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Direct multiplication of the gamma matrices in the chiral representation givesFor example ${\sigma }^1{\sigma }^2{\sigma }^3=i{I}_2$ by the Pauli matrix multiplication law, and the block multiplications give the displayed signs. Every ${\gamma }^{\mu }$ is off-diagonal in this representation, whereas the chirality matrix has opposite diagonal blocks. Therefore ${\gamma }^5{\gamma }^{\mu }+{\gamma }^{\mu }{\gamma }^5=0$ for all $\mu$.

The left-handed chirality condition makes the plane-wave coefficient $\lambda (p)=\left(\genfrac{}{}{0ex}{}{\chi }{0}\right)$. With $p\cdot x=E{x}^0-\mathbf{p}\cdot \mathbf{x}$, the Dirac equation givesSince $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^2{I}_2$, its determinant is $E^2-|\mathbf{p}{|}^2$. A nonzero spinor therefore exists exactly whenFor nonzero momentum, put $r=|\mathbf{p}|$ and write its direction as $(\sin \vartheta \cos \phi ,\sin \vartheta \sin \phi ,\cos \vartheta )$. Normalized eigenvectors of $\mathbf{\sigma }\cdot \hat{\mathbf{p}}$ areThe general nonzero solution is $\lambda (p)=C\left(\genfrac{}{}{0ex}{}{{\chi }_{-}}{0}\right)$ for $E=r$ and $\lambda (p)=C\left(\genfrac{}{}{0ex}{}{{\chi }_{+}}{0}\right)$ for $E=-r$, with arbitrary nonzero complex normalization $C$. At $p=0$ any constant two-component $\chi$ solves the equation, but the momentum direction and helicity are undefined. For the physical positive-energy branch this is the left-handed massless plane-wave spinor.

A rotation by angle $\alpha$ about $\hat{\mathbf{p}}$ acts on the two-component spinor asThe plane-wave argument is unchanged because the rotation fixes $\mathbf{p}$, and $S(R){\chi }_{-}=e^{i\alpha /2}{\chi }_{-}$. Thus the positive-energy solution has helicity $h=-1/2$, where a helicity-$h$ state has rotation phase $e^{-i\alpha h}$. The negative-energy coefficient has the opposite spinor phase.

In a quantized complex left-handed Weyl field, the positive-frequency term annihilates particles and the negative-frequency term creates antiparticles. Reinterpreting negative energy also reverses its spatial momentum; for the positive physical momentum the two spinor coefficients have the same left-chiral kernel. Because one multiplies an annihilation operator and the other a creation operator, the corresponding particle and antiparticle state phases are conjugate. The helicity content of a quantized Weyl field is consequentlyThere is one particle spin state at each nonzero momentum, rather than the two helicities supplied by a full massless Dirac field. The right-handed Weyl field interchanges these signs. This is the massless relation between chirality and helicity; it does not identify them for a massive field.

At momentum transfers small compared with $M_W$, the two weak charged current vertices each carry $g/(2\sqrt{2})$ when written with $(1-{\gamma }_5)$, and the W boson propagator reduces to a local factor proportional to $1/M_W^2$. Matching their product to the Fermi interaction gives, at tree level,The Fermi constant is thus determined by the electroweak coupling and symmetry-breaking scale; the four-fermion description is a low-energy effective interaction.

Only the cross term $J_L^{†}{J}_H$ can annihilate the pion and create the stated charged-lepton/antineutrino pair. Its leptonic matrix element is ${\bar{u}}_e(k){\gamma }^{\mu }(1-{\gamma }_5)v_{\nu }(q)$. Lorentz covariance permits a vacuum-to-spin-zero current matrix element only proportional to $p^{\mu }$. At rest this has only a temporal component. The vacuum is parity-even and the pion is parity-odd, while $V_H^0$ is parity-even, so $⟨0|V_H^0|\pi ⟩$ equals its own negative and vanishes. Lorentz covariance then makes the full vector matrix element vanish. The temporal axial current is parity-odd and is allowed: this is the vacuum-to-pseudoscalar current selection rule.

The minus sign in $J_H=V_H-A_H$ cancels the interaction's minus sign, up to an irrelevant common S-matrix phase. ThereforeThe pion decay constant convention here includes the given $\sqrt{2}$ in its current matrix element. Momentum conservation and the external Dirac equations give $p=k+q$, ${\bar{u}}_e/k=m_e{\bar{u}}_e$, $/q{v}_{\nu }=0$. Since $/q(1-{\gamma }_5)=(1+{\gamma }_5)/q$, the neutrino term vanishes andThis is helicity suppression: the amplitude vanishes in the zero charged-lepton-mass limit despite the larger available phase space.

