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Use mostly-minus metric signature, $\eta =\mathrm{diag}(1,-1,-1,-1)$. The Weyl sigma matrices are ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, with the three Pauli matricesThe Pauli matrix multiplication law ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ immediately gives ${\sigma }^{\mu }{\bar{\sigma }}^{\nu }+{\sigma }^{\nu }{\bar{\sigma }}^{\mu }=2{\eta }^{\mu \nu }I$: the $00$ case is $2I$, the mixed cases cancel, and the spatial cases give $-2{\delta }_{ij}I$. The spinor indices of ${\sigma }_{\alpha \dot{\beta }}^{\mu }$ express the correspondence between a Lorentz four-vector and a Hermitian $2\times 2$ matrix. For $X=x^{0}I+x^i{\sigma }_i$, $\det X=(x^0{)}^2-|\mathbf{x}{|}^2$; $X\mapsto MX{M}^{†}$ preserves this determinant for $M\in \mathrm{SL}(2,\mathbb{C})$, yielding the Lorentz spinor double cover.

In four-dimensional N=1 supersymmetry, the odd supercharges $Q_{\alpha }$ and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$ transform as conjugate Weyl spinors. Their Super-Poincaré algebra isThe ordinary translation and Lorentz generators satisfy the Poincare algebra, and the Lorentz generators act on the supercharges by the corresponding group representations on Weyl spinors. Tracing the mixed anticommutator gives $P_0=\frac{1}{4}{\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}$, proving energy positivity in global supersymmetry. For a massive rest-frame state, $P_{\mu }=(M,0,0,0)$, the rescaled charges $Q_{\alpha }/\sqrt{2M}$ obey the canonical anticommutation relations. Acting with their adjoints builds a supermultiplet containing states of alternating fermion parity.

Use the printed interaction, so the Dyson series contributes $-ig$ per vertex before its spinor matrix. The momentum-space Feynman rules areEach interaction vertex has one pseudoscalar line and one incoming and one outgoing fermion-number line. Conserve four-momentum at every vertex, integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$, and divide by any Feynman-diagram symmetry factor. A closed fermion loop contributes an extra minus sign. The external wavefunctions are $u$ for incoming fermions, $\bar{u}$ for outgoing fermions, $\bar{v}$ for incoming antifermions and $v$ for outgoing antifermions; spinor matrices are multiplied in order along the fermion line. An external pseudoscalar has unit wavefunction factor in standard relativistic normalization. Relative fermionic signs also occur when the external contractions are permuted.

The leading scattering amplitudes are of order $g^2$. Label the incoming momenta $p_1,p_2$ and outgoing momenta $p_3,p_4$, all with positive energy and $p_i^2=m^2$. Use the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. The tree scattering with pseudoscalar exchange is shown below; dashed edges are pseudoscalar propagators, and arrows indicate fermion-number flow, opposite to the motion of an antifermion.

Define the scattering amplitude by $⟨f|S-1|i⟩=i(2\pi {)}^4{\delta }^4(p_3+p_4-p_1-p_2)\mathcal{M}$. Use standard relativistic normalization for external states and spinors. To fix even the overall signs, suppress the positive state-normalization factors and write the creator order of the two-fermion states as $|i⟩=b_1^{†}{b}_2^{†}|0⟩$, $|f⟩=b_3^{†}{b}_4^{†}|0⟩$, and the fermion-antifermion states as $|i⟩=b_1^{†}{d}_2^{†}|0⟩$, $|f⟩=b_3^{†}{d}_4^{†}|0⟩$. Spin labels on the $u_i,v_i$ are implicit.

For two fermions, the direct $t$ graph connects $1\to 3$ and $2\to 4$, while the exchanged $u$ graph connects $1\to 4$ and $2\to 3$. The canonical anticommutation relations give signs $+$ and $-$ for these contractions in the chosen state order. Multiplication of the two vertices and the scalar propagator gives $(-ig{)}^{2}i=-i{g}^2$. HenceInterchanging the labels of either pair of identical external fermions reverses this amplitude, as required for the antisymmetric states of fermionic Fock space.

