Gauged Flavor Symmetries

Transcription

1 Gauged Flavor Symmetries NOW 2012 Julian Heeck Max-Planck-Institut für Kernphysik, Heidelberg based on J.H., Werner Rodejohann, PRD 84 (2011), PRD 85 (2012); Takeshi Araki, J.H., Jisuke Kubo, JHEP 07 (2012).

2 Outline for U(1) symmetries in type-i seesaw Neutrino hierarchies from U(1) Texture zeros and vanishing minors Summary Julian Heeck Gauged Flavor Symmetries 1 / 18

3 Flavor Symmetries Most popular way: discrete non-abelian global symmetries (A 4,... ) Drawbacks: complicated, many parameters and fields, unknown scalar sector at unreachable energies... Here: continuous abelian local symmetries, i.e. U(1) Very simple, i.e. few new particles/parameters Testable outside of neutrino sector, e.g. Z at LHC How to choose the U(1): Same charge for all quarks (FCNC): Y (q L ) = Y (u R ) = Y (d R ). Add three N i with charges that allow m D : Y (l Li ) = Y (e Ri ) = Y (N Ri ), Y (l Li ) Y (l Lj ). Anomaly cancellation: U(1) U(1) U(1) Julian Heeck Gauged Flavor Symmetries 2 / 18

4 Maximal Abelian Gauge Group Only one constraint on the charges: Two cases: 9 Y (q L ) + Y (l L1 ) + Y (l L2 ) + Y (l L3 ) = 0. B x α L α with x α = 3 [E. Ma, (1998)], α α y α L α with y α = 0. α α Famous examples: B L (GUT?), L α L β (anomaly free in the SM) [R. Foot, MPL A6 (1991)] More general: SU(3) l U(1) B L U(1) Le L µ U(1) Lµ L τ U(1) B L can be added to G SM without anomalies (l Li 3 l, e Ri 3 l, N Ri 3 l ). Julian Heeck Gauged Flavor Symmetries 3 / 18

5 Now what? Three interesting zeroth order approximations: M Le ν , 0 0 M L ν 0 0, 0 0 M Lµ Lτ ν [G. Branco, W. Grimus, L. Lavoura, NPB (1989); S. Choubey, W. Rodejohann, EPJC 40 (2005)] Conserve L e, L L e L µ L τ and L µ L τ, and lead to NH, IH and QD, respectively. L often used as global symmetry, gives bimaximal mixing. L µ L τ is special: M Lµ Lτ ν type-i seesaw. is invertible can be obtained from Julian Heeck Gauged Flavor Symmetries 4 / 18

6 L µ L τ Dirac matrices diagonal due to symmetry, M R of the L µ L τ symmetric form Seesaw gives L µ L τ symmetric M ν Add one or two scalars S that couple to N c i N j and get a VEV VEV fills zeros in M R and M ν and gives mass to Z boson M Z /g S For mixing angles: S 10 2 M R connection of seesaw-scale and M Z LHC can probe low seesaw-scale Julian Heeck Gauged Flavor Symmetries 5 / 18

7 L µ L τ Two scalars, ε = v S / M R = 0.02: 0.36 sin 2 Θ sin 2 Θ sin 2 Θ 13 sin 2 Θ 13 Julian Heeck Gauged Flavor Symmetries 6 / 18

8 L µ L τ at LHC Signal at LHC [S. Baek et al, PRD 64 (2001)] e, p, p µ +, τ + µ +, τ + Z, γ e +, p, p µ, τ Z µ, τ Full run (L = 100 fb 1 ) with g = 1: Up to M Z = 350 GeV pp, pp Z, γ 2µ Z 4µ 100 LHC (14 TeV) LHC (7 TeV) Tevatron (1.96 TeV) 10 σ [fb] MZ [GeV] Julian Heeck Gauged Flavor Symmetries 7 / 18

9 L µ L τ at low energies Magnetic moment of muon: µ Z µ γ For M Z m µ : a µ 1 12π 2 g 2 M 2 Z m 2 µ To explain 3.6σ deviation: M Z /g 220 GeV [E. Ma et al, PLB 525 (2002)] Julian Heeck Gauged Flavor Symmetries 8 / 18

10 L e and L Try B 3L e and B + 3L as anomaly free symmetries with the right lepton structure. Same procedure as for L µ L τ does not work, M Le R and ML R are not invertible. Weird coincidence: M R ε 1 1 ε ε M ν ε2 ε ε 1 1. ε 1 Low B + 3L breaking gives approximate L e symmetry! Broken B + 3L (say ε = 0.05) gives NH structure. Julian Heeck Gauged Flavor Symmetries 9 / 18

