- #1
kontejnjer
- 72
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Homework Statement
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(Note: this is not strictly homework, but it is related to one course I'm doing, and I can't find a useful solution anywhere)
Find the analytical expression for the scattering cross section of two longitudinally polarized electrons at tree level.
Homework Equations
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[itex]\displaystyle d\sigma=\left|\mathcal{M}_{fi}\right|^2\frac{d\Phi}{4I}[/itex]
Feynman rules
The Attempt at a Solution
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There's two diagrams at tree level since the particles are identical: the [itex]t[/itex] channel and the [itex]u[/itex] channel. There is a relative minus sign between the channels due to the fact the particles are identical fermions. The total scattering amplitude is:
[itex]\displaystyle i\mathcal{M}_{fi}=i\mathcal{M}_{fi}^t+i\mathcal{M}_{fi}^u=\frac{e^2}{t}\bar u^{s'}(p_4)\gamma^\mu u^s (p_2) \bar u^{r'} (p_3)\gamma_\mu u^r (p_1)-\frac{e^2}{u}\bar u^{r'}(p_3)\gamma^\mu u^s (p_2) \bar u^{s'} (p_4)\gamma_\mu u^r(p_1) [/itex]
Here [itex]t=(p_3-p_1)^2[/itex], [itex]u=(p_4-p_1)^2[/itex] are the Mandelstam variables, [itex]p_1[/itex] and [itex]p_2[/itex] are inital momenta with spins [itex]r[/itex] and [itex]s[/itex] respectively, and final momenta [itex]p_3[/itex] and [itex]p_4[/itex] with spins [itex]r'[/itex] and [itex]s'[/itex] respectively. As the initial electrons are longitudinally polarized, we know their spin states, however we still need to sum over the final spins, so the square of the amplitude is:
[itex] \left|\mathcal{M}_{fi}\right|^2=\sum\limits_{s',r'} \left(|\mathcal{M}_{fi}^t|^2+|\mathcal{M}_{fi}^u|^2-2(\mathcal{M}_{fi}^t)^* \mathcal{M}_{fi}^u\right)[/itex]
After some manipulation, and using the identity [itex]\sum\limits_{s}u^s(p)\bar u^s(p)=\gamma\cdot p+m[/itex], we can write down the square of each component as:
[itex]
\sum\limits_{s',r'}|\mathcal{M}_{fi}^t|^2=\frac{e^4}{t^2}{Tr}\left[\gamma^\mu u^s(p_2) \bar u^s(p_2)\gamma^\nu (\gamma\cdot{p}_4+m)\right]\cdot {Tr}\left[\gamma_\mu u^r(p_1)\bar u^r (p_1) \gamma_\nu (\gamma\cdot{p}_3+m)\right]\\
\sum\limits_{s',r'}|\mathcal{M}_{fi}^u|^2=\frac{e^4}{u^2}{Tr}\left[\gamma^\mu u^s(p_2) \bar u^s(p_2)\gamma^\nu (\gamma\cdot{p}_3+m)\right]\cdot {Tr}\left[\gamma_\mu u^r(p_1)\bar u^r (p_1) \gamma_\nu (\gamma\cdot{p}_4+m)\right]\\
\sum\limits_{s',r'}(\mathcal{M}_{fi}^t)^* \mathcal{M}_{fi}^u=\frac{e^4}{ut}{Tr} \left[\gamma^\mu u^s(p_2) \bar u^s(p_2)\gamma^\nu(\gamma\cdot{p}_3+m) \gamma_\mu u^r(p_1)\bar u^r (p_1) \gamma_\nu (\gamma\cdot{p}_4+m)\right][/itex]
This is the part that I get stuck on: the initial spins are predefined so I can't just sum over them, and the previously mentioned identity doesn't work in this case, all I know is that the initial spins should be helicity eigenstates. There's a reference paper for this exact topic by Adam M. Bincer - Scattering of longitudinally polarized fermions (DOI: 10.1103/PhysRev.107.1434), but the notation is somewhat outdated and I think it omits some crucial steps.
I know I could just pick a representation and a frame and do the whole trace calculation using some symbolic software, but the end result doesn't give me much qualitative info, what I'd like is to just have a frame-independent square of the scattering amplitude in terms of 4-momenta, so I can just pick any frame (say, COM or lab) and get the analytic result. Any advice or hints would be greatly appreciated.