IFIC/09-03 Tri-Bimaximal lepton mixing with A ⋉ (Z )3 4 2 9 Stefano Morisi a)1 0 0 a) AHEP Group, Institut de F´ısica Corpuscular – C.S.I.C./Universitat de Val`encia 2 Edificio Institutos de Paterna, Apt 22085, E–46071 Valencia, Spain n a J Abstract 8 2 We consider a model based on A ⋉(Z Z Z ) flavor symmetry for Tri- 4 e µ τ ] Bimaximal lepton mixing with two scalar A×-tripl×ets whose vacuum expectation h 4 p valueshavethesamealignment. Neutrinomassesaregenerated througheffective - dimension five Weinberg operators. Charged leptons mass hierarchies are gen- p e erated with an additional Froggatt-Nielsen U(1)F flavor symmetry under which h right-handed leptons are charged. [ 2 PACS numbers: 14,60.Pq, 11.30.Hv v 0 8 0 1 . 1 0 1 Introduction 9 0 : A successful phenomenological ansatz for leptons has been proposed by Harrison, v i Perkins and Scott (HPS) and is given by [3] X r a 2/3 1/√3 0  p  U = 1/√6 1/√3 1/√2 (1) HPS − −  1/√6 1/√3 1/√2  − which corresponds to tan2θ = 1, sin2θ = 0 and tan2θ = 0.5, providing a atm Chooz sol good first approximation to the values indicated by current neutrino oscillation data [1, 2]. There are many studies to derive the so called Tri-Bimaximal (TB) lepton mixing in eq.(1) by means of non-abelian groups like A [4, 5], T′[6, 7], S [8, 9, 10] and 4 4 ∆(27)[11]. However all models with non-abelian flavor symmetry G giving TB lepton f mixing, need at least two scalar fields whose vacuum expectation values (vevs) break G into distinct subgroups of G . In particular A -based models need two A -triplets, f f 4 4 for instances ϕ and χ, whose vevs are ϕ = ϕ = ϕ and χ = χ = χ = 0. 1 2 3 1 2 3 h i h i h i h i 6 h i h i We say that ϕ and χ have distinct vevs alignments. When ϕ and χ take vevs A 4 breaks respectively into Z and Z . Different papers [12, 13, 14, 15, 16] have recently 2 3 1 e-mail address: morisi@ific.uv.es 1 C = I C = T C = T2 C = S 1 2 3 4 { } { } { } { } 1 1 1 1 1 1′ 1 ω ω2 1 1′′ 1 ω2 ω 1 3 3 0 0 1 − Table 1: Character table of A where C are the different classes and ω3 1. 4 i ≡ emphasized the difficulty to have distinct vevs alignments1. In order to account for such alignments, extra-dimensions [17, 18], supersymmetry [19, 20] or Wilson lines [21] have been invoked. Recently has been studied models for TB mixing without vevs alignments [12, 16, 22] where the flavor symmetry is softly broken. In this paper we study the possibility to generate TB mixing in the lepton sector with two A scalar triplets. Their vevs break spontaneously A into Z and have 4 4 2 the same alignment. We do not have A scalar triplets whose vevs are misaligned 4 to get TB mixing. We need to enlarge the flavor group A to the group defined as 4 G = A ⋉ (Z Z Z ), that is the semidirect product 2 of A with the group f 4 e µ τ 4 × × (Z )3 (Z Z Z ). As in [14], we would like to be as minimal as possible 2 e µ τ ≡ × × generating charged lepton masses by dimension four operators and constructing the neutrino masses by dimension five Weinberg operators [23], without enter into the details of the particular dynamical model that generates such a operators. In the next section we introduce the A group, in section3 we give the main feature 4 of the model based on G flavor symmetry, in section 4 we study the Higgs potential f and then in section 5 we give the conclusions. 