Neutrino oscillations and Leptogenesis

Neutrino oscillations and Leptogenesis Anupam Yadav*, Sabeeha Naaz*, Jyotsna Singh* and R.B. Singh* arXiv:1901.06127v2 [hep-ph] 8 Apr 2019 Abstract : The symmetry breaking of left right symmetric model around few TeV range permits the existence of massive right handed neutrinos or gauge bosons. In this work the decay of lightest right handed neutrinos in a class of minimal left right symmetric model is analysed for the generation of adequate lepton asymmetry. An analytical expression for the lepton asymmetry is developed and the Boltzmann equation are solved. The effect of decay parameter and efficiency factor on the generation of leptogenesis is checked. Lower bound on right handed neutrino mass is imposed. The electroweak sphaleron processes converts the induced lepton asymmetry into baryon asymmetry. In an attempt to achieve the required baryogenesis, we have imposed certain constrains on the parameter space corresponding to low energy neutrino oscillation parameters (especially θ13 ) and the three phases ( CP, majorana and higher energy phase). Key words: Leptogenesis, Baryogenesis, Majorana Neutrino, Efficiency factor, Sphaleron Processes, CP Violation, Minimal left right Supersymmetric Model. 1 Introduction Standard model fails to explain the origin of baryonic asymmetry present in the observable universe. The Big-Bang nucleosynthesis shows that the density of baryons compared to that of photons in the universe is very low; η ≡ nB /nγ = (2.6 − 6.3) × 10−10. This ratio η can be related to the observed matter-antimatter or baryon-antibaryon asymmetry of the universe as, YB ≈ nB /s ≈ η /7.04 = (3.7 − 8.9) × 10−11 [1], where s denotes the entropy density. In order to produce a net baryon asymmetry in the standard Big-Bang model, three Sakharov necessary conditions must be satisfied :- (a) Baryon number violation, (b) C and CP violation, and (c) The departure from thermal equilibrium [2] [3]. The departure from thermal equilibrium has to be permanent else the baryon asymmetry will be washed out if the thermal equilibrium is restored. The first two conditions can be investigated only after a particle physics model is specified, whereas the third condition can be discussed in a more general way. Among several interesting and viable baryogenesis scenarios proposed in the literature, Fukugita and Yanagida’s leptogenesis mechanism has attracted a lot of attention due to the fact that neutrino physics is entering a flourishing era [4][5]. Further, amongst numerous neutrino mass models under consideration, seesaw mechanism appears as the most attractive model [6] for generation of the tiny neutrino masses reported by the neutrino oscillation experiments [7] [8]. The seesaw mechanism introduces a massive right handed neutrinos to the model. Since the right handed neutrinos are very heavy, usually in neutrino oscillation experiments, only the left handed neutrino mass or the low energy effective theory is considered for the study of neutrinos [9]. A more careful analysis of seesaw mechanism reveals that if the massive right handed neutrinos are completely ignored in the neutrino analysis or when seesaw mechanism is implemented in context of low energy effective theory, one can miss many essential ingredients of seesaw model required to explain the leptogenesis [10][11][12][13]. In general, without loss of generality one can work with both masses (tiny and massive), in a basis where the charged lepton mass matrix and right-handed neutrino Majorana mass matrix are diagAnupam Yadav*, Sabeeha Naaz*, Dr. Jyotsna Singh* and Dr. R.B. Singh* *Department of Physics, University of Lucknow, Lucknow-226007,India e-mail: [email protected],sabeehanaaz0786@gmail. 