Construction and Destruction of Horizontal Symmetry Not… · Lam Construction and Destruction of...
Transcript of Construction and Destruction of Horizontal Symmetry Not… · Lam Construction and Destruction of...
Construction and Destruction ofHorizontal Symmetry
C.S. Lam
McGill/UBC, Canada
Lam Construction and Destruction of Horizontal Symmetry
A Lesson from Archaeology
horizontal symmetry
¡
residual symmetry
Lam Construction and Destruction of Horizontal Symmetry
A Lesson from Archaeology
horizontal symmetry
¡
residual symmetry
Lam Construction and Destruction of Horizontal Symmetry
A Lesson from Archaeology
horizontal symmetry
¡
residual symmetry
Lam Construction and Destruction of Horizontal Symmetry
A Lesson from Archaeology
horizontal symmetry
¡
residual symmetry
Lam Construction and Destruction of Horizontal 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)
Lam Construction and Destruction of Horizontal 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)
Lam Construction and Destruction of Horizontal 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)
Lam Construction and Destruction of Horizontal Symmetry
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)
Lam Construction and Destruction of Horizontal Symmetry
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)
Lam Construction and Destruction of Horizontal Symmetry
A Lesson from Archaeology
horizontal symmetry
¡
residual symmetry
Lam Construction and Destruction of Horizontal 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
Lam Construction and Destruction of Horizontal 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
Lam Construction and Destruction of Horizontal 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
Lam Construction and Destruction of Horizontal 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
Lam Construction and Destruction of Horizontal 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
Lam Construction and Destruction of Horizontal 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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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√2 0
−1√2√3
−1√2 −
√3
(TBM)
TBM gives G1,G2,G3
if θ13 6= 0, assume G2 the same(accidental symmetry stillallowed)
(or, Me slightly non-diagonal)
Lam Construction and Destruction of Horizontal Symmetry
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√2 0
−1√2√3
−1√2 −
√3
(TBM)
TBM gives G1,G2,G3
if θ13 6= 0, assume G2 the same(accidental symmetry stillallowed)
(or, Me slightly non-diagonal)
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
Constructing the Unbroken Symmetry GL
{F ,G} ⊂ GLif 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
Lam Construction and Destruction of Horizontal Symmetry
Constructing the Unbroken Symmetry GL
{F ,G} ⊂ GLif 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
Lam Construction and Destruction of Horizontal Symmetry
Constructing the Unbroken Symmetry GL
{F ,G} ⊂ GLif 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
Lam Construction and Destruction of Horizontal Symmetry
CONTENTS H = RφL+ V (φ) G → G ′
Assumption: L = 3, 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
unbroken symmetryGL = {G′L
↑, G′L
↓}
G′↑, G′↓ from G & V (φ)
what φ’s?
demands a generic V •
leptons, quarks, quarks and leptons
Lam Construction and Destruction of Horizontal Symmetry
CONTENTS H = RφL+ V (φ) G → G ′
Assumption: L = 3, 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
unbroken symmetryGL = {G′L
↑, G′L
↓}
G′↑, G′↓ from G & V (φ)
what φ’s?
demands a generic V •
leptons, quarks, quarks and leptons
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
V (φ) for G = A4 = {F ,G2} f RφL+ hc
φ taken from 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 required
two-phase strategyb
a ‘generic’ (U(1)-inv) potentialV (φ3, φ1, φ1′ , φ1′′)
similar procedure works forother G
aAltarelli & Feruglio, NPB 741 (2006) 215bLam, PRD 83 (2011) 113002
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
−to build a model, need to specify howR transforms under A4. Not unique
−A4 promotes to S4 if f1′ = f1′′
Lam Construction and Destruction of Horizontal Symmetry
V (φ) for G = A4 = {F ,G2} f RφL+ hc
φ taken from 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 required
two-phase strategyb
a ‘generic’ (U(1)-inv) potentialV (φ3, φ1, φ1′ , φ1′′)
similar procedure works forother G
aAltarelli & Feruglio, NPB 741 (2006) 215bLam, PRD 83 (2011) 113002
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
−to build a model, need to specify howR transforms under A4. Not unique
−A4 promotes to S4 if f1′ = f1′′
Lam Construction and Destruction of Horizontal Symmetry
Two-Phase Analogy Water and Ice
H2O
0C < T < 100C T < 0C
Lam Construction and Destruction of Horizontal 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)
Lam Construction and Destruction of Horizontal Symmetry
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?)
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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
Lam Construction and Destruction of Horizontal Symmetry
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′
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′
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
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 •
Lam Construction and Destruction of Horizontal Symmetry