In collaboration withJ.-M. Frere (ULB), F.-S. Ling (ULB),E. Nugaev (INR), S. Troitsky (INR)

Flavour puzzle orWhy Neutrinos Are Different?

Maxim LibanovINR RAS, Moscow

Dual year Russia-Spain,11--11--11

Flavour Puzzle Maxim Libanov

The Flavour Puzzle In A Nutshell

XWhy three families in the SM?

Hierarchical masses + small mixing angles

XWhy massive neutrinos?

Tiny masses + two large mixing angles

XWhy very suppressed FCNC?

Strong limits on a TeV scale extension of the SM

Proposed solution:A model of family replication in 6D

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3 Families from 1 Maxim Libanov

3 Families In 4D From 1 Family In 6D

Core of the vortex --

Our 3D World

ϕ

r

Our 3D World is a core of Abrikosov-Nielsen-Olesen vortex:

scalarΦ(r,ϕ) = F(r)eiϕ

Ug(1)gaugefieldAr = A(r)

2

3 Families from 1 Maxim Libanov

We do not consider gravity. Extra dimensions can be flat (and infinitelylarge?) or compact, e.g., spherical.

◦ A vortex on a sphere is in fact like a magnetic monopole configurationin 3D

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3 Families from 1 Maxim Libanov

6D FermionIn general, 6D spinor contains 4 Weyl spinors

Ψ =

ψ+R

ψ+L

ψ−L

ψ−R

where

± ⇔1±Γ7

2

L, R⇔ Left, Right 4D chirality

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3 Families from 1 Maxim Libanov

Fermion zero modes.Chiral fermionic zero modes are trapped in the core due to specificinteraction with the A and Φ. Specific choise of Ug(1) fermionic gaugecharges ⇒

axial(3,0) ⇒ Φ3Ψ1−Γ7

2Ψ ⇒ 3 L zero modes

0ψ+L

ψ−L

0

axial(0,3) ⇒ Φ3Ψ1+Γ7

2Ψ ⇒ 3 R zero modes

ψ+R

00ψ−R

Zero modes⇔ three 4D fermionic families

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3 Families from 1 Maxim Libanov

Field Content

Fields Profiles Charges RepresentationsUg(1) UY(1) SUW(2) SUC(3)

scalar Φ F(r)eiϕ +1 0 1 1F(0) = 0, F(∞) = v

vector Aϕ A(r)/e 0 0 0 0A(0) = 0, A(∞) = 1

scalar X X(r) +1 0 1 1X(0) = vX, X(∞) = 0

scalar H H(r) --1 +1/2 2 1Hi(0) = δ2ivH, Hi(∞) = 0

fermion Q 3 L zero modes axial (3, 0) +1/6 2 3fermion U 3 R zero modes axial (0, 3) +2/3 1 3fermion D 3 R zero modes axial (0, 3) −1/3 1 3fermion L 3 L zero modes axial (3, 0) −1/2 2 1fermion E 3 R zero modes axial (0, 3) −1 1 1fermion N Kaluza-Klein 0 0 1 1

spectrum

r

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3 Families from 1. Hierarchy. Maxim Libanov

4D Fermion masses. Hierarchy.

r

ei·0·ϕei·1·ϕ

ei·2·ϕ

H(r) F(r)(Φ = F(r)eiϕ)

3 zero modes have different shapes, anddifferent angular momenta n = 1, 2, 3

JΨn ≡ −(i∂ϕ + 3

1 + Γ7

2

)Ψn = (n − 1)Ψn

Ψn(r→ 0) ∼ rn−1

Explicitly

Qn ∼

0

e−iϕ(n−3)f2(n, r)qn(xµ)e−iϕ(n−1)f3(n, r)qn(xµ)

0

Dm ∼

e−iϕ(m−1)f3(m, r)dm(xµ)

00

e−iϕ(m−3)f2(m, r)dm(xµ)

←

ψ+R

ψ+L

ψ−L

ψ−R

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3 Families from 1. Hierarchy. Maxim Libanov

r

ei·0·ϕei·1·ϕ

ei·2·ϕ

H(r) F(r)(Φ = F(r)eiϕ)

Allowed by symmetries couplings

YX · HXQ1 − Γ7

2D + YΦ · HΦQ

1 − Γ7

2D

mnm ∝2π∫0

dϕR∫

0

drQnDmHX(or F(r)eiϕ)

