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Gennaro Miele » 5.Baryogenesis

Baryon/anti-Baryon asymmetry

In the observed Universe matter is by far much more abundant than antimatter. In the nearby Universe all structures we see, stars, galaxies and clusters are made by nucleons and electrons, and one can only expects that the same holds for far structures, otherwise at the surface of separation between a matter and an antimatter bubble we would see a very intense γ ray emission, originating from the annihilation processes like p + anti-p → π’s, which including π₀ then decay into γ’s. Such activity has not been detected till now.

Moreover, we have also some estimate of the antimatter content of the Universe from the cosmic rays composition, which shows a ratio anti-p/p ≈ 10^-4and even smaller for Helium nuclei.

A baryon / anti-baryon asymmetry is a measurement of not vanishing baryon number B per comoving volume. Such parameter is the only free parameter in BBN and its value is well fixed by both primordial light elements like Deuterium and by CMB anisotropies. The value found is

$\eta_B \equiv \frac{n_B}{n_\gamma} \sim 273.97 \cdot 10^{-10}\, \Omega_B h^2 \sim 6 \cdot 10^{-10}$

A convenient way to express the baryon number per comoving volume is to normalize the asymmetry density to the entropy density for relativistic species, namely

$B = \frac{n_B-n_{\overline{B}}}{s_R}$

The equilibrium predictions

Let us assume B=0,at very high temperatures, namely T >> m_p, protons and neutrons are kept in thermal equilibrium with other species by strong and e.m. interactions. As the temperature drops below m_p, nucleon/anti-nucleon annihilation processes strongly dilute the nucleon density and their equilibrium distribution in phase space develops the exponential suppression factor e^-m_N/T. The nucleon relic aboundance can be obtained in terms of the freezing temperature, T_F, which is given by equating the interaction rate with the Hubble parameter. In particular at low energies the annihilation cross section is of the order of σ ≈ m_π^-2 , and the average relative velocity is given by v ≈ 1.6 (T/m_N)^1/2, thus we get

$m_\pi^{-2} \sqrt{\frac{T_F}{m_N}} 2 \left(\frac{m_N T_F}{2 \pi} \right)^{3/2} e^{-m_N/T_F} \sim \sqrt{\frac{8 \pi }{3 m_{Pl}^2} g_*\frac{\pi^2}{30}} \,T_F^2$

Solving this equation we get T_F = 21 MeV. Plugging this value into the Boltzmann distribution for nucleons we get

$\frac{n_B}{n_\gamma} = \frac{n_{\overline{B}}}{n_\gamma} \sim 10^{-17}$

which is much smaller than the B value inferred by both BBN and CMB.

Sakharov conditions

Since from the astrophysical observation we do not see enough antimatter we are led to the conclusion that the Universe has a non zero and positive baryonic charge B. For high temperatures the nucleon and antinucleon densities are comparable, once the temperature drops below m_N, almost all antinucleons disappear due to annihilations and we are left with a relic baryon density given by

$\frac{n_B-n_{\overline{B}}}{s_R} \sim \frac{n_B}{s_R}$

hence, the value of B today is related to η_B by the expression

$B \sim \left(\frac{n_{\gamma,0}}{s_{R,0}}\right)\, \eta_B \sim \, 0.28 \, \eta_B$

As will be clarified in the following η_Bcan be estimated by primordial nucleosynthesis and CMB anysotropies and taken at this level as a free parameter of standard cosmology. Nevertheless a much more appealing possibility is on the table, namely that B be generated during the evolution of the Universe starting from symmetric conditions. In a very famous work, A. Sakharov discussed the necessary and sufficient conditions for the production of a non zero baryon asymmetry, which are

The existence of B violating interactions;
The violation of C and CP by these interactions;
The occurrence of some departure from thermodynamical equilibrium.

