In first order of perturbation theory the S-matrix amplitude for electron scattering in the Coulomb field will be (up to normalization factors) $$ S_{fi} = \frac{iZ q^2}{\sqrt{2E_{f}2E_{i}}}\bar {u}(p_{f}, s_{f})\gamma_{0}u (p_{i}, s_{i}) \int \frac{d^{4}xe^{i(p_{f} - p_{i})x}}{|\mathbf x|}, $$ where $f, i$-indices mark correspondingly final and initial waves, $s_{i, f}$ marks the polarization.

What is the difference for the case of positron scattering (the amplitude of scattering has different expression)?

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    $\begingroup$ A difference in the Feynman rules for positrons is that they carry a $\bar{v}$ spinor instead of $u$ if they are incoming. They satisfy $\sum_{\mathrm{spins}} v(p)\bar{v}(p)=\gamma^{\mu}p_{\mu} -m$ rather than $\sum_{\mathrm{spins}} u(p)\bar{u}(p) = \gamma^{\mu}p_{\mu} +m$, neglecting spin indices. $\endgroup$ – user32361 Mar 13 '14 at 21:55

For positrons there is an overall minus sign and the spinors are the positron spinors. But to calculate the cross section you need to take the square of the S- matrix, so the sign does not matter. If you are going to calculate the unpolarized cross section you need to sum over the spins. In this case spinor parts are going to be the same too. Thus the unpolarized cross section for electrons and positrons turn out the be the same. But only for the first order S-matrix. If you write the S-matrix with the higher orders the they differ. Because of the signs of the charges, the second order term contributions have different signs for electrons and positrons.


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