For the spin sum, the Dirac adjoint of $1-{\gamma }_5$ is $1+{\gamma }_5$, and the completeness relations giveThe chiral projectors automatically eliminate the unused neutrino helicity when the full massless completeness relation is inserted. There is no initial spin average for a spin-zero pion.

In the pion rest frame, let $r=|\mathbf{k}|=|\mathbf{q}|$, $E=\sqrt{r^2+m_e^2}$. After integrating the spatial delta function and angles, the two-body Lorentz-invariant phase space becomesThe root is $r_{\ast }=(m_{\pi }^2-m_e^2)/(2m_{\pi })$ and $E_{\ast }=(m_{\pi }^2+m_e^2)/(2m_{\pi })$. The absolute derivative of the delta-function argument is $r_{\ast }/E_{\ast }+1=m_{\pi }/E_{\ast }$. Thus ${\Phi }_2=r_{\ast }/(4\pi m_{\pi })$. Multiplying by the spin sum and the initial factor $1/(2m_{\pi })$ yields the leptonic pseudoscalar decay widthIf the decay constant were instead defined by a matrix element $i{f}_{\pi }{p}^{\mu }$ without $\sqrt{2}$, then $f_{\pi }=\sqrt{2}{F}_{\pi }$ and the width coefficient would be $f_{\pi }^2/(8\pi )$; mixing these conventions would introduce a factor-of-two error.

The same derivation for a muon cancels the common hadronic and weak factors in the pion electron-to-muon decay ratio:The numerical value uses approximate masses $0.511,105.66,139.57$ MeV. Agreement with the corresponding Standard Model prediction tests lepton universality and the chiral, helicity-suppressing structure of the weak charged current, rather than the poorly calculable hadronic decay constant. A difference in the electron and muon weak couplings would multiply the ratio by their squared coupling ratio. Precision agreement must be assessed after radiative corrections, with matching photon-inclusive experimental definitions; the displayed tree-level ratio is not itself the radiatively corrected precision prediction.

The Polyakov action has independent worldsheet metric ${\gamma }_{ab}$ and embedding fields $X^{\mu }$. Varying the metric before imposing a gauge givesWorldsheet diffeomorphisms and Weyl transformations allow conformal gauge ${\gamma }_{ab}=e^{2\omega }{\eta }_{ab}$ locally. The Weyl factor then cancels out of the classical action, leavingThese are free two-dimensional scalar fields, but the surviving metric equations requireEquivalently, ${\dot{X}}^2+X^{\prime 2}=0$ and $\dot{X}\cdot X^{\prime }=0$. Gauge fixing does not license dropping these Virasoro constraints.

For an ordinary open string, the boundary conditions relate the two chiral sectors. With ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{k}^{\mu }$, their independent Fourier generators areThe classical oscillator Poisson brackets $\{{\alpha }_m^{\mu },{\alpha }_n^{\nu }\}=-im{\eta }^{\mu \nu }{\delta }_{m+n,0}$ give the classical Virasoro constraint algebraThus they are first-class constraints. In the quantum theory one applies normal ordering to the product and uses $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$. The oscillator commutator givesFor example, for $m>0$ the extra vacuum term is $(D/2){\sum }_{r=1}^{m-1}r(m-r)=D(m^3-m)/12$. This exhibits the Virasoro central extension rather than concealing it in a classical constraint equation. The matter central charge is $D$; the worldsheet ghost fields have charge $-26$. Quantum consistency sets $D=26$ and the string intercept to one.

In old covariant string quantization the physical-state Virasoro conditions for an open string arePhysical string states are then identified modulo physical null string states. One does not impose both positive and negative generators as annihilation conditions: their central extension would obstruct that prescription. The resulting null-state quotient of a string agrees with the transverse light-cone spectrum at nonzero momentum.

At $N=0$, the oscillator vacuum $|k⟩$ is a single spacetime scalar field with ${\alpha }^{\prime }{M}^2=-1$: the bosonic tachyon. At $N=1$, write $|\Psi ⟩={ϵ}_{\mu }{\alpha }_{-1}^{\mu }|k⟩$. The zero-mode equation gives $k^2=0$, and $L_1$ gives $k\cdot ϵ=0$. The null descendant $L_{-1}|k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|k⟩$ identifies $ϵ\sim ϵ+\lambda k$. This leaves 24 physical polarizations of a massless gauge vector.

