Leptogenesis due to oscillating Higgs field

We propose a new leptogenesis scenario in which the lepton asymmetry and matter particles are simultaneously generated due to the coherent oscillating Higgs background. To demonstrate the possibility of our scenario, we consider the type-I seesaw model as an illuminating example and show the numerical analysis. In order to generate the required lepton number $|n_L/s| = 2.4 \times 10^{-10}$, we find that the scales of the Higgs background oscillation and the lightest right-handed neutrinos mass are required to be higher than $10^{14-15}$ GeV


Introduction
The observation of the cosmic microwave background (CMB) supports two important cosmological events [1,2]. One is the cosmic inflation that causes the exponential expansion of the Universe. The evidence is shown in the scale invariance of the power spectrum. Another important event is the big bang nucleosynthesis (BBN) in which the abundance of baryons (hydrogen, deuterium, helium, etc.) is fixed from a second to 3 minutes after inflation. According to the theory of the BBN, the required initial conditions to realize the current Universe are a temperature higher than a few MeV and the baryon number to photons number ratio η = n B /n γ ∼ 6 × 10 −10 . Despite the CMB information is originated after 380,000 years later from the BBN, the observation shows η = (6.10 ± 0.04) × 10 −10 [2,3]. However, these two events cannot connect directly because the temperature of the Universe after inflation might go to zero due to the extreme dilution. On the other hand, the BBN must start with a high-temperature scale at least more than a few MeV. Therefore, the Universe must be heated due to some mechanism after inflation.
The (re)heating theory has been developed by many authors (for review, see e.g. [4][5][6]). Typically it is described by the decay of the inflation field into other particles after inflation, but a picture of the perturbative decay is not correct. Taking into account a particle coupled to the oscillating inflation field, the non-perturbative particle production occurs and the produced particle number can grow exponentially by parametric resonance [4,[7][8][9]. This process happens before the Universe is thermalized, so-called preheating era.
On the other hand, inflation makes our Universe no particle state because any particles are extremely diluted. Thus, the Universe must evolve from no baryon number state to non-zero baryon number state. This scenario is called baryogenesis that requires a small asymmetry between baryons and anti-baryons. The Standard Model (SM) cannot produce enough baryon asymmetry from a symmetric Universe. At present, one of the hopeful scenarios is the thermal leptogenesis [10] in which superheavy right-handed neutrinos and their interactions are added to the SM. The decay of right-handed neutrinos can generate lepton number. This generated lepton number can be converted into the baryon number through the sphaleron process [11] after the decay of the right-handed neutrinos.
In this paper we propose a new scenario of the baryogenesis that occurs in the preheating era. There are several baryogenesis scenarios associated with preheating [12][13][14] in which certain particles heavier than the inflation field can be produced through the nonperturbative process and they decay with baryon asymmetry later. In contrast to such scenarios, our scenario generates the baryons or leptons and their asymmetry simultaneously due to an oscillating background field. In earlier studies, it was pointed out the possibility of the asymmetry production in the preheating era with simple scalar cases that have charge violating interactions and C-and CP-violating parameters [15,16]. But it is still not clear that there exist realistic models that can generate enough baryon or lepton asymmetry.
To show the story with a concrete model, we consider the type-I seesaw model as an example. Then our scenario corresponds to a new-type of non-thermal leptogenesis. A sketch of our scenario is that the lepton asymmetry is produced by the non-perturbative particle production of the left-handed neutrinos due to the coherently oscillating Higgs background. For example, let us consider a lepton number violating operator C AB (H A )(H B ) first proposed by Weinberg [17], where C AB is a coupling and H is the Higgs doublet, and A is the lepton doublet in the generation A. This interaction also represents mass terms of the left-handed neutrinos in the case that the Higgs has a vacuum expectation value. If the Higgs vacuum expectation value varies non-adiabatically, the left-handed neutrinos would be produced. An important point is that this neutrino production violates the lepton number. Therefore, if the theory has C-and CP-violating parameters, then the lepton number asymmetry is also generated at the same moment. However, this process must happen at a lower energy scale than the scale of the coupling C AB . Otherwise, the backreactions or the neutrino-Higgs scatterings happen, erasing the generated asymmetry. Similar situations are investigated [18], but the interaction contents in that study are different from ours. The model in [18] was considered the type-I seesaw and a dimension six operator that is proportional to B + L current. Then, the lepton asymmetry can be generated by a single generation. In our model, we only include the type-I seesaw with three generations of the left-and the right-handed neutrinos.
Since it is difficult to obtain the analytic behavior, we will demonstrate the formulation and show the results by the numerical calculation. For simplicity, we neglect the spatial expanding effect in the later calculation. Furthermore, we assume that the energy scale of the produced left-handed neutrinos is much lower than the mass scale of the heavy righthanded neutrinos in order to avoid backreactions that would erase the generated asymmetry. Then the production of the right-handed neutrinos associated with the Higgs is forbidden, and the scattering of the left-handed neutrinos and the Higgs intermediated with the heavy right-handed neutrinos is also suppressed.
The paper is organized as follows. In section 2, we derive the basic operator equations and construct the effective theory in which the right-handed neutrinos do not appear. Using these results, we construct the equations of motion for two-point functions and the Higgs background to evaluate the lepton asymmetry in section 3. The numerical results are also shown in this section. We summarize our conclusion and discuss in section 4.

