Bhabha scattering: Deriving unpolarized cross section

Matrix elements

Both diagrams contribute to the transition matrix element. By letting k and k' represent the four-momentum of the positron, while letting p and p' represent the four-momentum of the electron, and by using Feynman rules one can show the following diagrams give these matrix elements:

			Where we use: $\gamma^\mu \,$ are the Gamma matrices, $u, \ \mathrm{and} \ \bar{u}\,$ are the four-component spinors for fermions, while $v, \ \mathrm{and} \ \bar{v}\,$ are the four-component spinors for anti-fermions (see Four spinors).
	(scattering)	(annihilation)
$\mathcal{M} = \,$	$-e^2 \left( \bar{v}_{k} \gamma^\mu v_{k'} \right) \frac{1}{(k-k')^2} \left( \bar{u}_{p'} \gamma_\mu u_p \right)$	$+e^2 \left( \bar{v}_{k} \gamma^\nu u_p \right) \frac{1}{(k+p)^2} \left( \bar{u}_{p'} \gamma_\nu v_{k'} \right)$

Notice that there is a relative sign difference between the two diagrams.

Square of matrix element

To calculate the unpolarized cross section, one must average over the spins of the incoming particles (s_e- and s_e+ possible values) and sum over the spins of the outgoing particles. That is,

$\overline{\|\mathcal{M}\|^2} \,$	$= \frac{1}{(2s_{e-} + 1)(2 s_{e+} + 1)} \sum_{\mathrm{spins}} \|\mathcal{M}\|^2 \,$
	$= \frac{1}{4} \sum_{s=1}^2 \sum_{s'=1}^2 \sum_{r=1}^2 \sum_{r'=1}^2 \|\mathcal{M}\|^2 \,$

First, calculate $|\mathcal{M}|^2 \,$ :

$\|\mathcal{M}\|^2 \,$ =	$e^4 \left\| \frac{(\bar{v}_{k} \gamma^\mu v_{k'} )( \bar{u}_{p'} \gamma_\mu u_p)}{(k-k')^2} \right\|^2 \,$	(scattering)
	${}- e^4 \left( \frac{ (\bar{v}_{k} \gamma^\mu v_{k'} )( \bar{u}_{p'} \gamma_\mu u_p)}{(k-k')^2} \right)^* \left( \frac{ (\bar{v}_{k} \gamma^\nu u_p )( \bar{u}_{p'} \gamma_\nu v_{k'}) }{(k+p)^2} \right) \,$	(interference)
	${}- e^4 \left( \frac{ (\bar{v}_{k} \gamma^\mu v_{k'} )( \bar{u}_{p'} \gamma_\mu u_p)}{(k-k')^2} \right) \left( \frac{ (\bar{v}_{k} \gamma^\nu u_p )( \bar{u}_{p'} \gamma_\nu v_{k'}) }{(k+p)^2} \right)^* \,$	(interference)
	${}+ e^4 \left\| \frac{(\bar{v}_{k} \gamma^\nu u_p )( \bar{u}_{p'} \gamma_\nu v_{k'} )}{(k+p)^2} \right\|^2 \,$	(annihilation)

Scattering term

Magnitude squared of M

$\|\mathcal{M}\|^2 \,$	$= \frac{e^4}{(k-k')^4} \Big( (\bar{v}_{k} \gamma^\mu v_{k'} )( \bar{u}_{p'} \gamma_\mu u_p) \Big)^* \Big( (\bar{v}_{k} \gamma^\mu v_{k'})( \bar{u}_{p'} \gamma_\mu u_p) \Big) \,$	$(1) \,$
	$= \frac{e^4}{(k-k')^4} \Big( (\bar{v}_{k} \gamma^\mu v_{k'} )^* ( \bar{u}_{p'} \gamma_\mu u_p)^* \Big) \Big( (\bar{v}_{k} \gamma^\mu v_{k'})( \bar{u}_{p'} \gamma_\mu u_p) \Big) \,$	$(2) \,$
	(complex conjugate will flip order)
	$= \frac{e^4}{(k-k')^4} \Big( \left(\bar{v}_{k'} \gamma^\mu v_{k} \right) \left( \bar{u}_{p} \gamma_\mu u_{p'} \right) \Big) \Big( \left( \bar{v}_{k} \gamma^\mu v_{k'} \right) \left( \bar{u}_{p'} \gamma_\mu u_p \right) \Big) \,$	$(3) \,$
	(move terms that depend on same momentum to be next to each other)
	$= \frac{e^4}{(k-k')^4} \left( \bar{v}_{k'} \gamma^\mu v_{k} \right) \left( \bar{v}_{k} \gamma^\mu v_{k'} \right) \left( \bar{u}_{p} \gamma_\mu u_{p'} \right) \left( \bar{u}_{p'} \gamma_\mu u_p \right) \,$	$(4) \,$

