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Choose a local orthonormal coframe $e^A$ with $g={\eta }_{AB}{e}^A{e}^B$. The Levi-Civita connection is represented by the connection 1-forms satisfyingThe gamma matrices satisfy the Clifford algebra relations $\{{\gamma }^A,{\gamma }^B\}=2{\eta }^{AB}I$. The matrices $\frac{1}{4}[{\gamma }^A,{\gamma }^B]$ represent the infinitesimal orthogonal transformations on spinor fields; lifting the frame connection to this representation defines the spinor covariant derivativeBoth frame indices are summed, so each antisymmetric pair occurs twice. Under a local Lorentz transformation with spin lift $S$, the spin connection transforms as ${\Omega }_{\mu }\mapsto S{\Omega }_{\mu }{S}^{-1}-({\partial }_{\mu }S)S^{-1}$ and $ϵ\mapsto Sϵ$; consequently ${\nabla }_{\mu }ϵ\mapsto S{\nabla }_{\mu }ϵ$. This establishes covariance. Global construction requires a compatible spin structure (and the appropriate orientation and time orientation); the formulas themselves work on a local orthonormal frame patch.

The supplied gamma matrices have squares $-I,I,I$ and anticommute pairwise, so $\eta =\mathrm{diag}(-1,1,1)$. Work on a connected regular patch with $r>0$ and $V\ne 0$. Since the metric depends only on $V^2$, we may use its positive smooth square root $V>0$. TakeTheir exterior derivatives are $d{e}^0=-V^{\prime }{e}^0\wedge e^1$, $d{e}^1=0$, and $d{e}^2=(V/r)e^1\wedge e^2$. Solving Cartan's first structure equation and imposing Lorentzian connection-form antisymmetry givesIn particular ${\omega }_{01}=-V{V}^{\prime }dt$ and ${\omega }_{12}=-Vd\theta$. The parallel spinor equations are thereforeDifferentiate the angular equation with respect to $r$ and use ${\partial }_rϵ=0$. This gives $V^{\prime }{\gamma }^1{\gamma }^2ϵ=0$. Since $({\gamma }^1{\gamma }^2{)}^2=-I$, this matrix is invertible. A parallel spinor that vanishes at any point vanishes everywhere on the connected domain, by uniqueness of parallel transport. Thus a nonzero parallel spinor implies $V^{\prime }=0$ throughout the patch. There are no nonzero parallel spinors unless $V$ is constant. The zero spinor field is always a solution, so “nonzero” is essential in interpreting the printed assertion. A smooth original $V$ of either sign is constant on such a patch exactly when $|V|$ is constant; $V=0$ is excluded because the given metric is not defined there.

Write the constant as $c=|V|>0$, and introduce $T=ct$, $\rho =r/c$, and $\varphi =c\theta$. The metric becomesA circle at small $\rho$ has circumference $cP\rho$ if $\theta$ has period $P$, whereas its proper radius is $\rho$. Smoothness requires circumference divided by radius to tend to $2\pi$, not a multiple of $2\pi$. Conversely, when $cP=2\pi$, the coordinates $X=\rho \cos \varphi$, $Y=\rho \sin \varphi$ exhibit a smooth flat spatial disk. HenceFor a prescribed period $2\pi$, this requires $c=1$; otherwise there is a conical singularity at the apex.

For constant $V=c$, the time and radial equations make the components independent of $t,r$. The angular equation has all solutionswhere ${ϵ}_0$ is an arbitrary constant two-component spinor field and ${\gamma }^1{\gamma }^2=-{\gamma }^0$. There are two independent real solutions for real spinor fields, or two independent complex solutions if complex spinors are used. At the smooth angular period, these polar-frame components change sign. This is exactly the polar-frame parallel spinor transformation under the spin lift of a full rotation: the rotating polar orthonormal frame does not extend through the origin. The spin structure extending over the disk uses this antiperiodic polar trivialization; in a Cartesian orthonormal frame the same parallel spinors have constant components and extend smoothly. Requiring periodic components in the rotating frame would select the other spin structure on the punctured disk, which does not extend over its center.

The supplied gamma matrices use the mostly-minus metric signature. The Pauli matrix multiplication law gives ${\sigma }^1{\sigma }^2{\sigma }^3=i{I}_2$, hence block multiplication yieldsThe off-diagonal blocks of every ${\gamma }^{\mu }$ connect opposite eigenspaces of this diagonal matrix. Explicitly,Thus $\{{\gamma }^5,{\gamma }^{\mu }\}=0$, and $({\gamma }^5{)}^2=I_4$. The chiral projectors $(1\mp {\gamma }^5)/2$ select the upper and lower two-component blocks, respectively.

