Resonant leptogenesis at TeV-scale and neutrinoless double beta decay

We investigate a resonant leptogenesis scenario by quasi-degenerate right-handed neutrinos which have TeV-scale masses. Especially, we consider the case when two right-handed neutrinos are responsible to leptogenesis and the seesaw mechanism for active neutrino masses, and assume that the CP violation occurs only in the mixing matrix of active neutrinos. In this case the sign of the baryon asymmetry depends on the Dirac and Majorana CP phases as well as the mixing angle of the right-handed neutrinos. It is shown how the yield of the baryon asymmetry correlates with these parameters. In addition, we find that the effective neutrino mass in the neutrinoless double beta decay receives an additional constraint in order to account the observed baryon asymmetry depending on the masses and mixing angle of right-handed neutrinos.


Introduction
Leptogenesis [1] is an attractive mechanism accounting for the baryon asymmetry of the Universe (BAU). See, for example, reviews [2,3]. In the canonical scenario the out-of-equilibrium decays of right-handed neutrinos ν R 's generate a lepton asymmetry, which is partially converted into the baryon asymmetry through the sphaleron effect at high temperatures [4]. When the their masses are hierarchical, the observed BAU [5] Y B OBS = n B s obs = (0.870 ± 0.006) × 10 −10 , where Y B is the ratio between the baryon number density n B and entropy density s, can be explained if their masses are heavier than O (10 9 ) GeV [6]. #1 It should be noted that such superheavy particles can also give a significant impact on neutrino masses. The various oscillation experiments have shown that neutrinos have the very suppressed but non-zero masses. The smallness of the masses can be naturally explained by the seesaw mechanism with superheavy ν R 's [8].
Leptogenesis can operate even if ν R 's masses are much smaller than the above value, which is resonant leptogenesis [9]. The mass degeneracy of ν R 's enhances the CP violating effects, which leads to the resonant production of lepton asymmetry by their decay. #2 The required mass degeneracy may be a consequence of the symmetry of the model.
In resonant leptogenesis scenario, since it can occur at relatively lower temperatures, the flavor effects of leptogenesis [14,15,16,17,18,19] can be essential. In such a case, the yield of the BAU depends on the mixing matrix of active neutrinos U called as the Pontecorvo-Maki-Nakagawa-Sakata (PMNS) matrix [20], and hence the CP violation due to the Dirac and/or Majorana phases in U can be an origin of the BAU.
We consider here resonant leptogenesis by right-handed neutrinos with TeV-scale masses in the framework of the seesaw mechanism. Especially, it is investigated the case in which the CP violation occurs only in the mixing matrix U of active neutrinos. We will show that how the yield of the BAU depends on the Dirac and/or Majorana phases.
Majorana masses of ν R 's break the lepton number by two units, which is necessary for leptogenesis. Then, various processes, which are absent in the Standard Model, are predicted by the models with the seesaw mechanism. One important example is the neutrinoless double beta (0νββ) decay [21]. The rates of such decays are parameterized by the so-called effective masses of neutrinos m eff , which depends on the CP violating parameters of active neutrinos. We then discuss the possible relation between the yield of BAU and m eff . #1 The lower bound on masses can be reduced as O (10 6 ) GeV if one considers the non-thermal production of righthanded neutrinos via the inflaton decays [7]. #2 A sufficient amount of the BAU can be generated by right-handed neutrinos even with masses O (1) MeV-O (10 2 ) GeV if one uses the flavor oscillations of ν R 's [10,11,12,13].
The present article is organized as follows: In Section 2 we explain the framework of the analysis in which the properties of right-handed neutrinos are specified. In Section 3 we present the method of estimating the BAU through resonant leptogenesis including the flavor effects. We then show how Y B depends on the CP violating parameters of active neutrinos. It is discussed in Section 4 that how the conditions accounting for the observed BAU give the impacts on the 0νββ decay. We will show that the BAU provides the upper and/or lower bound of the effective mass m eff in some cases, especially when the mass difference of ν R 's becomes larger. The final Section is devoted to conclusions. We add Appendix A to present the Boltzmann equations used in the analysis.

