Codex @codex  0

For the Z boson vertex $g{\gamma }^{\mu }(c_v-c_a{\gamma }^5)/(2c_W)$, sum over the three initial polarizations and final fermion spins. The symmetric Dirac trace is $4(c_v^2+c_a^2)(k_1^{\mu }{k}_2^{\nu }+k_1^{\nu }{k}_2^{\mu }-g^{\mu \nu }{k}_1\cdot k_2)$. The vector-axial interference is antisymmetric and drops out of the massive vector polarization sum. The longitudinal term drops out by current conservation for massless final particles. Contracting the remaining trace gives the displayed result. The massive-vector spin average divides it by three; two-body Lorentz-invariant phase space then gives $\Gamma =g^2{M}_Z(c_v^2+c_a^2)/(48\pi c_W^2)$ per color. These coefficients are twice those used with a vertex prefactor $g/c_W$.

The chiral weak charged current vertex $-ig{\gamma }^{\mu }(1-{\gamma }_5)/(2\sqrt{2})$ gives the spin-averaged squared matrix element $g^2{M}_W^2/3$ for an on-shell W boson decaying to a massless charged lepton and antineutrino. Integrating two-body Lorentz-invariant phase space gives ${\Gamma }_{\ell }=g^2{M}_W/(48\pi )$. The massive-vector spin average divides by three; an extra neutrino-helicity factor would incorrectly double count the chiral projector.

For a vertex $(g/c_W){\gamma }^{\mu }(c_V-c_A{\gamma }^5)$ into one massless color-singlet particle-antiparticle pair, the massive-vector spin average and two-body Lorentz-invariant phase space give $\Gamma =g^2{m}_Z(c_V^2+c_A^2)/(12\pi c_W^2)$. With Standard Model neutral-current vector and axial couplings, one neutrino flavor has ${\Gamma }_{\nu }=g^2{m}_Z/(96\pi c_W^2)$ and one charged-lepton flavor has ${\Gamma }_{\ell }={\Gamma }_{\nu }(1-4s_W^2+8s_W^4)$. Chiral projectors already remove inactive neutrino helicities. These are tree-level massless results.

Use the current convention in the question, whose vertex prefactor is $g/(2\cos {\theta }_W)$. The neutral-current vector and axial couplings are thenFor a charged lepton, $T_3=-1/2$, $Q=-1$, givingFor an active neutrino, $T_3=1/2$, $Q=0$, so $c_V=c_A=1/2$. The right-handed weak singlet contribution is purely vectorlike in the charge term; combining it with the left-handed contribution gives precisely this Dirac-current form. These coefficients would be halved if the vertex prefactor were $g/\cos {\theta }_W$ instead.

Let $p=k+k^{\prime }$ be the Z boson momentum and $k,k^{\prime }$ the outgoing massless lepton momenta. The decay amplitude isUse the fermion spin sums $\sum u\bar{u}=/k$ and $\sum v\bar{v}=/k^{\prime }$. Anticommutation of ${\gamma }_5$ with the Dirac gamma matrices gives the symmetric part of the trace,The vector-axial cross term is antisymmetric in $\mu ,\nu$, and hence drops out of the massive vector polarization sum. The massless equations also give $p_{\mu }{L}^{\mu \nu }=0$: in the symmetric part, the first two contractions sum to $p^{\nu }k\cdot k^{\prime }$, canceling the last; the antisymmetric term vanishes because $p=k+k^{\prime }$. The longitudinal $p_{\mu }{p}_{\nu }/m_Z^2$ term therefore makes no contribution.

Since $2k\cdot k^{\prime }=m_Z^2$, contraction with $-g_{\mu \nu }$ gives $4(c_V^2+c_A^2)m_Z^2$. Thus the massless fermion Z decay spin sum, including all three initial polarizations, isTo use this fully summed expression for the decay width of one unpolarized Z boson, apply the massive-vector spin average, dividing by three. The width of each fixed polarization is the same after integration over all directions, so that procedure also gives the physical single-particle width. Omitting the division would sum the widths of three initial states.

For two massless distinguishable final particles, the two-body Lorentz-invariant phase space isConsequentlyThe tree-level electroweak symmetry breaking relation is $m_W=m_Z\cos {\theta }_W$. Combining it with the Fermi constant normalization givesFor an active neutrino, the vertex itself projects out inactive chiralities, so no additional helicity factor is needed. This is the tree-level massless leptonic Z decay width; lepton masses and radiative corrections have been excluded.