2 Anupam Yadav*, Sabeeha Naaz*, Jyotsna Singh* and R.B. Singh* onal with real eigenvalues [14]. In this case, there will be a total of 18 parameters in neutrino sector and the lepton asymmetry will depend on all of these 18 parameters, where as low energy observables will depend on 9 parameters only. In this paper, lepton asymmetry ηB is estimated from light neutrino mass and mixing parameters by implementing the seesaw mechanism in the context of a class of supersymmetric left−right models. By considering a minimal Higgs sector, these models can predict the relation for the Dirac neutrino mass matrix, in a basis where the charged lepton mass matrix is diagonal; where c ≃ mt /mb is determined from the quark sector, and this assumption leaves only the Majorana mass matrix MR to be arbitrary. The three phases of MR can now be removed, leaving a total of 9 parameters which determine both the low energy neutrino masses and mixings as well as the baryon asymmetry. Various attempts have been made by different groups to establish a relationship between leptogenesis and low energy parameters that can be determined by the low energy neutrino experiments. 2 Matter-Antimatter Asymmatry : Numerical Approach: 2.1 CP Asymmatry : Conection between Low and High Energy Parameters: In an attempt to explain matter-antimatter asymmetry, leptogenesis is one of the simplest and well motivated mechanism. In this framework new heavy particles are introduced in the theory. The considered framework must naturally fullfill the three Sakharov conditions. The new heavy particles considered in Minimal Supersymmetric Model (MSSM) [14] are majorana neutrinos Ni , these particles have a hierarchical mass spectrum M1 <<M2 <M3 , so that the study of evolution of the number density of N1 suffices for the study of baryon asymmetry. One important factor that determines the baryon asymmetry, produced by thermal leptogenesis is CP asymmetry (εi ) as discussed above, which can be decomposed as CP asymmetry via N1 decays (and similarly N2 and N3 decays). ε1 = ΓN1 → lH − ΓN1 → lH ΓN1 → lH + ΓN1 → lH (1) (m†D mD )11 M1 8 π v2 (2) ΓN1 → lH = Where mD is dirac neutrino mass matrix and v = 174 GeV is the VEV (vacuum expectation value) of the Higgs doublet responsible for breaking the electroweak symmetry. For thermal leptogenesis the CP asymmetry produced by decay of N1 can be expressed as the sum of a vertex contribution and self energy contribution [9]. In the basis, where right handed neutrino mass matrix is diagonal and real, the CP asymmetry parameter can be written as [15] [16] , ε1 = 1 Mj 2 Mj 2 ∑ [Im(mD † mD )21 j [ fv ( M1 ) ] + [ fs( M1 ) ]] 8π v2(m†D mD )11 j=2,3 (3) Where fv (x) and fs (x) are the contribution arising from vertex and self energy corrections respectively [16]. For the MSSM case, fv (x) = p (x)log(1 + 1x ) ; fs (x) = √ 2 x x−1 M 3 fv (x) + fs (x) = − 2√ ; Here x = ( M1j )2 x Neutrino oscillations and Leptogenesis 3 In a model consisting of three heavy neutrino masses, the CP asymmetry parameter arising from the decay of