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Higher-order weak interaction decays

  1. Nov 3, 2006 #1
    Considering the decay of certain exotic mesons, such as the following:

    [tex]B^{+} --> D^{0} \pi^{+}[/tex]

    Apparently the decay proceeds via a weak interaction where multiple W boson exchanges occur.

    I was trying to nut out how this actually occurs, and draw up a sensible Feynman diagram representation of the process, but i can't quite nut it out.

    Recall that the B+ meson is comprised of a u and anti-b quark, and the D0 is a c and anti-up, and the Pi+ is an up and anti-down, FYR.

    Could someone offer me any pointers as to how such decay process usually work? First-order weak process such as beta decay are pretty straightforward, but this idea seems a bit more tricky.
  2. jcsd
  3. Nov 3, 2006 #2

    Meir Achuz

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    bbar-->cbar + W^+.
    W^+--> u + dbar.
    That u + cbar--> D^0.
    The incoming u + dbar--> pi^+.
  4. Nov 3, 2006 #3
    But doesn't the D^0 meson contain a c, not a cbar ?
  5. Nov 5, 2006 #4

    Meir Achuz

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    Yes. You fooled me with your first equation. It should be
    If you really meant D^0, then the diagram would be a mess and the decay never seen unless you have some completely new theory.
  6. Nov 5, 2006 #5
    Could you tell me how the decay could proceed anyway, even though it might never occur in practice?
  7. Nov 6, 2006 #6

    Meir Achuz

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    The final state you want, D_0 + pi^+ has the quarks [c ubar] and
    [u dbar]. You could get that from [u bbar] by the steps:
    1. bbar -->W^+ + cbar.
    2. W^+ -->[c] + [dbar].
    3. cbar --> W^- + dbar.
    4. W^- + u --> d.
    5. dbar +d --> [ubar] + (via strong intgeraction).
    This leaves you with the right quarks for D_0 + pi^+.
    The decay is doubly weak, so I cannot see how it could be observed.
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