Baryon wavefunctions

Question

The textbook I am currently using states that when all quarks have the same flavor, there are no $J=1/2$ baryon wavefunctions for the ground state $l=0$.

Is this an experimental result or is there a theoretical reason for this?

Answer 1

Both. It's fermi statistics.

The overall wavefunction must be totally antisymmetric under any (fermionic) quark interchange. The color part of the wavefunction is antisymmetric, so the l and spin wavefucntions must be symmetric. (There is no play in the flavor part of the wavefunction: take the quark to be s, so that part is sss.)

• So the spin wavefunction is the symmetric part of $1/2\otimes 1/2\otimes 1/2$, which is then necessarily spin 3/2, added to the l singlet part, yielding J=3/2.

The two spin 1/2 components of the triple spin product are both of mixed symmetry, so cannot be symmetric as required. It is thus impossible to contrive J=1/2 states!

You appreciate the actual $\Omega^-$ is part of the J=3/2 decuplet, and there is no such triply strange J=1/2 baryon.