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An asymptotically free theory has a running coupling that tends to zero at arbitrarily high energy. Its negative weak-coupling beta function makes short-distance perturbation theory increasingly accurate.

In Quantum chromodynamics, this means that a short-distance probe can resolve quarks with a small QCD coupling, even though isolated quarks are absent from the observed hadron spectrum. This made asymptotic freedom a bridge between the approximately free constituents of the parton model and an interacting quantum field theory. Asymptotic freedom supplies control at large momentum transfer; it does not itself prove confinement.^[ref]

Use the four-dimensional QCD coupling convention ${\alpha }_s=g_s^2/(4\pi )$. For the SU(3) gauge group and $n_f$ active quark flavors in the fundamental representation, the one-loop order beta function isHere $n_f$ counts quark flavors, not the three values of color charge. The chain rule converts the first equation to the last; changing from $\ln \mu$ to $\ln {\mu }^2$ supplies a factor of two. For ${\beta }_0>0$, the negative sign makes $g_s=0$ an attractive high-energy renormalization-group fixed point.^[ref]

For ${\beta }_0>0$, set the renormalization scale $\mu =Q$, where $Q$ is a positive energy scale. The one-loop running of the strong coupling then givesThe strong-coupling scale $\Lambda$ has the same units of measurement as $Q$, so the logarithm has a dimensionless argument. This is a fixed-$n_f$ approximation on the positive branch $Q>\Lambda$; controlled perturbative quantum field theory requires a small QCD coupling and suitably small omitted corrections, not merely a positive denominator. Heavy quark masses require matching the QCD scale across a quark threshold, changing $n_f$ and the associated $\Lambda$. Both the renormalization scheme and the loop order matter when quoting a numerical $\Lambda$.

At leading order in the naive parton model, dimensionless deep-inelastic structure functions depend only on Bjorken x. Quantum chromodynamics evolution introduces scale dependence.

In deep inelastic scattering, the scaling variable is $x=Q^2/(2P\cdot q)$ with $Q^2=-q^2$. In the massless collinear parton model, the outgoing on-shell condition $(xP+q{)}^2=0$ identifies $x$ with the struck parton's incoming longitudinal momentum fraction.

In the massless-target convention, the Callan-Gross relation sets this coefficient to zero at leading order in the parton model.

First fix the photon momentum convention. With $Q^2=-q^2>0$, $\nu =P\cdot q>0$ and $x=Q^2/(2\nu )>0$, an absorbed photon gives final momentum $p_X=P+q$. This is the physical momentum support of a deep-inelastic hadronic tensor. The original PDF instead prints ${\delta }^{(4)}(P-p_X-q)$ while keeping that positive-$x$ convention. Taken literally, it would requirein the Bjorken scaling limit, leaving no physical final states. The sign is therefore inconsistent with the requested nonzero deep-inelastic structure functions. Below, use the physically consistent absorption delta function ${\delta }^{(4)}(P+q-p_X)$. Alternatively all photon-momentum and scaling-variable conventions would have to be reversed together.

Work in the leading parton model: the target is unpolarized; scattering is incoherent from massless collinear quarks and antiquarks; intrinsic transverse momentum and target-mass effects are neglected. The parton distribution functions count constituents and include their colour multiplicity. At this order, neglect strong radiative corrections and the resulting logarithmic scale evolution.

Let a quark carry momentum $p=\xi P$, using a lightlike reference $P$ at leading order, and let $r=p+q$. The spin-averaged electromagnetic trace isIntegrating its one-particle final Lorentz-invariant phase space, with the $1/(4\pi )$ tensor normalization, supplies ${\delta }_{+}(r^2)/2$. Thus the spin-averaged electromagnetic parton tensor isOn physical support, $r^2=0$ gives $\xi =x$ and $p\cdot q=Q^2/2$. Define the transverse tensorsThe bracket in $w_f$ becomesThis follows by writing $p=\xi \tilde{P}-q/2$ on shell and expanding; it also verifies transversality to $q$.

