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Electron-neutrino scattering has a charged-current contribution absent for muon neutrinos. Fierz rearrangement combined with the external fermionic sign adds a left-electron current: $c_V\to c_V+1$, $c_A\to c_A+1$. The right-electron coupling stays unchanged; the left-electron coupling increases by two.

A fermionic operator is odd under fermion parity: if $P=(-1{)}^N$ is the parity operator, then $PAP=-A$. A fermionic creation operator or fermionic annihilation operator is an example. Interchanging odd insertions in a time-ordered product contributes the corresponding fermionic sign. Odd parity alone does not imply that every pair of such operators has zero anticommutator.

For a free Dirac field with action $\int \bar{\psi }K\psi$ and independent odd sources coupled as $\bar{\psi }J+\bar{J}\psi$, translation invariance of the Berezin integral and completion of the square give $Z[J,\bar{J}]=Z[0,0]\exp (-i\bar{J}{K}_F^{-1}J)$. Here $K_F^{-1}$ has Feynman i-epsilon prescription boundary conditions, and each bilinear includes its spacetime integrations. A left Grassmann derivative with respect to $\bar{J}$, followed by a right Grassmann derivative with respect to $J$, has normalized value $-i{K}_F^{-1}$ at zero sources. Multiplying by $-1$ removes the two insertion factors $i^2$ and yields the Dirac propagator $S_F=i{K}_F^{-1}$. Keeping source order fixed prevents a spurious fermionic sign.

A Grassmann derivative is an odd differentiation operation on a Grassmann algebra. A left Grassmann derivative obeys ${\partial }_{\theta }^L(ab)=({\partial }_{\theta }^{L}a)b+(-1{)}^{|a|}a({\partial }_{\theta }^{L}b)$ for homogeneous $a$. Right derivatives use the corresponding right-sided product rule. Their placement must be fixed when differentiating fermionic sources in a generating functional, since commuting ordinary derivatives would lose fermionic signs.

A normal-ordered product of free-field oscillator operators places all creation operators to the left of all annihilation operators, including the fermionic sign required to reorder fermions. Its vacuum expectation vanishes if any nontrivial oscillator factors remain. The Wick theorem expresses a free-field time-ordered product through these products and all Wick contractions.

Use the Berezin integral orientation $\int {\prod }_{i}d{\bar{\theta }}_{i}d{\theta }_i{\theta }_1{\bar{\theta }}_1\cdots {\theta }_n{\bar{\theta }}_n=1$. For an invertible matrix $B$, the normalized weight $e^{-\bar{\theta }B\theta }$ has $⟨{\theta }_i{\bar{\theta }}_j⟩=(B^{-1}{)}_{ij}$. Differentiate its determinant with respect to $B_{ji}$ to obtain this formula; two derivatives give the displayed four-point formula. The second pairing carries a fermionic sign. These are ordered contractions, so exchanging two odd factors changes their sign.

In the two-body centre-of-mass frame, spatial reflection reverses the relative coordinate of a quark-antiquark pair. Its orbital state of angular momentum $L$ has orbital parity $(-1{)}^L$, as follows from $Y_{Lm}(-\hat{\mathbf{r}})=(-1{)}^L{Y}_{Lm}(\hat{\mathbf{r}})$. Opposite constituent intrinsic parities multiply to minus one. Thus the meson parity and charge conjugation rules begin withTo obtain charge conjugation, consider a self-conjugate flavor state and write its spin-orbital wavefunction using ordered creation operators $b_{\sigma }^{†}(\mathbf{p})d_{\tau }^{†}(-\mathbf{p})$. Charge conjugation interchanges the quark and antiquark operators. Reordering them back to quark-first order produces a fermionic sign minus one. Interchanging the two spin-$1/2$ labels has eigenvalue $-1$ in the spin singlet state ($S=0$) and $+1$ in the spin triplet ($S=1$), namely $(-1{)}^{S+1}$. Reversing the relative momentum contributes $(-1{)}^L$. ThereforeReciprocal phases for the two conjugate creation operators cancel, so the result is independent of that convention. This applies to a state that is mapped to itself by charge conjugation, not to every electrically neutral flavor state. For example a neutral meson with unequal quark and antiquark flavors can be exchanged with a different antimeson. Charged states have no individual $C$ eigenvalue.

