Baryogenesis From Flavon Decays

Many popular attempts to explain the observed patterns of fermion masses involve a flavon field. Such weakly coupled scalar fields tend to dominate the energy density of the universe before they decay. If the flavon decay happens close to the electroweak transition, the right-handed electrons stay out of equilibrium until the sphalerons shut off. We show that an asymmetry in the right-handed charged leptons produced in the decay of a flavon can explain the baryon asymmetry of the universe.


I. INTRODUCTION
There is more matter than antimatter in the universe. This baryon asymmetry of the universe (BAU) is measured to be [1,2] where n B(B) is the (anti-)baryon number density and s is the entropy density.
In order to explain the BAU, the three so called Sakharov conditions [3] must be satisfied: (i) Baryon (or lepton) number must be violated, (ii) C and CP must be violated and (iii) There must be departure from thermal equilibrium.
Baryon number and CP are violated in the electroweak (EW) sector of the Standard Model (SM). However the CP violation in the SM is too small to account for the BAU. In addition, there is no out of equilibrium process in the SM. In order to produce the BAU, beyond the SM physics with additional degrees of freedom is needed. These degrees of freedom can also solve other deficiencies in the SM. Most notably example is in leptogenesis [4], where the right handed neutrinos are the only required additional degrees of freedom. These right neutrinos are part of the see saw mechanism [5], and thus providing an explanation for the smallness of neutrino masses. Through their out of equilibrium decays, the observed BAU can be obtained.
In this Letter, we show that another degree of freedom, the flavon, that conceivably plays an important role in understanding fermion masses, can also generate the observed BAU. In general, a flavon is a scalar particle whose vacuum expectation value (VEV) dictates the (Yukawa) couplings of the SM fermions, with the most popular example being the Froggatt-Nielsen field [6]. We show that the out of equilibrium decays of such a flavon can be the necessary beyond the SM physics that generates the observed BAU. The flavon decays may create a left right charged lepton asymmetry. In a universe where the flavon dominates the energy density, the right handed electrons may not come into chemical equilibrium, and an asymmetry in these right handed electron asymmetry can be converted into a baryon number asymmetry by sphalerons.

II. BASIC MECHANISM
Consider the couplings of the SM charged leptons to a SM singlet scalar, the flavon S, charged under the flavor symmetry U(1) FN , Here, φ denotes the electroweak Higgs doublet, f L is the f th generation left handed lepton doublet, and e g R represents the g th generation right handed charged lepton. v S is the flavon VEV, with Λ being the cutoff scale of the flavor symmetry, y 0 an O(1) coupling constant and n f g integers related to the Froggatt-Nielsen charges under U(1) FN . In the following, we will restrict ourselves to the simplest version of the scenario, in which the flavon only couples to leptons. As a benchmark model, we will take The key observation in our scenario is that the flavon decays are left right violating, e.g. an S particle only decays into L antiparticle and e R particle. Crucially, if there is an initial flavon asymmetry, say an excess of S over S * , after the flavons decay, there will be more left handed antileptons than left handed leptons. Note that the total lepton number, L, is conserved because the asymmetry in the left handed leptons will be balanced by an equal and opposite asymmetry in the right handed leptons. However, sphalerons will only act on left handed antileptons and partially convert this asymmetry into a baryon (B) asymmetry. This is similar to Dirac leptogenesis [7], in which nonzero baryon number asymmetry is possible with B−L conservation, as it utilizes the features that sphalerons only couple to left handed fields. In the scenario discussed here, the left right asymmetry originates from a primordial flavon antiflavon asymmetry, which we will discuss below in more detail. Our scenario requires the flavon to have a large asymmetry when it decays. This means that an asymmetry must be created and that flavon number must be conserved during the flavon oscillations. It has been pointed out in the context of the Affleck-Dine scenario [8] that scalar field dynamics in the early universe can exhibit these features. As explained in [9, Section III.C], in supersymmetric theories the mass terms of scalar fields S preserve S number while interaction terms do not. These interactions may provide an "initial kick" (in the terminology of [10]) which leads to an angular momentum of the S field in the complex plane, i.e. an S number asymmetry. These terms become unimportant later when the S field performs coherent oscillations, such that the asymmetry persists until the flavon decays.
A reason for concern might be the origin of the U(1) symmetry that ensures flavon number conservation. At first sight, one may conclude that if U(1) FN gets broken spontaneously, one is left with two options. (i) U(1) FN is local, in which case the would be Goldstone mode gets eaten, and one is left with a real scalar and no U(1) S that allows us to define the S number. (ii) U(1) FN is global, in which case there is a Goldstone mode, and the situation is even worse. However, in explicit models the situation is often richer than that. For instance, in supersymmetric scenarios, the scalar fields are complex. In such models, one starts with more degrees of freedom and the flavon is a complex linear combination of these fields, whose mass term preserves a U(1) symmetry. There are also non supersymmetric examples that have additional custodial symmetries (see e.g. [11,12] for recent applications of these symmetries in dark matter model building). We will assume that for energy scales far below Λ an approximate U(1) S is preserved by the flavon potential, i.e. V S = m 2 |S| 2 + S number violating terms suppressed by powers of Λ . (5)