Sum over spins

Next, we'd like to sum over spins of all four particles. Let s and s' be the spin of the electron and r and r' be the spin of the positron.

$\frac{(k-k')^4}{e^4} \sum_{\mathrm{spins}} \|\mathcal{M}\|^2 \,$	$= \left(\sum_{r'} \bar{v}_{k'} \gamma^\mu (\sum_{r}v_{k} \bar{v}_{k}) \gamma^\nu v_{k'} \right) \left(\sum_{s} \bar{u}_{p} \gamma_\mu (\sum_{s'}{u_{p'} \bar{u}_{p'}}) \gamma_\nu u_p \right) \,$	$(5) \,$
	$= \operatorname{Tr}\left( \Big(\sum_{r'} v_{k'} \bar{v}_{k'} \Big) \gamma^\mu \Big(\sum_{r}v_{k} \bar{v}_{k} \Big) \gamma^\nu \right) \operatorname{Tr} \left( \Big(\sum_{s} u_p \bar{u}_{p} \Big) \gamma_\mu \Big( \sum_{s'}{u_{p'} \bar{u}_{p'}} \Big) \gamma_\nu \right) \,$	$(6) \,$
	(now use Completeness relations)
	$=\operatorname{Tr}\left( (k\!\!\!/' - m) \gamma^\mu (k\!\!\!/ - m) \gamma^\nu \right) \cdot \operatorname{Tr}\left( (p\!\!\!/' + m) \gamma_\mu (p\!\!\!/ + m) \gamma_\nu \right) \,$	$(7) \,$
	(now use Trace identities)
	$=\left(4 \left( {k'}^\mu k^\nu - \mathbf{k' \cdot k}\eta^{\mu\nu} + k'^\nu k^\mu \right) + 4 m^2 \eta^{\mu\nu} \right) \left( 4 \left( {p'}_\mu p_\nu - \mathbf{p' \cdot p}\eta_{\mu\nu} + p'_\nu p_\mu \right) + 4 m^2 \eta_{\mu\nu} \right) \,$	$(8) \,$
	$=32\left( (k' \cdot p') (k \cdot p) + (k' \cdot p) (k \cdot p') -m^2 p' \cdot p - m^2 k' \cdot k + 2m^4 \right) \,$	$(9) \,$

Now that is the exact form, in the case of electrons one is usually interested in energy scales that far exceed the electron mass. Neglecting the electron mass yields the simplified form:

$\frac{1}{4} \sum_{\mathrm{spins}} \|\mathcal{M}\|^2 \,$	$= \frac{32e^4}{4(k-k')^4} \left( (k' \cdot p') (k \cdot p) + (k' \cdot p) (k \cdot p') \right) \,$
	(use the Mandelstam variables in this relativistic limit)
	$=\frac{8e^4}{t^2} \left(\tfrac{1}{2} s \tfrac{1}{2}s + \tfrac{1}{2}u \tfrac{1}{2} u \right) \,$
	$= 2 e^4 \frac{s^2 +u^2}{t^2} \,$

Annihilation term

The process for finding the annihilation term is similar to the above. Since the two diagrams are related by crossing symmetry, and the initial and final state particles are the same, it is sufficient to permute the momenta, yielding

$\frac{1}{4} \sum_{\mathrm{spins}} \|\mathcal{M}\|^2 \,$	$= \frac{32e^4}{4(k+p)^4} \left( (k \cdot k') (p \cdot p') + (k' \cdot p) (k \cdot p') \right) \,$
	$=\frac{8e^4}{s^2} \left(\tfrac{1}{2} t \tfrac{1}{2}t + \tfrac{1}{2}u \tfrac{1}{2} u \right) \,$
	$= 2 e^4 \frac{t^2 +u^2}{s^2} \,$