Past exam of the particle physics course of the University of Oxford / 2015 / 2

At energies large compared with fermion masses, the gauge interactions preserve chirality on each fermion line. A photon couples to charged fermions with equal left and right coefficients $e{Q}_f$. A gluon couples to either quark chirality with coefficient $g_s{T}^a$. A Z boson couples to both, with coefficients $(g/\cos {\theta }_W)(T_3-Q_f{\sin }^2{\theta }_W)$ on left-chiral fields and $(g/\cos {\theta }_W)(-Q_f{\sin }^2{\theta }_W)$ on right-chiral fields. A W boson couples only to left-chiral fermions and the associated right-helicity antifermions: $W^{-}\to {\ell }_L^{-}{\bar{\nu }}_R$ or $d_L{\bar{u}}_R$, and the charge-conjugate $W^{+}$ channels. For quarks the coefficient includes the CKM matrix. There is no Standard Model right-chiral neutrino gauge vertex.

For outgoing pairs at a neutral vertex, the allowed helicities are $f_L^{-}{\bar{f}}_R^{+}$ and $f_R^{-}{\bar{f}}_L^{+}$; the latter labels helicity, not the chirality of the corresponding antifield. Only the $s$-channel photon and Z boson diagrams contribute to $e^{+}{e}^{-}\to {\mu }^{+}{\mu }^{-}$ at tree level among these gauge exchanges. A W boson changes charged leptons into neutrinos, and gluons do not couple to leptons.

Assume unpolarized beams, $s\gg m_{\mu }^2$, a single exchanged photon, full azimuthal coverage and no radiative corrections. Take $c_{\ast }=\cos {\theta }^{\ast }$ between incoming $e^{-}$ and outgoing ${\mu }^{-}$. The four nonzero helicity amplitudes have magnitudes $e^2(1+c_{\ast })$ for equal incoming/outgoing negative-particle helicities, and $e^2(1-c_{\ast })$ for opposite ones. Averaging four initial spin states and summing the final ones givesThe massless two-body phase space givesThe even distribution reflects the equal left/right photon couplings and is symmetric under interchange of forward and backward directions.

Include the Z boson amplitude and its quantum interference with the photon amplitude. Define $s_W=\sin {\theta }_W$, $c_W=\cos {\theta }_W$, $e=g{s}_W$, $g_L^f=T_3^f-Q_f{s}_W^2$, $g_R^f=-Q_f{s}_W^2$, andFor the same unpolarized and massless approximations as b, the complete tree-level angular distribution isThe absolute squares include the pure terms and their interference. Put $A=|C_{LL}{|}^2+|C_{RR}{|}^2$ and $B=|C_{LR}{|}^2+|C_{RL}{|}^2$. The even part is $(A+B)(1+c_{\ast }^2)$ and the odd part is $2(A-B)c_{\ast }$. Thus the forward-backward asymmetry is $A_{\mathrm{FB}}=3(A-B)/(4(A+B))$.

At the Z boson pole, neglect the small photon contribution and use real tree-level couplings. Thenwhere $g_V=g_L+g_R$, $g_A=g_L-g_R$. For charged leptons, $g_V/g_A=1-4{\sin }^2{\theta }_W$. Fitting the forward/backward counts, after detector and radiation corrections, therefore measures the weak mixing angle, conventionally an effective leptonic angle once electroweak corrections are included. For illustration ${\sin }^2{\theta }_W=0.231$ gives ${\mathcal{A}}_{\ell }\simeq 0.151$ and $A_{\mathrm{FB}}^0\simeq 0.0171$. This illustrative parameter choice is not extracted from this paper's data. The real tree-level photon-Z boson interference vanishes exactly at the pole but matters away from it.