N1 can be written as [16], ε1 = −3 M1 1 M1 [Im(mD † mD )212 + Im(mD † mD )213 ] 16π v2 (m†D mD )11 M2 M3 (4) Here ε1 depends on the (1,1), (1,2) and (1,3) entries of (m†D mD ) . By the see-saw mechanism light neutrino mass matrix can be connected to heavy neutrino mass matrix by the expression [17], mν = −mD MN−1 mTD (5) mD = cml = c ∗ diag(me, mµ , mτ ) (6) where mD is, here ml is charged lepton mass matrix and c is defined as c = mt /mb . In the case of Neutrino, flavor eigen states (νe , νµ , ντ ) and mass eigen states (ν1 , ν2 , ν3 ) can be connected as [18] , † mν = Umdiag ν U (7) Where mdiag = diag(m1 , m2 , m3 ) and U is a 3 × 3 mixing matrix consisting of, majorana phases and dirac ν phase [19]. The solution for majorana mass matrix can be expressed as, T MN = c2 ml m−1 ν ml    m e 1 0 0 mτ 0 0  m    T   m m  0 mµτ 0  0 mµτ 0 UPMNS P2 0 m12 0  UPMNS m1 0 0 m 0 0 1 0 0 1 3 m e = c2 mτ2 m1 mτ 0 0  (8) (9) The matrix P contains two majorana phases P = (eiα , eiβ , 1). In an attempt to get proper form of MN , the values of different neutrino parameters are expressed in a small expansion parameter [15] [20] [21] γ , which is defined as, γ= mµ mτ ≃ 0.059 In terms of small expansion parameters γ , different parameters can be expressed as , me = a γ 3 mτ , m1 m3 = a13 γ , m1 m2 = tan2 θ12 + a12 γ , θ23 = π 4 + t23 γ , β = α + π /2 + bγ and sin θ13 = θ13 , since θ13 is very small. Where a, a13 , t13 , t23 are parameters of O(1) , with a = 1.400, a13 = 1.3, a12 = 1 and b = 1. Now the right handed mass matrix can be written as, 4 Anupam Yadav*, Sabeeha Naaz*, Jyotsna Singh* and R.B. Singh*   P γ 5 Qγ 3 R γ 2   MN =  Sγ 3 T γ 2 U γ  V γ2 W γ X (10) The unitary matrices of MN can be difined as (KU3U2U1 );  | M1 | 0  T (KU3U2U1 )MN (KU3U2U1 ) =  0 | M2 | 0 0 0   0  | M3 | (11) Where K = diag(k1 , k2 , k3 ) with k1 = e−iφ1 /2 , where φ1 , φ2 , φ3 are phase factors which make each right handed neutrino masses real. The values of masses m1 , m2 , and m3 considered in this work are , m1 = 0.0027 × 10−9GeV m2 = 0.0068 × 10−9GeV m3 = 0.0380 × 10−9GeV After the expansion in terms of small parameter, the values of (m†D mD )11 , (m†D mD )212 and (m†D mD )213 , which are connected to matrix MN takes the form, (m†D mD )11 = 2 tan2 θ 8a2 c2 m2τ γ 2 cos2 θ12 sin2 θ12 θ13 12 a13 cos4 θ12 [4a13(a212 − b2 ) cos 2θ12 ) + a13(a212 + 4b2)(3 + cos4θ12 )] (m†D mD )212 = (m†D mD )213 = 2 cos2θ + 4θ 2 ]2 2a2 c4 mτ4 γ 2 tan2 θ12 e−i(φ1 −φ2 ) e−2i(2α +δ )[−4θ13 12 13 [3a12 a13 − 2ia13b + 4 cos2 θ12 a12 a13 + a13(a12 + 2ib) cos4θ12 ]2 (12) (13) 2 sin2 θ γ 4 + 2a cos θ e−i(2α +δ ) θ sin θ γ 5 ] 2a2 c4 m4τ sin2 θ12 e−i(φ1 −φ3 ) [a213 cos2 θ12 γ 6 + e−2i(2α +δ )θ13 12 12 13 12 13 4 4 [a13 γ cos θ12 (a12 − 2ib + (a12 + 2ib) cos2θ12 )]2 (14) Now the Eq. (4), representing the CP asymmetry can be expressed as, ε1 = −3 1 M1 M1 [2ABC + D(A2 − B2)/(A2 + B2 )2 × ( )+2EFG+H(E 2 −F 2 )/(E 2 +F 2 )2 ×( )] 16π v2 (MD† MD )11 M2 M3 (15) Neutrino oscillations and Leptogenesis 5 Different terms used in the above expression are, A = (3a12a13 + 4 cos2 θ12 a12 a13 + a12a13 cos4θ12 ) B = i(2ba13 cos 4θ12 − 2a13b) C = 2a2 c4 mτ4 γ 2 tan2 θ12 [cos(φ1 − φ2 ) cos(4α + 2δ )− 2 cos 2θ + 4θ 2 ]2 sin(φ1 − φ2 ) sin(4α + 2δ )][−4θ13 12 13 D = 2a2 c4 m4τ γ 2 tan2 θ12 [sin(φ1 − φ2 ) cos(4α + 2δ )+ 2 cos 2θ + 4θ 2 ]2 cos(φ1 − φ2 ) sin(4α + 2δ )][−4θ13 12 13 p E = 2 (a13 )γ 2 cos2 θ12 b(cos 2θ12 − 1) p F = i[ (a13 )γ 2 cos2 θ12 a12 (1 + cos2θ12 )] G = 2a2 c4 m4τ sin2 θ12 cos(φ1 − φ3 ) 2 sin2 θ γ 4 cos(4α + 2δ ) + 2a cos θ θ sin θ γ 5 cos(2α + δ )] [a213 cos2 θ12 γ 6 + θ13 12 12 13 12 13 −2a2c4 mτ4 sin2 θ12 sin(φ1 − φ3 ) 2 sin2 θ γ 4 sin(4α + 2δ ) + 2a cos θ θ sin θ γ 5 sin(2α + δ )] [θ13 12 12 13 12 13 H = 2a2 c4 m4τ sin2 θ12 sin(φ1 − φ3 ) 2 sin2 θ12 γ 4 cos(4α + 2δ ) + 2a13 cos θ12 θ13 sin θ12 γ 5 cos(2α + δ )] [a213 cos2 θ12 γ 6 + θ13 +2a2c4 mτ4 sin2 θ12 cos(φ1 − φ3 ) 2 sin2 θ γ 4 sin(4α + 2δ ) + 2a cos θ θ sin θ γ 5 sin(2α + δ )] [θ13 12 12 13 12 13 The hightest contributing factor in the CP asymmetry ε1 can be stated as, 2 2 2 cos(φ1 − φ2 ) cos 2(2α + δ )− θ13 sin(φ1 − φ2 ) sin 2(2α + δ )+[cos(φ1 − φ3 )+sin(φ1 − φ3 )]/θ13 +....... ε 1 ∝ θ13 (16) The Davidson-Ibarra (DI) bound on a CP asymmetry is given as, |ε1 (M1d=5 , m̄)| ≤ |ε1max,d=5 (M1 , m̄)| = 3 M1 (m3 − m1 ) 16π v2 (17) 6 Anupam Yadav*, Sabeeha Naaz*, Jyotsna Singh* and R.B. Singh* 2.2 Quntitative Calculation of the Abundance of Right handed Neutrino: With in the minimal framework where we assume that the initial temprature Ti is larger than M1 , is the mass of the lightest heavy neutrino N1 we neglect the decays of two heavier neutrinos N2 and N3 assuming that it does not influence the final value of B-L asymmetry. Let us consider NN1 indicates the right handed neueq trino abundance and NN1 indicates the thermal equilibrium values of right handed neutrino. The Boltzmann equation for NN1 can be given as dNN1 = −(D + S)(NN1 − NNeq1 ) dz (18) From the above equation where z = M1 /T we notice that the lepton asymmetry will be generated when eq eq lightest right handed neutrino is ’out-of-thermal equilibrium’ (NN1 6= NN1 ). As NN1 drpos with universe temprature T, the out-of-thermal equilibrium can be satisfied, when the universe is cooling down. If the ΓD Hubble expansion rate is denoted by H then, D = Hz accounts for decays and inverse decays where as S = ΓS represents ∆ L = 1 scattering term. The Hubble expansion rate is given by, Hz H≃ q 8π 3 g∗ M12 1 90 MPI z2 where, g∗ = gSM = 106.75 is the total number of degrees of freedom, and MPI = 1.22 × 1019 GeV , is the Planck mass. The two terms D and S depend on effective neutrino mass (m̃1 ), which is expressed as, m̃1 = as, (m†D mD )11 M1 (19) Effective neutrino mass has to be compared with the equilibrium neutrino mass (m∗ ), which is expressed √ 16π 5/2 g∗ v2 √ m∗ = ≃ 1.08 × 10−3eV 3 5M pl (20) The deacy parameter K= ΓD (z = ∞) m̃1 = H(z = 1) m∗ (21) Neutrino oscillations and Leptogenesis 7 1 10 −1 10−2 10−3 1 NN (z) 10−4 10−5 10−6 10−7 10−8 10−9 10−10 10−11 −2 10 10−1 1 10 z = M1/T 102 Fig. 1: Result of heavy neutrino production in the case of zero initial abundance at K = 10−2 (green line), K = 100 (blue line) and thermal initial abundance (red line) at zeq . For the calculation of the abundance of right handed neutrino the decay parameter (K) plays a key role. For K>>1, the life time of right handed neutrino is much shorter than the age of the universe, t = H −1 /2 (z = 1) and the right handed neutrino decays and inverse decays many times before they become nonrelativistic. In this case the abundance of right handed neutrino resembles very closely to the equilibrium distribution as shown in Fig:1. On the other hand when K<<1, when right handed neutrino are already fully non-relativistic and the bulk of right handed neutrinos decay