Meaning of Sakharov conditions: B and C,CP violating processes

The condition 1) simply states the principle that if all interactions preserve the baryon number then the Hamiltonian H would commute with B. In this case if B=0 at some initial time, it will remain fixed at that value for ever. The condition 2) concerning the violation of both C and CP can be also easily understood. Let us consider some generic process from an initial state “i” to a final state “f” where some baryonic number is produced. If C is a symmetry for this process, then the rate Γ(i → f) will be equal to Γ(anti-i → anti-f). Hence no baryonic number can be produced. Let us remind that CPT theorem guarantees that, on the basis of quite general assumptions for a local relativistic quantum field theory, the combined transformation of C, P and time reversal T is an exact symmetry. This implies that CP violation (symmetry) is equivalent to T violation(symmetry). Hence by denoting with p_i (p_f) and s_i (s_f) are the initial (final) particle momenta and spins, respectively then CP symmetry would imply

$\Gamma(i \rightarrow f; p_i, s_i; p_f, s_f) = \Gamma(f \rightarrow i; -p_f, -s_f; -p_i, -s_i)$

Then since the baryonic number produced in the process i → f is opposite to the one produced in the reaction f → i, once we integrate over both momenta and spins we get no total baryonic number produced. For this reason, the violation of CP is a necessary requirement. However, as it is well known the CP violation in fundamental interactions has already been observed. The third condition states the necessity to have an out of equilibrium phase in order to produce a total baryonic number.

To prove that let us remind that in general the mean value of B, like for any physical quantity, when calculated in a system which is a statistical mixture of states must be computed in terms of the density matrix ρ, namely

$\langle \mbox{B} \rangle = \frac{\mbox{Tr} \left( \varrho \mbox{B}\right)}{\mbox{Tr} \left(\varrho \right)}$

Meaning of Sakharov conditions: Out of equilibrium condition

The third condition states the necessity to have an out of equilibrium phase in order to produce a total baryonic number. To prove that let us remind that in general the mean value of B, like for any physical quantity, when calculated in a system which is a statistical mixture of states must be computed in terms of the density matrix ρ, namely

$\langle \mbox{B} \rangle = \frac{\mbox{Tr} \left( \varrho \mbox{B}\right)}{\mbox{Tr} \left(\varrho \right)}$

Notice that at equilibrium ρ = e^{– β H}, with H commuting with the CPT operator Θ whereas B Θ = – Θ B. From these observations we get

$\langle \mbox{B} \rangle = \frac{\mbox{Tr} \left( \Theta \Theta^{-1} e^{-\beta \mbox{\small H}} \mbox{B} \right)}{\mbox{Tr} \left(\varrho \right)}=\frac{\mbox{Tr} \left( \Theta e^{-\beta \mbox{\small H}} \Theta^{-1} \mbox{B} \right)}{\mbox{Tr} \left(\varrho \right)}$ $= -\frac{Tr \left( \Theta e^{-\beta \mbox{\small H}} \mbox{B} \, \Theta^{-1} \right)}{Tr \left(\varrho \right)} =- \langle \mbox{B} \rangle$

and hence <B>=0. This result can be simply explained reminding that if there are baryon violating interactions the chemical potential associated to baryon number vanishes. In this case both baryons and anti-baryons at equilibrium have the same distribution function (due to CPT)

$f(p) = \left[\exp(\beta E(p)) \pm 1\right]^{-1}$

and therefore the asymmetry can only be zero.

C and CP violation and the need for new physics

In the electroweak SM C is maximally violated since we are working with a chiral gauge theory. Left-handed particles interacts with SU(2)_L gauge bosons, whereas right-handed fermions are singlets with respect to the same group. The same is true for the weak hypercharges. Moreover we know from the physics of K_L and K_S that CP symmetry is also broken in the SM. This evidence has been sequently corroborated by similar results obtained using B meson decays. As we know, in the electroweak SM the CP violation in the quark sector depends on a single parameter, namely the Jarlskog parameter that results to be of the order of 10^-5, and it is invariant under arbitrary phase redefinitions of quark fields. Such quantity depends on all mixing angles of Cabibbo-Kobayashi-Maskawa matrix as well as on its complex phase. Unfortunately, such amount of CP violation is too small to account for the observed value of η_B. This can be easily understood reminding that any CP violating effect, denoted for simplicity with ε_CP, has to be vanishing in the limit of quarks degenerate in masses (CKM matrix can be safely riabsorbed in the redefinition of quark fields). This means that ε_CP should be proportional to quark mass differences, actually to squared mass differences, since we can always change the overall sign of the fermion mass term by independently redefining the right and left components

$\epsilon_{{\mbox{\scriptsize CP}}} \sim J \, \frac{(m_t^2-m_c^2)(m_t^2-m_u^2)(m_c^2-m_u^2)(m_b^2-m_s^2)(m_b^2-m_d^2)(m_s^2-m_d^2)}{E^{12}}$

with E some energy scale. Since B violating processes in the framework of the SM are only effective at high energies, of the order of the electroweak breaking scale E_EW ≈ v ≈ 100 GeV or above this implies ε_CP≈ 10^-20 , which is orders of magnitude smaller than η_B.

