PHYS08200604017 Manimala Mitra - Homi Bhabha National Institute

Using the same arguments as above, it is not hard to see that the deviation from u 1 = u 2
is also mild. Again, u 1 = u 2 is satisfied when h ′ 6 = 0 and h ′′
6 = 0. Since h 6 = 0 is
also required for |v 1 | = |v 2 | to be satisfied, we conclude that exact µ − τ symmetry for
neutrinos demands that h 6 = 0, h ′ 6 = 0 and h′′ 6 = 0 simultaneously.
Finally, from the last minimization condition (6.56) we get the solution,
|v c | 2 = 1
4h 3
[
2h 6 (|v 2 | 2 −|v 1 | 2 )+2h ′′
6 (u2 2 −u2 1 )−2l′′ 1 (u2 1 +u2 2 )+2h 1
−2h 4 (|v 1 | 2 +|v 2 | 2 )−8u 1 u 2 l 1 −2l 4 v 2 ]
. (6.67)
We next use use the condition (6.67) to estimate the cut-off scale Λ. Since h 1 define
in Eq. (6.48) gives the square of the mass of the φ e fields, it could be large. The other
couplings h 3 , l 1 , l 4 , h 4 , h ′′
6 , l′′ 1 andh 6 aredimensionless andcanbeassumed tohave roughly
the same order of magnitude which should be much much smaller than h 1 . Dividing both
sides of Eq. (6.67) by Λ 2 and using |v 1 | 2 ≃ |v 2 | 2 = 10 −3 eV 2 , vc ≃ Λ 2×10−2 , u 1,2
≃ Λ 10−1
and hence ( v 1,2
Λ )2 ≪ ( vc
Λ )2 < ( u2 2 −u2 1) < ( u 1,2
Λ 2 Λ )2 , we get 9
Λ 2 ≃
h 1
×10 2 GeV 2 . (6.68)
4l 1 +2l 1
′′
The coupling h 1 has mass dimension 2 and in principle could be large. As an example, if
wetakeh 1 inTeVrange, forexample ifwetake √ h 1 = 10TeV, thenthecut-offscaleofthe
theory is fixed as 10 2 TeV, where we have taken l 1 and l ′′
1 ≃ O(1). From u 1
Λ
= u 2
Λ
∼ 10 −1
and vc ∼ Λ 10−2 , we then obtain u 1,2 = 10 TeV and v c = 1 TeV. The constraints from
the lepton masses themselves do not impose any restriction on the cut-off scale and the
VEVs. One can obtain estimates on them only through limits on the masses of the Higgs.
For instance, from Eqs. (6.38) and (6.39) one could in principle estimate u by measuring
the difference between doubly charged Higgs masses. This could then be combined with
the neutrino data to get Λ, and finally use the charged lepton masses to get v c .
Since we consider a model with triplet Higgs to generate Majorana neutrino masses,
it is pertinent to make some comments regarding breaking of lepton number and possible
creation of a massless goldstone called Majoron [26]. It is possible to break lepton number
explicitly by giving a lepton number to the fields ξ. In that case one would not break
lepton number spontaneously and there would be no Majoron.
These VEV alignments have been obtained by assuming no effect of renormalization
group running. However, it is understood that the running from the high scale where
S 3 is broken to the electroweak scale where the masses are generated, will modify the
9 h 6 ,h ′′
6=0 is motivated from µ−τ symmetry, which together with ( v1
Λ )2 ≪ ( vc
Λ )2 < ( u1
Λ )2 can also lead
to Eq. 6.68. Even for a mild breaking of µ−τ which leads to (u 2 2 −u2 1 ) ≪ u2 1,2 , the equation is valid.
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