Natural Vacuum Alignment from Group Theory Martin Holthausen - PowerPoint PPT Presentation

Natural Vacuum Alignment from Group Theory Martin Holthausen based on MH, Michael A. Schmidt JHEP 1201 (2012) 126 , arXiv: 1111.1730

Why Flavour Symmetry? in SM(+Majorana neutrinos) there are a total of 28 Parameters 1 1 12 2 10 2

Why Flavour Symmetry? in SM(+Majorana neutrinos) there are a total of 28 Parameters 1 1 12 2 10 2 most of them stem from interactions with the Higgs field, other interactions tightly constrained by symmetry principles

Why Flavour Symmetry? in SM(+Majorana neutrinos) there are a total of 28 Parameters 1 1 12 2 10 2 most of them stem from interactions with the Higgs field, other interactions tightly constrained by symmetry principles in quark sector: small mixing angles and hierarchical masses can be explained by Frogatt-Nielsen symmetry

Why Flavour Symmetry? in SM(+Majorana neutrinos) there are a total of 28 Parameters 1 1 12 2 10 2 most of them stem from interactions with the Higgs field, other interactions tightly constrained by symmetry principles in quark sector: small mixing angles and hierarchical masses can be explained by Frogatt-Nielsen symmetry in lepton sector: two large and one small mixing angle suggestive of non-abelian discrete symmetry

Lepton mixing from discrete groups G f residual symmetry of (me + me) residual symmetry of m ν complete flavour group G e =Z 3 G ν =Z 2 xZ 2 Sm ν S T = m ν Um ν U T = m ν Tm e m † e T † = m e m † e     1 0 0 1 0 0   0 1 0 S = 0 − 1 0 U = 0 0 1     T = 0 0 1   0 0 − 1 0 1 0 1 0 0 (smallest choice, but can also be continuous) (most general choice if mixing angles do not depend on masses & Majorana ν s) misaligned non-communting symmetries lead to 0 1 q 2 1 0 [He, Keum, Volkas ‘06; √ 3 3 „tri-bimaximal B C − 1 1 − 1 U P MNS = U T BM = Lam’07,‘08; B C √ √ √ mixing“(TBM) gives 6 3 2 @ A Altarelli,Feruglio‘05] 1 1 1 good LO description of √ √ √ 6 3 2 lepton mixing sin 2 θ 12 = 1 sin 2 θ 23 = 1 sin 2 θ 13 = 0 3 , 2 , sin 2 θ 12 = 0 . 312 +0 . 017 sin 2 θ 23 = 0 . 52 +0 . 06 sin 2 θ 13 = 0 . 013 +0 . 007 − 0 . 015 , − 0 . 07 , − 0 . 006

Candidate Groups SU � 3 � � � 216 Φ � � � 72 Φ � � � 360 Φ � � � 168 � T 7 ∼ = Z 7 o Z 3 � � 36 Φ � � � 60 � S 4 T 7 S 4 ∼ = ( Z 2 × Z 2 ) o S 3 � � 6 � 3 2 � A 4 A 4 ∼ = ( Z 2 × Z 2 ) o Z 3 T 0 ∼ = Z 2 .A 4 � � 3 � 3 2 � [Merle,Zwicky 1110.4891] ∆ (27) ∼ = ( Z 3 × Z 3 ) o Z 3

A 4 Symmetry Group A 4 is the smallest symmetry group that can lead to TBM mixing: = h S, T | S 2 = T 3 = ( ST ) 3 = 1 i A 4 ⇠ = ( Z 2 ⇥ Z 2 ) o Z 3 ⇠ (a) Character Table S T T 2 1 T S 1 1 11 1 1 1 1 11 1-d reps. 1 12 ! correspond to ! 2 1 1 ! 2 ! 12 1 13 reps. of Z 3 0 1 0 1 1 0 0 0 1 0 ! 2 1 1 13 ! 0 � 1 0 0 0 1 31 B C B C @ A @ A 3 0 0 -1 3 0 0 � 1 1 0 0 re ! = e i2 π / 3 0 1 0 1 3 × 3 = 11 + 12 + 13 + 3 S + 3 A 1 ( ab ) 11 = 3 ( a 1 b 1 + a 2 b 2 + a 3 b 3 ) √ 1 1 a 1 b 1 + ω 2 a 2 b 2 + ω a 3 b 3 a 1 b 1 + ω a 2 b 2 + ω 2 a 3 b 3 � � � � ( ab ) 12 = ( ab ) 13 = √ √ 3 3 @ A 0 1 0 1 a 2 b 3 + a 3 b 2 a 2 b 3 − a 3 b 2 ( ab ) A, 3 = 1 ( ab ) S, 3 = 1 a 3 b 1 − a 1 b 3 a 3 b 1 + a 1 b 3 B C B C where ( a 1 , a 2 , a 3 ) , ( b 1 , b 2 , b 3 ) ∼ 3 . 2 2 @ A @ A a 1 b 2 + a 2 b 1 a 1 b 2 − a 2 b 1

