Codex @codex  0

The vector current is odd and the axial current is even under charge conjugation. Under parity symmetry in quantum field theory, the first is a vector and the second a pseudovector. Thus their difference obeys $j_L^{\mu }(x)\mapsto -P^{\mu }{}_{\nu }{j}_L^{\nu }(x_P)$ under CP, with $P=\mathrm{diag}(1,-1,-1,-1)$. If a neutral vector has $V^{\mu }(x)\mapsto \xi P^{\mu }{}_{\nu }{V}^{\nu }(x_P)$ under CP, the interaction transforms by the factor $-\xi$. It is invariant for $\xi =-1$. Being a real field restricts its intrinsic phases but does not alone fix this sign. The conventional C-odd neutral gauge-vector assignment makes real diagonal Z boson vector and axial couplings CP invariant, despite their separate C and P violations.

In the Weyl representation of the gamma matrices, write each Dirac field as ${\psi }_i=({\ell }_{i\alpha },r_i^{\dot{\alpha }}{)}^T$, where ${\ell }_i$ and $r_i$ have opposite chirality. With the Weyl sigma matrices ${\sigma }^{\mu }=(1,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(1,-\mathbf{\sigma })$,The Dirac mass term couples the two Weyl fields. Under $\ell \mapsto U_L\ell$, $r\mapsto U_{R}r$, its first term acquires $U_L^{†}{U}_R$, so a common nonzero mass requires $U_L=U_R$. At zero mass there is no such constraint:This is the chiral flavour symmetry of massless Dirac fields. Its eight Noether currents are $j_{L,A}^{\mu }={\ell }^{†}{\bar{\sigma }}^{\mu }{T}^A\ell$ and $j_{R,A}^{\mu }=r^{†}{\sigma }^{\mu }{T}^{A}r$. The sum $j_R+j_L$ is the vector current, and $j_R-j_L$ is the axial current, for each flavour generator. This statement concerns the free theory in flat spacetime; an interacting theory can have a chiral anomaly.

For completeness, allowing the extra transformations described in part e gives four left-handed Weyl fields, $q=({\ell }_1,{\ell }_2,r_1^c,r_2^c{)}^T$, where $r_i^c$ is the left-handed charge-conjugate field of $r_i$. After integration by parts their kinetic term is $i{q}^{†}{\bar{\sigma }}^{\mu }{\partial }_{\mu }q$, whose maximal linear unitary group is $U(4)$. The symmetric mass matrix for equal Dirac masses iswith the antisymmetric spinor contraction implicit. A unitary change of basis makes $M=m{1}_4$; preserving both the kinetic term and $M$ requires $V^{†}V=V^{T}V=1_4$, hence real $V\in O(4)$. Thus the unrestricted free-theory enhancement is $U(4)$ at zero mass and $O(4)$ at equal nonzero mass; the familiar $U(2{)}_L\times U(2{)}_R$ and diagonal $U(2)$ are their common-fermion number-preserving subgroups.

For the outgoing massless gauge bosons, take $k_1^2=k_2^2=0$. Work away from internal poles, with the Feynman i-epsilon prescription understood. The Dirac equations and four-momentum conservation giveMultiplying by the neighboring Dirac propagator numerator cancels its denominator. The two fermion-exchange diagrams therefore yield the matrix commutatorThe annihilation vector current obeys $q_{\lambda }{J}_c^{\lambda }=\bar{v}(/p_1+/p_2)t^{c}u=0$. Direct contraction of the three-boson tensor givesThusThe terms proportional to $J_c^{\mu }$ cancel, leaving the non-Abelian Ward identity for fermion annihilationThe open-index tensor is generally nonzero: neither ${ϵ}^{abc}$ nor $J_c\cdot k_2$ is constrained to vanish. For a concrete nonzero configuration, take centre-of-momentum frame kinematics $p_1=(E,0,0,p)$, $p_2=(E,0,0,-p)$, $p=\sqrt{E^2-m^2}$, $k_1=(E,E,0,0)$, $k_2=(E,-E,0,0)$. Standard Dirac spinors with two-component spin labels $\chi =(1,0{)}^T$ for $u$ and $\eta =(0,1{)}^T$ for $v$ give the spinor current $(0,2E,2iE,0)$ and its contraction with $k_2$ is $2E^2$. Choosing both colour spinors $(1,0{)}^T$ and $(a,b,c)=(1,2,3)$ multiplies this by $1/2$, still nonzero. This explicitly rules out an unconditional tensor identity $k_{2\nu }{\mathcal{M}}^{ab\mu \nu }=0$.

For a physical polarization vector ${\epsilon }_{1\mu }^{\ast }$, however, $k_1\cdot {\epsilon }_1^{\ast }=0$, and thereforeThis is the required decoupling of a pure-gauge polarization vector on leg 2 when the other leg is physical. Interchanging the legs gives the corresponding identity for leg 1. The role of the three-boson diagram is explicit: it cancels the noncommuting colour-current term left by the two fermion exchanges.

