content_image/333650.jpg content_image/333649.jpg content_image/333648.jpg content_image/333647.jpg content_image/333646.jpg content_image/333645.jpg
written as, 

where M$_{R}$ = Diag( M$_{r}$$_{1}$, M$_{r}$$_{2}$, M$_{r}$$_{3}$ ) and M$_{ri}$ can be simplified as, 

How could we obtain the tri-bimaximal mixing matrix ? In the limit of x → 0 in Eq. (8), the light neutrino mass matrix in a basis where the charged lepton and heavy Majorana neutrino mass matrices are diagonal, that is, Y$_{ν}$ is replaced by ˜ Y$_{ν}$ ≡ $_{V}$† $_{L}$Y$_{ν}$ V$_{R}$ , can be obtained as follows 

<img src='content_image/333650.jpg'>

<img src='content_image/333649.jpg'>

<img src='content_image/333648.jpg'>

where ω$_{i}$ = z$_{i}$ 1 − $_{z$_{i}$}$[1 + z$_{i}$ ln z$_{i}$ 1 − z$_{i}$ ]. And the overall scale of neutrino mass matrix m$_{0}$ is given as 

<img src='content_image/333647.jpg'>

This mass matrix can be diagonalized by the so-called tribimaximal mixing matrix U$_{T B}$ with mixing angles given in Eq. (1), 

<img src='content_image/333646.jpg'>

we denote m$_{i}$ with i = 1 − 3 which are the eigenvalues of m$_{ν}$ , and the matrices U$_{TB}$ and P$_{ν}$ are 

<img src='content_image/333645.jpg'>

## A. Deviation from Tri-Bimaximal 

In order to achieve the deviation from the TBM matrix in neutrino sector we need to break µ − τ symmetry, that is, x ̸ = 0 in matrix Y$_{ν}$ (see Eq.(8)) where the VEVs are aligned 