Write the Dirac spinor as $\psi =({\chi }_L,{\chi }_R{)}^T$. Substituting the block matrices into the massless Dirac equation gives two independent Weyl spinor equations:They are decoupled because the mass term, which would join the two chirality components, is absent.

For a left-handed plane wave, use $p\cdot x=p^{0}t-\mathbf{p}\cdot \mathbf{x}$. Its coefficient $\xi$ obeysSince $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^2{I}_2$, the determinant is $(p^0{)}^2-|\mathbf{p}{|}^2$. A nonzero coefficient therefore requires the massless mass shell $\boxed{p^2=0}$. For positive energy and nonzero $\mathbf{p}$, put $P=|\mathbf{p}|=p^0$ and parameterize its direction by angles $\vartheta ,\phi$. ThenMultiplication gives $(\mathbf{\sigma }\cdot \hat{\mathbf{p}}){\chi }_{-}=-{\chi }_{-}$ and ${\chi }_{-}^{†}{\chi }_{-}=1$. Hence the left-handed massless plane-wave spinor can be writtenHere $N(p)$ is an arbitrary complex normalization; $|N|=\sqrt{2P}$ gives ${\lambda }_L^{†}{\lambda }_L=2P$. Away from the negative $z$ axis one can equivalently takeAt $\mathbf{p}=-P\hat{\mathbf{z}}$ choose $(1,0{)}^T$ instead. These are phase patches for the same one-dimensional kernel. The other frequency branch $p^0=-P$ uses the positive eigenvector ${\chi }_{+}=(\cos (\vartheta /2),e^{i\phi }\sin (\vartheta /2){)}^T$ for the direction of $\mathbf{p}$. At $p^{\mu }=0$, the algebraic equation vanishes and any two-component coefficient is possible, but no momentum direction or helicity is defined there.

A spatial rotation by $\alpha$ about $\hat{\mathbf{p}}$ acts on the two-component coefficient by $\exp (-i\alpha \mathbf{\sigma }\cdot \hat{\mathbf{p}}/2)$. The negative eigenvector therefore acquires the phase $e^{i\alpha /2}$. Since a state with helicity $h$ acquires $e^{-i\alpha h}$, the positive-energy left-handed particle has $\boxed{h=-1/2}$. There is only one such spin state at a fixed nonzero momentum, rather than two independent polarizations of the same chiral particle.

The helicity content of a quantized Weyl field also includes an antiparticle with opposite helicity. In the field expansion, the negative-frequency term multiplies an antiparticle fermionic creation operator, while the positive-frequency term multiplies a particle fermionic annihilation operator. Their adjoint roles give conjugate phases for the corresponding created states, so a left-handed field creates particles with $h=-1/2$ and antiparticles with $h=+1/2$. The right-handed Weyl spinor has the opposite assignments. Quantizing the full massless Dirac field, rather than one chiral component, therefore gives both particle helicities and both antiparticle helicities; every excitation has spin one-half, while chirality determines the helicity of each massless sector.

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$. Transposing a product reverses its order but does not conjugate the factor $i$. Thus the charge-conjugation matrix identity giveswhere reversing the four distinct gamma matrices requires six exchanges. Similarly,

To fix the transpose sign, work in the Weyl representation of the gamma matrices:The Pauli identity checks $C_0{\gamma }^{\mu T}{C}_0^{-1}=-{\gamma }^{\mu }$ directly, and ${\epsilon }^T=-\epsilon$ gives $C_0^T=-C_0$, with $C_0^{†}{C}_0=I$. Any other intertwiner differs from $C_0$ by a scalar: its ratio with $C_0$ commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. A matrix commuting with all matrix units is scalar. Under a spinor basis change $S$, the intertwiner becomes $S{C}_0{S}^T$, whose transpose is still its negative. Therefore in every such four-component representationThis proves the antisymmetry of the four-dimensional charge-conjugation matrix, rather than leaving the sign undetermined by the general transpose argument.

Hermitian conjugating the charged Dirac equation, using real $A_{\mu },e,m$ and ${\gamma }^0{\gamma }^{\mu †}{\gamma }^0={\gamma }^{\mu }$, gives the adjoint Dirac equationTranspose it and multiply by $C$. Since $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$, the field ${\psi }^c=C{\bar{\psi }}^T$ obtained by charge conjugation obeysThus algebraic charge conjugation reverses the electromagnetic coupling.

