Construction and Destruction of Horizontal Symmetry Not… · Lam Construction and Destruction of Horizontal Symmetry. ... no GUT and no SUSY ... Lam Construction and Destruction

Construction and Destruction ofHorizontal Symmetry

C.S. Lam

McGill/UBC, Canada

Lam Construction and Destruction of Horizontal Symmetry

Horizontal Symmetry?

badly broken


Horizontal Symmetry?

badly broken


A Lesson from Archaeology

horizontal symmetry

¡

residual symmetry



horizontal symmetry

¡

residual symmetry



horizontal symmetry

¡

residual symmetry



horizontal symmetry

¡

residual symmetry


The Fossils

masses and mixings

quark mixing verydifferent from neutrinomixing – hierarchy

(Wolfenstein) vs regularity

additional Higgs(valons) needed forsymmetry breaking

affect the SM Higgsmass and production?

many nice models, mostwith some ‘ad hoc’assumptions

Introduce as few newingredients and ad hocassumptions as possible

|Uquark | =

0.974 0.225 0.0030.225 0.973 0.040.008 0.04 0.9991

Ulepton = 1√6

2√

2 0

−1√

2√

3

−1√

2 −√

3

(TBM)

T2K (2.5σ), D. Chooz (1.5 σ)

MH ' 125GeV (∼ 3σ)


The Fossils

masses and mixings







|Uquark | =

0.974 0.225 0.0030.225 0.973 0.040.008 0.04 0.9991

Ulepton = 1√6

2√

2 0

−1√

2√

3

−1√

2 −√

3

(TBM)

T2K (2.5σ), D. Chooz (1.5 σ)

MH ' 125GeV (∼ 3σ)


The Fossils

masses and mixings







|Uquark | =

0.974 0.225 0.0030.225 0.973 0.040.008 0.04 0.9991

Ulepton = 1√6

2√

2 0

−1√

2√

3

−1√

2 −√

3

(TBM)

T2K (2.5σ), D. Chooz (1.5 σ)

MH ' 125GeV (∼ 3σ)


Hierarchy of Charged Fermions ξ = ρ− iη

|Uquark | =

0.974 0.225 0.0030.225 0.973 0.040.008 0.04 0.9991

= |U†upUdown|

U =

1− λ2/2 λ Aλ3ξ−λ 1− λ2/2 Aλ2

Aλ3(1− ξ∗) Aλ2 1

⇒

M := U

m1 0 00 −m2 00 0 m3

U† =

0 m −Aλ3ξm3

m m2 −Aλ2m3

−Aλ3ξ∗m3 −Aλ2m3 m3

= M†

'

0 m 0m m2 −Aλ2m3

0 −Aλ2m3 m3

(Fritzsch Texture)

if λ =√m1/m2 and m3 � m2 � m1

(m ≡ √m1m2)


Hierarchy of Charged Fermions ξ = ρ− iη

|Uquark | =

0.974 0.225 0.0030.225 0.973 0.040.008 0.04 0.9991

= |U†upUdown|

U =

1− λ2/2 λ Aλ3ξ−λ 1− λ2/2 Aλ2

Aλ3(1− ξ∗) Aλ2 1

⇒

M := U

m1 0 00 −m2 00 0 m3

U† =

0 m −Aλ3ξm3

m m2 −Aλ2m3

−Aλ3ξ∗m3 −Aλ2m3 m3

= M†

'

0 m 0m m2 −Aλ2m3

0 −Aλ2m3 m3

(Fritzsch Texture)

if λ =√m1/m2 and m3 � m2 � m1

(m ≡ √m1m2)



horizontal symmetry

¡

residual symmetry


CONTENTS H = RφL+ V (φ) G → G ′

Demands: no extra fermions, no Higgs driving field •at EW scale, no GUT and no SUSY •

Construction of G ′L & GL Destruction of G(= GL)

residual symmetries G′L↑

and

G′L↓

from the mixing data

minimal unbroken symmetryGL = {G′L

↑, G′L

↓}

G′↑, G′↓ from G & V (φ)

what φ’s?

demands a generic V •

leptons, quarks, quarks and leptons






and

G′L↓



↑, G′L

↓}

G′↑, G′↓ from G & V (φ)

what φ’s?








and

G′L↓



↑, G′L

↓}

G′↑, G′↓ from G & V (φ)

what φ’s?








and

G′L↓



↑, G′L

↓}

G′↑, G′↓ from G & V (φ)

what φ’s?








and

G′L↓



↑, G′L

↓}

G′↑, G′↓ from G & V (φ)

what φ’s?








and

G′L↓



↑, G′L

↓}

G′↑, G′↓ from G & V (φ)

what φ’s?




