Muon $g-2$ and non-thermal leptogenesis in $U(1)_{L_{\mu}-L_{\tau}}$ model

The gauged $U(1)_{L_{\mu}-L_{\tau}}$ symmetry is the simplest possibility to explain the observed muon $g-2$, while being consistent with the neutrino oscillations through the seesaw mechanism. In this paper, we investigate if leptogenesis can work at the same time. At first glance, leptogenesis seems challenging because the right-handed neutrino masses are related to the $U(1)_{L_{\mu}-L_{\tau}}$ breaking scale of $10\,$-$\,100\,$GeV as required from the muon $g-2$. Contrary to this expectation, we find that non-thermal leptogenesis with the right-handed neutrino masses of $\mathcal{O}(10^{7})\,$GeV is possible. The successful scenario results in strict predictions on the neutrino oscillation parameters, which will be tested in future experiments.


I. INTRODUCTION
The measurements of muon anomalous dipole moment at the Brookhaven National Laboratory [1] and at the Fermilab [2,3] have reported the 4.2 σ deviation from the Standard Model (SM) prediction [4]. 1 One of the solutions to this anomaly is to introduce an extra, neutral gauge boson Z associated with a gauged U (1) symmetry. Its contribution through the gauge interaction can enlarge the muon g − 2. Taking into account the results in search experiments of Z , the simplest remaining possibility is the U (1) Lµ−Lτ symmetry [27][28][29][30][31][32][33].
See Ref. [34] for the parameter space.
The gauge interaction of the U (1) Lµ−Lτ symmetry with leptons is given by where g Z is the gauge coupling constant, and Q e,µ,τ ≡ (0, +1, −1) are the charge of the U (1) Lµ−Lτ symmetry for each flavor. Here, L α ≡ (ν Lα , l Lα ) is the lepton doublet, andl Rα is the singlet charged lepton with α = e, µ, τ . Note that all the fermions are described by left-handed Weyl fermions. We follow the conventions of the spinor indices in Ref. [35]. The deviation of the muon g − 2 can be explained for g Z ≈ (3 -10) × 10 −4 and the mass of Z , m Z ≈ (1 -20) × 10 MeV, while avoiding other experimental constraints [34]. This region corresponds to the U (1) Lµ−Lτ breaking scale of 10 -100 GeV.
In this paper, we further investigate if leptogenesis can work while explaining the above two phenomena simultaneously. One can naively expect some obstacles to leptogenesis in this model. Firstly, masses of the right-handed neutrinos tend to be of the U (1) Lµ−Lτ breaking scale, 10 -100 GeV, to reproduce the neutrino oscillations by the seesaw mechanism. With such light right-handed neutrinos, thermal leptogenesis [47], for example, cannot be achieved.
Secondly, in analogy to the electroweak symmetry breaking, the U (1) Lµ−Lτ symmetry seems to be restored in the early universe, and thus the universe is in the U (1) Lµ−Lτ symmetric phase before the freeze-out of sphaleron processes. As we will see, leptogenesis does not work in the symmetric phase. Therefore, to find a successful scenario of leptogenesis, we have to seek a setup satisfying the following conditions: • Right-handed neutrinos have masses much larger than 10 -100 GeV.
• The U (1) Lµ−Lτ symmetry is broken even in the early universe.
We find that the first condition can be satisfied by a specific choice of Yukawa couplings of right-handed neutrinos to the U (1) Lµ−Lτ breaking fields, which in turn results in strict predictions on the neutrino oscillation parameters. The second condition can also be satisfied by choosing certain couplings between the U (1) Lµ−Lτ breaking scalar fields and the SM Higgs boson. Based on these outcomes, we demonstrate that non-thermal leptogenesis [48,49] can generate a sufficient amount of baryon asymmetry while explaining the muon g − 2 and the neutrino oscillations at the same time.
The rest of the paper is organized as follows. In Sec. II, we introduce a model with the gauged U (1) Lµ−Lτ symmetry and review the seesaw mechanism. In Sec. III, we discuss how heavy the right-handed neutrinos can be. In Sec. IV, we discuss restoration and breaking of the U (1) Lµ−Lτ symmetry in the early universe. In Sec. V, we discuss a possibility of leptogenesis in this model. Finally, we conclude this paper in Sec. VI.

II. MODEL WITH GAUGED U (1) Lµ−Lτ SYMMETRY
Let us start with the setup of the gauged U (1) Lµ−Lτ model which reproduces the active neutrino mass parameters [36,37] (see also Refs. [39,40] for earlier works). The U (1) Lµ−Lτ charge assignment for the doublet and the singlet leptons are given below Eq. (1). Three right-handed neutrinos are introduced to account for the neutrino oscillations via the type-I seesaw mechanism. They have the U (1) Lµ−Lτ charge as (N e ,N µ ,N τ ) = (0, −1, +1) in a natural way. Note again that all the fermions are described by left-handed Weyl fermions.
To break the U (1) Lµ−Lτ symmetry, we introduce two SM singlet scalar bosons σ 1,2 with the U (1) Lµ−Lτ charge +1 and +2, respectively. Note that the observed mixing angles among neutrinos can be reproduced by only σ 1 [36]. As we will see, however, σ 2 plays an important role in successful leptogenesis for the parameter region explaining the muon g − 2.
Hereafter, we call the sector consisting of Z and σ 1,2 the U (1) Lµ−Lτ sector. We summarize the phenomenological properties of the symmetry breaking sector in the Appendix A.
The Lagrangian relating to the neutrino masses is given by where we choose y α , λ α , and M αβ are real and positive by rotating the phases of L's,l R 's, andN 's. We call h αβ Majorana Yukawa coupling constants.
When the U (1) Lµ−Lτ charged scalar fields obtain non-vanishing expectation values, σ 1,2 , the U (1) Lµ−Lτ symmetry is spontaneously broken. In the present model, both σ 1,2 can be real positive (see the Appendix. A). In this case, Eq. (2) leads to the mass matrix ofN α , The corresponding mass of the Z boson is given by, To explain the deviation of the muon g − 2, we require, at the vacuum, (see Fig. 1). As we will discuss later, the temperature-dependent expectation values play important roles in successful leptogenesis. Thus, we put the subscript 0 on the vacuum expectation value (VEV) to distinguish it from the temperature-dependent expectation value.
The complex scalar field σ 2 is absent in the minimal model in Refs. [36][37][38]. In that case, the mass matrix is reduced to the so-called two-zero minor [50,51].
The active neutrino masses are given by the seesaw mechanism; where v EW denotes the electroweak symmetry breaking scale, Φ 0 = v EW 174 GeV. Note that σ 1,2 0 = 0 contribution is crucial to explain the observed neutrino oscillations since otherwise M R,eff induces only a mixing angle θ 23 between the active neutrinos. As we will see in the next section, even for σ 1,2 0 = 0, the mass parameters M ee and M µτ are severely constrained to reproduce the observed neutrino oscillations. Figure 1. Parameter region that explains the muon g − 2 within the 1σ (red) and the 2σ (pink) ranges [2]. We show m Z /g Z by the purple lines, which indicate the size of the VEVs of the scalar fields. The gray shaded region is excluded by the neutrino trident production experiment [52], the neutrino-electron scattering experiments [53,54], and the experiment searching for e − e + → µ − µ + Z [55]. In the blue shaded region, the ratio m 2 fd /µ 2 σ < 1 for T = T th in Eq. (40) (see Sec. IV A).
For later purposes, it is useful to consider λ −1 ν M R,eff λ −1 ν , which is related to the low energy observables via, where M d ν is a diagonal mass matrix defined by M d ν ≡ U T M ν U with the Pontecorvo-Maki-Nakagawa-Sakata (PMNS) matrix U. Here, M ν is invertible, and thus all the active neutrinos become massive. The PMNS matrix is represented by the mixing angles θ 12 , θ 23 , θ 13 , the Dirac CP phase δ, and the Majorana CP phases η 1 , η 2 as where s ij and c ij denote sin θ ij and cos θ ij , respectively. 2 The domains of the mixing angles are in [0, π/2), while those of the CP phases are in [0, 2π). We should note that, in this model, there are no additional CP phases in the lepton sector other than δ, η 1 , and η 2 .

