Møller scattering

The Møller scattering can be calculated from the QED point-of-view, at the tree-level, with the help of the two diagrams showed on this page. This two diagrams are contributing at leading order from the QED point-of-view. If we are taking in account the weak force, which is unified with the electromagnetic force at high energy, then we have to add two tree-level diagram for the exchange of a ${\displaystyle Z^{0}}$ boson. Here we will focus our attention on a strict tree-level QED computation of the cross section, which is rather instructive but maybe not the most accurate description from a physical point-of-view.

Before the derivation, we write the 4-momenta as (${\displaystyle p_{1}}$and ${\displaystyle p_{2}}$for incoming electrons, ${\displaystyle p_{3}}$and ${\displaystyle p_{4}}$for outgoing electrons, and ${\displaystyle m=m_{e}}$):

${\displaystyle p_{1}=(E,0,0,p),~p_{2}=(E,0,0,-p),}$

${\displaystyle p_{3}=(E,p\sin \theta ,0,p\cos \theta ),~p_{4}=(E,-p\sin \theta ,0,-p\cos \theta ).}$

The Mandelstam variables are:

${\displaystyle s=(p_{1}+p_{2})^{2}=(p_{3}+p_{4})^{2}\,}$

${\displaystyle t=(p_{1}-p_{3})^{2}=(p_{4}-p_{2})^{2}\,}$

${\displaystyle u=(p_{1}-p_{4})^{2}=(p_{3}-p_{2})^{2}\,}$

These Mandelstam variables satisfy the identity: ${\displaystyle s+t+u\equiv \sum m_{j}^{2}=4m^{2}}$.

According to the two diagrams on this page, the matrix element of t-channel is

${\displaystyle i{\mathcal {M}}_{t}=(-ie)^{2}{\bar {u}}(p_{3})\gamma ^{\mu }u(p_{1}){\frac {-i}{t}}{\bar {u}}(p_{4})\gamma _{\mu }u(p_{2})}$,

the matrix element of u-channel is

${\displaystyle i{\mathcal {M}}_{u}=(-ie)^{2}{\bar {u}}(p_{3})\gamma ^{\mu }u(p_{2}){\frac {-i}{u}}{\bar {u}}(p_{4})\gamma _{\mu }u(p_{1})}$.

So the sum is

${\displaystyle {\begin{aligned}i{\mathcal {M}}&=i({\mathcal {M}}_{t}-{\mathcal {M}}_{u})\\&=-i(-ie)^{2}[{\frac {1}{t}}{\bar {u}}(p_{3})\gamma ^{\mu }u(p_{1}){\bar {u}}(p_{4})\gamma _{\mu }u(p_{2})-{\frac {1}{u}}{\bar {u}}(p_{3})\gamma ^{\mu }u(p_{2}){\bar {u}}(p_{4})\gamma _{\mu }u(p_{1})].\end{aligned}}}$

Therefore,

${\displaystyle {\begin{aligned}|{\mathcal {M}}|^{2}&=e^{4}\{{\frac {1}{t^{2}}}[{\bar {u}}(p_{3})\gamma ^{\mu }u(p_{1})][{\bar {u}}(p_{1})\gamma ^{\nu }u(p_{3})][{\bar {u}}(p_{4})\gamma _{\mu }u(p_{2})][{\bar {u}}(p_{2})\gamma _{\nu }u(p_{4})]\\&~~+{\frac {1}{u^{2}}}[{\bar {u}}(p_{3})\gamma ^{\mu }u(p_{2})][{\bar {u}}(p_{2})\gamma ^{\nu }u(p_{3})][{\bar {u}}(p_{4})\gamma _{\mu }u(p_{1})][{\bar {u}}(p_{1})\gamma _{\nu }u(p_{4})]\\&~~-{\frac {1}{tu}}[{\bar {u}}(p_{3})\gamma ^{\mu }u(p_{1})][{\bar {u}}(p_{2})\gamma ^{\nu }u(p_{3})][{\bar {u}}(p_{4})\gamma _{\mu }u(p_{2})][{\bar {u}}(p_{1})\gamma _{\nu }u(p_{4})]\\&~~-{\frac {1}{tu}}[{\bar {u}}(p_{3})\gamma ^{\mu }u(p_{2})][{\bar {u}}(p_{1})\gamma ^{\nu }u(p_{3})][{\bar {u}}(p_{4})\gamma _{\mu }u(p_{1})][{\bar {u}}(p_{2})\gamma _{\nu }u(p_{4})]\}.\end{aligned}}}$

To calculate the unpolarized cross section, we average over initial spins and sum over final spins, with the factor 1/4 (1/2 for each incoming electron):

${\displaystyle {\begin{aligned}{\frac {1}{4}}\sum _{\text{spins}}|{\mathcal {M}}|^{2}&={\frac {e^{4}}{4}}\{{\frac {1}{t^{2}}}\mathrm {Tr} [\gamma ^{\mu }(\not p_{1}+m)\gamma ^{\nu }(\not p_{3}+m)]\mathrm {Tr} [\gamma _{\mu }(\not p_{2}+m)\gamma _{\nu }(\not p_{4}+m)]\\&~~+{\frac {1}{u^{2}}}\mathrm {Tr} [\gamma ^{\mu }(\not p_{2}+m)\gamma ^{\nu }(\not p_{3}+m)]\mathrm {Tr} [\gamma _{\mu }(\not p_{1}+m)\gamma _{\nu }(\not p_{4}+m)]\\&~~-{\frac {2}{tu}}\mathrm {Tr} [(\not p_{3}+m)\gamma ^{\mu }(\not p_{1}+m)\gamma ^{\nu }(\not p_{4}+m)\gamma _{\mu }(\not p_{2}+m)\gamma _{\nu }]\}\end{aligned}}}$

where we have used the relation ${\displaystyle \sum _{s}u^{s}(p){\bar {u}}^{s}(p)=\not p+m=\gamma ^{\mu }p_{\mu }+m}$. We would next calculate the traces.

The first term in the braces is

${\displaystyle {\begin{aligned}&~~{\frac {1}{t^{2}}}\mathrm {Tr} [\gamma ^{\mu }(\not p_{1}+m)\gamma ^{\nu }(\not p_{3}+m)]\mathrm {Tr} [\gamma _{\mu }(\not p_{2}+m)\gamma _{\nu }(\not p_{4}+m)]\\&={\frac {16}{t^{2}}}(p_{1}^{\mu }p_{3}^{\nu }+p_{3}^{\mu }p_{1}^{\nu }+(-p_{13}+m^{2})g^{\mu \nu })(p_{2}^{\mu }p_{4}^{\nu }+p_{4}^{\mu }p_{2}^{\nu }+(-p_{24}+m^{2})g^{\mu \nu })\\&={\frac {32}{t^{2}}}{\big (}p_{12}p_{34}+p_{23}p_{14}-m^{2}p_{13}-m^{2}p_{24}+2m^{4}{\big )}\\&={\frac {32}{t^{2}}}{\big (}p_{12}^{2}+p_{14}^{2}+2m^{2}(p_{14}-p_{12}){\big )}\\&={\frac {8}{t^{2}}}(s^{2}+u^{2}-8m^{2}(s+u)+24m^{4})\end{aligned}}}$

Here ${\displaystyle p_{ij}\equiv p_{i}\cdot p_{j}}$, and we have used the ${\displaystyle \gamma }$-matrix identity

${\displaystyle \mathrm {Tr} [\gamma ^{\mu }\gamma ^{\nu }\gamma ^{\rho }\gamma ^{\sigma }]=4\left(\eta ^{\mu \nu }\eta ^{\rho \sigma }-\eta ^{\mu \rho }\eta ^{\nu \sigma }+\eta ^{\mu \sigma }\eta ^{\nu \rho }\right)}$

and that trace of any product of an odd number of ${\displaystyle \gamma ^{\mu }}$ is zero.

Similarly, the second term is

${\displaystyle {\begin{aligned}&~~{\frac {1}{u^{2}}}\mathrm {Tr} [\gamma ^{\mu }(\not p_{2}+m)\gamma ^{\nu }(\not p_{3}+m)]\mathrm {Tr} [\gamma _{\mu }(\not p_{1}+m)\gamma _{\nu }(\not p_{4}+m)]\\&={\frac {32}{u^{2}}}{\big (}p_{12}p_{34}+p_{13}p_{24}-m^{2}p_{23}-m^{2}p_{14}+2m^{4}{\big )}\\&={\frac {8}{u^{2}}}(s^{2}+t^{2}-8m^{2}(s+t)+24m^{4})\end{aligned}}}$

Using the ${\displaystyle \gamma }$-matrix identities

${\displaystyle \mathrm {Tr} [\gamma ^{\mu }\gamma ^{\nu }\gamma _{\mu }\gamma _{\nu }]=-32}$,

${\displaystyle \mathrm {Tr} [\gamma ^{\rho }\gamma ^{\mu }\gamma ^{\sigma }\gamma ^{\nu }\gamma _{\mu }\gamma _{\nu }]=\mathrm {Tr} [\gamma ^{\rho }\gamma ^{\mu }\gamma ^{\nu }\gamma ^{\sigma }\gamma _{\mu }\gamma _{\nu }]=16g^{\rho \sigma }}$,