Restrict flavor to the $u,d$ isospin doublet. Its product with the conjugate doublet decomposes as $2\otimes \overline{2}=3\oplus 1$. Convenient normalized flavor states, with irrelevant overall phase choices, areOrbital excitation normally costs energy, so the simplest low-lying quark model states have $L=0$. Coupling the two constituent spins gives $S=0$ or $S=1$, and then $J=S$. The isotriplet with $S=0$ is the pion triplet, while the isotriplet with $S=1$ is the rho meson triplet. The isosinglet with $S=1$ is the nonstrange component of the omega meson. Their quantum numbers areThe charged pions have $J^P=0^{-}$ and the charged rho mesons have $J^P=1^{-}$; charge conjugation exchanges the positive and negative members. The same tensor product also allows an $S=0$, $I=0$ pseudoscalar state, so symmetry alone does not assert that only the listed mesons exist or predict their masses. The physical isosinglet spectrum involves further dynamics and, when strange flavor is included, mixing.

For a vector-meson decay to two spinless pions, conservation of angular momentum forces their relative orbital value to be $L=1$. Their combined parity is $(-1{)}^2(-1{)}^L=-1$, and the neutral charge combination has $C=(-1{)}^L=-1$, matching both ${\rho }^0$ and $\omega$. Thus $C$ and $P$ alone do not distinguish these two decays. The decisive restriction is isospin together with Bose statistics.

Two isotriplet pions have $3\otimes 3=1\oplus 3\oplus 5$, or total $I=0,1,2$. Under exchange the isospin-$I$ state has sign $(-1{)}^{2-I}$: the $I=2$ highest state and its lowered partners are symmetric, the $I=1$ highest combination $(|+0⟩-|0+⟩)/\sqrt{2}$ is antisymmetric, and the singlet $(|+-⟩+|-+⟩-|00⟩)/\sqrt{3}$ is symmetric. The $L=1$ orbital state is antisymmetric. Since pions are spin-zero bosons, their total state must be symmetric, requiring the antisymmetric isospin sector $I=1$. This allows ${\rho }^0\to {\pi }^{+}{\pi }^{-}$ but excludes an isospin-conserving $\omega \to {\pi }^{+}{\pi }^{-}$ amplitude.

There is no analogous obstruction for three pions. In a Cartesian isospin basis $a,b,c$, their totally antisymmetric combination ${ϵ}_{abc}|{\pi }^a{\pi }^b{\pi }^c⟩$ is an $I=0$ state. It can multiply a totally antisymmetric momentum amplitude proportional in the rest frame toMomentum conservation ${\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3=0$ makes this amplitude change sign under every exchange of two pion labels. The product of the flavor and momentum factors therefore obeys Bose statistics. The cross product supplies $J=1$ and an even orbital parity; multiplying by the three negative pion intrinsic parities gives total $P=-1$. In the charged component, charge conjugation exchanges ${\pi }^{+}$ and ${\pi }^{-}$ and changes the momentum factor's sign, giving $C=-1$. Thus $\omega \to {\pi }^{+}{\pi }^{-}{\pi }^0$ is compatible with $I=0$ and $J^{PC}=1^{--}$.

The same distinction is encoded in G parity: $G=C(-1{)}^I$ for the neutral member gives $G_{\pi }=-1$, $G_{\rho }=+1$, $G_{\omega }=-1$. An $n$-pion state has $G=(-1{)}^n$, so the G-parity selection rule for pion multiplicities allows two pions for the rho and three for the omega. The printed prohibition of the omega two-pion mode is an exact-isospin strong-interaction selection rule. In the physical theory, isospin-breaking effects allow a small two-pion mode; it is not an absolute prohibition. The scope of C parity is also stated in the Particle Data Group quark-model review, and the physical two-pion mode is recorded in its omega listing.

Use the Feynman vertex $-i\lambda$ for the scalar Yukawa coupling, the internal scalar propagator $i/(k^2-{\mu }^2+i0)$, incoming factors $u_r(p)$ and outgoing Dirac adjoints ${\bar{u}}_{r^{\prime }}(p^{\prime })$. Define $i\mathcal{M}$ as the coefficient in the connected S-matrix element after removing its overall momentum-conserving Dirac delta function. A direct pairing contributesThe crossed pairing interchanges the identical outgoing fermions and therefore has a relative fermionic sign of minus. With $t=(p-p^{\prime }{)}^2$ and $u=(p-q^{\prime }{)}^2$, the result isIt changes sign under exchange of the two outgoing momenta and spin labels, as required for identical fermions. No fermion spin sum or identical-particle phase-space factor belongs in this amplitude; those enter a relativistic scattering cross-section calculation.