III. HOW HEAVY CAN THE RIGHT-HANDED NEUTRINOS BE?
In this section, we consider possible ranges of the mass parameters of the right-handed neutrinos that reproduce the neutrino oscillations. As a reference, we summarize the observed oscillation parameters in Tab  First, let us consider the case where the active neutrino masses are not degenerate. Especially, we assume that the lightest active neutrino mass m l is much smaller than ∆m 2 sol/atm . In this case, the right hand side of Eq. (8) is given by where l is the index corresponding to the lightest active neutrino. For the central values of the neutrino mixing parameters in both of the normal and inverted orderings, the sizes of U αl U βl are at least of O(0.01). In the limit of m l ∆m 2 sol/atm , the first term in Eq. (10) is dominant. Then, we can obtain a constraint on the mass parameters by comparing the product of (1, 1) and (2, 3) elements and that of (1, 2) and (1, 3) elements of Eq. (8). We can also obtain another constraint by comparing the product of (2, 2) and (3,3) elements and that of (2, 3) and (3, 2) elements of Eq. (8). As a result, we find where we roughly take all the elements of U αl U βl ∼ 1. These constraints are independent of the Dirac Yukawa couplings, λ α 's. It can be read that these constraints are valid for m l 0.01 ∆m 2 sol/atm . From the second constraint, we find that M µτ is at most of O( σ 2 0 ). On the other hand, M ee can be as large as σ 1 2 0 /M µτ , which is much larger than σ 1 0 when M µτ is much smaller than σ 1 0 .
B. Quasi-degenerate active neutrino masses in the normal ordering From the CMB observations, the sum of the active neutrino masses is constrained as m i < 0.26 eV at the 95% C.L. [58], which means m l 0.082 eV. (When the CMB lensing and the BAO are included, it becomes m i < 0.12 eV, which means m l 0.030 eV.) As seen above, M ee and M µτ are tied to σ 1,2 0 if m l ∆m 2 sol/atm . To liberate M ee and M µτ from σ 1,2 0 , we consider m l comparable to ∆m 2 atm . As an example, we fix m l = 0.06 eV and adopt the center values for ∆m 2 sol/atm and the following mixing angles: ∆m 2 sol = 7.42 × 10 −5 eV 2 , ∆m 2 atm = 2.510 × 10 −3 eV 2 , sin 2 θ 12 = 0.304 , sin 2 θ 13 = 0.02246 .

(14)
We also take the rest of the oscillation parameters as where θ 23 and δ are in the 2σ ranges of the observational data [57]. In this case, the (1,2) and (2, 2) elements of Eq.
This structure corresponds to h eµ = h µµ = 0. Note also that this zero texture can be achieved with the opposite CP phases (mod 2π), although the corresponding Dirac CP phase is disfavored by observations [57].
With this structure of Eq. (16), the mass parameters and σ 1,2 0 are related as From this relation, we obtain 3 Here, we numerically find a parameter set in Eq. (15) where the (1, 2) and (2, 2) elements of Eq. (8) are suppressed by a factor of O 10 −13 compared with the other elements by varying (θ 23 , δ, η 1 , η 2 ). Thus, strictly speaking, the (1, 2) and (2, 2) elements might not vanish exactly. In this case, the constraints on the mass parameters in Eqs. (11) and (12) are relaxed by this factor, and the mass spectrum discussed in the following can be realized. This is also true for the case of the inverted ordering discussed below.
If λ µ 1, h eτ 1, σ 1 0 100 GeV, the mass parameters takes the maximum values as which is a rough estimate ignoring coefficients of O(0.01) in Eq. (16). Note that, even if one of λ α 's is unity, these choices satisfy the constraints from the charged flavor violation such as µ → e + γ [59] or τ → e + γ [60] due to the smallness of the other λ α 's.
For the normal ordering, it is also possible that both (2, 2) and (3, 3) elements in Eq. (8) vanish with specific choices of the lightest active neutrino mass and mixing parameters. This case is nothing but the minimal gauged U (1) Lµ−Lτ model without σ 2 studied in Refs. [37,38].
For this structure, however, the mass parameters are still constrained as Eq. (11) and either of M ee and M µτ is smaller than the scale of σ 1 0 .
At the end of this subsection, we comment on the constraints on the effective neutrino mass for the neutrinoless double beta (0νββ) decay, m ββ is bounded through the upper bound on the lifetime of the 0νββ decay from the KamLAND-Zen [61] and GERDA [62] experiments as respectively. The uncertainties of the upper bound come from the variety of nuclear matrix element calculations. For our parameter choice, we obtain which is consistent with the current constraints of the 0νββ decay experiments.
In this case, we find that the (1, 3) and (3, 3) elements of Eq. for where θ 23 and δ are in the 1σ ranges of the observational data [57]. This structure corresponds to h eτ = h τ τ = 0. As in the case of the normal ordering, this zero texture can be achieved with the opposite CP phases in spite of a disfavored Dirac CP phase.
Since the structure of Eq. (26) is the same as that of Eq. (16) except for the replacement of µ ↔ τ , we obtain the estimate of Thus, for λ τ 1, h eµ 1, σ 1 0 100 GeV, the mass parameters take the maximum values as in Eq. (19).
For this parameter choice, the effective neutrino mass of 0νββ decay is which is consistent with the current constraints of the 0νββ decay experiments.

