Duality in left-right symmetric seesaw

Transcription

1 Duality in left-right symmetric seesaw Evgeny Akhmedov KTH, Stockholm & Kurchatov Institute, Moscow In collaboration with Michele Frigerio Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 1

2 Why are neutrinos so light? A simple and elegant explanation The seesaw mechanism (Minkowski, 1977; Gell-Mann, Ramond & Slansky, 1979; Yanagida, 1979; Glashow, 1979; Mohapatra & Senjanović, 1980) In addition: Seesaw has a built-in mechanism for generating the baryon asymmetry of the Universe Baryogenesis via Leptogenesis (Fukugita & Yanagida, 1986; Luty, 1992; Covi et al., 1996; Buchmüller & Plümacher, 1996;...) The simplest version: Add 3 RH ν s N Ri yo the minimal SM Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 2

3 Heavy N Ri s make ν Li s light : L Y +m = Y ν ll N R H M RN R N R + h.c., Neutrino mass matrix in the (ν L, (N R ) c ) basis: M ν = 0 m D m T D M R N Ri are EW singlets M R M GUT (M I ) m D v. Block diagonalization: M N M R, m νl m D M 1 R mt D m ν (174 GeV)2 M R For m ν 0.05 ev M R GeV M GUT GeV! Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 3

4 Type II seesaw Type I seesaw: Neutrino mass generated through exchage of a heavy RH ν s N R ; in type II through exchage of a heavy SU(2) L - triplet scalars L : In the SM, RH neutrinos N R are singlets of the gauge group aliens. They are more natural in Left-Right symmetric extensions of the SM: SU(2) L SU(2) R U(1) B L, SU(2) L SU(2) R SU(4) PS, SO(10),... LR symmetric models explain in a nice way maximal P- and C-violation in low-e weak interactions as a spontaneous symmetry breaking phenomenon likely to be present in the final theory Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 4

5 In LR - symmetric models: In general, both type I and type II seesaw contributions to m ν are present : M ν = Block diagonalization of M ν : m L m T D m D M R m ν m L m D M 1 R mt D m D comes from Yukawa interactions with the bi-doublet Higgs Φ: Y 1 ll Φ l R + Y 2 ll Φ l R, Φ τ 2 Φ τ 2 m D = yv m L and M R from Yukawa couplings with triplet Higgses L and R m ν f L v L v 2 y (f R v R ) 1 y T In general: m ν, f L and f R complex symmetric, y complex matrix Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 5

6 Parity symmetry Discrete LR symmetry (P) in LR-symm. models two possible implementations: (1) l Li l Ri, L R, Φ Φ Yields f L = f R, y = y, (2) l Li l c Li (l Ri ) c, L R, Φ Φ T Yields f L = f R f, y = y T. Both implementations possible. Implem. (2) is more natural in SO(10) GUTs (is an automatic gauge symmetry) Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 6

7 Adhere to the second implementation m ν = fv L v 2 y (v R f) 1 y For realization (1) of parity: type I term contains (f ) 1 instead of f 1. N R, R, L are all at very high scale ( GeV) no direct way of probing this sector of the theory. Neutrinos may provide a low-energy window into new physics at very high energy scales! Take m ν from experiment Bottom-up approach: Take y from data + theoretical assumptions (quark-lepton symm., GUTs) Solve the seesaw relation for f E.A. and M. Frigerio, PRL 96 (2006) Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 7

8 What would we like to know? Try to learn as much as possible about the heavy neutrino sector What are the masses of heavy RH neutrinos? What are the mixing and CP violation in the heavy RH neutrino sector? Can all this give us a hint of the underlying theory? Is the solution that we found unique? Is type I or type II contribution in the light neutrino mass dominant or are they equally important? Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 8

9 Inverting the seesaw formula Pure type I (c 0 v 2 /v R ): m ν = c 0 ( y f 1 y T) f = c 0 ( y T m 1 ν y ) Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 9

10 Inverting the seesaw formula Pure type I (c 0 v 2 /v R ): Pure type II: m ν = c 0 ( y f 1 y T) f = c 0 ( y T m 1 ν y ) m ν = f v L f = m ν /v L Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 9

11 Inverting the seesaw formula Pure type I (c 0 v 2 /v R ): Pure type II: m ν = c 0 ( y f 1 y T) f = c 0 ( y T m 1 ν y ) m ν = f v L f = m ν /v L General type I + II seesaw: m ν = fv L v2 v R ( y f 1 y ) a nonlinear matrix equation for f. Yet can be readily solved analytically! Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 9