For a fermion and an antifermion, the graphs are $t$ exchange and $s$ annihilation. Their signs can be derived without guessing an analogy with the first process. The normal-ordered product of the pseudoscalar bilinear contains terms of the formsThe minus in the second term is the antifermion sign of a normal-ordered bilinear. Between the chosen external states, exchange has sign $-$ and pair annihilation followed by pair creation has sign $+$. ThereforeChanging an external-state phase can change the common overall sign; it cannot change the relative minus between the two channels. Both amplitudes use the coefficient exactly as printed. For the Hermitian convention $g=i{g}_P$ discussed in part (a), substitute $g^2=-g_P^2$ throughout, or equivalently use vertex $g_P{\gamma }^5$ from the outset.

Work initially in units $G=c=\hslash =k_B=1$ and consider gravitational collapse settling to a nonextremal Schwarzschild black hole. Classically, outgoing rays sufficiently close to the forming event horizon suffer an arbitrarily large redshift. Quantum mechanically, the meaning of positive frequency at past null infinity differs from that at future null infinity; a Bogoliubov transformation between these two mode descriptions can therefore mix creation operators and annihilation operators. Hawking's argument computes that mixing and obtains a thermal occupation, even when the incoming state has no particles.

To see why the frequency mixing is universal, use final Schwarzschild time $t$, the Schwarzschild tortoise coordinate $r_{\ast }$, and retarded time $u=t-r_{\ast }$. Near the final event horizon, $r_{\ast }\sim (2\kappa {)}^{-1}\log (r-2M)$, with Schwarzschild surface gravity $\kappa =1/(4M)$. On a fixed advanced-time section, $u=v-2r_{\ast }$, so $r-2M$ is proportional to $e^{-\kappa u}$. Tracing a ray back through the regular collapsing region maps this small separation smoothly to its separation from the last escaping incoming ray. If $U$ is an affine incoming null coordinate and $U_H$ labels that limiting ray, the Hawking exponential ray map isThe logarithm at a simple horizon root fixes the exponent. Smooth propagation through the earlier collapse changes $A$ and phases but not the leading exponential redshift.

For a massless bosonic test scalar field, an outgoing mode of frequency $\omega >0$ at future null infinity behaves as $e^{-i\omega u}$. Trace it backward in the high-frequency geometric optics approximation. Its relevant incoming dependence is, up to normalization and a phase,Here $\Theta$ is the Heaviside step function. Incoming rays with $U>U_H$ do not escape. This truncated logarithmic phase is not purely positive frequency with respect to $U$. Decomposing it into incoming positive-frequency solutions $e^{-i{\omega }^{\prime }U}$ and negative modes $e^{+i{\omega }^{\prime }U}$ gives the Bogoliubov transformation. With $x=U_H-U$, the two relevant Fourier transform integrals areThe former is the positive-frequency coefficient and the latter is the negative-frequency coefficient; the Klein-Gordon inner product introduces the same frequency normalization into their modulus ratio. By the defining integral of the gamma function and analytic continuation in its Laplace parameter,For a complex number $z$ in the right half-plane, $|z^{-1-ia}|=|z{|}^{-1}{e}^{a\arg z}$. As $ϵ\downarrow 0$, the two arguments tend to $+\pi /2$ and $-\pi /2$. This gives the thermal ratio of Hawking Bogoliubov coefficients:The regulator fixes the branches and is essential to the sign of the thermal exponent.