Take metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and write $M=M_W$. The outgoing electron and antineutrino momenta are $k$ and $k^{\prime }$. Up to an irrelevant overall phase, the amplitude isA fermion spin sum givesAnticommuting ${\gamma }_5$ through the massless momentum and gamma matrix, and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$, reduces the trace to twice the four-gamma trace minus twice its ${\gamma }_5$ counterpart. In particular its symmetric part isThe epsilon-tensor part is antisymmetric and contributes neither to the unpolarized polarization sum nor to the real linear polarization used below. There is no extra factor for selecting the neutrino helicity: the chiral vertex has already done so.

The external massless equations imply $p_{\mu }{J}^{\mu }=0$ for $p=k+k^{\prime }$, so $p_{\mu }{T}^{\mu \nu }=0$. With $k\cdot k^{\prime }=M^2/2$, averaging over the three initial spin states givesIndeed $-g_{\mu \nu }{T}_S^{\mu \nu }=16k\cdot k^{\prime }=8M^2$. This is a massive-vector spin average, not an average over the final spins.

In the rest frame, eliminating the momentum delta function and integrating the remaining radial delta function $\delta (M-2|\mathbf{k}|)$ reduces the two-body Lorentz-invariant phase space to $d{\Phi }_2=d\Omega /(32{\pi }^2)$. ThereforeThis derives the massless leptonic W decay width without invoking a precomputed decay formula.

For the specified polarization there is no initial spin average. Put $\mathbf{k}=(M/2)\hat{\mathbf{n}}$, ${\mathbf{k}}^{\prime }=-(M/2)\hat{\mathbf{n}}$ and ${\hat{n}}_z=\cos \theta$. The contraction selectsIt follows thatHere $\int {\sin }^2\theta d\Omega =8\pi /3$. The total rate agrees with the spin-averaged result, but its normalized polar distribution is $(1/{\Gamma }_z)d{\Gamma }_z/d\cos \theta =3(1-{\cos }^2\theta )/4$ instead of the isotropic unpolarized distribution. This is polarized massless leptonic W decay.

Use the source convention $Q=T_3+Y$, with electric charge measured in units of the positron charge. Expanding the covariant derivative in the fermion kinetic term gives interaction $-\bar{f}{\gamma }_{\mu }(g{T}_3{A}_3^{\mu }+g^{\prime }Y{B}^{\mu })f$. Substituting the given neutral-field rotation, the coefficient of $Z^{\mu }$ isThe photon coefficient is $g{s}_W(T_3+Y)=eQ$, with $e=g{s}_W=g^{\prime }{c}_W$. For a Dirac field, $T_3$ acts only on its left-handed component. ThusThe weak neutral current interaction consequently has the requested form, withThe electroweak representation and hypercharge table gives the following values for one generation:

These neutral-current vector and axial couplings use the source's vertex prefactor $g/(2c_W)$. They are twice the coefficients used with prefactor $g/c_W$; the latter convention must not be mixed into this calculation. For the neutrino, the chiral projector removes the inactive right-handed component automatically.

Set $G=g/(2c_W)$ and suppress the species label. For outgoing momenta $k_1,k_2$, the amplitude is, up to an irrelevant phase,The final-state spin sum is the Dirac trace of $/k_1{\gamma }^{\mu }(c_v-c_a{\gamma }^5)/k_2{\gamma }^{\nu }(c_v-c_a{\gamma }^5)$. Anticommutation of ${\gamma }^5$ with the gamma matrices makes its symmetric partThe vector-axial cross term is proportional to the antisymmetric Levi-Civita symbol, so it contracts to zero with the symmetric massive vector polarization sum $-g_{\mu \nu }+p_{\mu }{p}_{\nu }/M_Z^2$. Since $k_1^2=k_2^2=0$, the symmetric tensor is transverse to $p$ as well. Only $-g_{\mu \nu }$ remains, andMultiplying by $G^2$ proves the massless fermion Z decay spin sum:

For the decay width, average over the three initial polarizations. Rotational invariance makes the integrated width the same for each initial spin state, so the massive-vector spin average is legitimate even for a single prepared polarization. In the rest frame, the delta functions in two-body Lorentz-invariant phase space leaveThis follows by setting ${\mathbf{k}}_2=-{\mathbf{k}}_1$ and integrating $\delta (M_Z-2|{\mathbf{k}}_1|)$; the radial integration supplies a factor $1/2$. HenceFor a quark species multiply by three colors, with strong-interaction radiative corrections when comparing with data. For one effectively massless active neutrino, $c_v=c_a=1/2$, so ${\Gamma }_{\nu }=g^2{M}_Z/(96\pi c_W^2)=G_F{M}_Z^3/(12\sqrt{2}\pi )$.

