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D-meson decay

  1. Nov 4, 2012 #1
    Hi all,
    I have a question regarding to the branching ratio of neutral D-meson decay.
    Give three decays:
    1. D --> negative kaon + positive pion
    2. D --> positive pion + negative pion
    3. D --> positive kaon + negative pion

    The one with highest branching ratio is 1. due to its non-cross generation mixing. However, my question is: why isn't 2 more likely since D-meson can decay into two photons and split into two pairs of quarks? It's because Intermediate photons are involved with EM interactions, which dominate over weak interactions (1 and 3 b/c they don't conserve strangeness).

    Thanks :\

  2. jcsd
  3. Nov 4, 2012 #2


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    The first thing to notice is you MUST have a weak transition. So the leading order will be a weak-tree diagram, with NO photons (as that would be next order in QED).

    These are the weak transitions right?

    c-> s + dbar u
    ubar-> ubar

    = K- pi+

    c->d + dbar u
    ubar-> ubar
    = pi- pi+

    c->d + sbar u
    ubar-> ubar
    = K+ pi-

    c->s + sbar u
    ubar -> ubar
    = K+ K-

    So in this order we have:
    Vcs Vud ~ 1 * 1
    Vcd Vud ~ (-lambda) * 1
    Vcd Vus ~ (-lambda) * (lambda)
    Vcs Vus ~ 1 * (lambda)

    where lambda ~ 0.2257

    I imagine this is the first reason why they're ordered thusly.
  4. Nov 4, 2012 #3
    Yea, I see how this works like that.
    But as I mentioned before, since 'positive pion + negative pion' can be produced through two photons why wouldn't it be produced more often than the ones from weak interaction?

    Also a stupid question, when do you take the Vxx to be negative? Vxx = any element in KM matrix
  5. Nov 5, 2012 #4

    Vanadium 50

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    If you want #2 to proceed via two virtual photons, why would you not have an even larger branching fraction into two real photons?
  6. Nov 5, 2012 #5
    Vanadium 50, it's probably because of some CP conservation; so we have pions as products. This is why the intermediate photons are only virtual and we need quarks to be produced.
  7. Nov 5, 2012 #6

    Vanadium 50

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    No, that's not your problem. A D0 is not in a CP eigenstate.
  8. Nov 5, 2012 #7


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    ##D^0 \to \gamma \gamma## is not observed yet, with 2.5*10-5 as upper limit on the branching fraction (~1/5 of the WS ##D^0 \to K^+ \pi^-## BF). It needs weak and electromagnetic interaction at the same time. Why do you want photons in the diagram? Gluons would do the same, just with a stronger coupling. It cannot compete with the singly cabibbo suppressed tree diagram, however.

    If you look at the -> KK, Kpi, pipi branching fractions (and similar -> K pi pi pi, -> K K K pi and so on), powers of lambda dominate.
    At least roughly:
    BF ##D^0 \to \pi^+ \pi^-## is 0.14%
    BF ##D^0 \to K^+ K^-## is 0.40%

    And of course there is mixing... HCP2012 will have some news about that (14. November).
    Last edited: Nov 5, 2012
  9. Nov 5, 2012 #8
    mfb, it does make whole lot more sense now; thanks alot

    btw Vanadium 50, (correct me if I'm wrong) you don't need a particle decay to be in CP eigenstate to conserve charge conjugation/parity in any non-weak interactions.
  10. Nov 5, 2012 #9

    Vanadium 50

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    I was trying to point you in the right direction - or at least away from the wrong one.
  11. Nov 7, 2012 #10


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