Expand the outgoing annihilator as $b=\int d{\omega }^{\prime }({\alpha }_{\omega {\omega }^{\prime }}^{\ast }{a}_{{\omega }^{\prime }}-{\beta }_{\omega {\omega }^{\prime }}^{\ast }{a}_{{\omega }^{\prime }}^{†})$, including a complete set of incoming channels. The canonical identities for a bosonic Bogoliubov transformation express $[b,b^{†}]=1$. For normalized outgoing wave packets, they give $\int (|\alpha {|}^2-|\beta {|}^2)d{\omega }^{\prime }=1$. In the incoming Fock vacuum, the outgoing mean of the particle number operator is $N_{\omega }=\int |\beta {|}^{2}d{\omega }^{\prime }$. Combining the two identities with the thermal ratio, in the late-time narrow-frequency packet limit, yieldsWave packets avoid interpreting the divergent normalization of infinitely long continuum modes as a finite particle count. The temperature is the Hawking temperature, and the occupation is the zero-chemical-potential Bose-Einstein distribution. Restoring constants, the Schwarzschild black hole temperature is $T_H=\hslash c^3/(8\pi GM{k}_B)$ for physical mass $M$.

The preceding thermal occupation describes the horizon-originating channel before exterior scattering. The exterior angular and curvature potential partly reflects it. For a scalar field, the number flux at future null infinity is consequentlywhere the greybody factor ${\Gamma }_{\ell }$ is the transmission probability of a partial wave. Multiplication by $\omega$ gives its energy flux. Thus the asymptotic spectrum has the thermal denominator but is not an exact blackbody spectrum. For fermionic fields the canonical anticommutation relations replace the minus sign in the occupation denominator by a plus sign, at the same Hawking temperature.

The relevant state is the collapse vacuum for Hawking radiation: no incoming thermal bath at past null infinity, and regular short-distance behavior in freely falling coordinates at the forming event horizon. It predicts an outgoing flux, unlike equilibrium with a thermal bath on an eternal black hole. The derivation uses quantum field theory in curved spacetime on a prescribed classical geometry; it does not require outgoing particles to follow forbidden classical paths out of the interior. Positive energy carried to infinity is accompanied, in the semiclassical description, by negative Killing-energy flux into the black hole. Including that flux in slow backreaction decreases its mass, giving black-hole evaporation.

The approximation applies to the late radiation of a large nonextremal black hole while its surface gravity changes slowly on a time scale ${\kappa }^{-1}$. Backward propagation involves very high locally measured frequencies, so the calculation assumes the usual regular short-distance quantum-field state. It determines the leading radiation and temperature, not the Planck-scale endpoint of black-hole evaporation. The exponential horizon redshift forces positive/negative frequency mixing with a thermal ratio, giving outgoing Hawking radiation at $T_H=\kappa /(2\pi )$.

A supercharge is an odd Weyl spinor generator. The mixed anticommutator has one undotted and one dotted index, hence transforms as a Lorentz four-vector. In the minimal Super-Poincaré algebra, its only spacetime generator is the four-momentum, so covariance fixes it to $c{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$. Normalizing the charges canonically fixes $c=2$:The number two is a normalization, not something Lorentz covariance alone could fix under $Q\mapsto aQ$.

For an explicit closure calculation, use left Grassmann derivatives in superspace and the differential translation convention $P_{\mu }=i{\partial }_{\mu }$:Each of the two derivative-coordinate cross terms contributes $i{\sigma }^{\mu }{\partial }_{\mu }$; the other terms cancel by anticommutativity. This proves the displayed coefficient. The same calculation gives $\{Q,Q\}=\{\bar{Q},\bar{Q}\}=0$ and $[P_{\mu },Q]=[P_{\mu },\bar{Q}]=0$. There is no scalar central charge for a single supersymmetry label: symmetry of the anticommutator is incompatible with a term proportional only to the antisymmetric ${ϵ}_{\alpha \beta }$.

In the massive rest frame $P_{\mu }=(m,0,0,0)$, the mixed relation is $\{Q_{\alpha },Q_{\beta }^{†}\}=2m{\delta }_{\alpha \beta }$. These positive canonical anticommutation relations supply the input for the normalization and finite fermionic Fock space in the question-level continuation.