IV. BREAKING OF THE U (1) Lµ−Lτ SYMMETRY IN THE EARLY UNIVERSE
As we will see in the next section, successful leptogenesis requires non-vanishing expectation values of σ 1,2 , and hence, it is important to clarify the aspects of the U (1) Lµ−Lτ symmetry breaking in the early universe.
In section II, we introduced two U (1) Lµ−Lτ charged scalars. To discuss the nature of symmetry breaking, it is enough to consider a single U (1) Lµ−Lτ charged scalar σ. A treelevel Lagrangian for σ and the SM Higgs doublet Φ is given by where µ 2 σ , µ 2 Φ express mass parameters for each scalar field, λ σ , λ Φ are quartic self-couplings, and λ Φσ is a Higgs-σ coupling. In the following, we take µ 2 Φ , λ σ and λ Φ positive. The covariant derivative on σ is given by, At low energy, the electroweak symmetry is broken by the VEV of Φ, and hence, the mass term of σ around σ = 0 is given by, Since we require that the U (1) Lµ−Lτ symmetry is broken at the vacuum, we take µ 2 0 > 0.
The resultant VEV of σ is given by, where we take σ 0 real and positive without loss of generality.

A. Symmetry restoration by thermal/finite density effects
Let us consider the symmetry restoration in the early universe. In this subsection, we neglect λ Φσ for a while, and hence, µ 2 0 = µ 2 σ > 0. In the early universe, σ obtains an effective mass term as V eff = m 2 eff |σ| 2 , which depends on environment such as thermal bath. For m 2 eff > µ 2 0 , the U (1) Lµ−Lτ symmetry breaking does not occur. Therefore, we find the cosmic temperature T = T c at which the phase of the U (1) Lµ−Lτ symmetry changes, that is, m 2 eff | Tc = µ 2 0 . Note that µ 0 is bounded from above to explain the muon g − 2 as µ 0 = √ 2λ σ σ 0 100 GeV. First, let us evaluate the effective mass squared in the environment where the U (1) Lµ−Lτ sector is thermalized by the U (1) Lµ−Lτ gauge interaction with SM particles in the thermal bath.
The rate of the gauge interaction is given by Γ g ≈ (g 4 Z /4π)T for T m Z . Then, we obtain the thermalization temperature as where the interaction rate Γ g becomes equal to the Hubble expansion rate H. For T < T th , the U (1) Lµ−Lτ sector is thermalized. The Hubble rate is given by where M P is the reduced Planck scale and g * ∼ 100 is the effective degrees of freedom of the relativistic species.
When the U (1) Lµ−Lτ sector is thermalized, the thermal mass is given by At high temperatures where the thermal mass is larger than µ 2 0 , the U (1) Lµ−Lτ symmetry is in the symmetric phase. Decreasing the temperature, the symmetry breaking takes place at the breaking temperature, Therefore, the U (1) Lµ−Lτ symmetry is in the symmetric phase for at least T bre < T < T th .
Next, let us move on to higher temperatures, T > T th , where the U (1) Lµ−Lτ sector is not thermalized and σ does not obtain the thermal mass. Even in this case, σ has an effective mass squared, m 2 eff = m 2 fd , due to the finite density effect of the non-thermalized U (1) Lµ−Lτ sector particles, which are produced from the SM thermal bath. As derived in the Appendix B, it is given by where n X and p X are the number density and momentum of a particle X that interacts with σ, respectively, and C X is a coefficient depending on the interaction. Here, expresses the averaged value over the particle distribution.
To evaluate m 2 fd from the self-interaction of σ, we estimate the number density of σ, n σ . For the production of σ, e.g., µµ → σσ * , n σ follows the Boltzmann equation, where the cross section is given by σv ∼ g 4 Z /(4πT 2 ), and n µ = (3ζ(3)/2π 2 )T 3 . For T > T th , n σ is given by where κ is a numerical O(1) factor. For the self-interaction of σ in Eq. (30), the effective mass squared can be read as where we use p −1 σ T −1 since σ is produced from the SM thermal bath.
Compared with the tree-level mass squared µ 2 σ , for T > T th . In Fig. 1, we show the parameter region where m 2 fd (T th )/µ 2 σ < 1 as a blue shaded region. 4 The figure shows that m 2 fd (T th )/µ 2 σ > 1 for the parameter region that explains the muon g − 2, and hence, we find that the U (1) Lµ−Lτ symmetry is restored even at the temperature above T th . As a result, we find that the symmetry is preserved at T > T bre by combining the results from the thermal and finite density effects.
We should discuss here how restoration of the symmetry proceeds when the initial condition of the universe is in the U (1) Lµ−Lτ broken phase. Let us suppose that the U (1) Lµ−Lτ charged scalar field is initially settled at σ ∼ σ 0 and the scalar potential arises as V (σ) ∼ m 2 fd |σ| 2 just after reheating of the universe. In this situation, we can consider that σ rapidly moves to σ = 0 if By using Eq. (39), it turns out that the symmetry is restored immediately when the following condition is satisfied; Thus, even if the U (1) Lµ−Lτ symmetry is initially broken with σ ∼ σ 0 , it is restored at the temperature relevant to the following discussion. 4 We assume that λ σ g 2 Z = O 10 −6 , and then the effective mass squared is mainly contributed by the self-interaction of σ. In the figure, we take λ Φσ = 0, and hence µ σ / √

