By D. S. Mitrinovic
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Extra resources for Elementary Matrices - Tutorial Text No. 3
1) where τη is the neutron mean lifetime and we have set V = 1 since it always cancels. Now the mass difference of neutron and proton is mn - mp = A = 1-3 MeV. Thus the final particles will have maximum momenta of about 1 MeV and with a comparable maximum energy for the neutrino and electron. 30 TRANSITION AMPLITUDES AND PROBABILITIES [Ch. dpedQv 2 pv (mn - Ee) - pe · pv£v then since Ep ~ mn ~ 1000 MeV, we can drop the last term in the denominator and make the approximation N3= j dj>e düvEvPv and so Γ„ = — = — l — Σ Σ Wfi\2 dp.
The argument is probably more easily seen in modern language. Consider a particle (say the electron) and antiparticle (positron) with opposite properties particle — e — p — E —s antiparticle +e +p +E +s Now apply this situation to muon decay and a leptonic interaction c - μ+ -» e+vv β~μ+ -> vv If the electron has all the opposite properties to the positron then the kinematic conditions of the vv pair are unaltered. Thus it is immaterial whether we consider the creation of a particle or the annihilation of an antiparticle with opposite properties.
Now let us apply our result to muon decay, μ -> evv. 5) suggests an expression 1 G2 5 — ~ ^L 60π3 μ τμ where Αμ = πιμ - me - 100 MeV, thus τβ = (-j-Y τΛ ~ 10-6 sec. 8) This is satisfactorily close to the measured value. 2(d)]. The same basic expression may be used to evaluate the branching ratio of the pion in the mode π+ -> n°e+v. The total decay rate is ■* π 2a NBP = = ·*■ π-+μν "·" ■* n-*ev * ■* π->π°βν 32 TRANSITION AMPLITUDES AND PROBABILITIES [Ch. 2 where the measured value of τ„ ~ 10"8 sec.