Write $q_1={\bar{Q}}^{\dot{1}}$ and $q_2={\bar{Q}}^{\dot{2}}$. In the Jacobi-consistent convention just fixed,Applying the commutator product rule twice givesHere ${\sum }_i{\sigma }_i^2=3I$ produces the constant term; moving the $J_i$ to the right accounts for it. ConsequentlyThus the physically consistent numerical coefficients areIf one mechanically retains the literal positive-sign column transformation instead, the coefficient tuple would be $(\frac{1}{2},\frac{3}{4},1,-i,1,\frac{3}{4},1,i,-1)$. It cannot describe a unitary multiplet with the supplied rotation algebra, by the explicit Jacobi counterexample in part (a).

Take the spin-zero Clifford vacuum $|\Omega ⟩$, normalized to one and annihilated by the two annihilation charges. This is a lowest Fock state of a positive-mass representation, not a zero-energy vacuum annihilated by every charge. For the first one-charge state,The ladder commutators are $[J_{+},q_1]=-q_2$, $[J_{-},q_1]=0$, $[J_{+},q_2]=0$, $[J_{-},q_2]=-q_1$. Therefore these two states are the $j=1/2$ pair. A phase choice $|\frac{1}{2},+\frac{1}{2}⟩=-N{q}_2|\Omega ⟩$ makes the usual positive ladder coefficient hold.

Raising the dotted index only changes the oscillator basis by the unitary antisymmetric epsilon matrix, so $\{q_a,q_b^{†}\}=2m{\delta }_{ab}$. Since $q_a^{†}|\Omega ⟩=0$,The positive convention would formally label this state with $j_3=+1/2$ while $J_{+}$ acts nontrivially on it, contradicting the highest-weight ladder rule. This supplies an independent state-level check of the needed sign correction.

The two-charge state is a rotation singlet. Indeed, using $q_1^2=q_2^2=0$ and $q_2{q}_1=-q_1{q}_2$, the product rule gives $[J_i,q_1{q}_2]=0$ for every $i$: only the sum of the two diagonal spinor coefficients survives, and their trace is zero. Hence the requested commutators and quantum numbers areIts squared norm is $(2m{)}^2$ before multiplication by $N^4$, by the two oscillator anticommutators.

The spin-zero massive N=1 supermultiplet therefore consists of $|\Omega ⟩$, $N{q}_1|\Omega ⟩$, $N{q}_2|\Omega ⟩$ and $N^2{q}_1{q}_2|\Omega ⟩$. Their spins are $0,1/2,1/2,0$, all with the same mass $m$. Starting with an even vacuum, the two spin-zero states are bosonic and the two one-charge states fermionic. They are the two real scalar degrees of freedom and the two on-shell fermion polarizations of a massive chiral multiplet. Any product of three creation charges repeats an index and vanishes; annihilation charges reduce to these four states by the canonical anticommutation relations. There are exactly four independent states.

Put $n_n=c_n^{†}{c}_n$ and $P_n={\prod }_{m<n}(1-2n_m)$. The Jordan–Wigner transformation is $S_n^{+}=P_n{c}_n$, $S_n^{-}=P_n{c}_n^{†}$ and $S_n^z=1/2-n_n$. Since ${\sigma }_n^x=P_n(c_n+c_n^{†})$, $P_{n+1}=P_n(1-2n_n)$ and $(c_n+c_n^{†})(1-2n_n)=c_n^{†}-c_n$, adjacent bonds becomeThus, apart from the end bond,The end bond contains the global fermion parity and sets the sector-dependent periodic or antiperiodic fermion modes. It contributes an order-one boundary term, which is negligible for the thermodynamic energy density; it is not identically zero for the finite spin chain.

