Leptogenesis with heavy neutrino flavours: from density matrix to Boltzmann equations
Steve Blanchet, David A. Jones, Pasquale Di Bari, Luca Marzola
LLeptogenesis with heavy neutrino flavours: from density matrix to Boltzmann equations
Steve Blanchet a , Pasquale Di Bari b , David A. Jones b , Luca Marzola b a Institut de Th´eorie des Ph´enom`enes Physiques, ´Ecole Polytechnique F´ed´erale de Lausanne,CH-1015 Lausanne, Switzerland b School of Physics and Astronomy , University of Southampton, Southampton, SO17 1BJ, U.K.
October 16, 2018
Abstract
Leptogenesis with heavy neutrino flavours is discussed within a density matrixformalism. We write the density matrix equation, describing the generation of thematter-antimatter asymmetry, for an arbitrary choice of the right-handed (RH)neutrino masses. For hierarchical RH neutrino masses lying in the fully flavouredregimes, this reduces to multiple-stage Boltzmann equations. In this case we recoverand extend results previously derived within a quantum state collapse description.We confirm the generic existence of phantom terms. However, taking into accountthe effect of gauge interactions, we show that they are washed out at the produc-tion with a wash-out rate that is halved compared to that one acting on the totalasymmetry. In the N -dominated scenario they cancel without contributing to thefinal baryon asymmetry. In other scenarios they do not in general and they haveto be taken into account. We also confirm that there is a (orthogonal) componentin the asymmetry produced by the heavier RH neutrinos which completely escapesthe washout from the lighter RH neutrinos and show that phantom terms addition-ally contribute to it. The other (parallel) component is washed out with the usualexponential factor, even for weak washout. Finally, as an illustration, we study thetwo RH neutrino model in the light of the above findings, showing that phantomterms can contribute to the final asymmetry also in this case. a r X i v : . [ h e p - ph ] J a n Introduction
Leptogenesis [1] is a direct cosmological application of the see-saw mechanism [2] for theexplanation of neutrino masses and mixing and it realises a highly non trivial link betweencosmology and neutrino physics. The discovery of neutrino masses and mixing in neutrinooscillation experiments [3] has drawn great attention on leptogenesis that became one ofthe most attractive models of baryogenesis for the explanation of the matter-antimatterasymmetry of the Universe.In most cases, classical Boltzmann equations are sufficient for the calculation ofthe final asymmetry [4, 5, 6, 7, 8, 9]. However, when lepton flavour effects are takeninto account [6, 10, 11], different sets of classical Boltzmann equations apply depend-ing whether the asymmetry is generated in the one-flavour regime, when the mass ofthe decaying RH neutrinos M i is much above 10 GeV, in the two-flavour regime, for10 GeV (cid:29) M i (cid:29) GeV, or in the three-flavour regime for M i (cid:28) GeV. More-over classical Boltzmann equations fail in reproducing the correct result in the transitionregimes for M i ∼ GeV and for M i ∼ GeV. However, in the case that onlythe lightest RH neutrino species is assumed to be responsible for the generation of theasymmetry, classical Boltzmann equations provide quite a convenient description in phe-nomenological investigations, since the final asymmetry can be expressed in terms ofsimple analytical expressions that well approximate the numerical solutions [9, 12].On the other hand, the contribution from heavier RH neutrinos can also relevantlycontribute to the final asymmetry (heavy neutrino flavour effects) and has, therefore,consistently to be taken into account in general [13]. When lepton flavour effects arejointly considered [6, 14], a reliable calculation of the asymmetry cannot neglect thecontribution from the heavier RH neutrinos even in the two RH neutrino model [15]usually considered as a paradigmatic case for the validity of the traditional N -dominatedscenario. It has also been shown that a successful N -dominated scenario is naturallyrealised in the interesting class of SO (10) inspired models [16].When heavier RH neutrinos are included, one has to distinguish quite a large number ofpossible mass patterns with different corresponding sets of classical Boltzmann equationsfor the calculation of the final asymmetry. For example, in the typical case of three RHneutrinos one has ten different possible mass patterns [17] shown in Fig. 1. In addition,the requirement that all RH neutrino masses do not fall in a transition regime becomesclearly much more restrictive.Moreover new effects arise when heavy neutrino flavours are taken into account. First,part of the asymmetry generated by a heavier RH neutrino species, the flavour orthogonal2 GeV GeV GeV GeV M i M i Figure 1: The ten different three RH neutrino mass patterns requiring 10 different sets ofBoltzmann equations for the calculation of the asymmetry [17].component, escapes the washout from a lighter RH neutrino species [6]. Second, parts ofthe flavour asymmetries (phantom terms) produced in the one or two flavour regimes donot contribute to the total asymmetry at the production but can contribute to the finalasymmetry [18].Therefore, it is necessary to extend the density matrix formalism beyond the tradi-tional N -dominated scenario [6, 11, 19] and account for heavy neutrino flavours effects inorder to calculate the final asymmetry for an arbitrary choice of the RH neutrino masses.This is the main objective of this paper. At the same time we want to show how Boltz-mann equations can be recovered from the density matrix equations for the hierarchicalRH neutrino mass patterns shown in Fig. 1 allowing an explicit analytic calculation of thefinal asymmetry. In this way we will confirm and extend results that were obtained withina simple quantum state collapse description. For illustrative purposes, we will proceed ina modular way, first discussing the specific effects in isolation within simplified cases andthen discussing the most general case that includes all effects. The paper is organised inthe following way.In Section 2 we discuss the derivation of the kinetic equations for the N -dominatedscenario in the absence of heavy neutrino flavours. This is useful both to show theextension from classical Boltzmann to density matrix equations and to highlight some3eatures that will prove to be quite important when, in the following Sections, we willinclude heavy neutrino flavour effects. In particular we show the existence of phantomterms and how the expression for the CP asymmetry matrix can be unambiguously derivedfrom the flavoured CP asymmetries, taking into account the different flavour compositionsof the lepton and anti-lepton quantum states produced in RH neutrino decays.In Section 3 we start discussing the case where two heavy RH neutrino flavours areinvolved directly in the generation of the asymmetry, considering a simplified case withonly two charged lepton flavours. In this way we simplify the notation and we betterhighlight the main results. In this Section we are particularly interested to show twoeffects that specifically arise when the interplay between heavy neutrino and chargedlepton flavours is considered. The first one is phantom leptogenesis [18]. The second,that we call projection effect, is how part of the asymmetry generated by a heavy RHneutrino, the component orthogonal to the heavy neutrino flavour associated to a lighterRH neutrino, is not washed out by the inverse processes of the latter [6, 20]. We also showthat these two effects, phantom leptogenesis and projection effect, in general combine witheach other.In Section 4 we extend the discussion to the general case with three heavy neutrinoflavours and three charged lepton flavours. In this section we finally obtain general densitymatrix equations for the calculation of the asymmetry for an arbitrary choice of the RHneutrino masses. From these equations we derive the classical Boltzmann equations for aparticularly interesting case: the two RH neutrino model. The derivation can be easilyextended to all ten hierarchical RH neutrino mass patterns shown in Fig. 1. In Section 5we draw the conclusions. N dominated scenario We discuss leptogenesis within a minimal type I seesaw mechanism with three RH neutrinospecies, N , N and N , with masses M ≤ M ≤ M respectively where one adds right-handed neutrinos N iR to the SM lagrangian with Yukawa couplings h and a Majoranamass term that violates lepton number L = L SM + i N iR γ µ ∂ µ N iR − (cid:96) αL h αi N iR ˜Φ − M i N ciR N iR + h . c . . (1)After spontaneous symmetry breaking, a neutrino Dirac mass term m D = v h is generatedby the Higgs vev v . In the seesaw limit, M (cid:29) m D , the spectrum of neutrino masses splits4nto a light set given by the eigenvalues m < m < m of the neutrino mass matrix m ν = − m D M m TD , (2)and into a heavy set M < M < M coinciding to a good approximation with theeigenvalues of the Majorana mass matrix corresponding to eigenstates N i (cid:39) N iR + N ciR .In this section we review the main steps underlying the derivation of the kinetic equa-tions in leptogenesis when heavy neutrino flavours are neglected, assuming that only thelightest RH neutrino decays and inverse processes contribute to the final asymmetry: thetraditional N -dominated scenario.We first derive the Boltzmann (rate) equations and then we extend them writing thedensity matrix equations, accounting for quantum decoherence, flavour oscillations andgauge interactions. This discussion will prove to be useful not only to setup the notationbut also to highlight some basic features of the kinetic equations in leptogenesis that willbe relevant when we will include heavy neutrino flavour effects in the next section.We will neglect different effects, processes and corrections that have been studiedduring the last years and that will not play a relevant role in our discussion. These includefor example ∆ L = 2 washout [4, 8], ∆ L = 1 scatterings [5], momentum dependence[21], thermal corrections [8, 22], flavour coupling from the Higgs and quark asymmetries[6, 23, 10, 24, 18], quantum kinetic effects [25].We will moreover always assume vanishing pre-existing asymmetry though notice thatthe results that we will obtain in Section 3 on the projection effect, are also important inorder to describe the evolution of a non-vanishing pre-existing asymmetry [20, 17]. If we indicate with Γ the decay rate of the lightest RH neutrinos into leptons, N → (cid:96) + Φ † , and with ¯Γ the decay rate into anti-leptons, N → ¯ (cid:96) + Φ, we can introduce thedecay term D and the washout term W given respectively by D ( z ) ≡ Γ + ¯Γ H z = K z (cid:28) γ (cid:29) and W ( z ) ≡