B. Symmetry breaking by thermal effects
So far, we have concentrated on the effects of the U (1) Lµ−Lτ gauge interaction and the self-interaction of σ, which restores the U (1) Lµ−Lτ symmetry at T > T bre . The aspects of the U (1) Lµ−Lτ symmetry breaking, however, drastically change when the Higgs-σ coupling λ Φσ takes a certain negative value. 5 In the presence of a sizable λ Φσ , σ is thermalized through the interaction with Φ. Then, the effective scalar potential of σ obtains a contribution from the Higgs-σ interaction, where we omit a numerical coefficient. For |λ Φσ | λ σ , g 2 Z , the thermal mass is dominated by this contribution. As a result, σ acquires a non-zero expectation value, and thus the U (1) Lµ−Lτ symmetry is broken even at high temperatures.
It is subdominant compared with the top Yukawa contributions, and thus the back-reaction does not affect the dynamics of the electroweak sector significantly. Incidentally, we also find the thermal mass of Φ through the σ-loop is negligible due to |λ Φσ | λ Φ for |λ Φσ | λ σ .
Secondly, note that σ and Z do not affect the standard cosmology below T < O(m Z ).
To see this, we denote the mass eigenstates of the physical components in Φ and σ by H and S. The modulus of the Higgs boson Φ 0 contains H and S as Φ 0 = (H cos θ + S sin θ)/ √ 2 with a mixing angle θ. We take H to be the observed Higgs boson so that m H 125 GeV, while the mass of S, m S , is smaller. 6 We assume λ σ g 2 Z and then m S m Z . In this 5 In the following, we will take |λ Φσ | < O 10 −2 . Then, the VEVs of Φ and σ do not significantly affect each other while λ Φσ < 0 plays an important role at high temperatures. Moreover, µ 0 µ σ is justified for this range of λ Φσ , and the blue shaded region in Fig. 1, m 2 fd /µ 2 σ < 1, is still valid. 6 In the model discussed in the previous section, we have two complex scalars σ 1 and σ 2 . The discussion in the present section is given in the Appendix A.
limit, S decays into a pair of the longitudinal mode of Z immediately (see the Appendix A).
As Z also decays immediately into a pair of neutrinos, Z and σ do not cause cosmological Finally, we also comment on the experimental constraints on |λ Φσ |. In the model discussed in the previous section, we introduced two scalars, σ 1 and σ 2 , that couple to Φ through λ Φσ 1 and λ Φσ 2 , respectively. As we see in the Appendix A, the upper limit on the branching fraction of Higgs invisible decay mode, Br(H → invisible) < 0.11 [64], leads to (see also Ref. [65]).
We first consider a leptogenesis scenario with a negligible λ Φσ . In this case, as seen in the previous section, the U (1) Lµ−Lτ symmetry is not broken at T > T bre = 10 -100 GeV.
This temperature is lower than the temperature where sphaleron processes freeze out, T sph ≈ 130 GeV. 7 Since leptogenesis requires that the sphaleron processes convert the lepton asymmetry into the baryon asymmetry, we need to consider leptogenesis in the U (1) Lµ−Lτ symmetric phase. As we will see shortly, however, leptogenesis does not occur in the U (1) Lµ−Lτ symmetric phase.
On the other hand, when we consider a sizable λ Φσ < 0, the U (1) Lµ−Lτ symmetry is broken in the early universe. In this case, we find that the non-thermal leptogenesis [48,49] can generate sufficient lepton asymmetry to explain the observed baryon asymmetry of the universe.