IV. FLAVON COSMOLOY
By assumption, the flavon we consider is a weakly coupled scalar field. Hence it can perform coherent oscillations around the T = 0 minimum of its potential. In fact, thermal corrections to the potential will push the flavon away from its expectation value at T = 0 [13]. The energy density stored in these oscillations, ρ S , only drops as a −3 whereas the energy density of radiation, ρ rad , drops as a −4 , where a is the scale factor. It is expected that at a time t * , corresponding to a temperature T * , the energy density stored in flavon oscillations starts dominating over the radiation contribution. We define this point as The flavon decays to SM leptons with a decay rate Most of the decay products thermalize with the radiation and contribute to the radiation density. However, as we will show below, the right handed electrons might not come into chemical equilibrium before the sphalerons have switched off. The evolution of the relevant energy densities is given by where the Hubble rate is determined by the Friedmann equation with M Pl 1.2 × 10 19 GeV being the Planck mass.
We solve this set of equations numerically and show a benchmark scenario in Figure 1. Analytical approximations for the two energy densities can be given at a time t before the flavon decays, t * < t < τ ∼ Γ −1 S , as The first term in ρ rad is the initial radiation energy density and the second term is generated from flavon decays. This second term starts taking over ρ rad long before Γ −1 S . We will show in the next section that BAU can be produced via flavon decays for Γ S ∼ O(10 −13 − 10 −17 GeV). Thus, using Equation (7) we will require the flavon mass to be such that Λ is below the Planck scale.

V. GENERATION AND NON-EQUILIBRATION OF RIGHT-HANDED ELECTRONS
Among other leptons, flavon decays into right handed electrons with a branching fraction of B e ∼ ne ye nτ yτ 2 ∼ 7.5 × 10 −7 . Through its decays, the flavon asymmetry will get partially converted into an asymmetry in right handed electrons. Like in leptogenesis [4], this asymmetry is turned into a baryon asymmetry by sphalerons. Similarly to Dirac leptogenesis [7], our scenario does not require B − L violation. However, our scenario works both for Dirac and Majorana neutrinos. Right handed electrons equilibrate with the SM plasma mainly through their interactions with the Higgs boson and 2 → 2 scatterings. This equilibration rate has been recently calculated to be [14] Γ LR 10 −2 y 2 e T , which is larger by almost an order of magnitude than the initial estimate [15] and by a factor of a few than a refined estimate [16]. Comparing this rate to the Hubble rate for a radiation dominated universe, one finds that the right handed electrons come into equilibrium at T ∼ 10 5 GeV. Hence, in standard cosmology, any asymmetry in the right handed electrons would be washed out long before the EW transition at T ∼ 160 GeV. However, in a universe that is dominated by a flavon until temperatures around the electroweak scale, right handed electrons may not equilibrate. (See Figure 2.) We find the asymmetry in the number density of right handed electrons after the flavon decays by solving the Boltzman equation for n R = n eR − nē R , where B e is the flavon branching ratio to electrons The asymmetry in the flavon number density, n S , is parametrized through an initial flavon asymmetry, η S , as We solve for n R together with Equation (8). In Figure 3 we show the dimensionless quantity η R = nR s , where s is the entropy density, for a benchmark scenario with m S = 1 TeV , η S = 1 , T * = 100 TeV .
As usual, the temperature is defined as the temperature of radiation, which may or may not dominate the energy density of the universe. The behavior of η R can be understood qualitatively as follows. A right handed electron asymmetry is produced through flavon decays and is proportional to the flavon number density. When the Hubble rate drops below the left right equilibration rate, the asymmetry is washed out by SM interactions. The time when this washout starts depends on the flavon lifetime, τ = Γ −1 S , and is generally before τ . Smaller Γ S can delay the equilibration until after sphalerons shut off at T EW ∼ 160 GeV. However, this also means that a smaller number of flavons decay before T EW . Hence, in general, there is balancing between the flavon lifetime and Γ LR that describes the region where the right amount of right handed electron asymmetry is produced before the sphalerons shut off.