TeV-scale right-handed neutrinos
First of all, let us explain the framework of the present analysis. We consider the Standard Model (SM) extended by three right-handed neutrinos ν R I (I = 1, 2, 3) with the Lagrangian where L SM is the SM Lagrangian, and α (α = e, µ, τ) and H are lepton and Higgs doublets, respectively.    Table 1: The mixing angles and the mass squared differences from the global analysis of three flavor neutrino oscillations [22].
In order to induce the mixing angles and masses of neutrino oscillation observations, the Yukawa coupling matrix takes the form [23] where Ω is the arbitrary 3 × 3 complex orthogonal matrix, which is parameterized as where ω I J are complex mixing parameters.
In this analysis we consider the case when only two right-handed neutrinos are responsible to the seesaw mechanism as well as leptogenesis for simplicity. The mass and Yukawa couplings of the rest right-handed neutrino are taken to be sufficiently heavy and small, respectively. For the NH case with m 3 > m 2 > m 1 = 0 we consider the two right-handed neutrinos ν R2 and ν R3 and mixing parameters are taken to be On the other hand, for the IH case with m 2 > m 1 > m 3 = 0 we consider ν R1 and ν R2 and There are three CP violating parameters which can be a source of BAU under this situation. They are the Dirac phase δ CP , one combination of Majorana phases, i.e. , (α 21 − α 31 ) or α 21 for the NH or IH case, and the imaginary part of the mixing parameter, i.e. , Imω 23 or Imω 12 for the NH or IH case. In the present analysis we assume that the CP violation occurs only in the PMNS matrix and all the mixing parameters ω I J are real. It is then discussed whether two right-handed neutrinos which masses are TeV-scale can produce a sufficient amount of the BAU or not.

Baryon asymmetry and CP violations in leptonic sector
In this section we investigate the baryogensis scenario by TeV-scale right-handed neutrinos through resonant leptogenesis [9]. The BAU in Eq. (1) can be explained if their masses are quasi-degenerate.
We thus take M 3 = M N + ∆M /2 and M 2 = M N − ∆M /2 for the NH case while M 2 = M N + ∆M /2 and In this case the generation of the BAU is effective at TeV-scale temperatures and then the flavor effects of leptogenesis [15,16] must be taken into account. This is crucial to produce the baryon asymmetry by the CP violation in the PMNS matrix, since such an effect disappears for unflavored leptogenesis. From now on we will estimate the yield of the BAU by TeV-scale right-handed neutrinos and show how it depends on the low energy CP violating parameters in the PMNS matrix.
In this work, we use the Boltzmann equations for estimating the amount of the produced baryon asymmetry. In the N I decay the interference between tree and one-loop diagrams of vertex and selfenergy corrections induces the lepton asymmetry due to the CP violation the neutrino Yukawa coupling constants. It is characterized by the CP asymmetry parameter ε αI which is defined by where Γ(N I → α + Φ) is the partial decay width for N I → α + Φ. Now we consider the case with ∆M M N , and then the contribution from the self-energy correction dominates over that from the vertex correction. In this case ε αI is given by [9] ε αI In this equation A denotes a regulator that is estimated as A = M I Γ I + M J Γ J [24,25] where Γ I is the total decay rate of N I . It is then found that |ε αI | takes the maximal value when the condition A = |M 2 I −M 2 J | is satisfied. This means that the maximal value of |ε αI | is achieved when the mass difference is ∆M = ∆M * ≡ A/(2M N ). From now on, we call the yield of the BAU with ∆M = ∆M * as Y MAX B .
We estimate the yield of the BAU by using the Bolzmann equations for the yield of N I (Y N I ) and the charges (X α = B /3−L α ) associated with the baryon number B and the lepton flavor number L α . #3 The explicit equations are presented in Appendix A. The initial conditions are Y N I = Y eq N I and X α = 0, where Y eq N I is the equilibrium value of Y N I . We then solve the equations from the initial temperature T i M N #4 to the final temperature T f = T sph and calculate the yield of the BAU. Here T sph is the sphaleron freeze-out temperature and T sph = 131.7 GeV [28] for the observed Higgs boson mass.
We take M N = 1 TeV as a representative value and evaluate the maximal value Y MAX B by setting the mass difference as ∆M = ∆M * . In addition, as explained in the previous section, we consider the case when the CP violation occurs only in the mixing matrix U of active neutrinos, i.e. , we set Imω I J = 0. We take the central values of the mixing angle θ i j and the mass squared differences ∆m i j #3 The estimation based on the Kadanoff-Baym equation is found in Refs. [26,27]. #4 We take T i /M N = 100 for the numerical study. phase, e.g. , from accelerator neutrinos [30,31] is crucial for determining the sign of the BAU. It should be noted that the dependence on the CP violating phases is approximately given by which is found from the parameter dependence in ε αI as well as the strength of the wash-out effects, i.e. , the structures in the partial decay rates Γ N I → α + Φ . On the other hand, the right panel in Fig. 1 shows the contour in the mixing angle Reω 23 and δ CP plane when α 21 − α 31 = π. It is found #5 The dependence on Imω 23 is discussed in Ref. [29]. Next, we turn to consider the IH case. It is found from Fig. 2 Note that the subleading effect which disturbs the above dependence is larger than that in the NH case. The observed BAU cannot be produced for the vanishing mixing between ν R 's at Reω 12 = 0, π/2 similar to the NH case. In addition, the sign of the BAU correlates with the sign of Reω 12 . See also the below.
As described above, Y MAX In this case the sign of the BAU is determined by the Majorana phase and the mixing angle of ν R 's, which is represented in Fig. 3 in the Reω 12 -α 21 plane is shown in the right panel of Fig. 3. It is seen that the successful baryogenesis is realized for both Reω 12 < 0 and Reω 12 > 0. When Reω 12 < 0, α 21 2π is required and then the CP violation by δ CP is essential.
As explained above, the Majorana phase plays an important role for determining the sign of the BAU through leptogenesis scenario under consideration. It is therefore expected that the BAU may give an impact on the other phenomena in which the Majorana phase is essential. One such example is 0νββ decay, which will be discussed in the next section.
Before closing this section, we should mention the dependence on the averaged Majorana mass