CONSTRUCTION

Residual Symmetries G ′L↑ & G ′L

↓ from Mixing leptons

L mass matricesMe = M†eMe = M†e ,Mν = M

T

ν

Mixing matrix: UTMνU = diagthen U=mixing if Me is diag

residue symmetries G′L↑

and G′L↓

F †MeF = Me ,GTMνG = Mν

F 6= G or else U = 1

Zn symmetry for F(diagonal, non-degenerate)n = 3 well studied

Z2 × Z2 symmetry for GG 2i = 1, GiGj = Gk

eigenvectors of Gi are the threecolumns of Ueigenvalue of ith columns is +1for Gi and −1 for Gj ,Gk

Ulepton = 1√6

2√3 0

−1√3√2

−1√3√2

(TBM)

TBM gives G1,G2,G3

if θ13 6= 0, assume G2 the same(accidental symmetry stillallowed)

or, Me slightly non-diagonal

similar considerations for quarks





T

ν



and G′L↓






Ulepton = 1√6

2√3 0

−1√3√2

−1√3√2

(TBM)

TBM gives G1,G2,G3








T

ν



and G′L↓






Ulepton = 1√6

2√3 0

−1√3√2

−1√3√2

(TBM)

TBM gives G1,G2,G3








T

ν



and G′L↓






Ulepton = 1√6

2√2 0

−1√2√3

−1√2 −

√3

(TBM)

TBM gives G1,G2,G3


(or, Me slightly non-diagonal)





T

ν



and G′L↓






Ulepton = 1√6

2√2 0

−1√2√3

−1√2 −

√3

(TBM)

TBM gives G1,G2,G3


(or, Me slightly non-diagonal)


Constructing the Unbroken Symmetry GL

{F ,G} ⊂ GL

if F = diag(1, ω, ω2), then

{F ,G2,G3} = S4 ⊂ PSL2(7)

{F ,G2} = A4 ⊂ T ′ (Ma)

if F 6= {1, ω, ω2} thena

{F ,G2,G3} ⊃ S4

aC.S. Lam, PRL 101 (2008) 121602, PRD 78 (2008) 073015

F ,G of quarks do not generate a finite group



{F ,G} ⊂ GLif F = diag(1, ω, ω2), then

{F ,G2,G3} = S4 ⊂ PSL2(7)

{F ,G2} = A4 ⊂ T ′ (Ma)


{F ,G2,G3} ⊃ S4

aC.S. Lam, PRL 101 (2008) 121602, PRD 78 (2008) 073015





{F ,G2,G3} = S4 ⊂ PSL2(7)

{F ,G2} = A4 ⊂ T ′ (Ma)


{F ,G2,G3} ⊃ S4

aC.S. Lam, PRL 101 (2008) 121602, PRD 78 (2008) 073015





{F ,G2,G3} = S4 ⊂ PSL2(7)

{F ,G2} = A4 ⊂ T ′ (Ma)


{F ,G2,G3} ⊃ S4

aC.S. Lam, PRL 101 (2008) 121602, PRD 78 (2008) 073015




Assumption: L = 3, no extra fermions, no Higgs driving field •at EW scale, no GUT and no SUSY •



and

G′L↓


unbroken symmetryGL = {G′L

↑, G′L

↓}

G′↑, G′↓ from G & V (φ)

what φ’s?





Assumption: L = 3, no extra fermions, no Higgs driving field •at EW scale, no GUT and no SUSY •



and

G′L↓


unbroken symmetryGL = {G′L

↑, G′L

↓}

G′↑, G′↓ from G & V (φ)

what φ’s?




DYNAMICS of BREAKING

V (φ) for G = A4 = {F ,G2} f RφL+ hc

φ (valon) in IR 3, 1, 1′, 1′′

in F -diag representation, needsφe3 = (1, 0, 0) to render Me diagφν3 = (1, 1, 1) to get bimaximal

mixing for neutrinos (G2)

usual strategya

design a V (φe3, φν3 , φ1, ξ).

