Right-handed neutrino as a common progenitor of baryon number asymmetry and dark matter

Cosmological and astrophysical observations suggest that both energy densities of baryon and dark matter take the same order values in the present Universe. We propose a scenario to give an answer for this problem in a scotogenic model and its extension. The model naturally provides dark matter candidates as its crucial ingredients to explain the small neutrino mass. If a neutral component of an inert doublet scalar plays a role of dark matter and leptogenesis occurs at TeV scales, both dark matter abundance and baryon number asymmetry could be explained with a same mother particle. Coincidence between the order of their energy densities might be understood through such a background.


Introduction
Existence of dark matter (DM) suggested through various astrophysical and cosmological observations is one of unsolved problems in the standard model (SM) [1].Since there is no corresponding particle in the SM, it can be a crucial clue to study new physics beyond the SM.A promising candidate is a weakly interacting massive particle called a WIMP.
If it is in the thermal equilibrium in the early Universe and is frozen out at a certain stage, it is known that the expected DM abundance could be naturally realized as its relic.However, such a particle has not yet been discovered in various kind of experiments, but a lot of proposed models for it have been ruled out or constrained by them [2,3].
Another mysterious point on DM is cosmological coincidence of its abundance with baryon number asymmetry, that is, why their energy densities take the same order values in the present Universe.The WIMP scenario cannot answer this problem since its abundance is usually considered to be explained through physics irrelevant to the baryon number asymmetry.Asymmetric DM model [4] is a promising scenario which is motivated to give an answer for it.In this scenario, the DM abundance is explained as asymmetry of a certain conserved charge like the baryon number asymmetry.If we consider an alternative possibility to explain it, a scenario may be constructed by assuming that both the baryon number asymmetry and the DM are caused from a same mother particle.In that case, the same physics could be relevant to them and then it could give an answer for the problem.As such we study a class of model for the neutrino mass here.
The scotogenic model [5] has been proposed to explain the small neutrino mass and the existence of DM, simultaneously.In this model, the DM abundance is usually explained following the WIMP scenario.The model includes DM candidates as its important ingredients to generate the small neutrino mass.They are a lightest neutral component of an inert doublet scalar η and a lightest right-handed neutrino.If we choose the right-handed neutrino as the DM, a serious lepton-flavor violating problem appears [6].On the other hand, if we adopt a neutral component of the η as the DM, any serious phenomenological problem occurs, and then it can be considered as a promising DM candidate in the model.Its relic abundance and various features have been extensively studied in a lot of papers [7,8].
In this paper, we reconsider the DM abundance in this model from a view point of its relation to the baryon number asymmetry.In the scotogenic model, leptogenesis is known to generate baryon number asymmetry successfully [9,10].Since model parameters relevant to DM and leptogenesis could be almost independent in usually supposed cases, they have been studied separately.However, there could be an exceptional situation where they are closely related; this situation seems not to have been noticed and has not been studied.In the present study, we focus our attention on a case where the model allows successful leptogenesis at TeV scales as such a situation.There, we find that the usual estimation of the relic abundance of the DM should be modified.The DM abundance is considered to be explained as a cosmological relic which is neither a pure thermal relic nor an asymmetry of any conserved charge.It might give a new viewpoint for the cosmological coincidence between the DM and the baryon number asymmetry.
Following parts are organized as follows.In section 2 we briefly review some features of the scotogenic model relevant to the present study.In section 3, after we overview leptogenesis and the DM physics in the model, we discuss a scenario in which the baryon number asymmetry and the DM abundance can be closely correlated.We show its realization in the model which can give the origin of the CP phases in the PMNS matrix [11].
We summarize the paper in section 4. In the Appendix, we present a rough sketch of the derivation of the CP phases in the PMNS matrix in the model.

