written as, where M$_{R}$ = Diag( M$_{r}$$_{1}$, M$_{r}$$_{2}$, M$_{r}$$_{3}$ ) and M$_{ri}$ can be simplified as, How could we obtain the tri-bimaximal mixing matrix ? In the limit of x → 0 in Eq. (8), the light neutrino mass matrix in a basis where the charged lepton and heavy Majorana neutrino mass matrices are diagonal, that is, Y$_{ν}$ is replaced by ˜ Y$_{ν}$ ≡ $_{V}$† $_{L}$Y$_{ν}$ V$_{R}$ , can be obtained as follows where ω$_{i}$ = z$_{i}$ 1 − $_{z$_{i}$}$[1 + z$_{i}$ ln z$_{i}$ 1 − z$_{i}$ ]. And the overall scale of neutrino mass matrix m$_{0}$ is given as This mass matrix can be diagonalized by the so-called tribimaximal mixing matrix U$_{T B}$ with mixing angles given in Eq. (1), we denote m$_{i}$ with i = 1 − 3 which are the eigenvalues of m$_{ν}$ , and the matrices U$_{TB}$ and P$_{ν}$ are In order to achieve the deviation from the TBM matrix in neutrino sector we need to break µ − τ symmetry, that is, x = 0 in matrix Y$_{ν}$ (see Eq.(8)) where the VEVs are aligned 9 A. Deviation from Tri-Bimaximal U$_{TB}$ =    − √ 2 3 √ 1 3 0 1 6 1 3 1 √$_{2}$ √ 1 6 √ 1 3 − 1 √$_{2}$    , P$_{ν}$ =    e i ϕ$_{1}$ 2 0 0 0 1 0 0 0 e i ϕ$_{2}$ + π 2    . (21) m$_{ν}$ = U$_{TB}$ Diag . ( m$_{1}$, m$_{2}$, m$_{3}$ ) U T TB = m$_{0}$U$_{TB}$P$_{ν}$ Diag . ( ω$_{1}$ a , ω$_{2}$, ω$_{3}$ b ) P T ν U T TB , (20) m$_{0}$ = Δ m 2 η 16 π 2 g 2 ν M . (19) m$_{ν}$ = m$_{0}$ 3    2 ω 1 e iϕ$_{1}$ a + ω$_{2}$ ω$_{2}$ − ω 1 e iϕ$_{1}$ a ω$_{2}$ − ω 1 e iϕ$_{1}$ a ω$_{2}$ − ω 1 e iϕ$_{1}$ a − 3 ω 3 e iϕ$_{2}$ 2 b + ω 1 e iϕ$_{1}$ 2 a + ω$_{2}$ 3 ω 3 e iϕ$_{2}$ 2 b + ω 1 e iϕ$_{1}$ 2 a + ω$_{2}$ ω$_{2}$ − ω 1 e iϕ$_{1}$ a 3 ω 3 e iϕ$_{2}$ 2 b + ω 1 e iϕ$_{1}$ 2 a + ω$_{2}$ − 3 ω 3 e iϕ$_{2}$ 2 b + ω 1 e iϕ$_{1}$ 2 a + ω$_{2}$    , (18) m$_{ν}$ = Δ m 2 η 16 π 2 ˜ Y$_{ν}$$_{M}$− 1 R ˜ Y T ν , (16) M$_{ri}$ ≃    2 M$_{i}$, for z$_{i}$ → 1 M$_{i}$ [ ln z$_{i}$ − 1 ] − $^{1}$, for z$_{i}$ ≫ 1. (17)