∼ σ2n(−1) · δ(n − m( + 1)) ∼

σ4 σ3

σ2 σ1

1

Q∗n ∼

0

eiϕ(n−3)f2(n, r)qn(xµ)eiϕ(n−1)f3(n, r)qn(xµ)

0

Dm ∼

e−iϕ(m−1)f3(m, r)dm(xµ)

00

e−iϕ(m−3)f2(m, r)dm(xµ)

←

ψ+R

ψ+L

ψ−L

ψ−R

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3 Families from 1. Hierarchy. Maxim Libanov

r

ei·0·ϕei·1·ϕ

ei·2·ϕ

H(r) F(r)(Φ = F(r)eiϕ)

σ depends on the parameters of themodel. Hierarchy arises at σ ∼ 0.1

m2 : m1 : m0 ∼ σ4 : σ2 : 1 ∼ 10−4 : 10−2 : 1

UCKM∼

1 σ σ4

σ 1 σσ2 σ 1

Generation number⇔ Angular momentum

X The scheme is very constrained, as the profiles are dictated by the equations

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Why Neutrinos Are Different? Maxim Libanov

Neutrinos masses. Why is it different?N -- additional neutral spinor

⇒ Free propagating in the extra dim (up to dist. R ∼ (10 ÷ 100TeV)−1).⇒ Majorano-like 6D mass term

M2NcN + h.c.

⇒ Kaluza-Klein tower in 4D (no zero mode)⇒ Effective 6D couplings with leptons allowed by symmetries∑

S+

HS+L1 + Γ7

2N +

∑S−

HS−L1 − Γ7

2N + h.c.

S+ = X∗, Φ∗, X∗2Φ, . . .S− = X2, XΦ, Φ2, . . .

Non-zero windings ⇒more composite structure ofthe mass matrix

⇒ 4D Majorano neutrinos masses are generated by See-saw mechanism

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Why Neutrinos Are Different? Maxim Libanov

Neutrinos:

mνmn ∼

2π∫0

dϕR∫

0

drF(r,ϕ) ·[LcL ∝ LL

]

∼2π∫0

dϕei(4−n−m+...)ϕ ∼ δ(4 − m − n + . . .)

∼

· · 1· σ2 ·1 · ·

0eiϕ(3−n)

eiϕ(1−n)

0

0

eiϕ(3−m)

eiϕ(1−m)

0

←ψ+R

ψ+L

ψ−L

ψ−R

→

Charged fermions:

mchargedmn ∼

2π∫0

dϕR∫

0

drF(r,ϕ) ·[ΨΨ ∝ Ψ∗Ψ

]

∼2π∫0

dϕei(n−m+...)ϕ ∼ δ(n − m − . . .)

∼

σ4 · ·· σ2 ·· · 1

0eiϕ(3−n)

eiϕ(1−n)

0

e−iϕ(1−m)

00

e−iϕ(3−m)

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Why Neutrinos Are Different? Maxim Libanov

Neutrinos:

mνmn ∼

· · 1· σ2 ·1 · ·

U†νmνU

∗ν ∼ diag(−m,m, mσ2)

Uν ∼

1/√2 1/√2 σ

σ σ 1−1/√2 1/√2 σ

Charged fermions:

mchargedmn ∼

σ4 · ·· σ2 ·· · 1

mdiag

charged ∼ diag(µσ4, µσ2, µ)

UCKM∼

1 σ σ4

σ 1 σσ2 σ 1

m�µ due to See-Saw

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Why Neutrinos Are Different? Maxim Libanov

Consequences of this structure

mν ∼

· · 1· σ2 ·1 · ·

mdiagν ∼

−m 0 00 m 00 0 mσ2

Inverted hierarchy:

∆m2� = ∆m2

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∆m212

∆m213∼ σ2 ∼ 10−2

Solar angle automaticaly large

Small reactor angle

Ue3 ∼ σ ∼ 0.1

Uν ∼

1/√2 1/√2 σ

σ σ 1−1/√2 1/√2 σ

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Why Neutrinos Are Different? Maxim Libanov

Consequences of this structure

mν ∼

· · 1· σ2 ·1 · ·

mdiagν ∼

−m 0 00 m 00 0 mσ2

Pseudo-Dirac structure ⇒The − sign may be absorbed inthe mixing matrix, but contributesdestructively to the effective massfor neutrinoless double beta decay(Pseudo-Dirac structure when fullCabibbo-like mixing is introduced)0νββ decay

partial suppression

|〈mββ〉| '13

√∆m2⊕

•

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Why Neutrinos Are Different? Maxim Libanov

Semi-realistic numerical example

mdiagν =

−50.03 0 0

0 50.79 00 0 0.7089

[meV], UMNS =

0.808 0.559 0.186−0.286 0.660 −0.693−0.514 0.502 0.696

∆m2

12 = 7.63 × 10−5eV2

∆m213 = 2.50 × 10−3eV2 =⇒

∆m212

∆m213

= 3.05%

tan2 θ12 = 0.471(0.47+0.14

−0.10

)tan2 θ23 = 0.997

(0.9+1.0

−0.4

)sin2 θ13 = 3.46 · 10−2 (≤ 0.036)

Consequence for 0νββ decay

|〈mββ〉| =∣∣∣∑

imiU2

ei

∣∣∣ = 17.0 meV

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FCNC Maxim Libanov

Flavour Violation

Like in the UED, vector bosons can travel in the bulk of space. From the4D point of view:

1 massless vector boson in 6D=

1 massless vector bozon (zero mode)⇔ SM boson

+ KK tower of massive vector bosons Mn ∼nR

⇒ FCNC

+ KK tower of massive scalar bosons in 4D

⇒ KK scalar modes do not interact with fermion zeromodes

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FCNC Maxim Libanov

KK vector modes carry angular momentum = family number. In the absenceof fermion mixings, family number is an exactly conserved quantity ⇒processes with ∆G = ∆J ≠ 0 are suppressed by mixing.

K0L → µ±e∓

∆G = ∆J = 0

d

s

Jds = 1

Jµe = 1

∼κ2

M2Z′

JZ′ = 1

µ

e

µ→ eee

∆G = ∆J = 1

µ

JZ′ = 0

Jµ = 1

Je = 2

e

e

∼ σκ2

M2Z′

σ

Jee = 0

e

X κ = 1 for the particular model, but may be � 1 for extensions

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FCNC Maxim Libanov

Rare (forbidden) processes:

∆G = 0: K0L → µe, K+→ π+µ+e− P ∼ σ0κ4/M4

Z′

Bs→ µτ, Bd→ eτ∆G = 1: µ→ eee, µe-conversion, µ→ eγ P ∼ σ2κ4/M4

Z′

∆G = 2: mass difference KL − KS, CP-violation P ∼ σ4κ4/M4Z′

Bound onMZ′ & κ · 100 · TeV

A clear signature of the model would be an observation

K0L → µe

without observation other FCNC-processes at the sameprecision level

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FCNC Maxim Libanov

1 2 3 4 5M,TeV

0.050.1

0.51

510

Num

berof

even

ts

1

MZ′ = κ · 100TeV

L = 100 fb−1 √s = 14TeV

K0L → µe

forbidden

LHC beats fixed target

Tevatronlimit

Search at LHC

Search for an «ordinary»massive Z′(W′, g′, γ′)

Search for pp→ µ+e− + . . .

Search for pp→ µ−e+ + . . . ---one order below due to quarkcontent of protons

Search for pp→ t + c + . . . orpp→ b + s + . . . --- expect a few1000’s events, but must considerbackground!

LHC thus has the potential (in a specificmodel) to beat even the very sensitivefixed target K→ µe limit!

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Flavour puzzle or Why Neutrinos Are Different? Maxim LibanovConclusions

Family replication model in 6D: elegant solution to the flavour puzzleHierarchical Dirac masses + small mixing anglesNeutrinos are different: See-saw + Majorano-like mass for the bulkneutral fermion can fit neutrino dataFamily/lepton number violating FCNC suppressed by small fermionmixings

Predictions for neutrinosInverted hierarchyReactor angle ∼ 0.1Partially suppressed neutrinoless ββ decay

Other predictionsK→ µe will show up earlier than other FCNC-processesMassive gauge bosons with mass ∼ TeV or higherSearch for pp→ µ+e− at LHC can beat fixed targetConstraint on B-E-H (Higgs) boson: should be LIGHT

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