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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.

Higgsstrahlung produces a Higgs boson together with a massive electroweak gauge boson. At an electron-positron collider the process $e^{+}{e}^{-}\to Z^{\ast }\to ZH$ uses the Higgs boson coupling to Z bosons. The associated Z boson provides a recoil and identifiable decay products. This was the dominant Standard Model Higgs-production channel used at LEP2.

Each effectively massless active neutrino species contributes the same invisible Z boson partial decay width. Subtracting calculated or measured visible widths from the total width therefore measures the displayed number, assuming no additional invisible channels. The inference concerns particles lighter than $M_Z/2$ with ordinary weak couplings. It does not exclude heavier neutrinos or right-handed neutrinos that are gauge singlets. Early LEP resonance measurements were consistent with three light active species, with statistical uncertainty. See the OPAL report on its 1989 data for the original resonance-width comparison.

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 weak neutral current couples the Z boson to a fermion without changing its electric charge. Its vector and axial-vector coefficients depend on weak isospin, electric charge, and the weak mixing angle.

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.

Work with canonically normalized real components of the scalar field; a complex representation can first be written in terms of its real and imaginary components. At a vacuum expectation value ${\phi }_0$, expansion of the scalar potential givesThe linear term vanishes by stationarity. Thus the scalar mass matrix is the Hessian matrix of the scalar potential. Invariance under each Lie algebra generator gives the identity ${\partial }_{r}V(\phi )(i{T}_a\phi {)}_r=0$. Differentiating this identity with respect to ${\phi }_s$ and evaluating at the stationary point removes the term containing ${\partial }_{r}V$, leavingThe kernel of the map $T\mapsto iT{\phi }_0$ is precisely the Lie algebra of the stabilizer subgroup H. Consequently its restriction to a complementary space of broken Lie algebra generators is injective: a linear combination that annihilates ${\phi }_0$ would belong both to the stabilizer algebra and its complement, and hence would vanish. There are therefore $\dim G-\dim H$ independent Goldstone directions in the scalar mass matrix, tangent to the vacuum manifold. There are $\dim G-\dim H$ symmetry-protected massless scalars.

This Goldstone theorem guarantees at least that many zero quadratic masses. The claim that these are all the massless scalars additionally requires nonzero curvature transverse to the vacuum orbit. For example, $V=\kappa ({\phi }^T\phi -v^2{)}^4$, with $\kappa >0$, has an ordinary spherical vacuum manifold but its entire Hessian matrix vanishes there. These accidental massless radial modes beyond Goldstone modes are not excluded by the hypotheses. For the indicated orthogonal group breaking, the protected number is

For the Higgs doublet, put $r=\sqrt{2{\varphi }^{†}\varphi }=v+H$ and $z=\sqrt{2}\varphi /r$ wherever $r\ne 0$. If $z=(z_1,z_2{)}^T$, thenThis explicitly exhibits the three angular coordinates as an SU(2) transformation of a real radial field. In particular, the polar coordinates for the electroweak Higgs doublet can locally be chosen in the formAt the vacuum the three variations generated by $-i{\sigma }_1$, $-i{\sigma }_2$, and $i(I-{\sigma }_3)$ acting on $(0,1{)}^T$ are independent over the reals. They span the three broken directions; the remaining generator $Q=({\sigma }_3+I)/2$ annihilates the vacuum. This proves that the exponential supplies local angular coordinates. An inverse local gauge transformation, applied also to the gauge fields, removes these three fields in unitary gauge. They supply the longitudinal polarizations of the massive W bosons and Z boson through the Higgs mechanism. The parameterization is local about a nonzero vacuum, rather than a coordinate system through every possible zero of the Higgs doublet.

Use the metric $(+---)$ and define the charged W bosons and neutral gauge fields bywhere $s_W=\sin {\theta }_W$, $c_W=\cos {\theta }_W$, and $\tan {\theta }_W=g^{\prime }/g$. Direct substitution into the gauge-covariant kinetic term givesand henceThe electroweak doublet gauge-boson mass matrix in the $(A^3,B)$ basis isIts normalized zero direction is $(s_W,c_W)$ and its massive direction is $(c_W,-s_W)$. Thus the scalar-vector terms in physical fields arewith the tree-level electroweak gauge-boson massesThese are also charge eigenstates. The unbroken U(1) gauge symmetry has generator $Q=T_3+Y$; on the Higgs doublet, $Y=I/2$. Since $[Q,T_{\pm }]=\pm T_{\pm }$, the combinations $W^{\pm }$ have charges $\pm 1$, while Z, the photon, and H are neutral. The photon disappears from the displayed scalar-vector terms because $g{A}^3-g^{\prime }B=\sqrt{g^2+g^{\prime 2}}Z$. The photon is massless and has no direct tree-level coupling to the remaining radial scalar in this sector. This does not remove its interactions with charged W bosons, or exclude a scalar-to-photon interaction generated by loops.

