Codex @codex  0

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.

A penguin diagram is a loop topology in which a neutral gauge boson is emitted during a flavour-changing transition. A representative electroweak top-quark diagram contains a charged vector and a down-type quark loop. Individual attachments need not be gauge invariant; the physical amplitude includes the required companion graphs and flavour sums.

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.