Scotogenic model
The scotogenic model [5] is a simple extension of the SM with three right-handed neutrinos N j and an inert doublet scalar η.These new ingredients are assumed to have an odd parity under imposed Z 2 symmetry although all the SM contents have even parity.Relevant parts to these new ingredients in Lagrangian are given as where ℓ L i and ϕ stand for SM doublet leptons and a doublet Higgs scalar, respectively.The model can have a stable vacuum only if the potential V in eq. ( 2) satisfies the condition, where λ + is defined as Since η is assumed to have no vacuum expectation value (VEV), neutrino mass is forbidden at tree level but it can be generated by a VEV of the SM Higgs scalar through a one-loop diagram whose internal lines are composed of N j and η.Additionally, since η gets no VEV, the imposed Z 2 symmetry remains as an exact symmetry and it guarantees the stability of the lightest Z 2 odd particle, which could be a cold DM candidate as long as it is neutral.If λ 4 < 0 is satisfied, a lightest neutral component of η can be such a candidate.In the following study, λ 5 < 0 is assumed and then its real part η 0 R is supposed to be DM.Its mass m η 0 R has to satisfy m η 0 R < M N 1 where M N 1 is mass of the lightest right-handed neutrino N 1 .In that case, N 1 can decay to the lepton through the Yukawa coupling in eq. ( 1).This decay could generate lepton number asymmetry, which can be transformed to baryon number asymmetry through a sphaleron process [12,13].It should be also noted that η is also produced in this decay.Thus, the lightest right-handed neutrinos could be a common mother particle of both the baryon number asymmetry and the DM in this model.
The neutrino mass formula derived from the one-loop diagram is given as This formula can explain the small neutrino mass required by neutrino oscillation data [2] even for N j with the mass of TeV scales as long as |λ 5 | takes a sufficiently small value.a Moreover, if we note that only two right-handed neutrinos are necessary to derive two mass differences required to explain the neutrino oscillation data, we find that the N 1 could be irrelevant to the small neutrino mass generation.This suggests that the neutrino Yukawa coupling h ν i1 can take a much smaller value than others h ν ij (j = 2, 3), which should be fixed so as to satisfy the neutrino oscillation data through eq. ( 4).
Neutrino oscillation data suggest that the PMNS matrix which characterizes lepton flavor mixing could be described approximately through tribimaximal mixing [14].Ala A lower bound on |λ 5 | can be derived through a possible inelastic scattering of η 0 R with a nucleon in DM direct search experiments [10].
though it cannot cause nonzero mixing angle θ 13 , it can be modified by a mixing matrix for charged leptons even if the tribimaximal mixing matrix is assumed for the neutrino sector.b Tribimaximal mixing in the neutrino sector can be easily realized simply by assuming the neutrino Yukawa couplings in eq. ( 1) to satisfy [16], Using this assumption, we can present examples which explain the neutrino oscillation data well.Here, we take h 1 to be sufficiently small like O(10 −8 ) so that N 1 is irrelevant to the neutrino mass determination as mentioned above.Taking account of this, if we apply to eq. ( 4) the parameters c , (a) the mass differences required to explain the neutrino oscillation data are found to be realized for the couplings We use them in the following study.
3 Leptogenesis and dark matter abundance

Low-scale leptogenesis
This model makes leptogenesis [13] work well to generate the baryon number asymmetry in the same way as the type-I seesaw model [17].Sufficient lepton number asymmetry could be produced as a seed of the baryon number asymmetry through the out-of-equilibrium decay of N 1 [10].Moreover, it could happen at much lower scales than the type-I seesaw model if N 1 is in the thermal equilibrium [18].However, it is difficult to make it in the thermal equilibrium only by the neutrino Yukawa coupling h ν i1 in eq. ( 1) in a consistent way b A concrete example of it is presented in [15] and also in the Appendix of this article.In the following study, parameters given there are used.c We note that m η 0 R ≃ M η is satisfied for this small |λ 5 |.
with successful leptogenesis as suggested in [10].To overcome this difficulty, interactions, which can produce a sufficient amount of N 1 to reach its equilibrium value, are proposed in some extended frameworks of the scotogenic model in connection with various problems of the SM, for example, inflation [19,20], right-handed neutrino mass [21,22], and CP violation [15,23].For a while, we assume that N 1 is in the thermal equilibrium through certain interactions.Study of the coincidence problem based on a concrete interaction is given later.