For one Grassmann variable, define the Berezin integral by $\int d\theta 1=0$ and $\int d\theta \theta =1$, extending linearly. Since ${\theta }^2=0$, these rules determine it on every polynomial. With several Grassmann variables, integrate from the rightmost differential first. Equivalently, $\int d{\theta }_n\cdots d{\theta }_1$ extracts the coefficient of ${\theta }_1\cdots {\theta }_n$ in that order. Repeated indices give zero, while permuting distinct generators gives the fermionic sign of that permutation. Hencewhere $\epsilon$ is the Levi-Civita symbol.

For the doubled variables, fix the paired measure orientation byThe exponential in the Grassmann Gaussian integral has a finite Taylor expansion. Only its term of degree $2n$ survives integration. Each barred and unbarred index must occur exactly once; the $n!$ orders of the even bilinears cancel the Taylor factorial. Reordering the unbarred indices gives the determinant permutation sign. ThereforeThis identity holds even for a singular matrix. The normalized expectations below require $B$ invertible.

Put $C=B^{-1}$. Differentiating with respect to $B_{ji}$ inserts $-{\bar{\theta }}_j{\theta }_i={\theta }_i{\bar{\theta }}_j$. The derivative of a determinant then gives the ordered Grassmann Gaussian contractionsDifferentiate once more, using $\partial C_{ij}/\partial B_{lk}=-C_{il}{C}_{kj}$. The result isThe minus sign is the interchange sign in the fermionic Wick theorem. In particular $⟨\bar{\theta }M\theta ⟩=-\mathrm{tr}(MC)$. Reordering each barred-unbarred pair in the product and contracting the indices givesThe covariance of Grassmann bilinears is the connected second term. Its Feynman diagram is one oriented fermion loop through both insertions; the first term has two separate fermion loops, one at each insertion. Each closed fermion loop contributes a minus sign, so two loops give a plus and one gives a minus.

For the Dirac field, write $D=\gamma \cdot \partial +m$ and use signature $(-,+,\dots ,+)$, consistent with the denominator in the question. A finite regulator turns the free fermionic path integral into the same Gaussian problem with $B=iD$, because $e^{iS}=e^{-i\bar{\psi }D\psi }$. The time-ordered product covariance is consequently $(i{D}_F{)}^{-1}=-i{D}_F^{-1}$, where the subscript specifies the Feynman i-epsilon prescription.

With the Fourier transform convention $\psi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\tilde{\psi }(p)$, the derivative kernel is $i\gamma \cdot p+m$. The gamma matrices obeysince their anticommutator makes $(\gamma \cdot p{)}^2=p^{2}I$. Thus the mostly-plus Dirac propagator with a real derivative kernel isFor $p^2=-p_0^2+{\mathbf{p}}^2$, the denominator is $p_0^2-{\mathbf{p}}^2-m^2+i0$. Its positive-energy pole is below and negative-energy pole above the real axis, giving time ordering. Here $S_F$ is defined without the leading factor $i$, as in the question; conventions that include that factor in the name Dirac propagator give the same covariance.

Label the incoming momenta and spin states $(p_1,r_1),(p_2,r_2)$ and the outgoing ones $(p_3,r_3),(p_4,r_4)$, with $p_1+p_2=p_3+p_4$. Write $u_i=u^{r_i}({\mathbf{p}}_i)$ and $v_i=v^{r_i}({\mathbf{p}}_i)$. The Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. At lowest order, each connected Feynman diagram has two pseudoscalar Yukawa interaction vertices and one internal scalar propagator.

Define the scattering amplitude by $S_{fi}^{\mathrm{connected}}=i(2\pi {)}^4{\delta }^4(p_1+p_2-p_3-p_4)\mathcal{M}$. Fix the external Fock state ordering as $b_1^{†}{b}_2^{†}|0⟩$ to $b_3^{†}{b}_4^{†}|0⟩$ for the two-fermion process. The $t$ exchange connects $1$ to $3$ and $2$ to $4$; the $u$ exchange connects $1$ to $4$ and $2$ to $3$. The two Feynman vertices contribute $(-i\lambda {)}^2$, and the internal scalar propagator contributes $i$ divided by its denominator. Exchanging the two identical final fermions supplies a relative fermionic sign. ThusThis changes sign when the two outgoing labels are exchanged, as identical-fermion antisymmetry requires. There is no $s$ annihilation diagram for two incoming fermions: the interaction preserves Dirac fermion number conservation.

