Solutions to Problem Set 12 - Quantum Field Theory | PHY 396K, Assignments of Physics

Download Solutions to Problem Set 12 - Quantum Field Theory | PHY 396K and more Assignments Physics in PDF only on Docsity! PHY–396 K/L. Solutions for problem set #12. Problem 1: Note the correct muon decay amplitude M(µ− → e−νµν̄e) = GF√ 2 [ ū(νµ)(1− γ5)γαu(µ−) ] × [ ū(e−)(1− γ5)γαv(ν̄e) ] . (1) The complex conjugate of this amplitude M∗ = GF√ 2 [ ū(µ−)γβ(1 + γ5)u(νµ) ] × [ v̄(ν̄e)γβ(1 + γ 5)u(e−) ] (S.1) has the opposite sign of the γ5 because γ̄5 ≡ γ0(γ5)†γ0 = −γ5. Consequently, |M|2 = 12G 2 F [ ū(νµ)(1− γ5)γαu(µ−)ū(µ−)γβ(1 + γ5)u(νµ) ] (S.2) × [ ū(e−)(1− γ5)γαv(ν̄e)v̄(ν̄e)γβ(1 + γ5)u(e−) ] and hence 1 2 ∑ all spins |M|2 = 14G 2 F tr ( (1− γ5)γα(6pµ +Mµ)γβ(1 + γ5)(6pνµ +mνµ) ) × tr ( (1− γ5)γα(6pν̄e −mνe)γβ(1 + γ5)(6pe +me) ) ≈ 14G 2 F tr ( (1− γ5)γα(6pµ +Mµ)γβ(1 + γ5) 6pν ) × tr ( (1− γ5)γα 6pν̄γβ(1 + γ5) 6pe ) (S.3) where the approximation is me ≈ 0; we also make use of mνe = mνµ = 0 (which may be exactly true or just a very good approximation, future data will tell) and simplify notations: pν ≡ pνµ and pν̄ ≡ pν̄e . Please note that here and henceforth the indices µ, ν, ν̄, e denote the particles to which respective momenta belong and have nothing to do with the Lorentz indices of those momenta. For the Lorentz indices, I use here α, β and later also γ, δ, σ and ρ. Thus, pµα is the α’s component of the muon’s 4–momentum, etc., etc. 1 Having derived eq. (S.3), we now need to evaluate the traces. For the first trace, we have tr ( (1− γ5)γα(6pµ +Mµ)γβ(1 + γ5) 6pν ) = tr ( (1− γ5)γα(6pµ +Mµ)γβ 6pν(1− γ5) ) = tr ( (1− γ5)2γα(6pµ +Mµ)γβ 6pν ) = 2 tr ( (1− γ5)γα(6pµ +Mµ)γβ 6pν ) 〈〈using tr(γαγβ 6pν) = tr(γ5γαγβ 6pν) = 0〉〉 (S.4) = 2 tr ( γα 6pµγβ 6pν ) − 2 tr ( γ5γα 6pµγβ 6pν ) = 8 [ pαµp β ν + p β µp α ν − gαβ(pµ · pν) ] + 8iαγβδpµγpνδ. Similarly, the second trace evaluates to tr ( (1− γ5)γα 6peγβ(1 + γ5) 6pν̄ ) = 8 [ (peαpν̄β + peβpν̄α − gαβ(pe · pν̄) ] + 8iαρβσp ρ ν̄p σ e . (S.5) It remains to substitute the trace formulæ (S.4) and (S.5) back into eq. (S.3) and contract the Lorentz indices. Thus, 1 2 ∑ all spins |M|2 = 16G2F ([ pαµp β ν + p β µp α ν − gαβ(pµ · pν) ] + iαγβδpµγpνδ ) × ([ peαpν̄β + peβpν̄α − gαβ(pe · pν̄) ] + iαρβσp ρ ν̄p σ e ) 〈〈using symmetry/antisymmetry of factors under α↔ β 〉〉 = 16G2F ([ pαµp β ν + p β µp α ν − gαβ(pµ · pν) ] × [ peαpν̄β + peβpν̄α − gαβ(pe · pν̄) ] − αγβδpµγpνδ × αρβσpρν̄pσe ) = 16G2F ([ 2(pµ · pe)(pν · pν̄) + 2(pµ · pν̄)(pν · pe) − 2(pµ · pν)(pe · pν̄) − 2(pµ · pν)(pe · pν̄) + 4(pµ · pν)(pe · pν̄) ] + 2 [ 2(pµ · pν̄)(pν · pe) − 2(pµ · pe)(pν · pν̄) ]) = 64G2F (pµ · pν̄)(pν · pe) . (S.6) Q.E .D. 2 Problem 2: We are now ready to calculate the muon decay rate and the electons’ spectrum. According to the general rules of decay dΓ = |M|2 2M dP , (S.16) thus in light of eqs. (2) and (S.8), dΓ(µ− → e−νµν̄e) = G2F 8π5Mµ (pµ ·pν̄)(pe ·pν)×dEe dEν dEν̄ d3Ω δ(Ee+Eν+Eν̄−Mµ). (S.17) In the muon’s frame, (pµ · pν̄) = MµEν̄ (S.18) while (pe · pe) = EeEν − pepν cos θeν = EeEν + 12p 2 e + 1 2p 2 ν − 12p 2 ν̄ ; (S.19) neglecting the electron and neutrino’s masses, we may rewrite (S.19) as 1 2(Ee + Eν) 2 − 12E 2 ν̄ = 1 2Mµ(Mµ − 2Eν̄). (S.20) Consequently, dΓ(µ− → e−νµν̄e) = G2F 16π5 MµEν̄(Mµ−2Eν̄)×dEe dEν dEν̄ d3Ω δ(Ee+Eν+Eν̄−Mµ). (S.21) At this point we are ready to integrate over the final-state variables. In light of ∫ d3Ω = 8π2 and the kinematic limits (S.15), we immediately obtain Γ(µ− → e−νµν̄e) = G2FMµ 2π3 1 2 Mµ∫∫∫ 0 dEe dEν̄ dEν Eν̄(Mµ − 2Eν̄)δ(Ee + Eν + Eν̄ −Mµ) = G2FMµ 2π3 1 2 Mµ∫ 0 dEe 1 2 Mµ∫ 1 2 Mµ−Ee dEν̄ Eν̄(Mµ − 2Eν̄) = G2FMµ 2π3 1 2 Mµ∫ 0 dEeE 2 e ( 1 2Mµ − 2 3Ee). (S.22) In other words, the partial muon decay rate with respect to the final electron’s energy is given 5 by dΓ dEe = G2FMµ 12π3 × E2e (3Mµ − 4Ee) (S.23) or rather dΓ dEe ≈ { G2F 12π3 MµE 2 e (3Mµ − 4Ee) for Ee < 12Mµ, 0 for Ee > 1 2Mµ. (S.24) Graphically, ............................................................................................................................................................................................................................................................................................................................................................................................................................... ....... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .. ..... ..... dΓ/dEe Ee 1 2Mµ ......................... ............... ............ ........... .......... .......... ......... ......... ......... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ......... ......... ......... ......... .......... .......... ........... ............ ............... ........................ ................ Note how this curve smoothly reaches its maximim at Ee = 1 2Mµ and then abruptly falls down to zero. It remains to calculate the total decay rate of the muon by integrating the partial rate (S.24) over the electron’s energy. The result is Γtot(µ→ eνν̄) = G2FM 5 µ 192π3 . (S.25) 6