[QUE/QFT-12015] QFT-PROBLEM

$\newcommand{\Msc}{\mathscr M}$
For a decay process \(A (q) \to B(p) + C(\ell)\), where \(A\) and \(B\) are spin \(1/2\) fermions and \(C\) is a spin zero particle, the transition amplitude involves matrix element of the form \begin{equation} \Msc = \bar{u}^{(r)}_B(p)\big(g - g{'}\gamma_5\big) u^{(s)}_A(q) \end{equation} where \(u_B, u_A\) etc are Dirac spinors for the particles \(B\) and \(A\). We need to take absolute square of the matrix element and sum over all spins. Taking the adjoint, \(\dagger\), carefully gives the complex conjugate of \(\Msc\) \begin{eqnarray} \Msc^* &=& \{\bar{u}^{(r)}_B(p)\big(g - g{'}\gamma_5\big) u^{(s)}_A(q)\}^\dagger \\ &=& \{{u}^{(r)\dagger}_B(p)\gamma_0\big(g - g{'}\gamma_5\big) u^{(s)}_A(q)\}^\dagger\\ &=& \{{u}^{(s)\dagger}_A(q)\big(g^* - g^{{'}*}\gamma_5\big) \gamma_0 u^{(r)}_B(p)\} \\ &=& \{\bar{u}^{(s)}_A(q)\gamma_0\big(g^* - g^{{'}*}\gamma_5\big) \gamma_0 u^{(r)}_B(q)\}\\ &=& \{\bar{u}^{(s)}_A(q)\big(g^* + g^{{'}*}\gamma_5\big) u^{(r)}_B(p)\} \end{eqnarray} 

Therefore, the absolute square is given by \begin{eqnarray} \sum_{r,s} |\Msc|^2 &=& \Msc^* \Msc\\ &=& \sum_{r,s} \{\bar{u}^{(s)}_A(q)\big(g^* + g^{{'} *}\gamma_5\big) u^{(r)}_B(p) \} \times \big\{\bar{u}^{(r)}_B(p)\big(g - g{'}\gamma_5\big) u^{(s)}_A(q) \big\} \end{eqnarray}

Using projection operators, \begin{equation} \sum_r u^{(r)}(k) \bar{u}^{(r)}(k) = \frac{n\!\!\!{k}+ M }{2M} \end{equation} this expression can be written as trace of gamma matrices. To see this we write \(\Msc\) and \(\Msc^*\) terms explicitly in terms of Dirac indices as \begin{eqnarray} \Msc &=&\sum_{\alpha\beta} [\bar{u}^{(r)}_B(p)]_\alpha\,[\big(g -g{'}\gamma_5]_{\alpha\beta}\, [u^{(s)}_A(q)]_\beta\\ \Msc^*&=& \sum_{\sigma\rho}[\bar{u}^{(s)}_A(q)]_\sigma\,[\big(g^* + g^{{'}^*}\gamma_5\big)]_{\sigma\rho}\, [u^{(r)}_B(p)]_\rho \end{eqnarray} 

Therefore \(|\Msc|^2\), summed over all spins becomes
\begin{eqnarray} \sum_{r,s} |\Msc|^2 &=& \sum_{r,s}\sum_{\alpha\beta\sigma \rho} [\bar{u}^{(r)}_B(p)]_\alpha\,[\big(g-g{'}\gamma_5]_{\alpha\beta}\, [u^{(s)}_A(q)]_\beta [\bar{u}^{(s)}_A(q)]_\sigma\,[\big(g^* + g^{{'}^*}\gamma_5\big)]_{\sigma\rho}\, [u^{(r)}_B(p)]_\rho \end{eqnarray}

Now put \(u^{(r)}_A(p) \bar{u}^{(r)}_A(p)\) terms next to each other to get and use projection operators and complete spin sums over \(r,s\)
\begin{eqnarray} {\sum_{rs}|\Msc^2|} &=& \sum_{r,s}\sum_{\alpha\beta\sigma \rho} \Big(u^{(r)}_{B\rho}(p) \bar{u}^{(r)}_{B\alpha}(p)\Big)\,[\big(g -g{'}\gamma_5)]_{\alpha\beta}  \times \Big( u^{(s)}_{A\beta}(q) \bar{u}^{(s)}_{A\sigma} (Q)\Big) \,[\big(g^* + g^{{'}*}\gamma_5\big)]_{\sigma\rho}\\ &=& \sum_{\alpha\beta\sigma\rho} \left(\frac{n\!\!\!{p}+ M_B}{2M_B}\right)_{\rho\alpha} \Big(g -g{'}\gamma_5\Big)_{\alpha\beta}\, \left(\frac{n\!\!\!{q}+M_A}{2M_A}\right)_{\beta\sigma}\,\Big(g^* + g^{{'}*}\gamma_5\Big)_{\sigma\rho}\\ &=& \text{tr}\left\{\Big(\frac{n\!\!\!{p}+ M_B}{2M_B}\Big) \Big(g -g{'}\gamma_5\Big)\, \Big(\frac{n\!\!\!{q}+M_A}{2M_A}\Big)\Big(g^* + g^{{'}*}\gamma_5\Big)\right\} \end{eqnarray}
Thus problem of squaring the matrix element and summing over all spins has been reduced to computing trace of products of gamma matrices. \paragraph*{Question for You} Derive an general formula for spin sums taking \(|\Msc|^2\) to be of the form \begin{equation} \Msc= \bar{u}^{(s)}_B(p) \, \Gamma \, u^{(r)}_A(q) \end{equation} where \(\Gamma\) is an expression involving Dirac \(\gamma\) matrices. \paragraph*{Note} With this preparation you are now ready to compute life times and cross sections for processes involving fermions. \noindent Please go through the attached scan of computation of lifetime of \\ \(\Lambda^0 \to p + \pi^-\). \noindent Check all steps and factors carefully. Let me know if I have missed anything.