completely out of equilibrium hence in this case their equilibrium abundance is exponentially supressed by Boltzmann factor. This can be checked in Fig:1. Where we can see that the pattern for the production of the right handed neutrino at K = 100 overlaps the pattern produced for the same at zeq , after z = 1. Where as the production pattern of right handed neutrino for K<1 with stands for larger value of z. 2.3 Calculation of Baryon, Lepton Asymmetry and Efficiency factor: The Boltzmann equation for NB−L can be given as, dNB−L = −ε1 D(NN1 − NNeq1 ) − W NB−L dz (22) Γw where, W = Hz contributes two washout term, W = W0 + ∆ W ; the first term depends only on m̃1 , while the second term depends on the product M1 m̄2 , where m̄2 = m21 + m22 + m23 is sum of the light neutrino masses squared. The washout term is the term that tends to re-equilibrate the number of leptons and antileptons destroying the asymmetry generated by the CP violating term. It is simply a statistical re-equilibrating term that has to be present in order to respect the Sakharov third condition. Solution of above Boltzmann equation for NB−L is the sum of two term, i NB−L (z) = NB−L e R − zz dz′W (z′ ) i 3 − ε1 k(z; m̃1 , M1 m̄2 ) 4 (23) 8 Anupam Yadav*, Sabeeha Naaz*, Jyotsna Singh* and R.B. Singh* First term of the above equation accounts for possible generation of B − L asymmetry before N1 decays and the second part of the above equation expressed in terms of the efficiency factor (k) and CP asymmetry (ε1 ), describes B − L generation from N1 decays. In our analysis we have neglected the contribution arising from the first term. The efficiency factor does not depend on CP asymmetry ε1 . A global expression for the efficiency factor can be expressed as a sum of positive contribution k+f (K) when K>>1 and negative contribution k−f (K) when K<<1. k f (K) = k+f (K) + k−f (K) 2 3 2 k−f = −2e− 3 (N(K)+ 4 K αs ) (e 3 Ñ(K) − 1) k+f = Here, zB = M1 TB , 2 2 −1 2 (1 − e− 3 zB (K)K j(zB ) NN1 (zeq ) j(zeq ) ) zB (K)K j(zB )2 (24) (25) where TB is baryogenesis temperature and zB (K) is defined as, 3125π K 2 5 π K2 1 [ln( )] ) zB (K) ≃ 1 + ln(1 + 2 1024 1024 (26) D+S 1 a Ks 15 ≈ [ ln(1 + ) + ](1 + ) D a z Kz 8z (27) j(z) = a= 8π 2 K = Ks ln(M1 /Mh ) 9 ln(M1 /Mh ) (28) m̃1 ms∗ (29) 4π 2 gN1 v2 m∗ ≃ 10m∗ 9 mt2 (30) 2Ks 15 + 3K 8 (31) Scattering parameter, Ks = ms∗ = αs = Ñ(K) = 2N(K)z3eq ((9π )c + (2z3eq )c )1/c (32) zeq = ( K6 )1/3 ; c = 0.7; N(K) = 9π K/16 In this efficiency factor both decay and scattering are considered. The case without scattering can be recovered by substituting αs = 0 in k−f and j = 1 in k+f . Fig:2 illustrates the NB−L asymmetry produced for two different values of K (K<<1 and K>>1). Electroweak sphaleron processes is responsible for the conversion of lepton asymmetry into baryon asymmetry. The baryon asymmetry nB produced through the sphaleron transition of lepton asymmetry YL , while the quantum number B-L remains conserve, is given by, YB = nB = CYB−L = CYL s (33) Neutrino oscillations and Leptogenesis Where C = 8N f +4NH 22N f +13NH ; 9 N f is the number of fermionic family, NH is the number of Higgs doublets and s = 7.04 nγ . YL = (nL − nL ) 3 εi ki = ∑i=1 s g∗i (34) After substituting the eq(34) in eq(33), we get, YB = C ε1 k1 g∗1 (35) In our work we have to used k f in place of k1 and g∗ in place of g∗1 . 