I'm learning about particle physics at the moment, and have read that the J=3/2 multiplet contains baryons comprised of uuu, ddd, sss quarks. But the J=1/2 multiplet contains no baryons consisting of three quarks of the same flavour. Is there a reason for this? Is it something to do with quantum numbers the three quarks can take? (That doesn't really make sense to me, as the J=3/2 requires all spins of quarks aligned, so has even less "freedom" than than the J=1/2...)
Because the 3 quarks are antisymmetric in the color degree of freedom, they obey effective Bose statistics. Three identical quarks cannot be in a spin 1/2+1/2+1/2=1/2 state because this spin state is not fully symmetric.
It arises from Fermi-Dirac statistics. The wavefunction of the three quarks must be totally antisymmetric. It consists of parts: space, spin, flavor and color. The space part, assuming L=0, is symmetric. The flavor part, assuming three identical quarks, is symmetric. The color part, assuming the baryon will be colorless is totally antisymmetric. That leaves just the spin, and it must be totally symmetric. The totally symmetric coupling of three spin-1/2's is J=3/2.