The correct parton convolution normalization for a hadronic tensor isThe $1/\xi$ converts number weighting into covariant target-state normalization: a parton response per constituent is normalized by $2p^0$, whereas the hadron tensor uses $2P^0$, and $P^0/p^0=1/\xi$. Antiquark charge changes sign but its square and symmetric spin trace are the same.

Use the parton on-shell delta identity,Matching the two tensor coefficients now givesIn particular $F_2=2x{F}_1$, the Callan-Gross relation for spin-one-half partons. Scaling here means the naive leading approximation; full Quantum chromodynamics produces scale-dependent distributions and corrections.

Each quark contributes $+1$ and each antiquark contributes $-1$ to its flavour-vector charge. Sea pairs therefore cancel in the quark number sum rule:Here $N_f$ is the net flavour number, not the total number including sea pairs. This condition by itself fixes no antiquark distribution.

Under the separately imposed valence-only approximation, ${\bar{q}}_f=0$ and only up/down distributions remain. The proton has $(N_u,N_d)=(2,1)$ and the neutron has $(1,2)$. With $Q_u=2/3$, $Q_d=-1/3$, the valence-only proton and neutron structure-function moments areNo extra factor of three belongs here: colour is already included in the constituent distributions and the flavour counts. These are conditional valence-model moments, not unrestricted asymptotic statements in a theory whose evolution generates a sea.

The last panel of the preceding process figure shows deep inelastic scattering through a virtual photon. The blob represents the electromagnetic hadronic tensor response, with all final hadronic states included. With charges factored into $e$ and the hadronic current as in the question, a suitable scattering amplitude, up to an irrelevant overall phase, isThe electron fermion spin sum, before averaging the two initial spin states, gives the following normalization of the leptonic tensor:The initial electron spin average is a separate factor $1/2$. Equivalently one can absorb it into a tensor $L/2$, but the prefactor in the requested formula uses the unaveraged trace $L$. The massless Dirac equations imply $q^{\mu }{L}_{\mu \nu }=q^{\nu }{L}_{\mu \nu }=0$, the relevant Ward identity.

In the target rest frame the incident flux is $4p\cdot P=4ME$. This remains valid for a massive target because the electron is massless: the general flux $4\sqrt{(p\cdot P{)}^2-p^2{P}^2}$ reduces to that expression. The outgoing electron contributes $d^3{p}^{\prime }/((2\pi {)}^{3}2E^{\prime })$, and the inclusive final-state sum is $4\pi W_H^{\mu \nu }$. Including the electron spin average therefore givesThe implicit hadron spin average in $W_H$ must not be inserted a second time.

For an unpolarized target, Lorentz invariance permits tensor structures formed from $P$, $q$ and the metric. Hermiticity relates the two index orders. The electromagnetic response is parity even, excluding the pseudotensor $i{ϵ}^{\mu \nu \rho \sigma }{P}_{\rho }{q}_{\sigma }$. A possible ordinary antisymmetric $P^{\mu }{q}^{\nu }-q^{\mu }{P}^{\nu }$ term is excluded by current conservation. The symmetric part initially has four structures, $g^{\mu \nu }$, $P^{\mu }{P}^{\nu }$, $q^{\mu }{q}^{\nu }$ and $P^{\mu }{q}^{\nu }+q^{\mu }{P}^{\nu }$. Imposing both current-conservation conditions leaves the transverse decomposition of an unpolarized hadronic tensor:For example, contracting $q$ into the first structure gives $(1+A)q^{\mu }$ in the printed parametrization, and contracting into either parenthesis gives $(1+B)\nu$ or $(1+C)\nu$. Since the two independent deep-inelastic structure functions are not generally linked by symmetry, transversality requiresThere are only two invariant variables once $P^2=M^2$ is fixed; they may be chosen as $Q^2$ and Bjorken x $x=Q^2/(2\nu )$.

