Møller scattering

Feynman diagrams
t-channel
u-channel

Møller scattering is the name given to electron-electron scattering in quantum field theory, named after the Danish physicist Christian Møller. The electron interaction that is idealized in Møller scattering forms the theoretical basis of many familiar phenomena such as the repulsion of electrons in the helium atom. While formerly many particle colliders were designed specifically for electron-electron collisions, more recently electron-positron colliders have become more common. Nevertheless, Møller scattering remains a paradigmatic process within the theory of particle interactions.

We can express this process in the usual notation, often used in particle physics:

$${\displaystyle e^{-}e^{-}\longrightarrow e^{-}e^{-},}$$

In

The asymmetry in Møller scattering is

$${\displaystyle A_{\rm {PV}}=-m_{e}E{\frac {G_{\rm {F}}}{{\sqrt {2}}\pi \alpha }}{\frac {16\sin ^{2}\Theta _{\text{cm}}}{\left(3+\cos ^{2}\Theta _{\text{cm}}\right)^{2}}}\left({\frac {1}{4}}-\sin ^{2}\theta _{\rm {W}}\right),}$$

where m_e is the electron mass, E the energy of the incoming electron (in the reference frame of the other electron), ${\displaystyle G_{\rm {F}}}$ is Fermi's constant, ${\displaystyle \alpha }$ is the

${\displaystyle \Theta _{\text{cm}}}$

${\displaystyle \theta _{\rm {W}}}$

QED computation

The Møller scattering can be calculated from the QED point-of-view, at the tree-level, with the help of the two diagrams shown on this page. These 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

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

Therefore,

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

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):

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

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

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

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, ${\displaystyle m\ll p}$,

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

References

External links

• SLAC E158: Measuring the Electron's WEAK Charge

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