PHYS08200604017 Manimala Mitra - Homi Bhabha National Institute

where we have defined B = (b+f 1 +f 2 ), F = f 3 −2f 2 and we have used α 2 = α 1 . Using
Eq. (6.53) we obtain,
−2a|v 2 |+4B(|v 2 | 2 +|v 1 | 2 )|v 2 |+2|v 2 |[4e 1 u 1 u 2 +e ′ 1(u 2 1 +u 2 2)]+2|v 1 |[e 2 (u 2 2 +u 2 1)+e ′ 2u 1 u 2 ]
+4F|v 2 ||v 1 | 2 +2h 4 |v 2 ||v c | 2 +2a 1 |v 2 |v 2 −2h 6 |v c | 2 |v 2 |+2h ′ 6 |v 1|(u 2 2 −u2 1 ) = 0 . (6.60)
Multiplying Eq. (6.59) by |v 2 | and Eq. (6.60) by |v 1 | and subtracting one from the other
we obtain,
(2e 2 (u 2 1 +u 2 2)+2e ′ 2u 1 u 2 +4F|v 1 ||v 2 |)(|v 1 | 2 −|v 2 | 2 ) = 4|v 1 ||v 2 |[h 6 |v c | 2 −h ′ 6(u 2 2 −u 2 1)] (6.61) .
In the limit h 6 = 0 and h ′ 6 = 0 we get |v 2 | = |v 1 | (if e 2 , e ′ 2 and F ≠ 0 simultaneously),
which is required for exact µ-τ symmetry in the neutrino sector. However, there is no
a priori reason to assume that h 6 , and h ′ 6 are zero. In the most general case keeping
non-zero h 6 and h ′ 6, we obtain
|v 1 | 2 = |v 2 | 2 + 4|v 1||v 2 |[h 6 |v c | 2 −h ′ 6 (u2 2 −u2 1 )]
2e 2 (u 2 1 +u 2 2)+2e ′ 2u 1 u 2 +4F|v 1 ||v 2 | . (6.62)
Since |v 1 ||v 2 | ≪ u 1 u 8 2 and (u 2 1 +u2 2 ) we neglect the 4F|v 1||v 2 | term from the denominator.
For u 1 ≃ u 2 = u and e 2 ≃ e ′ 2 , one obtains
|v 1 | 2 = |v 2 | 2 + 2|v 1||v 2 |h 6 |v c | 2
3e 2 u 2 . (6.63)
For a fixed v 2 , this is a quadratic equation in v 1 which allows the solution v 1 ≃ v 2 (1+ǫ)
where ǫ = h 6vc
2
3e 2
. For h
u 2 6 and e 2 of the same order and u = Λ 10−1 , vc = Λ 10−2 we obtain
ǫ ≃ 10 −2 ≪ 1 . This would give rise to a very mild breaking of the µ−τ symmetry. We
have discussed this case in section 6.3.2.
Using Eqs. (6.54) and (6.55) and repeating the same exercise we get the deviation
from u 1 = u 2 as
u 2 1 = u2 2 + A B
(6.64)
where A and B are
A = 4u 1 u 2 [h ′′
6 |v c| 2 −h ′ 6 (|v 2| 2 −|v 1 | 2 )] (6.65)
B = (−4c+16du 1 u 2 −4d ′ u 1 u 2 +4d ′′ (u 2 1 +u 2 2)+4e 1 (|v 1 | 2 +|v 2 | 2 )+2e ′ 2|v 1 ||v 2 |+4l 1 |v c | 2
+4a 2 v 2 ) . (6.66)
8 v1,2 2 ∼ 10 −3 /10 −4 eV 2 , u 1,2 /Λ ∼ 10 −1 from neutrino phenomenology. Hence the hierarchy between
the VEV’s is justified.
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