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XYZ spectroscopy and the existence of possible 4-quark states

  1. May 27, 2014 #1
    Hi everyone, I've been studying the so-called XYZ spectroscopy and the existence of possible 4-quark states.

    The LHCb collaboration recently confirmed the existence of a particle called [itex]Z(4430)^-[/itex]. This particle is the unambiguous evidence for the existence of 4-quark states. From what I understood the reason is that this particle decays as [itex]Z(4430)^-\to \psi' \pi^-[/itex]. Since it decays in a [itex]c\bar c[/itex] it must contain such quarks as valence quarks. Moreover it is charged and therefore its minimal quark content can only be [itex]c\bar c d\bar u[/itex].

    My question is: some time ago was also discovered another particle, the [itex]Z_c(3900)^+[/itex] decaying in [itex]J/\psi \pi^+[/itex]. Why this is not considered as an evidence for 4-quark states? It seems to me that it follows the same criteria as the [itex]Z(4430)^-[/itex].

    Does anyone know something about it?

    Thanks
     
  2. jcsd
  3. May 27, 2014 #2

    mfb

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    It is an unambiguous evidence for something beyond the established categories.

    The Zc(3900)+ has the same features, but the discovery of this particle was not as clear as the observation of the Z(4430).
     
  4. May 27, 2014 #3
    What do you mean exactly? If I remember correctly the Z(3900) was safely observed in two different decay channels with significantly more than 5 sigma. What other factors define the "significance" of the discovery?
     
  5. May 27, 2014 #4

    mfb

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    I don't have the time to check the original references now, but systematic uncertainties do not follow a normal distribution. See the Opera neutrino measurement for example - they could have observed 50 sigma and still be wrong because of an unaccounted measurement error. It is easy to get a model wrong (and therefore see a particle that does not exist), especially in areas with so many open question as in XYZ states. Two different experiments don't make that less likely if they use the same models.
     
  6. May 27, 2014 #5
    Understood. So, at the end of the day what makes the LHCb discovery so important is that it is more solid that the others right?
     
  7. May 27, 2014 #6

    mfb

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    Right.
     
  8. May 27, 2014 #7

    Vanadium 50

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    I disagree - there's nothing "better" about the 4430 discovery than the 3900. Both have an unambiguous peak. What the LHCb result adds is something called a Argand diagram showing the phase at the mass pole: this is highly technical, but it demonstrates that the peak in the mass plot behaves as a particle and not some sort of bizarre background.
     
  9. May 27, 2014 #8
    Do you know if there is any particular reason that prevent the other experiment to perform the same phase analysis for the Z(3900)? I mean, it was discovered a few years ago and no one did that. Is that because of it was impossible for the other experiments or what?
     
  10. May 27, 2014 #9

    Vanadium 50

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    Among other reasons, it takes a lot of events.
     
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