12 Inverting the seesaw formula Pure type I (c 0 v 2 /v R ): Pure type II: m ν = c 0 ( y f 1 y T) f = c 0 ( y T m 1 ν y ) m ν = f v L f = m ν /v L General type I + II seesaw: m ν = fv L v2 v R ( y f 1 y ) a nonlinear matrix equation for f. Yet can be readily solved analytically! Important point: A duality property of LR-symmetric seesaw If f is a solution, so is ˆf (m ν /v L f) Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 9

13 Inverting the seesaw formula Pure type I (c 0 v 2 /v R ): Pure type II: m ν = c 0 ( y f 1 y T) f = c 0 ( y T m 1 ν y ) m ν = f v L f = m ν /v L General type I + II seesaw: m ν = fv L v2 v R ( y f 1 y ) a nonlinear matrix equation for f. Yet can be readily solved analytically! Important point: A duality property of LR-symmetric seesaw If f is a solution, so is ˆf (m ν /v L f) There is always an even number of solutions! (#sol. = 2 n ) Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 9

14 LR symmetry important for duality! It was important that f L = f R and y T = y. For the other realization of parity symmetry (f L = f R and y = y ) exactly the same duality holds! f 1 = 1 F Two-generation case f 22 f 12 f 12 f 11, y = y e2 y e1 y µ1 y µ2, F detf, y µ1 = y e2. Define: x v L v R /v 2 and m m ν /v L xf(f 11 m ee ) = f 22 ye1 2 2f 12 y e1 y e2 + f 11 ye2 2, xf(f 22 m µµ ) = f 22 ye2 2 2f 12 y e2 y µ2 + f 11 yµ2 2, xf(f 12 m eµ ) = f 22 y e1 y e2 f 12 (y e1 y µ2 + ye2) 2 + f 11 y e2 y µ2, Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 10

15 A system of coupled 3rd order equations Admits a simple exact analytic solution! Go to the basis where y is diagonal: y = diag(y 1, y 2 ) (no loss of generality) xf(f 11 m ee ) = f 22 y 2 1 xf(f 22 m µµ ) = f 11 y 2 2 xf(f 12 m eµ ) = f 12 y 1 y 2 Rescaling: f ij = λf ij, m ij = λm ij, y ij = λy ij, λ - arbitrary; fix it by requiring F detf = 1. The system of eqs. for f ij becomes linear Express f ij back through unprimed m ij, y 1,2 and subst. into F 4th order characteristic equation for λ = 1 Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 11

16 The solution: f = xλ (xλ) 2 y 2 1 y2 2 xλm ee + y1m 2 µµ m eµ (xλ y 1 y 2 )... xλm µµ + y2m 2 ee λ has to be found from [ (xλ) 2 y1y ] 2 [ x detm(xλ y1 y 2 ) 2 xλ +(m ee y 2 + m µµ y 1 ) 2 (xλ) 2] = 0 Has 4 complex solutions 4 solutions for f which form 2 dual pairs Determinant of the seesaw relation x ˆf = yf 1 y x 2 F ˆF = y 2 1y 2 2 F = 1 F = λ; therefore x 2 λˆλ = y 2 1y 2 2 Allows to express the four solutions for λ in a simple closed form Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 12

17 Which seesaw type dominates? If m αβ m γδ 4 y i y j /x, one of the solutions λ i satisfies xλ 1 y i y j f 1 m (type II seesaw) Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 13

18 Which seesaw type dominates? If m αβ m γδ 4 y i y j /x, one of the solutions λ i satisfies xλ 1 y i y j f 1 m (type II seesaw) Then for the dual solution λ 2 : xλ 2 y i y j f 2 = ˆf 1 corresponds to type I seesaw The remaining 2 solutions are of mixed type. Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 13

19 Which seesaw type dominates? If m αβ m γδ 4 y i y j /x, one of the solutions λ i satisfies xλ 1 y i y j f 1 m (type II seesaw) Then for the dual solution λ 2 : xλ 2 y i y j f 2 = ˆf 1 corresponds to type I seesaw The remaining 2 solutions are of mixed type. If m αβ m γδ 4 y i y j /x, all 4 solutions are of mixed type. Similar situation in 3-generation case Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 13

20 Which seesaw type dominates? If m αβ m γδ 4 y i y j /x, one of the solutions λ i satisfies xλ 1 y i y j f 1 m (type II seesaw) Then for the dual solution λ 2 : xλ 2 y i y j f 2 = ˆf 1 corresponds to type I seesaw The remaining 2 solutions are of mixed type. If m αβ m γδ 4 y i y j /x, all 4 solutions are of mixed type. Similar situation in 3-generation case In general, if there is a solution with seesaw type I dominance, there is always also a solution with type II sessaw dominance and vice versa. Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 13