1 10−1 10−2 10−3 |NB-L(z)| 10−4 10−5 10−6 10−7 10−8 10−9 10−10 10−11 10−2 10−1 1 z = M1/T 10 102 Fig. 2: Result of baryon and lepton asymmetry in the case of thermal initial abundance at K = 10−2 (blue line) and K = 100 (red line). If we assume that the universe reheats to a thermal bath composed of particles with case interaction after inflation, the final baryon to photon number ratio ηB can be estimated as the product of three suppresion factor 1) Leptonic CP asymmetry εi . 2) An efficiency factor k f ,arising due to washout processes and scattering, decays and inverse decays. 3) A reduction factor due to chemical equilibrium charge conservation and the redistribution of asymmetry among different particle species. 3 Bounds on the Mass of Right handed Neutrino: The light neutrino masses can be either quasi degenerate or hierarchical, with m2 − m1 <<m3 − m2 as normal hierarchy and m2 − m1 >>m3 − m2 as inverted hierarchy. The maximal CP asymmetry ε1max depends on M1 and m̃1 and via the light neutrino masses mi , on absolute neutrino mass scale m̄. For given light neutrino 10 Anupam Yadav*, Sabeeha Naaz*, Jyotsna Singh* and R.B. Singh* 1 masses, ε1 is maximized in the limit m m̃1 tending to zero. The upper bound on CP asymmetry ε1 as a function of M1 and m̃1 is expressed in eq(17). Eq(17) reaches its maximum value for fully hierarchical neutrinos with p m1 = 0 and m3 = matm = ∆ m2atm and can be written as, ε1max (M1 , m̄) = 3 M1 m3 16π v2 (36) From the latest neutrino oscillation experimental results the values of ∆ m2atm and ∆ m2sol are as follows [29], ∆ m2atm = (2.5 ± 0.03) × 10−3eV 2 (NH) −3 2 ∆ m2atm = 2.45+0.03 −0.04 × 10 eV (IH) −5 2 ∆ m2sol = 7.55+0.2 −0.16 × 10 eV matm = (0.05 ± 0.003)eV A recent combined analysis of baryon to photon ratio [27] is, −10 ηBCMB = 6.0+0.8 −1.1 × 10 (37) The CP asymmetry in terms of ηBCMB can be written as, [28], ε1CMB ≃ 6.3 × 10−8( ηBCMB )k−1 6 × 10−10 f (38) For maximal baryon asymmetry ηBmax we get maximal CP asymmetry, which is evident from eq(38). By CMB constraint we get ηBmax ≥ ηBCMB . Since matm is fixed quantity then from eq(36), maximal value of CP asymmetry will depends only on M1 , ε1max (M1 ) = 3 M1 M1 matm m3 ≃ 10−6( 10 )( ) 2 16π v 10 GeV 0.05eV (39) ηBCMB 0.05eV −1 )( )k f −10 6 × 10 matm (40) M1 >M1min ≃ 6.4 × 108GeV ( Bounds on M1 depends on the combination ηBCMB /matm , 8 −1 M1min (m̃1 ) = (6.4 ± 0.6) × 108GeV k−1 f ≥ 4 × 10 k f (m̃1 ) (41) Neutrino oscillations and Leptogenesis 11 1016 1015 1014 M1 (GeV) 1013 1012 1011 1010 109 10−6 10−5 10−4 −3 ~ 10(eV) m 1 10−2 10−1 1 Fig. 3: Lower bounds on M1 for thermal initial abundance (blue line) and for zero initial abundance (green line). The observed baryon asymmetry ηB ≈ 10−10 sets a lower bound limit on ε1 and therefore on M1 . If right handed neutrinos are produced thermally then ηB ≤ 10−2ε1 and M1 >108 GeV. From Fig:3 the lower bound on right handed neutrino mass M1 in thermal initial abundance case is ∼ 5 × 108 GeV and in zero initial abundance case is ∼ 2 × 109 GeV. 4 Results and Discussion: In developing the above formulation, we have assumed low energy supersymmetry, where the Dirac neutrino mass matrix has a determined structure. As a result, we have connected the lepton asymmetry with measurable low energy neutrino parameters. Here right handed neutrino masses are not independent of CP asymmetry parameter. Our study is restricted to the case where the baryon asymmetry is generated