Contents Flavour permutational symmetry A minimal S 3 -invariant - PowerPoint PPT Presentation

The flavour permutational symmetry S 3 A. Mondrag´ on Instituto de F´ ısica, UNAM Prog. of Theoretical Physics 109 , 795 (2003) Rev. Mex. Fis , S52, N4 , 67-73 (2006) Phys. Rev. D, 76 , 076003, (2007) J. Phys. A: Mathematical and Theoretical 41 , 304035 (2008) Phys. of Atomic Nuclei 74 , 1075-1083 (2011). XIII Mexican Workshop on Particles and Fields 20 - 26 October 2011, Le´ on Guanajuato, M´ exico

Contents • Flavour permutational symmetry • A minimal S 3 -invariant extension of the Standard Model • Masses and mixings in the quark sector • Masses and mixings in the leptonic sector • The neutrino mass spectrum • FCNCs • The anomaly of the moun’s magnetic moment • Summary and conclusions 1

The Group S 3 The group S 3 of permutations of three objects Permutations Rotations 3 � 1 � 2 3 a 120 ◦ − rotation around the ⇐ ⇒ 3 1 2 V 1 invariant vector V 1 V2A � 1 � 2 3 a 180 ◦ rotation around the ⇐ ⇒ 2 2 1 3 invariant vector V 2 s V2s 1 Symmetry adapted basis       1 1 1 1 1 1  , | v 2 s > =  , | v 1 > = | v 2 A > = √ − 1 √ 1 √ 1     2 6 3 0 − 2 1 2

Irreducible representations of S 3 The group S 3 has two one-dimensional irreps (singlets ) and one two-dimensional irrep (doublet) • one dimensional: 1 A antisymmetric singlet, 1 s symmetric singlet • Two - dimensional: 2 doublet Direct product of irreps of S 3 1 s ⊗ 1 s = 1 s , 1 s ⊗ 1 A = 1 A , 1 A ⊗ 1 A = 1 s , 1 s ⊗ 2 = 2 , 1 A ⊗ 2 = 2 2 ⊗ 2 = 1 s ⊕ 1 A ⊕ 2 the direct (tensor) product of two doublets � p D 1 � � q D 1 � p D = and q D = p D 2 q D 2 has two singlets, r s and r A , and one doublet r T D r s = p D 1 q D 1 + p D 2 q D 2 is invariant , r A = p D 1 q D 2 − p D 2 q D 1 is not invariant � p D 1 q D 2 + p D 2 q D 1 � r T D = p D 1 q D 1 − p D 2 q D 2 3

A Minimal S 3 invariant extension of the SM The Higgs sector is modified, (Φ 1 , Φ 2 , Φ 3 ) T Φ → H = H is a reducible 1 s ⊕ 2 rep. of S 3 1 � � H s = √ Φ 1 + Φ 2 + Φ 3 3  1  2 (Φ 1 − Φ 2 ) √ H D =     1 6 (Φ 1 + Φ 2 − 2Φ 3 ) √ Quark, lepton and Higgs fields are Q T = ( u L , d L ) , u R , d R , L † = ( ν L , e L ) , e R , ν R , H All these fields have three species (flavours) and belong to a reducible 1 ⊕ 2 rep. of S 3 4

Leptons’ Yukawa interactions Leptons − Y e 1 L I H S e IR − Y e 3 L 3 H S e 3 R − Y e L YE = 2 [ L I κ IJ H 1 e JR + L I η IJ H 2 e JR ] Y e 4 L 3 H I e IR − Y e − 5 L I H I e 3 R + h.c., − Y ν 1 L I ( iσ 2 ) H ∗ S ν IR − Y ν 3 L 3 ( iσ 2 ) H ∗ L Yν = S ν 3 R Y ν 2 [ L I κ IJ ( iσ 2 ) H ∗ 1 ν JR + L I η IJ ( iσ 2 ) H ∗ − 2 ν JR ] Y ν 4 L 3 ( iσ 2 ) H ∗ I ν IR − Y ν 5 L I ( iσ 2 ) H ∗ − I ν 3 R + h.c. � 0 � � 1 � 1 0 κ = ; η = I, J = 1 , 2 1 0 0 − 1 Furthermore, the Majorana mass terms for the right handed neutrinos are − ν T IR C M I ν IR − M 3 ν T L M = 3 R Cν 3 R , C is the charge conjugation matrix. 5

