S 3 Symmetry as the Origin of CKM Matrix

S 3 Symmetry as the Origin of CKM Matrix Ujjal Kumar Dey Physical Research Laboratory October 25, 2015 Based on: PRD 89, 095025 and arxiv:1507.06509 Collaborators: D. Das and P. B. Pal 1 / 25

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Discrete symmetries in particle physics In SM fermions (both leptons and quarks) come in three generations There are inter-generational differences, in contrast to their uniformity in gauge interactions Two types of hierarchies in the flavor sector: Large hierarchy within the charged fermion sector and enormous hierarchy between charged fermion and neutrino masses Mixing information in quark and lepton sector Finite discrete symmetry groups (e.g., S 3, S 4, D 4, A 4 etc.) provide an effective way of explaining some of these flavor issues We will consider the case of S 3 symmetric model 4 / 25

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S 3 is the permutation group of three objects The order of S 3 is 3! = 6 The six elements correspond to the following transformations e : (x 1, x 2, x 3) (x 1, x 2, x 3), a 1 : (x 1, x 2, x 3) (x 2, x 1, x 3), a 2 : (x 1, x 2, x 3) (x 3, x 2, x 1), a 3 : (x 1, x 2, x 3) (x 1, x 3, x 2), a 4 : (x 1, x 2, x 3) (x 3, x 1, x 2), a 5 : (x 1, x 2, x 3) (x 2, x 3, x 1). S 3 can also be thought of as the symmetry of an equilateral triangle with a 1 and a 1a 2 being the reflection and the 2π/3 rotation respectively 6 / 25

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Scalar Potential The scalar sector consists of three SU(2) L scalar doublets φ i (i = 1, 2, 3) (φ 1, φ 2) T transform as an S 3 doublet and φ 3 is an S 3 singlet Most general S 3-symmetric scalar potential with these fields can be written as, V (φ i ) =µ 2 1(φ 1 φ1 + φ 2 φ2) + µ2 3φ 3 φ3 + λ1(φ 1 φ1 + φ 2 { φ2)2 + λ 2(φ 1 φ2 φ 2 φ1)2 + λ 3 (φ 1 φ2 + φ 2 φ1)2 + (φ 1 φ1 φ 2 φ2)2} { } + λ 4 (φ 3 φ1)(φ 1 φ2 + φ 2 φ1) + (φ 3 φ2)(φ 1 φ1 φ 2 φ2) + h.c. + λ 5(φ 3 φ3)(φ 1 φ1 + φ 2 φ2) + λ6 { (φ 3 φ1)(φ 1 φ3) + (φ 3 φ2)(φ 2 φ3) } + λ 7 { (φ 3 φ1)(φ 3 φ1) + (φ 3 φ2)(φ 3 φ2) + h.c. } + λ 8(φ 3 φ3)2 8 / 25

Scalar Potential The minimization conditions for the potential are: 2µ 2 1 = 2λ 1(v 2 1 + v 2 2 ) 2λ 3(v 2 1 + v 2 2 ) 6λ 4v 2v 3 (λ 5 + λ 6 + 2λ 7)v 2 3, 2µ 2 1 = 2λ 1(v 2 1 + v 2 2 ) 2λ 3(v 2 1 + v 2 2 ) 3v3 v 2 λ 4(v 2 1 v 2 2 ) (λ 5 + λ 6 + 2λ 7)v 2 3, 2µ 2 3 = λ 4 v 2 v 3 (v 2 2 3v 2 1 ) (λ 5 + λ 6 + 2λ 7)(v 2 1 + v 2 2 ) 2λ 8v 2 3. There exists three nontrivial conditions for consistency, λ 4= 0 a massless scalar v 2= 3v 1 and v 3 = 0 interesting for DM aspect (?) v 1= 3v 2 and v 3 arbitrary present case 9 / 25