The CP asymmetry ε in the decay N 1 → ℓ L i η † is expressed as [24] ε where . Since the dependence on h ν i1 in this asymmetry ε can be suppressed under a suitable flavor structure, h ν i1 can be assigned a much smaller value compared with h ν i2 and h ν i3 keeping a value of ε to be a magnitude required for successful leptogenesis.On the other hand, a small value of h ν i1 makes the decay of N 1 delay so that the washout of generated lepton number asymmetry could be ineffective when its substantial decay starts.These could make successful leptogenesis possible for the N 1 with the TeV scale mass in the scotogenic model [15,18,20,22,23].
The decay of N 1 is expected to start around the temperature T L , which satisfies a condition Γ D N 1 = H(T L ), where Γ D N 1 and H(T ) are the decay width of N 1 and the Hubble parameter at the temperature T , respectively.They are expressed as where g * is relativistic degrees of freedom at T and M pl is a reduced Planck mass.If we use the tribimaximal assumption given in eq. ( 5) for the neutrino Yukawa couplings, the temperature T L can be estimated as where g * = 116 is used.
Here, we define Y i as Y i ≡ n i s by using the number density n i of a particle species i and the entropy density s.Lepton number asymmetry is expressed by Y L ≡ n ℓ −nl s , where Fig. 1 Evolution of the lepton number asymmetry Y L , the lightest right-handed neutrino number density "eq" stands for its equilibrium value.In the left panel for the case (a), n ℓ and nl are the number density of leptons and antileptons, respectively.If washout processes of the lepton number asymmetry Y L decouple at a temperature T F and T L < T F is satisfied, the Y L generated through the N 1 decay is not affected by the washout effect which is caused by the inverse decay of the right-handed neutrinos and 2-2 scatterings mediated by N 2,3 .d In such a case, the lepton number asymmetry at T L can be roughly estimated by using ε in eq. ( 8) as Y L = εY eq N 1 (T L ) where Y eq N 1 (T L ) takes the equilibrium number density of the relativistic particle as Y eq N 1 (T L ) = O(10 −3 ).Since T L < M N 1 has to be satisfied at least, eq. ( 10) shows The required baryon number asymmetry can be generated for Y L = O(10 −10 ) and then ε should have a value of O(10 −7 ).Eq. ( 8) shows that such an ε is consistent with the values of h 2,3 given in eq. ( 7) as long as a maximal CP violation is assumed.
The generated lepton number asymmetry is converted to the baryon number asymmed If the lepton asymmetry is generated at the temperature lower than the η mass as a result of a small h 1 , it can escape the washout by these processes since they could be sufficiently suppressed by the Boltzmann factor.It is confirmed through the numerical study shown in Fig. 3 which shows that T L and try through the sphaleron process which is considered to be in the thermal equilibrium at temperature higher than 100 GeV [12].Since the lepton number violating N 1 decay has to occur at a higher temperature than it, this imposes that Γ D N 1 > H(T ) should be satisfied at T = 100 GeV.This condition can be expressed as If we impose that leptogenesis occurs successfully at TeV regions, we find from eqs. (11) and ( 12) that h 1 should take a value of O(10 −8 ) for M N 1 , which is larger than m η 0 R assumed in eq. ( 6).
To examine these qualitative arguments, we solve the Boltzmann equations for Y L and Y N 1 given in [10] numerically by assuming that N 1 is initially in the thermal equilibrium and using the the parameters given in eqs.( 6) and (7).The results for both cases (a) and (b) are given in the left and right panels of Fig. 1, respectively.Both panels show that the N 1 decay delays and Y N 1 keeps its relativistic value until the temperature T reaches the value which satisfies T ≪ M N 1 .Since Y L is found to be realized as εY eq N 1 (T L ) from it, the above discussion can be justified quantitatively.The sufficient lepton number is found to be produced in both cases.