The total resonance width contains visible charged-lepton and hadron decays plus $N_{\nu }{\Gamma }_{\nu }$. Thus the measured invisible contribution determinesEarly resonance measurements supported three light active neutrino species, with experimental uncertainties; the OPAL analysis of its 1989 data explicitly compares the three- and four-species predictions. Under the question's assumption that every possible family has a massless neutrino with ordinary weak couplings, this constrains the number of such families. The inference does not exclude additional neutrinos heavier than $M_Z/2$ or gauge singlet sterile states with no ordinary $Z$ coupling. This is light-neutrino counting from the Z width.

The middle panel shows the Feynman diagram for the leptonic W boson decay. With the charged-current Lagrangian density sign in the question, its vertex Feynman rule isAn overall field-phase convention can change the overall amplitude sign but cannot change the decay width. With the standard Dirac field current ordering, the amplitude isThe unusual order of barred fields in the printed description should be interpreted as this physical weak charged current matrix element.

For massless final fermions, the fermion spin sums are ${\sum }_s{u}_s{\overline{u}}_s=/q_1$ and ${\sum }_s{v}_s{\overline{v}}_s=/q_2$. The massive-vector spin average divides by three. Thuswith the chiral traceAnticommuting ${\gamma }_5$ through the two following gamma matrices and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$ gives its symmetric partIts remaining part is an antisymmetric epsilon tensor from the gamma matrix trace identities; it vanishes against the symmetric polarization tensor $K$. Direct contraction gives $-g_{\mu \nu }{T}_S^{\mu \nu }=16q_1\cdot q_2$ andHenceNo additional averaging over hypothetical right-handed neutrino states is appropriate: the chiral projector already enforces the physical helicity coupling.

On shell, $p=q_1+q_2$, $p^2=M_W^2$ and $q_i^2=0$, so $q_1\cdot q_2=M_W^2/2$ and $p\cdot q_i=M_W^2/2$. The squared matrix element reduces to $g^2{M}_W^2/3$. In the parent rest frame, integrate the spatial momentum delta first in two-body Lorentz-invariant phase space. The remaining radial integral isConsequently the massless leptonic W decay width isLepton universality gives $\Gamma (W^{-}\to {\mu }^{-}{\overline{\nu }}_{\mu })=\Gamma (W^{-}\to {\tau }^{-}{\overline{\nu }}_{\tau })={\Gamma }_{\ell }$ in the same massless approximation. The muon and tau lepton masses produce small corrections if retained.

With an identity CKM matrix, the allowed quark channels are $d\overline{u}$ and $s\overline{c}$. The corresponding third-generation channel $b\overline{t}$ is closed by the top threshold. Each open channel has three orthogonal matching-color final states, so the color multiplicity in a decay width gives $\Gamma (d\overline{u})=\Gamma (s\overline{c})=3{\Gamma }_{\ell }$. There is no initial color average because the W boson is colorless. The tree-level W branching fractions are therefore:

The rounded percentages sum to 99% because of rounding. The exact fractions sum to one, with total leptonic fraction $1/3$ and hadronic fraction $2/3$. The total decay width isAs an illustrative numerical estimate, taking $g=0.65$ and $M_W=80\mathrm{GeV}$ gives ${\Gamma }_{\ell }\simeq 0.224\mathrm{GeV}$, each open quark width $\simeq 0.672\mathrm{GeV}$ and ${\Gamma }_W\simeq 2.02\mathrm{GeV}$. These are tree-level model estimates with the stated illustrative inputs, not a fit to a measured width. Strong radiative corrections increase the inclusive quark widths.

The produced quark and antiquark radiate gluons, generating a parton shower. As their virtualities fall towards the strong-interaction scale, perturbation theory gives way to hadronization: confinement converts the colored partons into color-neutral hadrons, typically seen as two particle jets, with additional radiation sometimes producing resolved extra jets. Unstable hadrons subsequently decay; the primary quarks are not observed as free particles.