21 Which seesaw type dominates? If m αβ m γδ 4 y i y j /x, one of the solutions λ i satisfies xλ 1 y i y j f 1 m (type II seesaw) Then for the dual solution λ 2 : xλ 2 y i y j f 2 = ˆf 1 corresponds to type I seesaw The remaining 2 solutions are of mixed type. If m αβ m γδ 4 y i y j /x, all 4 solutions are of mixed type. Similar situation in 3-generation case In general, if there is a solution with seesaw type I dominance, there is always also a solution with type II sessaw dominance and vice versa. Using low-energy data only, it is impossible to say which seesaw type dominataes (if any) additional criteria necessary (leptogenesis?) Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 13

22 Three lepton generations A system of six coupled 4th order equations can be readily solved analytically by making use of the same trick as in 2-g case plus duality. Characteristic equation for the rescaling factor λ is of 8th order 8 solutions λ i 8 solutions f i (i = 1,...,8). Form 4 pairs of mutually dual solutions For given y (i.e. m D ), all 8 solutions result in exactly the same mass matrix of light neutrinos m ν! Analysis of solutions under way Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 14

23 Conclusions In LR - symmetric models (minimal LR, Pati-Salam, SO(10). E 6,...) one naturally has type I + II seesaw mechanism of neutrino mass generation Discrete LR symmetry (parity) leads to a relation between type I and type II contributions to m ν, which results in a duality property of the seesaw formula: f m ν /v L f For given y, there are 2 n (8 for three lepton generations) matrices f which result in exactly the same mass matrix of light neutrinos m ν A simple analytic method developed for solving the seesaw nonlinear matrix equation. Allows bottom-up reconstraction of the Yukawa coupling matrix f of heavy RH neutrinos The results may be used for neutrino mass model building and studies of baryogenesis via leptogenesis. Allow to analytically explore the interplay of type I and II contributions to m ν Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 15

24 Backup slides Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 16

25 RGE issues The discrete LR (parity) symmetry of the underlying theory must be broken at some scale v LR renormalization group evolution below this scale may result in a violation of conditions f L = f R f, y = y T at lower energies. Corrections to matrix elements of y and f depend logarithmically on ratios of masses of RH neutrinos and Higgs triplets; are suppressed by loop factors and possibly by small couplings. Numerically checked: If LR-violating corrections to the matr. elements are of the order of percent, reconstruction of the matrix f remains accurate at a percent level. The stability may be lost if small matrix elements receive corrections proportional to the large elements a dedicated study necessary Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 17

26 Mass matrix of light neutrinos m ν Symmetric mass matrix diagonalized as m ν = U T m diag U, m diag = diag(m 1, m 2, m 3 ) The masses m i known up to the overall scale parameterized by m 1 : m 2 = m m2 21, m 3 = m m2 31 Leptonic mixing matrix U: U = c 23 s 23 0 s 23 c 23 c 13 0 s 13 e iδ s 13 e iδ 0 c 13 c 12 s 12 0 s 12 c e iα e iβ c ij cos θ ij, s ij sinθ ij Experiment: θ 12 34, θ 23 45, θ 13 < 10. CP violating phases can be varied between 0 and 2π. Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 18

27 The solutions in 2-generation case xλ i = 1 4 [xdetm + r ± ± 2(detm) 2 x 2 + 4kx + 2xr ± detm], where r ± = ± (detm) 2 x 2 + 4kx + 16y 2 1 y2 2, k = m 2 eey m 2 eµy 1 y 2 + m 2 µµy 2 1. When λ 1 satisfies xλ 1 y i y j f 1 m (type II seesaw case) The dual solution xλ 2 = xˆλ 1 = y1y 2 2/(xλ 2 1 ) has modulus y i y j f 2 = ˆf 1 takes type I seesaw form Condition xλ 1 y i y j can only be satisfied when m αβ m γδ 4 y i y j /x. This ensures the existence of solutions with one seesaw type dominance. But: in general not all 4 solutions correspond to one seesaw type dominance in this limit: if detm 4 y i y j /x, the remaining 2 dual solutions f 3 and f 4 are of mixed type. When m αβ m γδ 4 y i y j /x, all 4 sols. are of mixed type. Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 19

28 Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 20

29 Three lepton generations Once again go to the basis where y = diag(y 1, y 2, y 3 ). f 1 = 1 f 22 f 33 f F System of 6 coupled 4th order equations: xf (m 11 f 11 ) = y 2 1 (f 22 f 33 f 2 23)..... Simple rescaling would not do! Use similar equations for duals: x ˆF (m 11 ˆf 11 ) x ˆF f 11 = y 2 1 ( ˆf 22 ˆf33 ˆf 2 23) and ( ˆf 22 ˆf33 ˆf 2 23) = (m 22 f 22 )(m 33 f 33 ) (m 23 f 23 ) 2 Allows to express quadratic in f terms through linear and f-indep. terms and ˆF Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 21