C and CP violation and the need for new physics

This simple calculation suggests that the production of a large enough baryon asymmetry requires some physics beyond the SM.

Very popular models for baryogenesis are represented by Grand Unified Theories (GUT), which naturally predict B violating processes, like proton decay. Furthermore in these theories there is room for large CP–violating effects obtained due to the enlarged number of Higgs scalars and gauge boson fields.

An alternative possibility is provided by lepton physics and in particular by neutrinos. As will be clarified in the following, leptons result to be relevant in producing a baryon asymmetry even though they do not carry any baryon number. This is possible due to sphalerons that reshuffle baryon and lepton numbers but keeping unchanged their difference B-L. This means that if we have some finite value for L, somehow produced in the early Universe, it will give rise to a baryon asymmetry too. Such scheme is generally known as Baryogenesis through Leptogenesis and will discuss in the following with some detail.

It is worth reminding that lepton number L is not a symmetry if we introduce both Dirac and Majorana neutrino mass terms, as it occurs in the see-saw models. Moreover, if we have some sterile states, N_s , one can introduce Yukawa terms in the Lagrangian density which couple these states to left-handed neutrinos and Higgs field Φ. After spontaneous symmetry breaking such terms will produce the neutrino Dirac mass term

${\cal L}_{Y}^R = Y^{\nu}_{ \alpha s} \, \overline{L_\alpha} \, i \tau_2 \, \Phi^* \, N_{s}\, + h.c.$

Since the coupling Y_αs^νis generally complex, this provides a source for CP violation effects.

B and L violation

The electroweak SM Lagrangian is invariant under the two global symmetries that lead to the conservation of baryon and lepton numbers, respectively. Such property is guaranteed by considering any renormalizable terms which are invariant under the electroweak SM group. Hence, there are no processes computed in the perturbative approach changing B or L. Nevertheless, the same could be not true for non perturbative effects which may give rise to some violation. This is just the case of instanton and sphaleron configurations of the gauge and Higgs fields which will be discussed later.

The situation obviously changes if one considers extensions of the electroweak SM gauge symmetry, namely a Grand Unified Theory (GUT).

The concept of GUT is quite simple. One can imagine that for energies larger than some energy threshold M_GUT the symmetry group of fundamental interactions is a group G larger than the electroweak SU(3)_C x SU(2)_L x U(1)_Y .

There are some famous examples proposed in literature like the simple groups SU(5) or SO(10), or semi-simple like SU(4) x SU(2)_L x SU(2)_R, or finally containing an U(1) factor as SU(3)_C x SU(2)_L x SU(2)_R x U(1)_B-L. All of these contain the electroweak SM gauge group as a proper subgroup.

B and L violation

Differently from the SM, in a GUT theory both quarks and leptons of a given generation can be accomodated in the same irreducible representation of G. This is for example the case of SU(5) where a single generation of fermions is allocated into

$\overline{\bf 5} \oplus {\bf 10}$

In particular we have

$\overline{\bf 5} \equiv (q_{\alpha R}^{D C} , L_{\alpha L}) \,\,\,\, {\mbox and} \,\,\,\,\, {\bf 10} = (Q_{\alpha L}, q_{\alpha R}^{U C}, l_{\alpha R}^C)$

The case of SO(10) is even more elegant, since all left-handed quarks and leptons and their conjugate states are accomodated in a single representation, the 16-dimensional spinorial representation of SO(10), which also naturally include the SM sterile state $\nu^C_R$

The Higgs mechanism leading from G to SU(3)_C x SU(2)_L x U(1)_Yneeds a chain of several intemediate breaking steps at decreasing energy scales

$G \rightarrow G' \rightarrow G'' \rightarrow... \rightarrow SU(3)_C \times SU(2)_L \times U(1)_Y$

