Lepton mixing angle $θ_{13} = 0$ with a horizontal symmetry $D_4$

Details Lepton mixing angle $θ_{13} = 0$ with a horizontal symmetry $D_4$ Lepton mixing angle $θ_{13} = 0$ with a horizontal symmetry $D_4$

2004-07-09

arxiv.org/abs/hep-ph/0407112v1

JHEP 0407 (2004) 078

Physics

High Energy Physics

High Energy Physics - Phenomenology

13 pages, one figure

Scientific paper

10.1088/1126-6708/2004/07/078

We discuss a model for the lepton sector based on the seesaw mechanism and on a $D_4$ family symmetry. The model predicts the mixing angle $\theta_{13}$ to vanish. The solar mixing angle $\theta_{12}$ is free--it will in general be large if one does not invoke finetuning. The model has an enlarged scalar sector with three Higgs doublets, together with two real scalar gauge singlets $\chi_i$ ($ i = 1, 2$) which have vacuum expectation values < \chi_i >_0$ at the seesaw scale. The atmospheric mixing angle $\theta_{23}$ is given by $\tan \theta_{23} = <\chi_2>_0$ /<\chi_1>_0$, and it is maximal if the Lagrangian is $D_4$-invariant; but $D_4$ may be broken softly, by a term of dimension two in the scalar potential, and then < \chi_2_0$ becomes different from < \chi_1_0$. Thus, the strength of the soft $D_4$ breaking controls the deviation of $\theta_{23}$ from $\pi / 4$. The model predicts a normal neutrino mass spectrum ($m_3 > m_2 > m_1$) and allows successful leptogenesis if $m_1 \sim 4 \times 10^{-3} \mathrm{eV}$; these properties of the model are independent of the presence and strength of the soft $D_4$ breaking.

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