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In the incoherent massless collinear parton approximation, $f_f(\xi )d\xi$ is the number of quarks in a momentum-fraction interval, not a covariantly normalized external state weight. A single-particle response divided by its external-state normalization is proportional to $w/(2k^0)$; the hadron normalization is $2P^0$. Thus converting a number-weighted parton sum to the target tensor contributes $P^0/k^0=1/\xi$. This gives the displayed convolution, including antiquarks. The spin-averaged electromagnetic parton tensor and $\delta (2\xi P\cdot q-Q^2)=\delta (\xi -x)/(2P\cdot q)$ then give $F_1=\frac{1}{2}{\sum }_f{Q}_f^2(f_f+{\bar{f}}_f)$ and $F_2=x{\sum }_f{Q}_f^2(f_f+{\bar{f}}_f)$, proving the Callan-Gross relation at this order. Target masses, transverse momentum and radiative effects are beyond this leading approximation.

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.

Use an incoming spacelike momentum transfer $q$, so the physical absorption process has $p_X=P+q$, $Q^2=-q^2>0$, and $P\cdot q>0$. There is a sign inconsistency in the original PDF: its delta function has $P-p_X-q$, while its positive Bjorken x definition and target formulas use incoming $q$. With that literal delta function and $x>0$, the final-state invariant would bewhich becomes negative in the stated high-energy limit and cannot represent a physical hadronic final state. We therefore use ${\delta }^4(P+q-p_X)$, preserving the printed $x=Q^2/(2P\cdot q)$ and structure-function conventions. Equivalently, keeping an outgoing momentum variable requires reversing the signs in those definitions as well.

Assume an unpolarized target, incoherent scattering from effectively free spin-one-half partons, and the massless collinear parton approximation. In the Bjorken scaling limit, take $Q^2,P\cdot q\to \infty$ at fixed $0<x<1$, neglecting target and quark masses and intrinsic transverse momentum relative to $Q$. Write the incoming struck-quark momentum as $k=\xi P$ in the leading lightlike-reference approximation. Let $f_f(\xi )$ and ${\bar{f}}_f(\xi )$ be quark and antiquark parton distribution functions, counting all colours. For the naive scaling result also neglect radiative scaling violations; in full QCD these distributions depend logarithmically on the hard scale.

The outgoing struck quark has momentum $k^{\prime }=k+q$. Its on-shell condition givesAverage its initial spin and sum its final spin. The spin-averaged electromagnetic parton tensor has numeratorThe supplied three-gamma identity gives this directly: contract its middle index with $(k+q{)}_{\lambda }$ and trace with $/k$; the ${\gamma }_5$ trace vanishes. Thus no antisymmetric term survives the unpolarized spin average.

Define transverse tensorsOn the cut, $2k\cdot q=Q^2$, so $\tilde{k}=k+q/2=\xi \tilde{P}$. Expanding the two transverse terms givesThis also explicitly verifies electromagnetic current conservation, $q_{\mu }{S}^{\mu \nu }=q_{\nu }{S}^{\mu \nu }=0$.

The one-particle final-state Lorentz-invariant phase-space measure obeysIncluding the current's charge factor and the given $1/(4\pi )$ normalization, a single quark therefore contributesAn antiquark has the same unpolarized tensor and the same squared charge $Q_f^2$.

The parton convolution normalization for a hadronic tensor isThe $1/\xi$ is important: the distributions count parton numbers, whereas covariantly normalized one-particle responses carry the state normalization $2k^0$. The incoherent sum of normalized responses is $W/(2P^0)={\sum }_f\int d\xi (f_f+{\bar{f}}_f)w_f/(2k^0)$; in the collinear frame $k^0=\xi P^0$, giving the factor above. There is no extra colour multiplicity because it is already included in the distributions.

On the physical cut,Comparing the two transverse coefficients in the hadronic tensor gives, with $n_f=f_f+{\bar{f}}_f$,Therefore the requested deep-inelastic structure functions areThe notation $f_f$ here is the question's $f$ with its flavour index made explicit. The calculation proves the Callan-Gross relation, $F_2=2x{F}_1$, rather than merely citing it. Finite masses and radiative corrections modify this leading approximation; at leading QCD order the same formulas use scale-dependent parton distribution functions.

The quark number sum rule follows because a distribution is a number density. Integrating $f_f$ counts quarks and integrating ${\bar{f}}_f$ counts antiquarks; their difference is the net conserved flavour charge:Every sea quark-antiquark pair contributes zero to this difference. More formally the flavour-vector charge counts quarks minus antiquarks and has a fixed eigenvalue on the hadron state. Thus $N_f$ is the valence or net quark number, not the total number including the sea.

Use both final assumptions printed in the question: only up/down flavours and no antiquarks. The second does not follow just from restricting the flavour list. Then $\int f_u=N_u$, $\int f_d=N_d$, with squared charges $Q_u^2=4/9$ and $Q_d^2=1/9$. The proton has $(N_u,N_d)=(2,1)$ and the neutron $(1,2)$. Hence the valence-only proton and neutron structure-function moments areThe integral limits assume the scaling approximation is valid for the indicated weighted moments, for example by dominated convergence of the weighted integrands. The same net-quark sum rule alone would not fix these individual moments in the presence of antiquarks, since $F_2$ involves the sum of quark and antiquark distributions.