11 Mixing Angles and Collider Since µ and τ have same Y charge, m D not diagonal θ 23 random (i.e. large). One scalar S with charge Y (S) = 6 and VEV 1 10 TeV < v S ε M R gives: sin 2 Θ sin 2 Θ 13 LEP-II constraint: v S > 2.3 TeV, LHC prospects in [H. S. Lee and E. Ma, PLB 688 (2010)]. Julian Heeck Gauged Flavor Symmetries 10 / 18

12 How to get Inverted Hierarchy Problem: anomaly cancellation demands odd number of RHN, but then M L R is not invertible, which destroys symmetry! Solution: Decouple one N i with a Z 2. Make N 3 odd: L N3 = in 3 γ µ ( µ i( 3)g Z µ) c N3 Y χ S N 3 N3 + h.c. = i 2 χt Cγ µ µ χ 3 ( 2 g Z µχ T Cγ µ v S γ 5 χ Y χ χ T Cχ 1 + s ). 2 v S Majorana fermion χ will be dark matter candidate. Neutrino mass: a 0 ( ) 1 ( ) a 2 B abx acx M ν 0 b A X a 0 0 b X B 0 b c 2 A bca. 0 c c 2 A Julian Heeck Gauged Flavor Symmetries 11 / 18

13 Mixing Angles Only two neutrinos massive, M ν exhibits scaling θ 13 = 0. Solution: Add five N i instead of three (still anomaly free) and decouple one of them: ) a M ν a b 0 0 ( 0 0 c d A X X 0 0 e f T B Need larger breaking than before. ε = 0.1: 0.36 b c e. 0 d f 0.34 sin 2 Θ sin 2 Θ 13 Julian Heeck Gauged Flavor Symmetries 12 / 18

14 Scalar Sector Scalar sector identical to U(1) B L models: [L. Basso et al, PRD 80 (2009)] V (H, S) = µ 2 1 H 2 + λ 1 H 4 µ 2 2 S 2 + λ 2 S 4 + δ S 2 H 2, In unitary gauge, mixing via δ: ( ) ( φ1 cos α sin α = φ 2 sin α cos α Masses: ) ( h s), tan 2α = δvv S λ 2 v 2 S λ 1v 2. M Z = 6g v S, m 2 m s 2λ 2 v S, M χ = 2Y χ v S. Julian Heeck Gauged Flavor Symmetries 13 / 18

15 Dark Matter Dark matter sector identical to U(1) B L Z 2 sector: [N. Okada and O. Seto, PRD 82 (2010)] L χ = i 2 χt Cγ µ µ χ 3 ( 2 g Z µχ T Cγ µ v S γ 5 χ Y χ χ T Cχ 1 + s ). 2 v S Relic density can be obtained around the scalar resonances or the Z resonance (m 1 = 125 GeV, m 2 = 500 GeV, M Z = 3.5 TeV): 100 Χ h sin Α 0.1 sin Α 0.3 sin Α 0.5 WMAP MΧ GeV Julian Heeck Gauged Flavor Symmetries 14 / 18

16 Direct Detection Direct detection cross section via Z are suppressed by Lorentz structure: χγ µ γ 5 χ f γ µ f. XENON1T only sensitive to scalar exchange: 10 6 Σp pb XENON sin Α 0.5 sin Α 0.3 sin Α M Χ GeV Julian Heeck Gauged Flavor Symmetries 15 / 18

17 And now for something completely different... B α x αl α or α y αl α symmetries forbid entries in M R texture zeros! Y (N c 2xe xe xµ xµ 3 i N j ) = x e x µ 2x µ x e 3 x µ 3 x e 3 2x e + 2x µ 6 Choose x i so that some entries are allowed, others filled by S, e.g. B L e + L µ 3L τ with Y (S) = 2: M R = M S M 1 ν. 0 Dirac matrices diagonal by symmetry, so M R M 1 ν. Julian Heeck Gauged Flavor Symmetries 16 / 18

18 One Scalar With just one scalar S, we can get five of the seven valid patterns: Symmetry generator Y Y (S) Texture zeros in M R Texture zeros in M ν L µ L τ 1 (M R ) 33, (M R ) 22 (C R ) B L e + L µ 3L τ 2 (M R ) 33, (M R ) 13 (B R 4 ) (Mν) 12, (M ν) 22 (B ν 3 ) B L e 3L µ + L τ 2 (M R ) 22, (M R ) 12 (B R 3 ) (Mν) 13, (M ν) 33 (B ν 4 ) B + L e L µ 3L τ 2 (M R ) 33, (M R ) 23 (D R 2 ) (Mν) 12, (M ν) 11 (A ν 1 ) B + L e 3L µ L τ 2 (M R ) 22, (M R ) 23 (D R 1 ) (Mν) 13, (M ν) 11 (A ν 2 ) Many patterns are hard to distinguish via neutrino experiments, e.g. D R 1 and D R 2, but the symmetries B + L e 3L µ L τ and B + L e L µ 3L τ are very different new possibilities to disentangle texture zeros. More scalars or discrete Z N subgroups can generate other allowed patterns. Julian Heeck Gauged Flavor Symmetries 17 / 18