If ${\gamma }_5\psi =\eta \psi$, $\eta =\pm 1$, Hermiticity of ${\gamma }_5$ and its anticommutation with ${\gamma }^0$ implyUsing the first transpose identity,Hence $\boxed{\bar{\psi }{\gamma }_5=-\eta \bar{\psi },{\gamma }_5{\psi }^c=-\eta {\psi }^c}$: algebraic charge conjugation reverses chirality. This concerns conjugating the already projected spinor, not the distinct operation of keeping a chiral projector outside an operator conjugation.

For the Grassmann variation of a chiral Majorana mass term, variations of odd fields must be moved past odd fields with a minus sign. If $A^T=-A$, componentwise reordering givesBoth $C$ and $C^{-1}$ are antisymmetric. Treating $\psi ,\bar{\psi }$ independently, integrating the kinetic variation by parts, and placing varied Grassmann fields on the left givesThe factors two from variation cancel the mass-term factors $1/2$. The first coefficient gives one equation; multiplying the second by $C$ gives the other:The variations can be restricted to their chiral subspaces: the two terms in each equation have the same opposite output chirality, so no inactive component equation has been imposed.

Applying $i/\partial$ to the first equation and substituting the second, using $(i/\partial {)}^2=-\square$, givesA phase rotation removes the phase of $m$, and combining the chiral field with its conjugate then gives a Majorana fermion of this mass, not two independent chiral species.

Let $\chi ={\varphi }^{\prime †}\Psi$ and $\bar{\chi }=\bar{\Psi }{\varphi }^{\prime }$. Question 1 shows they are gauge singlets, so both terms in ${\mathcal{L}}_G=G{\chi }^T{C}^{-1}\chi -G^{\ast }\bar{\chi }C{\bar{\chi }}^T$ preserve electroweak gauge invariance. The original PDF has $C^{-1}$ in the first term; the TeX aid incorrectly replaces it by $G^{-1}$. At the Higgs field vacuum,soComparing with the just-derived chiral mass equations yields the Majorana neutrino mass from the Weinberg operator:For nonzero $G$ this generates a mass without introducing a right-handed neutrino. Under lepton number rotation $\Psi \mapsto e^{i\alpha }\Psi$, the first operator acquires $e^{2i\alpha }$ and its conjugate acquires $e^{-2i\alpha }$. Thus the interaction violates lepton number by $\boxed{\Delta L=2}$. Each composite contains one scalar of mass dimension one and one fermion of dimension $3/2$, so the interaction is dimension five and $G$ has dimension $-1$. This is the Weinberg operator, an effective extension of the minimal renormalizable Standard Model, consistent with the absence of bare quadratic masses in Question 1.

Use coordinate indices $\mu ,\nu$ and Lorentz-frame indices $A,B$, and take ${\eta }_{AB}=\mathrm{diag}(-1,1,1,1)$. Locally choose four independent one-forms $e^A=e^A{}_{\mu }d{x}^{\mu }$ satisfyingOne construction begins with a smooth timelike vector field in a sufficiently small neighborhood, normalizes it, and applies the Gram-Schmidt process to its spacelike orthogonal complement. The dual orthonormal coframe in spacetime gives the vierbein; equivalently its matrix factorizes the metric as $g=e^T\eta e$. The inverse matrix gives the frame $e_A=e_A{}^{\mu }{\partial }_{\mu }$ with $e_A{}^{\mu }{e}^B{}_{\mu }={\delta }_A^B$. This is a local construction: a global frame is not guaranteed by a metric, and a globally defined spinor field requires a compatible spin structure.

The metric does not determine a unique vierbein. A local Lorentz transformation changes the coframe components byThe coordinate index is unchanged in this frame transformation. Restrict to the proper orthochronous Lorentz group for its connected spin lift; orientation-reversing transformations require further structure and are not part of this local spin-group discussion.

Let constant gamma matrices obey the Clifford algebra $\{{\gamma }^A,{\gamma }^B\}=2{\eta }^{AB}$, and write ${\gamma }^{AB}=\frac{1}{2}[{\gamma }^A,{\gamma }^B]$. A spinor field transforms by the spin lift of the Lorentz transformation, not by its vector matrix:For $\Lambda =1+\alpha +O({\alpha }^2)$, with ${\alpha }_{AB}=-{\alpha }_{BA}$, the Clifford algebra identityshows explicitly thatIndeed the commutator of ${\gamma }^C$ with this infinitesimal spin matrix is ${\alpha }^C{}_B{\gamma }^B$. A finite transformation connected to the identity is obtained by exponentiating these generators along a path. Its lift has two possible signs, so spinors are not ordinary Lorentz vectors: a $2\pi$ spatial rotation acts by $-1$, and $4\pi$ by $+1$. Consistent transition lifts define the spin structure; the metric alone does not select it.

