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.

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    $\begingroup$ More on tetraquarks: physics.stackexchange.com/q/107570/2451 $\endgroup$
    – Qmechanic
    Apr 11, 2014 at 21:04
  • $\begingroup$ In principle, it's not hard to construct states with zero color out of four quarks, say if you have a bound state of a $\pi^{0}$ and an $\eta$ particle. $\endgroup$ May 12, 2014 at 3:43

1 Answer 1


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.).

  • $\begingroup$ So even in a bound-state we can say that the quark and an anti-quark are distinguishable? So problem would start if you had more than 3 quarks in the bound-state? $\endgroup$
    – user6818
    Apr 11, 2014 at 21:11
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    $\begingroup$ Yes, for example, they always have different electric charges. If you had more than 3 quarks, it would be like having more than 2 electrons in an atom. You can't put them all in the same orbital but you can give them different energies and angular momentum to make room. However a quark bound state has to be color neutral, and this not always possible, for example it's not possible to make a colorless $qq$ state. I don't know if there's a general condition on the number of quarks and antiquarks that ensures you can make it colorless. $\endgroup$ Apr 11, 2014 at 21:21
  • $\begingroup$ So to make a bound-state colour neutral we would need a notion of "anti-colour" for the anti-quarks - right? $\endgroup$
    – user6818
    Apr 11, 2014 at 21:25
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    $\begingroup$ Yes, antiquarks have anticolor. If $r, g, b$ (red, green, blue) represent the color charges, and $\overline{r},\overline{g},\overline{r}$ are the anti-colors (anti-red, anti-green, anti-blue) the color neutral state is proportional to $r\overline{r} + b\overline{b} + g\overline{g}$. $\endgroup$ Apr 11, 2014 at 21:29

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