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The valence quark distributions describe net quark content: $\int u_v(x)dx=2$ and $\int d_v(x)dx=1$ for a proton. Their momentum densities $x{u}_v,x{d}_v$ are small at very small $x$, peak at moderate momentum fractions, and vanish toward $x=1$, where one parton carrying almost all proton momentum is unlikely. The larger $u$ distribution reflects two valence up quarks and its comparatively harder shape; its pointwise value need not be exactly twice the down distribution.

The sea quark distribution is generated largely by gluon radiation and $g\to q\bar{q}$ splitting. It grows strongly toward small $x$ and falls at large $x$. The plotted sea curve is scaled by $0.05$, so it must be multiplied by twenty before numerical use. The total up-quark PDF isusing the valence-up distribution and the up-sea distribution. The summed sea curve alone does not fix $\bar{u}$ without a flavour assumption. Under the later stated four-flavour symmetry, $\bar{u}=\Sigma /8$.

A proton contains up quarks on which the charged current can act:through $W^{-}$ exchange, weighted by $V_{ud}$. Spectator quarks complete the hadronic final state. At the exclusive nucleon level the charge assignment is $e^{-}p\to {\nu }_{e}n$; at the high momentum transfers specified, the generally dominant description is inclusive $e^{-}p\to {\nu }_{e}X$ with proton structure and hadronic breakup, not an unsuppressed pointlike proton-to-neutron transition. The quark-level diagram in the figure displays this allowed charge flow. The electron's left-chiral selection is the same as in part d.

Let ${\theta }^{\star }$ be the angle from the incoming up quark, along the proton direction, to the outgoing charged antilepton in the W boson rest frame. The initial and final spin projections in ii are $-1$ and $+1$. The Wigner small-d matrix therefore givesAfter squaring and normalizing over $c=\cos {\theta }^{\star }$, the charged-lepton angular distribution isIts integral is one since ${\int }_{-1}^1(1-c{)}^{2}dc=8/3$. The charged antilepton is preferentially emitted toward the antiproton. Measuring the angle from the incoming down antiquark instead replaces $c$ by $-c$ and produces $(1+c{)}^2$; the axis convention is part of the answer. Per unit solid angle the normalized density is $3(1-\cos {\theta }^{\star }{)}^2/(16\pi )$.

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.

A leading valence-level process is $u\bar{d}\to W^{+}\to e^{+}{\nu }_e$, with the up quark supplied by the proton and the down antiquark by the antiproton. The same Feynman diagram works with $e^{+}{\nu }_e$ replaced by ${\mu }^{+}{\nu }_{\mu }$. Both vertices are weak charged current vertices; the production scattering amplitude contains the CKM matrix element $V_{ud}$ or its conjugate according to the chosen fermion-flow convention. The intermediate W boson is produced near its mass shell.

Assume approximate charge symmetry of nucleon parton distributions, the interchange of up quarks and down quarks under the isospin rotation carrying the proton into the neutron:Corrections from the up quark-down quark mass difference and the electromagnetic interaction are neglected. With the proton-only shorthand in d,If charm is active, add the same $4x(c^p+{\bar{c}}^p)/9$ to each; heavier flavors are treated similarly. These formulae use the same $Q^2$ and factorization convention for all parton distribution functions.

Subtracting the deep-inelastic structure functions in e givesThus at moderate and large $x$, the charge-weighted up quark content of the proton exceeds its down quark content. In the valence quark dominated regime this is consistent with the $uud$ valence structure, compared with $udd$ for the neutron. The larger $2/3$ up quark electric charge makes the proton scatter more strongly. This observation alone does not establish $u^p(x)=2d^p(x)$ pointwise: the valence-number sum rules constrain integrals, not every $x$ bin.

If the small-$x$ observation means equality to good relative accuracy, it supports dominance by approximately charge-symmetric sea quarks, with $U\simeq D$ and negligible relative valence excess. Gluon splitting creates quark-antiquark pairs, and the leading strong interaction treats light quark flavors similarly. It does not follow that every sea quark flavor has exactly the same distribution at every $x$.