An A 4 Prototype model (A 4, Z 4 ) charge assignments: l ∼ (3,i), e c ∼ (1 1, -i), 휇 c ∼ (1 2, -i) , 휏 c ∼ (1 3, -i) , 휒 ∼ (3,1), Φ ∼ (3,-1), 휉 ∼ (1,-1) auxiliary Z 4 separates neutral and charged lepton sectors at LO A 4 (L 휒 )e c (ll) 3 Φ +(ll) 1 휉 〈 휒 〉 ∼ (1,1,1) 〈 Φ 〉 ∼ (1,0,0) Z 3 = 〈 T 〉 Z 2 = 〈 S 〉 0 1 0 1 ˜ a 0 0 y e y µ y τ m 휈 ∼ TBM v m e ∼ ˜ ! 2 y τ 0 a ˜ d A , y e ! y µ B C B C @ @ A Λ ˜ ! 2 y µ 0 d a ˜ y e ! y τ (Z 2 xZ 2 symmetry accidental) Vacuum alignment crucial! [e.g. Ma,Rajasekaran’01, Babu, Ma, Valle ’03, Altarelli,Feruglio, ’05,’06]

Can Vacuum Alignment be realized?

Can Vacuum Alignment be realized? SSB: Minimum Energy State exhibits less symmetry than full Lagrangian

Can Vacuum Alignment be realized? SSB: Minimum Energy State exhibits less symmetry than full Lagrangian Most straightforward case: scalar potential in 4D DOES NOT work: V χ = m 2 0 ( χχ ) 11 + λ 1 ( χχ ) 11 ( χχ ) 11 + λ 2 ( χχ ) 12 ( χχ ) 13 has minima (1,1,1) and (1,0,0). Effect of breaking to Z 2 in another sector can be included by adding: V soft,Z 2 = m 2 A χ 2 1 + m 2 B χ 2 2 + m 2 C χ 2 χ 3

Can Vacuum Alignment be realized? SSB: Minimum Energy State exhibits less symmetry than full Lagrangian Most straightforward case: scalar potential in 4D DOES NOT work: V χ = m 2 0 ( χχ ) 11 + λ 1 ( χχ ) 11 ( χχ ) 11 + λ 2 ( χχ ) 12 ( χχ ) 13 has minima (1,1,1) and (1,0,0). Effect of breaking to Z 2 in another sector can be included by adding: V soft,Z 2 = m 2 A χ 2 1 + m 2 B χ 2 2 + m 2 C χ 2 χ 3 Minimization conditions then give:  ∂ V � 2 ⇣ √ ⌘ v 0 + 4 λ 1 v 0 3 m 2 3 m 2 0 = χ i = v 0 = 0 + √ A ∂χ 1 3  ∂ � ∂ = 2 m 2 B v 0 0 = V − V ∂χ 2 ∂χ 3 χ i = v 0  ∂ � ∂ 2 m 2 A − m 2 � � v 0 V − V 0 = χ i = v 0 = C ∂χ 1 ∂χ 3