For the metric connection, take the Levi-Civita connection and determine the spin connection from the tetrad postulateIt is equivalently determined by $d{e}^A+{\omega }^A{}_B\wedge e^B=0$ and ${\omega }_{\mu AB}=-{\omega }_{\mu BA}$. Explicitly, in terms of the inverse frame,Thus both the coordinate connection and the Lorentz connection 1-form are determined by the metric and chosen frame; no independent torsion has been supplied.

An ordinary derivative fails covariance because ${\partial }_{\mu }(S\psi )=S{\partial }_{\mu }\psi +({\partial }_{\mu }S)\psi$. Seek ${\nabla }_{\mu }\psi =({\partial }_{\mu }+{\Omega }_{\mu })\psi$. Covariance requiresThe coframe's connection obeys ${\omega }^{\prime }=\Lambda \omega {\Lambda }^{-1}-d\Lambda {\Lambda }^{-1}$. Applying the spin representation of the same Lorentz algebra, whose infinitesimal generator was just determined, suppliesThis proves the transformation law rather than choosing the coefficient arbitrarily. It is the spinor covariant derivative induced by the spin connection. With curved matrices ${\gamma }^{\nu }(x)=e_A{}^{\nu }{\gamma }^A$, the same coefficient and tetrad postulate giveso differentiation is compatible with Clifford multiplication and $\{{\gamma }^{\mu }(x),{\gamma }^{\nu }(x)\}=2g^{\mu \nu }(x)$.

Finally, expand the commutator of these derivatives:The spin representation preserves the Lorentz algebra bracket. Consequently $d\Omega +\Omega \wedge \Omega =\frac{1}{4}{R}_{AB}{\gamma }^{AB}$, where $R^A{}_B=d{\omega }^A{}_B+{\omega }^A{}_C\wedge {\omega }^C{}_B$. In components the spinor curvature identity isFor the full second derivative, ${\nabla }_{\mu }({\nabla }_{\nu }\psi )$ also includes $-{\Gamma }_{\mu \nu }^{\rho }{\nabla }_{\rho }\psi$, because its first derivative has a covector index. This term cancels on antisymmetrization for the torsion-free connection. With unit-weight brackets, $X_{[\mu \nu ]}=(X_{\mu \nu }-X_{\nu \mu })/2$, the requested expression is thereforeThis convention distinguishes the antisymmetrized second derivative from the full commutator by its factor of two. In a noncoordinate orthonormal frame, an operator commutator must additionally subtract the derivative along the frame-vector commutator; using the full tensorial second derivative incorporates that correction.

Use signature $\eta =\mathrm{diag}(1,-1,-1,-1)$, the numerical matrices ${\sigma }_{\mu }=(I,{\sigma }_i)$ and ${\bar{\sigma }}_{\mu }=(I,-{\sigma }_i)$ given in the question, and raise genuine tensor indices with $\eta$. A real Minkowski spacetime vector corresponds to the Hermitian matrixFor SL(2,C) matrices, $X^{\prime }=AX{A}^{†}$ is Hermitian and has the same determinant. It therefore induces a real linear Lorentz transformation; composition of the congruence actions agrees with matrix multiplication. The kernel consists of $\pm I$: preservation of $X=I$ makes a kernel element unitary, and preservation of every Hermitian $X$ makes it scalar.

Polar decomposition of an invertible complex matrix connects every determinant-one complex matrix to its unitary factor, so the group is connected. The action gives the Lorentz spinor double cover of the proper orthochronous Lorentz group. It covers rotations via $A_R=e^{-i\vartheta \mathbf{n}\cdot \mathbf{\sigma }/2}$ and boosts via the positive Hermitian matrices below. Every proper orthochronous transformation is a boost followed by a rotation, since one can first match its image of the future unit time vector and then use its rotation stabilizer. The congruence construction does not cover spatial parity or time reversal. In particular parity has determinant $-1$ as a four-vector transformation; all transformations continuously produced from SL(2,C) have determinant $+1$.

Set $c_{\theta }=\cosh (\theta /2)$, $s_{\theta }=\sinh (\theta /2)$ and $N=\mathbf{n}\cdot \mathbf{\sigma }$. The Pauli matrix multiplication law gives $N^2=I$. The proposed boost matrix is $A_B=c_{\theta }I+s_{\theta }N$, with eigenvalues $e^{\pm \theta /2}$, determinant one and $A_B^{†}=A_B$. Split $\mathbf{x}=x_{\parallel }\mathbf{n}+{\mathbf{x}}_{\perp }$. The perpendicular Pauli part anticommutes with $N$, so multiplication givesThis is an active Lorentz boost. The image of a rest worldline has velocity $\mathbf{v}=\tanh \theta \mathbf{n}$, fixing the velocity sign convention.