Baryon number violation and proton decay

B violating terms lead to proton instability, via processes like $p \rightarrow e^+ \pi^0$ From the experimental point of view it is available an extremely high lower limit on the lifetime in these decay channels like $\tau_{p \rightarrow e^+ \pi^0} \geq 8.2 \cdot 10^{33}\,\,\, years$ that implies that the typical mass of these gauge bosons must be very heavy. GUT models underwent spontaneous symmetry breaking in the very early Universe, when the temperature was of the order of M_GUT ≥ 10¹⁵ GeV. The interactions violating B and L are a good news for baryogenesis. In fact, in the early stages of the expansion, where the typical temperature of the plasma was very high, these interactions are not suppressed by the large values of leptoquark masses, so they might lead to some baryon or lepton asymmetry. Such scheme, called GUT baryogenesis, represents the most simple scenario proposed to explain the value of η_B. It is essentially based on the out of equilibrium decay scenario of gauge or heavy Higgs scalars. Its basic structure will be discussed in the following. The main problem of GUT like SU(5) is that despite B and L are violated, nevertheless their combination B-L is preserved. Thus even though some initial B asymmetry is produced at very early times, it will be completely erased by the lower energy sphaleron processes, which wash out any initial asymmetry in B+L, while they cannot change the value of their difference B-L. If B-L is initially zero, no baryon and lepton asymmetry is left over today. This is the case for SO(10) or left-right symmetric models. In this case B-L is associated to a gauge symmetry which is spontaneously broken at some intermediate scale. At the B-L breaking scale the $\nu^C_R$ take a Majorana mass term so we cannot assign a definite value of L to these states if they also have standard Dirac mass coupling with their left-handed partners. This is the main starting point of Leptogenesis.

Sphalerons

It was first noted by ‘t Hooft that due to the non trivial vacuum structure of non abelian gauge theories, and to the anomalous violation of baryon and lepton currents, in the electroweak SM one has B and L violating processes at the non perturbative level, mediated by instantons. It is worth reminding that for massless fermions both left-handed and right-handed currents are classically conserved, nevertheless the triangle diagram leads to an quantum anomaly. If we consider quantum electrodynamics we have

$\partial_\mu J_L^\mu = \partial_\mu \overline{\psi}_L \gamma^\mu \psi_L = -\frac{e^2}{16 \pi^2} F^{\rho \sigma} \tilde{F}_{\rho \sigma}$ $\partial_\mu J_R^\mu = \partial_\mu \overline{\psi}_R \gamma^\mu \psi_R = \frac{e^2}{16 \pi^2} F^{\rho \sigma} \tilde{F}_{\rho \sigma}$

This means that the vector current J^μ= J_L^μ+J_R^μ is still conserved while the axial one is anomalous

$\partial_\mu J_5^\mu = \partial_\mu \overline{\psi} \gamma^\mu \gamma_5 \psi = \frac{e^2}{8 \pi^2} F^{\rho \sigma} \tilde{F}_{\rho \sigma}$

The same result would hold for non abelian gauge theories but the electroweak SM is a chiral gauge theory, namely left- and right-handed couple differently to gauge fields. If we consider SU(2)_L , since only left-handed fermion couple to gauge fields in this case $\partial_\mu J_R^\mu = 0$ and hence the vector current is also anomalous

$\partial_\mu \overline{\psi} \gamma^\mu \psi = -\frac{g^2 \kappa}{16 \pi^2} F_a^{\rho \sigma} \tilde{F}_{a\rho \sigma}$

where k is a coefficient depending on the number and type of virtual particles in the triangle Feynman diagram.

Sphalerons

The charge associated

$Q = \int d^3 x \, J^0$ counts the number of particles minus antiparticles (summed over the two possible chiralities). The Gauss theorem gives

$Q(t_f) - Q(t_i) = -\frac{g^2 \kappa }{16 \pi^2 } \int_{t_i}^{t_f} dt \int d^3 x F_a^{\rho \sigma} \tilde{F}_{a\rho \sigma} \neq 0$ which means that the charge is not conserved. From the previous considerations we get that both baryon and lepton vector currents are not conserved

$\partial_\mu J^{(B) \mu}=\partial_\mu J^{(L) \mu} = \frac{N_f}{32 \pi^2} \left( -g^2 F_a^{\rho \sigma} \tilde{F}_{a\rho \sigma} +g'^2 B^{\rho \sigma} \tilde{B}_{\rho \sigma} \right)$

where N_f is the number of generation and $F_a^{\rho \sigma} \,\,\,\,and \,\,\,\, B^{\rho \sigma}$ are the SU(2)_Land U(1)_Y field strength, respectively. Since both lepton and baryon currents have the same anomaly (for the same number of quark and lepton generation), while B+L can change, the difference B-L is still a conserved charge. The processes where B+L is not preserved are related to the non trivial structure of vacua in non abelian gauge theories. If we integrate the previous equation we can link the variation of B with the Chern-Simons numbers

$B(t_f)-B(t_i) = N_f \left( N_{CS}(t_f)-N_{CS}(t_i)$ where

$N_{CS} = - \frac{g^2}{16 \pi^2} \int d^3 x \, 2 \epsilon^{ijk} \mbox{Tr} \left( \partial_i A_j A_k + \frac{2}{3} g A_i A_j A_k \right)$ with A_i the SU(2)_L gauge fields.