Formulation of operator equations
The goal of this paper is to demonstrate that the lepton asymmetry can be generated by the oscillating Higgs background. At first, we explain what we must calculate to know the generated lepton asymmetry. As mentioned in the previous section, we consider our model with the type-I seesaw model that includes the SM and the three generations of the right-handed neutrinos. Furthermore, we neglect the spacial expanding effect in the later calculation for simplicity and assume that the energy scale of the produced neutrinos is much lower than the heavy right-handed neutrino scale in order to avoid backreactions erasing the generated asymmetry. In our formulation, we use the notation of the metric as g µν = diag(+1, −1, −1, −1), and use the two-component spinors as the representation of fermions.
The net lepton number can be defined by a vacuum expectation value of U (1) Noether charge of the leptons as where the superscript A runs the generation, and ν L , e L and e c R are the left-handed neutrino, the left-handed electron and the charge conjugate of the right-handed electron, respectively. The lepton number density n L can be obtained by n L = L/V where V ≡ d 3 x is a spatial volume of the system. As shown in (2.1), we need to follow every leptonic two-point function. To derive the equations of motion for each two point function, at first we derive the operator equations for leptons.
In later calculation, we derive the equations of motion for each operator field and construct the differential equations for all the required two-point functions using the operator equations. Then, we solve them numerically, and we follow the time evolution of the generated lepton asymmetry referring to the solved two-point functions.

Lagrangian
The Lagrangian relating to the lepton sector is given by where superscripts A, B run 3 generations (flavor basis) and a, b run SU (2) L components, and mass matrix of the right-handed neutrinos M R is chosen to be diagonal and real components. We choose the unitary gauge as H 1 = 0, H 2 = h/ √ 2, 1 = ν L , 2 = e L , then the Lagrangian by each component is shown by where B µ and W a µ for a = 1, 2, 3 are the gauge boson fields, g Y and g W are the gauge couplings corresponding to U (1) Y and SU (2) L gauge symmetries, respectively.

Approximate solution for right-handed neutrinos
At first, we derive the equations of motion for the right-handed neutrinos. From (2.3), the operator equations for the right-handed neutrinos can be obtained as Since we assume that the scale of the non-perturbative particle production is enough lower than the right-handed neutrino mass scale, we can construct an effective equation in which the right-handed neutrinos do not appear. Assuming M −1 R ∂ 1 in (2.4) and (2.5), we can obtain the approximate solution as Note that the first term in the above equation is of the leading order, and the second and third terms are of the next-to-leading order.