For the fermion-antifermion process take initial state $b_1^{†}{d}_2^{†}|0⟩$ and final state $b_3^{†}{d}_4^{†}|0⟩$. The $t$ exchange uses bilinears $({\bar{u}}_3{\gamma }^5{u}_1)({\bar{v}}_2{\gamma }^5{v}_4)$, and the $s$ annihilation uses $({\bar{v}}_2{\gamma }^5{u}_1)({\bar{u}}_3{\gamma }^5{v}_4)$. To fix the sign without a guess, expand the current $J=:\bar{\psi }{\gamma }^5\psi :$ in creation and annihilation operators. Its relevant terms have the signsThe antifermion sign of a normal-ordered bilinear makes the exchange contraction negative, whereas the product of annihilation and creation contractions is positive for the specified external ordering. Multiplying by $(-i\lambda {)}^{2}i=-i{\lambda }^2$ therefore givesA different common phase convention for an external state can reverse the overall amplitude, but never the relative sign of its two contributions. Both results are tree scattering with pseudoscalar exchange and are of order ${\lambda }^2$.

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.

Use the free-boson operator product expansionWrite $E_p=:e^{ip\cdot X}:$ and $ϵ=z-w$. The needed Wick contractions are $\partial X^{\mu }(z)\partial X^{\nu }(w)\sim -{\alpha }^{\prime }{\eta }^{\mu \nu }/(2{ϵ}^2)$ and $\partial X^{\mu }(z)E_p(w)\sim -i{\alpha }^{\prime }{p}^{\mu }{E}_p(w)/(2ϵ)$. For $V_{\zeta }={\zeta }_{\mu }:\partial X^{\mu }{E}_p:$, the cross double contraction in the worldsheet stress tensor has a cubic pole: either of its two derivatives can contract with the derivative in $V_{\zeta }$, while the other contracts with $E_p$. Including the factor $-1/{\alpha }^{\prime }$ givesA primary operator has no cubic pole. Hence the first vertex is primary precisely whenThe antiholomorphic conformal weight is $\bar{h}={\alpha }^{\prime }{p}^2/4$. Primarity alone puts no mass-shell condition on $p$.

For $W_{\zeta }={\zeta }_{\mu \nu }:\partial X^{\mu }\bar{\partial }{X}^{\nu }{E}_p:$, the holomorphic cubic pole is $-i{\alpha }^{\prime }{p}^{\mu }{\zeta }_{\mu \nu }:\bar{\partial }{X}^{\nu }{E}_p:/(2{ϵ}^3)$. The antiholomorphic operator product expansion gives the analogous pole involving $p^{\nu }{\zeta }_{\mu \nu }$. Thus the transversality of derivative-exponential primary vertices requiresThere is no trace-free requirement for primarity. A physical integrated string vertex operator must additionally have conformal weights $(1,1)$; for this vertex that gives $p^2=0$.

Now keep the fermionic signs in the real chiral Majorana fermion calculation. Contracting the external $\psi (w)$ through $:\psi (z)\partial \psi (z):$ givesMultiplying by $-1/2$ and Taylor expanding $\psi (z)$ at $w$ producesThis proves that $\psi$ is a primary operator with weight $h_{\psi }=1/2$. To find the central charge, in $T(z)T(w)$ the two double-contraction pairings contribute, before the overall factor $1/4$,The single Wick contractions give $2T(w)/{ϵ}^2+\partial T(w)/ϵ$, soComparing the leading term with $c/(2{ϵ}^4)$ proves $c_{\psi }=1/2$, as in the free chiral Majorana fermion conformal field theory.

With four embedding bosons and the ordinary reparameterization bc ghost system, cancellation of the worldsheet Weyl anomaly in each chirality requiresThus there are 44 real left-moving and 44 real right-moving fermions, equivalently 44 two-dimensional nonchiral Majorana fermions. These are internal free fermions in a four-dimensional bosonic string. Adding internal matter to the bosonic string does not introduce local worldsheet supersymmetry or its superghosts.