2 The group A 4 A is the finite group of the even permutations of four objects (for a short introduction 4 to A , see for instance [20] and references therein). All the 12 elements of the group 4 can be generated from two elements S and T that satisfy the following defining rules S2 = T3 = (ST)3 = . (2) I Since A group has four equivalence classes, there are four irreducible representations, 4 one triplet 3 and three singlets 1,1′,1′′, see table(1). The product of the singlets is equal to the Z product, then 1′ 1′′ 1, 1′ 1′ 1′′ and so on. The product of two 3 × ∼ × ∼ triplet 3 3 contains the three singlets and two triplets. The form of the irreducible × representations contained into the product of two triplets, depends from the choice of the structure of the generators S and T. Here we take the basis where T is diagonal 1 0 0 1 2 2   1  −  T = 0 ω 0 , S = 2 1 2 . (3) 3 −  0 0 ω2   2 2 1  − 1 Also called the “misalignment problem”[6]. 2 See for instance [21] for a short introduction. 2 L ec µc τc ξ h ϕ φ φ φ 1 2 3 A 3 1 1′ 1′′ 1 1 3 3 4 Z + + + + + + + + 2e − − Z + + + + + + + + 2µ − − Z + + + + + + + + 2τ − − Z ω ω2 ω2 ω2 1 ω 1 1 1 1 3 Table 2: Lepton and scalar multiplet structure of our model, see text. In such a basis the product of two triplets χ and ϕ is given by 1 (χϕ) = (χ ϕ +χ ϕ +χ ϕ ), 1 1 2 3 3 2 ∼ 1′ (χϕ)′ = (χ ϕ +χ ϕ +χ ϕ ), 3 3 1 2 2 1 ∼ 1′′ (χϕ)′′ = (χ ϕ +χ ϕ +χ ϕ ), 2 2 1 3 3 1 ∼ (4) 2χ ϕ χ ϕ χ ϕ χ ϕ χ ϕ 1 1 2 3 3 2 2 3 3 2  − −   −  3 2χ ϕ χ ϕ χ ϕ , 3 χ ϕ χ ϕ . s 3 3 1 2 2 1 a 1 2 2 1 ∼ − − ∼ −  2χ ϕ χ ϕ χ ϕ   χ ϕ χ ϕ  2 2 1 3 3 1 1 3 3 1 − − − We observe that in the basis of eq.(3) the product 3 3 is different from 3 ¯3 since T × × is complex. For instance, the product of χ and ϕ¯ is given by eq.(4) with ϕ ϕ . 2 3 ↔ 3 The model Consider the model defined in Table(2). The left-handed doublets transform as a triplet of A , namely L = (L ,L ,L ) and the right-handed fields ec, µc, τc as different 4 e µ τ singlets of A . Each right-handed field lc is charged under the corresponding Z with 4 a 2a a = e,µ,τ. We have two Higgs doublets h and ξ that are singlets of A and they differ 4 for an extra abelian symmetry Z . There is an A -triplet ϕ = (ϕ ,ϕ ,ϕ ) of Higgs 3 4 1 2 3 doublets. We have also one more A -triplet of Higgs doublets φ = (φ ,φ ,φ ) where 4 1 2 3 each scalar field φ transforms with respect to Z . As in [24, 25] the (Z )3 symmetries a 2a 2 glue each lc with the corresponding φ . As we will show below, (Z )3 remove off- i i 2 diagonal terms in the charged lepton sector, and as a consequence the charged lepton mass matrix is diagonal. The G Z invariant Lagrangian reads f 3 × y y = y L ecφ +y L µcφ +y L τcφ + a(LL)ξh+ b(LLϕ)h, (5) e e 1 µ µ 2 τ τ 3 L Λ Λ where (LL) means the A -singlet contained into 3 3, (LLφ) the A -singlet contained 4 4 × into 3 3 3 and Λ is the cut-off scale. According to the A symmetry also the 4 × × following terms are allowed (see eq.