DM abundance
We consider the η 0 R abundance as DM in this model.The relic abundance of η 0 R is determined by solving Boltzmann equation for it.If we define a dimensionless parameter where H(m η 0 R ) and s(m η 0 R ) represent the Hubble parameter and the entropy density at T = m η 0 R , respectively.⟨σv⟩ is a thermally averaged annihilation cross section of η 0 R and Γ D N 1 is the decay width of N 1 given in eq. ( 9).We take account of an effect of the N 1 decay to η as the second term in the right-hand side of eq. ( 13).Here, we define as the solution of eq. ( 13) with the second term and the one without it, respectively.Since this term can be safely neglected for the case there and DM is considered as thermal relic.We consider such a case first.
In that case, the present abundance of η 0 R is fixed through its equilibrium density at temperature T D where its annihilation processes are frozen out [25,26].We can estimate T D by using its thermally averaged annihilation cross section ⟨σv⟩ and Hubble parameter H(T ) through a condition 2n eq η 0 R (T D )⟨σv⟩ = H(T D )x D as found from eq. ( 13).Using H(T ) given in eq. ( 9), x D can be determined by solving the above mentioned condition as where g is an internal degree of freedom of , the present DM abundance can be expressed as s 0 /ρ 0 where ρ 0 and s 0 are the energy density and the entropy density in the present Universe.They are given as ρ 0 = 3M 2 pl H 2 0 and s 0 = 2.27 × 10 −38 GeV 3 .If we solve eq. ( 13) by taking account of , we find that Ω η 0 R can be approximately expressed as where we use H 0 = 2.13 × 10 −42 h GeV.Applying it to the present observational result Ω DM h 2 = 0.12, we find that Y ∞ In Fig. 2, we draw contours of Ω η 0 R h 2 = 0.12 for typical values of m η 0 R in the (λ + , λ 3 ) plane.In this plane, we have to take account of both the stability condition given in eq. ( 3) and λ 4 = λ + − λ 3 − λ 5 < 0 which is a necessary condition for η 0 R to be lighter than the charged components.Combining them, we find that only points on the contours included in a triangle region of the upper-right quadrant in the (λ + , λ 3 ) plane are allowed.
On the other hand, the same parameters are also constrained by the present results of DM direct search experiments.In this model, η 0 R -nucleon elastic scattering is caused by the t-channel Higgs exchange.Its cross section can be expressed as where fN is a coupling between the Higgs scalar and a nucleon.Masses of the nucleon and the Higgs boson are represented by m N and m h , respectively.A present most stringent bound on σ N through the direct search experiment is given by the XENONnT experiment [27].Since it gives an upper bound on |λ + |, as found from eq. ( 17), a region included in a band sandwiched with dotted vertical lines fixed for each m η 0 R is allowed.These suggest that only restricted points in the (λ + , λ 3 ) plane can be consistent with the present data for the DM abundance.Future direct detection experiments and collider experiments might find a η 0 R signature in this parameter region.

Coincidence of DM and baryon number asymmetry
We reconsider realization of the DM abundance in relation to the baryon number asymmetry generated through leptogenesis.Since the required relic abundance is known to be realized for x D ∼ 25 in the WIMP scenario, T D < 100 GeV < T L can be satisfied for η 0 R whose mass is less than 3 TeV. .After that, Y η 0 R converges to a constant value which gives Ω η 0 R h 2 = 0.12.It suggests that Y η 0 R is determined irrelevantly from the determination of Y L .It is generally expected for the case where N 1 is much heavier than η 0 R .Next, we consider a case where Y ∞ η 0 R is determined through a different way from the above case.It can happen in a situation where h 1 takes a small value and also N 1 has the similar mass to η 0 R .The condition for h 1 required by the leptogenesis is given by eqs.( 11) and (12).If the Yukawa coupling h 1 takes a value such as H(T D ) > ∼ Γ D N 1 , the N 1 abundance could be larger than Y eq η 0 R (x D ) which saturates the required DM abundance.