Mass matrices We will assume that < H D 1 > = < H D 2 > � = 0 and < H 3 > � = 0 and � 246 � 2 < H 3 > 2 + < H D 1 > 2 + < H D 2 > 2 ≈ 2 GeV Then, the Yukawa interactions yield mass matrices of the general form   µ 1 + µ 2 µ 2 µ 5 M = µ 2 µ 1 − µ 2 µ 5   µ 4 µ 4 µ 3 The Majorana masses for ν L are obtained from the see-saw mechanism − 1 ( M νD ) T M ν = M νD ˜ ˜ M with M = diag ( M 1 , M 2 , M 3 ) 6

Mixing matrices The mass matrices are diagonalized by unitary matrices � � U † = diag m d ( u,e ) m s ( c,µ ) m b ( t,τ ) d ( u,e ) L M d ( u,e ) U d ( u,e ) R and � � U T ν M ν U ν = diag m ν 1 , m ν 2 , m ν 3 The masses can be complex, and so, U eL is such that � | m e | 2 , | m µ | 2 , | m τ | 2 � U † eL M e M † e U eL = diag , etc. The quark mixing matrix is U † V CKM = uL U dL and, the neutrino mixing matrix is U † = eL U ν V MNS 7

Masses and mixings in the quark sector The mass matrices for the quark sector take the general form  µ u ( d ) + µ u ( d ) µ u ( d ) µ u ( d )  1 2 2 5 µ u ( d ) µ u ( d ) − µ u ( d ) µ u ( d ) M u ( d ) =   2 1 2 5   µ u ( d ) µ u ( d ) µ u ( d ) 4 4 3 � | m u ( d ) | 2 , | m c ( s ) | 2 , | m t ( b ) | 2 � U † u ( d ) L M u ( d ) M † u ( d ) U u ( d ) L = diag , U † V CKM = uL U dL The set of dimensionless parameters µ u 1 /µ u µ u 3 /µ u µ u 2 /µ u 0 = − 0 . 000293 , 0 = − 0 . 00028 , 0 = 1 , µ u 4 /µ u µ u 5 /µ u 0 = 0 . 031 , 0 = 0 . 0386 , µ d 1 /µ d µ d 3 /µ d µ d 2 /µ d 0 = 0 . 0004 , 0 = 0 . 00275 , 0 = 1 + 1 . 2 I, µ d 4 /µ d µ d 5 /µ d 0 = 0 . 283 , 0 = 0 . 058 , 8

The quark mixing matrix Yields the mass hierarchy and the mixing matrix m u /m t = 2 . 5469 × 10 − 5 , m c /m t = 3 . 9918 × 10 − 3 , m d /m b = 1 . 5261 × 10 − 3 , m s /m b = 3 . 2319 × 10 − 2 , The computed mixing matrix is   0 . 968 + 0 . 117 I 0 . 198 + 0 . 0974 I − 0 . 00253 − 0 . 00354 I V CKM = − 0 . 198 + 0 . 0969 I 0 . 968 − 0 . 115 I − 0 . 0222 − 0 . 0376 I   0 . 00211 + 0 . 00648 I 0 . 0179 − 0 . 0395 I 0 . 999 − 0 . 00206 I   0 . 975 0 . 221 0 . 00435 | V th CKM | = 0 . 221 0 . 974 0 . 0437   0 . 00682 0 . 0434 0 . 999 which should be compared with  (3 . 96 ± 0 . 09) × 10 − 3  0 . 97383 ± 0 . 00024 0 . 2272 ± 0 . 0010 | V exp (42 . 21 ± 0 . 45 ± 0 . 09) × 10 − 3 CKM | = 0 . 2271 ± 0 . 010 0 . 97296 ± 0 . 00024     (8 . 14 ± 0 . 5) × 10 − 3 (41 . 61 ± 0 . 12) × 10 − 3 0 . 9991 ± 0 . 000034 The Jarlskog invariant is J = Im [( V CKM ) 11 ( V CKM ) 22 ( V ∗ CKM ) 12 ( V ∗ CKM ) 21 ] = 2 . 9 × 10 − 5 , J exp = (3 . 0 ± 0 . 3) × 10 − 5 9

The leptonic sector To achieve a further reduction of the number of parameters, in the leptonic sector, we introduce an additional discrete Z 2 symmetry − + H I , ν 3 R H S , L 3 , L I , e 3 R , e IR , ν IR then, Y e 1 = Y e 3 = Y ν 1 = Y ν 5 = 0 Hence, the leptonic mass matrices are µ e µ e µ e µ ν µ ν     0 2 2 5 2 2 µ e − µ e µ e µ ν − µ ν M e = and M νD = 0  2 2 5   2 2  µ e µ e µ ν µ ν µ ν 0 4 4 4 4 3 10