VI. BARYON ASYMMETRY
In the SM B +L number is violated by EW processes called sphalerons [17,18]. Sphalerons are unsuppressed above the EW transition temperature T EW , with a thermal rate Γ sph (α 2 T ) 4 , while they are exponentially suppressed due to finite gauge boson masses after EW symmetry breaking. While sphalerons violate B+L, they conserve B−L. Hence a lepton asymmetry will be turned into a baryon asymmetry [4]. In our scenario this baryon asymmetry is given by [16] η B ≡ n B s 198 481 We numerically solve Equations (8) and (13) to find the electron asymmetry and the entropy of the universe at T EW 160 GeV. Our results are shown in Figures 3  and 4. As can be seen in Figure 4, there is a large parameter space where asymmetric flavon decays produce the observed BAU. We comment on the dependence of η B on the parameters of the model.
• Our scenario is rather sensititve to the flavon decay rate. If the flavon decays too early, T d T EW , right handed electrons equilibrate and (essentially) no baryon asymmetry is generated. If the flavon decays too late, T d T EW , the right handed electron asymmetry is produced when sphalerons are inoperative, and again no baryon asymmetry emerges. We find that the observed baryon asymmetry is produced for 10 −16 GeV Γ S 10 −13 GeV.
• The baryon asymmetry is virtually independent of the time t * when the flavon domination starts as long as T * T d , T EW .
• As the flavon mass increases, its number density drops. This is because each flavon decay produces entropy, the dilution factor being m S /T d . Hence the baryon asymmetry is inversely proportional to m S . The flavon decay rate also depends on its mass. For given values of m S and Γ S , the appropriate flavon scale Λ can be found from Equation (11). We find that a flavon with O(TeV) mass can produce the observed BAU. This requires Λ ∼ O(10 9 GeV). We note that for such a large flavon scale, constraints on the flavon mass is O(100 GeV) [19].
• The baryon asymmetry is directly proportional to the primordial flavon asymmetry η S . We show our results in Figure 4 for the maximal value of η S = 1. If we allow for a flavon as light as 200 GeV, we only require η S 10 −2 .

VII. CONCLUSIONS
In this Letter, we have shown that the observed baryon asymmetry can be produced in flavon decays. This scenario requires the flavon to decay around the time of the electroweak transition and that the flavon carries an O(10 −2 − 1) primordial asymmetry. The flavon mass is O(1 − 10 TeV). The viable parameter ranges are illustrated in Figure 4. The role of the flavon is twofold: 1. Its decays produce a left right asymmetry in the lepton sector, the left handed part of which is converted to a baryon asymmetry by sphalerons.
2. It dominates the universe before the EW scale, thus increasing the Hubble rate and preventing the right handed electrons from equilibrating.
Our scenario is rather constrained and thus predictive. We "lose" a significant amount of efficiency because our scalar is a flavon, i.e. it predominantly decays into the heavier generations. A more "efficient" version of this scenario would be to couple a scalar to light fermions only, in which case one may also consider the decay into up or even down quarks. However, it is arguably an appealing aspect of our scenario that we do not have to introduce a new degree of freedom in order to explain the observed baryon asymmetry. Rather, a traditional flavon can do that without being tweaked. work was performed in part at Aspen Center for Physics, which is supported by National Science Foundation grant PHY 1607611.