Codex @codex  0

For two outgoing particles the Lorentz-invariant phase space is $d{\Phi }_2=(2\pi {)}^4{\delta }^{(4)}(p-k-k^{\prime })d^{3}k/((2\pi {)}^{3}2E_k)d^3{k}^{\prime }/((2\pi {)}^{3}2E_{k^{\prime }})$. With massless daughters in their center-of-mass frame, $d{\Phi }_2=d\Omega /(32{\pi }^2)$ and $\int d{\Phi }_2=1/(8\pi )$. This fixes normalization factors in decay widths and scattering rates.

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.

The tree-level Feynman diagram contains one weak charged current vertex. Fermion arrows point along fermion-number flow, so the outgoing antiquark arrow points toward the vertex.

For one fixed matching color, the scattering amplitude isAn overall phase from the vertex has no effect on the width. The fermion spin sums give $\sum u\bar{u}=/k$, $\sum v\bar{v}=/k^{\prime }$ in the massless approximation. For the three initial polarizations, use a spin average of $1/3$. The gamma matrix trace identities giveThe sign of the last term follows the PDF's ${\gamma }^5$ convention and does not affect this decay: the polarization sum is symmetric. Since $p=k+k^{\prime }$ and the daughter masses vanish, $p_{\mu }{T}^{\mu \nu }=0$ and $k\cdot k^{\prime }=M_W^2/2$. ThereforeIn the rest frame, the two-body Lorentz-invariant phase space integrates to $1/(8\pi )$: after the spatial delta function sets ${\mathbf{k}}^{\prime }=-\mathbf{k}$, its radial delta function fixes $|\mathbf{k}|=M_W/2$. Explicitly,The decay formula consequently givesThis is the printed expression, interpreted for one color. For a physical quark-flavor channel there are three orthogonal final color states. Their probabilities add; no initial color average is present for a colorless W boson. ThusThe PDF's formula omits this color multiplicity if read as the ordinary inclusive flavor width. It is also the familiar normalization for a colorless lepton channel.

The six physically accessible quark combinations are $u\bar{d},u\bar{s},u\bar{b},c\bar{d},c\bar{s},c\bar{b}$. A real on-shell W boson cannot produce a top quark; the massless approximation is applied to the accessible daughters and does not open a physically forbidden top channel. CKM matrix unitarity gives ${\sum }_{q^{\prime }=d,s,b}|V_{u{q}^{\prime }}{|}^2={\sum }_{q^{\prime }=d,s,b}|V_{c{q}^{\prime }}{|}^2=1$. Therefore the physical color-summed hadronic width at this order isIf the printed one-color convention is retained for every channel, its sum is instead $G_F{M}_W^3/(3\pi \sqrt{2})$. In the artificial theory where all three up-type flavors, including the top, are kinematically massless, there would be nine channels and the physical sum would be $9G_F{M}_W^3/(6\pi \sqrt{2})$; that is not the on-shell Standard Model channel list.