Obtaining a nonzero 13 in lepton models based on SO(3) A 4 Yuval - PowerPoint PPT Presentation

Obtaining a nonzero  13 in lepton models based on SO(3)  A 4 Yuval Grossman and Wee Hao Ng , Cornell University arXiv:1404.1413 [hep-ph] Phenomenology 2014 1

Outline • Tri-bimaximal mixing found and lost; how do A 4 models cope? • Basics of SO(3)  A 4 model • Nonzero  13 in SO(3)  A 4 model • Summary 2

Tri-bimaximal mixing and A 4 models • For long time, PMNS matrix thought to be consistent with TBM   i δ sin θ e   13 2 1 13   0 3  3    1 1 1     U TBM 6 3 2     1 1 1     6 3 2  • Highly specific pattern… discrete lepton flavor symmetries? 3

Tri-bimaximal mixing and A 4 models • Simplest A 4 lepton models – Components • SM Higgs and leptons • Right handed neutrinos Representations of A 4 • Scalar flavons  ,  ’ – Features • Additional Z 2 symmetry so flavons sectorized • Flavons gain VEV  = (v, v, v),  ’  = (v’, 0, 0)  TBM! 4

Tri-bimaximal mixing… lost • Recently,  13 found to be much larger! • Daya Bay (2012):      2 sin 2 θ 0.092 0.016(stat) 0.005(syst) 13 • RENO (2012):      2 sin 2 θ 0.113 0.013(stat) 0.019(syst) 13 • TBM ruled out! 5

How do A 4 models cope? • Actually many A 4 models already allow this! • Approaches: – Higher dimension operators, e.g. [Altarelli and Meloni (2009)] – More flavons, e.g. [Chen et al (2013)] – Perturb flavon alignments, e.g. [King (2011)] – Radiative corrections, e.g. [Antusch et al (2003)] 6

Continuous symmetry  A 4 • Class of models based on continuous symmetry  A 4 . • Originally motivated to explain origin of A 4 . • Specific example: SO(3)  A 4 [Berger and Grossman (2009)] • Turns out  13 already nonzero at tree level! A4 scale • Even better:  13 related to SO(3) scale 7

Basics of SO(3)  A 4 model Left-handed leptons Right-handed leptons Scalars 8

Basics of SO(3)  A 4 model • Most general Lagrangian for leptons H H H H         a a a b ab T a abc bc 5 ò abc a bd cd L y ψ ψ y ψ ψ y ψ T ψ y ψ ψ l e l e m l m m l m m l 5 m Λ Λ Λ Λ            a a a b ab T a abc bc 5 ò abc a bd cd y ψ ψ y ψ ψ y ψ T ψ y ψ ψ e f e m f m m f m m f 5 m x       ca a ca b c abc a a ν L M ψ ψ ψ ψ T y ψ Hψ ν n n n n ν l n Λ • Flavons gain VEV        v 0 v v v 5 5                   (a b c) T v x y z , v , v 0 v , 0       T 5 5 5         v  v v 0   0  5 5 3D coordinate basis vectors  6  6 mass matrices 9

Obtaining U PMNS • U PMNS characterize charged-current interactions between light eigenstates. • 6  6 mass matrices: – Block-diagonalise 6  6 mass matrices. – Then diagonalise 3  3 mass matrices. • Neutrinos:   2 2 y v    ν H 0 0   M   a 0 0    2 2 2 2 2   y Mv Λ y x v v v Λ    3 3   ν H ν ν H T M 0 , of form 0 b c     ν   2 2 2 2 2 2 2 2 2 2  M Λ x v v M Λ x v v    ν T ν T 0 c b      2 2 2 2 2 y x v v v Λ y Mv Λ    ν ν H T ν H 0       2 2 2 2 2 2 2 2 2 2  M Λ x v v M Λ x v v  ν T ν T 10

Obtaining U PMNS   v v 2 π • Charged leptons: H H A B   i 3     6 6 M ω e Λ Λ  l    C D            5 2 5 2 y v y v y v ( ω ω) y v y v (ω ω )       a b c  e m m 5 m m 5               5 2 5 2 2 2 A y v y v y v ( ω ω) ω y v y v (ω ω ) ω , ofform a b ω c ω         e m m 5 m m 5   2    a b ω c ω          5 2 2 5 2 y v y v y v ( ω ω) ω y v y v (ω ω ) ω       e m m 5 m m 5      T 5 5 y v 2y v y v y v y v m m T m m 5 m 5     T B y v 2y v y v  m m T m      5 T 5 T y v y v y v y v y v y v   m m 5 m T m m 5 m T                5 2 5 2 y v y v y v ( ω ω) y v y v (ω ω )       a b c  e m m 5 m m 5                    5 2 5 2 2 2 C y v y v y v ( ω ω) ω y v y v (ω ω ) ω , ofform a b ω c ω         e m m 5 m m 5   2    a b ω c ω               5 2 2 5 2 y v y v y v ( ω ω) ω y v y v (ω ω ) ω       e m m 5 m m 5           T 5 5 y v 2y v y v y v y v m m T m m 5 m 5        T D y v 2y v y v  m m T m            5 T 5 T y v y v y v y v y v y v   m m 5 m T m m 5 m T 11

Nonzero  13 • Block diagonalize   6 6 6 6 † M (M ) l l 2 v            3 3 3 3 † † † † † † † 1 † † H M (M ) AA BB (AC BD )(CC DD ) (CA DB )   l l 2 Λ • Useful to define    T T E B (y / y )D m m • Then A,C,E~O(v,v ), B,D~O(v ) 5 T • Expanding in small parameter   ò O(v / v )~O(v / v ) T T   – LO: T v y  13 still 0!    3 3 H  m  M A C  l T Λ  y  m   T – NLO: v y Suggests A4 scale      3 3 1 H m M  A C ED C   l T Λ y    13 ~  ! m SO(3) scale 12

Nonzero  13 • Actual contribution to  13 from deviation turns out to be ~ (m  /m  )  ~ 10  • Random simulation results In agreement with predictions! 13

Summary and further work • TBM ruled out, but many A 4 models allow for this scenario • In particular, SO(3)  A 4 model allow nonzero tree-level  13 of size m A4 scale τ ~ m SO(3) scale μ • % level separation of A 4 and SO(3) breaking scales 14

Thank you! 15