The Mass Matrix of the charged leptons as function of its eigenvalues The mass matrix of the charged leptons is �  1+ x 2 − ˜ m 2  mµ ˜ mµ ˜ µ 1 1 1 √ √ √ √ √ 1+ x 2 1+ x 2 1+ x 2 2 2 2         � m 2 1+ x 2 − ˜   mµ ˜ mµ ˜ µ 1 − 1 1 M e ≈ m τ √ √ . √ √ √   1+ x 2 1+ x 2 1+ x 2  2 2 2        me (1+ x 2) me (1+ x 2)   ˜ ˜ e iδe e iδe 0   � � m 2 m 2 1+ x 2 − ˜ 1+ x 2 − ˜ µ µ x = m e /m µ , m µ = m µ /m τ and ˜ ˜ m e = m e /m τ This expression is accurate to order 10 − 9 in units of the τ mass There are no free parameters in M e other than the Dirac Phase δ !! 11

The Unitary Matrix U eL The unitary matrix U eL is calculated from U † eL M e M † m 2 e , m 2 µ , m 2 � � eL U eL = diag τ We find 1 , 1 , e iδD � � U eL = Φ eL O eL , Φ eL = diag and m 2 µ +4 x 2+ ˜ m 4 m 2 m 2 m 4 m 2   (1+2 ˜ µ +2 ˜ e ) (1 − 2 ˜ µ + ˜ µ − 2 ˜ e ) 1 − 1 1 2 x √ √ √ � � m 2 m 4 m 6 m 2 2 m 2 m 4 m 6 m 2 2 µ +5 x 2 − ˜ e +12 x 4 µ + x 2+6 ˜ 1+ ˜ µ − ˜ µ + ˜ 1 − 4 ˜ µ − 4 ˜ µ − 5 ˜  e        (1+4 x 2 − ˜ m 4 m 2 m 2 m 4 µ − 2 ˜ e ) (1 − 2 ˜ µ + ˜ µ )   − 1 1 1 2 x √ √ √ O eL ≈ ,   � � m 2 m 4 m 6 m 2 2 m 2 m 4 m 6 m 2 2 µ +5 x 2 − ˜ e +12 x 4 µ + x 2+6 ˜ 1+ ˜ µ − ˜ µ + ˜ 1 − 4 ˜ µ − 4 ˜ µ − 5 ˜   e     √   � m 2 µ + x 2 − 2 ˜ m 2 (1+ x 2 − ˜ m 2 m 2 � 1+2 x 2 − ˜ m 2 m 2 1+2 x 2 − ˜ m 2 m 2   1+ x 2 ˜ µ − ˜ e (1+ ˜ e ) µ − 2 ˜ e ) µ − ˜ e me ˜ mµ − − x   � � � m 2 m 4 m 6 m 2 m 2 m 4 m 6 m 2 m 2 µ +5 x 2 − ˜ e +12 x 4 µ + x 2+6 ˜ 1+ x 2 − ˜ 1+ ˜ µ − ˜ µ + ˜ 1 − 4 ˜ µ − 4 ˜ µ − 5 ˜ e µ x = m e /m µ , m µ = m µ /m τ and ˜ ˜ m e = m e /m τ 12

The neutrino mass matrix I The Majorana masses for ν L are obtained from the see-saw mechanism M − 1 R M T M ν = M νD ˜ νD with ˜ � � M R = diag M 1 , M 2 , M 3 M 1 � = M 2 � = M 3 and   µ 2 µ 2 0 = µ 2 − µ 2 0 M ν   µ 4 µ 4 µ 3 Then  2 µ 2  2 µ 2 µ 4 2 λµ 2 2 ¯ ¯ 2 M M  2 µ 2  M ν =  . 2 λµ 2  2  2 λ 2 µ 2 µ 4 ¯  2  M 2 µ 2 µ 2  2 µ 2 µ 4 4 3 2 λµ 2 µ 4 M + ¯ ¯ ¯ M M � 1 � 1 M = 1 1 1 λ = 1 1 � � + and − ¯ 2 M 1 M 2 2 M 1 M 2 13

The neutrino mass matrix II M ( o ) is reparametrized in term of the neutrinomasses ν M ν = M (0) + δM µ ν �   ( m ν 3 − m ν 1 )( m ν 2 − m ν 3 ) e − iδν m ν 3 0 M (0)   = 0 m ν 3 0   ν   � ( m ν 3 − m ν 1 )( m ν 2 − m ν 3 ) e − iδν m ν 1 + m ν 2 − m ν 3 ) e − 2 iδν � 0   0 1 0 mν 1 mν 2 mν 1 � 1 0 (1 − mν 2 )( mν 3 − mν 3 ) δM ν = 2 λµ 2     2  mν 1 mν 2 mν 1  � 0 (1 − mν 2 )( mν 3 − mν 3 ) 0 2 λµ 2 = m ν 3 µ 2 14