6.3. SOLAR NEUTRINO PROBLEM 351
- There are three discrete eigenvaluesλjof (6.3.17) with eigenstates:
(6.3.18) Hˆνj=λjνj for 1≤j≤ 3 ,such thatνe,νμ,ντare some linear combinations of{νj| 1 ≤j≤ 3 }:(6.3.19)
νe
νμ
ντ
=A
ν 1
ν 2
ν 3
,
whereA∈SU( 3 )is a third-order complex matrix given by (6.3.26) below.Remark 6.16.The formulas (6.3.17)-(6.3.19) constitute the current model of neutrino os-
cillation, which requires the neutrinos being massive. However, the massive neutrino as-
sumption gives rise two serious problems. First, it is in conflict with the known fact that
the neutrinos violate the parity symmetry. Second, the handedness of neutrinos implies their
velocity being at the speed of light.
In fact, by using the Weyl equations as the neutrino oscillation model we can also deduce
the same conclusions and solve the two mentioned problems. Moreover, theνmediator
introduced by the authors in (Ma and Wang,2015b) leads to an alternate explanation to the
6.3 Solar Neutrino Problem. CONTENTS xi
Under the above three hypotheses (6.3.17)-(6.3.19), the oscillation betweenνe,νμand
ντare given in the following fashion. For simplicity we only consider two kinds neutrinos
νe,νμ, i.e.ντ=0. In this case, (6.3.19) becomes
(6.3.20)
ν 1 =cosθ νμ−sinθ νe,
ν 2 =sinθ νμ+cosθ νe.By the Dirac equations (6.2.15) and (6.3.18),ν 1 andν 2 satisfy
i ̄h∂ νk
∂t=λkνk fork= 1 , 2.The solutions of these equations read
(6.3.21) νk=νk( 0 )e−iλkt/h ̄, k= 1 , 2.
Assume that the initial state is atνe, i.e.νe( 0 ) = 1 , νμ( 0 ) = 0.Then we derive from (6.3.20) that
(6.3.22) ν 1 ( 0 ) =−sinθ, ν 2 ( 0 ) =cosθ.
It follows from (6.3.21) and (6.3.22) that
(6.3.23) ν 1 =−sinθe−iλ^1 t/ ̄h, ν 2 =cosθe−iλ^2 t/h ̄.