(4)) y (L ecφ +L ecφ )+y (L µcφ +L Hµcφ )+y (L Hτcφ +L Hτcφ ), (6) e µ 3 τ 2 µ e 3 τ 1 τ e 2 τ 1 but these terms are not invariant under the Z Z Z symmetry. e µ τ × × When the φ’s Higgs doublets take vevs as below φ = φ = φ = v, (7) 1 2 3 h i h i h i 3 the charged lepton mass matrix is diagonal with masses proportional to the yukawa couplings y , y and y . We can generate the charged lepton mass hierarchies assum- e µ τ ing that the right-handed charged fields transform with respect to an extra Froggatt- Nielsen (FN) symmetry U(1) . Assuming for instance the FN charges F to be F(ec) = F 4, F(µc) = 2, F(τc) = 0 and a flavon field θ with F(θ) = 1, then y λ4, y λ2 e µ − ∼ ∼ and y 1 where λ θ /Λ. We can reproduce the correct lepton mass hierarchies if τ ∼ ∼ h i λ 0.22. ≈ When the scalar Higgs fields ξ and h take vevs ξ = u and h = t and assuming h i h i ϕ = ϕ = ϕ = k, (8) 1 2 3 h i h i h i the Majorana neutrino mass matrix takes the form a+2b b b  − −  M = b 2b a b , (9) ν − −  b a b 2b  − − where a = y ut/Λ and b = y kt/Λ. We will show in the next section that the vevs a b alignments of eqs.(7) and (8) are natural in our model. The matrix M is µ τ ν ↔ invariant and M +M = M +M . Therefore the atmospheric angle is maximal, ν11 ν13 ν22 ν23 the reactor angle is zero and the solar angle trimaximal, see for instance [16]. We have three different eigenvalues m = a+3b, m = a and m = a+3b and it is possible to 1 2 3 − reproduce the ratio r = ∆m2 /∆m2 , see for instance [17]. sol atm 4 The scalar potential In our model there are 8 Higgs doublets that belong to different representations of G . We have assumed the vev of the two A -triplets to be equally aligned, that is f 4 φ = φ = φ and ϕ = ϕ = ϕ . We have showed that these alignments 1 2 3 1 2 3 h i h i h i h i h i h i together with the (Z )3 symmetry give TB mixing. In [17] has already been showed 2 that such a alignment is a solution of the most generic A invariant potential. Here 4 we study the Higgs potential and we show that the above alignment is one possible minimum. The scalar potential invariant under G Z is f 3 × V = µ h†h+µ ξ†ξ +µ (φ†φ)+µ (ϕ†ϕ)+ 1 2 3 4 + λ (φ†φ) (φ†φ) + λ′ (ϕ†ϕ) (ϕ†ϕ) + λ′′ (φ†φ) (ϕ†ϕ) + X ab a b X ab a b X ab a b a,b=r a,b=r a,b=r + λ (φ†φ)ξ†ξ +λ (φ†φ)h†h+λ ξ†ξξ†ξ +λ h†hh†h+λ ξ†ξh†h+ 1 2 3 4 5 + λ (ϕ†ϕ)ξ†ξ +λ (ϕ†ϕ)h†h+h.c. (10) 6 7 where r = 1,1′,1′′,3 ,3 . We assume the vevs of the Higgs doublets to point in the s a same direction of SU(2) and, for simplicity all the vevs to be real. We define ϕ = k , φ = v , ξ = u, h = t, i i i i h i h i h i h i 4 then the scalar potential reads V = µ t2 +µ u2 +µ (v2 +v2 +v2)+µ (k2 +k2 +k2)+ min 1 2 3 1 2 3 4 1 2 3 + λ11(v12 +v22 +v32)2 +λ1′1′′(v3v2 +v1v2 +v3v1)2 + + λ (2v2 v2 v2)2 +2(2v v v v v v )2 + 3s3s(cid:2) 1 − 2 − 3 3 2 − 1 3 − 2 1 (cid:3) + λ (v2 v2)2 2(v v v v )2 + 3a3a (cid:2) 3 − 2 − 1 3 − 1 2 (cid:3) + λ′ (k2 +k2 +k2)2 +λ′ (k k +k k +k k )2 + 11 1 2 3 1′1′′ 3 2 1 2 3 1 + λ′ (2k2 k2 k2)2 +2(2k k k k k k )2 + 3s3s(cid:2) 1 − 2 − 3 3 2 − 1 3 − 2 1 (cid:3) + λ′ (k2 k2)2 2(k k k k )2 + 3a3a (cid:2) 3 − 2 − 1 3 − 1 2 (cid:3) + λ′′ (v2 +v2 +v2)(k2 +k2 +k2)+λ′′ (v v +v v +v v )(k k +k k +k k )+ 11 1 2 3 1 2 3 1′1′′ 3 2 1 2 3 1 3 2 1 2 3 1 + λ′′ ((2v2 v2 v2)(2k2 k2 k2)+ 3s3s 1 − 2 − 3 1 − 2 − 3 +2(2v v v v v v )(2k k k