psi }\Psi$ is a Lorentz scalar. Infinitesimally the spin terms in $\delta (\bar{\Psi }\Psi )$ cancel; at fixed coordinates only the ordinary scalar orbital transformation remains.

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.

For this question use the mostly-plus Dirac convention: the Minkowski metric is ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex $4\times 4$ matrices representing the spacetime Clifford algebra with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where ${\sigma }^i$ are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.

In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$ changes the Hermitizing matrix to $H^{\prime }=M^{-†}{\gamma }^0{M}^{-1}$ and the charge-conjugation matrix to $C^{\prime }=MC{M}^T$. Then ${\gamma }^{\prime a†}=H^{\prime }{\gamma }^{\prime a}{H}^{\prime -1}$ and $C^{\prime -1}{\gamma }^{\prime a}{C}^{\prime }=-{\gamma }^{\prime aT}$. Thus the simple formula with ${\gamma }^{\prime 0}$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.

Applying $(i{\gamma }^a{\partial }_a-m)$ to the Dirac equation gives $(\square -m^2)\Psi =0$. Its mass shell is $p^2=-m^2$, so the frequencies are $p^0=\pm \sqrt{{\mathbf{p}}^2+m^2}$. The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the spin angular momentum matrices are $\mathrm{diag}({\sigma }^i,{\sigma }^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.

In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, $e^{ip\cdot x}=e^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ is positive frequency. The negative-frequency coefficient obeys $(/p+m)v_s(p)=0$. The fermionic annihilation operators $b_s$ and $d_s$ satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The $b_s^{†}$ excitations are particles; the $d_s^{†}$ excitations are antiparticles with the same positive energy, mass and spin-$1/2$ but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of $b_s^{†}{b}_s+d_s^{†}{d}_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.

For the printed $e^{ip\cdot x}$ wave, ${\partial }_a\Psi =i{p}_a\Psi$. Substitution givesThe spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, Hermitian conjugation and ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$ giveThese are right and left null-vector equations for the same on-shell matrix.

To obtain the Gordon identity, take both external Dirac spinors to have the same real mass $m$. Their two equations implyDefine ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$. The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by $-1$ proves the required formula exactly:For $m\ne 0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at $m=0$. The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(/p+m)u=0$; the corresponding identity uses $(p_b-p_b^{\prime })$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.

Take Euclidean gamma matrices with $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\delta }_{\mu \nu }$, and independent Grassmann fields $\psi ,\bar{\psi }$. The minimally coupled Dirac action isIts mass term is the Dirac mass term, and the interaction is $-ig{\bar{\psi }}^a{\gamma }_{\mu }{A}_{\mu }^b(T_{\mathrm{ad}}^b{)}_{ac}{\psi }^c$. With Fourier convention $e^{ip\cdot x}$, the free kinetic operator is $i/p+m$. The Euclidean adjoint Dirac Feynman rules are thereforeThe vertex sign follows by expanding $e^{-S_{\psi }}$, not $e^{i{S}_{\psi }}$. Momentum conservation supplies the usual Dirac delta function at each vertex. There is only a one-gluon fermion vertex, with no two-gluon seagull because the fermionic covariant derivative is first order. A closed fermion loop has the usual fermionic sign $-1$. The identity $(i/p+m)(-i/p+m)=(p^2+m^2)I$ checks the Dirac propagator.

The Dirac action isTo first order,The commutator found in part i givesTogether with $\det \Lambda =1$, the transformations in part ii therefore leave both the kinetic and mass terms invariant through first order. Thus $S_D^{\prime }=S_D+O({\Omega }^2)$.