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I am referring to this, http://home.web.cern.ch/about/updates/2014/04/lhcb-confirms-existence-exotic-hadron

So how does this work if we stick to keeping quarks in the 3 dimensional fundamental representation of $SU(3)$?

This bound-state seems to have 2 anti-quarks and 2 quarks. So with just 3 colours how do we make the whole thing anti-symmetric with respect to the colour quantum number?

Is there anything called "anti-colour" quantum number that an anti-quark can posses so that there are a total of $(3\times 2)^2$ colour options to choose from for the 2 quarks and 2 anti-quarks? I have never heard of such a thing!

The point is that unlike the $U(1)$ charge, the non-Abelian charge doesn't occur in the Lagrangian for the quarks. The Lagrangian only sees the different flavours, the gauge groups and the gauge coupling constant.

JamalS
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user6818
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1 Answers1

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Antiquarks can be distinguished from quarks, so you only need to antisymmetrize two at a time. That's no problem, and even if you had 3 quarks it wouldn't be. Furthermore, you only need the total state to be antisymmetric. You could have antisymmetry in space, symmetry in spin and symmetry in color, and the whole thing would be antisymmetric. (Like how you can put two electrons in each atomic orbital and both singlet and triplet are allowed.).

Robin Ekman
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