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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.

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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    $\begingroup$ Sorry, What is PEP here? Thanks $\endgroup$ – wonderich Sep 30 '17 at 17:11
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The valence quark content of a baryonic state is $qqq$, that is to say three identical quarks make up the content of the bound state amongst a sea of partons. Each quark therein carries a series of quantum numbers which serve to denote the different possible states of the quark and, collectively, must be such that a simultaneous permutations of the degrees of freedom yields an overall antisymmetric state, in accordance with the fact the quantum state must obey Fermi-Dirac statistics.

For non excited states (the so called lowest lying states), the spatial component of the direct product is always symmetric ( $S$-wave orbital angular momentum) and colour is antisymmetric so the combination $|\text{spin} \rangle \otimes |\text{flavour} \rangle$ must be symmetric. The observable states transform under irreducible multiplets of (approximate) $SU(3)$ flavour symmetry and so the '$|uds \rangle$' state is actually a flavour symmetric combination in the $10$ decuplet, a flavour antisymmetric combination in the $1$, and appears with mixed flavour symmetries in the remaining two octets in agreement with the group theoretic decomposition $3 \otimes 3 \otimes 3 = 1 \oplus 8 \oplus 8 \oplus 10$.

So. e.g for the totally flavour symmetric combination, the spin state of the wavefunction must be in one of the symmetric spin $3/2$ states.

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