PHY–396 K/L. Solutions for problem set #12.
Problem 1:
Note the correct muon decay amplitude
M(µ−→e−νµ¯νe) = GF
√2¯u(νµ)(1 −γ5)γαu(µ−)ׯu(e−)(1 −γ5)γαv(¯νe).(1)
The complex conjugate of this amplitude
M∗=GF
√2h¯u(µ−)γβ(1 + γ5)u(νµ)iׯv(¯νe)γβ(1 + γ5)u(e−)(S.1)
has the opposite sign of the γ5because ¯γ5≡γ0(γ5)†γ0=−γ5. Consequently,
|M|2=1
2G2
Fh¯u(νµ)(1 −γ5)γαu(µ−)¯u(µ−)γβ(1 + γ5)u(νµ)i(S.2)
ׯu(e−)(1 −γ5)γαv(¯νe)¯v(¯νe)γβ(1 + γ5)u(e−)
and hence
1
2X
all
spins
|M|2=1
4G2
Ftr(1 −γ5)γα(6pµ+Mµ)γβ(1 + γ5)(6pνµ+mνµ)
×tr(1 −γ5)γα(6p¯νe−mνe)γβ(1 + γ5)(6pe+me)
≈1
4G2
Ftr(1 −γ5)γα(6pµ+Mµ)γβ(1 + γ5)6pν
×tr(1 −γ5)γα6p¯νγβ(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 p¯ν≡p¯νe. Please note that here and henceforth the indices µ, ν, ¯ν , 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,βand later also γ,δ,σand ρ. Thus, pµα is
the α’s component of the muon’s 4–momentum, etc.,etc.
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