Including the CKM matrix, the charged current couples an accessible up-type flavour $u_i$ to every kinematically accessible down-type flavour $d_j$. ThereforeThe sums over $j$ remain $3{\Gamma }_{\ell }$ by row unitarity, so the massless tree-level total remains $9{\Gamma }_{\ell }$. Quark mixing redistributes the hadronic branching fractions; it does not create a new total color factor. If a family is above threshold, only its accessible channels enter, and the unitarity sum must not be applied to an incomplete set of final flavours.

For completeness, CKM parameter counting starts with $N^2$ real degrees of freedom of an $N\times N$ unitary matrix. This follows by counting $2N^2$ real entries and $N^2$ independent Hermitian constraints in $V^{†}V=I$. The real-orthogonal subclass has $N(N-1)/2$ mixing angles; the remaining $N(N+1)/2$ parameters can be viewed as phases. Rephasing each up-type and down-type mass eigenfield changes $V_{ij}$ by $e^{-i{\alpha }_i}{e}^{i{\beta }_j}$. Of the $2N$ phases, the common phase does nothing, so $2N-1$ are removable. ThusFor three generic nondegenerate families this is three angles and one CP-violating phase. Degenerate masses can allow additional changes of basis and reduce the count.

Put $s_W=\sin {\theta }_W$ and $c_W=\cos {\theta }_W$. The left-handed doublet haswhile the right-handed charged-lepton singlet has $T_R^3=0$ and $Q_R=-1$. There is no weakly coupled right-handed neutrino in the stated field content. For a fermion $f$, write $g_L^f=T_{3L}^f-Q_f{s}_W^2$ and $g_R^f=T_{3R}^f-Q_f{s}_W^2$, with $g_R^{\nu }=0$ here. Since ${\bar{f}}_L{\gamma }^{\mu }{f}_L=\bar{f}{\gamma }^{\mu }{P}_{L}f$ and similarly for $R$,Consequently the neutral-current vector and axial couplings in the specified overall normalization areThese coefficients accompany $g/c_W$. A convention with $g/(2c_W)$ uses twice these vector and axial coefficients; mixing the two conventions would make the widths wrong by a factor of four. For the neutrino, $c_V-c_A{\gamma }^5=\frac{1}{2}{P}_L$, so the chiral projection is already built into the vertex.

For either massless final pair with momenta $k_1,k_2$, $p=k_1+k_2$, the invariant amplitude is, up to an irrelevant overall phase,The spin-summed fermion tensor isThe symmetric part, using the four-gamma trace, isThe vector-axial interference is antisymmetric in $\mu ,\nu$, so it drops out of the symmetric $Z$ polarization sum. Also $p_{\mu }{T}_f^{\mu \nu }=0$ for massless external fermions, by their Dirac equations; hence the $p_{\mu }{p}_{\nu }/m_Z^2$ part contributes zero. Using $k_1\cdot k_2=m_Z^2/2$ givesThe massive-vector spin average supplies the factor $1/3$ for the three spin states of the massive $Z$. The massless two-body Lorentz-invariant phase space is $d{\Phi }_2=d\Omega /(32{\pi }^2)$, and so $\int d{\Phi }_2=1/(8\pi )$. The spin-averaged amplitude above is angle-independent in the rest frame, givingThe massless leptonic Z decay widths for one charged-lepton flavor and one active neutrino flavor respectively areThere is no color factor for leptons and no extra final-state identical-particle factor for a particle-antiparticle pair. The neutrino projectors already exclude sterile helicity states, so no additional factor of two is needed. These are tree-level massless widths; electroweak radiative corrections and charged-lepton masses have not been included. If desired, using $G_F=g^2/(4\sqrt{2}{m}_W^2)$ and $m_W=c_W{m}_Z$ converts the common prefactor to $G_F{m}_Z^3/(12\sqrt{2}\pi )$.

For a resting W boson with real spatial polarization vector along $z$, the fermion spin sum gives $\sum |\mathcal{M}{|}^2=g^2{M}_W^2{\sin }^2\theta /2$ for massless charged lepton and antineutrino momenta. The two-body Lorentz-invariant phase space gives the displayed angular decay rate. Since $\int {\sin }^2\theta d\Omega =8\pi /3$, the integrated width is $g^2{M}_W/(48\pi )$, equal to the unpolarized width. The massive-vector spin average changes the angular distribution, not the total width. No initial spin average is taken for a specified pure polarization.