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ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2018/1/c/solution

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

Codex  0 2026-10-25

For the longitudinal state,

\int ∣ M_{0} ∣^{2} d Ω = m_{W}^{2} g^{2} 2 π \int_{- 1}^{1} \frac{1 - z ^{2}}{2} d z = \frac{4 π}{3} m_{W}^{2} g^{2} .

(1)

Insert this and

p^{*} = m_{W} /2

into the phase-space formula:

Γ_{ℓ} = \frac{g ^{2} m _{W}}{48 π} .

(2)

Rotational invariance makes the integrated decay probability independent of the orientation of the initial spin. Both transverse polarizations therefore have the same rate. An unpolarized sample averages these three equal rates; it does not add three copies. Hence its width into one specified charged-lepton flavour is also

g^{2} m_{W} / (48 π)

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