Past exam of the particle physics course of the University of Oxford / 2018 / 6 / b / Solution

Let $z=\cos {\theta }^{\star }$. Rotating the spin-one intermediate state between initial and final axes gives probability amplitudes proportional to $d_{\pm 1,\pm 1}^1$ for equal fermion helicities and $d_{\mp 1,\pm 1}^1$ for opposite ones. Thus, apart from a common positive normalization,

The common factor $1/4$ from the rotation elements is absorbed into the normalization. Distinct external helicities are orthogonal, so an unpolarized rate adds their squared probability amplitudes incoherently.

The source interchanges the conventional left and right electron coupling labels. With $g_L=T_3-Q{\sin }^2{\theta }_W$ and $g_R=-Q{\sin }^2{\theta }_W$, the physical electron values areAt ${\sin }^2{\theta }_W=0$, for example, the left electron still carries weak isospin $-1/2$, while the right electron is a weak singlet; the printed assignment gives the opposite result. This fixes how the helicity-channel table should be interpreted.^[ref]

For a muon final state, lepton universality gives the same pair of couplings. Crucially, swapping both labels does not change the summed distribution. Put $L=g_L$, $R=g_R$; thenHere $\mathcal{N}$ includes the resonance propagator, initial spin average and common coupling normalization. Equivalently,The source's simultaneous label interchange reverses ${\mathcal{A}}_{\ell }$ but leaves its square unchanged.