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D meson decay

  1. Nov 30, 2011 #1
    I am trying to work out the Feynman diagram for the decay

    [tex]D^0\to K^++\pi^-[/tex]

    But I can't seem to get it unless the D meson is its own antiparticle. Could someone tell me if this is the case? Ie, is [tex]|\bar{u}c\rangle = |u\bar{c}\rangle[/tex]?

  2. jcsd
  3. Nov 30, 2011 #2


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    That's because the decay is actually D0 → K- + π+. Maybe you want to start with an anti-D0
  4. Nov 30, 2011 #3
    No, the question gives three decays, one of which is the one you posted and another is the one I posted.

    Could I change D to anti-D by having the two constituent quarks exchange a gluon, so that the decay I posted would then proceed from the antiparticle?

    See for example decay modes 42 and 213 here: http://pdg.lbl.gov/2008/listings/s032.pdf
    Last edited: Nov 30, 2011
  5. Nov 30, 2011 #4

    Vanadium 50

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    Do you know what Cabibbo suppression is? That decay is doubly Cabibbo suppressed. That should help you.
  6. Nov 30, 2011 #5
    Update: seem to work if the c decays into d and a W^+ meson, which then decays into u and anti-s

    any problems there?

    this decay is apparently the much less likely than the one you posted. Why is that?

    EDIT: I don't yet know what cabibbo suppression is. Do you know of a resource that explains it? Google turns up a lot of scholarly stuff that assumes I know it already.
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