Now make the leading massless collinear parton approximation. In the hard parton calculation, use a lightlike collinear reference for $P$, neglecting target-mass effects of relative order $M^2/Q^2$ as well as quark masses. Literally combining a massive $P$ with an exactly massless constituent momentum $\xi P$ would be inconsistent. The exact rest-frame flux derived above does not require this approximation; it is the following partonic response that does.

Let $r=\xi P$ and $k=r+q$. Strip the quark charge out of the current, since its square is already the factor $Q_f^2$ in the parton convolution. Averaging the initial quark spin and summing the final spin gives the spin-averaged electromagnetic parton tensorThe spatial momentum delta fixes $\mathbf{k}=\mathbf{q}+\xi \mathbf{P}$. The remaining one-particle Lorentz-invariant phase space, together with $1/(4\pi \xi )$ in the definition, suppliesSince $r^2=k^2=0$ and $r\cdot k=\xi P\cdot q$, the full parton response isTerms with a free $q$ index vanish when contracted with the conserved leptonic tensor. Thus the part of $\tilde{W}$ contributing to the cross-section iswhere $\doteq$ denotes equality in that contraction, rather than an equality of the full tensors. The omitted terms are necessary for the full hadronic tensor to be transverse; the wording about contributing terms in the PDF is important here.

To prove the delta identity with the correct Jacobian, set $\omega =E_q+\xi E_P$. On physical support $E_k>0$ and $\omega >0$. Factoring $E_k^2-{\omega }^2=(E_k-\omega )(E_k+\omega )$ gives the positive-energy on-shell delta function identityThis is restricted to the positive-energy branch; the other root must not be included. Using the spatial momentum constraint and $P^2=0$ in the leading hard approximation,The Dirac delta function change of variable therefore gives the parton on-shell delta identityHere $\nu >0$. This is the actual printed identity: the converted TeX's final $\delta (x)/E_k$ is a transcription error. If target mass is formally retained in the delta argument, it becomes $Q^2-2\xi \nu -{\xi }^2{M}^2$, so the stated simple root and Jacobian no longer apply.

Let $D(x)={\sum }_f{Q}_f^2(q_f(x)+{\overline{q}}_f(x))$. The parton integral collapses at $\xi =x$, givingThe transverse decomposition of an unpolarized hadronic tensor, in the same contraction, is $-g^{\mu \nu }{F}_1+P^{\mu }{P}^{\nu }{F}_2/\nu$. Comparing the two independent structures yieldsThis derives the Callan-Gross relation from the spin-one-half trace and on-shell constraint. Antiquarks give the same symmetric response and therefore the sum $q_f+{\overline{q}}_f$. The result applies at leading order in the massless parton model; target masses, intrinsic transverse momentum and Quantum chromodynamics radiative effects can generate a nonzero longitudinal deep-inelastic structure function and modify the relation.

Let $s=q^2>0$, neglect $m_e$, and average over the two spin states of each incoming particle. Only the single-photon channel is being considered. The spectral-density symbol ${\rho }_h$ is not defined in the question, so specify its normalization before calculating. A useful photon-field convention isHere $d{\Pi }_X$ is the product of final-particle measures $d^{3}k/[(2\pi {)}^{3}2E]$, including the appropriate sums and symmetry factors. Lorentz invariance and current conservation give the transverse tensor form. In this convention ${\rho }_h$ is the hadronic photon spectral density, with mass dimension $-2$, rather than a density of a stationary time series.

The electron annihilation amplitude can be written $e\bar{v}(p_{+}){\gamma }_{\mu }u(p_{-})⟨X|A^{\mu }|0⟩$, up to an overall phase. The leptonic tensor after the initial spin average isIt obeys $q^{\mu }{L}_{\mu \nu }=0$ and $g^{\mu \nu }{L}_{\mu \nu }=-s$. In the center-of-momentum frame, the flux denominator in the hint is $2s$, since the relative speed is 2 and each beam energy is $\sqrt{s}/2$. Contracting the tensors gives the inclusive relativistic cross-sectionHere $\alpha$ is the fine-structure constant. The $1/4$ in the initial spin average is essential because the hint's particles were spinless.