12 Γ ID + ¯Γ ID H z = 14 K K ( z ) z , (3)where z ≡ M /T , K ≡ (Γ + ¯Γ ) T =0 /H T = M is the decay parameter, H is the expan-sion rate and the averaged dilution factor, in terms of the Bessel functions, is given by (cid:104) /γ (cid:105) = K ( z ) / K ( z ). Under the fore-mentioned assumptions and approximations andconsidering the unflavoured regime, the calculation of the asymmetry is described by themost traditional set of kinetic equations for leptogenesis from the decays of the lightest5H neutrinos N [5, 6, 7, 8, 9] dN N dz = − D ( N N − N eq N ) , (4) dN B − L dz = ε D ( N N − N eq N ) − W N B − L , (5)where with N X we indicate any particle number or asymmetry X calculated in a portion ofco-moving volume containing one heavy neutrino in ultra-relativistic thermal equilibrium,in a way that N eq N i ( T (cid:29) M i ) = 1. In this way the baryon-to-photon number ratio atrecombination is related to the final B − L asymmetry by η B = a sph N f B − L N rec γ (cid:39) . N f B − L , (6)to be compared with the value measured from the CMB anisotropies observations [26] η CMB B = (6 . ± . × − . (7)Let us very shortly recall the basic steps for the derivation of the Eq. (5) for the B − L asymmetry. Ignoring the reprocessing action of sphalerons, we can write dN B − L dz = dN ¯ (cid:96) dz − dN (cid:96) dz . (8)The net production rate of leptons and anti-leptons is then given by the difference betweenthe production rate due to decays and the depletion rate due to inverse decays, for leptons dN (cid:96) dz = Γ H z N N − Γ ID H z N (cid:96) (9)and for anti-leptons dN ¯ (cid:96) dz = ¯Γ H z N N − ¯Γ ID H z N ¯ (cid:96) . (10)The inverse decay rates are related to the decay rates by [5] Γ ID = Γ N eq N N eq (cid:96) and ¯Γ ID = ¯Γ N eq N N eq¯ (cid:96) , (11) This expression directly accounts for the resonant ∆ L = 2 contribution that is needed not to violatethe Sakharov condition on the necessity of a departure from thermal equilibrium for the generation of anasymmetry [4]. Here we are interested in showing the separate Boltzmann equations for lepton anti-leptonnumbers that we will use in the next subsection to derive the CP violating term in the density matrixequations. N eq (cid:96) = N eq¯ (cid:96) ≡ N eq (cid:96) = 1 is the number of leptons (cid:96) and of anti-leptons ¯ (cid:96) in thermalequilibrium for vanishing asymmetry. The number of leptons and anti-leptons can thenbe recast as N (cid:96) = 12 (cid:0) N (cid:96) + N ¯ (cid:96) (cid:1) + 12 (cid:0) N (cid:96) − N ¯ (cid:96) (cid:1) = N eq (cid:96) − N B − L + O ( N B − L ) (12)and N ¯ (cid:96) = 12 (cid:0) N (cid:96) + N ¯ (cid:96) (cid:1) − (cid:0) N (cid:96) − N ¯ (cid:96) (cid:1) = N eq (cid:96) + 12 N B − L + O ( N B − L ) . (13)Inserting these last expressions into the Eq. (8) and neglecting terms O ( N B − L ), the Eq. (5)is obtained, with D and W given by the Eqs. (3).The solution for the final asymmetry has a very simple analytical expression [9] N f B − L = ε κ ( K ) , with κ ( x ) ≡ x z B ( x ) (cid:20) − exp (cid:18) − x z B ( x ) (cid:19)(cid:21) , (14)where κ ( K ) is the final efficiency factor that here we have written, for simplicity, in thecase of initial thermal N -abundance. For K (cid:38)
3, the strong wash-out regime favouredby neutrino oscillation experiments, the asymmetry is generated in quite a narrow intervalof temperatures around T B ≡ M /z B , where z B ≡ z B ( K ) = O (1 ÷
10) [9].The unflavoured assumption, underlying the Eqs. (4) and (5), proves to describe thecorrect final asymmetry only for masses M (cid:38) GeV [19, 27]. In this range of masses,during all the relevant period of the asymmetry production, the lepton and anti-leptonquantum states produced from the decays of the N , that we will indicate respectivelysimply with | (cid:105) and | ¯1 (cid:105) , can be treated, in flavour space, as pure states between theirproduction at decay and their absorption at a subsequent inverse decay. They can beexpressed as a linear combination of flavour eigenstates ( α = e, µ, τ ) , | (cid:105) = (cid:88) α C α | α (cid:105) , C α ≡ (cid:104) α | (cid:105) and | ¯1 (cid:105) = (cid:88) α ¯ C ¯1¯ α | ¯ α (cid:105) , ¯ C ¯1¯ α ≡ (cid:104) ¯ α | ¯1 (cid:105) . (15)Notice that in general, even though in order to simplify the notation we are indicating thefinal anti-leptons produced by the N decays with ¯ (cid:96) , they do not coincide with the CP conjugated of the final lepton states. This means that, introducing the CP conjugatedstates CP | ¯1 (cid:105) = ¯ C τ | τ (cid:105) + ¯ C τ ⊥ | τ ⊥ (cid:105) , with ¯ C α = ¯ C (cid:63) ¯1¯ α , (16)in general one has ¯ C α (cid:54) = C α [10]. 7t will prove useful to introduce the branching ratios p α ≡ |C α | and ¯ p α ≡ | ¯ C α | giving respectively the probabilities that a lepton (cid:96) or an anti-lepton ¯ (cid:96) is found eitherin a flavour eigenstate α or ¯ α in a flavour measurement process. It is also useful to recastthe branching ratios as p α = p α + δp α , ¯ p α = p α + δ ¯ p α , (17)where, in general, the tree level values p α (cid:54) = ( p α + ¯ p α ) / δp α (cid:54) = − δ ¯ p α . The deviations from the tree level values, δp α = p α − p α and δ ¯ p α =¯ p α − p α , originate from the CP violating contributions due to the interference with loopdiagrams (see discussion in the Appendix).If the charged lepton interactions are negligible during the period of generation ofthe asymmetry (one-flavour regime) , for z (cid:39) z B , the lepton flavour compositions donot play any role since the only other relevant interactions, the gauge interactions, areflavour blind and lepton and anti-lepton quantum states propagate coherently betweenproduction from decays and absorption from inverse decays. However, we will notice thatgauge interactions have some important interplay with lepton flavour compositions. Thissituation is realised for M (cid:38) GeV. On the other hand, for masses 10 GeV (cid:29) M (cid:29) GeV, the coherent evolution of the | (cid:105) and | ¯1 (cid:105) quantum states breaks downbefore they can inverse decay interacting with the Higgs bosons, due to collisions withright-handed tauons. At the inverse decays, lepton quantum states can then be describedas an incoherent mixture of tauon eigenstates | τ (cid:105) and of | τ ⊥ (cid:105) quantum states. Thesesecond ones are a coherent superposition of muon and electron eigenstates that can beregarded as the projection of the lepton quantum states | (cid:105) on the plane orthogonal to thetauon flavour (see Fig. 2). In this two-fully flavored regime, classical Boltzmann equationscan be still used as in the unflavored regime, with the difference, in general, that now theflavour compositions of leptons and anti-leptons do play a role in the generation of theasymmetry. In this case each single flavour asymmetry has to be tracked independentlyand the total final B − L asymmetry has to be calculated after freeze-out as the sum ofthe two flavoured asymmetries, a τ asymmetry and a τ ⊥ asymmetry. To this extent, wehave to introduce the flavoured CP asymmetries ε iα ≡ ¯ p iα Γ i − p iα Γ i Γ i + Γ i = ¯ p iα + p iα ε i − ∆ p iα , (18)where we defined ∆ p iα ≡ p iα − ¯ p iα and all other quantities are a straightforward gen-eralisation of the quantities previously defined for the lightest RH neutrino species N to the case of a generic RH neutrino species N i . Since sphaleron processes conserve the8 Τ (cid:166) e ΜΤ Figure 2: For 10 GeV (cid:29) M (cid:29) GeV, the lepton quantum states | (cid:105) can be treatedas an incoherent mixture of a τ and of a τ ⊥ component during the generation of theasymmetry and a two fully flavoured regime applies.quantities ∆ α ≡ B/ − L α ( α = e, µ, τ ) [6], these are the convenient independent variablesto be used in the set of Boltzmann equations that can be written as dN N dz = − D ( N N − N eq N ) , (19) dN ∆ τ dz = ε τ D ( N N − N eq N ) − p τ W N ∆ τ ,dN ∆ τ ⊥ dz = ε τ ⊥ D ( N N − N eq N ) − p τ ⊥ W N ∆ τ ⊥ , where p τ ⊥ ≡ p e + p µ and ε τ ⊥ ≡ ε e + ε µ and where we neglected terms O (∆ p N ∆ α ).Using the decomposition of the flavoured CP asymmetries in terms of p iα and ∆ p iα (cf. Eq. (18)) and assuming strong washout for both flavours (i.e. K τ , K τ ⊥ (cid:29) N f B − L (cid:39) ε κ ( K ) + ∆ p τ (cid:104) κ ( K τ ⊥ ) − κ ( K τ ) (cid:105) , (20)where K iα ≡ p iα K i . This approximated expression shows how, compared to the expres-sion obtained in the unflavoured case, large lepton flavour effects can arise only whenleptons and anti-leptons have a different flavour composition , for non-vanishing ∆ p τ . Notice that relaxing the assumption of strong washout for both flavours one can only get an asymme-try that is even closer to the unflavoured calculation. Indeed in the limit of no washout ( K τ , K τ ⊥ (cid:28)