Logical qualification. Equality of the limiting values alone is weaker than equality of their ratio. For example $U=2$, $D=1$, $S=0$ near $x=0$ gives $F_2^p=x$ and $F_2^n=2x/3$: both tend to zero, although $U\ne D$ and $F_2^n/F_2^p=2/3$. The usual inference about a symmetric sea needs $F_2^n/F_2^p\to 1$ or equivalent information about relative differences, rather than merely coincident zero limits.

The neutron has valence content $udd$ and total spin $J=1/2$; its ground-state parity is positive. The two down quarks contribute charge $-2/3$ and the up quark contributes $+2/3$, giving net charge zero. “Three valence quarks” specifies the net flavor quantum numbers, rather than excluding gluons or additional sea quark–antiquark components from the full Quantum chromodynamics state.

In the approximate light-flavor SU(3) classification, the neutron belongs to the ground-state baryon octet, containing eight hadrons:These are eight particle species, without counting their antiparticles again as extra members of the same octet. The corresponding antibaryons form the conjugate particle set. Within the smaller SU(2) isospin classification, the neutron instead belongs to the two-member nucleon doublet $(p,n)$. Stating the symmetry distinguishes the usual eight-member flavor multiplet answer from the also valid two-member isospin multiplet. The baryon decuplet has different ground-state spin and is not the neutron's octet.

The charged weak current changes one neutron down quark into an up quark:The other two valence quarks are spectators in this tree-level Feynman diagram. The intermediate W boson is highly virtual; the available neutron decay energy is far below its on-shell rest energy. Charge is conserved at each vertex, and the electron's lepton number is balanced by the electron antineutrino.

The reversed quark-level charged current is allowed as an interaction, but a free proton cannot decay as $p\to n+e^{+}+{\nu }_e$ because $m_p<m_n+m_e$. Four-momentum conservation leaves no accessible final-state Lorentz-invariant phase space. The absence of this free-proton beta-plus decay is therefore kinematic, not an absence of a $u\to d$ weak vertex. Binding-energy differences in a nucleus, or incoming energy in a collision, can change whether the corresponding conversion is possible. Exotic baryon-number-violating proton decay is a separate question and is not generated by this ordinary charged-current diagram.

For a three-light-flavor example, take up quarks, down quarks and strange quarks with the following additive quantum numbers. Here $Q$ denotes electric charge in units of the positive elementary charge $e$, $B$ is baryon number, $S$ is strangeness, $I_3$ is the third component of isospin, and $Y=B+S$ is flavor hypercharge. An antiquark reverses all five displayed values. These are flavor coordinates, not spatial coordinates or electroweak hypercharge.

Adding these columns immediately explains familiar hadrons. A proton with valence quarks $uud$ has $Q=2/3+2/3-1/3=1$ and $B=1$; a neutron with $udd$ has $Q=0$ and $B=1$. A positive kaon with $u\bar{s}$ has $Q=2/3+1/3=1$, $B=0$ and $S=+1$. Thus the same constituent rules describe both baryons and mesons.

Two hadrons can have identical additive labels while differing in their internal state. The Lambda baryon and neutral Sigma baryon both have valence content $uds$, yet their total isospin differs. Likewise, the neutral pion flavor direction $(u\bar{u}-d\bar{d})/\sqrt{2}$ and the Eta octet state $(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$ both have $Q=Y=0$, but their inner product vanishes. Expanding that inner product gives $(1-1+0)/\sqrt{12}=0$. Constituent counting must therefore be supplemented by quantum superposition and the symmetry of the state.

For a colour singlet hadron built entirely from up quarks, down quarks, strange quarks, antiquarks and gluons, the net number of quarks is a multiple of three. The central SU(3) colour element multiplies the state by ${\omega }^{n_q-n_{\bar{q}}}$, while a singlet must be unchanged. The central flavour element supplies the same phase because all three light flavours form the fundamental representation. Every resulting flavour multiplet therefore has zero SU(3) triality. Heavy quarks are singlets of light-flavour SU(3), so hadrons containing them do not obey this argument. The zero-triality restriction allows higher light-quark multiplets as well as the familiar singlet, octet and decuplet.