only due to the decay of right handed neutrinos. Three right handed neutrinos, having hierarchical mass structure is considered in this work. In the model considered for the generation of baryon asymmetry it is assumed that in early universe, at temperature of order N1 , the main thermal process, which entered in the production of lepton asymmetry was the decay of lightest right handed neutrino. In order to estimate the baryon asymmetry arising due to the formulated analytical expression, the dilution factor, often referred as the efficiency factor k f , that takes into account the washout processes (inverse decays and lepton number violating scattering) has to be known a priori. Hence the solution of Boltzmann equation for the abundance of right handed neutrino and for the generation of NB−L is performed. While performing the solution of Boltzmann equation the effect of various parameters i.e. washout effect (W ), decay parameter (K), efficiency (k f ) and the ratio of right handed neutrino mass to the temprature (z = M1 /T ) are studied for the generation of excess N1 and NB−L , to achieve observed value of baryogenesis.The generated lepton asymmetry gets converted to the baryon asymmetry in the presence of the sphaleron induced anomalous B-L violating processes before the electroweak phase transition. The effect of the decay parameter K, z and efficiency factor k f on the value of NN1 and NB−L is shown in Fig:1 and Fig:2. The efficiency factor considered in our work takes into account scattering and decay both. The lowest bound on the right handed neutrino is also examined and is illustrated in Fig:3. The lowest bound achieved in this work is M1 ≥ 5 × 108. 12 Anupam Yadav*, Sabeeha Naaz*, Jyotsna Singh* and R.B. Singh* 1 10−1 10−2 ϵ1 10−3 10−4 10−5 10−6 10−7 10−8 10−9 1 2 3 4 θ013 5 6 7 8 Fig. 4: Evolution of CP asymmetry parameter ε1 using analytical results as a function of neutrino oscillation angle θ13 . 4 YB × 10 10 3 2 1 0 −1 0 1 2 3 4 5 6 α(radian) Fig. 5: Plot for Baryon asymmetry parameter YB as a function of majorana phase angle α . Neutrino oscillations and Leptogenesis 13 4 YB × 10 10 3 2 1 0 −1 0 1 2 3 4 5 6 δ(radian) Fig. 6: Evolution of Baryon asymmetry parameter YB as a function of dirac phase angle δ . The parameter space corresponding to the parameters θ13 , the CP phase δ and the Majorana phase α are scanned. The value of ε1 is checked for a given set of parameters and the current best fit value of θ13 [24] as expressed in table 1. In our analysis if ε1 <1.3 × 10−7 [15] , the induced baryon asymmetry would be too small to explain the experimental observations. In Fig.4 we observe that at current best fit value of oscillation angle θ13 , the value of ε1 is sufficient enough to generate observable the baryogengesis signals at low energy neutrino experiments. The value of leptogenesis change with the variation in input parameters. Table 1: The value of various selected parameters used for our analysis. a b c a12 a13 γ M1 (GeV) M2 (GeV) M3 (GeV) mτ (GeV) θ12 1.4 1.0 41.1 1.0 1.3 0.059 1×109 8.7×1011 2.6×1014 1.77 θ13 34.50 8.450 From Eq.