To escape such a disastrous situation, h 1 has to satisfy This is consistent with eq. ( 12).If these conditions are satisfied and Y eq η 0

R
(x D ) is smaller than the value required by the DM abundance, the decay of N 1 could give a crucial contribution to the relic density of We can estimate Y N 1 (x D ) by solving eq. ( 14) as ), the exponential factor in eq. ( 19) can be rewritten as where we use m η 0 R = 2000 GeV.The required DM abundance can be realized for Y η 0 R (x D ) = O(10 −12 ) as found from the left panel of Fig. 1.If we impose Y N 1 (x D ) = O(10 −12 ), eqs.(19) and (20) suggest that the η 0 R produced by the N 1 decay at x > x D could supply a substantial part of the DM abundance for This relation is consistent with conditions ( 11) and (12).
If h 1 and M N 1 take values following the condition (21), The N 1 decay which causes the lepton number asymmetry could also generate a dominant part of the relic η 0 R for parameters λ + and λ 3 which are ruled out in the ordinary estimation of the relic η 0 R .The present abundance of η 0 R should be estimated based on since all components of η produced through the N 1 decay finally come to η 0 R .As the Boltzmann equation for Y η 0 R , we have to use eq.( 13) which takes account of the N 1 decay to η.
To examine the above observation quantitatively, we solve Boltzmann equations ( 13) and ( 14) numerically assuming that N 1 is initially in the thermal equilibrium.Although the mass difference between η 0 R and N 1 is small, their coannihilation can be neglected in ⟨σv⟩ because their coupling h 1 is small enough.In the right panel of Fig. 1, the evolution obtained as the solutions of the Boltzmann equations in the case (b) is plotted.In this calculation, λ + and λ 3 are fixed to 0.66 and 2.14, respectively.These make Y (o) η 0 R coming from the thermal η 0 R smaller than the required value Ω η 0 R h 2 = 0.12.It occupies about 50 % of the total and a remaining part is supplied through the N 1 decay.The figure shows that both the sufficient baryon number asymmetry and the DM abundance are simultaneously realized through the N 1 decay.
Comparison of both panels in Fig. 1 clarifies features of the present case.In the left panel which corresponds to the usually supposed case, the η 0 R abundance is realized by the freeze-out of the thermal η 0 R and the yields from the N 1 decay is irrelevant.The baryon number asymmetry and the DM abundance is explained based on irrelevant physics.On the other hand, we can find that the N 1 decay plays a crucial role to determine the η 0 R abundance in the right panel.In this example, the contribution from the thermal η 0 R is only 50% and the remaining one is caused by the N 1 decay.It suggests that the η 0 R yielded through the N 1 decay could supply the substantial part of the η 0 R abundance.

The coincidence of the baryon number density and the dark matter density in the present
Universe could be explained naturally there since common parameters relevant to the N 1 decay control them simultaneously.

A feasible model for the scenario
Finally, in order to show that this scenario works in a realistic way, we adopt a well motivated model with possible interactions which can generate N 1 in the thermal equilibrium even if the Yukawa coupling h 1 is too small to produce it effectively.The model has been proposed to give a prospect to the CP issues in the SM [15].It can solve the strong CP problem [28] through the Nelson-Barr mechanism [29,30] and give the origin of CP phases in the PMNS and CKM matrices [11,31].