Chern-Simons numbers

The Chern-Simons are relative integer numbers, hence equal to 0, ± 1, ±2,… etc., which label the infinite set of possible vacua of a non abelian gauge theory.

In a classical gauge theory, the ground state should be time independent and the minimum of the energy. If we choose the gauge A₀=0, this implies that A_i should correspond to a pure gauge, i.e.

$A_i = \frac{i}{g}\left( \partial_i U(x) \right) U(x)^{-1}$

with U(x) denoting a generic gauge transformation. At the same time the Higgs field should be at the minimum of the potential, namely

$\Phi = U(x)\left( \begin{tabular}{c} 0 \\ $v$ \end{tabular} \right)$

Choosing the trivial transformation U(x) proportional to the identity matrix I, we get A_i=0, and in this case the Chern-Simons number is vanishing. This would be the same for each time-independent gauge transformation that can be continuously transformed into the trivial one. In this case all vacua would result identical. This is not true for non abelian gauge theories since the U(x) can be decomposed in homotopy classes labelled by integer (positive or negative) “winding numbers” n.

Chern-Simons numbers

Two generic transformations U(x) and U’(x) are in the same class if they can be continuously deformed one into the other. Let us take a representative U⁽ⁿ⁾(x) in each of these classes we can then write down an infinite set of topologically inequivalent vacua

$A^{n}_i = \frac{i}{g}\left( \partial_i U^{(n)}(x) \right) \left(U(x)^{(n)} \right)^{-1}, \,\,\,\, \Phi^{(n)} = U^{(n)}(x) \left( \begin{tabular}{c} 0 \\ $v$ \end{tabular} \right)$

which have different Chern-Simons numbers N_CS=n.

Baryon and lepton number violation is related to the probability of transition from one configuration of gauge fields to another one with different values of N_CS. This probability is negligible at low energies. This can be easily proved by reminding that in the SM vacuum states are separated by a potential barrier of the order of 8 π v/g ≈ 10 TeV.

For vanishing temperature this transition can occur only by a tunnel effect. This is exactly the role played by the instanton configuration studied by ‘t Hooft, who found that the probability of a variation by one unit of N_CS and thus of B and L violation processes is exponentially suppressed by the factor exp(-16 π²/g²) ≈ 10^-160.

For this reason at our energy baryon and lepton violating processes never occur. Different was the situation in the early Universe when the temperature and density conditions were extreme. The potential barrier between states with different N_CS could be classically overcome due to the kinetic energy that was sufficiently large.

Chern-Simons numbers

The transition rate in this case is determined by a particular solution of classical gauge and Higgs equation of motion better known as sphaleron. It corresponds to an unstable static solution whose mass M_sph is given by the height of the potential barrier and which interpolates between two contiguous vacua. The probability of exciting such a configuration is proportional to the factor exp(- M_sph/T). Reminding that the Higgs vev v decreases with the increasing of the temperature, and eventually vanishes above the electroweak phase transition, the probability per unit time and volume for a ΔN_CS=1 process is

$\Gamma \propto \exp \left( -\frac{8 \pi m_W(T)}{g^2 T} \right)$

Above the electroweak transition at T_EW≈ 100 GeV, v vanishes and the gauge bosons are massless, in this case the barrier disappears. There are no more sphalerons and the transitions can occur much more efficiently. A rough estimate of the transition rate gives

$\Gamma \sim \left(\frac{g^2}{4 \pi} T \right)^4$

This leads to the presence of very fast B and L violating processes till very high energies. Using the standard criterium we have efficient processes till

$\frac{\Gamma}{T^3} \gtrsim H(T) =\left( \frac{8 \pi}{3} g_* \frac{\pi^2}{30}\right)^{1/2} \frac{T^2}{m_{Pl}}$

which means up to T ≈ 10¹² GeV.