(16), we observe that the CP asymmetry factor depends strongly on parameters θ13 , α , cos(φ1 − φ3 ) and cos 2(2α + δ ). In an attempt to observe the signatures of baryogenesis at neutrino oscillation experiments, we have imposed constraints on (φ1 − φ2 ) , α , δ and (2α + δ ) ( dirac and majorana phases) parameters. In Fig:8 baryon asymmetry YB is plotted as a function of α and from this plot we can observe that the disfavoured range of α is 100o − 170o and 280o − 340o . From Fig:6 we observe that the al- lowed range of dirac phase δ lies in the range 114o − 220o. The currently constrained value of δ (or δCP ) by low energy neutrino oscillation experiment lies in the above mentioned range, which motivates the search of leptogenesis signatures at low energy neutrino oscillation experiments. From Eq. 16 we can observe that the dependent phase (2α + δ ) contributes significantly in CP asymmetry. 14 Anupam Yadav*, Sabeeha Naaz*, Jyotsna Singh* and R.B. Singh* 4 YB × 10 10 3 2 1 0 −1 0 1 2 3 4 5 6 2α + δ(radian) Fig. 7: Evolution of Baryon asymmetry parameter YB as a function of the dependent phase angle (2α + δ ). Fig:7 illustrates, that the baryon asymmetry (originating from CP asymmetry) depends on (2α + δ ), a low energy dependent phase. The allowed values of this dependent phase to generate observed baryon asymmetry are 0o − 57o and 292o − 360o. 2.5 2 1.5 YB × 10 10 1 0.5 0 −0.5 −1 −1.5 −2 −2.5 0 1 2 3 4 5 6 ϕ1 − ϕ 3(radian) Fig. 8: Evolution of Baryon asymmetry parameter YB as a function of the dependent phase angle (φ1 − φ3 ) . Fig:8. shows the dependance of baryogenesis on phases φ1 and φ3 , which are the phases used for the digonalization of the right handed mass matrix as shown in Eq. (11). The allowed range of (φ1 - φ3 ) is 30o − 110o. Neutrino oscillations and Leptogenesis 15 5 Conclusions: In this work, we have performed a study of thermal leptogenesis which is considered as mechanism responsible for the generation of baryon asymmetry. The analytical expression for ε1 (CP asymmetry) at low temperature is derived in terms of small expansion parameter γ . This expression is derived for non zero or the present value of oscillation angle, θ13 . In an attempt to generate baryon asymmetry from CP asymmetry Boltzmann equation for NN1 (excess right handed neutrino) and NB−L are solved. Fig:1 illustrates that the pattern for the production of the right handed neutrino at K = 100 overlaps the pattern produced for the same by zeq , after z = 1. Hence we can say that to produce baryon or lepton asymmetry the value of K should be less 100 or K<1 will be preferred. This preference for the value of K is observed by green line (K = 10−2 ) of Fig:1. The Fig:2 also indicates that K<1 is preferred for the generation of NB−L . The final efficiency expression used for the generation of the baryon asymmetry is sum of positive contribution k+f (K) when K>>1 and negative contribution k−f (K) when K<<1 in which both scattering and decay processes are considered. The observed baryon asymmetry ηB ≈ 10−10 sets a lower bound limit on ε1 and therefore on M1 . For thermally genereted right handed neutrinos we get ηB ≤ 10−2ε1 and M1 >108 GeV. From Fig:3 the lower bound on right handed neutrino mass M1 in thermal initial abundance case is ∼ 5 × 108 GeV and in zero initial abundance case is ∼ 2 × 109. The results illustrated in Fig:4,5,6,7 and 8 show that with the present value of θ13 and present bounds im- posed on δ by NoVA and T2K [25], the neutrino experiments can be considered as one of potential source to compute the baryon asymmetry. However, a few relevant parameters are presently unknown. 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