 Quark-model baryon-decuplet construction

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.