k k k ))+ 3 2 1 3 2 1 3 2 1 3 2 1 − − − − + λ′′ (v2 v2)(k2 k2) 2(v v v v )(k k k k ) + 3a3a (cid:2) 2 − 3 2 − 3 − 1 3 − 1 2 1 3 − 1 2 (cid:3) + λ (v2 +v2 +v2)u2 +λ (v2 +v2 +v2)t2 +λ u4 +λ t4 +λ u2t2 + 1 1 2 3 2 1 2 3 3 4 5 + λ (k2 +k2 +k2)u2 +λ (k2 +k2 +k2)t2. (11) 6 1 2 3 7 1 2 3 A possible solution of the system ∂V/∂x = 0 with x = v , k , u, t is i i i i v = v = v = v = 0, k = k = k = k = 0, t = 0, u = 0, (12) 1 2 3 1 2 3 6 6 6 6 with v = λ λ λ′′µ 2λ λ λ′′µ 2λ λ λ′′µ +λ λ λ′′µ +6λ λ2µ 6λ λ λ µ + { 5 6 1 − 3 7 1 − 4 6 2 5 7 2 4 6 3 − 5 6 7 3 + 6λ λ2µ 24λ λ λ′µ +6λ2λ′µ +4λ λ λ′′µ λ2λ′′µ + 3 7 3 − 3 4 3 5 3 3 4 4 − 5 4 + λ ( λ2µ +4λ λ′µ +λ λ µ 2λ λ′µ +λ λ µ λ λ µ )+ 2 − 6 1 3 1 6 7 2 − 5 2 5 6 4 − 3 7 4 + 3λ (λ λ µ 2λ λ′µ λ2µ +4λ λ′µ 2λ λ µ +λ λ µ ) / 1 6 7 1 − 5 1 − 7 2 4 2 − 4 6 4 5 7 4 } 9λ2λ2 36λ2λ λ′ +λ2(λ2 4λ λ′) 4λ(λ λ2 λ λ λ +λ λ2 4λ λ λ′ +λ2λ′)+ { 1 7 − 1 4 2 6 − 3 − 4 6 − 5 6 7 3 7 − 3 4 5 + 12λ λ λ λ′′ 6λ λ λ λ′′ 4λ λ λ′′2 +λ2λ′′2 + 1 4 6 − 1 5 7 − 3 4 5 2λ (3λ λ λ 6λ λ λ′ +λ λ λ′′ 2λ λ λ′′) , (13) 2 1 6 7 1 5 5 6 3 7 − − − } with λ = 9(λ11 +λ1′1′′), λ′ = 9(λ′11 +λ′1′1′′) and λ′′ = 9(λ′1′1 +λ′1′′1′′). Similar relations can be found for k, u and t. We have verified that there is a non-vanishing portion of the parameter space where the Hessian matrix ∂2V/∂x ∂x has positive eigenvalues for i j the solution in eq.(12). 5 Conclusions We have studied a model for TB lepton mixing. The model is based on the flavor symmetry A ⋉(Z )3. Most of the models based on A need two scalar flavor-triplets 4 2 4 that take vevs with different alignments in order to give TB mixing. The reason is that one of the two scalar triplet breaks A into Z in the neutrino sector, while the 4 2 second scalar triplet breaks A into Z in the charged lepton sector. The misalignment 4 3 between the two sectors yields to the large TB mixing. However it has been shown that vevs misalignment is not a natural solution of the scalar potential without introducing supersymmetry or extra-dimensions. 5 In our model the vevs of the two scalar flavor-triplets interacting respectively with the neutrino sector and the charged lepton sector, have the same alignment. Unwanted off-diagonal terms in the charged lepton mass matrix are removed by means of the (Z )3 symmetry. Since the groups A and (Z )3 do not commute, the flavor group 2 4 2 is bigger than A and is given by their semidirect product, namely A ⋉ (Z )3. This 4 4 2 method to remove off-diagonal terms in the charged sector has already been used in [13, 22, 24, 25] where the basic groups are S and ∆(27). In [13, 22, 24, 25] the fermion 3 masshierarchies arerelatedtothedifferent vevs ofφ ,φ andφ andithasbeenshowed 1 2 3 that φ φ φ is possible by assuming soft-breaking terms of dimension two 1 2 3 in thhe scia≪larhpoite≪ntihal ilike φ†φ and/or φ†φ . Our approach is completely different i i i j because we have A then the fermion masses are proportional to the yukawa couplings. 4 We can generate fermion mass hierarchies by adding an extra FN symmetry and we do not require to softly break the flavor symmetry. We have studied the G Z invariant f 3 × Higgs potential. 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