At $N=2$, use the level-two open-string polarization decompositionThe mass is $M^2=1/{\alpha }^{\prime }$. The oscillator commutators give all the nontrivial positive-mode conditions:Higher positive modes annihilate this level. To see the quotient explicitly, the 25 transverse-parameter null states $L_{-1}(\xi \cdot {\alpha }_{-1}|k⟩)$, with $k\cdot \xi =0$, shiftThe additional level-two scalar Virasoro null state $(L_{-2}+\frac{3}{2}{L}_{-1}^2)|k⟩$ shiftsSubstitution into the two physical equations verifies the null-state conditions; the second residual is $(D-26)/2$. Because $k$ is timelike, these null shifts span all components of $b$, so choose a representative with $b=0$. The physical conditions then say $k^{\mu }{h}_{\mu \nu }=0$ and $h^{\mu }{}_{\mu }=0$. In the rest frame this is a traceless second-rank tensor in 25 spatial dimensions, with $25\cdot 26/2-1=324$ polarizations. The light-cone basis independently gives $24$ states ${\alpha }_{-2}^i|k⟩$ and $24\cdot 25/2=300$ symmetric states ${\alpha }_{-1}^i{\alpha }_{-1}^j|k⟩$. Hence

For the fermionic extension, an explicit convention prevents ambiguity in the signs. Keep worldsheet signature $(-,+)$ and useTake $\psi =({\psi }_1,{\psi }_2{)}^T$ and the constant parameter $ϵ=({ϵ}_1,{ϵ}_2{)}^T$ to be real Grassmann-odd Majorana components, with $\bar{\psi }={\psi }^T{\rho }^0$. Put ${\partial }_{\pm }={\partial }_{\tau }\pm {\partial }_{\sigma }$. Apart from the common factor $1/(4\pi {\alpha }^{\prime })$, the combined Lagrangian isThe rigid transformations can be writtenor, in components,For the ${ϵ}_2$ part, the bosonic variation is $i{ϵ}_2({\partial }_{+}{\psi }_1\cdot {\partial }_{-}X+{\partial }_{+}X\cdot {\partial }_{-}{\psi }_1)$, while the fermionic variation is $-i{ϵ}_2{\partial }_{-}X\cdot {\partial }_{+}{\psi }_1+i{ϵ}_2{\psi }_1\cdot {\partial }_{+}{\partial }_{-}X$. Anticommuting the constant parameter past the fermion is essential in the second term. The first terms cancel. The other chiral component works in the same way, giving the chiral proof of rigid worldsheet supersymmetryThus the action is invariant off shell up to a boundary term. The transformation is a genuine supersymmetry: on $X$ its commutator for parameters $ϵ,\zeta$ is the translationOn $\psi$ the same translation follows using ${\partial }_{+}{\psi }_1=0$ and ${\partial }_{-}{\psi }_2=0$; the remaining cross terms are proportional to these Dirac equations. This is on-shell closure of rigid worldsheet supersymmetry. On a surface with boundary, compatible endpoint conditions must make the boundary flux vanish. On a compact worldsheet, a constant parameter must preserve the spin structure; the local identity alone does not make a constant transformation compatible with every antiperiodic sector.

The fermions contribute to the worldsheet stress-energy tensor and produce a fermionic superconformal current. In holomorphic normalization with ${\psi }^{\mu }(z){\psi }^{\nu }(w)\sim {\eta }^{\mu \nu }/(z-w)$,The free-field contractions give $G(z)G(w)\sim (2c/3)/(z-w{)}^3+2T(w)/(z-w)$ and conformal weight $3/2$ for $G$, where $c=D+D/2=3D/2$. Consequently the modes obey the N=1 super-Virasoro algebraDropping the central terms gives the classical graded constraint algebra. Integer $r$ gives the Ramond sector; half-integer $r$ gives the Neveu–Schwarz sector. In the displayed conformal-plane convention $G_0^2=L_0-c/24$ in the Ramond sector. Shifting to the Ramond vacuum-normal-ordered zero mode gives the usual Ramond supercurrent zero-mode square $G_0^2=L_0^{(R)}$.

Rigid symmetry of matter already supplies these currents, but promoting their vanishing to new gauge constraints requires coupling to worldsheet supergravity and varying its worldsheet gravitino. In superconformal gauge one retains $T=G=0$. The fermionic current constraints and their physical-state conditions extend the bosonic Virasoro conditions. The ghost charges are $-26+11=-15$, so cancellation with $3D/2$ gives the critical dimension of the RNS superstring $D=10$; the NS and Ramond intercepts are respectively $1/2$ and zero. This distinguishes the global supersymmetry established above from its locally gauged string completion.