9n the Appendix we further discuss some interesting aspects and consequences of thispoint that is crucial for flavour effects to have a strong impact on the final asymmetryand that, as we will see, will play a very important role in the results discussed in thispaper. The most extreme case is realized when ε = 0 [10]. In the unflavoured casethis would imply a vanishing final asymmetry but in the flavoured case it does not. Itshould be indeed noticed that when flavour effects are considered, B − L violation is nota necessary condition for the generation of a baryon asymmetry via leptogenesis, it is suf-ficient to have a ∆ α violation accompanied by an asymmetric washout between the twoflavours, that in this context corresponds to the requirement of departure from thermalequilibrium.For M (cid:28) GeV muon interactions are able to break the coherent evolution alsoof the | τ ⊥ (cid:105) quantum states between decays and inverse decays during the period of thegeneration of the asymmetry. In this case a three-fully flavoured regime is realised andthe set of classical Boltzmann equations is a straightforward generalisation of that onewritten in the two fully flavoured regime. In the N -dominated scenario with hierarchicalRH neutrinos, because of the lower bound M (cid:38) GeV for successful leptogenesis [28, 7],a three fully flavoured regime is not relevant for the calculation of the final asymmetry. Onthe other hand, in a N -dominated scenario, a three flavoured regime has to be consideredin the calculation of the lightest RH neutrino washout [14]. Within a density matrix formalism [6, 11, 29, 19], the description of leptogenesis is moregeneral than with classical Boltzmann equations, since it makes possible to calculate theasymmetry in those intermediate regimes where lepton quantum states interact with thethermal bath via charged lepton interactions between decays and inverse decays thoughnot so efficiently that a quantum collapse approximation can be applied in a statisticaldescription. In this case the ensemble of lepton quantum states cannot be describedneither in terms of pure states nor as an incoherent mixture. Yukawa interactions andcharged lepton interactions compete with each other in the determination of the averageproperties of the lepton quantum states. We will show that also gauge interactions playan active, though indirect, role. A statistical quantum-mechanical description of leptonflavour cannot treat leptons as decoupled from the thermal bath. Therefore, the conceptof lepton quantum states itself is blurred since one should consistently describe togetherleptons and thermal bath. A density matrix formalism [30] is then particularly convenientsince it still allows to describe the leptonic subsystem in a separate way, neglecting back-10eaction effects and encoding the coupling with the thermal bath in the evolution of theoff-diagonal terms of the lepton density matrices.Let us see how a density matrix equation for the B − L asymmetry can be obtainedstarting first from the case where charged lepton interactions are negligible. In this casewe just expect to reproduce the Eq. (5).Let us consider a simple two lepton flavour case able to describe the intermediateregime between the unflavoured case and the two fully flavoured regime that are recoveredas asymptotic limits. The two relevant flavours are then τ and τ ⊥ (see Fig. 2). In thistwo flavour space the flavour composition of the lepton quantum states produced by the N decays can be written as ( α = τ, τ ⊥ ) | (cid:105) = C τ | τ (cid:105) + C τ ⊥ | τ ⊥ (cid:105) , C α ≡ (cid:104) α | (cid:105) , (21) CP | ¯1 (cid:105) = ¯ C τ | τ (cid:105) + ¯ C τ ⊥ | τ ⊥ (cid:105) , ¯ C α ≡ (cid:104) α | CP | ¯1 (cid:105) . (22)This definition can be straightforwardly generalised to the lepton quantum states pro-duced by a generic RH neutrino species N i that can be written in terms of amplitudes C iα and ¯ C iα . At tree level, the amplitudes C iα and ¯ C iα are given by ( i = 1 , , C iα = h αi (cid:112) ( h † h ) ii and ¯ C iα = h αi (cid:112) ( h † h ) ii . (23)Including one-loop CP -violating corrections, these amplitudes become C iα = 1 (cid:112) ( h † h ) ii − h † h ξ u ) ii ( h αi − ( h ξ u ) αi ) , (24)¯ C iα = 1 (cid:112) ( h † h ) ii − h † h ξ (cid:63)v ) ii ( h αi − ( h ξ (cid:63)v ) αi ) . (25)This shows explicitly that the flavour compositions of leptons (cid:96) i and of the ( CP conju-gated) anti-leptons ¯ (cid:96) i are different, provided ξ (cid:63)v (cid:54) = ξ u , which has to be the case for CP violation to be non-zero, as we show below. We are following here the notation and for-malism introduced in [31] and more recently in [32]. The one-loop corrections are includedin the ξ u and ξ v functions, which are given by (cid:2) ξ u ( M i ) (cid:3) ki ≡ (cid:2) u ( M i ) + M b ( M i )( h † h ) T M (cid:3) ki , (26) (cid:2) ξ v ( M i ) (cid:3) ki ≡ (cid:2) v ( M i ) + M b ( M i )( h † h ) M (cid:3) ki . The first term on the right-hand side describes the self-energy correction, whereas thesecond one is the vertex correction. Note that the mass matrix being diagonal, we simply11ave M ki = M i δ ki . The u and v terms in Eq. (26) are given by u ki ( M i ) = ω ki ( M i ) (cid:2) M i Σ N,ki ( M i ) + M k Σ N,ik ( M i ) (cid:3) , (27) v ki ( M i ) = ω ki ( M i ) (cid:2) M i Σ N,ik ( M i ) + M k Σ N,ki ( M i ) (cid:3) . They depend on the propagator ω ik and self-energy Σ N,ki ( M i ) = a ( M i )( h † h ) ki , where a is a loop factor, both evaluated on mass-shell for the RH neutrino N i .It can be easily checked that the difference of branching ratios, ∆ p iα ≡ |C iα | − | ¯ C iα | ,does not vanish in general, implying different flavour compositions of leptons and anti-leptons. This can be indeed expressed as |C iα | − | ¯ C iα | = 1( h † h ) ii (cid:88) k (cid:8) M i M k Im (cid:2) b ki ( M i ) (cid:3) Im (cid:2) h (cid:63)αi h αk ( h † h ) ik (cid:3) (28)+4 M k Re (cid:2) ω ki ( M i ) (cid:3) Im (cid:2) a ( M i ) (cid:3) Im (cid:2) h (cid:63)αi h αk ( h † h ) ik (cid:3) +4 M i Re (cid:2) ω ki ( M i ) (cid:3) Im (cid:2) a ( M i ) (cid:3) Im (cid:2) h (cid:63)αi h αk ( h † h ) ki (cid:3) − | h αi | ( h † h ) ii M k (cid:0) M i Im (cid:2) b ki ( M i ) (cid:3) + Re (cid:2) ω ki ( M i ) (cid:3) Im (cid:2) a ( M i ) (cid:3)(cid:1) Im (cid:2) ( h † h ) ik (cid:3)(cid:27) , where Im [ a ( M i )] = − / (16 π ), and the imaginary part of the other loop factor b ( M i )evaluated on mass shell for the RH neutrino N i is given byIm (cid:2) b ki ( M i ) (cid:3) = 116 πM i M k f ( x k /x i ) , (29)where x i ≡ M i /M and f ( x ) = √ x (cid:0) − (1 + x ) log (cid:0) xx (cid:1)(cid:1) . Lastly, the real part of thepropagator ω , evaluated on shell, is found to beRe (cid:2) ω ki ( M i ) (cid:3) = M i ( M k − M i )( M k − M i ) + ( M k Γ i − M i Γ k ) . (30)It can now be easily checked that the expression Eq. (28) consistently satisfies the decom-position Eq. (18), explicitly |C iα | − | ¯ C iα | ≡ ∆ p iα = ( p iα + ¯ p iα ) ε i − ε iα , (31)where ( p iα + ¯ p iα ) (cid:39) | h αi | / ( h † h ) ii , the flavoured CP asymmetries [33] ε iα = 316 π ( h † h ) ii (cid:88) j (cid:54) = i (cid:40) Im (cid:2) h (cid:63)αi h αj ( h † h ) ij (cid:3) ξ ( x j /x i ) (cid:112) x j /x i + 23( x j /x i −
1) Im (cid:2) h (cid:63)αi h αj ( h † h ) ji (cid:3)(cid:41) , (32)with ξ ( x ) = 23 x (cid:20) (1 + x ) ln (cid:18) xx (cid:19) − − x − x (cid:21) , (33)12nd, finally, the total CP asymmetries ε i = 316 π ( h † h ) ii (cid:88) j (cid:54) = i Im (cid:2) ( h † h ) ij (cid:3) ξ ( x j /x i ) (cid:112) x j /x i . (34)Let us now focus again on the N -dominated scenario. We can introduce the quantumstates | ⊥ (cid:105) and CP | ¯1 ⊥ (cid:105) orthogonal, in flavour space, respectively to the lepton quantumstates | (cid:105) and CP | ¯1 (cid:105) and with flavour compositions | ⊥ (cid:105) = −C (cid:63) τ ⊥ | τ (cid:105) + C (cid:63) τ | τ ⊥ (cid:105) and CP | ¯1 ⊥ (cid:105) = − ¯ C (cid:63) τ ⊥ | τ (cid:105) + ¯ C (cid:63) τ | τ ⊥ (cid:105) . (35)In the two flavour bases (cid:96) – (cid:96) ⊥ and CP (¯ (cid:96) –¯ (cid:96) ⊥ ), the lepton and anti-lepton density matricesare respectively simply given by the projectors ρ (cid:96)ij = P (1) ij = diag(1 ,
0) and ρ ¯ (cid:96)ij = P (1) ij =diag(1 , i, j = 1 , ⊥ if, for the time being, we assume that there are no otherleptons beyond the (cid:96) ’s produced by the RH neutrino decays and that they are thermalisedjust by the Yukawa interactions. This is clearly not true either if one starts from vanishingRH neutrino abundances or if Yukawa interactions are weak or both. We will discuss ina moment how gauge interactions are able to thermalise the leptons and will play a role,affecting the results on the asymmetries. Notice that, for matrices, we indicate the heavyneutrino flavour index with a superscript in round brackets. Since we are dealing with CP conjugated anti-lepton states, we can use the same flavour indices for the matrix entries ofleptons and anti-leptons. However, it is important to notice that, because of the differentflavour composition of leptons and anti-leptons, the two bases do not coincide.If we introduce the lepton and anti-lepton number density matrices, respectively N (cid:96)ij ≡ N (cid:96) ρ (cid:96)ij and N ¯ (cid:96)ij ≡ N ¯ (cid:96) ρ ¯ (cid:96)ij , their evolution at T ∼ T L is given by dN (cid:96)ij dz = (cid:18) Γ H z N N − Γ ID H z N (cid:96) (cid:19) ρ (cid:96)ij , dN ¯ (cid:96)ij dz = (cid:18) ¯Γ H z N N − ¯Γ ID H z N ¯ (cid:96) (cid:19) ρ ¯ (cid:96)ij . (36)In order to obtain an equation for the total B − L asymmetry matrix N B − L ≡ N ¯ (cid:96) − N (cid:96) ,we have first to write these two equations in the same flavour basis, for convenience thelepton flavour basis τ – τ ⊥ , and then subtract them. The rotation matrices are then givenby R (1) αi = (cid:32) C τ −C (cid:63) τ ⊥ C τ ⊥ C (cid:63) τ (cid:33) and ¯ R (1) αi = (cid:32) ¯ C τ − ¯ C (cid:63) τ ⊥ ¯ C τ ⊥ ¯ C (cid:63) τ (cid:33) , (37)for leptons and anti-leptons respectively. Also notice that at tree level, corresponding toneglect CP violation, they simply coincide, i.e. R (1)0 αi = (cid:32) C τ −C (cid:63) τ ⊥ C τ ⊥ C (cid:63) τ (cid:33) = ¯ R (1)0 αi . (38)13n the charged lepton flavour basis one can finally write the equation for the B − L asymmetry matrix as dN B − Lαβ dz = ¯ R (1) αi dN ¯ (cid:96)ij dz ¯ R (1) † jβ − R (1) αi dN (cid:96)ij dz R (1) † jβ , (39)whose trace gives the B − L asymmetry N B − L . In the charged lepton flavour basis thetwo projectors become P (1) αβ ≡ R (1) αi (cid:32) (cid:33) R (1) † jβ = (cid:32) p τ C τ C (cid:63) τ ⊥ C (cid:63) τ C τ ⊥ p τ ⊥ (cid:33) , (40) P (1) αβ ≡ ¯ R (1) αi (cid:32) (cid:33) ¯ R (1) † jβ = (cid:32) ¯ p τ ¯ C τ ¯ C (cid:63) τ ⊥ ¯1 ¯ C (cid:63) τ ¯ C τ ⊥ ¯1 ¯ p τ ⊥ ¯1 (cid:33) , (41)which, at tree level, simply coincide and are given by P (1)0 αβ = R (1)0 αi (cid:32) (cid:33) R (1)0 † jβ = (cid:32) p τ C τ C (cid:63) τ ⊥ C (cid:63) τ C τ ⊥ p τ ⊥ (cid:33) = 1( h † h ) (cid:32) | h τ | h τ h (cid:63)τ ⊥ h (cid:63)τ h τ ⊥ | h τ ⊥ | (cid:33) . (42)Using these results, we can now rewrite Eq. (39) as dN B − Lαβ dz = (cid:18) ¯Γ H z N N − ¯Γ ID H z N ¯ (cid:96) (cid:19) P (1) αβ − (cid:18) Γ H z N N − Γ ID H z N (cid:96) (cid:19) P (1) αβ , (43)that can be recast, using eqs. (12) and (13) assuming thermal abundances, first as dN B − Lαβ dz = ε (1) αβ D (cid:0) N N − N eq N (cid:1) − W N B − L P (1) αβ Γ + P (1) αβ ¯Γ Γ + ¯Γ (44)and then, neglecting terms O ( ε N B − L ) and O (∆ p N B − L ), as dN B − Lαβ dz = ε (1) αβ D ( N N − N eq N ) − W N B − L P (1)0 αβ . (45)Notice that this result has been obtained assuming that there are only (cid:96) leptons and¯ (cid:96) anti-leptons. Notice that we defined the CP asymmetry matrix for the lightest RHneutrino N as ε (1) = P (1) Γ − P (1) Γ Γ + Γ = ε P (1) + P (1) − ∆ P (1) , (46)where ∆ P (1) ≡ P (1) − P (1) . This expression [6] generalises the eq. (18) that is obtained forthe diagonal terms in the charged lepton flavour basis where the diagonal terms simply14orrespond to the flavoured CP asymmetries, ε (1) αα = ε α , while the off-diagonal termsobey ε (1) αβ = ( ε (1) βα ) (cid:63) and are not necessarily real. This expression can be generalised to the CP asymmetry matrix ε ( i ) αβ , of any RH neutrino species N i that in terms of the Yukawacouplings can be written as ε ( i ) αβ = 332 π ( h † h ) ii (cid:88) j (cid:54) = i (cid:40) i (cid:2) h αi h (cid:63)βj ( h † h ) ji − h (cid:63)βi h αj ( h † h ) ij (cid:3) ξ ( x j /x i ) (cid:112) x j /x i +i 23( x j /x i − (cid:2) h αi h (cid:63)βj ( h † h ) ij − h (cid:63)βi h αj ( h † h ) ji (cid:3)(cid:27) , (47)where the ξ function was defined in Eq. (33). This expression slightly differs from thatone in [11, 19] (simply, there, the first term is minus the imaginary part of the first termwritten here, so that the off-diagonal terms are real) while it agrees with the expressiongiven in [27]. The diagonal components of the Eq. (39) can be explicitly written as dN B − Lττ dz = ε (1) ττ D ( N N − N eq N ) − p τ W N B − L , (48) dN B − Lτ ⊥ τ ⊥ dz = ε (1) τ ⊥ τ ⊥ D ( N N − N eq N ) − p τ ⊥ W N B − L . (49)Summing these two equations, one finally recovers the usual Eq. (5) for the total B − L asymmetry N B − L = Tr[ N B − Lαβ ], which is washed out in the usual way at the production.On the other hand, from Eqs. (48) and (49), one finds the relation1 p τ dN B − Lττ dz − p τ ⊥ dN B − Lτ ⊥ τ ⊥ dz = − ∆ p τ (cid:32) p τ + 1 p τ ⊥ (cid:33) D ( N N − N eq N ) (50)which, together with Eq. (4), forms a system of equations that can be solved analytically.At low temperatures T (cid:28) T B = M /z B (cid:28) M , the final values are then found to be N B − L, f ττ (cid:39) p τ N f B − L − ∆ p τ N in N , (51) N B − L, f τ ⊥ τ ⊥ (cid:39) p τ ⊥ N f B − L + ∆ p τ N in N . This solution shows that the flavoured asymmetries contain terms that escape the washoutat the production and are proportional to the initial abundance of RH neutrinos: theseare the phantom terms [18]. If one only considers the one-flavour regime, where chargedlepton interactions can be neglected as we have done so far, the flavoured asymmetriesare not themselves measured and the phantom terms cannot give any physical effect, inparticular they cannot affect the baryon asymmetry.15e want now to consider the effect of charged lepton interactions and of gauge interac-tions. When charged lepton interactions become effective, at T ∼ GeV, tauon leptoninteractions start to be in equilibrium breaking the coherence of the lepton quantumstates.Charged lepton interactions and gauge interactions are described by additional termsin Eqs. (36), which then generalise into [34, 11, 35, 27] dN (cid:96)αβ dz = Γ H z N N P (1) αβ −
12 Γ ID H z (cid:8) P (1) , N (cid:96) (cid:9) αβ + Λ αβ + G αβ , (52) dN ¯ (cid:96)αβ dz = ¯Γ H z N N P (1) αβ −
12 ¯Γ ID H z (cid:110) P (1) , N ¯ (cid:96) (cid:111) αβ + ¯Λ αβ + ¯ G αβ . where Λ αβ and ¯Λ αβ are the terms describing the effect of charged lepton interactions,Λ αβ = − i Re(Λ τ ) H z (cid:34)(cid:32) (cid:33) , N (cid:96) (cid:35) αβ − Im(Λ τ ) H z (cid:32) N (cid:96)ττ ⊥ N (cid:96)τ ⊥ τ (cid:33) , (53)¯Λ αβ = +i Re(Λ τ ) H z (cid:34)(cid:32) (cid:33) , N ¯ (cid:96) (cid:35) αβ − Im(Λ τ ) H z (cid:32) N ¯ (cid:96)ττ ⊥ N ¯ (cid:96)τ ⊥ τ (cid:33) . (54)The real and imaginary parts of the tau-lepton self-energy are respectively given by[36, 37] Re(Λ τ ) (cid:39) f τ T and Im(Λ τ ) (cid:39) × − f τ T , (55)where f τ is the tauon Yukawa coupling. The commutator structure in the third term onthe RHS of Eq. (52) accounts for oscillations in flavor space driven by the real part of theself energy, and the second terms damp of the off-diagonal terms driven by the imaginarypart of the self energy. The terms G αβ and ¯ G αβ describe the gauge interactions and havethe effect to thermalise leptons and anti-leptons so that kinetic and chemical equilibriumcan be assumed during all the transition from the unflavoured regime to the two fullyflavoured regime. Since they are CP conserving, they cannot change the total and flavourasymmetries while thermalising the asymmetries but, as we are going to discuss, theyplay an active, though indirect, role in the final values of the asymmetries.Let us show how, from the set of density matrix equations (52), one can derive correctlyboth the one-flavour (cf. eq. (45)) and the (two) fully flavoured regime (cf. eq. (19)).In the one-flavour case we have seen that neglecting gauge interactions correspondsto have N (cid:96) = N (cid:96) P (1) and N ¯ (cid:96) = N ¯ (cid:96) P (1) , where we had to assume that N (cid:96) and N ¯ (cid:96) arethermalised by the same Yukawa interactions, an assumption that does not describe thecase either when Yukawa interactions are weak or if one starts from a non-thermal RHneutrino abundance. 16f we now take into account the effect of gauge interactions, these will thermalise notonly the abundances of the leptons (cid:96) and of the anti-leptons ¯ (cid:96) , independently of thestrength of the Yukawa interactions and of the RH neutrino abundance, but also theabundances of their orthogonal states (cid:96) ⊥ and ¯ (cid:96) ⊥ . Since they are flavour blind and CP conserving, their presence is described by an additional unflavoured term in the leptonand anti-lepton abundance matrices that in this way get generalised as N (cid:96) = N eq (cid:96) I + N (cid:96) P (1) − N (cid:96) + N ¯ (cid:96) P (1)0 , (56) N ¯ (cid:96) = N eq (cid:96) I + N ¯ (cid:96) P (1) − N (cid:96) + N ¯ (cid:96) P (1)0 . The third terms in the right-hand side describe how annihilations mediated by gaugeinteractions drag out of the (cid:96) and ¯ (cid:96) their tree-level components, CP conjugated of eachother, that are thermalised. In this way the gauge interactions annihilations act as a sortof detector of the differences of flavour compositions of leptons and anti-leptons, thoughthey cannot measure the flavour compositions themselves, as implied by the term N eq (cid:96) I that is invariant under rotations in flavour space. If we linearise N (cid:96) and N ¯ (cid:96) using theeqs. (12) and (13) respectively, they can be recast as N (cid:96) = N eq (cid:96) I + (cid:18) N (cid:96) + N ¯ (cid:96) (cid:19) δ P (1) − N B − L P (1) , (59) N ¯ (cid:96) = N eq (cid:96) I + (cid:18) N (cid:96) + N ¯ (cid:96) (cid:19) δ P (1) + 12 N B − L P (1) , Notice that now these equations also describe consistently the case of vanishing initial RH neutrinoabundance that would yield seemingly unphysical negative values of N (cid:96) + N ¯ (cid:96) . Indeed now negativevalues correspond to the production of orthogonal states (cid:96) ⊥ and ¯ (cid:96) ⊥ , considering that P (1) = I − P (1) ⊥ , P (1)0 = I − P (1)0 ⊥ and analogously for the anti-leptons. One could wonder whether instead of terms proportional to the tree level components P (1)0 , oneshould subtract in the eqs. (56) terms proportional to the average components (cid:16) P (1) + P (1) (cid:17) /
2. However,one can verify that one would anyway obtain in the end the same result eq. (64) unless O (∆ P ) terms.Notice also that with this modification (and neglecting terms O ( N B − L ∆ P (1) )) the expressions (59) canbe written as N (cid:96) = N eq (cid:96) I + (cid:18) N (cid:96) + N ¯ (cid:96) (cid:19) ∆ P (1) − N B − L P (1) + P (1) , (57) N ¯ (cid:96) = N eq (cid:96) I − (cid:18) N (cid:96) + N ¯ (cid:96) (cid:19) ∆ P (1) N B − L P (1) + P (1) , respecting the constraint [27] N (cid:96) − N eq (cid:96) I = − ( N ¯ (cid:96) − N eq (cid:96) I ) , (58)that would be a matrix generalisation of the thermal equilibrium conditions Eqs. (12) and (13). δ P (1) ≡ P (1) − P (1)0 and δ P (1) ≡ P (1) − P (1)0 . From these equationsone can find an expression for the asymmetry matrix, N B − L = N B − L P (1) + P (1) − (cid:0) N (cid:96) + N ¯ (cid:96) (cid:1) ∆ P (1) , (60)that has to be compared with the eq. (46) for the CP asymmetry matrix: the first termis the usual contribution proportional to the total asymmetry, while the second term isthe contribution to the flavour asymmetry matrix coming from the difference in flavourcompositions yielding the phantom terms. Notice that the quantity ( N (cid:96) + N ¯ (cid:96) ) / N B − L . We can also writean expression for the sum N (cid:96) +¯ (cid:96) ≡ N (cid:96) + N ¯ (cid:96) = 2 N (cid:96) eq I + N (cid:96) + N ¯ (cid:96) (cid:16) δ P (1) + δ P (1) (cid:17) − N B − L P (1) . (61)Considering that in the tree level basis one has ( i , j = 1 , ⊥ ) δ P (1) i j = (cid:32) δp (cid:63) δp (cid:33) , (62)with δp = −C τ ⊥ δ C τ + C τ δ C τ ⊥ and δ C α ≡ C α − C α , one obtains the equalities (cid:8) P (1) , δ P (1) (cid:9) = δ P (1) + O ( δ P ) , (cid:110) P (1) , δ P (1) (cid:111) = δ P (1) + O ( δ P ) , (63)and neglecting terms O ( ε ∆ P ) and O ( δ P ), one arrives at the following equation dN B − L dz = ε (1) D ( N N − N eq N ) − W N (cid:96) + N ¯ (cid:96) (cid:16) P (1) − P (1) (cid:17) − W N B − L (cid:16) P (1) + P (1) (cid:17) . (64)Using the eqs. (63), it can be also recast more compactly as dN B − L dz = ε (1) D ( N N − N eq N ) − W (cid:8) P (1)0 , N B − L (cid:9) . (65)The Eq. (64) implies that, having accounted for the unflavoured thermal bath fromgauge interactions, phantom terms are washed out contrarily to the previous calculationwhere it was neglected. However, non trivially, the wash-out term acting on phantomterms is half compared to that one acting on the total asymmetry. Let us show this resultexplicitly, finding the solutions for the diagonal components in the charged lepton flavour We wish to thank M. Herranen and B. Garbrecht for pointing out to us that the eq. (65) implies somewash-out of the phantom terms and that, therefore, is not equivalent to the eq. (45) when the differencesbetween lepton and anti-lepton flavour compositions are taken into account. N B − Lττ and N B − Lτ ⊥ τ ⊥ . If one first considers the eq. (65) in the tree level basis, in thisbasis the decomposition of ε (1) in the right-hand side of eq. (46) specialises into ε (1) i j = (cid:32) ε