For the other common convention, define the hadronic electromagnetic current with the coupling omitted, $J_h^{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }^{\mu }{q}_f$, and write its inclusive tensor asEquivalently, for $i\int d^{4}x{e}^{iqx}⟨0|T{J}_h^{\mu }(x)J_h^{\nu }(0)|0⟩=(q^{\mu }{q}^{\nu }-q^2{g}^{\mu \nu }){\Pi }_J(q^2)$, one has ${\rho }_J=\mathrm{Im}{\Pi }_J/\pi$. At leading order in the electromagnetic interaction, $⟨X|A^{\mu }|0⟩$ is proportional to $e⟨X|J_h^{\mu }|0⟩/s$. ThusIf the symbol ${\rho }_h$ is instead used for this dimensionless hadronic electromagnetic-current spectral density, the last formula applies with ${\rho }_J$ renamed ${\rho }_h$. The normalization must not be silently switched between these formulas.

At $\sqrt{s}$ well above the strong-coupling scale of Quantum chromodynamics, asymptotic freedom makes production over distances of order $1/\sqrt{s}$ perturbative. The electromagnetic current initially creates a quark-antiquark pair. Subsequent strong interactions produce hadrons, but an inclusive sum over all hadronic final states is much less sensitive to this rearrangement than an exclusive channel. This is the regime in which quark-hadron duality motivates a leading parton model calculation, with radiative and power-suppressed corrections. It is not a pointwise theorem at individual resonances or near thresholds; the comparison is most reliable for sufficiently inclusive or suitably averaged high-energy observables. This argument remains restricted to photon exchange, even where additional electroweak channels could also contribute.

For one active massless quark flavor with charge $Q_{f}e$, the tree-level scattering amplitude isThe two gamma-matrix traces, the initial spin average, and the final color charge sum yieldwhere $t=-s(1-\cos \theta )/2$ and $u=-s(1+\cos \theta )/2$. The massless two-body Lorentz-invariant phase space then givesSumming the distinct final flavors, rather than interfering amplitudes for them, givesAccording to the permitted approximation in the paper, active flavors satisfy $m_f^2<s$ and are treated as massless; the others are omitted. This is the stipulated step approximation, not the exact pair-production threshold $s\ge 4m_f^2$.

The hadronic R ratio is $R=N_c\sum Q_f^2$, relative to the massless muon-pair relativistic cross-section $4\pi {\alpha }^2/(3s)$. If $N_u$ and $N_d$ active flavors have charges $2/3$ and $-1/3$, respectively,For $N_c=3$, the usual sets of three, four, five and six active flavors give $R=2,10/3,11/3,5$. In the two spectral conventions the same leading calculation gives

In the parton model, the incoming parton has $k=\xi P_H$ and the outgoing massless parton has $k^{\prime }=\xi P_H+q$. Neglecting $P_H^2$ and imposing its on-shell condition givesThus the Bjorken scaling variable measures the struck parton's longitudinal momentum fraction in this approximation. The inelasticity is unchanged when the hadron momentum is replaced by its parton's collinear momentum:Let $S_H=(p+P_H{)}^2\simeq 2p\cdot P_H$. The partonic invariant in part (b) is $s(\xi )=\xi S_H$, and $Q^2=-q^2=xy{S}_H$. Insert its two parton cross sections into the parton distribution function convolution:Since the event's measured $x$ fixes $\xi =x$, the integrand is the cross-section density in $x$. No additional Jacobian from $y_q$ is needed because $y_q=y$. ThereforeEquivalently $S_H=2E_{\nu }{E}_H(1-\cos {\vartheta }_{\nu H})$ in the chosen massless-hadron frame. The explicit factor $x$ comes from the partonic energy $s=x{S}_H$, not from redefining the parton distribution functions. This expression uses the stated two-flavor and low-energy approximations, with masses and generation mixing omitted.

