# Identifying the flavour singlet baryon

I would like to ask how to identify the lowest lying flavour-singlet baryon in the data on known states of particles. So far I managed to derive the following constraints:

1. Its quark content must be $uds$ by full antisymmetry in flavour indices. Thus it goes by the name $\Lambda$. As always, it is also fully antisymmetric in colour.

2. If its spatial wavefunction was an $s$-wave then we would need to make it fully antisymmetric in spin indices to satisfy Pauli's principle. This is impossible for three doublets. Hence it must be a p-wave. Therefore parity $P=1$. Then to make the whole thing fully antisymmetric I need $S= \frac{3}{2}$.

3. By angular momenta addition it has $J=\frac{1}{2}, \frac{3}{2}, \frac{5}{2}$. However I have no intuition whatsoever which of these is prefered.

Of course there is a lot of states in PDG which satisfy these requirements. The questions are how do I decide which one of these is the lowest-lying flavour singlet? Moreover, can we infer from the theory any further constraints on its quantum numbers and other characterstics?

• What does "flavor singlet" means ? I guess you have some flavour symmetry transformations in mind but you are not stating them ... Commented Nov 17, 2017 at 22:59
• In any case, maybe I am rusty on Pauli, but are you sure that it is relevant to consider exclusion principle when the flavor of the quarks is different? Your state has three quarks all of different flavors, so I think they qualify as "distinguishable" particles and, please check!, Pauli poses no constraints on their wave-function. Commented Nov 17, 2017 at 23:06
• I mean $SU(3)$ flavour symmetry mixing the $u$, $d$ and $s$ quarks. It would be a symmetry of the strongly interacting spectrum if quarks were massles. Commented Nov 17, 2017 at 23:14

I found the answer to this question in the paper "SU(3) systematization of baryons" by V. Guzey and M. V. Polyakov (arXiv:hep-ph/0512355). Apparently the lightest flavour singlet is the state $\Lambda(1520)$ with $J^P=\frac{3}{2}^-$.