Scaling in the Neutrino Mass Matrix and the See-Saw Mechanism Werner Rodejohann (MPIK, Heidelberg) Erice, 20/09/09 1
A. S. Joshipura, W.R., Phys. Lett. B 678, 276 (2009) [arxiv:0905.2126 [hep-ph]] A. Blum, R. N. Mohapatra, W.R., Phys. Rev. D 76, 053003 (2007) [arxiv:0706.3801 [hep-ph]] R. N. Mohapatra, W.R., Phys. Lett. B 644, 59 (2007) [arxiv:hep-ph/0608111] 2
What is Scaling? Ansatz for the neutrino mass matrix m ν obtainable in many scenarios/models leads to inverted hierarchy alternative to L e L µ L τ alternative to tri-bimaximal, trimaximal, µ τ symmetry etc. U e3 = 0 stable under RG predictive for low energy phenomenology and for see-saw parameters 3
m ν = Ansatz for m ν A B B/c B D D/c B/c D/c D/c 2 rank(m ν ) = 2 one eigenvalue zero 0 m ν 1 = 0 c so the predictions are inverted hierarchy (m 3 = 0) with U e3 = 0 and tan 2 θ 23 = 1/c 2 in general: θ 23 non-maximal! 4
Predictivity 5 physical parameters for θ 12, θ 23, m 2, m 2 A, m ee only one (Majorana) phase: m ee = m 2 A 1 sin 2 2θ 12 sin 2 α <m> [ev] 0.055 0.05 0.045 0.04.035.03.025 0.02 0.015 0.01 0.005 0.24 0.26 0.28 0.3 0.32 0.34 sin 2 θ 12 0.36 0.38 0.4 5
Renormalization take RG into effect by multiplying (m ν ) αβ with m ν = I K = (1 + δ α ) (1 + δ β ) with δ α = C m2 α 16π v 2 ln M X M Z with m ν obeying scaling: A B B/c (1 + δ τ ) B D D/c (1 + δ τ ) B/c (1 + δ τ ) D/c (1 + δ τ ) D/c 2 (1 + δ τ ) 2 Ã B B/ c B D D/ c with c = c (1 + δ τ) B/ c D/ c D/ c 2 m 3 = U e3 = 0 not modified! (Grimus, Lavoura, J. Phys. G 31, 683 (2005)) 6
Comparison with L e L µ L τ usual Ansatz for inverted hierarchy (Petcov, Phys. Lett. B 110, 245 (1982)) 0 A B m ν = A 0 0 B 0 0 also gives m 3 = U e3 = 0 and θ 23 π/4 but predicts m 2 = 0 and θ 12 = π/4 highly tuned perturbations required: m 2 / m 2 A π/4 θ 12 perturbations must be of order 30-40 % effective mass small: m ee cos 2 θ 13 m 2 A cos2θ 12 7
Charged Lepton Corrections U = U l U ν with U ν from scaling: if U l is only 23-rotation: c c (c cos θ l 23 sinθ l 23)/(cosθ l 23 + c sinθ l 23) from µ τ symmetry in general: U e3 1 2 sin θ12 l sinθ13 l e iφ 1 sin 2 θ 23 1 2 + ( sinθl 23 cosφ 2 1 4 sin 2 θ12 l sin 2 ) θ13 l + 1 2 cosφ 1 sinθ12 l sin θ13 l rather tuned to have U e3 = 0 and θ 23 π/4 8
A Model Field D 4 Z 2 quantum number L µ L e 1 + 1 e R, N e, φ 1 1 1 L τ N µ, φ 2 1 + 2 N τ 1 2 φ 3 1 4, µ R τ R φ 4 φ 5 2 + 2 Mohapatra, W.R., Phys. Lett. B 644, 59 (2007) 9
A Model charged leptons and heavy neutrinos M R are diagonal a e iϕ 0 0 m D = v b d e 0 0 0 m ν = v2 M 2 and low energy mass matrix M 2 M 1 a 2 e 2iϕ + b 2 b d b e b d d 2 d e b e d e e 2 gives scaling with tan 2 θ 23 = e 2 /d 2 10
A Model 1 10-8 1 10-9 Y B 1 10-10 1 10-11 0.01 0.02 0.03 0.04 0.05 0.06 <m> [ev] Blum, Mohapatra, W.R., Phys. Rev. D 76, 053003 (2007) 11