Can Vacuum Alignment be realized? SSB: Minimum Energy State exhibits less symmetry than full Lagrangian Most straightforward case: scalar potential in 4D DOES NOT work: V χ = m 2 0 ( χχ ) 11 + λ 1 ( χχ ) 11 ( χχ ) 11 + λ 2 ( χχ ) 12 ( χχ ) 13 has minima (1,1,1) and (1,0,0). Effect of breaking to Z 2 in another sector can be included by adding: V soft,Z 2 = m 2 A χ 2 1 + m 2 B χ 2 2 + m 2 C χ 2 χ 3 Minimization conditions then give:  ∂ V � 2 ⇣ √ ⌘ v 0 + 4 λ 1 v 0 3 m 2 3 m 2 0 = χ i = v 0 = 0 + This thus requires m A = m B = m C =0, √ A ∂χ 1 3  ∂ i.e. all non-trivial contractions � ∂ = 2 m 2 B v 0 0 = V − V between Φ and 휒 have to vanish ∂χ 2 ∂χ 3 χ i = v 0  ∂ in the potential. � ∂ 2 m 2 A − m 2 � � v 0 V − V 0 = χ i = v 0 = C ∂χ 1 ∂χ 3

Can Vacuum Alignment be realized? SSB: Minimum Energy State exhibits less symmetry than full Lagrangian Most straightforward case: scalar potential in 4D DOES NOT work: V χ = m 2 0 ( χχ ) 11 + λ 1 ( χχ ) 11 ( χχ ) 11 + λ 2 ( χχ ) 12 ( χχ ) 13 has minima (1,1,1) and (1,0,0). Effect of breaking to Z 2 in another sector can be included by adding: V soft,Z 2 = m 2 A χ 2 1 + m 2 B χ 2 2 + m 2 C χ 2 χ 3 Minimization conditions then give:  ∂ V � 2 ⇣ √ ⌘ v 0 + 4 λ 1 v 0 3 m 2 3 m 2 0 = χ i = v 0 = 0 + This thus requires m A = m B = m C =0, √ A ∂χ 1 3  ∂ i.e. all non-trivial contractions � ∂ = 2 m 2 B v 0 0 = V − V between Φ and 휒 have to vanish ∂χ 2 ∂χ 3 χ i = v 0  ∂ in the potential. � ∂ 2 m 2 A − m 2 � � v 0 V − V 0 = χ i = v 0 = C ∂χ 1 ∂χ 3 Breaking to the same subgroup of A 4 can be realized. The non-trivial couplings, i.e. ( Φ Φ ) 3 ( 휒휒 ) 3 thus force breaking of group to the same subgroup. the couplings cannot be forbidden by an internal symmetry that commutes with A 4 , as e.g.( Φ † Φ ) 3 is invariant under the commuting symmetry.

Solutions in the Literature In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. Altarelli, Feruglio 2005

Solutions in the Literature In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. Altarelli, Feruglio 2005 In SUSY, one has to introduce a continuous R-symmetry and additional fields with R- charge 2(driving fields). These fields enter the superpotential only linearly and allow the vacuum alignment. Altarelli, Feruglio 2006

Solutions in the Literature In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. Altarelli, Feruglio 2005 In SUSY, one has to introduce a continuous R-symmetry and additional fields with R- charge 2(driving fields). These fields enter the superpotential only linearly and allow the vacuum alignment. Altarelli, Feruglio 2006 Babu and Gabriel(2010) proposed the flavour group (S 3 ) 4 ⋊ A 4 , which has the properties leptons transform only under A 4 subgroup if one takes Φ ∼ 16, vacuum alignment possible as V=V( Φ )+V( 휒 )+( Φ Φ ) 1 ( 휒휒 ) 1 neutrino masses then generated by coupling to 〈 Φ 4 〉 ∼ (1,0,0)

Problems with the Solutions In models with extra dimensions(ED), it is possible to locate the various fields at different locations in the ED, thereby forbidding the cross-couplings. Altarelli, Feruglio 2005 In SUSY, one has to introduce a continuous R-symmetry and additional fields with R- charge 2(driving fields). These fields enter the superpotential only linearly and allow the vacuum alignment. Altarelli, Feruglio 2006 Babu and Gabriel(2010) proposed the flavour group (S 3 ) 4 ⋊ A 4 , which has the properties leptons transform only under A 4 subgroup if one takes Φ ∼ 16, vacuum alignment possible as V=V( Φ )+V( 휒 )+( Φ Φ ) 1 ( 휒휒 ) 1 neutrino masses then generated by coupling to 〈 Φ 4 〉 ∼ (1,0,0)