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Decay D0-> K+ K-

  1. Oct 21, 2014 #1
    Hey community!

    Since a couple of hours I've been dealing with the question why the decay $$D^0 \rightarrow K^+ K^-$$ is suppressed.
    If I draw the Feynman diagram for the first order decay, everything seems pretty alright as the charm quark can couple with a strange quark. The resulting W+ can then produce an up and anti-strange quark which gives above K mesons.
    What piece am I missing?
  2. jcsd
  3. Oct 21, 2014 #2


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    up and anti-strange are two different generations, there you get the Cabibbo suppression (compared to ##D^0 \rightarrow K^- \pi^+##).

    It is not a rare decay. It's just not as frequent as the decay to ##K^- \pi^+##.
  4. Oct 21, 2014 #3
    Aaah, somehow I thought the Cabibbo suppression just applies in case of a quark-quark decay. But it holds whenever any kind of quark coupling is involved (independent of the "direction" I go in the diagram), right? (Sorry in advance for these basic questions..)
  5. Oct 22, 2014 #4


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    Correct, whenever you have a coupling of a W with quarks belonging to different generations, you get a suppression (an even larger suppression if the generations are 1,3 or 2,3).
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