On the physical positive-energy branch, put $\omega =q^0+\xi E_P$. The massless outgoing parton obeysSince $E_{k^{\prime }}=\omega >0$ on the support and $\nu >0$, the positive-energy on-shell delta function identity givesInsert this into the parton model sum, and writeThe parton distribution functions $q_f,{\bar{q}}_f$ are number densities in momentum fraction. The contributing tensor becomesComparing with part (b), under the same leptonic tensor contraction, givesThis is the Callan-Gross relation for massless partons of spin angular momentum $1/2$ at leading order. The longitudinal deep-inelastic structure function is $F_L=F_2-2x{F}_1=0$ in this approximation. Target-mass effects and radiative Quantum chromodynamics corrections can change the relation. The leading parton model also gives Bjorken scaling: at this level the deep-inelastic structure functions depend on $x$ alone.

Use the massless collinear parton approximation in a high-energy frame: $P^2=0$, $k=\xi P$, and $k^{\prime }=\xi P+q$ with $k^{\prime 2}=0$. This neglects target-mass corrections to the parton model; it does not literally set a stationary massive target to a massless particle in the earlier flux formula.

For a quark of dimensionless charge $Q_f$, the electromagnetic vector current matrix element is $Q_f\bar{u}(k^{\prime }){\gamma }^{\mu }u(k)$. The spin average and gamma-matrix trace giveIntegrating the three-momentum Dirac delta function in the parton hadronic tensor leavesSince $k\cdot k^{\prime }=\xi \nu$, this isFor the massless electron momenta, $p^2=p^{\prime 2}=0$ and $q=p-p^{\prime }$. Substitution into the leptonic tensor givesand likewise $q^{\nu }{L}_{\mu \nu }=0$. These Ward identities eliminate every term with an exposed $q$ index in the contraction. ThereforeHere $\doteq$ means equality after contraction with the leptonic tensor. The shortened tensor is not itself conserved; the omitted terms restore current conservation in the full hadronic tensor.

Evidence for quarks comes from several kinds of observation that test different aspects of their interpretation. The light meson and baryon spectra organize into approximate flavour multiplets with the charges and strangeness expected for $q\bar{q}$ and $qqq$. In particular, the baryon octet and baryon decuplet encode the three light quark flavours and their spin couplings. This is an economical account of hadron quantum numbers, but spectroscopy alone would not establish that the constituents behave as localized scattering centers.

Deep inelastic scattering supplies that second test. In energetic electron–proton scattering the outgoing lepton measures the energy and momentum transferred through a virtual photon, while the broken-up target is summed over. Approximate Bjorken scaling means the structure functions depend primarily on $x=Q^2/(2P\cdot q)$ rather than varying rapidly with the resolving momentum scale $Q^2$. It indicates nearly pointlike constituents. In the leading spin-half parton model,The charge-squared weights and the second relation test the constituents' fractional charges and spin-half nature. Scaling violations reflect quark and gluon radiation rather than a failure of the constituent picture.^[ref]

Particle jets give complementary final-state evidence. A high-energy quark produces a collimated spray of hadrons because confinement prevents it from emerging as an isolated particle. Back-to-back jets fit quark–antiquark production; three-jet events show hard gluon radiation. Their angular correlations test a spin-one radiated parton. This distinguishes constituents of hadrons from a literal observation of free fractional-charge particles.

Three independent arguments identify an additional threefold colour degree of freedom. First, the spin-aligned $uuu$ state of a ground-state ${\Delta }^{++}$ would have symmetric space, spin and flavour factors. The Pauli exclusion principle requires an antisymmetric additional factor. The three-quark colour singlet supplies it. The same construction accommodates the spin-aligned $sss$ state of the ${\Omega }^{-}$ without altering its measured quantum numbers.

