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At tree level, the Feynman diagram has an electron-positron current and a muon-antimuon current joined by one virtual photon. In the massless limit, the spin average and final-state spin sum give $\overline{|\mathcal{M}{|}^2}=e^4(1+{\cos }^2\theta )$, and the integrated relativistic scattering cross-section is $4\pi {\alpha }^2/(3s)$.

The product of lepton-current matrix elements with a specified spin sum or spin average. Its normalization must be stated when quoting a differential scattering cross-section.

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.

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.

Use metric signature $(+---)$ and average over the four initial spin states, summing over final spins and the $N_c=3$ quark color charge states. The Electrons carry no color charge, so their initial color charge average is trivial. A final color charge average would not give the inclusive relativistic cross-section requested.

Photon exchange gives the tree amplitude, up to an irrelevant overall phase,Massless spin sums give $\sum u\bar{u}=/p$ and $\sum v\bar{v}=/p$. ThereforeThe color charge multiplicity is ${\sum }_{a,b}|{\delta }_{ab}{|}^2=N_c$, not $N_c^2$. Each trace equals $4(a^{\mu }{b}^{\nu }+a^{\nu }{b}^{\mu }-g^{\mu \nu }a\cdot b)$. Contracting them, the terms proportional to $(p_1\cdot p_2)(q_1\cdot q_2)$ cancel, leaving $32[(p_1\cdot q_1)(p_2\cdot q_2)+(p_1\cdot q_2)(p_2\cdot q_1)]$. Thus

In the centre-of-momentum frame every energy is $\sqrt{s}/2$. Let $\theta$ be the angle between $p_1$ and $q_2$. ThenConsequently $\overline{|\mathcal{M}{|}^2}=N_c{e}^4{Q}_q^2(1+{\cos }^2\theta )$.

The supplied two-body Lorentz-invariant phase space formula has a factor-of-two error. Deriving the normalization explicitly,so the massless two-body scattering flux normalization isFor example, integrating the spatial Dirac delta in $d{\Phi }_2$ leaves $[4(2\pi {)}^2{]}^{-1}dkd\Omega \delta (\sqrt{s}-2k)$; its radial integral supplies the factor $1/2$. The printed denominator $64{\pi }^2{p}_1\cdot p_2$ would double the relativistic cross-section and cannot reproduce the requested result with the standard scattering amplitude.

The correctly normalized differential and total results for electron-positron annihilation into a quark pair areand hence

At energies where photon exchange and the parton description apply, sum over kinematically accessible quark flavors. The produced quarks undergo hadronization, so the leading inclusive hadronic rate isThis is the hadronic R ratio. Perturbative QCD adds a leading relative correction ${\alpha }_s(s)/\pi$; thresholds, narrow resonances and, near the Z boson scale, electroweak exchange require additional treatment. The calculation is a photon-exchange, massless result from a tree-level Feynman diagram.

Use natural units and the Minkowski metric $(+---)$, and work at tree level with on-shell final particles. Write $M=m_H$ and, for a final particle of mass $m$, ${\beta }_m=\sqrt{1-4m^2/M^2}$. In the Higgs boson rest frame the Lorentz-invariant phase-space measure reduces toTo obtain this, integrate the momentum delta function to set ${\mathbf{k}}_2=-{\mathbf{k}}_1$; the energy delta function is $\delta (M-2\sqrt{k^2+m^2})$, whose radial Jacobian is $E/(2k)$. Hence for an angle-independent final-state spin sum $S=\sum |\mathcal{M}{|}^2$,There is no initial-state spin average for a scalar, and neither charged-particle pair here requires an identical-particle factor of $1/2$.

For Higgs decay to two W bosons, the Feynman vertex gives $\mathcal{M}=(2m_W^2/v){\epsilon }_1^{\ast }\cdot {\epsilon }_2^{\ast }$, up to an overall phase. Contracting the two massive vector polarization sums givesThe constant 2 follows from $4-1-1$ in the contraction, with the last term coming from the two momentum projectors. Putting $x_W=m_W^2/M^2$, the full massive result isThe on-shell two-body width is zero below this threshold; decays through off-shell W bosons into more particles are different channels.

For Higgs decay to a fermion pair, put $y_b=m_b/v$. The amplitude is $\mathcal{M}=-y_b\bar{u}(k_1)v(k_2)$, up to an overall phase and a color Kronecker delta. The fermion spin sums and gamma-matrix trace giveThe color multiplicity in a decay width is $N_c=3$: only a quark and antiquark with matching colors contribute, so the factor is three rather than nine. ThereforeAgain the on-shell two-body width is zero below threshold. This is a partonic tree-level answer with every mass retained; hadronization is outside the specified calculation.

The W boson channel dominates for large $M$ within the tree-level comparison. The widths scale as $M^3/v^2$ and $m_b^{2}M/v^2$, respectively, andThe enhancement comes from longitudinal polarization of a massive vector boson: its polarization vector grows as momentum divided by $m_W$. Thus the longitudinal pair survives the apparently small $m_W^2$ factor in the interaction. At masses so large that the scalar sector is strongly coupled, the tree-level extrapolation itself needs corrections.

At this order there is one annihilation tree-level Feynman diagram. The squared-amplitude diagram is its product with its complex conjugate, with the same external momenta and spin labels in both factors:

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o
 e+(q,r) <---/          \<--- mu+(q',r')

                     times

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o                       *
 e+(q,r) <---/          \<--- mu+(q',r')

A wavy line is the virtual photon; the arrows show fermion-number flow. The star means complex conjugation of the entire second graph, including its external spinors. This represents $|\mathcal{M}{|}^2=\mathcal{M}{\mathcal{M}}^{\ast }$ for fixed spins, without adding a new physical loop to the amplitude.

For the spin sum in part (iv), identify each external spin label between the two copies and sum it. The fermion spin sum sews each spinor chain into a gamma-matrix trace. The two incoming unpolarized spin states are then handled by a spin average, giving the factor $1/4$. In particular, the diagrammatic contraction becomes

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.

The spin sum becomes a gamma-matrix trace:Since $(p+q{)}^2=M^2$, $p\cdot q=(M^2-2m^2)/2$. Therefore the normalization requested in the paper gives