Baryon Octet/Decuplets

https://www.physicsforums.com/showthread.php?t=291801

 

The short answer is the exclusion principle. But to see that you need to know a bit about groups and their representations. Specifically that the direct product of 3 fundamental '3's of SU(3) decomposes like 3 * 3 * 3 = 10 + 8 + 8 + 1. The decuplet (which contains uuu) is necessarily spin 3/2 and completely symmetric.

 

The attempt to construct a spin 1/2 state |uuu> results in a violation of the Pauli exclusion principle.

As far as I remember from my QCD lectures (~ two decades ago!) the explanation in the other thread misses the spatial part of the wave function. One must take into account
|color> * |spin> * |isospin> * |space>
Ofcourse this doesn't make things easier ...

 

yes the most basic answer is to combine the two of haelfix and tom.stoer

 

The ground state L=0 is space-symmetric in the constituent quark model.

 

The answer is simpler than group theory, but is based on the Pauli principle (Fermi-Dirac statistics) and the addition of spins. The combined state of (color)(space)(spin) must be completely antisymmetric for three identical u quarks. The wave function is antisymmetric in color. The spin addition 1/2+1/2+1/2=1/2 is of mixed symmetry, and so cannot combine with the presumed symmetry spatial ground state and antisymmetric color state to form a completely antisymmetric state. Spin 1/2 for three quarks is only possible if they are not all identical, for instance uud.