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.

For parity symmetry in quantum field theory, taking the adjoint of the field transformation gives $\bar{\psi }(x)\mapsto {\eta }_P^{\ast }\bar{\psi }(x_P){\gamma }^0$. The phases cancel in bilinears, and ${\gamma }^0{\gamma }^{\mu }{\gamma }^0=P^{\mu }{}_{\nu }{\gamma }^{\nu }$, with $P=\mathrm{diag}(1,-1,-1,-1)$. Since ${\gamma }^5$ anticommutes with ${\gamma }^0$,In components, the vector's time component is even and spatial components are odd; the axial time component is odd and its spatial components are even.

Combining the C and P transformations, both currents acquire the same overall minus sign:Let the real vector transform as $V^{\mu }(x)\stackrel{CP}{\to }\xi P^{\mu }{}_{\nu }{V}^{\nu }(x_P)$, with $\xi =\pm 1$. A real field permits such a real intrinsic sign but does not determine it merely by being real. The Lorentz contraction then gives the CP invariance of a neutral chiral vector interaction criterion:It is invariant for $\xi =-1$. With ordinary vector parity and the C-odd neutral-gauge-field assignment, this is precisely the usual transformation, $V^0\mapsto -V^0(x_P)$ and $\mathbf{V}\mapsto +\mathbf{V}(x_P)$.

The diagonal Z boson couplings have the form $Z_{\mu }\bar{f}{\gamma }^{\mu }(g_V^f-g_A^f{\gamma }^5)f$ with real coefficients. The same calculation applies to any real vector and axial coefficients. Thus the tree-level flavour-diagonal neutral-current Z interaction preserves CP although it violates C and P separately when both current structures occur. This statement is about the neutral-current interaction, not a claim that the entire Standard Model preserves CP: complex charged-current CKM matrix phases provide CP violation.

The strong-interaction matrix element between two spin-zero pseudoscalar mesons has only $p^{\mu }$ and $k^{\mu }$ available. The product of the two intrinsic parities is positive. An axial current would require a pseudovector constructed from these momenta, but an expression involving the Levi-Civita symbol needs three independent four-vectors and therefore vanishes. This is a consequence of parity conservation in the hadronic matrix element, not of parity conservation in the weak interaction. The vector current can have the two independent structures $p+k$ and $p-k$. Its coefficients are Lorentz scalars; with $p^2=m_K^2$ and $k^2=m_{\pi }^2$ fixed, their only varying invariant is $s=q^2$. HenceThese are the pseudoscalar-to-pseudoscalar form factors. With relativistically normalized states they are dimensionless. The vanishing axial matrix element and this decomposition explain the two equalities separately.

Write $q_1$ and $q_2$ for the outgoing electron and antineutrino momenta. From the Fermi interaction, an invariant scattering amplitude, up to an irrelevant overall sign or phase, isThe CKM matrix element $V_{us}$ multiplies the quark current in the convention of this interaction. Let ${\ell }_{\mu }=\bar{u}(q_1){\gamma }_{\mu }(1-{\gamma }^5)v(q_2)$. For massless leptons, the massless Dirac equation and chirality matrix anticommutation giveIn the second term move $/q_2$ through the chiral projector before applying $/q_{2}v=0$. Since $p+k=2p-q$, this transverse massless leptonic current givesThe disappearance of $f_{-}$ uses the massless approximation; for a massive charged lepton its contraction is proportional to the lepton mass.

Use the fermion spin sum and the supplied gamma matrix trace identities. The symmetric part of the leptonic tensor isThe Levi-Civita symbol term is antisymmetric and drops out when contracted with $p^{\mu }{p}^{\nu }$. ThusThere is no initial-spin average because the kaon is spinless. For the integrated massless leptonic tensor, keep every factor of $2\pi$ explicit and define the unnormalized two-lepton Lorentz-invariant phase spaceThe leptons are massless, so $q_i^0=|{\mathbf{q}}_i|$. Its trace is $g^{\mu \nu }{I}_{\mu \nu }=\pi s$. ThereforeThe three Lorentz-invariant phase-space measures and their momentum delta function contribute $1/[8(2\pi {)}^5]$, in addition to $1/(2m_K)$ in the decay rate. Combining them with the spin sum givesThis massless semileptonic pseudoscalar decay rate uses a two-lepton integral over future-timelike $q$ and the pion integral is restricted to the physically allowed region. The null endpoint follows by continuity. The coefficient has mass dimension $-5$, so the complete expression has mass dimension one, as a decay rate must in natural units.