19 Summary With three right-handed neutrinos, the SM gauge group can be extended by U(1) B L U(1) Le L µ U(1) Lµ L τ. Can use subgroups to influence neutrino mixing by introducing just one or two scalars. Applications: Enforcing neutrino hierarchies, two-zero textures, two vanishing minors, maybe other stuff? Z and s can be searched for outside the neutrino sector testable models! Julian Heeck Gauged Flavor Symmetries 18 / 18

20 Remark Easy way to see anomaly freedom: Choose different basis in group space: L e L µ = diag(1, 1, 0), (L e L µ ) + 2 (L µ L τ ) = diag(1, 1, 2). Cartan subalgebra of SU(3) l with l Li 3 l, e Ri 3 l, N Ri 3 l. Real reducible rep [SU(3) l ] 3 anomaly vanishes (like quarks). Anomalies: SU(3) l SU(3) l U(1) Y : Y = 3 (2 Y (L e ) Y (e R )) = 0, 3 l SU(3) l SU(3) l U(1) B L : (B L) = 3 (2 ( 1) + (+1) + (+1)) = 0. 3 l G SM SU(3) l U(1) B L G SM G max is anomaly free. Julian Heeck Gauged Flavor Symmetries 19 / 18

21 One more remark For U(1), field strength tensor F µν is already gauge invariant Every gauge group with U(1) 1 U(1) 2 part allows for kinetic mixing : [B. Holdom, PLB 166 (1986)] L mix = c mix F µν 1 F 2 µν Diagonalize kinetic terms: A 1 couples to j µ 2 new effects Can be extended to U(1) 1 U(1) 2 U(1) 3 [J.H. and W.R., PLB 705 (2011)] Julian Heeck Gauged Flavor Symmetries 20 / 18

22 Take M ν and set two independent entries to zero four constraints on the nine low-energy parameters (m 1, m 2, m 3 ), (θ 23, θ 12, θ 13 ) and (δ, α, β) (CP violating phases) 15 two-zero textures possible, only 7 allowed at 3σ: [H. Fritzsch et al, JHEP 1109 (2011)] A ν 1 : 0 0 0, A ν 2 : 0 0, B ν 1 : 0 0, B ν 2 : 0 0, B ν 3 : 0 0 0, B ν 4 : 0, C ν : Julian Heeck Gauged Flavor Symmetries 21 / 18

23 Vanishing Minors Same idea, but with M 1 ν instead of M ν. Seven patterns for M 1 ν allowed: D R 1 : 0 0, D R 2 : 0, B R 3 : 0 0, B R 4 : 0, 0 0 B R 1 : 0 0, B R 2 : 0, C R : Two-zero texture in M 1 ν M ν. E.g. M 1 = corresponds to two vanishing minors in M 11 M 13 M 31 M 33 = 0 = M 21 M 22 M 31 M 32 Julian Heeck Gauged Flavor Symmetries 22 / 18

24 Two Scalars M R pattern Symmetry generator Y Y (S i ) D R 1 B al e 3L µ + al τ, a / { 9, 3, 0, 1, 3} 2 a, 3 + a B 2L µ L τ 1, 2 B Le 9 2 Lµ 3, 3 2 B Le 7 7 Lµ 3 7 Lτ 18 7, 6 7 B Le 7 3 Lµ Lτ 2, 2 3 D R 2 D R 1 with Lµ Lτ B R 3 D R 1 with Le Lτ B R 4 B R 3 with Lµ Lτ B R 1 B + 3L µ 6L τ 3, 12 B 2L µ L τ 2, 3 B 9 2 Lµ Lτ 3, 9 2 B 6L e + 3L µ 3, 12 B Le 9 2 Lµ 3, 9 2 B L e 2L µ 2, 3 B L e + 3L µ 5L τ 2, 10 B 5L e + 3L µ L τ 2, 10 B R 2 B R 1 with Lµ Lτ C R B + 3L e al µ (6 a)l τ, a / { 3, 0, 1, 3, 5, 6, 9} 5, 3 a B 6L µ + 3L τ 3, 6 B 3L e ± 9L µ 9L τ 6, 12 Julian Heeck Gauged Flavor Symmetries 23 / 18