 Solution

ID: 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 becomes

Γ = \frac{1}{2 M} \int ∣ M ∣^{2} (2 π)^{4} δ^{(4)} (P - p_{1} - p_{2}) \frac{d ^{3} p _{1}}{( 2 π ) ^{3} 2 E _{1}} \frac{d ^{3} p _{2}}{( 2 π ) ^{3} 2 E _{2}} .

p_{2} = - p_{1}

. The energy delta function fixes the common magnitude

p^{*}

; its radial Jacobian converts the remaining phase-space factor into

Γ = \frac{p ^{*}}{32 π ^{2} M ^{2}} \int ∣ M ∣^{2} d Ω.

Only one momentum magnitude remains because the two final momenta are constrained, not independent. Explicitly,

p^{*} = \frac{[ M ^{2} - ( m _{1} + m _{2} ) ^{2} ] [ M ^{2} - ( m _{1} - m _{2} ) ^{2} ]}{2 M} .

For leptonic W boson decay both final masses are small compared with

m_{W}

, including the tau mass at the precision requested, so

p^{*} ≃ m_{W} /2

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