The relation between baryon and lepton numbers

Sphaleron processes reshuffle the baryon and lepton numbers via transition which change B+L, but keep constant the value of B-L. For simplicity one can imagine that sphaleron transitions if in equilibrium completely cancel ony initial value of B+L. Under this assumption if we denote by B₀ and L₀ the initial values for B and L at some high energy scale one gets

$B_0 = \frac{B_0+L_0}{2}+\frac{B_0-L_0}{2} , \,\,\,\,\, L_0 = \frac{B_0+L_0}{2}-\frac{B_0-L_0}{2}$

if the initial B₀ + L₀is completely washed out we get

$B = \frac{B_0-L_0}{2} = -L$

However, such prediction results to be correct only at the level of orders of magnitude. In fact, a precise calculation must consider the relation between the temperature T and the value of Higgs vev. Let us assume that we are at a value of T >> v, in order to perform a better calculation we can consider also quarks as freely propagating particles. In this case for μ_i<<T the asymmetry for the i-th relativistic particle specie results

$n_i -n_{ \overline{i}} \simeq \left\{ \begin{array}{ll} \frac{g}{6} T^2 \mu_i & {\mbox fermion} \\ & \\ \frac{g}{3} T^2 \mu_i & {\mbox boson} \end{array} \right.$

The relation between baryon and lepton numbers

Using the previous relation concerning the sphaleron processes, one finds that for three generations (N_f=3) and for a ΔN_CS=1 the transition involves 12-left handed fermions, namely

$\emptyset \leftrightarrow 2u + d+2c+s+t+2b+ e^- + \nu_\mu + \tau^-$

Such reaction satisfies ΔB = ΔL = N_f=3, and preserves the neutrality.

Let us remind that all interaction processes at equilibrium and conserved charges Q imply relations on the chemical potentials of particles of the SM.

If some particles are involved in a process at equilibrium or possess a non zero Q charge we have a linear relation among the corresponding chemical potentials.

The relation between baryon and lepton numbers

The constraints on μ_i we get are the following:

1. For very high temperatures all the SU(3)_C x SU(2)_L x U(1)_Y gauge interactions are in equilibrium, this implies that all particles in the same group representation share the same chemical potential, and gauge bosons have vanishing chemical potential.

2. The “efficient” 12–particle interactions due to sphaleron processes together with condition 1) lead to

$\sum_{\alpha} \left( 3 \mu_{Q_{\alpha L}} + \mu_{L_{\alpha L}} \right)=0$

where the sum is over the generation index α.

3. Left- and right-handed quarks interact via SU(3)_C instanton processes, hence we have a corresponding relation

$\sum_{\alpha} \left( 2 \mu_{Q_{\alpha L}} - \mu_{q^U_{\alpha R}} -\mu_{q^D_{\alpha R}}\right)=0$

4. The Yukawa interactions link the left-handed and right-handed fermion and Higgs chemical potentials

$\mu_{Q_{\alpha L}} - \mu_\Phi -\mu_{q^D_{\alpha R}}=0;$ $\mu_{Q_{\alpha L}} + \mu_\Phi -\mu_{q^U_{\alpha R}}=0;$ $\mu_{L_{\alpha L}} - \mu_\Phi -\mu_{l_{\alpha R}}=0.$

5. Hypercharge neutrality implies

$\sum_{\alpha} \left( \mu_{Q_{\alpha L}} + 2 \mu_{q^U_{\alpha R}}-\mu_{q^D_{\alpha R}}-\mu_{L_{\alpha L}}-\mu_{l_{\alpha R}} \right) + 2 \mu_\Phi =0$

The relation between baryon and lepton numbers

Assuming for simplicity that there is no dependence on the family index, we can express all chemical potentials in terms of one of them only. Let us take for example all expressions as function of the left-handed lepton chemical potential, hence we get

$\mu_\Phi =\frac{4 N_f}{6 N_f+3}\mu_{L_{L}},$ $\mu_{q^U_R} = \frac{2 N_f-1}{6 N_f+3}\mu_{L_{L}},$ $\mu_{q^D_R} = -\frac{6 N_f+1}{6 N_f+3}\mu_{L_{L}},$

$\mu_{Q_{L}} = - \frac{1}{3}\mu_{L_{L}},$ $\mu_{l_{R}} = \frac{2 N_f+3}{6 N_f+3}\mu_{L_{L}}$

Hence the Baryon and Lepton numbers result to be for a generic number of generations N_f given by

$B \propto N_f \left( 2 \mu_{Q_{L}}+\mu_{q^U_R}+\mu_{q^D_R} \right)= -\frac{4}{3} N_f \mu_{L_{L}},$

$L \propto N_f \left( 2 \mu_{L_{L}}+\mu_{l_{R}} \right) = \frac{14 N_f+9}{6 N_f+3} N_f \mu_{L_{L}}$

or equivalently

$B = \frac{8 N_f+4}{22 N_f+13} (B-L),$ $L = -\frac{14 N_f+9}{8 N_f+4} B.$

These relations give for T >> v and for N_f=3 the relation L = – 1.8 B. Such expression is almost valid also for smaller T.