For two nonzero boosts, write $A^{\prime }=c^{\prime }I+s^{\prime }{\mathbf{n}}^{\prime }\cdot \mathbf{\sigma }$ and $A=cI+s\mathbf{n}\cdot \mathbf{\sigma }$. Their product isThe last term is anti-Hermitian. Thus $A^{\prime }A$ is not Hermitian unless the axes are collinear; a pure boost has Hermitian lifts $\pm A_B$, so it cannot give the same Lorentz transformation. The noncollinear boost obstruction from Pauli products is thereforeThat rotation is a Wigner rotation. A zero-rapidity factor is the trivial exception, regardless of the arbitrary axis assigned to it.

Define $J_i=\frac{1}{2}{ϵ}_{ijk}{M}_{jk}$ and $K_i=M_{0i}$. The printed Lorentz algebra brackets giveSet $L_i=(J_i+i{K}_i)/2$, $R_i=(J_i-i{K}_i)/2$. ThenThis is the chiral decomposition of the complex Lorentz algebra. The two copies are the complexifications of the SU(2) algebras conventionally labelled left and right. They are not two independent compact real subalgebras of the real Lorentz algebra: $L,R$ are complex linear combinations of the real generators. The distinction is needed for noncompact boosts.

To verify the infinitesimal two-component action, putIts trace vanishes by antisymmetry, so $\det A=1+O({\omega }^2)$. The Pauli matrix multiplication law impliesSince $a^{†}=\frac{1}{4}{\omega }^{\mu \nu }{\bar{\sigma }}_{\nu }{\sigma }_{\mu }$, antisymmetrizing this identity yieldsConsequently $AX{A}^{†}=X+{\sigma }_{\mu }{\omega }^{\mu }{}_{\nu }{x}^{\nu }+O({\omega }^2)$, exactly the claimed infinitesimal coordinate transformation with ${\omega }^{\mu }{}_{\nu }={\omega }^{\mu \rho }{\eta }_{\rho \nu }$.

The two Weyl spinor representations transform asBoth maps preserve group multiplication. For $\Lambda =e^{\omega }$, choose its spin lift and write the finite matrices asFor rotations these coincide as the SU(2) doublet; for boosts their generator signs are opposite. Any complex-linear intertwiner would commute with all rotations and hence be scalar by the Schur lemma, but a nonzero scalar cannot intertwine the opposite boost matrices. They are therefore inequivalent. Infinitesimally their $K_i$ matrices are $-i{\sigma }_i/2$ and $+i{\sigma }_i/2$, while their $J_i$ matrices are both ${\sigma }_i/2$. Thus they have chiral labels $(1/2,0)$ and $(0,1/2)$. The finite matrices are representations of the spin cover; choosing $A$ or $-A$ matters for spinors even though their four-vector transformations coincide.

Using the supplied block gamma matrices, the generator of the Dirac spinor isThe antisymmetry of $\omega$ removes the symmetric Clifford part. Hence

There is a conjugation-order error in the last displayed gamma identity in the PDF. The Clifford algebra gives, with ${\gamma }^{\rho }={\eta }^{\rho \sigma }{\gamma }_{\sigma }$,Exponentiating this linear commutator action proves the inverse Lorentz action on gamma matricesThe second identity is the order required for the requested bilinears when $\psi \mapsto S\psi$. An explicit countercheck to the printed order is a positive boost along the third axis: ${\omega }^{03}=-\theta$ giveswhereas the printed right side has a plus sign. This cannot be repaired by dropping index raising; the same raising convention is needed in the preceding coordinate transformation.

The adjoint relation supplied in the question gives ${\Omega }^{†}{\gamma }^0=-{\gamma }^0\Omega$, hence the Dirac spinor pseudo-unitarity identity $S^{†}{\gamma }^{0}S={\gamma }^0$. The Dirac adjoint therefore transforms as $\bar{\psi }\mapsto \bar{\psi }{S}^{-1}$. It now follows that the Dirac scalar bilinear isthe vector isand the antisymmetric second-rank tensor isThese are respectively a Lorentz scalar, Lorentz four-vector and Lorentz tensor. The source's inconsistent gamma identity is replaced by its correct inverse/order pair; all three transformation laws then follow with the stated spinor transformation.

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: