# About the recent discovery of tetraquark boundstates

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.

• More on tetraquarks: physics.stackexchange.com/q/107570/2451 – Qmechanic Apr 11 '14 at 21:04
• 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. – Jerry Schirmer May 12 '14 at 3:43

• 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. – Robin Ekman Apr 11 '14 at 21:21
• 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 $r\overline{r} + b\overline{b} + g\overline{g}$. – Robin Ekman Apr 11 '14 at 21:29