The out of equilibrium decay scenario

Let us discuss a scheme to produce out of equilibrium conditions which was first proposed in the framework of GUT baryogenesis.

We start considering some particle specie X with mass m_X and its antiparticle which are coupled to light degrees of freedom, like fermionic particles of the SM or the scalar Higgs, via baryon (or lepton) violating number violating interaction terms. To describe the main characteristics of the model let us take the simplest case, where the X particle can decay into two different channels simply denoted as f₁ (with baryon number B₁) and f₂ (with baryon number B₂). The decay rates in the two exclusive channels can be written as

$\Gamma(X \rightarrow f_1) = \frac{1}{2} (1+ \epsilon) \Gamma_D,$ $\Gamma(\overline{X} \rightarrow \overline{f}_1) = \frac{1}{2} (1 + \overline{\epsilon}) \Gamma_D$

$\Gamma(X \rightarrow f_2) = \frac{1}{2} (1- \epsilon) \Gamma_D,$ $\Gamma(\overline{X} \rightarrow \overline{f}_2) = \frac{1}{2} (1- \overline{\epsilon}) \Gamma_D$

where Γ_D is the total decay rate that due to CPT and unitarityis the same for X and its antiparticle. The quantities $\epsilon-\overline{\epsilon}$ parametrize the strength of C and CP violation.

At high temperatures T >> m_X, X and their antiparticles are in thermodynamical equilibrium with other relativistic species due to scattering processes, $X \, \overline{X}$ pair annihilations and pair productions, decays and inverse decays $f_{1,2} \rightarrow X, \overline{X}$ .

At this temperature the rates are of the order of α T, with some coupling constant depending on the particular process.

Since there are baryon violating processes and equilibrium holds no baryon asymmetry is initially present, thus $n_X=n_{\overline{X}}$

The out of equilibrium decay scenario

When the temperature drops below m_X, the densities of X and their antiparticles exponentially decrease as exp(-m_X/T) whenever the interactions are strong enough to keep them in equilibrium. Typically the main processes regulating their abundance are decay and inverse decay, whose rates for a renormalizable theory reads

$\Gamma_D = g^2 m_X$

$\Gamma_{ID} \sim \Gamma_D \left( \frac{m_X}{T} \right)^{3/2} e^{-m_X/T}$

In case Γ_Dis larger than H for T ≈ m_X, decays are able to keep X and their antiparticles in equilibrium, so their density rapidly decrease. On the contrary if decay processes are quite slow

$\Gamma_D \lesssim H(T\sim m_X)$ (a)

the above densities per comoving volume remain frozen, and become much larger than the equilibrium value. As T continues to decrease, the inverse decays become quite negligible, since the process is kinematically suppressed. By construction the decay of X particles will occur for a temperature T_Dfor which the universe is at least as old as the particle lifetime, which implies Γ_D≈ H(T_D). Such decays will occur in out of equilibrium conditions, since

$n_{X,\overline{X}}>> n^{eq}_{X,\overline{X}}$

and this is by virtue of Sakharov theorem a necessary condition for a production of a baryon asymmetry.

The out of equilibrium decay scenario

The equation (a) of previous trasparency translates into a lower bound on the X particle mass. Assuming that the Universe is radiation dominated we have

$m_X \gtrsim \frac{g^2}{\sqrt{g_*}} m_{Pl}$

In the case for which X is a gauge boson of a GUT theory, we have as a typical value g²≈ 10^-2 – 10^-1, and so

$m_X \gtrsim (10^{17}-10^{18}) /\sqrt{g_*} \, {\mbox GeV}$

If X is a heavy neutrinos coupled to the Higgs and the left handed lepton doublet, we have g²≈ m_D²/v², with m_D of the order of the charged lepton Dirac masses, hence

$m_X \gtrsim (10^{7} - 10^{15}) /\sqrt{g_*} \, {\mbox GeV}.$

In both cases the X particles should be much heavier than the electroweak breaking scale. This supports the scenario of a GUT baryogenesis, since the GUT breaking scale is typically of the order of 10¹⁶ GeV, and are of the same order tha masses of leptoquarks.