Higgs Potential 3X 3X V = µ 2 i φ i φ i µ 2 4(φ 4φ 4 + φ 5φ 5 ) + λ i (φ i φ i) 2 + λ 4 (φ 4φ 4 + φ 5φ 5 ) 2 i=1 i=1 + λ 12 (φ 1φ 1 )(φ 2φ 2 ) + λ 13 (φ 1φ 1 )(φ 3φ 3 ) + λ 23 (φ 2φ 2 )(φ 3φ 3 ) 3X + κ i (φ i φ i)(φ 4 φ 4 + φ 5 φ 5) + α 1 [(φ 1 φ 2) 2 + h.c.] + α 2 φ 1 φ 2 2 i=1 + α 3 [(φ 2 φ 3) 2 + h.c.] + α 4 φ 2 φ 3 2 + α 5 (φ 4 φ 5 + φ 5 φ 4) 2 + α 6 (φ 4 φ 5 φ 5 φ 4) 2 + α 7 (φ 4 φ 4 φ 5 φ 5) 2 + α 8 (φ 4 φ 4 φ 5 φ 5)(φ 1 φ 3 + h.c.) + α 9 [(φ 2φ 4 ) 2 + (φ 2φ 5 ) 2 + h.c.] + α 10 ( φ 2φ 4 2 + φ 2φ 5 2 ) + α 11 [(φ 1φ 4 ) 2 + (φ 1φ 5 ) 2 + h.c.] + α 12 ( φ 1φ 4 2 + φ 1φ 5 2 ) + α 13 [(φ 3φ 4 ) 2 + (φ 3φ 5 ) 2 + h.c.] + α 14 ( φ 3φ 4 2 + φ 3φ 5 2 ) + α 15 [(φ 1φ 4 )(φ 3φ 4 ) (φ 1φ 5 )(φ 3φ 5 ) + h.c.] + α 16 [(φ 1 φ 4)(φ 4 φ 3) (φ 1 φ 5)(φ 5 φ 3) + h.c.] 12
Higgs Potential 5 Higgs doublets 5 real scalars 4 pseudoscalars 4 charged scalars potential can be minimized by choozing Φ 1,2,4,5 = 0 v/ and Φ 3 = 5 0 v/ 5 with masses of scalars above experimental limits 13
0 B @ 0 B @ 0 B @ 0 B @ 0 B @ 0 B @ m D Leptogenesis tan 2 θ 23 1 0 0 0 a 2 0 0 C A I e 23 = a2 2 a2 3 sin2α c 2 3 3 b 2 a 3 e iα 3 3 b 3 c 3 1 0 0 0 a 2 b 2 c 2 C A I e 23 = a2 2 a2 3 sin2α c 2 2 3 b 2 a 3 e iα 2 3 0 0 1 a 1 0 0 0 0 0 C A I e 13 = a2 1 a2 3 sin2α c 2 3 3 b 2 a 3 e iα 3 3 b 3 c 3 1 a 1 b 1 c 1 0 0 0 C A I e 13 = a2 1 a2 3 sin2α c 2 1 3 b 2 a 3 e iα 1 3 0 0 1 a 1 0 0 a 2 e iα 2 b 2 c 2 C A I e 12 = a2 1 a2 2 sin2α c 2 2 2 b 2 2 0 0 0 1 a 1 b 1 c 1 a 2 e iα 2 0 0 C A I e 12 = a2 1 a2 2 sin2α c 2 1 2 b 2 1 0 0 0 Goswami, Khan, W.R., Phys. Lett. B 680, 255 (2009) 14
Scaling and See-Saw m ν = m T D M 1 R usually: m D = i M R R m D m diag ν U infinite number of m D are allowed and hence no predictions for LFV or leptogenesis possible In contrast, scaling constraints m D to have the form: m D = a 1 b b/c a 2 d d/c a 3 e e/c no matter what M R is!! 15
Proof m ν ψ = 0 where ψ = 0 1 c det(m ν ) = 0 and therefore (M R can t be singular) det(m D ) = 0 = m D χ = 0 for one of its eigenvectors χ. With m ν = m T D M 1 R m D it follows m ν χ = 0 Hence, χ is proportional to ψ, which means m D ψ = 0 solution for m D is as given above 16
Scaling and Z 2 Suppose Z 2 generated by ν L S(θ) ν L 1 0 0 S(θ) = 0 cos 2θ sin 2θ 0 sin 2θ cos2θ acts on low energy mass matrix from L = 1 2 νc L m ν ν L this implies θ 23 = θ and U e3 = 0 generalized µ τ symmetry (Grimus et al., Nucl. Phys. B 713, 152 (2005)) scaling can be special case of this S(θ) when cos2θ = c2 1 1+c 2 17
Scaling and Z 2 Now assume see-saw L = 1 2 N R M R N c R + N R m D ν L and Z 2 with ν L S(θ) ν L implies that m D S(θ) = m D and thus a 1 b b/c m D = a 2 d d/c a 3 e e/c scaling in m ν!! 18
Scaling and Lepton Flavor Violation SUSY see-saw BR(l i l j γ) ( m D L m 2 D) ij with L ij = δ ij log M i /M X Note: basis in which M R diagonal: m D = V T R m D If m D obeys scaling, then ( m D L m 2 D) 12 ( m D L m 2 = c2 = cot 2 θ 23 D) 13 and τ eγ too rare to be observable 19
m D = Lepton Flavor Violation Note: if µ τ symmetric see-saw a 1 d 1 d 1 W X X a 2 d 2 d 3 and M R = X Y Z a 2 d 3 d 2 X Y Y then LFV prediction is ( m D m 2 D) 12 ( m D m 2 = 1 D) 13 and logarithmic corrections due to L 20