00 0 (cid:33) + (cid:32) ε (cid:63) ∆ ε (cid:33) , (66)where ∆ ε = ( δ ¯ p − δp ) / ≡ − ∆ p/
2. In this way in this basis the 1 term is just thetotal asymmetry N B − L that gets washed out by W . Instead the off-diagonal terms, uponrotation to the charged lepton flavour basis, give the phantom terms that are washed by W /
2. In this way, in the charged lepton flavour basis, one finds N B − L, f ττ (cid:39) p τ N f B − L − ∆ p τ κ ( K / , (67) N B − L, f τ ⊥ τ ⊥ (cid:39) p τ ⊥ N f B − L + ∆ p τ κ ( K / . This result confirms the presence of phantom terms but it also clearly shows how theeffect of the gauge interactions annihilations in detecting the differences between leptonand anti-lepton flavour compositions results into a wash-out of the phantom terms, thoughwith a wash-out rate that is halved compared to the wash-out rate acting on the totalasymmetry.Let us now consider the (two) fully flavoured regime. This can be recovered moreconveniently considering that in the eqs. (52) the off-diagonal terms are damped bythe charged lepton interactions [6]. Therefore, one has that N (cid:96) and N ¯ (cid:96) are diago-nal in the charged lepton flavour basis, so that N (cid:96)αβ = diag( N (cid:96)ττ , N (cid:96)τ ⊥ τ ⊥ ) and N ¯ (cid:96)αβ =diag( N ¯ (cid:96)ττ , N ¯ (cid:96)τ ⊥ τ ⊥ ). The gauge interactions thermalise the τ and the τ ⊥ abundances. Inthis way, taking the diagonal components, one straightforwardly recovers the eqs. (19).Notice that one could also try to get this result from a closed differential equation for N B − Lαβ . Subtracting the two equations (52) one obtains dN B − Lαβ dz = ε (1) αβ D N N − D (cid:34) Γ ID1 Γ + Γ (cid:110) P (1) , N ¯ (cid:96) (cid:111) αβ − Γ ID1 Γ + Γ (cid:8) P (1) , N (cid:96) (cid:9) αβ (cid:35) (68)+∆Λ αβ + ∆ G αβ . Recasting then N (cid:96) = N (cid:96) + N ¯ (cid:96) − N B − L N ¯ (cid:96) = N (cid:96) + N ¯ (cid:96) N B − L , (69)one obtains first dN B − Lαβ dz = ε (1) αβ D N N − D N eq N N eq (cid:96) (cid:110) ε (1) αβ , N (cid:96) +¯ (cid:96) (cid:111) αβ (70)19 D (cid:34) Γ ID1 Γ + Γ (cid:110) P (1) , N B − L (cid:111) αβ + Γ ID1 Γ + Γ (cid:8) P (1) , N B − L (cid:9) αβ (cid:35) +∆Λ αβ + ∆ G αβ . and then, neglecting terms O (∆ P N B − L ), dN B − Lαβ dz = ε (1) αβ D N N − D N eq N N eq (cid:96) (cid:110) ε (1) αβ , N (cid:96) +¯ (cid:96) (cid:111) αβ − W (cid:8) P , N B − L (cid:9) αβ (71)+i Re(Λ τ ) H z (cid:34)(cid:32) (cid:33) , N (cid:96) +¯ (cid:96) (cid:35) αβ − Im(Λ τ ) H z (cid:32) N B − Lττ ⊥ N B − Lτ ⊥ τ (cid:33) + ∆ G αβ . A Boltzmann equation for the quantity N (cid:96) +¯ (cid:96) is given by dN (cid:96) +¯ (cid:96)αβ dz (cid:39) − Re(Λ τ ) H z ( σ ) αβ N B − Lαβ − S g ( N (cid:96) +¯ (cid:96)αβ − N eq (cid:96) δ αβ ) , (72)where S g ≡ Γ g / ( Hz ) accounts for gauge interactions. As shown in [27], this term hasthe effect of damping the flavour oscillations. This is because the gauge interactions force N (cid:96) +¯ (cid:96)αβ (cid:39) N eq (cid:96) δ αβ [6, 35, 27], as it can be seen explicitly from eq. (61). This in turn makesin a way that the oscillatory term becomes negligible and that the second term on theright-hand side can be approximated with the usual inverse decay CP violating term,obtaining in the end dN B − Lαβ dz = ε (1) αβ D ( N N − N eq N ) − W (cid:8) P , N B − L (cid:9) αβ − Im(Λ τ ) H z ( σ ) αβ N B − Lαβ , (73)that generalises the Eq. (65). When the off-diagonal terms are fully damped, one againcorrectly obtains the Eqs. (19) in the fully flavoured regime and the usual Eq. (5) forthe total asymmetry in the unflavoured regime. Notice that we have now shown that theeq. (73) holds also starting from the Eqs. (56), taking into account differences betweenlepton and anti-lepton flavour compositions. This is because they anyway respect theapproximation N (cid:96) +¯ (cid:96)αβ (cid:39) N eq (cid:96) δ αβ (cf. 61).Suppose now that the asymmetry was generated in the unflavoured regime at temper-atures T (cid:29) GeV. Let us indicate with T (cid:63) (cid:28) GeV that value of the temperaturebelow which one can approximate, with the desired precision, the lepton quantum statesas a fully incoherent mixture of | τ (cid:105) and | τ ⊥ (cid:105) quantum states corresponding to a com-plete damping of the off-diagonal terms in the lepton density matrix (analogously foranti-leptons). This means that for T (cid:46) T (cid:63) the τ and τ ⊥ lepton asymmetries, givenby the diagonal entries of N B − Lαβ , are fully measured by the thermal bath and repro-cessed by sphaleron processes conserving the ∆ τ and ∆ τ ⊥ asymmetries, so that at T (cid:63) N T (cid:63) ∆ τ = N B − Lττ ( z B ) and N T (cid:63) ∆ τ ⊥ = N B − Lτ ⊥ τ ⊥ ( z B ). However, notice that since in thecase M (cid:29) GeV the total B − L asymmetry got already produced and frozen in theunflavoured regime, this fully flavoured regime stage does not affect the final total B − L asymmetry. Therefore, phantom terms do not contribute to the final asymmetry becausethey cancel with each other. In other words, within the N -dominated scenario, phantomterms have no effect on the final asymmetry . Therefore, phantom terms do not haveany consequence on the final asymmetry in the N -dominated scenario, in the absence ofheavy neutrino flavour effects.Our results show explicitly the presence of the phantom terms extending previousresults where this had not been noticed [19, 27]. In particular, in [19], the lepton andanti-lepton density matrices were assumed to be diagonalisable in bases that are CP conjugated of each other precluding the derivation of the phantom terms. However, aswe have seen, as far as the N -dominated scenario is concerned, phantom terms canbe safely neglected in the calculation of the final asymmetry, and therefore there is nocontradiction between our results and previous ones where phantom terms have not beenidentified [19, 27].On the other hand, we are interested in accounting for heavy neutrino flavour effects.On this case we cannot neglect the phantom terms since in this case, as we discuss inthe next Section, they can contribute to the total final asymmetry and even dominate(phantom leptogenesis [18]). In this section we account for heavy neutrino flavour effects considering a simplified twocharged lepton flavour case. This will greatly simplify the notation making the new resultsmore easily readable. It will be then quite straightforward in the next Section to generaliseall the equations to a realistic three lepton flavour case.For definiteness we consider masses M i (cid:29) GeV, when only tauon lepton inter-actions have to be taken into account. We also assume that the heaviest RH neutrinos On the other hand, as discussed, if one considers 10 GeV (cid:28) M (cid:28) GeV, the two fully flavouredregime holds during the period of leptogenesis and the density matrix equations reduce to the set ofclassical Boltzmann equations Eqs. (19). The terms in the flavoured asymmetries coming from CP violating terms due to a different flavour composition of leptons and anti-leptons are still present butthey are not phantom, since they are measured directly at production and undergo washout. Therefore,if there is a flavour-asymmetric production, they contribute to the final asymmetry, yielding the secondterm in the Eq. (20), and can even dominate. Τ (cid:166) e ΜΤ Figure 3: Flavour configuration of the two heavy neutrino lepton flavours, (cid:96) and (cid:96) , lead-ing to the simplified two charged lepton neutrino flavour case considered in this section. N do not contribute to the final asymmetry. This is in any case a valid assumption if M (cid:29) T RH (cid:29) M , since in this way the N ’s would not thermalise. As for lepton flavours,we will extend the results to the three heavy neutrino flavour case in the next Section.Notice that with these assumptions, the two charged lepton flavour case can be re-garded as a special case where the two heavy neutrino lepton flavours, (cid:96) and (cid:96) , lie onthe same plane orthogonal to the e − µ plane and therefore τ ⊥ = τ ⊥ = τ ⊥ (see Fig. 3).Correspondingly the two anti-lepton flavours, ¯ (cid:96) and ¯ (cid:96) , also lie on the same plane or-thogonal to the e − µ plane and therefore ¯ τ ⊥ = ¯ τ ⊥ ≡ ¯ τ ⊥ with ¯ τ ⊥ that is now assumed tobe CP conjugated of τ ⊥ . In this way, in the whole following discussion in this Section, wewill have only two charged lepton flavours, τ and τ ⊥ .The density matrix equation Eq. (73), valid for N leptogenesis, gets then generalisedinto ( α, β = τ, τ ⊥ ) dN B − Lαβ dz = ε (1) αβ D ( N N − N eq N ) − W (cid:8) P (1)0 , N B − L (cid:9) αβ (74)+ ε (2) αβ D ( N N − N eq N ) − W (cid:8) P (2)0 , N B − L (cid:9) αβ − Im(Λ τ ) H z ( σ ) αβ N B − Lαβ , where N N is described, as N N , by an analogous eq. (4). We now discuss the threeasymptotic cases, the first one for M (cid:29) GeV (cid:29) M , the second for M , M (cid:29) GeV and the third for M , M (cid:28) GeV, where Boltzmann equations are recovered.In this way we will derive, within a density matrix formalism, results that were alreadyobtained within an instantaneous collapse of the quantum state formalism: the first isphantom leptogenesis [18], the second is the heavy neutrino flavour projection [6, 20]. M (cid:29) GeV (cid:29) M : three stages phantom leptogene-sis Let us consider the asymmetry production from the N ’s at T ∼ M . This is basically de-scribed by the same equations that we wrote in the previous section for the N -dominatedscenario where now simply all quantities have to be relabelled in a way that 1 →