Second, the inclusive hadronic R ratio, away from thresholds and narrow resonances, is at leading electromagnetic orderEach quark flavour has $N_c$ distinguishable colour final states. With $N_c=3$, the predicted plateaux for three, four and five accessible flavours are $2$, $10/3$ and $11/3$; without the factor three they would be a third as large. The measured inclusive normalization is therefore a colour-counting test, with radiative and threshold corrections included in precision comparisons.^[ref]

Third, the neutral pion's two-photon amplitude contains a quark-loop colour sum, proportional to $N_c(e_u^2-e_d^2)$; its rate and the other evidence are consistent with three colours. Quantum chromodynamics identifies this degree of freedom as a gauge charge, with eight gluons. Observed hadrons are colour singlets; the evidence for colour is its quantitative and symmetry consequences, not visible colours or isolated coloured quarks.

Use the parton model, neglect antiquarks, binding and target mass corrections, quark masses, muon mass, QCD scaling violation and the W boson propagator correction. The struck quark is effectively free during the short interaction and carries parton momentum fraction $x$, giving $\hat{s}\simeq xs$. The charged current reaction is ${\bar{\nu }}_{\mu }{u}_i\to {\mu }^{+}{d}_j$. Its helicity amplitude has the same $(1-y)$ factor as in a. Define the effective parton distribution function by $q(x)={\sum }_{i,j}|V_{ij}{|}^2{u}_i(x)$, summing only kinematically accessible final quarks; for valence-only protons and accessible $d,s$ this is approximately $u(x)$. It is not the unweighted sum $u+d$.Here $y=p\cdot (k-k^{\prime })/(p\cdot k)$ is the inclusive deep inelastic scattering variable, and $x=Q^2/(2p\cdot (k-k^{\prime }))$ in the scaling limit. Integrating $0<y<1$ gives $1/3$, henceThe moment measures the momentum fraction in the active quarks. Approximate linearity is evidence for scattering from pointlike spin-one-half constituents with approximately scale-independent parton distribution functions, rather than coherent scattering from an extended proton. It is an intermediate-energy prediction: thresholds, experimental deep inelastic scattering cuts, QCD scaling violation and eventually the finite W boson propagator modify it.

The basic Feynman diagrams have ${\tau }^{-}\to {\nu }_{\tau }{W}^{-\ast }$ followed by $W^{-\ast }\to e^{-}{\bar{\nu }}_e$, ${\mu }^{-}{\bar{\nu }}_{\mu }$, or $d\bar{u},s\bar{u}$, with the latter pairs undergoing hadronization. See the charged-current diagrams in a. Neglect all daughter masses, QCD corrections and inaccessible charm final states, and assume lepton universality. Each leptonic mode then has one unit of decay width; the two accessible quark combinations have three color charge states and total CKM weight $|V_{ud}{|}^2+|V_{us}{|}^2\simeq 1$. ThereforeThe result is an inclusive parton model estimate. Perturbative QCD corrections increase the hadronic contribution, while muon mass, nonperturbative dynamics and radiative corrections also matter; the physical fraction is roughly $0.65$.

Every ordinary hadronic mode already contains a tau neutrino, so it contains a lepton as well as hadrons if invisible particles are counted. For an additional visible charged lepton, a virtual photon can internally convert in a hadronic mode, for example ${\tau }^{-}\to {\nu }_{\tau }{\pi }^{-}{\gamma }^{\ast }\to {\nu }_{\tau }{\pi }^{-}{e}^{+}{e}^{-}$. The extra electromagnetic vertices suppress such modes; they are not a large fraction of the inclusive width. Alternatively a daughter pion or kaon can decay semileptonically or leptonically in flight, for example ${\tau }^{-}\to {\nu }_{\tau }{\pi }^{-}{K}^0$ followed by a semileptonic neutral-kaon decay. Those are secondary decay chains, not independent primary tau lepton branching fractions. Lepton-flavour-violating neutrinoless modes such as ${\tau }^{-}\to {\mu }^{-}{\pi }^{+}{\pi }^{-}$ do not contribute appreciably in the Standard Model.

