 Solution

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.