Left-handed neutrinos
The operator equation for the left-handed neutrinos is given by Substituting the approximate solution (2.6) to the above equation, we obtain an approximate equation without the right-handed neutrinos as where · · · means higher order terms. In the above operator equation, we impose the following approximations: These indicate that the Higgs has a homogeneous background but the gauge fields do not have backgrounds 1 . Then (2.8) becomes a simpler equation as Although there are no backgrounds of gauge fields (one-point functions), two-point functions that correspond to their number density are not negligible. In our scenario, the gauge bosons can be produced by the coherent oscillation of the Higgs background. We will see in section 3.2 that the bosonic two-point functions appear in the Higgs sector.
In (2.10), a matrix y ν M −1 R y T ν · h 2 means the masses of the left-handed neutrinos but it is not diagonalized in general. This mass matrix can be diagonalized by Pontecorvo-Maki-Nakagawa-Sakata matrix U PMNS as where m ν is a diagonal matrix and every component is real. Since U T PMNS y ν M −1 R y T ν U PMNS is a constant matrix, it can be represented by the present Higgs vacuum expectation value and the present left-handed neutrino masses: and m 1 , m 2 , m 3 are the masses of the left-handed neutrinos in mass basis. Equation (2.12) can also be rewritten as [19] where is an orthogonal complex matrix. Using this matrix, the Yukawa matrix y ν can be represented by Hence, we can obtain Finally, the operator equation (2.10) written by the mass eigenstate can be shown as where the superscripts I, J run mass eigenstate indices, is a mass eigenstate of the left-handed neutrinos, and Note that Z is a non-diagonal Hermitian matrix in general. The CP-violating parameters are included by the non-diagonal complex elements.
Since it is convenient to treat the equations by the Fourier transformed representation, we represent (2.18) by each Fourier mode. We expand ν L (x) by a plane wave and each helicity mode as where (e s k ) α is an eigen-spinor for the helicity operator that satisfies 2 : Then (2.18) in the Fourier space can be represented by (2.28) Later, we will use this differential equation to derive each two-point functions.

Left-and right-handed electrons
Since the electron sector does not have interaction with right-handed neutrinos, each equation for electrons is the same as in the SM. Moreover, as long as we use the approximation (2.9), the electron sector is separated from the left-handed neutrinos. Therefore, even if the lepton asymmetry is generated in the neutrino sector, the effect does not influence the electron sector. For this reason, we ignore the electron sector in this work.

Analysis of lepton asymmetry
As shown in the previous section, the net lepton number is defined by (2.1). We expect that the lepton asymmetry from the electron sector can be neglected. Thus, we only consider the neutrino sector and obtain In this paper, we use the following representation: is a phase defined by the x-and y-components of the momentum. Useful properties: Therefore, we need to follow the values of each two-point functions in order to know the net lepton number.
According to the CMB observation [1], the baryon number in the Universe is given by where n B is the baryon density and s is the entropy density. The lepton number required for achieving this observation can be estimated as follows. After the lepton number is generated in our scenario, a part of it converts to the baryon number through the sphaleron process which is in equilibrium at the temperature T 10 12 GeV. The amount can be estimated by [11] Thus the required lepton number is

Differential equations for two-point functions
We can construct differential equations for the two-point functions using the operator equation (2.28). The relevant differential equations are given by Note that correlation functions of three or more points that provide the interaction effects do not appear in the above equations because we approximate the bosonic operators by background fields in the previous section. Although it is still difficult to analytically solve these differential equations because of the time-dependent matrices m ν and Z, the numerical analysis can be performed in a relatively simple way. The initial conditions for each two-point function are chosen to be 3 where t = t 0 is an initial time, V ≡ d 3 x is a spatial volume of the system, and

