Past exam of the particle physics course of the University of Oxford / 2018 / 1 / b / Solution

With relativistically normalized one-particle states, Fermi's golden rule becomesIn the parent rest frame, the three-momentum delta function fixes ${\mathbf{p}}_2=-{\mathbf{p}}_1$. The energy delta function fixes the common magnitude $p^{\ast }$; its radial Jacobian converts the remaining phase-space factor intoOnly one momentum magnitude remains because the two final momenta are constrained, not independent. Explicitly,For leptonic W boson decay both final masses are small compared with $m_W$, including the tau mass at the precision requested, so $p^{\ast }\simeq m_W/2$.