Past exam of the nuclear and particle physics course of the University of Oxford / 2022 / 4 / b / Solution

The diagram refers to ground state, $L=0$, color singlet baryons. Their spatial wavefunction is symmetric and their color wavefunction is antisymmetric. The Pauli exclusion principle therefore requires a symmetric combined spin-and-flavor wavefunction. One cannot independently attach every three-spin coupling to each flavor assignment.

For three spin-$1/2$ particles, there are formally two ways to couple to total spin $1/2$. In the ground state quark model these are tied to the corresponding mixed-symmetry flavor functions. Their required symmetric combination yields one physical spin-$1/2$ flavor octet, not two independent species at each peripheral weight. In spin-flavor symmetry notation,under flavor $SU(3)$ and spin $SU(2)$. Thus there is one peripheral species in the baryon octet, each still possessing its two ordinary spin projections. The two abstract flavor-octet copies in $3\otimes 3\otimes 3$ do not double the physical ground state octet.

At the dotted positions, the flavor content would be $uuu$, $ddd$ or $sss$. These flavor wavefunctions are completely symmetric. The spin wavefunction must then also be completely symmetric, which means total spin $3/2$, not $1/2$. The corresponding baryon decuplet states are ${\Delta }^{++}$, ${\Delta }^{-}$ and ${\Omega }^{-}$. This exclusion is for the stated ground state spin-$1/2$ multiplet; it is not an exclusion of all excited baryons at those weights.