Evolution of Higgs background
The time dependence of the matrices m ν (t) and Z(t) is described by the dynamics of the Higgs background. Thus, we must follow the time evolution of the Higgs background in order to solve Eqs.(3.5)-(3.7). The Lagrangian relating to the Higgs sector is given by where L Yukawa is the Higgs interaction terms with fermions. We neglect the quadratic term of the Higgs because the scale we focus is much higher than the electro-weak scale.
Using the unitary gauge, H 1 = 0, H 2 = h/ √ 2, the Lagrangian in each component can be represented as

18)
3 Actually, the asymptotic solution of (2.28) corresponding to zero particle state is given by and a s k (a s † k ) is an annihilation (a creation) operator which satisfies The Lagrangian (3.17) leads the equation of motion for the Higgs as Taking the vacuum expectation value in the above equation, we can obtain the equation for the Higgs background as where J BR is a backreaction term that consists of This backreaction plays a quite important role. As we will see later, the lepton asymmetry cannot be fixed without the backreaction. Because the backreaction (3.23) includes too much information and thus the form is complicated, let us extract the relevant effect and approximate them to be a useful form. At first, we neglect three-point functions since we constructed the differential equations up to two-point functions. This approximation is valid as long as Z µ ∼ W ±µ ∼ 0. Moreover, we can expect that the bosonic two-point functions would be much more significant than the fermionic functions because each two-point function relates to the number density and the bosons would be exponentially produced due to the parametric resonance by the coherent oscillation of the Higgs background. Therefore, we can approximate the backreaction term as Speaking roughly, δm 2 h is the product of the couplings and the total number densities of the Higgs and the gauge bosons. Because it is complicated to follow the time evolution of the two-point functions of these species (3.25), we approximate them by a single real scalar field 4 as where χ is an artificial scalar field that satisfies is an effective degrees of freedom included in (3.25). The coefficients 1, 3, 6 in (3.30) correspond to the degrees of freedom of h, Z, W ± bosons, respectively. Finally, the differential equation for the Higgs background can be derived as 5 where u k is χ's time-dependent wave function which satisfies The above initial conditions for u k indicate zero-particle state as the initial state. Thus, this analysis is valid in the case that the thermal particle number described by the temperature of the Universe is negligible. We assume that the produced bosons due to the oscillating Higgs background are more than the thermal particles. Otherwise, the backreaction does not sufficiently affect to the Higgs background and the final lepton number would not be fixed. 5 The representation leads to but this term diverges. In order to renormalize, we add a counter term as by hand. The counter term 1/2ω χk corresponds to 0(t)|χ 2 |0(t) where |0(t) is a vacuum state defined at time t.

Scales of the particle production
In this section, we discuss what the momentum scale of the produced left-handed neutrinos is . Non-perturbative particle production occurs when the adiabatic condition is violated. The condition for producing the heaviest left-handed neutrino is given by where [ω k ] heaviest ≡ [ω k ] II and m heaviest ≡ [m ν,now ] II for the heaviest generation I of the lefthanded neutrinos. In order to obtain the production scale, we need to know the dynamics of the Higgs background. At the beginning of the particle production era, the backreaction J BR in (3.22) can be neglected. Then the time derivative of the Higgs background can be represented as In the derivation of the above equation, we assume ḣ = 0 when h = h max . Substituting (3.39) into (3.38), we obtain (3.41) Note that the parameter Q corresponds to the resonance parameter q known in the Mathieu equation 6 . In the case of Q 1 which corresponds to the broad resonance, the condition (3.40) can be simplified to This result shows us the production area of the left-handed neutrinos. Furthermore, applying the Tayler expansion to the Higgs background where t = t * is a time when h(t = t * ) = 0, we obtain the time scale of the particle production around h = 0 as (3.46) 6 The Mathieu equation for a function y = y(x) is given by where A and q are the resonance parameters. In this equation, the non-adiabatic condition is obtained as This time scale implies the momentum scale of the produced particles as In the case of Q 1 which corresponds to the narrow resonance, the condition (3.40) is satisfied for almost any value of h except the narrow area of h ∼ h max . Therefore, the time scale of the particle production is estimated by the oscillation time scale of the Higgs background and thus the momentum scale can be estimated by (3.49)

