# Could a tetraquark $q \bar{q} q \bar{q}$ be colorless?

CERN just posted this article where it informs that it was found an hadron which cannot be classified within the traditional quark model. What other models are there to explain this result? Or is it possible to introduce a correction to the quark model to explain such find?

• The particle they found is a Tetraquark. The traditional quark model only had particles with 3 quarks or a quark-antiquark pair. Apr 9, 2014 at 13:58
• The Z(4430) is believed to be made up of quarks, or at least no-one is suggesting differently. It's just that it appears to be made up of four quarks not three. Apr 9, 2014 at 13:58
• But then it can't be color neutral if it's composed by $c \bar{c}d \bar{u}$. I thought all hadrons had to be neutral.
– PML
Apr 9, 2014 at 14:07
• Why can't $c \bar c d \bar u$ be color neutral? Doesn't $3\times\bar3\times3\times\bar3$ contain a singlet? Apr 9, 2014 at 14:15
• $rgb$ isn't the only colorless combination. $r\bar{r} g \bar{g}$ and other such 4-quark states are colorless as well. Apr 9, 2014 at 15:50

There are three flavours of quarks in the fundamental $3$ representation of $SU(3)$, the QCD gauge group. Their antiparticles are in the conjugate representation $\bar3$ or $3^\star$.

QCD is confining; the quarks form bound, colorless states, which are singlets in $SU(3)$.

• Mesons are $q\bar q$. The general tensor $3\times\bar 3$ can be decomposed into irreducible represetations; $3\times\bar 3 = 1 +8$. Note that this contains a singlet.
• Tetra-quarks are $q\bar q q\bar q$. Since $3\times\bar 3 = 1 +8$, $3\times\bar 3 \times 3\times\bar 3$ clearly contains a singlet.

Decomposition of an arbitrary tensor doesn't always contain a singlet e.g. $3\times3=\bar6+3$.

The discovery of a tetra-quark does not require the model of quarks and their interactions to be adjusted.

A tetraquark (and pentaquark) had been previously predicted as extensions of known mesons (and baryons, respectively). There is no reason why they cannot be colour neutral. e.g. in RBG the quarks could be R 255 G 0 B 0 R 0 G 255 B 0 R 0 G 0 B 128 R 0 G 0 B 127

• This is nonsense... Quarks don't have RGB in the same way a computer does. There is no "half blue" charge. Apr 17, 2014 at 10:27