2. Sincecharged lepton interactions are negligible, we can use the Eq. (65) for the calculation of N B − Lαβ , that now, in the charged lepton flavoured basis, simply becomes dN B − Lαβ dz = ε (2) αβ D ( N N − N eq N ) − W (cid:8) P (2)0 , N B − L (cid:9) αβ , (75)with an obvious re-definition of all quantities that now refer to N . At T (cid:39) T B ≡ M /z B ,the τ and τ ⊥ asymmetries are described by the Eqs. (76) with 1 → N B − Lττ ( T (cid:39) T B ) (cid:39) p τ N T (cid:39) T B B − L − ∆ p τ κ ( K / , (76) N B − Lτ ⊥ τ ⊥ ( T (cid:39) T B ) (cid:39) p τ ⊥ N T (cid:39) T B B − L + ∆ p τ κ ( K / . Again, at temperatures below T (cid:63) (cid:28) GeV, the N B − Lαβ off-diagonal terms are fullydamped by the tauon charged interactions, so that the N ∆ τ and N ∆ τ ⊥ asymmetries,corresponding to the diagonal terms, can be treated as measured quantities.At T ∼ T B , the phantom terms in the eq. (76) cancel with each other and they do notcontribute to the total asymmetry. Therefore, so far, the description of the asymmetryevolution is completely analogous to that one discussed in the N -dominated scenario.However, there is still a third stage to be taken into account: the lightest RH neutrinowashout. For T ∼ M , the tauon and the τ ⊥ asymmetries are washed out by the lightestRH neutrino inverse processes. At T (cid:39) T B = M /z B , they get frozen to their finalvalues N f∆ τ (cid:39) (cid:20) p τ N T (cid:39) T B B − L − ∆ p τ κ ( K / (cid:21) e − π K τ , (77) N f∆ τ ⊥ (cid:39) (cid:20) p τ ⊥ N T (cid:39) T B B − L + ∆ p τ κ ( K / (cid:21) e − π K τ ⊥ , (78)23o that the final total asymmetry N f B − L (cid:39) N f∆ τ + N f∆ τ ⊥ . If for the flavour α = τ ( τ ⊥ ) onehas K α (cid:46)
1, while for the other flavour β = τ ⊥ ( τ ) one has K β (cid:29)
1, the final asymmetrywill be dominated by the α asymmetry, N f B − L (cid:39) p α N T (cid:39) T B B − L − ∆ p α κ ( K / . (79)Interestingly the phantom term is affected by a washout at the production that is half thatone acting on the final asymmetry. Since in the strong wash-out regime approximately κ ( K ) ∝ /K . , the phantom term contribution gets enhanced by a factor 3 comparedto that one proportional to the total asymmetry at the production and this could make itdominant. Having included the effect of gauge interactions, now in the strong wash-outregime ( K (cid:29)
1) the phantom terms are independent of the initial conditions. Therefore,phantom terms have to be included even in the case of initial vanishing abundance.Phantom leptogenesis was first discussed within an instantaneous quantum state col-lapse description without gauge interactions [18]. Here we have re-derived it within adensity matrix formalism showing the importance of gauge interactions that determine awash-out of the phantom terms, though halved. Notice that there are three well sepa-rated stages: N asymmetry production at T (cid:39) T B , decoherence at T ∼ T (cid:63) and flavourasymmetric N washout at T ∼ M .Notice also that phantom leptogenesis has some analogies with the scenario of N -leptogenesis with ε = 0 [10] that we discussed in the previous section. In both cases thefinal asymmetry originate from the CP violating terms ∝ ∆ p iα due to a different flavourcomposition of leptons and anti-leptons. In both cases a non-vanishing final asymme-try relies on an asymmetric washout acting on the two flavour asymmetries. There arehowever important differences. In the case of N leptogenesis with ε = 0 one has thatproduction, decoherence and washout occur simultaneously, while in the case of phan-tom leptogenesis they occur at different stages and between the production and the N washout stage the phantom terms they cancel in the final asymmetry. Another importantdifference is that in the case of phantom leptogenesis one has not to assume the specialassumption ε or ε = 0 ( B − L conservation): if the washout at the production is suffi-ciently strong phantom terms can potentially dominate because of the reduced wash-outcompared to the total asymmetry.As we are going to show, phantom leptogenesis is even more general and it doesnot necessarily require that the N production and the N washout stages occur in twodifferent fully flavoured regimes. 24 .2 Case M (cid:38) M (cid:29) GeV : heavy neutrino flavour projec-tion and two stages phantom leptogenesis
Let us now consider the case when both heavy neutrino masses M , M (cid:29) GeV andcharged lepton interactions do not affect the final asymmetry. This can be called the heavyflavoured scenario [17] since the only lepton flavours that affect the final asymmetry arethose produced from the heavy RH neutrinos. The density matrix equation (74) can thenbe recast simply as dN B − Lαβ dz = ε (1) αβ D ( N N − N eq N ) − W (cid:8) P (1)0 , N B − L (cid:9) αβ (80)+ ε (2) αβ D ( N N − N eq N ) − W (cid:8) P (2)0 , N B − L (cid:9) αβ . For illustrative purposes, we want first to describe just the N washout of the asymmetryproduced by the N decays, without any additional effect. Therefore, we first neglect thedifferent flavour compositions of leptons and anti-leptons assuming that ∆ p α = ∆ p α = 0.With such a simplifying assumption, the lepton quantum states are given by | (cid:105) = C τ | τ (cid:105) + C τ ⊥ | τ ⊥ (cid:105) and | ¯1 (cid:105) = C (cid:63) τ | ¯ τ (cid:105) + C (cid:63) τ ⊥ | ¯ τ ⊥ (cid:105) ( CP | ¯1 (cid:105) = | (cid:105) ) , (81) | (cid:105) = C τ | τ (cid:105) + C τ ⊥ | τ ⊥ (cid:105) and | ¯2 (cid:105) = C (cid:63) τ | ¯ τ (cid:105) + C (cid:63) τ ⊥ | ¯ τ ⊥ (cid:105) ( CP | ¯2 (cid:105) = | (cid:105) ) . (82)Assuming the hierarchical limit, M (cid:38) M [40], there are two well distinguished differentstages. In a first stage at T ∼ M , an asymmetry is produced from N decays. The leptondensity matrix is then given by ρ (cid:96)ij = diag(1 ,
0) in the basis (cid:96) − (cid:96) ⊥ . Analogously the anti-lepton density matrix is given by ρ ¯ (cid:96)ij = diag(1 ,
0) in the basis ¯ (cid:96) − ¯ (cid:96) ⊥ that at the momentwe are assuming to be CP conjugated of (cid:96) − (cid:96) ⊥ . As in the previous subsection, theasymmetry production from N decays is again described by the Eq. (75) with vanishingphantom terms so that we simply have N B − Lττ ( T (cid:39) T B ) (cid:39) p τ N T (cid:39) T B B − L , N B − Lτ ⊥ τ ⊥ ( T (cid:39) T B ) (cid:39) p τ ⊥ N T (cid:39) T B B − L , (83)where N T (cid:39) T B B − L (cid:39) ε κ ( K ). We have now to consider the N washout stage at T ∼ M .Since at the moment we are just interested in describing the N washout, we also neglectthe N asymmetry production assuming a vanishing ε (1) αβ . Moreover let us first furtherassume, just for simplicity, | (cid:105) = | τ (cid:105) and correspondingly | ¯1 (cid:105) = | ¯ τ (cid:105) .25n this way, at T ∼ M , the Eqs. (80) for the asymmetry evolution in the chargedlepton flavour basis can be simply rearranged as ( α, β = τ, τ ⊥ ) dN B − Lαβ dz = − W (cid:32) N B − Lττ N B − Lττ ⊥ N B − Lτ ⊥ τ (cid:33) , (84)and, at the end of the N -washout at T (cid:39) T B , one simply finds N B − Lττ ( T (cid:39) T B ) (cid:39) e − π K p τ N T (cid:39) T B B − L , N B − Lτ ⊥ τ ⊥ ( T (cid:39) T B ) (cid:39) p τ ⊥ N T (cid:39) T B B − L . (85)Finally, at T ∼ GeV, the charged lepton interactions damp the off-diagonal termsmeasuring the tauon and the ‘non-tauon’ (i.e. the τ ⊥ ) asymmetries.This result can be easily generalised. Let us, first of all, allow an arbitrary | (cid:105) flavourcomposition but continuing, for the time being, to neglect the N asymmetry production,at T ∼ T B . The Eq. (84) has now to be written in the basis (cid:96) − (cid:96) ⊥ , dN B − Li j dz = − W (cid:32) N B − L
11 12 N B − L ⊥ N B − L ⊥ (cid:33) ( i , j = 1 , ⊥ ) . (86)The solution is again quite trivial in this basis: the 11 term is washed out, N B − L ( T (cid:39) T B ) = e − π K N B − L ( T (cid:39) T B ) , (87)together with the off-diagonal terms, while the 1 ⊥ ⊥ term is unwashed. The asymmetrymatrix at T ∼ T B , in the (cid:96) − (cid:96) ⊥ basis, can now be calculated in terms of the rotationmatrices (cf. Eq. (38)) as N B − Li j ( T (cid:39) T B ) = N T (cid:39) T B B − L R (1)0 † i α R (2)0 αi (cid:32) (cid:33) R (2)0 † j β R (1)0 βj . (88)In a more compact way, considering that N B − L ( T (cid:39) T B ) = N T (cid:39) T B B − L | (cid:105)(cid:104) | , this can bemore conveniently written as N B − Li j ( T (cid:39) T B ) = N T (cid:39) T B B − L (cid:32) p (cid:104) | (cid:105)(cid:104) | ⊥ (cid:105)(cid:104) ⊥ | (cid:105)(cid:104) | (cid:105) − p (cid:33) , (89)where [17] p ≡ |(cid:104) (cid:96) | (cid:96) (cid:105)| = (cid:12)(cid:12) ( h † h ) (cid:12)(cid:12) ( h † h ) ( h † h ) . (90)The final asymmetry can then be calculated as N f B − L = Tr[ N B − Li j ( T (cid:39) T B )] = e − π K p N T (cid:39) T B B − L + (1 − p ) N T (cid:39) T B B − L . (91)26he asymmetry can be also rotated in the charged lepton flavour basis, N B − Lαβ ( T (cid:39) T B ) = R (1)0 αi N B − Li j ( T (cid:39) T B ) R (1)0 † j β . (92)At T (cid:39) GeV the charged lepton interactions just damp the off-diagonal terms withoutaffecting the total asymmetry given by the trace and for this reason we could directly writethe Eq. (91).This result fully confirms what one expects within an instantaneous quantum statecollapse description. It is not only confirmed that just the (cid:96) -parallel component of theasymmetry undergoes the N washout while the orthogonal component completely escapesit [6, 20], but also that the washout of the parallel component is exactly described bythe factor exp[ − (3 πK / K [17]. Notice that in anintermediate regime K ∼
