 #1
JD_PM
 1,128
 158
 TL;DR Summary

I want to understand how to get (as an example)
##X_{aa} = \operatorname{Tr}\{\gamma^{\beta} (\not{\!p} + \not{\!k} +m) \gamma^{\alpha} (\not{\!p} + m) \gamma_{\alpha} (\not{\!p} + \not{\!k} +m) \gamma_{\beta} (\not{\!p'} +m)\}##
Given the Feynman amplitude (Compton Scattering by electrons) ##\mathscr{M}=\mathscr{M}_a+\mathscr{M}_b##, where
$$\mathscr{M}_a = i e^2 \frac{\bar u' \not{\!\epsilon'} (\not{\!p}+\not{\!k}+m) \not{\!\epsilon} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = i e^2 \frac{\bar u' \not{\!\epsilon} (\not{\!p}\not{\!k'}+m) \not{\!\epsilon'} u}{2(pk')} \tag 1$$
Given also the formula to sum over spin and polarization states.
$$Y:= \frac 1 4 \sum_{r} \sum_{s} \mathscr{M}^2=\frac 1 4 \sum_{r} \sum_{s} \Big\{\mathscr{M_a}^2+\mathscr{M_b}^2+\mathscr{M_a}\mathscr{M_b}^{\dagger}+\mathscr{M_b}\mathscr{M_a}^{\dagger}\Big\}$$ $$=\frac{e^4}{64 m^2} \Big\{ \frac{X_{aa}}{(pk)^2} + \frac{X_{bb}}{(pk')^2}  \frac{X_{ab}+X_{ba}}{(pk)(pk')}\Big\} \tag 2$$
Where ##r## represents polarization states and ##s## spin states.
My issue is understanding how to get the ##X_{..}## expressions, which are traces.
My idea is work out ##X_{aa}## as an example, so that I understand the process. The answer is
$$X_{aa} = \operatorname{Tr}\{\gamma^{\beta} (\not{\!p} + \not{\!k} +m) \gamma^{\alpha} (\not{\!p} + m) \gamma_{\alpha} (\not{\!p} + \not{\!k} +m) \gamma_{\beta} (\not{\!p'} +m)\} \tag 3$$
Useful trick: setting ##\mathscr{M}:=\epsilon_{\alpha}\epsilon_{\beta}'\mathscr{M}^{\alpha \beta}## allows us to write ##Y=\frac 1 4 \sum_{r} \sum_{s} \mathscr{M}^2=\frac 1 4 \sum_{s} \mathscr{M^{\alpha \beta}}\mathscr{M_{\alpha \beta}}^{\dagger}##
Note: from here on ##\mathscr{M}## really means ##\mathscr{M_a}##
$$\frac 1 4 \sum_{r} \sum_{s} \mathscr{M}^2=\frac 1 4 \sum_{r} \sum_{s}
(\bar{u_s} (\vec p') \Gamma u_r (\vec p))( \bar{u_r} (\vec p) \tilde \Gamma u_s (\vec p')) \tag 4$$
Where I've defined ##\Gamma:=\not{\!p}+\not{\!k}+m##, ##\tilde \Gamma = \gamma^0 \Gamma^{\dagger} \gamma^0##
But this leads to a trace of the form
$$\frac{1}{4} \operatorname{Tr}\Big\{\frac{\not{\!p'}+m}{2m}\Gamma\frac{\not{\!p}+m}{2m}\tilde \Gamma\Big\} \tag 5$$
Which is not the provided answer.
Source: Mandl & Shaw second edition, pages 143 & 144.
Any help is appreciated.
$$\mathscr{M}_a = i e^2 \frac{\bar u' \not{\!\epsilon'} (\not{\!p}+\not{\!k}+m) \not{\!\epsilon} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = i e^2 \frac{\bar u' \not{\!\epsilon} (\not{\!p}\not{\!k'}+m) \not{\!\epsilon'} u}{2(pk')} \tag 1$$
Given also the formula to sum over spin and polarization states.
$$Y:= \frac 1 4 \sum_{r} \sum_{s} \mathscr{M}^2=\frac 1 4 \sum_{r} \sum_{s} \Big\{\mathscr{M_a}^2+\mathscr{M_b}^2+\mathscr{M_a}\mathscr{M_b}^{\dagger}+\mathscr{M_b}\mathscr{M_a}^{\dagger}\Big\}$$ $$=\frac{e^4}{64 m^2} \Big\{ \frac{X_{aa}}{(pk)^2} + \frac{X_{bb}}{(pk')^2}  \frac{X_{ab}+X_{ba}}{(pk)(pk')}\Big\} \tag 2$$
Where ##r## represents polarization states and ##s## spin states.
My issue is understanding how to get the ##X_{..}## expressions, which are traces.
My idea is work out ##X_{aa}## as an example, so that I understand the process. The answer is
$$X_{aa} = \operatorname{Tr}\{\gamma^{\beta} (\not{\!p} + \not{\!k} +m) \gamma^{\alpha} (\not{\!p} + m) \gamma_{\alpha} (\not{\!p} + \not{\!k} +m) \gamma_{\beta} (\not{\!p'} +m)\} \tag 3$$
Useful trick: setting ##\mathscr{M}:=\epsilon_{\alpha}\epsilon_{\beta}'\mathscr{M}^{\alpha \beta}## allows us to write ##Y=\frac 1 4 \sum_{r} \sum_{s} \mathscr{M}^2=\frac 1 4 \sum_{s} \mathscr{M^{\alpha \beta}}\mathscr{M_{\alpha \beta}}^{\dagger}##
 My attempt to get ##X_{aa} = \operatorname{Tr}\{\gamma^{\beta} (\not{\!p} + \not{\!k} +m) \gamma^{\alpha} (\not{\!p} + m) \gamma_{\alpha} (\not{\!p} + \not{\!k} +m) \gamma_{\beta} (\not{\!p'} +m)\}##
Note: from here on ##\mathscr{M}## really means ##\mathscr{M_a}##
$$\frac 1 4 \sum_{r} \sum_{s} \mathscr{M}^2=\frac 1 4 \sum_{r} \sum_{s}
(\bar{u_s} (\vec p') \Gamma u_r (\vec p))( \bar{u_r} (\vec p) \tilde \Gamma u_s (\vec p')) \tag 4$$
Where I've defined ##\Gamma:=\not{\!p}+\not{\!k}+m##, ##\tilde \Gamma = \gamma^0 \Gamma^{\dagger} \gamma^0##
But this leads to a trace of the form
$$\frac{1}{4} \operatorname{Tr}\Big\{\frac{\not{\!p'}+m}{2m}\Gamma\frac{\not{\!p}+m}{2m}\tilde \Gamma\Big\} \tag 5$$
Which is not the provided answer.
Source: Mandl & Shaw second edition, pages 143 & 144.
Any help is appreciated.