Quark-model constituent bookkeeping ($Q={\sum }_i{Q}_i,B={\sum }_i{B}_i,S={\sum }_i{S}_i$)

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.

A particularly economical example is the baryon decuplet. Restrict to fully symmetric three-quark flavor states. The nonnegative occupation numbers satisfy $n_u+n_d+n_s=3$, giving $\left(\genfrac{}{}{0ex}{}{3+3-1}{3}\right)=10$ possibilities by stars and bars. Their additive labels areThese formulas follow directly from the constituent table. With fixed $n_s$, there are $4-n_s$ possible values of $n_u$, so the weight diagram has rows of four, three, two and one states. Its outer triangle is therefore a consequence of the constituent construction, rather than an arbitrary drawing convention.

The bottom corner has $n_s=3$, hence $I_3=0$, $Y=-2$ and $Q=-1$: the Omega baryon position. The top-right corner has $uuu$ and $Q=2$, corresponding to the doubly charged Delta baryon. Every point satisfies the Gell-Mann--Nishijima formula $Q=I_3+Y/2$.

The flavor construction is only part of the state. The Pauli constraint on three-quark flavour multiplets also involves spin angular momentum, the spatial wavefunction and color charge. In particular, three identical fermions cannot be assigned a completely symmetric total state merely because the flavor diagram is symmetric. Nor does the same flavor diagram determine the hadron masses: an approximate flavor symmetry can organize a spectrum without making all its members degenerate.