Calculation of B to l nu decay width

[Safinaz]

Hi,

1. Homework Statement

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.

Homework Equations

[/B]
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 ..

The Attempt at a Solution

[/B]
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.

 

[Safinaz]

Hi, yes I got an answer. Thanks.