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Neutrino masses and lepton asymmetry in the 3-3-1 model with right-handed neutrinos

N. T. ThuyYonsei University

Jindo Workshop, Sep 20~23, 2012

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Outline

I. IntroductionII. A review of the modelIII. Higgs sextet and neutrino massesIV. Lepton asymmetryV. Summary

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I. Introduction

• Traditionally, the 3-3-1 model with right-handed neutrinos works with the three Higgs triplets. At the tree level the neutrino mass spectrum is degenerated with the mass 0, -m, m. -> This spectrum is not realistic.

By introducing an addition Higg sextet, the neutrino mass problem can be solved.

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• All cosmological models agree that the Universe started with the same amount of baryon and anti-baryon. However from Big-Bang Nucleosynthesis (BBN) and WMAP ,

where are the number densities of baryons and antibaryons. ->The baryon asymmetry must be generated dynamically.• Sakharov: requires Baryon number violation C and CP violation Out of equilibrium dynamics• Consider The in the quark mixing matrix, leads to much too small

a asymmetry -> We study lepton asymmetry.

B B B Bn n n n= Þ ?

CPCP

B B-

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II. A review of the model The particle content in this model is given as follow:

where a= 1, 2, 3 and = 2, 3 are generation indices. The electric charges of U (D) are the same as of u (d).

a

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The three Higgs with VEVs are introduced:

01

203

102

301

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0(3,3, 1/ 3), 0 ,

/ 20

(3,3, 2/ 3), / 2 ,0

(3,3, 1/ 3

v

cc c c

c w

rr r r

r

hh h

h

-

+

+

-

æ öæ ö ÷ç÷ç ÷ç÷ç ÷÷ çç ÷÷ ç ÷ç= - =÷ ç ÷ç ÷ ç ÷ç ÷ ç ÷ç ÷ ÷ç÷ç ÷ ÷ç ç ÷çè ø è øæ öæ ö ÷÷ çç ÷÷ çç ÷÷ çç ÷÷ çç ÷= + =÷ çç ÷÷ ç ÷ç ÷ ç ÷ç ÷ ÷ç÷ç ÷ç ÷çè ø è ø

æ ö÷ç ÷ç ÷ç ÷ç= -÷ç ÷ç ÷ç ÷÷ç ÷çè ø

:

:

:/ 2

), 0 .0

uh

æ ö÷ç ÷ç ÷ç ÷ç ÷=ç ÷ç ÷ç ÷÷ç ÷çè ø

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The Yukawa interactions can be written in the most general form as follows

where the subscripts LNC and LNV, respectively indicate to the lepton number conserving and lepton number violating ones.

where a, b, and c stand for SU(3) L indices.

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where

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The gauge group is broken via two steps:

The VEV w is responsible for the first step of symmetry breaking giving mass for the exotic quarks, while the second step is due to u and v giving mass for the usual quarks and all ordinary leptons. Therefore, they have to satisfy the constraints:

,(3) (1) (2) (1) (1)w vuL X L Y QSU U SU U UÄ ¾¾® Ä ¾¾¾®

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III. Higgs sextet and neutrino masses The Lagrangian for the tree-level neutrino masses is obtained as

where

The tree-level spectrum consists of only Dirac neutrinos with one particle massless and two others degenerate in mass: 0, -m, m.

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• The sextet Higgs is introduced to solve neutrino masses problem, defined by

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• Impose that the active neutrinos gain mass via a type II seesaw:

• It turns out that the neutrino masses in this model naturally small because of suppression of a large scale and a small constrained from the -parameter, where

• The sterile neutrinos have large masses in the scale:

• Taking we get

Lk r

L2 2

11 2

22 ,Mv f vM

M

n

L L: :

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IV. Lepton asymmetry• There must have the non conservation of the lepton

number. Any lepton asymmetry generated by A would be neutralized by the decays of A*, unless CP conservation is also violated and the decays are out of thermal equilibrium in the universe.

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• The most general potential for the three Higgs triplets and sextet which are invariant under the is given as

V=

+

(3) (3) (1)C L XSU SU UÄ Ä

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• Under the discrete symmetry the potential can be written as following:

, ,i i S iSc c r r® ® ®

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The constraints of the parameters can be calculated from the minimum condition of the potential. We get

11 15 17 0,l l l= = =

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We assume that there is no hierarchy between the couplings of Higgs sextet with Higgs triplets, .

On the other hand, S gives mass to right handed neutrinos, so the Higgs sextet is much heavier than the three Higgs triplets.

=> We can ignore the mixing of Higgs sextet with Higgs triplets.

1 0, 2,3,4,6i il ® =

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• The relevant Lagrangian is

If we assign the lepton number L=0 to the sextet S, the Yukawa couplings explicitly break L.

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• We consider neutrinos as Majorana particles

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The heavy particles in this model are N and S.• At tree-level,

• Contributions to the CP violation is due to the interference between the tree-level and the one-loop diagrams.

• The asymmetry can be generated through the decays

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Example of Feynman diagrams

In general, the CP asymmetry is given by

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What are we going to do?• We are going to calculate the asymmetry of all possible

channels of N and S.• Whether the decay of N or S is important for the

asymmetry. • We can calculate the baryon asymmetry after we get the

lepton asymmetry.• Another idea: can be consider as inflaton with

the potential

• All the above works can be considered in the economical 331 model.

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• This idea is from Chia-Min Lin: Phy. Lett B 659 (2011)

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V. Summary

• Neutrino masses problem has been solved by introducing one sextet S in 3-3-1 model with right handed neutrinos. In our model, Majorana neutrinos gain mass via like-type II seesaw.

• The asymmetry can be generated through the decays of N and S.

• Work is in process

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Thanks for Your Attentions!