1, the quantum states at T (cid:39) T B are left in a sort of partiallyincoherent mixture, with some residual flavour oscillations that however do not affect thetotal asymmetry.Notice that this result also applies to a possible pre-existing asymmetry produced bysome other external mechanism [20, 17]. Therefore, the conclusions of [17], employingthis result in various situations, are also confirmed.One can then easily further generalise this result accounting also for a possible N asymmetry generation, simply obtaining for the final asymmetry N f B − L = ε κ ( K ) + (cid:16) e − π K p + 1 − p (cid:17) ε κ ( K ) . (93) We still miss a last step. We have so far assumed that the flavour compositions of the (cid:96) and ( CP conjugated) ¯ (cid:96) quantum states are the same. We want now to show that, whenthis additional flavoured CP violation contribution is taken into account, phantom termscontribute to the final asymmetry and the eq. (93) gets generalised. Notice that this timethe role played by the charged lepton flavour basis in the previous subsection, is replacedby the heavy neutrino lepton basis (cid:96) – (cid:96) ⊥ . Notice that in general now also the basis (cid:96) − (cid:96) ⊥ does not coincide with ¯ (cid:96) − ¯ (cid:96) ⊥ and therefore there can be an ambiguity about the basison which one should project. However, one can calculate the wash-out in the tree-levelbasis 1 − ⊥ , so that the eq. (45) can be still used also in this case.Therefore, the quantum states | (cid:105) and | ¯2 (cid:105) have now to be projected, more generally,on the tree-level basis 1 − ⊥ so that they can be written as | (cid:105) = (cid:104) | (cid:105) | (cid:105) + (cid:104) ⊥ | (cid:105) | ⊥ (cid:105) and | ¯2 (cid:105) = (cid:104) ¯1 | ¯2 (cid:105) | ¯1 (cid:105) + (cid:104) ¯1 ⊥ | ¯2 (cid:105) | ¯1 ⊥ (cid:105) . (94)27herefore, writing the Eq. (75) in this basis, we have at the production ( i , j = 1 , ⊥ ) dN B − Li j dz = ε (2) i j D ( N N − N eq N ) − W (cid:8) P (2)0 , N B − L (cid:9) i j , (95)where as usual the superscript “0” indicates the tree level quantities that can be approx-imately fully employed in the calculation of the washout term. In this way we obtainexpressions for the heavy neutrino lepton flavour asymmetries, that are analogous to theeqs. (76-83) for the charged lepton flavoured asymmetries, N B − L ( T (cid:39) T B ) (cid:39) p ε κ ( K ) − ∆ p κ ( K / , (96) N B − L ⊥ ⊥ ( T (cid:39) T B ) (cid:39) (1 − p ) ε κ ( K ) + ∆ p κ ( K / . (97)The quantity ∆ p is defined analogously to the ∆ p iα ’s (cf. eqs. (17), (18)), explicitly∆ p ≡ |(cid:104) | (cid:105)| −|(cid:104) ¯1 | ¯2 (cid:105)| . Finally, taking into account the lightest RH neutrino washoutand asymmetry production, we obtain for the final asymmetry N f B − L = ε κ ( K ) + (cid:104) p e − π K + (1 − p ) (cid:105) ε κ ( K ) + (cid:16) − e − π K (cid:17) ∆ p κ ( K / . (98)Therefore, the phantom terms give an additional contribution to both components and inparticular to the orthogonal component. If K (cid:28)
1, both the parallel and the orthogonalcomponents are unwashed and the phantom terms cancel with each other. On the otherhand, in the opposite case, for K (cid:29)
1, the parallel component is completely washedout so that only the orthogonal one survives (together with the additional N -unwashedphantom term contribution).This result shows that phantom leptogenesis goes even beyond the case where the twoRH neutrino masses fall into two different flavour regimes [18].Finally, it should be clear that an account of the different flavour compositions of the (cid:96) and ¯ (cid:96) quantum states at the production from N , would lead to additional phantom terms.These, however, cancel with each other and do not contribute to the final asymmetry, asalready discussed in section 2. GeV (cid:29) M , M When M , M (cid:28) GeV both RH neutrinos produce their asymmetry in the two-flavourregime. The production from the heavier RH neutrinos is given by the usual result N B − Lττ ( T (cid:39) T B ) = ε τ κ ( K τ ) , N B − Lτ ⊥ τ ⊥ ( T (cid:39) T B ) = ε τ ⊥ κ ( K τ ⊥ ) . (99)28n the strong washout regime for both flavours, K τ ⊥ , K τ (cid:29)
1, the sum, i.e. the totalasymmetry, can be approximated by the Eq. (20) rewritten for the heavier RH neutrino.When the temperature drops down to T ∼ T B , the washout from the lighter RH neutrinostarts to act. Similarly to the previous cases, this washout factorizes from the generalexpression and can be expressed as a simple exponential pre-factor so that N B − Lττ ( T (cid:39) T B ) = ε τ κ ( K τ ) e − π K τ , (100) N B − Lτ ⊥ τ ⊥ ( T (cid:39) T B ) = ε τ ⊥ κ ( K τ ⊥ ) e − π K τ ⊥ . (101)The production of the asymmetry from the N decays is then added to what is left fromthe N production, so that we finally obtain N B − Lττ ( T (cid:39) T B ) = ε τ κ ( K τ ) e − π K τ + ε τ κ ( K τ ) , (102) N B − Lτ ⊥ τ ⊥ ( T (cid:39) T B ) = ε τ ⊥ κ ( K τ ⊥ ) e − π K τ ⊥ + ε τ ⊥ κ ( K τ ⊥ ) . (103)It should be noticed that there are no phantom terms in this case because of the assump-tion made at the beginning of the Section that τ ⊥ = τ ⊥ ≡ τ ⊥ . In this case, we have aneffective two-flavour problem and there are no phantom terms cancelling out. If we relaxthe two-flavour assumption allowing τ ⊥ (cid:54) = τ ⊥ , we have to work in a full three-flavour ba-sis, and, as we will see, phantom terms appear again in the final asymmetry. We discussthis more general case in the next Section. If we consider the general realistic case with three lepton flavours, the density matrixequations have to be written in terms of 3 × M i (cid:29) GeV, the density matrix equation eq. (74) further generalises into( α, β = τ, µ, e ) dN B − Lαβ dz = ε (1) αβ D ( N N − N eq N ) − W (cid:8) P (1)0 , N B − L (cid:9) αβ (104)+ ε (2) αβ D ( N N − N eq N ) − W (cid:8) P (2)0 , N B − L (cid:9) αβ + ε (3) αβ D ( N N − N eq N ) − W (cid:8) P (3)0 , N B − L (cid:9) αβ e ΜΤ Figure 4: A generic three heavy neutrino lepton flavour configuration. − Im(Λ τ ) , , N B − L αβ (105) − Im(Λ µ ) , , N B − L αβ . We have implied the effect of gauge interactions in setting the condition of thermal equi-librium on the lepton abundances.If one of the three masses is lower than ∼ GeV, electron flavour interactions termshave to be included as well, though they have no real impact, within this framework,on the final asymmetry. This is because the electron asymmetry is in any case alreadymeasured as a ‘neither-muon-nor-tauon’ asymmetry.This master equation can now be used to calculate the final asymmetry not only forall the ten mass patterns shown in Fig. 1, but also when the M i ’s fall in one of the flavourtransition regimes.Notice that, though in this paper we are only considering hierarchical RH neutrinomass patterns, this equation can also be used to calculate the asymmetry beyond thehierarchical limit [40] and even in the resonant case [41]. In this latter case, however,many different effects can become important and should be included [42].30olutions of this set of equations are particularly difficult when at least two of thefive kinds of interactions are simultaneously effective, something that goes beyond ourobjectives. Here, as an example with three flavours, we want to show a particularlyinteresting asymptotic limit that cannot be described within the simplified two-flavourcase discussed in the previous section: the two RH neutrino model [15]. We will showthat, even in this case, phantom terms have in general to be taken into account. We consider a two RH neutrino model [43] corresponding to a situation where M issufficiently large ( M (cid:29) GeV) to decouple in the seesaw formula for the calculationof the neutrino masses [44]. In order to reproduce the observed baryon asymmetry one hasto impose M (cid:38) GeV so that the muon interactions can be neglected in the Eq. (104)On the other hand, in order to have M and M as low as possible, it is interesting toconsider the case 10 GeV (cid:29) M (cid:38) M (cid:29) × GeV in a way to obtain a RH neutrinomass spectrum corresponding to the third panel (from upper left) in Fig. 1.This model has been recently revisited in [15]. We want here to re-derive, starting fromthe density matrix equation (104), the Boltzmann kinetic equations and the consequentformula for the final asymmetry that in [15] has been used to calculate the value of M necessary to reproduce the observed baryon asymmetry .Thanks to the hierarchical limit, we can again introduce different simplifications. Firstof all we can impose the complete damping of the τ α and ατ ( α (cid:54) = τ ) off-diagonal termsin the asymmetry matrix.Second, we can consider the N production at T (cid:39) T B . With these assumptions,only the N -terms can be considered in the Eq. (104) and the asymmetry matrix can betreated as a 2 × τ – τ ⊥ flavour space. In this way the density matrix equationreduce to a set of two Boltzmann equations in an effective two fully flavoured regime, dN B − Lττ dz = ε (2) ττ D ( N N − N eq N ) − p τ W N B − Lττ , (106) dN B − Lτ ⊥ τ ⊥ dz = ε (2) τ ⊥ τ ⊥ D ( N N − N eq N ) − p τ ⊥ W N B − Lτ ⊥ τ ⊥ . (107)As usual, assuming first for simplicity that the | τ ⊥ (cid:105) and the | ¯ τ ⊥ (cid:105) quantum states havethe same flavour compositions, one finds N B − Lττ ( T (cid:39) T B ) = ε τ κ ( K τ ) and N B − Lτ ⊥ τ ⊥ ( T (cid:39) T B ) = ε τ ⊥ κ ( K τ ⊥ ) . (108) It has been shown in [15] that even the N contribution to the asymmetry depends just on M andnot on M , provided that this is much smaller than 10 GeV . e ΜΤ Τ (cid:166) Τ (cid:166) Τ (cid:166) (cid:166) Figure 5: Relevant lepton flavours in the two RH neutrino model.These values of the asymmetries at the end of the N production