Finally, retain the printed normalization of the Higgs field potential:The kinetic term is $\frac{1}{2}(\partial H{)}^2$, so the Higgs radial mass for a half-normalized quartic potential is $\boxed{m_H^2=\lambda v^2}$. The extra factor of two familiar from a potential with prefactor $\lambda$, rather than $\lambda /2$, does not apply here.

Put $s=q^2>0$ and neglect electron and final-quark masses compared with $\sqrt{s}$. First fix a final color charge; its multiplicity will be restored for the inclusive rate. The photon-exchange scattering amplitude is, up to an overall phase,Averaging over the two spins of each incident particle and summing final spins givesBy the gamma matrix trace identities, the first trace is $4(p_2^{\mu }{p}_1^{\nu }+p_2^{\nu }{p}_1^{\mu }-g^{\mu \nu }{p}_1\cdot p_2)$, and likewise for the second. Contracting the two tensors cancels the terms proportional to $(p_1\cdot p_2)(k_1\cdot k_2)$, givingIn the centre-of-mass frame all four energies are $\sqrt{s}/2$. The first pair of scalar products equals $s(1-\cos \theta )/4$ each, and the second pair equals $s(1+\cos \theta )/4$ each. Hence $\overline{|\mathcal{M}{|}^2}=e^4{Q}_q^2(1+{\cos }^2\theta )$.

For clarity, the printed flux hint has a factor-of-four inconsistency: when $\mathcal{F}=4\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=2s$, the prefactor multiplying the spin-averaged amplitude is $1/\mathcal{F}$, not $1/(4\mathcal{F})$. Alternatively, $1/(4F_0)$ is correct if $F_0=\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=s/2$. Use this massless two-body scattering flux normalization, keeping the spin average separate.

To obtain the two-body Lorentz-invariant phase space directly, integrate the momentum-conserving spatial delta function, leaving ${\mathbf{k}}_2=-{\mathbf{k}}_1$. The remaining radial integral is $\int dk\delta (\sqrt{s}-2k)=1/2$, so $d{\Phi }_2/d\Omega =1/(32{\pi }^2)$. The relativistic scattering cross-section is consequentlySince $\int d\Omega (1+{\cos }^2\theta )=2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$, the integrated electron-positron annihilation into a quark pair result, for this fixed color, isThe first two printed formulas are thus color-stripped. Distinct final colors are orthogonal states and must be summed, producing $N_c=3$. At leading order each accessible quark flavor contributes independently. Away from thresholds and narrow resonances, quark-hadron duality relates the inclusive parton rate to the observed inclusive electron-positron annihilation into hadrons, givingThis formula concerns the photon channel, assumes $\sqrt{s}$ is large compared with the relevant quark masses and the strong-interaction scale, and does not include Z boson exchange.

Beyond leading order, an exactly exclusive pair of colored quarks is not an observable asymptotic state because of confinement and hadronization. Moreover its virtual corrections have infrared divergences; real soft or collinear gluon emission cannot be excluded if a finite rate is wanted. The inclusive sum includes both the corrected two-parton channel and channels such as $q\bar{q}g$, with their degenerate soft and collinear limits. Real and virtual infrared divergences then cancel for an infrared-safe observable. A specified jet observable can also be meaningful if it defines an infrared-safe observable; a bare two-quark final state is not.

A convenient inclusive calculation uses the optical theorem. For the electromagnetic hadronic current $j_{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }_{\mu }{q}_f$, defineThe hadronic electromagnetic-current spectral density is obtained from its discontinuity. In this convention the hadronic R ratio is $R(s)={\sigma }_{\mathrm{had}}/{\sigma }_{\mu \mu }=12\pi \mathrm{Im}\Pi (s+i0)$, where ${\sigma }_{\mu \mu }=4\pi {\alpha }^2/(3s)$ at this order. Cutting the forward amplitude sums all allowed hadronic intermediate states, so perturbative Quantum chromodynamics applied to this inclusive quantity implements the required sum of real and virtual contributions.