This model is an extension of the scotogenic model with a singlet scalar S and vectorlike fermions D L,R and E L,R , where D L,R and E L,R are down-type singlet quarks and charged leptons, respectively.Their Yukawa couplings are given as where d R k and e R k are the singlet down-type quark and the singlet charged lepton in the SM, respectively.Since CP invariance is assumed in the model, all coupling constants in eq. ( 23) are considered to be real.If the singlet scalar S gets a VEV as ⟨S⟩ = 1 √ 2 ue iρ 0 , spontaneous CP violation is caused.Complex phases due to this CP violation could be brought about in the PMNS and CKM matrices through the mixing between the SM fermions and vectorlike fermions which is caused by these interactions.e The last term in the first line of ( 23) induces the mass of the right-handed neutrino N k .If we redefine N k to make its mass real, M N k and Λ k in eq. ( 4) can be expressed as This θ k determines the CP violation in the N 1 decay and then it fixes the CP asymmetry ε given in eq. ( 8).
e A rough sketch of this scenario is given in the Appendix.In the context of this paper, we should especially note that these interactions can through the exchange of S. If the mass of the vectorlike fermions is smaller than the reheating temperature, they could be in the thermal equilibrium through the SM gauge interactions.Heavier right-handed neutrinos N 2,3 could also be in the thermal equilibrium effectively through their neutrino Yukawa interactions at the temperature less than O(10 8 ) GeV if their couplings take the values given in eq. ( 7) [21,22].In such a case, these processes can produce N 1 effectively to reach its equilibrium density.Thus, if its mass M N 1 takes a similar value to m η 0 R as in the case (b) shown in the right panel of Fig. 1, we can expect that the N 1 decay generates the lepton number asymmetry sufficiently and it also contributes to the relic abundance of η 0 R substantially.f For its quantitative check, we solve the Boltzmann equations for Y N 1 , Y η 0 R , and Y L by taking account of these scattering processes.Eq. ( 14) should be modified by introducing f We should note that a scalar interaction κ Sη S † Sη † η is not forbidden in the model, which could give an additional source of η 0 R through the scattering.However, since this process is effective at the temperature higher than m η 0 R , it does not affect the present calculation of the DM abundance.It should be also noted that any change in the neutrino mass formula (4) is not caused by this interaction since the one-loop neutrino mass depends on the squared mass difference of η 0 R and η 0 I which is not changed by this interaction.Fig. 4 The lightest right-handed neutrino mass M N1 dependence of the ratio of the relic abundance of η 0 R originated from the N 1 decay to the total relics which are composed of the thermal one Ω T and Ω D from the N 1 decay.The masses of N 2,3 and η 0 R are fixed to the ones given as (b) in eq. ( 6).
the right-hand side additional terms where the suffix α describes the above processes.We take Y N 1 = 0 as an initial value of Y N 1 .Couplings in eq. ( 23) and the mass of the vectorlike fermions, which are crucial for the determination of the PMNS matrix, are fixed to the ones used in [15].They are presented in the Appendix.We assume u = 10 6 GeV and ρ 0 = π 4 as a VEV of the singlet scalar S and then these couplings fix mass eigenvalues of the fourth charged lepton and N 2,3 as M E = 3165 GeV and M N 2,3 given in eq. ( 6), respectively.It is noticeable that these can realize the present experimental results for the PMNS matrix well through a framework presented in the Appendix as described in [15].We also note that the CP violation, which fixes the CP asymmetry ε, can take a maximal value.
The results of this calculation are given in Fig. 3.It shows that N 1 reaches its equilibrium number density through the introduced interactions and its decay produces both the sufficient lepton number asymmetry and a substantial part of the required relic η 0 R abundance as in the same way as the right panel of Fig. 1.In this example, the η 0 R yielded through the N 1 decay occupies about 70% of the required abundance Ω η 0 R h 2 = 0.12.The remaining part is supplied from η 0 R in the thermal equilibrium by setting the relevant parameters as λ + = 0.66 and λ 3 = 2.6.Since the N 1 decay starts at a larger x due to a smaller value of h 1 compared with the right panel of Fig. 1, its contribution to the relic η 0 R becomes larger.