A. Failures of leptogenesis in the U (1) Lµ−Lτ symmetric phase
To discuss leptogenesis in the U (1) Lµ−Lτ symmetric phase, it is convenient to construct a Dirac fermion from the two left-handed Weyl fermions,N µ andN τ , which has a U (1) Lµ−Lτ charge −1. In the symmetric phase, M µτ provides the Dirac mass term of Ψ µτ . The remaining right-handed neutrino forms a four-component Majorana fermion, whose Majorana mass is given by M ee . The four component lepton doublets are also given by, which satisfy P L Ψ Lα = Ψ Lα with P L being the projection operator on left-handed fermions.
In terms of them, the Lagrangian in Eq. (2) is rewritten as, Figure 2. Tree, vertex, and wave-function diagrams for the decay of right-handed neutrinos. The plain solid lines, solid lines with arrows, and dashed lines correspond to the right-handed neutrinos, left-handed leptons, and scalar bosons, respectively.
Here, the bar over the fermion denotes the Dirac conjugate, the superscript c denotes the charge conjugation, 8 and P R is the projection operator on right-handed fermions. The usual lepton number 9 is violated by the mass terms and Majorana Yukawa interactions in Eqs. (50) and (51).
As discussed in section III, both M µτ and M ee can be as large as O( 10 7 ) GeV. Then, leptogenesis by the decay of the heavy right-handed neutrinos can be considered. Let us first focus on the decay of Ψ e . In this case, the asymmetry is generated through the interference of the decay amplitudes given by Here, c e0 denotes the tree-level amplitude, c e1 is the one-loop amplitude where the kinematical loop-integration function F e is factored out. The tree-level and one-loop diagrams contributing to the decay are shown in Fig. 2. Note that Ψ e does not decay into Ψ (c) Lµ,τ at the tree level, and hence, we do not consider those modes. The lepton asymmetry is proportional The imaginary part of F e appears from the on-shell singularities, while the imaginary part of c e0 c * e1 depends on the interaction coefficients appearing in the diagrams. In the U (1) Lµ−Lτ symmetric phase, the flavor-changing (Ψ e ↔ Ψ µτ ) wave-function diagrams in Fig. 2 do not appear. Thus, such diagrams do not contribute to the asymmetry.
and hence, Im[c e0 c * e1 ] = 0. Therefore, no asymmetry is generated by the decay of Ψ e in the U (1) Lµ−Lτ symmetric phase.
Next, we focus on the decay of Ψ µτ . The asymmetries are obtained from the following amplitudes, Here, c µ0 and c τ 0 are tree-level amplitudes, and c µ1 and c τ 1 are one-loop amplitudes where the loop-integration functions F µ,τ are factored out. In this case, the lepton asymmetry is proportional to the sum of, As in the case of the decay of Ψ e , we look at the vertex diagrams with only Dirac Yukawa couplings, and we find Note that, in the one-loop vertex diagrams of the decay mode into Ψ Lµ , only Ψ c Lτ appears as a virtual state, and vice versa. From Eqs. (62) and (63), we find that the decay of Ψ µτ does not generate the lepton asymmetry. As a result, the decay of right-handed neutrinos in the U (1) Lµ−Lτ symmetric phase does not generate the lepton asymmetry. This consequence coincides with that in previous work [66]. 10 For M µτ and M ee = O(10) GeV leptogenesis via right-handed neutrino oscillations can be considered [67,68]. In the U (1) Lµ−Lτ symmetric phase, however, the right-handed neutrinos have different U (1) Lµ−Lτ charges, and hence, the oscillations among them do not occur.
Thus, leptogenesis via right-handed neutrino oscillations cannot work.
Hereafter, we assume that both σ 1 and σ 2 obtain non-vanishing expectation values in the early universe. In the broken phase, U (1) Lµ−Lτ charge is no longer conserved, and hence, leptogenesis by the right-handed neutrinos may take place. To discuss leptogenesis in the broken phase, we take the Majorana mass eigenstates of the right-handed neutrinos Ψ I (I = 1, 2, 3) with the mass eigenvalues M I (see Eqs. (C3) and (C14)). Due to the nonzero σ 1,2 , one degree of freedom in σ 1,2 is absorbed into the gauge boson Z , and the rest remains as physical degrees of freedom, S 1 , S 2 , and P (see the Appendix A).
As the simplest possibility, let us discuss the non-thermal leptogenesis where the inflaton mainly decays into the right-handed neutrinos [48,49]. The lepton asymmetry is gener-ated by the subsequent decay of the right-handed neutrinos into the SM leptons and the Higgs bosons. In this scenario, the reheating temperature after inflation is much lower than all three right-handed neutrino masses, T R M I . This condition can be satisfied for M ee , M µτ T R . Such large mass parameters are consistent with the observed neutrino oscillations in the cases in Secs. III B and III C.
We also assume that the decay rates of the right-handed neutrinos are much larger than that of the inflaton, i.e., Γ D,I Γ inf . In this case, since the Hubble expansion rate at the reheating is also much smaller than the decay rates of the right-handed neutrinos, we may approximate that σ 1,2 ∝ T is time-independent to discuss leptogenesis.
The yields of the right-handed neutrinos from the inflaton decay are given by Here, s is the entropy density, ρ R,inf are the energy densities of the radiation and the inflaton, respectively, and n inf,I are the number densities of the inflaton and each right-handed neutrino Ψ I . The parameter f I is the number of the right-handed neutrinos expected at the decay of one inflaton. Here, we assume that at least one of f I is of O(1). In the second equality, we have used ρ R ρ inf at the reheating time. By noting ρ R /s T R , n inf /ρ inf 1/m inf with m inf being the inflaton mass, the yields amount to As the right-handed neutrinos immediately decay after the production, the yield of the lepton asymmetry is given by where ∆n L is the difference between the number densities of (L α ,l † Rα ) and (L † α ,l Rα ), and˜ I denotes the asymmetry parameters of each right-handed neutrino defined bỹ We decompose˜ I as where I denotes the asymmetry parameters of each right-handed neutrino when Ψ I decays into Ψ L +Φ and Ψ c L +Φ † with the branching ratio, Br(Ψ I → Ψ (c) L +Φ ( †) ). Here, we sum over the flavor of leptons for both I and the branching ratio.
To evaluate the lepton asymmetry, we present the relevant interactions of the right-handed neutrinos in the mass basis. The couplings to Z are given by the covariant derivative as where Q IJ can be read from Eq. (C7). The Dirac Yukawa matrix, λ ν , is given by The breaking scalars couple with the right-handed neutrinos through the Majorana Yukawa term as with for X = S 1 , S 2 , P . See the Appendix C for the explicit form of these couplings.
First, we evaluate the branching ratio, Br(Ψ I → Ψ (c) L +Φ ( †) ). In addition to the decay into the SM lepton and Higgs, the right-handed neutrino can also decay into another lighter right-handed neutrino and the breaking scalar: Ψ I → Ψ J + X, or another lighter righthanded neutrino and the U (1) Lµ−Lτ gauge field: Ψ I → Ψ J + Z . Here, we neglect 1 → 3 decay processes because their rates will be subdominant in the total decay rate due to the phase space suppression. The tree decay rates of Ψ I → Ψ (c) L +Φ ( †) , Ψ I → Ψ J + X, and Ψ I → Ψ J + Z are given by where r J ≡ M J /M I , r Z ≡ m Z /M I . Since T R M I , m Z and m X evaluated at T R satisfy r Z 1 and m X /M I 1 and we used these limits. From these decay rates, we obtain the branching ratio as where Γ tree,I is the sum of Eqs. (73), (74), and (75). Here, we used Γ D,I Γ tree,I . Note that L +Φ ( †) ) = 1 for the lightest right-handed neutrino, I = 1.
Next, we evaluate I . To this end, we consider the tree and one-loop diagrams of the decay of the right-handed neutrinos into Ψ (c) L +Φ ( †) . In our setup, three types of one-loop diagrams contribute to the asymmetry parameters. In Fig. 3, we show the relevant diagrams: where (a) The fractions with Γ 2 tree,J and Γ 2 tree,K come from the finite width of the right-handed neutrinos and regulates the asymmetry parameters in the resonant limit, M I → M J,K [69].
From these formulae, we evaluate the asymmetry parameters in the normal and inverted ordering of active neutrino masses assuming the parameters obtained in Secs. III B and III C.
The results are summarized in Tab. II. Here, we fix the expectation values of σ 1,2 in the  vacuum and at the reheating for both the orderings as which can be realized by In the normal ordering case, we also fix the parameters as 11 |M ee | = 5 × 10 6 GeV , λ µ = 1 , |h eτ | = 1 .
In this case, the mass eigenvalues of the right-handed neutrinos become Here, M 2 and M 3 are highly degenerate because they are dominated by M µτ and their 11 Here, we consider a large value of h eµ , whose effect was not discussed in the previous section. Since we investigate non-thermal leptogenesis, the right-handed neutrinos are not in the thermal bath. They decay promptly after being generated from the inflatons, and hence we expect that they do not alter the potential of σ. Even if finite density effects of the right-handed neutrinos exist, such effects are negligible because their number density is much smaller than that of other particles in the thermal bath. degeneracy is slightly broken by h τ τ σ 2 T R , which is much smaller than M µτ .
On the other hand, in the inverted ordering case, we fix which leads to where M 2 and M 3 are highly degenerate as in the normal ordering case.
Finally, we evaluate the resultant yield of the lepton asymmetry. Since the observed baryon asymmetry is n B /s 8.7 × 10 −11 [58] and the sphaleron processes convert the lepton asymmetry generated at high temperatures into baryon asymmetry as n B (28/79)×n B−L , the success of leptogenesis requires Y L −2.5 × 10 −10 .
We first consider the case where the inflaton mainly decays into Ψ 1 . In this case, we adopt f 1 = 2 and f 2 = f 3 = 0 as a typical value when the inflaton decays into a pair of the right-handed neutrinos. From Eq. (66), the lepton asymmetry becomes Since the decay of the inflaton into a pair of Ψ 1 requires m inf > 2M 1 , we obtain the maximum lepton asymmetry as This value in the inverted ordering is compatible with the observed baryon asymmetry in the universe. As we will see below, the washout effects reduce Y L by a factor of a few at most.
Next, we consider the case where the inflaton mainly decays into Ψ 2 and Ψ 3 . As a typical value, we adopt f 2 = f 3 = 1. While Ψ 2,3 mainly decays into the left-handed lepton and Higgs, a part of them generates Ψ 1 . Thus, even if the inflaton does not directly decay into Ψ 1 , f 1 is effectively nonzero and estimated as 12 for both the normal and inverted orderings. From m inf 2M 3 , we obtain the maximum lepton asymmetry as This value in the normal ordering is compatible with the observed baryon asymmetry of the universe.
Note that the signs of asymmetry parameters of leptogenesis above are reversed when we adopt the CP phases opposite to those in Secs. III B and III C. However, such values of the Dirac phase δ (∼ 90 • ) are disfavored in the current neutrino oscillation experiments.

