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About the recent discovery of 4-quark boundstates.

+ 7 like - 0 dislike
137 views

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 (3x2)^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.


This post imported from StackExchange Physics at 2014-04-13 14:30 (UCT), posted by SE-user user6818

asked Apr 11, 2014 in Phenomenology by user6818 (955 points) [ revision history ]
recategorized Apr 13, 2014 by dimension10
More on tetraquarks: physics.stackexchange.com/q/107570/2451

This post imported from StackExchange Physics at 2014-04-13 14:30 (UCT), posted by SE-user Qmechanic

The same ''problem'' already occurs for mesons (one quark-antiquark pair), and is solved by anti-color, as discussed below.

1 Answer

+ 2 like - 0 dislike

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.).

This post imported from StackExchange Physics at 2014-04-13 14:30 (UCT), posted by SE-user Robin Ekman
answered Apr 11, 2014 by Robin Ekman (215 points) [ no revision ]
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?

This post imported from StackExchange Physics at 2014-04-13 14:30 (UCT), posted by SE-user user6818
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.

This post imported from StackExchange Physics at 2014-04-13 14:30 (UCT), posted by SE-user Robin Ekman
So to make a bound-state colour neutral we would need a notion of "anti-colour" for the anti-quarks - right?

This post imported from StackExchange Physics at 2014-04-13 14:30 (UCT), posted by SE-user user6818
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 $r\overline{r} + b\overline{b} + g\overline{g}$.

This post imported from StackExchange Physics at 2014-04-13 14:30 (UCT), posted by SE-user Robin Ekman

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