30 Resulting system: x(f+ ˆF) f 11 = y1 2 [(m 22 m 33 m 2 23) (m 22 f 33 + m 33 f 22 2m 23 f 23 )] + xf m Now one can rescale: f ij = λ 1/3 f ij, m ij = λ 1/3 m ij, y ij = λ 1/3 y ij, Fix λ by requiring F detf = 1 Determinant condition: x 3 F ˆF = y 2 1 y 2 2 y 2 3 x 3 ˆF = (y 1 y 2 y 3) 2 System of 6 linear eqs. for f ij easily solved. Substitution into F = 1 (F = λ) 8th order characteristic eq. for λ. Yields 4 pairs of dual solutions for the matrix f. Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 22

31 3 generation case more details Equations for the matrix elements of f and its dual ˆf: xf(f ij m ij ) = y i y j F ij, (1) x ˆF( ˆf ij m ij ) = x ˆFf ij = y i y j ˆFij, (2) Here: F detf, ˆF det ˆf and F ij 1 2 ǫ iklǫ jmn f km f ln, ˆF ij 1 2 ǫ iklǫ jmn ˆfkm ˆfln = M ij T ij + F ij, M ij 1 2 ǫ iklǫ jmn m km m ln, T ij ǫ ikl ǫ jmn f km m ln. A system of 6 coupled quartic equations for f ij. RH sides are quadratic rather than linear in f ij a simple rescaling would not linearize the system. But: using the dual system of eqs. gives for these RH sides y i y j F ij = x ˆFf ij + y i y j (T ij M ij ) Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 23

32 Linearization: rescaling by a factor λ 1/3 Determinant condition: x 3 F ˆF = y 2 1 y 2 2 y 2 3. Fix λ by requiring F (λ) = 1; ˆF = (y 1y 2y 3) 2 /x 3. The linearized system: [x 3 (y 1y 2y 3) 2 ]f ij x 3 m ij = x 2 y iy j(t ij M ij). - Simplified case: y 1 0 (physically well motivated). The result: f 11 = m ee, f 12 = m eµ, f 13 = m eτ, ( f 23 = m µτ + y 2y 3 m eµ m ) eτ /d 2, xλ [ ( )] f 22 = m µµ + y2 2 M 22 y2 3m ee m 2 eµ xλ xλ [ ( f 33 = m ττ + y2 3 M 33 y2 2m ee m 2 eτ xλ xλ /d 1, )] /d 1, Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 24

33 Here: Characteristic equation for λ: d 1 = 1 y2 2y3m 2 2 ee (xλ) 2, d 2 = 1 + y 2y 3 m ee xλ. λ 4 { [(xλ) 2 m 2 eey 2 2y 2 3] 2 x [ detm(xλ m ee y 2 y 3 ) 2 xλ +(M 22 y 2 + M 33 y 3 ) 2 (xλ) 2]} = 0. If a solution λ of general charact. eq. 0 in the limit y 1 0, then ˆλ = y1y 2 2y 2 3/(x 2 3 λ) 0 determinant of ˆf vanishes. The dual of any solution that is finite for y 1 0 becomes singular and must be discarded. For y 1 0 there are only 4 (rather than 8) solutions with no duals. The corresp. values of λ are zeros of the factor in curly brackets (which is quartic in λ). Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 25

34 A strong connection with the pure 2-g case A different duality among the 4 remaining solutions: If λ satisfies the charact. equation, so does λ y 2 2y 2 3m 2 ee/(x 2 λ), and it corresponds to where f m f, m αβ = m αβ + m eα m eβ /m ee. There are two pairs of such solutions. Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 26

35 Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 27

36 Why is m ν = 0 in the Standard Model? No RH neutrinos N Ri Dirac mass terms cannot be introduced Operators of the kind l lhh, which could could produce Majorana neutrino mass after H H, are dimension 5 and so cannot be present at the Lagrangian level in a renormalizable theory These operators cannot be induced in higher orders either (even nonperturbatively) because they would break not only lepton number L but also B L, which is exactly conserved in the SM In the Standard Model: B and L are accidental symmetries at the Lagrangian level. Get broken at 1-loop level due the axial (triangle) anomaly. But: their difference B L is still conserved and is an exact symmetry of the model Evgeny Akhmedov SNOW 2006 Stockholm May 4, 2006 p. 28