stage have to be used asinitial values in the set of equations describing the evolution of the asymmetries duringthe N production, dN B − Lττ dz = ε (1) ττ D ( N N − N eq N ) − p τ W N B − Lττ , (109) dN B − Lτ ⊥ τ ⊥ dz = ε (1) τ ⊥ τ ⊥ D ( N N − N eq N ) − p τ ⊥ W N B − Lτ ⊥ τ ⊥ . (110)The τ ⊥ component of the asymmetry at the end of the N production has to be decom-posed into a τ ⊥ parallel component and into a τ ⊥ orthogonal component that we indicatewith the symbol τ ⊥ ⊥ . In this way one finds that the final asymmetry is the sum of threeflavour components (see Fig. 5), N f B − L = N B − Lττ ( T (cid:39) T B ) + N B − Lτ ⊥ τ ⊥ ( T (cid:39) T B ) + N B − Lτ ⊥ ⊥ τ ⊥ ⊥ ( T (cid:39) T B ) , (111)where N B − Lττ ( T (cid:39) T B ) = ε τ κ ( K τ ) + ε τ κ ( K τ ) e − π K τ , (112) N B − Lτ ⊥ τ ⊥ ( T (cid:39) T B ) = ε τ ⊥ κ ( K τ ⊥ ) + p τ ⊥ τ ⊥ ε τ ⊥ κ ( K τ ⊥ ) e − π K τ ⊥ , (113) N B − Lτ ⊥ ⊥ τ ⊥ ⊥ ( T (cid:39) T B ) = (cid:16) − p τ ⊥ τ ⊥ (cid:17) ε τ ⊥ κ ( K τ ⊥ ) . (114)32his expression coincides with the result found in [15] and is valid neglecting phantomterms. If one takes into account the different flavour compositions between the | τ ⊥ (cid:105) andthe | ¯ τ ⊥ (cid:105) quantum states, then phantom terms are, in general, present. The procedureis essentially the same discussed in Section 3.2, with the only difference that now thephantom terms will appear only in the τ ⊥ and τ ⊥ ⊥ components but not in the measuredtauon component. We can therefore directly write the final result, N f B − L = ε τ κ ( K τ ) + ε τ κ ( K τ ) e − π K τ (115)+ ε τ ⊥ κ ( K τ ⊥ ) + (cid:32) p τ ⊥ τ ⊥ ε τ ⊥ κ ( K τ ⊥ ) − ∆ p τ ⊥ τ ⊥ κ ( K τ ⊥ / (cid:33) e − π K τ ⊥ + (cid:16) − p τ ⊥ τ ⊥ (cid:17) ε τ ⊥ κ ( K τ ⊥ ) + ∆ p τ ⊥ τ ⊥ κ ( K τ ⊥ / , where each of the three lines corresponds respectively to the τ , τ ⊥ and τ ⊥ ⊥ componentsand where now ∆ p τ ⊥ τ ⊥ ≡ |(cid:104) τ ⊥ | τ ⊥ (cid:105)| − |(cid:104) ¯ τ ⊥ | ¯ τ ⊥ (cid:105)| . This last example shows, once more,how phantom terms are present whenever the production occurs either in one or in atwo flavour regime, though only those generated by the heavier RH neutrinos can beafterwards asymmetrically washed out by the lighter RH neutrinos and contribute to thefinal asymmetry without cancelling with each other. Within a Boltzmann classical kinetic formalism one has to distinguish the ten differentRH neutrino mass patterns shown in Fig. 1. These are obtained in the limits where themasses M i are hierarchical and do not fall in the transition regimes. We have extendedthe density matrix formalism for the calculation of the matter-anti matter asymmetry inleptogenesis including heavy neutrino flavours. In this way we obtained a density matrixequation for the calculation of the asymmetry for any choice of the RH neutrino masses,even beyond the hierarchical limit.Within this more general description, the ten hierarchical RH neutrino mass patternsof Fig. 1 correspond to those cases where the (five) different interactions are only oneby one effective within a given range of temperatures. In this way the evolution of theasymmetry can be described in well separated stages where the density matrix equationsgreatly simplify reducing to multiple sets of Boltzmann equations, one for each stage.In these cases we recovered or extended results that had already been derived within asimpler description based on an instantaneous collapse of lepton quantum states.33he flavour projection effect, where the orthogonal component of a previously pro-duced asymmetry escapes the RH neutrino washout, is fully confirmed. We have alsoshown that the washout of the parallel component is exactly described by the usual ex-ponential washout factor independently of the washout regime.Phantom terms emerge as quite a generic feature of flavoured leptogenesis and haveto be taken into account even for vanishing initial RH neutrino abundances. They cancontribute to the final asymmetry even if the production from an heavier RH neutrinospecies and the washout from a lighter RH neutrino species occur in the same fullyflavoured regime and so their presence goes beyond the N -dominated scenario wherethey were originally discussed [18]. However, we have shown that, when the effect ofgauge interactions in thermalising the lepton abundances is taken into account, phantomterms get washed-out at the production, though their wash-out rate is halved comparedto that one acting on the final asymmetry. In this way, in the strong wash-out regime,phantom terms give a contribution that is also independent of the initial conditions.Even though we have explicitly calculated the final asymmetry only in one of the tenasymptotic limits RH neutrino mass patterns shown in Fig. 1, in the case of the twoRH neutrino model, the procedure can be easily extended to all others neutrino masspatterns. For example one can easily show the expression for the final asymmetry in the N -dominated scenario, when M (cid:28) GeV [13, 14, 24, 18].It would be desirable in future to calculate the asymmetry beyond these ten asymp-totic limits, solving the full density matrix equation. In this way the calculation of thematter-anti matter asymmetry would be extended to a generic RH neutrino mass pat-tern, including the cases where the RH neutrino masses fall in the transition regimeswhere quantum decoherence from charged lepton interactions acts simultaneously withasymmetry generation and wash-out. This would make possible to interpolate betweenthe asymptotic limits, finding the exact conditions on the RH neutrino masses for thevalidity of the solutions that we have discussed here.
Acknowledgments
We wish to thank Stefan Antusch and Steve King for many useful discussions. PDB andLM acknowledge financial support from the NExT Institute and SEPnet. SB acknowl-edges support from the Swiss National Science Foundation under the Ambizione grantPZ00P2 136947. DAJ is thankful to the STFC for providing studentship funding.34 ppendix
In this Appendix we review and discuss some insightful aspects and properties of theheavy neutrino lepton and anti-lepton bases, respectively { (cid:96) , (cid:96) , (cid:96) } and { ¯ (cid:96) , ¯ (cid:96) , ¯ (cid:96) } .At tree level the two bases are CP conjugated of each other and the probabilities p ij ≡ |(cid:104) (cid:96) | (cid:96) (cid:105)| = |(cid:104) ¯ (cid:96) | ¯ (cid:96) (cid:105)| can be expressed as [17] p ij = (cid:12)(cid:12)(cid:12) ( m † D m D ) ij (cid:12)(cid:12)(cid:12) ( m † D m D ) ii ( m † D m D ) jj = | (cid:80) h m h Ω (cid:63)hi Ω hj | (cid:101) m i (cid:101) m j , (A.1)where (cid:101) m i ≡ ( m † D m D ) ii /M i . In the last expression we expressed the terms ( m † D m D ) ij through the orthogonal matrix Ω providing a useful parameterisation of the neutrinoDirac mass matrix given by m D = U √ D m Ω √ D M [45], where U is the leptonic mixingmatrix, D M ≡ diag( M , M , M ) and D m ≡ diag( m , m , m ).In general, the (tree level) bases { (cid:96) , (cid:96) , (cid:96) } and { ¯ (cid:96) , ¯ (cid:96) , ¯ (cid:96) } are not orthonormal (seeFig. 4) [10], i.e. in general p ij (cid:54) = δ ij . This case would indeed correspond to special forms ofthe Dirac mass matrices where the orthogonal matrix is either the identity (Ω ij = δ ij ), orone of the other five special forms obtained from the identity permuting rows or columns.These six special forms imply [13] that the see-saw formula reduces to the case whereeach light neutrino mass m j proportional to a different inverse RH neutrino mass M i ,so called form dominance models [46]. However, when one of these six special cases areexactly realised and the two bases are orthonormal, both the total and the flavoured CP asymmetries exactly vanish for the simple reason that in this case there is no interferencebetween the tree level and the one loop graphs, since this requires that in the decay of aRH neutrino N i a virtual RH neutrino N j (cid:54) = i couples to a lepton (cid:96) i while orthonormalityimplies that it does not.This means that for these six special forms, even including perturbative effects, theheavy neutrino lepton and anti-lepton bases remain equal to the tree level bases and,therefore, they are still orthonormal. Therefore, in order to have successful leptogenesis,the heavy neutrino lepton and anti-lepton bases have necessarily to be non-orthonormalto some level.When some interference between tree level and one loop graphs is turned on, implyingnon-orthonormality of the two bases, then in general this will induce both non-vanishingtotal CP asymmetries, with proportional contributions in the flavoured CP asymmetriesgiven by the first terms in the eq. (20), but also different flavour compositions betweenthe heavy neutrino lepton basis and heavy neutrino anti-lepton basis. This can be seeneasily from the expressions for the flavoured CP asymmetries recast in the orthogonal35arameterisation and for example in [12] it was noticed how ∆ p α (cid:54) = 0, with a strongenhancement of the asymmetry compared to the unflavoured case (cf. eq. (20)) can beinduced by the presence of low energy phases. As a matter of fact, it is well knownthat the total CP asymmetry depends only on a subset of phases (3) compared to theflavoured CP asymmetries. This difference can be precisely traced back to the presence ofthe ∆ p α in Eq. (20), which includes the dependence on the additional three phases. Onecan wonder why the flavour composition of the final leptons and anti-leptons is affectedby an account of the interference between tree level and one loop graphs. In particularthe neutrino Yukawa matrix can be always brought to a triangular form h = V † h ∆ , where V is a unitary matrix. One can then switch from the weak basis to another orthonormalbasis (cid:96) ∆ i = V iα (cid:96) α . In this basis one has that at tree level N → (cid:96) ∆1 = (cid:96) = V α (cid:96) α .However, this does not remain valid accounting for the interference with one loop graphsthat make now possible to have N → (cid:96) j (cid:54) =1 . 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