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Tetraquark found in Japan

  1. Nov 16, 2003 #1
    This people claims that is possible that they have discovered a particle made of 4 quarks, and I've read that there's also speculation about a particle called hexaquark, made of 6 quarks. Is there a limit to the number of quarks that can have a particle?
    Would be surprising if someone in the future discovers the "dodecaquark"
  2. jcsd
  3. Nov 16, 2003 #2


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    The restriction for hadrons is simply that they must be color neutral. There are a couple of ways this can happen:

    Two quarks:
    • [tex]red + \overline{red}[/tex]
    • [tex]green + \overline{green}[/tex]
    • [tex]blue + \overline{blue}[/tex]

    Three quarks:
    • [tex]red + green + blue[/tex]
    • [tex]\overline{red} + \overline{green} + \overline{blue}[/tex]

    A color-neutral tetraquark could be made of two color-neutral pairs.

    A color-neutral pentaquark could be made of one color-neutral pair and one color-neutral triplet, and so on.

    - Warren
  4. Nov 16, 2003 #3
    This is amazing. I've just read that there's a recent proposal that Theta+(1540) can be a heptaquark (7 quarks)
  5. Nov 17, 2003 #4

    The lightest potential members of the base scalar nonet, a0(980) and f0(980), have also been theorized as being tetraquarks or bound KKbar states. X(3872) is well above the DDbar threshold, after all. Ds+Ds- could be a possibility at that level, but DD* would be at 3871.2 MeV. That's just beautifully exact. Hope to hear about its quantum numbers soon, it would make great sense if it ends up being a scalar... no, never mind that (I would edit that part out, but that's too much momentary stupidity to just erase without feeling dishonest). If it is a DD* composite, and decays into pi+pi- + J/psi, then it has to be a vector particle. I'd be willing to bet (if I was a betting man) that its quantum numbers turn out to be either 0-(1--) or 1+(1--), the latter being rather unlikely. If the former, then it will probably be redesignated psi(3872) some time in the near future.

    This brings up the decay mode there. B+ --> K+ + pi+pi- + J/psi with pi+pi- + J/psi coming from X(3872) decay. If X(3872) is indeed a bound DD* state coming from B+ (composite of u + -b), then it should follow the currently known models for B-decay into D-mesons, which are via b --> c + -u + d and b --> c + -c + s. The latter could account for the production of the K+ and a DD* bound state as long as an extra up/anti-up quark pair can be added into the decay. This gives u + -b --> (u + -s) + (c + -u + -c + u), which is B+ --> K+ + X(3872), leading to K+ + pi+pi- + J/psi. *(Just a note here, I had previously made a very dumb remark on how this transition could happen via a flavor-changing neutral current; after some thought I realised it was blatently wrong, and so I edited it out.)

    As for theta+(1540), I can kind of see the K+N idea, but adding in the pion on top of that has got to be really tricky. Pions are so common in many kinds of light meson decay. So anyway, since its a stranged particle would it temporarily fall under the category of strange mesons when it hits the Particle Listings of the PDG in 2004?
    Last edited: Nov 20, 2003
  6. Nov 20, 2003 #5


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    Re: tetraquark?

    Isn't an alpha particle a dodecaquark?
  7. Nov 21, 2003 #6
    In a sense, possibly. But probably not by the technical definition. In order to have an alpha particle be a true dodecaquark, the protons and neutrons would have to sort of meld together, rather than simply orbit each other in a bound state such as the nucleus. Even if X(3872) is a DD* bound state, it is very likely that there is a semi-homogeneous mixture so that the two D-mesons would no longer be distinguishable until decay. Probing of X(3872) would not likely show two separate states bound together. The alpha particle, on the other hand, can still have the separate protons and neutrons within it distinguished easily if one was to probe it.
  8. Nov 21, 2003 #7


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    What keeps the protons and neutrons all together? Isn't there an extremely strong electrostatic repulsion?
  9. Nov 21, 2003 #8
    The residual strong force is responsible for keeping them together. The residual strong force, unlike the strong force as mediated by gluons, acts on color neutral particles via meson exchange. Protons contain three valence quarks (uud) and a "sea" of quark-antiquark pairs, which can interact with other protons and neutrons. The neutrons are the same way, except that their valence quarks are (udd). Hyperons with valence quarks (uds) also experience the same behavior in hypernucleii. The residual strong force, mediated by mesons, has sufficient strength to hold nucleons of all kinds togther.
    Last edited: Nov 21, 2003
  10. Dec 2, 2003 #9


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    By the way, is f0(600) still considered as the potential candidate for the scalar sigma-meson? In the nuclear theory I have read recently, most authors discard it, as they prefer to think in terms of "effective mesons", unrelated to the quark model.
  11. Dec 2, 2003 #10
    sigma meson

    It's still in the listings. Although they keep changing their minds on the pole mass, and hence changing the name every few years. I would say that in general it is simply ignored; many physicists have simply decided to move on. Personally, I find it to be quite possibly an important factor in my own work. f0(600) may well be a member of a light scalar SU(3) nonet, as opposed to replacing f0(1370) as the lighter isosinglet in the 1(3)P0 multiplet. The possibility of having two scalar multiplets in the ground state is very interesting, and essentially indicates that the scalars occur at two poles. The well established members of the 1(3)P0 nonet in SU(3) are;

    I = 0 f0(1370), f0(1710)
    I = 1 a0(1450)
    I = 1/2 K*0(1430)

    The other potential members at this time are;

    I = 0 f0(600), f0(980)
    I = 1 a0(980)
    I = 1/2 k(800)

    It is interesting to note several things here:
    1) f0(980) and a0(980) represent one of the largest breakings of isospin symmetry ever found (until this latest find of X(3872) out of KEK);
    2) by taking the mass-squared pole positions of the isospin groups (i.e. f0(600) and f0(1370), f0(980) and f0(1710), a0(980) and a0(1450), k(800) and K*0(1430)), you find mass poles that fit in the standard resonance pattern of the 1P spin triplets, i.e. 1(3)P0 mesons (poles) are lighter than 1(3)P1 mesons, which are lighter than 1(3)P2 mesons, all in appropriate sized steps;
    3) with a quick and crude calculation, the poles for the isoscalar members (I=0) occur just above 980 MeV (within about 78 MeV) and just above 1370 MeV (within about 24 MeV), possibly indicating that the f0(980) and f0(1370) are fairly pure states as they are, and placing them as the light and heavy isoscalars respectively of a scalar pole multiplet;
    4) f0(980) and a0(980) have been postulated as possible candidates for KKbar bound states in the past, and now with the find of X(3872) as a possible D*Dbar candidate this possibility is all the more realistic.

    Alot of the answers to the mystery of the scalars may in fact hinge on the interpretation of f0(600).
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