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In the Murray Gell-Mann model, particles are brought together as a function of their angular momentum. The classification diagrams can be seen as irreducible representations of $SU(3)$, following multiplication rules $3\otimes 3\otimes 3$=$10\oplus 8\oplus 8\oplus 1$.

My question is: must $J^p$ be equal between each of these decuplet, octets, and singlet? (Is there a $J^p=1/2^+$ baryon decuplet?)

I know it is the case for the mesons, but I am not sure if this is by chance or it is demanded.

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The answer to your questions is no , and for your parenthetical specific question, yes--but it may be a weirdo one.

First, review your generic, flavor-blind quark assignments. The total J comes as a combination of the three s=1/2 quarks and the orbital L, which enforces its even or odd parity P to the baryon. The total w.f. must be fermion interchange antisymmetric, is color antisymmetric, so the combination of flavor SU(3) and spin/angular momentum part must be over-all symmetric. Since the 10 is flavor-symmetric, its combination J w.f. must be symmetric.

The decuplet that contains the Ω-, which is just 3 strange quarks, must be spin symmetric for L=0, so then S=J=3/2+ . This is a fact of nature, the confirmation linchpin of SU(3), historically, which provides a counterexample to your first question/mis-conjecture.

The trailing parenthetical question of (is there...) is answerable in the affirmative, in principle, for S=3/2, L=2, resulting in a 1/2+, but it would take enormous and improbable baryon-spectroscopic gumption to isolate and confirm a state like that...

As for your oddly flawed rule for the mesons... you have a pseudoscalar meson octet and a vector meson octet, see the two figures here. How would your rule work? In this article, you see the analogous spin/angular momentum composition rules for the mesons.

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