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Calculation of B to l nu decay width

  1. Apr 16, 2015 #1
    Hi,

    1. The problem statement, all variables and given/known data


    In the calculation of the matrix element amplitude of ## B \to l \nu ##, I got a factor ## \bar{l} (1-\gamma_5)\nu ## as in [hep-ph/0306037v2]. For ##|M|^2## I made :

    ##| \bar{l} (1-\gamma_5)\nu|^2 = (/\!\!\! p_l+m_l) (1-\gamma_5) /\!\!\! p_\nu(1+\gamma_5) = 2(/\!\!\! p_l+m_l) /\!\!\! p_\nu(1+\gamma_5) = 8 p_l. p_\nu ~##(1) , after taking the trace.


    2. Relevant equations

    My question about the value of (1 )which depends on the kinematics of the process.I got a factor

    ##\frac{m_B^2}{2} ( 1 - \frac{2m_l^2}{m_B^2}) ##, while in [hep-ph/0306037v2] equ. 5, (1) gave a factor ## m_B^2( 1 - \frac{m_l^2}{m_B^2}) ## instead ..

    3. The attempt at a solution

    I defined

    ## p_l = ( E_l,\bf{p_l}) ## and ## p_\nu= ( E_\nu,- E_\nu) ##, where

    ## -E_\nu = -\bf{p_\nu} \equiv - \bf{p_l}, ~ \bf{p_l}^2= E_l^2 - m_l^2, ~ E_l =E_\nu = m_B/2##, then


    ## p_l. p_\nu = E_l E_\nu +\bf{p_l}^2 = 2E_l^2 - m_l^2 = m_B^2/ 2 - m_l^2 =\frac{m_B^2}{2} ( 1 - \frac{2m_l^2}{m_B^2}) ##!

    Bests.
     
  2. jcsd
  3. Apr 21, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Apr 22, 2015 #3
    Hi, yes I got an answer. Thanks.
     
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