The baryon wavefunction is comprised of the direct product of contributions forming different Hilbert spaces such that $$|\Psi \rangle = |\text{spin} \rangle \otimes |\text{flavour}\rangle \otimes | \text{colour}\rangle \otimes |\text{space}\rangle. $$ The necessity for a colour degree of freedom is usually motivated in the literature from the delta(++) spin 3/2 containing quark content ##uuu##. It is flavour symmetric by inspection, is symmetric in the spin quantum numbers and for lowest lying states, has symmetric space state. The state thus contains identical fermions but is overall symmetric under interchange of any of the quarks. This is in violation of the PEP – the resolution was of course the addition of the colour degree of freedom which is necessarily antisymmetric so as to conform to the principle.(adsbygoogle = window.adsbygoogle || []).push({});

It’s then said that the generic wavefunction for a baryon is overall antisymmetric. Is this only the case where the baryon wavefunction is of the form ##\epsilon_{ijk} \psi^{(1)}_i \psi^{(1)}_j \psi^{(1)}_k## or ##\epsilon_{ijk} \psi^{(1)}_i \psi^{(1)}_j \psi^{(2)}_k## where in the former case we have all three qqq the same and in the latter only two are identical?

I say this because if we consider one of the baryons with no flavour symmetry e.g ##|uds \rangle## this is a state with no identical fermions so does the the PEP (i.e total wavefunction antisymmetry) have to hold for this case? (I guess analogously to the fact that the mesons have no requirement of antisymmetry because the content of the valence quarks is ##q \bar q## and this is never two identical quarks)

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# I Is baryon antisymmetry always true?

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