Scaling and Leptogenesis m D = if µ τ symmetric see-saw a 1 d 1 d 1 a 2 d 2 d 3 and M R = a 2 d 3 d 2 W X X X Y Z X Y Y then 2 RH neutrinos: no leptogenesis, neither flavored nor unflavored 3 RH neutrinos: unflavored Y B m 2 Mohapatra, Nasri, Phys. Rev. D 71, 033001 (2005); Mohapatra, Nasri, Yu, Phys. Lett. B 615, 231 (2005) 21
m D = Scaling and Leptogenesis 2 RH neutrinos and scaling A 1 B B/c and M R = VR D R V R A 2 D D/c gives decay asymmetry ε 1 = 3 16π v 2 M 1 m 1 r m 2 1 + r e iρ 2 sinρ just as for µ τ symmetry and 3 RH neutrinos! wash-out parameter: m 1 = m ( ) 1 + r m 2 1 + r e iρ = O m 2 A strong wash-out (flavored leptogenesis: ε α i determined by m 2 A ) 22
Scaling and Leptogenesis add additional Z 2 N R S(θ) N R which implies S(θ) m D = m D and S(θ) M R S(θ) = M R m D = 0 B @ A 1 B B s 23 /c 23 A 2 c 23 D c 23 D s 23 A 2 s 23 D s 23 D s 2 23/c 23 This Z 2L Z 2R gives 1 C A and M R = 0 B @ A B B/c B F (c 1/c) + G F B/c F G 1 C A One heavy RH neutrino decouples and formulae are as above 23
Scaling and Non-Standard Neutrino Physics Dirac neutrinos: m ν = a 1 b b/c a 2 d d/c a 3 e e/c sterile neutrinos: a 1 b 1 b 1 /c d 1 e 1 b 1 b 2 b 2 /c d 2 e 2 m ν = b 1 /c b 2 /c b 2 /c 2 d 2 /c e 2 /c d 1 d 2 d 2 /c d 2 /c 2 e 2 /c 2 e 1 e 2 e 2 /c e 2 /c 2 e 5 0 1 1+c 2 c 1+c 2 0 0 = U e3 U µ3 U τ3 U s1 3 U s2 3 24
m ν = Summary A B B/c B D D/c B/c D/c D/c 2 highly predictive and reconstructible scenario inverted hierarchy and U e3 = 0, stable under RG obtainable in flavor models, see-saw texture analyses,... determines m D irrespective of M R interesting differences to µ τ symmetry 25
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Scaling and Leptogenesis m D m D = ZZ 0 0 0 m 1 e iα 1 = Z2 11 M 1 m 2 e iα 2 = Z2 22 M 2 and ( 1 + Z2 21 ( Z11 2 1 + Z2 21 Z11 2 M 1 M 2 M 1 M 2 ) ) Z 12 Z 22 = Z 21 Z 11 M 1 M 2 = r Z2 11 M 1 ( 1 + r e iρ ) Z2 ( 22 ) 1 + r e iρ, M 2 M1 e iρ/2 M 2 27
Scaling and Leptogenesis The individual flavored decay asymmetries read ε µ 1 = c2 23 (ε 1 ε e 1) ε τ 1 = s 2 23 (ε 1 ε e 1) and ε e 3 M 1 ( ( ) 1 = r m 2 16π v 2 m 1 1 + r e iρ 2 2 s 2 12 m 2 1 c 2 12 sinρ+ c 12 s 12 m1 m 2 r ((m 1 m 2 r) sin(α 1 α 2 4β ρ)/2+ (m 1 r m 2 ) sin(α 1 α 2 4β + ρ)/2)) 28
Normal hierarchy Normal hierarchy U e3 2 0.01 0.001 0.0001 1e-05 1e-06 1e-07 A AB # BB BM BO CM CY DMM DR GK + JLM < VR > YW > # 1e-08 0.94 0.95 0.96 0.97 0.98 0.99 1 < + U e3 2 0.01 0.001 0.0001 1e-05 1e-06 1e-07 > + < # A AB # BB BM BO CM CY DR GK + JLM < VR > YW 1e-08 0.25 0.3 0.35 0.4 sin 2 2θ 23 Inverted hierarchy sin 2 θ 12 Inverted hierarchy 0.01 0.01 0.001 0.001 U e3 2 1e-4 U e3 2 1e-4 1e-05 1e-05 1e-06 1e-06 1e-07 0.88 0.9 0.92 0.94 0.96 0.98 1 1e-07 0.25 0.3 0.35 0.4 sin 2 2θ 23 sin 2 θ 12 Albright, W.R., Phys. Lett. B 665, 378 (2008) 29
Comparison with µ τ Symmetry m µ τ ν = A B B B D E B E D obtained from Z 2 invariance ν L S µτ ν L leading to Sµτ 1 m ν S µτ = m ν with 1 0 0 S µτ = 0 0 1 0 1 0 (cf. with scaling and c = 1) Broken µ τ symmetry gives in general O( θ 23 π/4 ) = O( U e3 ) 30