‘driving fields’ ξ requiredV not generic

two-phase strategyb

a ‘generic’ (U(1)-inv) potential• V (φ3, φ1, φ1′ , φ1′′)

similar procedure works forother G

aAltarelli & Feruglio, NPB 741 (2006) 215bLam, PRD 83 (2011) 113002

Potential (may include 1, 1′, 1′′ in V )

V = 〈φ3φ3|g1P1 + g1′P1′ + g1′′P1′′+

g3P3|φ3φ3〉 − µ2〈φ3|φ3〉

Solutions

every nondegenerate eigenvector ofevery g ∈ A4 = {F ,G2} is a solution

F

100

=

100

, G2

111

=

111

different solution defines a differentphase with a different energy V anda different residual symmetry

Remarks

− gi and V depend on electric charge.

This enables e and ν to belong to

a different phase



φ (valon) in IR 3, 1, 1′, 1′′



usual strategya



two-phase strategyb





V = 〈φ3φ3|g1P1 + g1′P1′ + g1′′P1′′+

g3P3|φ3φ3〉 − µ2〈φ3|φ3〉

Solutions


F

100

=

100

, G2

111

=

111


Remarks



a different phase



φ (valon) in IR 3, 1, 1′, 1′′



usual strategya



two-phase strategyb





V = 〈φ3φ3|g1P1 + g1′P1′ + g1′′P1′′+

g3P3|φ3φ3〉 − µ2〈φ3|φ3〉

Solutions


F

100

=

100

, G2

111

=

111


Remarks



a different phase



φ (valon) in IR 3, 1, 1′, 1′′



usual strategya



two-phase strategyb





V = 〈φ3φ3|g1P1 + g1′P1′ + g1′′P1′′+

g3P3|φ3φ3〉 − µ2〈φ3|φ3〉

Solutions


F

100

=

100

, G2

111

=

111


Remarks



a different phase



φ (valon) in IR 3, 1, 1′, 1′′



usual strategya



two-phase strategyb





V = 〈φ3φ3|g1P1 + g1′P1′ + g1′′P1′′+

g3P3|φ3φ3〉 − µ2〈φ3|φ3〉

Solutions


F

100

=

100

, G2

111

=

111


Remarks



a different phase



φ (valon) in IR 3, 1, 1′, 1′′



usual strategya



two-phase strategyb





V = 〈φ3φ3|g1P1 + g1′P1′ + g1′′P1′′+

g3P3|φ3φ3〉 − µ2〈φ3|φ3〉

Solutions


F

100

=

100

, G2

111

=

111


Remarks



a different phase



φ taken from IR 3, 1, 1′, 1′′



usual strategya


‘driving fields’ ξ required

two-phase strategyb

a ‘generic’ (U(1)-inv) potentialV (φ3, φ1, φ1′ , φ1′′)



Solutions


F

100

=

100

, G2

111

=

111


Remarks



a different phase

−to build a model, need to specify howR transforms under A4. Not unique

−A4 promotes to S4 if f1′ = f1′′



φ taken from IR 3, 1, 1′, 1′′



usual strategya


‘driving fields’ ξ required

two-phase strategyb

a ‘generic’ (U(1)-inv) potentialV (φ3, φ1, φ1′ , φ1′′)



Solutions


F

100

=

100

, G2

111

=

111


Remarks



a different phase

−to build a model, need to specify howR transforms under A4. Not unique

−A4 promotes to S4 if f1′ = f1′′


Two-Phase Analogy Water and Ice

H2O

0C < T < 100C T < 0C


Quarks and Leptons

The Fossils

masses and mixings







|Uquark | =

0.974 0.225 0.0030.225 0.973 0.040.008 0.04 0.9991

Ulepton = 1√6

2√

2 0

−1√

2√

3

−1√

2 −√

3

(TBM)

T2K (2.5σ), D. Chooz (1.5 σ)

MH ' 125GeV (∼ 3σ)


A Common G

assume L quarks and leptons to belong to the same IR of G •Dirac and Majorana fermions may live in two different phases of abroken G •the Dirac phase has very little symmetry

the Majorana phase has a finite-group symmetry such as A4

a candidate for G is SO(3), • because it contains A5,S4,A4,Dn.It should also be able to decide which finite group to take (why A4?)


V (φ) of G = U(1)× SO(3) RφL+ hc

L = 3, R = 3 •• ⇒φ = 5, 3, 1

J = 2(t), 1(v), 0(s)

potentially 5+3+1=9unknowns

two solutions (phases):

nematic (N), tetrahedral (T)

N: {e−πiJx , e−πiJy , e−πiJz}t = (sin η, 0,

√2 cos η, 0, sin η) = 1

v = (0, 0, 0), s = 1

ends up with 2+1=3 parameters

V (φ) = −µ2〈t|t〉 − ν2〈v |v〉+〈tt|g4P4 + g2P2 + g0P0|tt〉+〈vv |f2P2 + f0P0|vv〉+ h〈ss|ss〉+〈tv |α3P3 + α2P2 + α1P1|tv〉+β〈ts|ts〉+ γ〈vs|vs〉 − λ2〈s|s〉14 non-zero couplings, 9 cubic eqs

more couplings ⇒ less solutions!