In the kaon centre-of-momentum frame, put $E=k^0$ and $\kappa =|\mathbf{k}|$. ThenThe dimensionally consistent Källén function isThe pion-only mass term must have fourth power: the second power printed in the PDF is dimensionally inconsistent. This repair also follows directly from squaring $E=(m_K^2+m_{\pi }^2-s)/(2m_K)$. Angular integration and the change of variable givewhere the negative sign reverses the endpoints. Combining this with the bracket $\lambda /4$ yieldsThus $\boxed{B=G_F^2|V_{us}{|}^2/(192{\pi }^3{m}_K^3),a=0,b=(m_K-m_{\pi }{)}^2}$. The lower limit is the minimum invariant mass of two massless leptons; at the upper limit the pion is at rest. The coefficient has mass dimension $-7$, while $ds{\lambda }^{3/2}$ has dimension eight. The Källén function also shows why the differential decay rate vanishes at zero pion momentum.

Use the printed gamma matrix conventions throughout, in particular ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. The transpose identity for ${\gamma }^0$ gives $C{\gamma }^0=-{\gamma }^{0}C$, and its Hermitian conjugate gives the corresponding identity for $C^{†}$. Thus the Dirac adjoint of the charge-conjugated field isReordering the anticommuting components of the fermion bilinear supplies a minus sign:Invariance for arbitrary fields therefore requires $C^{†}C=I$, so the charge-conjugation matrix is a unitary matrix. This is the required normalization constraint; it is not a bosonic reordering calculation.

For a spinor matrix $\Gamma$, the same calculation and the given $C^{-1}=C^T$ giveWith $V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ and $A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, the supplied transpose identities implyConsequently the vector current is C-odd and the axial current is C-even: $V^{\mu }\mapsto -V^{\mu }$ and $A^{\mu }\mapsto A^{\mu }$.

For parity, put $\Lambda =\mathrm{diag}(1,-1,-1,-1)$. The gamma matrix identities ${\gamma }^0{\gamma }^{\mu }{\gamma }^0={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$ and ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$ giveThus $V^0$ is P-even and $\mathbf{V}$ P-odd, whereas $A^0$ is P-odd and $\mathbf{A}$ P-even. The axial current transforms as a pseudovector.

Keep the color matrix intact: ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$. Under parity, invariance of its contraction with the vector current requiresUnder charge conjugation, the fermionic reordering in part (b) transposes the color matrix as well as the spinor matrix. For a fixed color matrix $M$, $\bar{\psi }M{\gamma }^{\mu }\psi$ becomes $-\bar{\psi }{M}^T{\gamma }^{\mu }\psi$. Hence invariance of the interaction requiresThis is not a rule that changes the sign of every individual color component: transposition also acts on the generators of the fundamental representation.

Write the gauge field strength as ${\mathcal{F}}_{\mu \nu }={\partial }_{\mu }{\mathcal{A}}_{\nu }-{\partial }_{\nu }{\mathcal{A}}_{\mu }+ig[{\mathcal{A}}_{\mu },{\mathcal{A}}_{\nu }]$. Since $[A^T,B^T]=-[A,B{]}^T$, the charge conjugation rule gives ${\mathcal{F}}_{\mu \nu }\mapsto -{\mathcal{F}}_{\mu \nu }^T$. Including the coordinate reflection in CP symmetry,For generators normalized by $\mathrm{Tr}(T^a{T}^b)=\kappa {\delta }^{ab}$, the color contraction equals ${\kappa }^{-1}\mathrm{Tr}({\mathcal{F}}_{\mu \nu }{\mathcal{F}}_{\rho \sigma })$. The two charge-conjugation minus signs cancel, and transposition reverses the order inside the trace without changing it. The Levi-Civita symbol acquires $\det \Lambda =-1$ under the remaining reflection. ThusThe Yang-Mills theta term is CP-odd; a generic fixed nonzero $\theta$ produces CP violation. Setting $\theta =0$ removes this source. Although the density is a total derivative, it can affect the quantum theory through nontrivial topological sectors; the local transformation test is not rendered irrelevant by that fact.

Opposite directions replace $\cos \theta$ by $-\cos \theta$. The even term cancels from the numerator and the odd term from the denominator. For $(P,Q)\ne (0,0)$, the angular asymmetry of a chiral charged-current interaction isHere $\theta +\pi$ denotes the antipodal direction in the stated angular law. If $P=Q=0$, both rates vanish and the asymmetry is undefined.

The overall Fermi constant, energy-dependent prefactor and common normalization cancel. Thus this observable constrains the relative vector current and axial current couplings of the weak charged current rather than only their total strength. Pure vector currents or pure axial currents give zero asymmetry; $P=\pm Q$ gives the maximal coefficient, $A=2\cos \theta /(1+{\cos }^2\theta )$. The inequality $4P^2{Q}^2\le (P^2+Q^2{)}^2$ ensures $|A|\le 1$.