For the requested polynomial string vertex operators, work in the unprojected Neveu–Schwarz sector. A full model also chooses spin structures and a consistent projection, which the central-charge condition alone does not specify. If the holomorphic and antiholomorphic oscillator contributions to the conformal weights are ${\ell }_L,{\ell }_R$, physicality requiresTherefore only equal levels at most one can contribute to the requested nonpositive bosonic string mass spectrum.

With zero fermion insertions, level zero gives the tachyon vertex operator $E_p$, a scalar with $M^2=-4/{\alpha }^{\prime }$. Level one gives the massless closed-string vertex operator ${\zeta }_{\mu \nu }:\partial X^{\mu }\bar{\partial }{X}^{\nu }{E}_p:$, with the transverse conditions above and the string-state gauge redundancy that removes longitudinal polarizations. The physical transverse polarization tensor decomposes into a symmetric trace-free graviton, an antisymmetric Kalb–Ramond field, and a scalar dilaton. In four spacetime dimensions these have respectively two, one, and one physical polarizations; the two-form can be dualized to a scalar.

With two fermion insertions there are two possibilities. One fermion in each chirality givesThese are $44^2=1936$ scalar tachyons before any projection. There is no spacetime derivative or spacetime polarization to impose a vector transversality condition.

Two different fermions in the same chirality form an antisymmetric fermion bilinear, a weight-one current $J^{ij}=:{\psi }^i{\psi }^j:$; equal-species undifferentiated bilinears vanish. Closed-string level matching then requires a weight-one bosonic derivative in the opposite chirality. The resulting string vertex operators areThey have $M^2=0$ and are gauge vectors, with $p\cdot A^{ij}=0$ and $A_{\mu }^{ij}\sim A_{\mu }^{ij}+p_{\mu }{\lambda }^{ij}$, and likewise for $\tilde{A}$. Each chirality supplies $\left(\genfrac{}{}{0ex}{}{44}{2}\right)=946$ gauge vectors. The fermion bilinears generate the two $\mathfrak{so}(44)$ current algebras associated with the Special orthogonal Lie algebra, giving the unprojected gauge group $SO(44{)}_L\times SO(44{)}_R$. The tempting vertex $:{\psi }^i{\psi }^j{E}_p:$ alone fails closed-string level matching because its weights are $(1,0)$ before the common momentum contribution. Additional derivatives raise the levels above the nonpositive-mass range; a product of a left and a right bilinear has four fermion insertions and is outside this requested sector. This completes the zero- and two-fermion vertices of an internally fermionized bosonic string; a specified global projection can remove some of the listed states.

Expanding the covariant derivative isolates the ghost-gluon vertex interactionThe derivative acts on the antighost. With all momenta incoming, let its momentum be $k$, the gluon's Lorentz index be $\mu$, and their color indices be $(a,b,c)$ in the order $(\bar{c},A,c)$. The convention $e^{-ikx}$ replaces ${\partial }^{\mu }$ by $-i{k}^{\mu }$, and multiplication by $i$ from $e^{iS}$ givesMomentum conservation is implicit. The structure constants reflect the Adjoint representation, and the antighost momentum follows directly from its differentiated field. Reversing the momentum convention changes the displayed vertex sign consistently; exchanging the ghost and antighost source order also carries a fermionic sign.

Keep the mostly-plus Minkowski metric and the Fourier transform $e^{ip\cdot x}$. Write $/p={\gamma }^a{p}_a$, with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$. To fix the otherwise unspecified phase of ${\gamma }^5$ and the Dirac adjoint, takewhere ${\Gamma }^a$ are ordinary mostly-minus gamma matrices. Then $({\gamma }^5{)}^2=-1$, $\{{\gamma }^5,{\gamma }^a\}=0$, and the fermionic time-derivative term is $i{\psi }^{†}{\partial }_t\psi$. In these conventions the interaction with real $g$ is Hermitian: in mostly-minus notation it is $ig{\bar{\psi }}_{\mathrm{std}}{\Gamma }^5\psi \varphi$. If one instead calls $i{\gamma }^5$ the square-one chirality matrix, its coefficient must be $-ig$ to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding $e^{i{S}_{\mathrm{int}}}$ yields the following Feynman rules with relativistically normalized external states:

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.

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.