${\displaystyle \mathrm {Tr} [\gamma ^{\rho }\gamma ^{\mu }\gamma ^{\sigma }\gamma ^{\nu }\gamma ^{\lambda }\gamma _{\mu }\gamma ^{\tau }\gamma _{\nu }]=-32g^{\rho \lambda }g^{\sigma \tau }}$,

and the identity of Mandelstam variables: ${\displaystyle s+t+u\equiv \sum m_{j}^{2}}$, we get the third term

${\displaystyle {\begin{aligned}&~~-{\frac {2}{tu}}\mathrm {Tr} [(\not p_{3}+m)\gamma ^{\mu }(\not p_{1}+m)\gamma ^{\nu }(\not p_{4}+m)\gamma _{\mu }(\not p_{2}+m)\gamma _{\nu }]\\&=-{\frac {32}{tu}}{\big (}-2p_{12}p_{34}+2m^{2}(p_{12}+p_{13}+p_{14})-2m^{4}{\big )}\\&={\frac {16}{tu}}(s^{2}-8m^{2}s+12m^{4})\end{aligned}}}$

Therefore,

${\displaystyle {\begin{aligned}{\overline {|{\mathcal {M}}|^{2}}}&\equiv {\frac {1}{4}}\sum _{\text{spins}}|{\mathcal {M}}|^{2}\\&=2e^{4}{\Big \{}{\frac {1}{t^{2}}}{\big (}s^{2}+u^{2}-8m^{2}(s+u)+24m^{4}{\big )}\\&~~+{\frac {1}{u^{2}}}{\big (}s^{2}+t^{2}-8m^{2}(s+t)+24m^{4}{\big )}\\&~~+{\frac {2}{tu}}{\big (}s^{2}-8m^{2}s+12m^{4}{\big )}{\Big \}}\end{aligned}}.}$

Substitute in the momentums we have set here, which are

${\displaystyle s=4E^{2}=E_{CM}^{2}}$,

${\displaystyle t=2p^{2}(\cos \theta -1)}$,

${\displaystyle u=2p^{2}(-\cos \theta -1)}$.

Finally we get the unpolarized cross section

${\displaystyle {\begin{aligned}{\frac {d\sigma }{d\Omega }}&={\frac {1}{64\pi ^{2}E_{CM}^{2}}}{\frac {|{\vec {p}}_{f}|}{|{\vec {p}}_{i}|}}{\overline {|{\mathcal {M}}|^{2}}}\\&={\frac {\alpha ^{2}}{2E_{CM}^{2}}}{\Big \{}{\frac {1}{t^{2}}}{\big (}s^{2}+u^{2}-8m^{2}(s+u)+24m^{4}{\big )}\\&~~+{\frac {1}{u^{2}}}{\big (}s^{2}+t^{2}-8m^{2}(s+t)+24m^{4}{\big )}\\&~~+{\frac {2}{tu}}{\big (}s^{2}-8m^{2}s+12m^{4}{\big )}{\Big \}}\\&={\frac {\alpha ^{2}}{E_{CM}^{2}p^{4}\sin ^{4}\theta }}{\Big [}4(m^{2}+2p^{2})^{2}+{\big (}4p^{4}-3(m^{2}+2p^{2})^{2}{\big )}\sin ^{2}\theta +p^{4}\sin ^{4}\theta {\Big ]}.\end{aligned}}}$

with ${\displaystyle E^{2}=m^{2}+p^{2}}$ and ${\displaystyle E_{CM}=2E}$.

In the nonrelativistic limit, ${\displaystyle m\gg p}$,

${\displaystyle {\begin{aligned}{\frac {d\sigma }{d\Omega }}&={\frac {m^{4}\alpha ^{2}}{E_{CM}^{2}p^{4}\sin ^{4}\theta }}{\Big (}4-3\sin ^{2}\theta {\Big )}\\&={\frac {m^{4}\alpha ^{2}}{E_{CM}^{2}p^{4}\sin ^{4}\theta }}{\Big (}1+3\cos ^{2}\theta {\Big )}.\end{aligned}}}$

In the ultrarelativistic limit, ${\displaystyle m\ll p}$,

${\displaystyle {\begin{aligned}{\frac {d\sigma }{d\Omega }}&={\frac {\alpha ^{2}}{E_{CM}^{2}p^{4}\sin ^{4}\theta }}{\Big (}16p^{4}-8p^{4}\sin ^{2}\theta +p^{4}\sin ^{4}\theta {\Big )}\\&={\frac {\alpha ^{2}}{E_{CM}^{2}\sin ^{4}\theta }}{\Big (}3+\cos ^{2}\theta {\Big )}^{2}.\end{aligned}}}$

• SLAC E158: Measuring the Electron's WEAK Charge