C. Wash-out effects in non-thermal leptogenesis
In the presence of lepton number violating processes, lepton asymmetry in the thermal bath is washed out due to the unbalance between the rates of the processes including leptons and anti-leptons. We evaluate the wash-out effects from the inverse decay of Ψ I and lepton number violating scatterings with ∆L = 1 and 2 in our scenario. It turns out that the effects are not significant with T R 4 × 10 4 GeV in the case of the benchmark point in the previous subsection.

Inverse decay
The inverse decay of Ψ I from the thermal bath violates lepton number by ∆L = 1. Since non-thermal leptogenesis requires M I T R , the inverse decay rate is suppressed. We here estimate the condition of z I ≡ M I /T R for the wash-out effect of the inverse decay to be inefficient. The inverse decay rate, Γ ID,I , is given by where g N = 2. 13 From the above equations, Γ ID,I is obtained with respect to z I as when M 1 5 × 10 6 GeV and M 2,3 1.4 × 10 7 GeV. 13 In the limit of ∆n L /n eq 1, the lepton asymmetry follows ∆ṅ L + 3H∆n L ⊃ −Γ ID,I ∆n L .

∆L = 1 scatterings
Scattering processes of Ψ I and left-handed leptons with ∆L = 1 also contribute to the wash-out effects, e.g., Ψ I + Ψ Lα → f +f , where f is a SM fermion. Here, the right-handed neutrinos do not appear in the final state since M I T R in non-thermal leptogenesis. The Boltzmann equation of the lepton asymmetry in terms of Y L is given by, where the second term expresses the wash-out effect from the ∆L = 1 scatterings. Here, Y L is at most I˜ I Y I (T R ) = I˜ I f I T R /m inf . Assuming that the inflaton mainly decays into one of Ψ I with favorable˜ I , it is sufficient to focus on such a Ψ I in the Boltzmann equation; This shows that the wash-out effect due to the ∆L = 1 scattering becomes negligible when Ref. [70]. We have also checked that Γ D,I Γ ∆L=1,I is satisfied when we consider interactions involving the U (1) Lµ−Lτ sector particles.

∆L = 2 scatterings
Finally, scattering processes between left-handed leptons and SM Higgs with a virtual Ψ I can violate the lepton number by ∆L = 2 like Ψ Lα Φ → Ψ c L β Φ † and Ψ Lα Ψ L β → ΦΦ. For T M I , the invariant amplitude squared is given by where N d = 10 is a numerical coefficient from relevant diagrams (N d = 8 comes from Ψ Lα Φ → Ψ c L β Φ † and N d = 2 comes from Ψ Lα Ψ L β → ΦΦ), and p Lα is the four-momentum of left-handed lepton with α flavor. We used the relations of Yukawa couplings and masses between the flavor and mass bases in the last equality (see Appendix C). From this amplitude squared, the cross-section of the ∆L = 2 scatterings is obtained as where, as a typical value, we took the energies of SM particles as T .
For the neutrino oscillation parameters discussed in Sec. III B for the normal ordering, For the benchmark point of non-thermal leptogenesis discussed in Sec. V B, the largest wash-out effect is for α = β = µ, and the scattering rate is given by Then, Γ ∆L=2 | µµ < H requires Since T R in our benchmark point is close to this bound, the generated lepton asymmetry estimated in the previous subsection may be reduced by about a factor of two. Therefore, the result for inverted ordering in Eq. (88) is barely consistent with the observed baryon asymmetry. 14 On the other hand, that for normal ordering in Eq. (90) is sufficiently large to explain the observed value.

VI. CONCLUSION AND DISCUSSION
An extension of the SM with the gauged U (1) Lµ−Lτ symmetry can explain the muon g − 2 anomaly. The gauged U (1) Lµ−Lτ symmetry is also consistent with the observed neutrino oscillations through the seesaw mechanism where three right-handed neutrinosN e ,N µ ,N τ have the U (1) Lµ−Lτ charges 0, −1, +1, respectively. In this paper, we investigated if leptogenesis can work while explaining the neutrino masses and muon g − 2 anomaly at the same time. 15 In our discussion, we sought the scenario where all the right-handed neutrinos are much heavier than the electroweak scale. Such a spectrum is highly non-trivial because righthanded neutrino masses are typically tied to the U (1) Lµ−Lτ breaking scale, 10-100 GeV.
Nevertheless, we found that it is possible that all the right-handed neutrinos can be as heavy as 10 7 GeV only when the Yukawa couplings have specific structures as in Eqs. (16) and (26).
We also found that leptogenesis requires U (1) Lµ−Lτ symmetry breaking in the early universe because the U (1) Lµ−Lτ symmetry prohibits the flavor mixing among the right-handed neutrinos. As we have seen, however, the gauge and self interactions of the breaking fields σ's tend to restore the U (1) Lµ−Lτ symmetry. Thus, to achieve the U (1) Lµ−Lτ broken phase in the early universe, we assumed a sizable negative value of Higgs-σ coupling λ Φσ , which leads to σ ∼ |λ Φσ |/λ σ T .
With these observations, we considered non-thermal leptogenesis and estimated the generated lepton asymmetry for both orderings of active neutrino masses. Even taking into account the wash-out effects on the asymmetry, we found the parameter points to gener- 14 In our analysis, we have not sought the optimal value of the lepton asymmetry for each ordering, and it is possible to achieve a larger lepton asymmetry by a factor of O(1). 15 In Ref. [71], leptogenesis has been discussed in a setup where the Z mass and the U (1) Lµ−Lτ breaking scale are highly separated by considering a hierarchical charge assignment between the U (1) Lµ−Lτ breaking fields and the SM leptons. ate a sufficient lepton asymmetry compatible with the observed baryon asymmetry of the universe. Therefore, we conclude that this model can explain the above three phenomena beyond the SM simultaneously.
In closing, we should mention that this scenario will be tested from various aspects in near future. Firstly, the extra neutral gauge boson Z explaining the muon g − 2 anomaly will be probed by COHERENT [72] and NA64µ at CERN [73,74]. Secondly, improvement of the upper bound on the sum of active neutrino masses from the CMB observations will probe this scenario. To have the right-handed neutrino masses much larger than σ 1,2 0 , our scenario requires m i 0.18 eV, which will be robustly tested in the future CMB observations such as CMB-S4 [75]. Finally, theoretical and experimental progress on the 0νββ decay will also be important for its test because relatively large values of the effective neutrino mass are suggested for this successful non-thermal leptogenesis as Eqs. (23) and (29). In this appendix, we summarize the symmetry breaking sector of the U (1) Lµ−Lτ model.