The parton model treats a fast proton as an ensemble of approximately collinear, nearly free quarks and antiquarks. A hard deep inelastic scattering collision resolves one parton on a time scale short compared with its binding motion. The probabilities are parton distribution functions; the simplest approximation neglects intrinsic transverse momentum, quark masses and the proton mass in the hard collision. Quantum chromodynamics introduces logarithmic violations of Bjorken scaling, so the parton distribution functions depend on a resolution scale as well as a parton momentum fraction.

Write the incident electron and proton four-momenta as $k,P$, the scattered electron four-momentum as $k^{\prime }$, and set $q=k-k^{\prime }$. With the $(+---)$ Minkowski metric, the Bjorken scaling variable and the inelasticity areAn incident parton with four-momentum $\xi P$ emerges on shell, giving $(\xi P+q{)}^2\simeq 2\xi P\cdot q-Q^2=0$. Thus $\xi =x_B$: the observable Bjorken scaling variable measures the struck parton momentum fraction in this approximation.

For clarity, measure the scattering angle $\vartheta$ from the incoming electron direction, and denote the outgoing electron energy by $E_e^{\prime }$. Direct four-momentum products giveThe expression uses the known beam energies and two measured quantities. An electromagnetic calorimeter measures $E_e^{\prime }$; a tracking detector, the collision interaction vertex and the position of the electromagnetic shower determine $\vartheta$. Coverage close to the incoming electron direction matters at small $Q^2$. A hadronic calorimeter supplies an independent consistency check and rejects backgrounds. If the angle is instead measured from the incoming proton, its sine and cosine half-angle factors interchange.

Neglect the participating quark and lepton masses, so that the weak charged current selects a left-helicity $u$ and a right-helicity $\bar{d}$. Choose $+z$ along the proton beam in the partonic centre-of-momentum frame. Then the up quark moves along $+z$ with spin projection $-1/2$, while the down antiquark moves along $-z$ with helicity $+1/2$ and also has spin projection $-1/2$ along $+z$. ThusThe collinear two-body initial state has no orbital projection about its collision axis. In this approximation the produced W boson therefore has the indicated transverse polarization; its longitudinal spin state is absent. The unpolarized hadron beams contain other constituent spin states, but these do not participate in this massless charged-current subprocess.

In the final state the Positron has helicity $+1/2$ and the electron neutrino has helicity $-1/2$. Along the outgoing charged-lepton direction their combined angular-momentum projection is $+1$, because the oppositely moving neutrino has spin projection $+1/2$ on that axis. Finite masses, noncollinear production and sea contributions modify the ideal configuration.

Use the leading-order parton model: scatter incoherently from an approximately on-shell collinear quark carrying nucleon momentum fraction $x$, neglect quark and lepton masses, use an unpolarized target, and neglect nuclear binding and higher-twist corrections. The condition $Q^2\gg m_p^2$ suppresses target-mass effects, while $Q^2\ll M_W^2$ permits the local Fermi interaction instead of the full W propagator. CKM-weighted accessible final flavours are approximated by their leading light-quark sums. The struck-quark centre-of-mass energy is $\hat{s}\simeq xs$, with $s=(k+p{)}^2\simeq 2m_N{E}_{\nu }$.

A neutrino scatters through $\nu d\to {\mu }^{-}u$ or $\nu s\to {\mu }^{-}c$, and through the corresponding crossed antiquark channels such as $\nu \bar{u}\to {\mu }^{-}\bar{d}$. The chiral charged current makes the massless quark contribution independent of $y$ and the antiquark contribution proportional to $(1-y{)}^2$. The scattering probability is weighted by the parton distribution functions; the explicit factor $x$ comes from $\hat{s}=xs$. In the usual convention $q(x)dx$ is a number probability density, while $xq(x)$ is a momentum density. Approximate Bjorken scaling leaves mainly $x$ dependence; QCD evolution adds a logarithmic $Q^2$ dependence.