Choose the discrete Fourier transform convention $c_n=N^{-1/2}{\sum }_k{e}^{-ikn}{c}_k$. Hopping gives $-2J\cos k{c}_k^{†}{c}_k$. Opposite-momentum pairing gives $iJ\sin k(c_k^{†}{c}_{-k}^{†}+c_k{c}_{-k})$, using the canonical anticommutation relations to antisymmetrize the coefficient. ThereforeFor the Ising-chain Nambu spinor ${\Psi }_k=(c_k,-i{c}_{-k}^{†}{)}^T$, expansion ofrecovers every term: the diagonal contributes $J(g-\cos k)(n_k+n_{-k}-1)$, and the two off-diagonal entries supply the pairing. Since ${\sum }_k\cos k=0$, the constant is $-JgN$. Reversing the Fourier sign changes the pairing convention; the specified sign makes the displayed matrix agree directly.

Write $a_k=g-\cos k$, $s_k=\sin k$ and $r_k=\sqrt{a_k^2+s_k^2}$. The Ising-chain Bogoliubov diagonalization uses the eigenvalues $\pm r_k$ of the two-by-two Nambu matrix. For distinct paired modes $k\ne -k$, choose $\cos {\theta }_k=a_k/r_k$, $\sin {\theta }_k=-s_k/r_k$ and ${\theta }_{-k}=-{\theta }_k$. The rotation $U_k=e^{-i{\theta }_k{\sigma }_y/2}$ satisfies $U_k^{†}{M}_k{U}_k=r_k{\sigma }_z$, and its first transformed component isThe opposite signs of the paired angles preserve the canonical anticommutation relations. Self-paired spinless fermion modes have zero pairing and are treated directly as occupied or empty number levels; they do not require this paired-angle formula.

Both $k$ and $-k$ occur in the sum, so the physical quasiparticle coefficient is $2J{r}_k$, not $J{r}_k$. ConsequentlyThis is the bulk quadratic result under the stipulated boundary simplification. Restoring the finite-chain parity sectors adjusts allowed modes and global excitation constraints. In the thermodynamic limit, the ground-state energy per site isFor $g\ge 0$, the bulk quasiparticle gap is $2J|g-1|$. The requested three regimes are shown in the original sketch and discussed separately below.

Use the metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and the usual Dirac adjoint $\bar{\psi }={\psi }^{†}{\gamma }^0$.

Field transformation. The mode expansion of a Dirac field isCharge conjugation exchanges particle and antiparticle operators. Choose their phases so thatThe conjugate relations for creation and annihilation operators preserve their canonical anticommutation relations. Thus the transformed expansion is ${\eta }_C$ times the same integral with $a_s{u}^s$ replaced by $b_s{u}^s$ and $b_s^{†}{v}^s$ replaced by $a_s^{†}{v}^s$. On the other hand, transposing the adjoint expansion gives $b_{s}C{\bar{v}}^{sT}{e}^{-ipx}+a_s^{†}C{\bar{u}}^{sT}{e}^{ipx}$. The given spinor identities identify these expressions, proving

Symmetry of the spin matrix. Transposing $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$ gives ${\gamma }^{\mu }{C}^T=-C^T{\gamma }^{\mu T}$. Substitute ${\gamma }^{\mu T}=-C^{-1}{\gamma }^{\mu }C$ and multiply on the right by $C^{-1}$:The complex four-dimensional Clifford algebra representation generated by the gamma matrices is irreducible; equivalently its sixteen independent gamma products span the full matrix algebra. The Schur lemma therefore gives $C^T{C}^{-1}=zI$. Hence $C^T=zC$. Transposing once more gives $C=z^{2}C$, so $z^2=1$ andThis is the transpose symmetry of a charge-conjugation matrix. The given antisymmetric choice is consistent with this conclusion.

Equation of motion. Taking the adjoint of the free Dirac equation yields $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Its transpose, multiplied by $C$, isThe defining matrix relation converts this intoThus the charge conjugation of a Dirac field maps free solutions to free solutions. With an external electromagnetic field, its charge would also reverse; no such background is present here.