Finally, write $a={\alpha }_s({\mu }^2)$, $C=1.99-0.11n_f$, $L=\ln (s/{\mu }^2)$, and ${\beta }_0=11-2n_f/3$. The given truncated correction isDifferentiate with respect to $t=\ln {\mu }^2$, using the stated one-loop beta function $da/dt=-{\beta }_0{a}^2/(4\pi )$ and $dL/dt=-1$. The derivative of $a/\pi$ cancels the explicit logarithmic derivative at order $a^2$. Keeping the residual derivative of the truncated expression givesThus the principle of minimal sensitivity, applied to this expression with nonzero coupling and ${\beta }_0\ne 0$, selects the one-loop stationary scale for the hadronic annihilation correctionThis is a prescription based on the residual order-$a^3$ scale dependence of an order-$a^2$ approximation. At strictly order $a^2$, the derivative already vanishes for every renormalization scale; the prescription does not assert exact scale dependence of the observable. It is useful only while the selected running coupling remains perturbative. If $a=0$ or ${\beta }_0=0$, the stated derivative imposes no unique scale.

A useful model of spontaneous breaking of a Z2 scalar symmetry is the four-dimensional real scalar theoryThe action is invariant under the Z2 symmetry $\varphi \mapsto -\varphi$. Its classical vacuum manifold consists of the two values $\varphi =\pm v$. In the broken quantum phase, choose a pure vacuum with $⟨\varphi ⟩=v$; the symmetry maps it to a different vacuum with opposite expectation value, although it leaves the action unchanged. To distinguish quantum symmetry breaking from simply minimizing a classical potential, introduce a small source $J\varphi$ and take the infinite-volume limit before $J\to 0^{+}$. A finite-volume symmetry-preserving ground state can remain an even superposition because of tunneling, whereas the selected infinite-volume pure vacua have nonzero vacuum expectation values. The parameter $v$ below is the tree-level expectation, with its quantum value determined by renormalized parameters.

Writing $\varphi =v+h$ givesThe original sign symmetry relates expansions about the two vacua; in one expansion it acts as $h\mapsto -2v-h$. It is not a symmetry that fixes the chosen vacuum. Since the broken group is discrete, there is no continuous broken generator and no required Goldstone boson.

With relativistically normalized external states, the general two-body partial decay width isHere the bar sums final spins and averages initial spins only if the initial ensemble is unpolarized. The identical final-state symmetry factor is $\mathcal{S}=2!$ for two identical daughters and one for distinguishable daughters. Integrating the three-momentum delta function in the parent rest frame leaves ${\mathbf{p}}_2=-{\mathbf{p}}_1$. The energy delta function fixes $k=|{\mathbf{p}}_1|$ and has radial Jacobian $k(1/E_1+1/E_2)$. Hence the two-body decay phase space isConsequentlywhen the spin-summed amplitude is angle independent.

For Higgs decay to two W bosons, the scalar parent has no spin average and the amplitude, up to an irrelevant phase, is $\mathcal{M}=g{M}_W{\epsilon }_1^{\ast }\cdot {\epsilon }_2^{\ast }$. Use the massive vector polarization sum with $p_i^2=M_W^2$:Momentum conservation gives $p_1\cdot p_2=(M_H^2-2M_W^2)/2$. With the PDF's ratio $x=M_W/M_H$, this becomesThe charged daughters are distinguishable. Combining the last two equations with $g^2/M_W^2=4\sqrt{2}{G}_F$ givesFor Higgs decay to two Z bosons, the Higgs boson coupling to Z bosons is $2i{M}_Z^2{g}_{\mu \nu }/v=ig{M}_Z^2{g}_{\mu \nu }/M_W$, using $v=2M_W/g$. This compensates the changed powers of $M_Z$ in the polarization contraction, giving the same normalization of the squared amplitude with $x$ replaced by $y=M_Z/M_H$. The two Z bosons are identical, so the phase-space symmetry factor supplies an additional half:Both expressions have mass dimension one, vanish at their two-body thresholds and approach the ratio $2:1$ for $M_H\gg M_Z$. The cubic large-mass behavior comes from longitudinal-vector polarizations. These are on-shell tree-level widths under the stated heavy-Higgs hypothesis; below threshold the physical off-shell multi-particle decays require a different calculation.

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.

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.