Model with a single scalar
For the sake of brevity, let us first consider the model with a single scalar, σ. The scalar potential of σ and the Higgs doublet Φ is given by, where µ 2 σ , µ 2 Φ express mass parameters for each scalar field, λ σ , λ Φ are quartic self-couplings, and λ Φσ is a Higgs-σ coupling. As we have discussed in Sec. IV B, we assume λ Φσ < 0, while λ σ and λ Φ are positive. Around the vacuum, we decompose Φ and σ as, where v EW and v are VEVs . The charged Higgs scalar H + and the CP-odd scalars a and A are would-be Goldstone modes, which are set to be zero in the unitary gauge.
From ∂V /∂Ĥ = ∂V /∂Ŝ = 0, we find The squared masses of the CP-even scalars are given by, in the (Ĥ,Ŝ) basis. The mass eigenstates are obtained by, where tan 2θ (A6) As we are interested in the parameter region where λ Φ |λ Φσ |, λ σ with v 10 -100 GeV, we approximate Eq. (A6) by, where the mass eigenvalues are given by, To reproduce the observed Higgs mass m H 125 GeV and v EW 174 GeV, we take λ Φ 0.13.
To the leading order of λ Φσ and λ σ , the scalar couplings relevant to H and S decays are given by, with Although the couplings to A are vanishing in the unitary gauge, they are useful to estimate the Higgs decay rate into Z through the Goldstone equivalence theorem. The scalar couplings to Z in the unitary gauge are also obtained from the kinetic term, |D µ σ| 2 , as where with Q σ being the U (1) Lµ−Lτ charge of σ.
Now, let us calculate the decay rates of H and S. The decay rates into a pair of Z 's are given by, These decay rates are in agreement with those into the Goldstone modes, Γ H→AA and Γ S→AA , which demonstrates the Goldstone equivalence theorem. The Higgs boson also decays into a pair of S's with the decay rate Γ H→SS = 1 32π where we have assumed m H m S in the second equality. In Fig. 4, we show the partial decay rates of S into a pair of Z 's and those into the SM particles. The figure shows that S dominantly decays into Z for sin θ 1. As Z mainly decays into a pair of ν's, the decays of S are virtually invisible.
Since decays of S and Z are invisible, both Γ H→Z Z and Γ H→SS contribute to the invisible decay mode of the Higgs boson as where Γ SM is the predicted value of the Higgs decay width into the SM particles, Γ SM 4.1 MeV [77]. For m H m Z , m S , we obtain Thus, from the upper limit on the branching fraction of Higgs invisible decay mode, Br(H → invisible) < 0.11 [64], we find the upper limit on |λ Φσ |,
Around the vacuum, the scalar fields are decomposed as In this case, H + , a, and a linear combination ofÂ 1,2 are would-be Goldstone modes, which are set to zero in the unitary gauge.
From ∂V /∂Ĥ = ∂V /∂Ŝ i = 0, we find The squared mass matrix of the CP-even scalars are given by, in the (Ĥ,Ŝ 1 ,Ŝ 2 ) basis. As we are interested in the case where λ Φ |λ Φσ i |, λ σ i (i = 1, 2), we can approximately diagonalize the matrix by, where the small mixing angles are given by, for i = 1, 2 with small multiplicative correction factors of O(λ Φσ i , cv i /m H , λ σ i ). To the leading order of λ Φσ i , cv i /m H , λ σ i , the squared mass matrix of (S 1 ,S 2 ) is given by, We define the mass eigenstates of M 2 S as (S 1 , S 2 ) whose masses are denoted by m S 1 and m S 2 .
The squared mass matrix of the CP-odd scalars is given by, in the (Â 1 ,Â 2 ) basis. The matrix can be diagonalized by Here, A corresponds to the massless would-be Goldstone mode, while the mass of P is given by, After eliminating the mixings toŜ i 's, the Higgs couplings to the scalar sector are given by, where The absence of HS 1S2 and HÂ 1Â2 is due to our simplification that we have omitted the |σ 1 | 2 |σ 2 | 2 term. The Higgs couplings to Z 's and P are, on the other hand, given by where s α ≡ sin α and c α ≡ cos α.
As in the case of the single scalar model, S 1,2 dominantly decay into a pair of Z 's. As we will see below, P can decay into a CP-even scalar and Z . Thus, the branching ratio of the invisible Higgs decay is given by, Br(H → invisible) = Γ H→Z Z + Γ H→SS + Γ H→P P + Γ H→Z P Γ SM + Γ H→Z Z + Γ H→SS + Γ H→P P + Γ H→Z P , where Γ H→SS denotes the sum of the decay rates into the CP-even scalars (S 1 , S 2 ). For m H m Z , m S 1 , m S 2 , m P , we find 16 Γ H→Z Z + Γ H→SS + Γ H→P P + Γ H→Z P (|λ As a result, the upper limit on the branching fraction of Higgs invisible decay mode, Br(H → invisible) < 0.11 [64], results in a constraint, for the two scalar models (see Eq. (A18)). 16 Note that Γ H→Z Z + Γ H→P P + Γ H→Z P = Γ H→Â1+Â1 + Γ H→Â2+Â2 , in the limit of m H m P , which is in agreement with the Goldstone equivalence theorem.
Finally, let us comment on the fate of the CP-odd scalar P . Through the interactions, P decays into Z and a CP-even scalar when kinematically allowed. In such a case, the decay rate of P is comparable to those of the CP-even scalars in Fig. 4 for m P = O(m S i ).
For example, for we obtain which allows P → Z + S 2 . Hence, P does not cause any cosmological problems for m P = O(1) GeV.
When P is lighter than both the CP-even scalars, P decays into Z Z Z through a virtual CP-even scalar. In this case, the decay rate of P is expected to be roughly suppressed by λ 2 σ i m 4 P /(8πm 4 S i ) compared to the two-body decay rate in Fig. 4. Even in such a case, P does not cause cosmological problems as long as its decay temperature, T D √ Γ P M P , is much higher than O(10) MeV.

