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D* -> D + gamma BR calculation

  1. Nov 21, 2009 #1

    Hepth

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    Can anyone walk me through this calculation? I'm not completely sure how to do it (merely a student).

    So is this considered a 1 body radiative decay? So the factorized amplitude would be something like
    [tex]M \propto <D^0|\bar{u}\gamma_\mu\left(1-\gamma_5\right)c|0>[A_1 <\gamma|\bar{u}\gamma_\mu\left(1-\gamma_5\right)c|D^{*0}>+A_2<\gamma|\bar{u}i \sigma_{\mu \nu} p_\nu \left(1+\gamma_5\right)c|D^{*0}> [/tex]

    And each matrix can be decomposed into some constants/vectors/polarizations etc.

    [tex]<D^0|\bar{u}\gamma_\mu\left(1-\gamma_5\right)c|0> = i f_{D^0} P_{D^{0}\mu}[/tex]

    and the second and third terms can be similarly decomposed into linear combinations of terms depending only on:
    Photon Polarization [tex]\epsilon^{*}_{\gamma \alpha}[/tex]
    [tex]D^{*0}[/tex] Polarization [tex]\epsilon^{*}_{D^{*0} \beta}[/tex]
    Momentums of the photon and D* (q and p respectively)
    And the constants can depend on p^2.

    So any sort of combination of these such that the end result yields one remaining index.

    Now my first question, for those that do these sort of things, am I handling the D* matrices correctly? Am I missing any effective terms/form factors?
     
  2. jcsd
  3. Nov 22, 2009 #2

    Vanadium 50

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    This is over my head, because I am an experimenter, but what are all the gamma_5's doing there? This is an electromagnetic decay, so why are you projecting out the left handed state?
     
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