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How to form a J^{PC} = 1^{-} state with only gluons?

  1. Sep 2, 2006 #1
    How to form a J^{PC} = 1^{--} state with only gluons?

    Two gluons? Three ?

    How do the P, C, J values work?
     
  2. jcsd
  3. Sep 2, 2006 #2

    Meir Achuz

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    In QCD, one gluon can convert into two gluons, so the gluon must have C=+1, and there is no way to get any number of gluons with C=-1.
    One photon cannot convert into two photons because c=-1 for a photon.

    Gluons, as vector particles, have intrinsic P=-1. For N gluons,
    P=(-1)^N times (-1)^L for each orbital angular momentum. This gets complicated for more than three gluons.

    J can get complicated. You have to add up the gluons spin and orbital angular momentum, much like many electron atoms, or many nucleon nuclei.
     
  4. Sep 8, 2006 #3
    Why in PDG booklet, gluon has J^{PC} value as 1^-, where C is abscent?
     
  5. Sep 8, 2006 #4

    Meir Achuz

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    The gluon has color charge, so it can't be its own antiparticle and can't be a C eigenstate.
     
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