Let the masses of $\Phi$ and $\varphi$ be $M$ and $\mu$, respectively. Write their external four-momenta as $P=k_1+k_2$, with $P^2=M^2$ and $k_1^2=k_2^2={\mu }^2$. The condition $M>2\mu$ opens the two-body decay phase space. Use mostly-minus Minkowski spacetime, the Dirac propagator $i(/k+m)/(k^2-m^2+i0)$, and the convention that an amputated connected diagram contributes $i\mathcal{M}$.

There is no tree-level three-scalar Feynman vertex in the displayed interaction. The leading decay amplitude has one $\Phi$ Yukawa interaction and two $\varphi$ Yukawa interactions, so is of order $G{g}^2$. The Yukawa fermion triangle is:

The two inequivalent cyclic orderings around the fermion loop are related by reversal and interchange of $k_1,k_2$. They give equal integrals: reversing all internal four-momenta leaves the masses and denominators unchanged, and the scalar gamma matrix trace identities below give the same numerator. Both must be included. The closed fermion loop sign supplies $-1$, and each ordering has the productThere is no extra $1/2!$ in the decay amplitude. Such a factor belongs to integration over the identical-particle final phase space when calculating a decay rate.

Route $a=\ell$, $b=\ell +k_1$, $c=\ell -k_2$ and define $D_a=a^2-m^2+i0$, and similarly for $D_b,D_c$. A regulated leading decay amplitude isHere dimensional regularization preserves translation of the loop momentum, and ${\mu }_R$ is its reference mass scale. Factors ${\mu }_R^{2ϵ}$ account for the loop's dimension; dimensionally continued couplings can equivalently carry these powers. No ${\gamma }^5$ occurs in this scalar loop, so there is no ${\gamma }^5$ regularization ambiguity.

Using gamma matrix trace identities, traces with an odd number of gamma matrices vanish, while $\mathrm{tr}1=4$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4g^{\mu \nu }$. Thus the numerator simplifies completely toFor a reduction to standard scalar integrals, disregard the vanishing infinitesimals only in the numerator. Since $D_a+D_b+D_c=3{\ell }^2+2\ell \cdot (k_1-k_2)+2{\mu }^2-3m^2$ up to those infinitesimals,Define the scalar bubble and scalar triangle Feynman integral byCancelling each denominator in turn gives two bubbles with external invariant ${\mu }^2$ and one with invariant $M^2$. ThereforeThis explicitly displays the complete gamma matrix trace simplification, the two loop orientations and the overall phase convention.

There is a renormalization qualification to this answer. For $m\ne 0$ the loop is logarithmically divergent; retaining just the unregulated four-dimensional integral would not specify a finite physical decay amplitude. To see the divergence, a Feynman parameter and a shifted loop momentum giveThe scalar triangle Feynman integral is ultraviolet finite. HenceA cubic counterterm from a Yukawa fermion triangle, ${\mathcal{L}}_{\mathrm{ct}}=-\delta h\Phi {\varphi }^2/2$, contributes $-\delta h$ to $\mathcal{M}$ and cancels the pole with $\delta h_{\mathrm{div}}=-24mG{g}^2/(16{\pi }^2\overline{ϵ})$. This is an instance of counterterm closure of a massive Yukawa theory. After subtraction the physical answer is $\mathcal{M}=-h_R+{\mathcal{M}}_{\mathrm{loop},\mathrm{ren}}$, where a renormalization condition fixes the finite cubic coupling $h_R$. The displayed interaction alone does not supply that condition. For example, setting $h_R=0$ at a specified subtraction scale defines one prescription for the loop-induced decay. The regulated expressions above remain the lowest-order loop answer without such extra data. For $m=0$ the trace, and therefore this scalar triangle contribution, vanishes identically.

Now take the Hermitian pseudoscalar Yukawa interaction with the sign printed in the PDF. Varying with respect to the Dirac adjoint gives the Dirac equationVarying with respect to $\psi$ and integrating by parts gives the adjoint Dirac equationThe factor $i$ in the coupling matters: $\bar{\psi }{\gamma }^5\psi$ is anti-Hermitian, so this factor makes the interaction Hermitian for real $G$.

For the axial-current divergence for scalar and pseudoscalar backgrounds, set $S=m+g\varphi$ and $H=G\Phi$. The two equations are equivalentlyIn the divergence of the axial current, use $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ and $({\gamma }^5{)}^2=1$:ThusThis is the classical field-equation identity requested here. It includes explicit breaking by both backgrounds; it is not a claim that a renormalized quantum axial current has no chiral anomaly when gauge interactions are present.

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

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

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

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

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

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

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

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

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

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

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

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

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

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