Entropy density
Since the cosmological observation of the baryon number density n B is normalized by the entropy density s as n B /s, we need to estimate not only the net lepton number density but also the entropy density. The entropy density can be evaluated by the distributions of all species as 7 by where N i and f (i) k are the degrees of freedom and the distribution function for species i particle, respectively.
In our scenario, sizable gauge bosons and the Higgs bosons are produced by the oscillating Higgs background. This process can be regarded as the entropy production. As we approximated in (3.27), we can also approximate (3.52) by the effective degrees of freedom N deg and the distribution of χ particle. Then, we can obtain where ω k = |k| 2 + m 2 is one-particle energy, µ is chemical potential and T is temperature, one can derive the familiar representation of the entropy density from (3.52) as where ρ is the energy density, p is the pressure and n is the number density.
where the distribution function can be represented by the wave function as (3.54)

Model parameters
Before we show our numerical results, we mention the required input parameters for our analysis. There are 17 model parameters required by the analysis of the lepton asymmetry: • The gauge couplings g Y , g W and the Higgs self-coupling λ • Complex orthogonal matrix O: 6 real parameters The SM parameters g Y , g W , and λ could be determined by the renormalization group running once the initial conditions for h(t 0 ) , ḣ (t 0 ) are determined. For simplicity, we assume that m 3 is the heaviest left-handed neutrino and that a non-degenerate mass spectrum to the left-handed neutrinos in later analysis 8 . Taking into account the observations of the neutrino oscillation [20], we set the heaviest mass of the left-handed neutrinos as The mass scale of the right-handed neutrinos is constrained in our analysis. The lightest right-handed neutrino must be much heavier than the heaviest left-handed neutrino because of the validity of the effective theory. Hence, the lightest mass of the right-handed neutrinos must be where h max is a maximal value of | h(t) |. Note that the typical scale in our scenario is characterized by h now 2 m 3 = 1.21 × 10 15 GeV. where c ij ≡ cos θ ij = cosh(Im θ ij ) · cos(Re θ ij ) − i sinh(Im θ ij ) · sin(Re θ ij ) (3.59) The complex parameters θ 12 , θ 23 , θ 13 correspond to 6 real parameters.

Numerical results
Finally, we show our numerical results with a set of specific parameters. In our analysis, we set the gauge couplings and the Higgs self-coupling at the Higgs oscillation scale as 9 As we mentioned, one can see that the net lepton number is produced with a certain magnitude but it continues to fluctuate between positive and negative. Therefore, the amount is not fixed. The case with backreaction is shown in Figure 2. Contrary to the case without backreaction, the evolution of the net lepton number freezes after several times of the oscillation of the Higgs background. The upper panel shows the time evolution of the lepton-entropy ratio, its absolute value and the amplitude of the Higgs background. The graph of n L /s seems that the generated net lepton number oscillates when the Higgs background reaches to the edge of the amplitude and that the final value is rather small. The graph of |n L /s| with the logarithmic scale shows us that the magnitude of the asymmetry is quite steady but its sign is flipping. This behavior is similar to the case without backreaction. At the last stage, however, the generated net number is fixed with non-zero value when the amplitude of the Higgs background decreases rapidly. The numerical result shows n L /s −6.5 × 10 −10 at t = 100∆t where ∆t is a time scale of the particle production defined in (3.46). The main reason for the amplitude reduction of the Higgs background is the resonant production of the gauge bosons and the Higgs boson. The production of the left-handed neutrinos with lepton number violation also occurs but the energy conversion from the Higgs background is much smaller because the interaction is suppressed by the right-handed neutrino mass scale. Once enough bosons are produced, their plasma behaves as an effective mass of the Higgs background. In consequence, the Higgs background loses its energy and the non-adiabatic condition, and hence the resonant particle production finally stops. On the other hand, one can see that the graph of |n L /s| seems to decrease in its evolution and to be fixed at the -1.5e-08 -1.0e-08 -5.0e-09 0.0e+00 5.0e-09 1.0e-08 1.5e-08  volume ∆t 3 . Actually, the magnitude of the net lepton number is almost fixed except its sign. But the entropy is generated exponentially by the parametric resonance. As a result, n L /s reduces at the early stage, and the magnitude is fixed because the entropy production becomes steady at the later stage. It is also interesting that the produced entropy is much smaller than the bosonic number density.
We show the comparison with different values for h(t 0 ) and M 1 in Figure 3. According to this result, larger scale of h ( t 0 ) and M 1 gives larger magnitude of n L /s. Although we fix h(t 0 ) /M 1 = 0.1 in this comparison, the case of h(t 0 ) /M 1 < 0.1 leads smaller magnitude of n L /s during the whole time evolution. This figure also shows that more than 10 14 GeV scales for h(t 0 ) and M 1 are required to generate |n L /s| ∼ 10 −10 in our scenario.