Finally, we examine how the relative share of η 0 R produced from the N 1 decay in the total relic depends on the N 1 mass.Results are shown in Fig. 4 where the abundance of the thermal η 0 R and the η 0 R produced from the N 1 decay is respectively represented as Ω T and Ω D , and (Ω T + Ω D )h 2 = 0.12 is imposed.The scalar couplings λ 3,4 are fixed to realize this value.In this study, m η 0 R and M N 2,3 are fixed to the ones given as (b) in eq.( 6), and then M N 1 is allowed in the range 2000 GeV< M N 1 < 6000 GeV.We also choose two values of h 1 which are used in the right panel of Fig. 1 and Fig. 3.The figure shows that the relic η 0 R can be entirely produced by the N 1 decay if N 1 takes a close value with the η 0 R mass.In that case, λ 3 and |λ 4 | have to take large values near their perturbative bound since the η 0 R abundance produced by the N 1 decay is large and its substantial part has to be reduced through the annihilation processes.We also find from the figure that Ω D h 2 can reach the order of the required DM abundance in a wide region of M N 1 .This result suggests that the coincidence between the baryon number density and the DM density in the present Universe can be recognized as a natural consequence of the model.If η 0 R is discovered as the DM, the origin of the relic η 0 R can be an interesting subject.In that case, the present study suggests that detailed experimental study of the scalar couplings λ 3,4 might tell us the ratio of Ω D to Ω T .

Summary
We study a new scenario for the DM abundance in the scotogenic model from a view point of the coincidence of the baryon number asymmetry and the DM abundance in the present Universe.In this model, the abundance of the inert doublet DM is usually considered to be explained as the thermal relic following the WIMP scenario.Since it is irrelevant to the baryon number asymmetry in that case, the model cannot give any answer for this problem.However, if we note that the decay of the lightest right-handed neutrino N 1 generates both the lepton number asymmetry and the DM candidate η 0 R , a correlation can be found between them in a certain situation such that N 1 takes the same order mass as η 0 R .The mass of the right-handed neutrinos can take a TeV scale value consistently with the neutrino oscillation data in the model.Moreover, the decay of such N 1 is known to generate the sufficient baryon number asymmetry through leptogenesis if the N 1 is in the thermal equilibrium through certain interactions.These suggest that the relic η 0 R as the DM could be supplied not only as the thermal relic but also as the yields of the decay of N 1 which also causes the lepton number asymmetry.If the latter gives a dominant part of the DM abundance, the baryon number asymmetry has a close relation to the DM abundance.The model could give an insight for the coincidence problem.
We examine this idea quantitatively by solving the Boltzmann equations under the assumption that N 1 is in the thermal equilibrium initially.The result shows that their coincidence can be explained well.We also propose a well-motivated model which contains the interactions to make N 1 reach the thermal equilibrium and show how the coincidence of the baryon number asymmetry and the DM abundance can happen in this extended model.An interesting point of the model is that these interactions can give an explanation for the CP issues in the SM.
If both the baryon number asymmetry and a substantial part of the DM are produced through the decay of a common mother particle, it could be a promising scenario to give an answer to the coincidence problem.In such a context, low-scale leptogenesis may provide an interesting possibility not only from a phenomenological viewpoint but also from a cosmological viewpoint.The extended model studied in this paper might be considered as a prototype model which can realize such a scenario naturally.

Appendix
In this Appendix, we briefly address how CP phases in the PMNS matrix can be induced through Yukawa interactions given in eq. ( 23) following [15].If the singlet scalar gets a VEV as ⟨S⟩ = 1 √ 2 ue iρ 0 , which causes spontaneous CP violation, the CP phase can appear in the PMNS matrix through the couplings of the singlet S with vectorlike charged leptons E L,R .These Yukawa interactions extend the SM charged lepton mass matrix m e ij to a 4 × 4 mass matrix M e such as ( lL i , ĒL ) where ℓ L i and e R i are the charged leptons in the SM.F e j , G i and µ E are defined as F e j = 1 The charged lepton effective mass matrix me is obtained as a result of the mixing between the light charged leptons and the extra heavy lepton.If ỹe j is not equal to y e j and |µ E | 2 < F e F e † is satisfied, the matrix Ã could have a large CP phase.