For the case v 1 = 3v 2, even after spontaneous symmetry breaking, a Z 2 subgroup of S 3 remains preserved To be precise, the vacuum in the (φ 1, φ 2) T basis is ( 3, 1) T and ( ) 1 3 ( 3 ) ( ) ( ) 2 2 2 3 1 3 = with 2 2 = 1 2 1 1 3 2 1 2 3 2 1 2 This remnant Z 2 will be used to construct the realistic model for quarks 10 / 25

Physical Scalars After electroweak symmetry breaking from three SU(2) L doublet scalars we will have nine physical scalars Physical States Transformation under Z 2 h 0, H ± 1, A1 Odd h, H, H ± 2, A2 Even Here h will have the SM-like couplings in the alignment limit i.e., sin(β α) = 1 where β = tan 1 (2v 2/v 3) and α is the mixing angle in the h H sector 11 / 25

λs and Masses The quartic couplings and the physical scalar masses can be connected via the following relations: { 1 ( λ 1 = 2v 2 sin 2 m 2 h β cos2 α + m 2 ) ( H sin2 α + m 2 1+ m2 2+ cos2 β 1 )} 9 m2 h0, 1 { λ 2 = 2v 2 sin 2 (m 2 } 1+ β m2 A1 ) (m2 2+ m2 A2 ) cos2 β, ( ) 1 4 λ 3 = 2v 2 sin 2 β 9 m2 h0 + m2 2+ cos2 β m 2 1+, λ 4 = 2 mh0 2 1 9 v 2 sin β cos β, λ 5 = 1 { sin α cos α ( v 2 m 2 ) H sin β cos β m2 h + 2m 2 2+ + 1 m 2 } h0 9 cos 2, β λ 6 = 1 ( 1 m 2 ) h0 v 2 9 cos 2 β + m2 A2 2m2 2+, λ 7 = 1 ( 1 m 2 ) h0 2v 2 9 cos 2 β m2 A2 { 1 ( λ 8 = 2v 2 cos 2 m 2 h β sin2 α + m 2 ) H cos2 α 1 } 9 m2 h0 tan2 β. 12 / 25

Constraints from unitarity and stability log 10 (tan β) log 10 (tan β) Unitarity and stability demands that tan β [0.3, 17] and the physical scalars are below 1 TeV 13 / 25

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Quark Sector In the gauge basis the S 3 transformation properties of the quarks, 1 : Q 3, u 3R, d 3R, ( ) ( ) Q1 u1r 2 :,, Q 2 u 2R ( d1r where Q i s are the usual left-handed SU(2) L quark doublets and u ir s (d ir s) are the right-handed up (down) type SU(2) L singlets The Yukawa Lagrangian is given by, L Yuk = L U Yuk + L D Yuk where ( L U Yuk = y1 u Q1 φ3u 1R + Q ) {( 2 φ3u 2R y2 u Q1 φ2 + Q ) 2 φ1 u 1R ( + Q1 φ1 Q } 2 φ2 )u 2R y3 u Q 3 φ3u 3R y4 u Q ( 3 φ1u 1R + φ ) 2u 2R ( ) y5 u Q 1 φ 1 + Q 2 φ 2 u 3R + h.c. L D Yuk =L U Yuk with (u ir d ir, y u i d 2R y d i, φ i φ i ) ) 15 / 25

Quark Mass Matrices The general form of the quark mass matrix (for brevity only up-type case is shown) is given by, y1 u v 3 + y2 u v 2 y2 u v 1 y5 u v 1 M u = y2 u v 1 y1 u v 3 y2 u v 2 y5 u v 2 y4 u v 1 y4 u v 2 y3 u v 3 This can be block-diagonalized by the unitary matrix, 1 3 0 2 2 X = 3 1 0 2 2 0 0 1 in the following way, y1 u v 3 2y2 u v 2 0 0 M block u = X M ux = 0 y1 u v 3 + 2y2 u v 2 2y5 u v 2 0 2y4 u v 2 y3 u v 3 Recall that the vacuum alignment v 1 = 3v 2 implies the breaking S 3 Z 2 16 / 25

