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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.

The weak charged current creates a left-handed neutrino and a right-handed outgoing positron in the massless limit. In the parent's rest frame their momenta are opposite, so these two fixed helicities give total angular-momentum projection of magnitude one along their flight axis. A spin-one W boson can supply that projection, and its leptonic width remains nonzero:A spin-zero pion cannot produce the same helicity configuration. The forbidden component is rescued only by a charged-lepton mass insertion that reverses its helicity, giving helicity suppression.

This can also be seen directly from the pion decay constant. Lorentz symmetry gives a vacuum-to-pion axial current matrix element proportional to $f_{\pi }{p}_{\pi }^{\mu }$. Contracting $p_{\pi }=k_e+k_{\nu }$ with the leptonic current givesusing the massless neutrino equation and $(/k_e+m_e)v_e=0$. Hence ${\mathcal{M}}_{\pi }\propto m_e$ and ${\Gamma }_{\pi }\propto m_e^2\to 0$. The W boson polarization vector is not constrained to be proportional to the total momentum, so its amplitude has no analogous compulsory mass factor.

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.

For a normal ordering of a fermion bilinear, the result of part (a) and the fermionic sign from interchanging the fields giveFor the Dirac current, $(C^{-1}{\gamma }^{\mu }C{)}^T=-{\gamma }^{\mu }$. For the axial current, the definition of ${\gamma }^5$ and the gamma matrix identities implyThus the charge conjugation of fermion bilinears givesThe Dirac current is odd under charge conjugation, and the axial current is even under charge conjugation. The normal ordering specifies the usual treatment of coincident-field contact terms.

Write $J_V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ for the Dirac current and $J_A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ for the axial current. Under parity symmetry in quantum field theory, the Dirac spinor phases cancel andThe extra sign for the axial current follows from ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$. Combining these transformations with part (c), both currents transform under CP symmetry as $J_{V,A}^{\mu }(x)\mapsto -{\mathbb{P}}^{\mu }{}_{\nu }{J}_{V,A}^{\nu }(x_P)$. The vector relativistic quantum field transforms as $V^{\mu }(x)\mapsto {\eta }_C{\mathbb{P}}^{\mu }{}_{\nu }{V}^{\nu †}(x_P)$. The two parity matrices cancel in the tensor contraction, givingFor a real vector relativistic quantum field with ${\eta }_C=-1$, the interaction is invariant under CP symmetry. A complex vector relativistic quantum field instead maps to the conjugate-field interaction, which must be compared with its Hermitian conjugation partner.

The flavour-diagonal weak neutral current coupled to the real Z boson has real vector and axial current couplings. With ${\eta }_C(Z)=-1$, part (e) shows that this tree-level interaction is invariant under CP symmetry, although the simultaneous presence of the Dirac current and axial current generally violates charge conjugation and parity symmetry in quantum field theory separately. This conclusion concerns the tree-level weak neutral current, rather than every interaction or radiative effect in the Standard Model.

The weak charged current couples distinct quark flavours and involves the Cabibbo-Kobayashi-Maskawa matrix:Here $u_{Li}$ and $d_{Lj}$ are left-handed quark fields. A CP symmetry transformation exchanges the two conjugate-field operators; because it acts through a unitary operator, it does not complex-conjugate their numerical coefficients. Equality with the original interaction therefore requires a Cabibbo-Kobayashi-Maskawa matrix that can be made real by quantum field rephasings. With three generations, a physical CP-violating phase remains in general, measured by the Jarlskog invariant. Therefore the W boson interactions can violate CP symmetry. Merely seeing a complex coefficient is insufficient: the phase must survive quantum field rephasing, unlike the real two-generation Cabibbo angle rotation.

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.

Under parity, $\overline{\psi }(t,\mathbf{x})\mapsto \overline{\psi }(t,-\mathbf{x}){\gamma }^0$. Since
${\gamma }^0{\gamma }^0{\gamma }^0={\gamma }^0$ and
${\gamma }^0{\gamma }^i{\gamma }^0=-{\gamma }^i$,Moreover ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$, so the axial current is a pseudovector:

For a massless Dirac fermion,The local gauge symmetry gives the vector current $J^{\mu }=e\bar{\psi }{\gamma }^{\mu }\psi$. The global chiral transformation $\psi \mapsto e^{i\beta {\gamma }^5}\psi$ gives the classical axial currentUsing the massless Dirac equation and $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ gives ${\partial }_{\mu }{J}_{\mathrm{ax}}^{\mu }=0$ classically.

Quantum mechanically, a gauge-invariant regulator for the fermion measure is not invariant under the axial rotation. In the Fujikawa form, its infinitesimal Jacobian containsThe first nonzero term in the heat kernel expansion is quadratic in $F_{\mu \nu }$. Usingand the Gaussian momentum integral gives the chiral anomalyThe overall sign follows the gamma-matrix, charge, and Levi-Civita conventions stated in the question. The same coefficient is obtained from the one-loop axial-vector-vector triangle diagram.

The matrix ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ anticommutes with every Dirac matrix. Moving ${\gamma }^{\mu }$ through each power in the exponential series givesUnder the chiral transformationHermiticity of ${\gamma }^5$ and its anticommutation with ${\gamma }^0$ implyConsequentlyso the action is invariant. The corresponding classical axial current is

A meson with zero spin angular momentum and negative parity. The pion and kaon are familiar examples. Between two pseudoscalar mesons, parity conservation in the strong interaction forces the axial current matrix element to vanish, while the vector current has pseudoscalar-to-pseudoscalar form factors. The weak decay as a whole need not conserve parity.

Between spin-zero pseudoscalars, Lorentz covariance decomposes a vector current matrix element into $(p+k{)}^{\mu }{f}_{+}(q^2)+(p-k{)}^{\mu }{f}_{-}(q^2)$, with $q=p-k$. Strong interaction parity symmetry in quantum field theory makes the corresponding axial current matrix element vanish.

A quark mass is the coefficient of its Dirac mass term. This term couples left- and right-handed quarks, so nonzero quark masses explicitly break independent rotations of the two chiralities. For two equal masses, the vector flavor symmetry survives, while the nonsinglet axial current obeys the flavor-nonsinglet axial Ward identity with equal quark masses. In the renormalized theory the mass parameter depends on the subtraction scale and scheme, while measurable hadron masses are physical observables.