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Why when people/textbooks talk about strong interaction, they talk only about bound states of 2 or 3 quarks to form baryons and mesons?

Does the strong interaction allow bound states of more than 3 quarks?

If so, how is the stability of a bound state of more than 3 quarks studied?

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up vote 9 down vote accepted

There is no known reason that you can't have bound states like $qq\bar{q}\bar{q}$ or $qqqq\bar{q}$ or higher number excitations, but none have been observed to date.

You do have to make a color-neutral state, of course.

In the mid-2000 some folks thought that they had of pentaquark states (that the $qqqq\bar{q}$) for a while, but it was eventually concluded that they were wrong.

Added June 2013: Looks like we may have good evidence of four-quark bound states, though the detailed structure is not yet understood, and in the comments Peter Kravchuk points out that pentaquarks have come back while I wasn't paying attention (and the same state, too). Seem some egg may have moved from face to face.

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How the stability of such bound states is studied theoretically then? – Revo Aug 6 '12 at 19:14
QCD is hard. Last time I heard, you could find theorists claiming either condition. – dmckee Aug 6 '12 at 19:16
That is weird. Don't string theorists work with nonabelian gauge theories all the time. Since QCD is just another nonabelian gauge theory, I thought the answer to such question must have been known for sometime now. – Revo Aug 6 '12 at 19:19
Being able to write down the theory and catalog it's properties is very different from being able to compute accurate solution to complicated problems. Full computations are very hard, which is why there is so much interest in solutions on the lattice. – dmckee Aug 6 '12 at 19:28
A recent article from DESY about tetraquarks can be found in the (freely distributed) magazine femto, first volume, pp.36-37 – glS Dec 24 '14 at 12:01

As a quick explanation: all bound states are color-neutral. The intuitive reason is that the strong interaction is so strong that it would pull any color-charged particles together. (Because the strong force increases with distance, you can't get around this by spreading out the charged particles, as you can with the EM interaction.)

Since there are 3 colors, you can either achieve a color-neutral state by combining one quark of each color, which gives you a baryon, or a quark and an antiquark of the same color (e.g. blue and antiblue), which gives you a meson. Any combination of more quarks or antiquarks that works out to being color-neutral, such as the hypothetical pentaquark, can be broken down into some combination of baryons and mesons, which means that such a particle would probably naturally decay in that way, if it could even exist (which there is no evidence for).

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If it is allowed by QCD then it must exist, right? – Revo Aug 6 '12 at 19:16
That's generally a good guideline, but not a rule. So no, not necessarily. – David Z Aug 6 '12 at 20:55
Are there any calculations within QCD regarding stability, lifetime (..etc) of any of high quark content states? – user10001 Aug 7 '12 at 3:10
@dushya People are trying to do such things, mostly within the framework of lattice QCD (if you'd like a buzzword to search for), but the calculations are extremely complicated and we're not up to the point of getting those results out of them yet. – David Z Aug 11 '12 at 5:51
@Revo Not necessarily, remember the gauge invariant $\theta$-term in QCD. Experimentally $\theta$ is consistent with zero whereas theoretically one would expect $\theta\approx 1.$ – Faq Jan 19 '15 at 12:33

In a sense every nucleus is a bound state of 3N quarks. After all, the nuclear force between nucleons (protons and neutrons) is a result of the leakage of the strong color force outside the "boundary" of the nucleon. So there are undoubtedly gluons and even quark exchanges between the nucleons of a nucleus.

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