The Z 2-odd combination of fermion will not mix with the Z 2-even counterparts in the fermion mass terms We can define the top quark as the Z 2-odd combination with mass m t = y u 1 v 3 2y u 2 v 2 = v y u 1 cos β y u 2 sin β The block-diagonal form of M block u implies that one can define the following intermediate basis in the up-sector separating the Z 2-odd and Z 2-even components, t c = X u u 1 u 2 u 3 Rotate further the 2 2 block to get the physical u and c quarks Similar treatment can be followed for down sector by identifying the b as the Z 2-odd combination 17 / 25

Getting the CKM Now one can get M diag u = U L Mblock u U R = diag(m t, m c, m u) where both U L and U R are block-diagonal One can take U L to be of the form 1 0 0 U L = 0 cos θ u sin θ u 0 sin θ u cos θ u Then M diag u = U L Mblock u U R = U L X M uxu R = U L MuU R (say); where U L,R = XU L,R Similarly in the down sector D L,R = XD L,R where 1 0 0 D L = 0 cos θ d sin θ d 0 sin θ d cos θ d 18 / 25

The CKM matrix is thus given by, V CKM = U L D L = U L D L = b s d t 1 0 0 c 0 cos θ C sin θ C where θ C = θ d θ u u 0 sin θ C cos θ C By choosing sin θ C = sin(θ d θ u) λ, one can reproduce the Cabibbo block of the CKM matrix The near block-diagonal structure of CKM matrix can be thought of as a direct consequence of the remnant Z 2 symmetry Only a mild breaking of this Z 2 can result in the exact structure of the CKM matrix 19 / 25

Mild Breaking of Z 2 Gives the Exact Form of CKM Matrix Introduce a soft S 3-breaking term (say, µ 2 13(φ 1 φ3 + φ 3 φ1)) This will slightly modify the VEV alignment, v 1 = 3v 2 + with v 1,2 Thus the mass matrix in the up sector (for example) will be, 0 y2 u y5 u 0 u M u = M u + y2 u 2 u 5 0 0 M u + u y4 u 2 0 0 0 0 u 4 0 0 Clearly, now we will have M diag u = U M L uu R where U L (and similarly U R ) can be defined as U L = U L U L and U L is close to unit matrix which takes care of the very small, 1 Aλ 2 Cλ 3 U L = Aλ 2 1 0 + O(λ 4 ) C λ 3 0 1 Note that D L can also be approximated as 1 3 20 / 25

Thus choosing the parameter C = A(ρ + iη), one can write, V CKM = U L U L D LD L b s d t 1 Aλ 2 Aλ 3 (1 ρ iη) = c Aλ 2 1 λ2 λ 2 + O(λ4 ) u Aλ 3 (ρ iη) λ 1 λ2 2 Even if we consider small departure of D L from 1 3 the general form of V CKM will not change 21 / 25

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The Z 2-odd particles (e.g., h 0, A 1, H ± 1 ) will only have off-diagonal couplings with quarks involving 3rd generation physical quarks (t or b). But the mild breaking of Z 2 will lead to tiny diagonal couplings For example, if h 0 is light enough, it can be looked for in the channel t ch 0 followed by h 0 µτ or eτ. But for heavy h 0 ILC is the better option to produce h 0 via the coupling with SM-like Higgs h, (h 0 h 0 h) The precise measurement of the decay width h γγ can also put stringent constraints (see arxiv:1408.6133) Like most of the extended scalar sector models here also the FCNC related issues are to be dealt with much care 23 / 25

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Starting from an S 3 symmetric theory and its spontaneous breaking to an approximate Z 2 answers the following puzzles Why the third generation of quarks are so different (massive) from the other two? As the third generation quarks are Z 2-odd Why the CKM matrix is nearly block-diagonal? Because of the misalignment of the mixing between the first two generations in the up and down sectors, both of which are Z 2-even The study of the lepton sector and a detailed exploration of the flavour constraints are yet to be done (future project) 25 / 25