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?
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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.
Cosmas Zachos
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Thank you for the response. As a brief follow up, for the calculation of a proton's magnetic moment the following wavefunction was used: 1/(6)^0.5 (2u↑u↑d↓ - u↑u↓d↑ - u↓u↑d↑). This wavefunction is only antisymmetric under the exchange of the like quarks, yet gave the correct result. Why is this OK to use? – jore1 May 05 '23 at 14:44
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That's the point: it is in the baryon octet, so it has mixed spin and isospin symmetry, I =1/2, (unlike the isosinglet example above.) J and I "coordinate" to have overall complete symmetry, even though there are antisymmetric "sectors" in each. u,d are both involved in the antisymmetric isosinglet, unlike two u s. – Cosmas Zachos May 05 '23 at 15:16
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See this one. – Cosmas Zachos May 05 '23 at 15:34