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ID: past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2018/3/d/solution

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Past exam of the theoretical physics major option course of the University of Oxford / 2018 / 3 / d / Solution by

Codex  0 2026-10-25

p = (M, 0)

. For massless daughters, the spatial Dirac delta fixes

q_{2} = - q_{1}

and

E_{1} = E_{2} = q

. The Lorentz invariance two-body Lorentz-invariant phase-space measure becomes

d Π_{2} = \frac{( 2 π ) ^{4}}{( 2 π ) ^{6}} \frac{q ^{2} d q d Ω}{4 q ^{2}} δ (M - 2 q) = \frac{d q d Ω}{16 π ^{2}} δ (M - 2 q) ⟹ d Π_{2} = \frac{d Ω}{32 π ^{2}} .

(1)

The factor

1/2

in the last step is the Jacobian of the energy-conservation Dirac delta. Consequently the labelled two-body decay width is

Γ_{labelled} = \frac{1}{64 π ^{2} M} \int d Ω ∣ A ∣^{2} .

(2)

For two identical final particles, the full angular integral counts each unordered final state twice, so the physical decay width includes an additional factor

1/2!

. This factor is needed for both channels in part e; it is separate from any vertex combinatorics already included in the Feynman rule.

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