10.3 Solutions 585
νμ=ν 1 cosθ+ν 2 sinθ (1)
andνe=−ν 1 sinθ+ν 2 cosθ (2)
which form a set of ortho normal states.νμandνeare the sort of states which
are produced in charged pion decay (π→μ+νμ) and neutron decay (n→
p+e−+νe); ν 1 andν 2 are the mass eigen states, corresponding to the
neutrino massesm 1 andm 2. In the matrix form
(
νμ
νe)
=
(
cosθ sinθ
−sinθcosθ)(
ν 1
ν 2)
(3)
The difference in the masses leads to difference in the characteristic fre-
quencies with which the neutrinos are propagated. Hereθis known as the
mixing angle which is analogous to the Cabibbo angleθc. Using natural units
(=c=1) at timet
=0,
ν 1 (t)=ν 1 (0)e 1 −iE^1 t
ν 2 (t)=ν 2 (0)e 2 −iE^2 t (4)
where the exponentials are the usual oscillating time factors associated with
any quantum mechanical stationary state. Since the momentum is conserved
the statesν 1 (t) andν 2 (t) must have the same momentump.Ifthemass
mi<<Ei(i= 1 ,2)Ei=(p^2 +m^2 i)^1 /^2 ∼=p+m^2 i
2 p(5)
Suppose att=0, we start with muon type of neutrinos so that,νμ(0)= 1
andνe(0)=0 then by (3),
ν 2 (0)=νμ(0) sinθ (6)
ν 1 (0)=νμ(0) cosθ
and νμ(t)=cosθν 1 (t)+sinθν 2 (t)(7)
Thus at timet
=0, the muon–neutrino beam is no longer pure but devel-
ops an electron neutrino component. Similarly, an electron–neutrino beam
would develop a muon–neutrino beam component.
Using (4) and (6) in (7)
νμ(t)
νμ(0)=cos^2 θ.e−iE^1 t+sin^2 θ.e−iE^2 tand the intensity
Iμ(t)
Iμ(0)=
∣
∣
∣
∣
νμ(t)
νμ(0)∣
∣
∣
∣
2
=cos^4 θ+sin^4 θ+sin^2 θcos^2 θ[
ei(E^2 −E^1 )t+e−i(E^2 −E^1 )t]
P(νμ→νe)= 1 −sin^22 θsin^2[
(E 2 −E 1 )
2
t]
(8)
wherewehaveused(cos^4 θ+sin^4 θ)=(cos^2 θ+sin^2 θ)^2 −2 cos^2 θsin^2 θ
in simplifying the above equation.