T: A4 = {G = ω2 e−2πiJy/3,

F = e−iβJy e−πiJz }, cos 12β =

√13

t = (1, 0, 0,√

2, 0) = 1

t ′ = (0,−√

2, 0, 0, 1) = 1′

v = (0, 0, 0), s = 1′′

ends up with 2 free parameters



L = 3, R = 3 •• ⇒φ = 5, 3, 1

J = 2(t), 1(v), 0(s)





√2 cos η, 0, sin η) = 1

v = (0, 0, 0), s = 1




T: A4 = {G = ω2 e−2πiJy/3,


√13

t = (1, 0, 0,√

2, 0) = 1

t ′ = (0,−√

2, 0, 0, 1) = 1′

v = (0, 0, 0), s = 1′′




L = 3, R = 3 •• ⇒φ = 5, 3, 1

J = 2(t), 1(v), 0(s)





√2 cos η, 0, sin η) = 1

v = (0, 0, 0), s = 1




T: A4 = {G = ω2 e−2πiJy/3,


√13

t = (1, 0, 0,√

2, 0) = 1

t ′ = (0,−√

2, 0, 0, 1) = 1′

v = (0, 0, 0), s = 1′′



Fermion Mass Matrices RφL

Cartesian basis: t → symmetric traceless mass matrix, s → unit matrix

N s′η − c ′η − σ′ 0 00 −s′η − c ′η − σ′ 00 0 2c ′η − σ′

s ′η = τ ′ sin η, c ′η = τ ′ cos η/√3

T

−σ + τ 0√2τ

0 −σ − τ 0√2τ 0 −σ

F -diagonbal basis

−c ′η − σ′ s′η 0s′η −c ′η − σ′ 00 0 2c ′η − σ′

−σ τ ττ −σ ττ τ −σ


Fermion Mass Matrices RφL

Cartesian basis: t → symmetric traceless mass matrix, s → unit matrix

N s′η − c ′η − σ′ 0 00 −s′η − c ′η − σ′ 00 0 2c ′η − σ′

s ′η = τ ′ sin η, c ′η = τ ′ cos η/√3

T

−σ + τ 0√2τ

0 −σ − τ 0√2τ 0 −σ

F -diagonbal basis

−c ′η − σ′ s′η 0s′η −c ′η − σ′ 00 0 2c ′η − σ′

−σ τ ττ −σ ττ τ −σ


F -diagonbal basis

N −c ′η − σ′ s′η 0s′η −c ′η − σ′ 00 0 2c ′η − σ′

diagonal if s ′η = 0, ⇒2 parameters

3 parameters if s ′η 6= 0,

but maximal 1-2 mixing

T −σ τ ττ −σ ττ τ −σ

magic and 2-3 symmetric.

TBM neutrino mixing in

Me-diagonal basis

leptons

Me ,Mν (s ′η = 0) MN

me = mµ in Me ; |c ′η| � σ′ in Mν mν OK, TBM neutrino mixing

quarks

Mu,Md (s ′η 6= 0)

masses OK, but no mixing

F -diagonbal basis

N −c ′η − σ′ s′η 0s′η −c ′η − σ′ 00 0 2c ′η − σ′







Me-diagonal basis

leptons



quarks

Mu,Md (s ′η 6= 0)


F -diagonbal basis

N −c ′η − σ′ s′η 0s′η −c ′η − σ′ 00 0 2c ′η − σ′







Me-diagonal basis

leptons



quarks

Mu,Md (s ′η 6= 0)


CONCLUSION

A different strategy to search for horizontal symmetry

Mixing can be used to construct residual symmetry and unbrokenhorizontal symmetry of leptons

A two-phase mechanism enables the breaking of a generic potential(for A4 and G) to retain the correct residual symmetries

The difference of quark & lepton mixings may reflect two phases ofa common horizontal symmetry such as SO(3)

To get a realistic model of quarks and leptons, we have to relaxsome of the assumptions in •


Documents

Construction and Destruction of Horizontal Symmetry Not… · Lam Construction and Destruction of Horizontal Symmetry. ... no GUT and no SUSY ... Lam Construction and Destruction