Conclusion and discussion
In this paper, we proposed a new leptogenesis scenario in which the lepton asymmetry is generated by the coherent oscillating Higgs background, and demonstrated that the type-I seesaw model as an illuminating example can generate enough lepton number. Although the analytic results are not derived because of the difficulty of the analysis, we showed the numerical results with some choices of parameters. We emphasize that in our scenario the particle production and the asymmetry generation occur simultaneously. Hence any perturbative decay processes do not need at all. This is a quite different point from the ordinary scenario. Although we discussed the type-I seesaw model, a similar scenario is possible to be constructed by other models that include the baryon or lepton number violating interaction, C-and CP-violating parameters, and the time-varying background fields. If such a model has the oscillating background field and baryon or lepton number violating interaction with C and CP violation, then the asymmetry could be generated.
Because we used some approximations in our analysis to avoid complicated formulations, we need to mention the validity and the condition. First, we constructed an effective theory in which the right-handed neutrinos do not appear. If one wants to apply to a case that the scale of the initial amplitude of the Higgs background is larger than the mass scale of the right-handed neutrinos, a complete calculation is needed. Secondary, we neglected the fermionic two-point functions and any correlation functions of more than two-points. In the case that the bosonic resonant production is not relevant, the fermionic two-point function could affect the backreaction in (3.23) as well as the bosonic terms. If the effects of the correlation functions of three and more points could be included in the analysis, the results would describe the effects of decay and scattering processes, which could provide secondary sources of the lepton asymmetry. Finally, we neglected the spatial expanding effect in the whole calculation. Although a realistic model must include the expansion effect, we ignored it for simplicity and to see a clear structure of the dynamics. If the Hubble parameter H is much larger than the oscillation scale of the Higgs background (H √ λ| h max |), the Higgs background maintains the initial value of its amplitude. When the expansion scale becomes smaller than the oscillation scale (H √ λ| h max |), the Higgs background can oscillate. Therefore, this motivates us to adopt the initial conditions of the Higgs background as h(t 0 ) = h max , ḣ (t 0 ) = 0. However, the expanding effect might change the whole dynamics seriously. Since the time scale of the particle production ∆t is much smaller than the Hubble inverse H −1 in many cases, the expanding effect at the moment of the particle production can be negligible. But the spatial expansion makes the amplitude of the Higgs background shrink, and thus the velocity of the background at the particle production area becomes smaller. Consequently, the amount of the produced left-handed neutrinos would be reduced, and thus there is a possibility that the lepton asymmetry would become smaller. On the other hand, entropy production would also be reduced. The result might strongly depend on the evolution of the Hubble parameter, i.e., the matter contents. We leave the analysis with the expanding effect to a future work.