We assume that the neutrino mass matrix M ν is diagonalized by a tribimaximal matrix U ν as U T ν M ν U ν = M diag ν , where the matrix U ν can be expressed as Majorana phases α 1 and α 2 are written by using eq.( 24) as The PMNS matrix is obtained as V P M N S = Ã † U ν where Ã is fixed through eq. ( 30).
Since the matrix Ã is expected to be almost diagonal from hierarchical masses of the charged leptons, the structure of V P M N S is considered to be mainly determined by U ν in the neutrino sector.Although tribimaximal mixing cannot realize a nonzero mixing angle θ 13 which is required by the neutrino oscillation data, the matrix Ã could compensate this fault and a desirable V P M N S may be derived as V P M N S = Ã † U ν .If we fix the relevant parameters in eq. ( 23) as y e = (0, 3 × 10 −3 , 0), ỹe = (0, 0, 10 −3 ), y E = ỹE = 3.3 × 10 −6 , y N = (2.2 × 10 −3 , 6 × 10 −3 , 7 × 10 −3 ), ỹN = (2.2 × 10 −3 , 0, 0), V P M N S obtained in this way is found to be rather good realization of the experimental results including nonzero θ 13 as shown in [15].
h 1 = 2.0 × 10 −8 and M N1 = 3.1 × 10 4 GeV are used.In the right panel for the case (b), h 1 = 5.4 × 10 −8 and M N1 = 3.1 × 10 3 GeV are used.Horizontal lines represent values of Y L (dotted) and Y η 0 R (solid) which are required to explain the abundance of the baryon number asymmetry and the DM abundance in the present universe, respectively.

Fig. 2 2 √ λ 1 λ 2 with λ 2
Fig. 2 Contours of the η 0 R relic density Ω η 0 R h 2 = 0.12 are plotted by colored solid lines in the (λ + , λ 3 ) plane.Each contour corresponds to the one with m η 0 R =2200 GeV (blue), 2000 GeV (red), and 1800 GeV (green), respectively.An upper bound on |λ + | based on the direct search of XENONnT is represented by vertical dashed lines with the same color as the one of the relic density for each mass.A diagonal black solid line and a vertical black dash-dotted line represent the condition λ 4 = 0 and λ + = −2 √ λ 1 λ 2 with λ 2 = 0.1, respectively.The allowed region is restricted into an upper-triangle region surrounded by these lines.

R.
the estimation of the relic abundance of η 0 R in the previous section can be justified.The solution of the Boltzmann equation for Y η 0 R added in the left panel of Fig. 1 proves it.In this calculation, λ + and λ 3 are fixed to 0.66 and 1.56, which are the consistent values with the contour for m η 0 R = 2 TeV shown in Fig. 2.There is no contribution to Y η 0 R from the N 1 decay and then Y The annihilation processes of η 0 R are frozen out at x D ∼ 26 where Y N 1 becomes much smaller than Y eq η 0 R

Fig. 3
Fig. 3 Left: Evolution of the lepton number asymmetry Y L and the η 0 R number density Y η 0 R .In the calculation, h 1 = 3.4 × 10 −8 and M 1 = 3.1 × 10 3 GeV for the case (b) are used.Horizontal lines represent the required values of Y L (dotted) and Y η 0 R (solid) to explain the baryon number asymmetry and the DM abundance, respectively.Right: Evolution of the relevant reaction rates included in the Boltzmann equations.Γ (ab) f stands for the reaction rate of the scattering ab → N 1 N 1 mediated by f .The washout of the lepton number asymmetry Y L is caused by Γ (ηℓ) N and Γ (ℓℓ) N .