Past exam of the particle physics course of the University of Oxford / 2022 / 1

An incoming fermion contributes a Dirac spinor $u_s(p)$ and an outgoing fermion contributes its Dirac adjoint ${\bar{u}}_s(p)=u_s^{†}(p){\gamma }^0$. The label $s$ specifies a spin state; $p$ is the on-shell four-momentum. Incoming and outgoing antifermions instead contribute ${\bar{v}}_s(p)$ and $v_s(p)$.

The exchanged boson is a virtual photon, represented in the Feynman rules by the photon propagator $-i{g}_{\mu \nu }/(q^2+iϵ)$ in Feynman gauge. Its momentum transfer is $q=p_1-p_3=p_4-p_2$; unlike an external real photon, it need not satisfy $q^2=0$. Each Feynman vertex contracts a gamma matrix with this propagator and contributes $-ie{\gamma }^{\mu }$ in a conventional QED sign convention. The coupling constant $e$ sets the interaction strength, with fine-structure constant $\alpha =e^2/(4\pi )$. One may reverse the sign of the photon field and both vertex signs without changing the result. The gamma matrices obey $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }I$, with Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$.

There is one tree-level Feynman diagram: the distinguishable electron and muon exchange a photon in the $t$ channel. There is no identical-particle exchange diagram.

Multiplying the external Dirac spinors, the two Feynman vertices and the photon propagator givesThe overall phase of this scattering amplitude is convention dependent. Four external spin labels with two choices each give $2^4=16$ possible configurations before taking a limit. For $E\gg m,M$, the vector QED interaction preserves fermion chirality, which becomes helicity for each external particle. Thus the surviving configurations, written as $(h_e,h_{\mu }{)}_{\mathrm{in}}\to (h_e,h_{\mu }{)}_{\mathrm{out}}$, areHere $L,R$ specify helicity along each particle's own momentum. The other configurations require at least one mass-suppressed flip and vanish in the strictly massless limit.

Define the Mandelstam variables for the scattering labels by $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2=q^2$ and $u=(p_1-p_4{)}^2$. Dropping the electron and muon masses givesThese identities follow by expanding the corresponding squared four-momenta, using $p_i^2=0$ and four-momentum conservation. Substitution into the supplied spin-averaged matrix element yieldsThe $1/4$ initial spin average is already included in the given expression; no further averaging factor belongs here.

To implement crossing symmetry with the annihilation labels, replace the old scattering four-momenta by $(p_1,-p_3,-p_2,p_4)$. The scattering Mandelstam variables then become $(s_{\mathrm{old}},t_{\mathrm{old}},u_{\mathrm{old}})=(t,s,u)$, where now $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. ConsequentlyIn the centre-of-momentum frame, take $\theta$ between the incoming electron and outgoing positive muon, $p_3$. For negligible masses, $t=-s(1-\cos \theta )/2$ and $u=-s(1+\cos \theta )/2$. ThereforeDefining $\theta$ using the outgoing negative muon interchanges $t$ and $u$ and leaves the result unchanged.

The intermediate photon carries spin $J=1$. In the massless limit, a back-to-back fermion-antifermion pair coupled to its vector current has opposite helicities, so the initial and final projections along their respective axes are $m,m^{\prime }=\pm 1$. A rotation between these axes contributes the Wigner D-matrix elements $d_{m^{\prime }m}^1(\theta )$. The relevant elements have magnitudes $(1+\cos \theta )/2$ and $(1-\cos \theta )/2$. Summing the four squared helicity amplitudes gives the angular factorThe common coupling and normalization factors recover the result in d. The $m=0$ terms do not contribute: the massless external vector currents eliminate the corresponding helicity configurations.