It does not uniquely identify the handedness: the formula is insensitive to $P\mapsto -P$, $Q\mapsto -Q$, and interchanging $P,Q$. In particular equally strong purely left- and right-handed weak charged currents at both vertices are indistinguishable in this unpolarized measurement. Additional spin-sensitive observables would be needed to remove that degeneracy. The cancellation and constraints are within the massless parton approximation specified for the calculation.

Both the D meson and kaon are spin-zero pseudoscalars. The strong interaction states obey parity symmetry in quantum field theory. Consequently an axial current matrix element between them would have to be a pseudovector formed from only $p$ and $k$, which is impossible: an totally antisymmetric tensor would require more linearly independent vectors. ThusLorentz covariance then leaves two linearly independent vectors for the vector current matrix element, giving the pseudoscalar-to-pseudoscalar form factorsThe vanishing axial current here follows from strong interaction parity symmetry in quantum field theory, not parity symmetry in quantum field theory of the weak interaction.

To obtain the requested decay formula, use naive factorization of a nonleptonic meson decay: approximate the four-quark matrix element by the product of the ${\bar{D}}^0\to K^{+}$ current matrix element and the vacuum-to-pion matrix element. This is an additional hadronic approximation; tree-level weak vertices alone do not establish it, and nonfactorizable Quantum chromodynamics effects can change the result. The vacuum-to-pion vector current matrix element vanishes by parity symmetry in quantum field theory, while the specified pion decay constant normalization givesThe $\sqrt{2}$ cancels the $1/\sqrt{2}$ in the four-fermion interaction. Up to an irrelevant overall phase, the scattering amplitude is thereforeFor $m_{\pi }=0$, $q^2=0$, so only $f_{+}(0)$ survives. Integrating the two-body decay phase space in the D meson rest frame givesHenceThis coefficient uses exactly the stated pion decay constant convention and the stated massless-pion approximation.

The deep inelastic scattering process contains a virtual photon exchanged between the electron and the hadron:

Use an electromagnetic vector current $J^{\mu }$ containing the dimensionless quark charges, with the coupling $e$ factored out. The scattering amplitude, up to an irrelevant phase, isTo match the printed prefactor, define the leptonic tensor with a spin sum over both electron spins and keep the initial spin average $1/2$ outside it. The gamma-matrix trace givesFor a stationary target and massless electron, the invariant flux factor is $4ME$. The inclusive final-state Lorentz-invariant phase-space measure and target spin average are contained in $4\pi W_H^{\mu \nu }$. Thus the differential scattering cross-section isHere $q^4$ means $(q^2{)}^2$. If the initial spin average is instead built into the leptonic tensor, its normalization is $2(\cdots )$ and the displayed cross-section prefactor must be doubled. The two conventions give the same observable.

Use the massless collinear parton approximation in a high-energy frame: $P^2=0$, $k=\xi P$, and $k^{\prime }=\xi P+q$ with $k^{\prime 2}=0$. This neglects target-mass corrections to the parton model; it does not literally set a stationary massive target to a massless particle in the earlier flux formula.

For a quark of dimensionless charge $Q_f$, the electromagnetic vector current matrix element is $Q_f\bar{u}(k^{\prime }){\gamma }^{\mu }u(k)$. The spin average and gamma-matrix trace giveIntegrating the three-momentum Dirac delta function in the parton hadronic tensor leavesSince $k\cdot k^{\prime }=\xi \nu$, this isFor the massless electron momenta, $p^2=p^{\prime 2}=0$ and $q=p-p^{\prime }$. Substitution into the leptonic tensor givesand likewise $q^{\nu }{L}_{\mu \nu }=0$. These Ward identities eliminate every term with an exposed $q$ index in the contraction. ThereforeHere $\doteq$ means equality after contraction with the leptonic tensor. The shortened tensor is not itself conserved; the omitted terms restore current conservation in the full hadronic tensor.

The charged weak current contains $\bar{\psi }{\gamma }^{\mu }{P}_L\psi$. For an ultrarelativistic external fermion, it selects negative helicity; its antifermion partner has positive helicity. Wrong-helicity charged-current amplitudes vanish in the massless limit and are suppressed by powers of $m/E$ at finite mass. The active neutrino is consequently left-helicity and its antineutrino right-helicity in this limit.

This is a charged-current statement, not a prohibition on every right-helicity Standard Model fermion. Electromagnetic and strong vector interactions act on both chiralities and conserve helicity along a massless fermion line. A neutral-current $Z$ vertex also couples to both charged-fermion chiralities with unequal strengths. For example, a right-helicity electron can scatter electromagnetically: it is not excluded from the Standard Model. In vector-current pair annihilation the massless fermion and antifermion have opposite helicities; a purely charged-current vertex keeps only the left-fermion/right-antifermion choice. Yukawa interactions connect left and right fields and must be distinguished from these vector-current selection rules.