Neutrinoless double beta decay
In the seesaw mechanism active neutrinos are Majorana fermions, which break the lepton number in contrast to the SM. One interesting example of the lepton number violating processes is the 0νββ decay: (A, Z ) → (A, Z + 2) + 2e − [21]. The decay rate is proportional to m 2 eff which is the effective neutrino mass given by Here we take into account the contribution from active neutrinos only, because the contribution from heavy neutral leptons is negligible in the considering situation.
We have assumed that only two right-handed neutrinos are responsible to the seesaw mechanism of neutrino masses, and then the lightest active neutrino is massless. In this case the effective mass is written as for the NH case and We have shown in the previous section that the successful scenario for resonant leptogenesis requires a certain range of the Majorana phase (i.e. , α 21 − α 31 or α 21 for the NH or IH case) as well as the mixing angles of ν R 's (i.e. , Reω 23 or Reω 12 for the NH or IH case). It is, therefore, expected that the predicted range of m eff is restricted for generating the sufficient amount of the BAU by leptogenesis.
It is important to note that Y MAX We have so far taken the Dirac phase in Eq. (12). The allowed region in Fig. 6 changes by the value of δ CP for the NH case, while it remains almost the same for the IH case. This difference comes from the δ CP dependence of Y B . The experimental determination of δ CP is thus important for the predictions of the BAU as well as the 0νββ decay for the NH case.

Conclusions
We have investigated the Standard Model extended by right-handed neutrinos, in which two ν R 's are quasi-degenerate with TeV-scale masses. We have studied the origins of the neutrino masses and the BAU in this setup. By assuming that the neutrino masses are generated by the seesaw mechanism, the production of the BAU through resonant leptogenesis has been studied by using the Boltzmann equations with the flavor effects.
We have considered the case when the CP violation occurs only in the active neutrino sector.
Since leptogenesis by TeV-scale right-handed neutrinos is considered, the flavor effects are essential

A Boltzmann equations
In this appendix we present the formulae to estimate the yield of the BAU by using the Boltzmann equations. The yield of the right-handed neutrino N I , Y N I , is given by where n N I is the number density of N I . The entropy density of the universe is s = (2π 2 /45)g * s T 3 with the cosmic temperature T . The effective degrees of freedom is taken as g * s = 106.75 throughout this analysis. When N I is in equilibrium, the yield is estimated as where the variable z is defined by z = M /T , in which M is the mass of lighter right-handed neutrinos, i.e. , M = M 2 or M 1 for the NH or IH case, respectively. K 2 (z) is the modified Bessel function of the second kind. When the sphaleron processes are in thermal equilibrium, there are three conserved charges X α = B /3 − L α (α = e, µ, τ) associated with baryon number B and lepton flavors L α . The yield of X α is denoted by Y X α = n X α /s. The Boltzmann equations for Y N I and Y X α used in this analysis are given by [33,34] (1)αγ N + γ (2)αγ N + I ,J C αβ −C γβ γ (1)αγ where H (M ) is the Hubble expansion rate for T = M and Y eq = 45/(2π 4 g * s ). #6 C and C Φ are given by [15,16] The reaction densities γ's are found in Refs. [33,34]. We have considered the non-supersymmetric case and taken into account the flavor effects of leptogenesis. Our notations of the reaction densities #6 We have applied the Bolzmann approximation.