Chirality. Define the chiral projectors $P_L=(1-{\gamma }^5)/2$ and $P_R=(1+{\gamma }^5)/2$. The operator $\hat{C}$ acts on the field operators and commutes with these numerical matrices:The transformed component is therefore left-chiral, as requested. There is a useful distinction: the algebraic charge conjugate of a left-chiral spinor is right-chiral. Indeed $C{\gamma }^{5T}{C}^{-1}={\gamma }^5$ and $\overline{P_L\psi }=\bar{\psi }{P}_R$ giveThis is the operator conjugation of a chiral field component; it avoids confusing projection after operator conjugation with charge conjugation of the already projected spinor.

Baryogenesis. Consider equal initial populations of a heavy particle $X$ and its antiparticle, with two baryon-number-violating decay channels carrying different final baryon numbers $b_1,b_2$. Let $r$ and $\bar{r}$ be the branching fractions into a channel and its antiparticle channel. The net baryon number produced per initial particle-antiparticle pair isExact charge conjugation symmetry equates the conjugate branching fractions and makes this vanish. Exact CP symmetry also equates the fully summed conjugate decay rates, because parity reverses momenta and helicities without changing baryon number. Thus both violation of charge conjugation and CP violation are necessary for generating an asymmetry from a symmetric initial state.

Violation of charge conjugation alone can generate a helicity asymmetry while leaving total baryon number zero. For example, with $L,R$ denoting helicities, CP symmetry may enforce $\Gamma (X\to q_L{q}_L)=\Gamma (\bar{X}\to {\bar{q}}_R{\bar{q}}_R)$ and the analogous relation with $L,R$ exchanged, even when same-helicity charge-conjugate rates differ. The CP rate cancellation in baryogenesis means that summing over helicities cancels the baryon asymmetry. These are the symmetry requirements in the Sakharov conditions; baryon-number violation and departure from thermal equilibrium are also needed in the conventional decay mechanism.

The canonical anticommutation relations are preserved precisely when the real matrix acting on $({\gamma }_{k\uparrow },{\gamma }_{-k\downarrow }^{†})$ is orthogonal. Directly,The other relations are identical or vanish between disjoint pairs. Hence $u_k^2+v_k^2=1$, and one may write $u_k=\cos {\theta }_k$, $v_k=\sin {\theta }_k$. This fermionic Bogoliubov transformation uses an ordinary rotation; bosonic mixing instead preserves a difference of squared coefficients.

The exponential acts on the Fock vacuum: $|\psi ⟩=e^{\zeta {\sum }_n{\psi }_n{a}_n^{†}}|0⟩$. For bosons, ${\psi }_n$ are commuting complex numbers. Since $[a_n,{\sum }_m{\psi }_m{a}_m^{†}]={\psi }_n$, commuting $a_n$ through the exponential gives $a_n|\psi ⟩={\psi }_n|\psi ⟩$. These are unnormalized bosonic coherent states.

For fermions, the labels are independent odd Grassmann variables, which anticommute with one another and with the fermionic operators. For one mode, the fermionic coherent state is $|\psi ⟩=(1-\psi a^{†})|0⟩$. The canonical anticommutation relations give $a(-\psi a^{†})|0⟩=\psi |0⟩=\psi |\psi ⟩$, using ${\psi }^2=0$. The even factors $1-{\psi }_n{a}_n^{†}$ commute between modes, so the same argument applies to every $n$. A Grassmann eigenvalue is a formal extension of the state space, not an ordinary complex eigenvalue of the nilpotent annihilation operator.

With dual state $⟨\psi |=⟨0|e^{\zeta {\sum }_n{a}_n{\bar{\psi }}_n}$, both cases obey $⟨\psi |{\psi }^{\prime }⟩=e^{{\sum }_n{\bar{\psi }}_n{\psi }_n^{\prime }}$.

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: