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Leptogenesis in the E6SSM: Flavour Dependent LeptonAsymmetries

S. F. King∗, R. Luo†, D. J. Miller† and R. Nevzorov†

∗School of Physics and Astronomy, University of Southampton, Southampton, SO17 1BJ, U.K.†Department of Physics and Astronomy, University of Glasgow, Glasgow, G12 8QQ, U.K.

Abstract. We discuss flavour dependent lepton asymmetries in the Exceptional Supersymmetric Standard Model (E6SSM). Inthe E6SSM, the right-handed neutrinos do not participate in gaugeinteractions, and they decay into leptons and leptoquarks.Their Majorana nature allows violation of lepton number. New particles and interactions can result in substantial leptonasymmetries, even for scales as low as 106 GeV.

Keywords: neutrino, leptogenesis, supersymmetryPACS: 98.80.Cq; 13.15.+g; 13.35.Hb; 11.30.Pb; 12.10.-g

INTRODUCTION

The generation of the Baryon Asymmetry of the Uni-verse (BAU) is a long-standing problem of particle cos-mology. One plausible solution is thermal leptogenesis,where the Majorana right-handed (RH) neutrinos de-cay out-of-equilibrium into left-handed lepton doublets.CP violation in these decays leads to a non-zero leptonasymmetry as the universe cools down.Sphaleron pro-cesses convert part of this lepton number asymmetry intoa baryon asymmetry in the universe before nucleosynthe-sis.

However, in the seesaw scenario, the Yukawa cou-plings of RH neutrinos influence the light neutrinomasses, constraining the generation of lepton asymme-try. In order to generate a sufficient lepton asymmetryin the Standard Model (SM) or the Minimal Supersym-metric Standard Model (MSSM), the mass of the lightestRH neutrino,M1, should be 109GeV or larger. On theother hand, the reheating temperatureTR should be of or-der ofM1 because RH neutrinos are produced thermally.The reheating temperature is constrained by the produc-tion of gravitinos in supergravity models, which requiresTR < 107GeV. This tension in the choice of the reheat-ing temperature is known as the gravitino overproductionproblem. Therefore it is interesting to consider leptogen-esis in extended supersymmetric models; here we inves-tigate the E6SSM [1].

E6SSM

The E6SSM [2][3] is based on gauge groupSU(3)c ×SU(2)W ×U(1)Y ×U(1)N at low energies, which canoriginate from the breaking of anE6 GUT theory. The

extraU(1)N is a combination ofU(1)χ andU(1)ψ

U(1)N =14

U(1)χ +

√154

U(1)ψ , (1)

whereU(1)χ andU(1)ψ appear from the decompositionof SO(10) and E6 respectively. RH neutrinos are sin-glets with respect toU(1)N and therefore may gain largemasses. The particle content of E6SSM includes threefundamental representations of E6. Each 27-plet containsa generation of SM leptons and quarks, the RH neutrinoNi, two SU(2) doubletsH1,i H2,i with quantum numbersof Higgs fields, a singlet field under the SM gauge groupsSi and triplets of exotic quarksDi andDi, wherei=1,2,3is the generation index. Here, the third generation ofSU(2) doubletsHd ≡ H1,3 andHu ≡ H2,3 play the roleof the down-type and up-type Higgs fields respectively.The exotic quarks can be either diquarks, carrying 2/3baryon number or leptoquarks carrying 1 lepton numberand 1/3 baryon number (refered to as model I and modelII respectively). The anomalies are canceled within eachgeneration. In addition, extra doublets (only one gener-ation) L4 and L4 with mass∼ 1TeV are introduced inorder to achieve gauge unification.L4 effectively car-ries one lepton number and couples to SM leptons viaYukawa couplings. In order to suppress Flavour Chang-ing Neutral Currents (FCNC) processes, aZH

2 symmetryis postulated, under whichH1,3, H2,3, S3 are even and allother fields are odd. The E6superpotential invariant undera ZH

2 symmetry is,

WE6SSM = λ S(HuHd)+λαβ S(H1αH2β )+κi jS(DiD j)

+ fαβ (HdH2α)Sβ + fαβ (H1α Hu)Sβ

+hUi j(HuQi)u

cj + hD

i j(HdQi)dcj

+hEi j(HdLi)e

cj + hN

i j(HuLi)Ncj

+12

Mi jNci Nc

j + µ ′(L4L4)

+hE4 j(HdL4)e

cj + hN

4 j(HuL4)Ncj . (2)

The ZH2 symmetry can only be approximate in order

to allow exotic quarks to decay; couplings which breakZH

2 should be suppressed. ViaZH2 symmetry breaking,

the RH neutrinos couple to all leptons andH2,i of allgenerations. In model II, couplings of RH neutrinos toleptoquarks and down-type quarks (three generations)are also allowed. The corresponding contribution to thesuperpotential reads:

WN = hNkx j(H

uk Lx)N

cj + gN

ki jDkdci Nc

j (3)

wherex=1,2,3,4 andk, i, j = 1,2,3, andHuk ≡H2,k with

hN3i j ≡ hN

i j andhN34j ≡ hN

4 j. The first term only is present inmodel I while both terms are present in model II.

LEPTON ASYMMETRIES

The first step in the study of leptogenesis is the calcula-tion of the lepton asymmetries. Let’s consider the situ-ation where theZH

2 symmetry is conserved. In this sce-nario, the RH neutrinos couple to the first three gener-ations of leptons andL4 with Hu. The decay channelsinclude

N1 → Lx +Huk , N1 → Lx + Hu

k ,

N1 → Lx + Hu

k , N1 → Lx +Huk , (4)

wherek = 3 only. The flavour dependent lepton asym-metries originate from the interference of tree-level de-cay amplitudes and one-loop corrections. Calculating theone-loop diagrams we find

ε31, ℓx

≃− 38π ∑

j=2,3

Im

[(hN†hN)1 jhN∗

x1 hNx j

]

(hN†hN)11

M1

M j, (5)

whereMi=1,2,3 are the masses of RH neutrinos. We as-sumeM1 ≪ M2,3 for all scenarios in this contribution.

When the effect ofZH2 symmetry breaking is consid-

ered, we should include the first and second generationsof Hu

k in the final states. The possible RH (s)neutrinosdecay channels are given by Eq.(4) withk = 1,2,3, re-sulting in,

εk1, ℓx

=1

8πA1∑

j=2,3Im

{2A jh

N∗kx1hN

kx jM1

M j

+∑m,y

hN∗my1hN

mx jhNky jh

N∗kx1

M1

M j

}, (6)

where A j = ∑m,y hN∗my1hN

my j and x,y = 1,2,3,4, i,k =1,2,3.

It is convenient to define thetotal lepton asymmetriesassociated with each lepton flavour,

εtot1, f = ∑

k

εk1, f . (7)

In model II, leptoquarks may also contribute to thegeneration of lepton asymmetries. The additional RHneutrino decays are,

N1 → Dk + dci, N1 → Dk + dc

i

N1 → Dk + di, N1 → Dk + dci . (8)

Notice that no baryon number is generated due to thepresence ofDk and dc

i in the final state, so only thelepton number violation need be considered. The leptonasymmetries with exotic quarks in the final states are,

ε i1,Dk

=1

8πA0∑

j=2,3

Im

{2A jg

Nki jg

N∗ki1

M1

M j

+∑m,n

gN∗mn1gN

mi jgNkn jg

N∗ki1

M1

M j

}, (9)

where

A j = A j +32 ∑m,n gN∗

mn1gNmn j , (10)

A0 = ∑k, i gNki1gN∗

ki1 . (11)

The new coupling of RH neutrinos and leptoquarks alsocontribute toεk

1, ℓx. This can be described by Eq.(6), but

with A2, A3 replaced byA2, A3 separately.To illustrate how lepton asymmetries can be enhanced

by these new couplings, we consider the generation oflepton decay asymmetries within see-saw models withsequential dominance [4]. The RH neutrino mass matrixand the Yukawa coupling matrix can be written as,

MN =

MA 0 00 MB 00 0 MC

, hN =

A1 B1 C1A2 B2 C2A3 B3 C3

.

(12)

According to sequential dominance|AiA j |

MA≫ |BiB j |

MB≫

|CiC j |MC

and|A1| ≪ |A2,3|. The left-handed neutrino massescan be obtained by diagonalizing the effective mass ma-trix in the basis of(νi, Ni)

M =

(0 hN vu

hN† vu MN

), (13)

which gives,

m3 ≃(|A2|2+ |A3|2)v2

u

MA, m2 ≃

|B1|2v2u

s212MB

, m1 ≃ 0,

(14)

-20-18-16-14-12-10-8-6-4-2 0 2

log

|hN H

u 3L4N

2|

log

|ε3 1,

µ| =

log

|ε3 1,

τ|

-5 -4 -3 -2 -1 0log |hN

Hu3L4N1

|

-5

-4

-3

-2

-1

0

1lo

g |h

N Hu 3L

4N2|

log

|ε3 1,

µ| =

log

|ε3 1,

τ|

-6

-8

-10

-12

-14

log |hNHu

3L4N1|

-20

-18

-16

-14

-12

-10

-8

-6

-4

-2

0

2

log

|hN H

u 3L4N

2|

log

|ε3 1,

L 4 |

-5 -4 -3 -2 -1 0log |hN

Hu3L4N1

|

-5

-4

-3

-2

-1

0

1

log

|hN H

u 3L4N

2|

log

|ε3 1,

L 4 |-4

-6

-8

-10

-2

-12

0

log |hNHu

3L4N1|

FIGURE 1. Logarithm of the maximal values of|ε31,µ | =

|ε31,τ | (left) and |ε3

1,L4| (right) with unbrokenZH

2 symmetry

versuslog|hNHu

3 L4N1| and log|hN

Hu3 L4N2

| for M1 = 106 GeV andM2 = 10·M1 .

-20-18-16-14-12-10-8-6-4-2 0 2

log

|hN H

u 2L3N

2|

log

|εto

t1,

µ|

-5 -4 -3 -2 -1 0log |hN

Hu2L3N1

|

-5

-4

-3

-2

-1

0

1

log

|hN H

u 2L3N

2|

log

|εto

t1,

µ|

-6

-8

-10

-12

-14

-16

log |hNHu

2L3N1|

-20-18-16-14-12-10-8-6-4-2 0 2

log

|hN H

u 2L3N

2|

log

|εto

t1,

τ|

-5 -4 -3 -2 -1 0log |hN

Hu2L3N1

|

-5

-4

-3

-2

-1

0

1

log

|hN H

u 2L3N

2|

log

|εto

t1,

τ|

0

-2

-4

-6

-8

-10

log |hNHu

2L3N1|

FIGURE 2. Logarithm of the maximal values of|ε tot1,µ | (left)

and|ε tot1,τ | (right) with brokenZH

2 symmetry versuslog|hNHu

2 L3N1|

andlog|hNHu

2 L3N2| for M1 = 106 GeV andM2 = 10·M1 .

wherevu is the vev of Higgs fieldHu ands12 ≡ sinθ12is the sine of the 1-2 neutrino mixing angle. We furtherassume thatA1 = 0, A2 = −A3 = |A|eiφA andB1 = B2 =B3 = |B|eiφB as in Constrained Sequential Dominance.Note thatMA,B,C ≃ M1,2,3, the RH neutrino masses, sincediagonalizing the mass matrix gives a tiny contribution tothe RH neutrino masses. Using the light neutrino massesm2 ≃ 8.7×10−3eV, m3 ≃ 4.9×10−2eV as in the case ofa normal hierarchy, the relations outlined above provideall relevant Yukawa couplings, dependent only on the RHneutrino masses.

We plot the maximal lepton decay asymmetries inFigs.(1-3). We adopt a notation where couplings of theform hN

Huk LxN j

= hNkx j andgN

D3d3Ni= g33i. We estimate that

in order to generate the observed baryon asymmetry withefficiency factor ofη ∼ 0.1, we require lepton decaysasymmetries of order 10−5.

In Fig.(1) we restrict our consideration to theZH2 sym-

metric case and ignore the couplings of the heaviest RHneutrino. One can see that a substantial CP asymme-try associated withL4 can be achieved even forM1 =106GeV if hN

Hu3L4N2

is of the order 10−1 – 10−2.

Figs.(2) and (3) illustrate scenarios with brokenZH2

symmetry. In this case, for simplicity, we ignore theYukawa couplings ofL4 to the RH neutrino. In Fig.(2) weexamine the dependence of the maximal CP asymmetry

-20-18-16-14-12-10-8-6-4-2 0 2

log

|gN D

3d3N

2|

log

|ε3 1,

µ|

-5 -4 -3 -2 -1 0log |gN

D3d3N1|

-5

-4

-3

-2

-1

0

1

log

|gN D

3d3N

2|

log

|ε3 1,

µ|

-2

-4

-6

-8

-10

log |gND3d3N1

|

-20-18-16-14-12-10-8-6-4-2 0 2

log

|gN D

3d3N

2|

log

|ε3 1,

D3|

-5 -4 -3 -2 -1 0log |gN

D3d3N1|

-5

-4

-3

-2

-1

0

1

log

|gN D

3d3N

2|

log

|ε3 1,

D3|

-2

-4

-6

-8

-10

0

log |gND3d3N1

|

FIGURE 3. Logarithm of the maximal values of|ε31,µ | =

|ε31,τ | (left) and|ε3

1,D3| (right) and with unbrokenZH

2 symmetry

versuslog|gND3d3N1

| and log|gND3d3N2

| for M1 = 106 GeV andM2 = 10·M1 .

in Model I, assuming that onlyhNHu

2L3N1andhN

Hu2L3N2

havenon-zero values. Once again we see a substantial CPasymmetry can be generated ifhN

Hu2L3N2

is of the order

10−1 – 10−2.Finally in Fig.(3) we consider the generation of CP

asymmetries in model II. For simplicity, we ignore thecouplings of all exotic particles except those of the thirdgeneration of leptoquarksD3, which predominantly cou-ples to the RHb-quarkd3 and the two lightest RH neutri-nos,N1 andN2. As in the previous two cases, an appro-priate amount of CP asymmetry can be generated by thecouplings of the exotic quarks.

CONCLUSIONS

We calculate the flavour dependent lepton asymmetriesin the E6SSM. New particles and new sources of CP vio-lation may result in the drastic enhancement of lepton de-cay asymmetries. We demostrate how the new couplingsincrease lepton CP asymmetries within seesaw modelswith Sequential Dominance. Successful leptogenesis canbe achieved for a wide range of parameters. To providea quantification of the resulting baryon asymmetry re-quires detailed study of the complete system of flavourdependent Boltzmann equations.

REFERENCES

1. S. F. King, R. Luo, D. J. Miller, and R. Nevrozov (2008),0806.0330.

2. S. F. King, S. Moretti, and R. Nevzorov,Phys. Lett. B634,278–284 (2006),hep-ph/0511256.

3. S. F. King, S. Moretti, and R. Nevzorov,Phys. Rev. D73,035009 (2006),hep-ph/0510419.

4. S. F. King, Nucl. Phys. B786, 52–83 (2007),hep-ph/0610239.