Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 2 / b

First derive the contracted gamma matrices identities from the Clifford algebra in four dimensions. Moving the outer gamma matrix through two factors givesMoving it through three factors, and using this two-factor result, givesConsequently ${\gamma }^{\nu }/b{\gamma }_{\mu }/c{\gamma }_{\nu }=-2/c{\gamma }_{\mu }/b$ and ${\gamma }_{\mu }/b/d{\gamma }^{\mu }=4(b\cdot d)1$. Apply these in succession:The final step uses $\mathrm{tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$. This is a genuinely four-dimensional instance of the gamma matrix trace identities; using it unchanged inside a $D$-dimensional loop calculation would require separate justification.

Label the four-momenta as $e^{-}(p_1){\mu }^{-}(p_2)\to e^{-}(p_3){\mu }^{-}(p_4)$. There is one tree-level Feynman diagram, the $t$-channel photon exchange shown in the lower-left panel. The two lepton species are distinct, so there is no exchanged-final-lepton diagram. With $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$, the Feynman rules giveThe massless fermion spin sum is ${\sum }_r{u}_r(p){\bar{u}}_r(p)=/p$. Averaging the two initial spins of each lepton supplies $1/4$, soThe gamma matrix trace identities imply, for arbitrary null external four-momenta,To see the contraction, each trace is $4(a^{\mu }{b}^{\nu }+a^{\nu }{b}^{\mu }-g^{\mu \nu }a\cdot b)$; the terms proportional to $(a\cdot b)(c\cdot d)$ cancel in four dimensions, leaving the displayed two products. Here $p_1\cdot p_2=p_3\cdot p_4=s/2$ and $p_1\cdot p_4=p_2\cdot p_3=-u/2$. ThereforeThe requested proportionality constant is $2e^4=32{\pi }^2{\alpha }_{\mathrm{em}}^2$, where ${\alpha }_{\mathrm{em}}$ is the fine-structure constant. Both numerator terms are required by the spin sum; no final-state statistical factor enters a squared scattering amplitude for distinct leptons.

For Møller scattering, both the $t$-channel and $u$-channel photon exchange tree-level Feynman diagrams contribute, as shown in the two remaining lower panels. Exchanging the identical final electrons reverses the sign of the scattering amplitude, so the fermionic sign fixesThe two direct squares follow from the preceding distinct-lepton calculation:The interference cannot be discarded because the final particles are identical. Summing their spins joins all four external Dirac spinors into a single trace:The result from part (i) was used with relabelled contracted indices. Its negative sign combines with the explicit subtraction of the two exchange scattering amplitudes, so the interference contribution to the averaged square is positive:ThusThis is symmetric under $t\leftrightarrow u$, as required by the identical final electrons. There is no additional $1/2!$ in the squared scattering amplitude. Such a factor is used when integrating the cross section over a final-state region that counts both orderings of the identical pair; alternatively one integrates over one representative half of that region.