Bhabha scattering

In this article, the Mandelstam variables are defined by

${\displaystyle s=\,}$	${\displaystyle (k+p)^{2}=\,}$	${\displaystyle (k'+p')^{2}\approx \,}$	${\displaystyle 2k\cdot p\approx \,}$	${\displaystyle 2k'\cdot p'\,}$
${\displaystyle t=\,}$	${\displaystyle (k-k')^{2}=\,}$	${\displaystyle (p-p')^{2}\approx \,}$	${\displaystyle -2k\cdot k'\approx \,}$	${\displaystyle -2p\cdot p'\,}$
${\displaystyle u=\,}$	${\displaystyle (k-p')^{2}=\,}$	${\displaystyle (p-k')^{2}\approx \,}$	${\displaystyle -2k\cdot p'\approx \,}$	${\displaystyle -2k'\cdot p\,}$

where the approximations are for the high-energy (relativistic) limit.

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

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

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,

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

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

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

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

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.

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

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:

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

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

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

(This is proportional to ${\displaystyle (1+\cos ^{2}\theta )}$ where ${\displaystyle \theta }$ is the scattering angle in the center-of-mass frame.)

Evaluating the interference term along the same lines and adding the three terms yields the final result

${\displaystyle {\frac {\overline {|{\mathcal {M}}|^{2}}}{2e^{4}}}={\frac {u^{2}+s^{2}}{t^{2}}}+{\frac {2u^{2}}{st}}+{\frac {u^{2}+t^{2}}{s^{2}}}\,}$

The completeness relations for the four-spinors u and v are

${\displaystyle \sum _{s=1,2}{u_{p}^{(s)}{\bar {u}}_{p}^{(s)}}=p\!\!\!/+m\,}$

${\displaystyle \sum _{s=1,2}{v_{p}^{(s)}{\bar {v}}_{p}^{(s)}}=p\!\!\!/-m\,}$

where

${\displaystyle p\!\!\!/=\gamma ^{\mu }p_{\mu }\,}$      (see Feynman slash notation)

${\displaystyle {\bar {u}}=u^{\dagger }\gamma ^{0}\,}$

To simplify the trace of the Dirac gamma matrices, one must use trace identities. Three used in this article are:

1. The Trace of any product of an odd number of ${\displaystyle \gamma _{\mu }\,}$'s is zero
2. ${\displaystyle \operatorname {Tr} (\gamma ^{\mu }\gamma ^{\nu })=4\eta ^{\mu \nu }}$
3. ${\displaystyle \operatorname {Tr} \left(\gamma _{\rho }\gamma _{\mu }\gamma _{\sigma }\gamma _{\nu }\right)=4\left(\eta _{\rho \mu }\eta _{\sigma \nu }-\eta _{\rho \sigma }\eta _{\mu \nu }+\eta _{\rho \nu }\eta _{\mu \sigma }\right)\,}$

Using these two one finds that, for example,

${\displaystyle \operatorname {Tr} \left((p\!\!\!/'+m)\gamma _{\mu }(p\!\!\!/+m)\gamma _{\nu }\right)\,}$	${\displaystyle =\operatorname {Tr} \left(p\!\!\!/'\gamma _{\mu }p\!\!\!/\gamma _{\nu }\right)+\operatorname {Tr} \left(m\gamma _{\mu }p\!\!\!/\gamma _{\nu }\right)\,}$
	${\displaystyle +\operatorname {Tr} \left(p\!\!\!/'\gamma _{\mu }m\gamma _{\nu }\right)+\operatorname {Tr} \left(m^{2}\gamma _{\mu }\gamma _{\nu }\right)\,}$
	(the two middle terms are zero because of (1))
	${\displaystyle =\operatorname {Tr} \left(p\!\!\!/'\gamma _{\mu }p\!\!\!/\gamma _{\nu }\right)+m^{2}\operatorname {Tr} \left(\gamma _{\mu }\gamma _{\nu }\right)\,}$
	(use identity (2) for the term on the right)
	${\displaystyle ={p'}^{\rho }p^{\sigma }\operatorname {Tr} \left(\gamma _{\rho }\gamma _{\mu }\gamma _{\sigma }\gamma _{\nu }\right)+m^{2}\cdot 4\eta _{\mu \nu }\,}$
	(now use identity (3) for the term on the left)
	${\displaystyle ={p'}^{\rho }p^{\sigma }4\left(\eta _{\rho \mu }\eta _{\sigma \nu }-\eta _{\rho \sigma }\eta _{\mu \nu }+\eta _{\rho \nu }\eta _{\mu \sigma }\right)+4m^{2}\eta _{\mu \nu }\,}$
	${\displaystyle =4\left({p'}_{\mu }p_{\nu }-(p'\cdot p)\eta _{\mu \nu }+p'_{\nu }p_{\mu }\right)+4m^{2}\eta _{\mu \nu }\,}$