Appendix B: Finite density effect
In this appendix, we derive an effective mass of a scalar field that couples to the particles with finite densities. For simplicity, let us consider a model where a complex scalar field σ couples to a real scalar field χ that has finite density. The Lagrangian density of them is assumed to be where λ σ and κ are dimensionless coupling constants, and v and m χ are the parameters with mass dimension one. We assume λ σ > 0 so that the scalar potential of σ is bounded from below. In general, the sign of κ can be positive or negative. In this model, σ obtains a non-vanishing expectation value at the vacuum, which breaks the global U(1) symmetry of σ.
Let us consider a system with a finite volume V 3D = L 3 , where L is a large length scale.
The mode expansion of χ is then given by, where n = (n 1 , n 2 , n 3 ) is a set of three integers, with which the 3D momentum of χ is given by, The creation and annihilation operators satisfy Let us assume that the number density of χ is n χ = 0, which is given by where N n is the particle number for each mode in the 3D volume V 3D . The corresponding particle state ofχ is given by, where In this state, the expectation value of χ 2 (x) is given by Note that we define the renormalized operator [χ 2 (x)] R by Now consider the effective mass of σ around its origin, Thus, for κ > 0, the effective mass of σ becomes positive for and hence, the U(1) symmetry of σ is restored due to the finite density effect. When χ is thermalized, for example, the above mass term reproduces the thermal mass Now, let us apply this result to the U (1) Lµ−Lτ model. In this case, σ couples to the U (1) Lµ−Lτ gauge boson, which is expected to have a finite density in the early universe. The relevant coupling is where we take the unitary gauge. Thus, σ couples to the finite density particle with κ = 2g 2 > 0, and hence, we find that the U(1) symmetry of σ can be restored by the finite density of the gauge boson.
Let us also comment that the finite density of σ itself can also contribute to the effective mass of σ. Since the quartic coupling constant λ σ is required to be positive to avoid the potential unbounded from below. Thus, the finite density of σ also can restore the symmetry.
In the case of the thermalized σ, the finite density effects on the mass reduce to the thermal mass, component fermions, the Lagrangian related to right-handed neutrinos is given by When the U (1) Lµ−Lτ symmetry is spontaneously broken due to nonzero σ 1,2 , the Majorana mass term receives additional contributions from the Majorana Yukawa terms as Thus, either ofN α is not the mass eigenstates in general. Since M R,eff is a complex symmetric matrix, it is diagonalized by a unitary matrix Ω using the Autonne-Takagi factorization as Here, M I is the mass eigenvalue with I = 1, 2, 3 and we choose Ω so that M I is real and M 1 ≤ M 2 ≤ M 3 without loss of generality. Thus, the Majorana mass term is diagonalized whereN I ≡ Ω † IαN α represents the mass eigenstates.
Using this mass basis, we rewrite the Lagrangian in Eq. (C1) as Here, and the covariant derivative is given by, with The Dirac Yukawa couplings are given by and the Majorana Yukawa couplings are represented by the matrices: with Next, we rewrite the Lagrangian in terms of four-component fermions: The mass eigenstates of right-handed neutrinos as four-component fermions are given by Then, the Lagrangian becomes L = L kin + L mass + L DY + L MY , with Here, For later convenience, we also define (C24) Figure 5. Momentum assignment in tree, wave-function, and vertex diagrams.
To evaluate the asymmetry parameters, we consider tree (Fig. 6) and one-loop (Figs. 7 -11) diagrams. The one-loop diagrams are classified into wave-function and vertex diagrams. The assignment of the momenta in the following calculations is shown in Fig. 5. Hereafter, we approximate that the particles other than the right-handed neutrinos are massless. We also take the leading order of λ Φσ i and cv i /m H . In other words, we ignore the mixings among the Higgs andŜ 1,2 and identifyŜ 1,2 with the mass eigenstates S 1,2 . The CP-odd scalarsÂ 1 andÂ 2 are related to the physical degree of freedom P bŷ where we neglect the would-be Goldstone mode, A (see also the Appendix A).
First, we consider the tree diagrams in Fig. 6. The amplitudes of these processes are given by iM tree =ū(−iλ ναI P L )u .
The decay rate of right-handed neutrinos through these processes is proportional to |M tree | 2 + |M tree | 2 = 2[λ † ν λ ν ] IIū P L uūP R u (D6) Figure 6. Tree diagrams for decays of right-handed neutrinos. Plain solid lines, solid lines with arrows, and dashed lines correspond to right-handed neutrinos, left-handed leptons, and scalar bosons, respectively.

Wave-function diagram with Ψ L Φ loop
The diagrams with Ψ L β loop in the upper side of Fig. 7 give where n w = 2 represents the degree of freedom of the SU (2) doublet in the loop.
The diagrams with Ψ c L β loop in the lower side of Fig. 7 give iM L c a =ū(−iλ * ναJ P R ) As discussed in Sec. V A, the asymmetry parameters are proportional to the imaginary part coming from loop integrals. Thus, we pick up the imaginary part in the loop integrals and obtain Then, the difference between |M| 2 and |M| 2 from diagram (a) is As a result, we obtain the asymmetric parameter as  where we introduce the regulator with Γ 2 tree,J reflecting the finite width of the right-handed neutrinos [69].

Vertex diagram with Ψ L ΦΨ loop
The vertex diagrams with Ψ L ΦΨ loop in Fig. 8 give (D18) Note that M b does not include n w .
As in the case of diagram (a), we pick the imaginary part in the loop integrals and obtain M b ⊃ ir J 16π λ * ναJ [λ † ν λ ν ] JI 1 + (1 + r 2 J ) log r 2 J − (1 + r 2 J ) log 1 + r 2 Figure 9. Wave-function diagrams with Ψ X loop for decays of right-handed neutrinos.

Vertex diagram with ΦΨ X loop
The vertex diagrams with ΦΨ X loop in Fig. 10 include the scalar coupling proportional to λ Φσ σ i . From the dimensional analysis, the contribution from this diagram is suppressed by σ i /M I compared with the contributions above. Moreover, λ Φσ i is bounded from above as λ Φσ i 7 × 10 −3 as shown in the Appendix A while λ ναI and F X IJ are O(1) at most. Thus, the contribution of these diagrams to the total asymmetric parameter can be safely neglected.

Vertex diagram with Ψ L Ψ Z loop
The vertex diagrams with Ψ L Ψ Z loop in Fig. 11 include two of the gauge couplings g Z .
Thus, the asymmetric parameter from these diagrams is suppressed by g 2 Z 10 −6 and can be safely neglected.