The value of g_* depends on the theoretical framework considered. In the SM, at very high energy scale g_* ≈ 10². This value is typically larger in theories which extend the SM group or in supersymmetric models. In any case, the main conclusions of the model do not depend very much on this detail.

Baryon asymmetry in the out of equilibrium decay scenario

In general, apart of decays one should also consider the inverse decay processes as well as baryon violating scatterings f₁ ↔ f₂ mediated by the X particles. These two contributions have to be taken into account in case the out of equilibrium condition are only partially satisfied, $\Gamma \sim H(m_X)$

Such additional processes tend to wash out the effect of decays and reduce the final baryon asymmetry. In this case the time evolution of baryon number B can only be found by solving a set of Boltzmann equations for the species involved. Let us assume that T_D < m_X in such a way that we can neglect the inverse decay processes and baryon violating scatterings. The average baryon number density produced in X decays is

$n_X(T_D) \frac{\Gamma(X \rightarrow f_1) \mbox{B}_1 + \Gamma(X \rightarrow f_2) \mbox{B}_2}{\Gamma_D} = n_X(T_D) \left( \frac{\mbox{B}_1+\mbox{B}_2}{2} + \epsilon \frac{\mbox{B}_1-\mbox{B}_2}{2} \right)$

wheras for the X antiparticle decay we have

$n_{\overline{X}} (T_D) \left( -\frac{\mbox{B}_1+\mbox{B}_2}{2} - \overline{\epsilon} \frac{\mbox{B}_1-\mbox{B}_2}{2} \right)$

Summing the two contributions, the final total baryon number density normalized to the specific entropy of relativistic species s_R after decays reads

$B = \frac{n_X(T_D)}{s_R} (\epsilon -\overline{\epsilon}) (\mbox{B}_1-\mbox{B}_2)$

Baryon asymmetry in the out of equilibrium decay scenario

It is worth noticing that B vanishes if $\epsilon -\overline{\epsilon}=0$ (C and CP symmetry) or if B₁=B₂ (B is a conserved quantum number). In case at decay the universe energy density is still dominated by relativistic d.o.f., we can neglect the entropy release from the decaying heavy X particles, we have $s_R \sim g_s T_D^3$ . By using the relation

$n_X(T_D) \sim n_\gamma (T_D) \sim T_D^3$

we get

$B \sim \frac{(\epsilon-\overline{\epsilon}) (\mbox{B}_1-\mbox{B}_2)}{g_s}$

The situationis alittle bit diferent if the Universe is dominated by X energy before its decay. In this case the value s_R must be carefully computed. In the limit of instantaneous decay at T_D, the reheating temperature T_RH of relativistic d.o.f. can be obtained from conservation of energy

$m_X n_X(T_D) = g_* \frac{\pi^2}{30} T_{RH}^4$

which implies

$s_R = \frac{4}{3} g_s \frac{\pi^2}{30} T_{RH}^3 = \frac{4}{3} n_X(T_D) \frac{m_X}{T_{RH}}$

Note that we have assumed that all relativistic d.o.f. share the same temperature.

Baryon asymmetry in the out of equilibrium decay scenario

The temperature T_RH is obtained requiring that at T_D the decay rate equals H

$\Gamma_D \sim \left( \frac{8 \pi}{3 m_{Pl}^2} \rho_X(T_D) \right)^{1/2} = \left( \frac{8 \pi}{3 m_{Pl}^2} g_* \frac{\pi^2}{30} T_{RH}^4 \right)^{1/2}$

From the previous expression we get

$T_{RH} \sim \frac{0.78}{g_*^{1/4}} \left( \Gamma_D \, m_{Pl} \right)^{1/2}$

Thus finally we can get

$B \sim \frac{3}{4}\frac{T_{RH}}{m_X} (\epsilon -\overline{\epsilon})(\mbox{B}_1-\mbox{B}_2) \sim \frac{1}{{g_*^{1/4} } } \left( \frac{\Gamma_D \, m_{Pl}}{m_X^2} \right)^{1/2} (\epsilon-\overline{\epsilon}) (\mbox{B}_1-\mbox{B}_2)$

Few elements of Leptogenesis

As discussed in the previous slides due to Baryogenesis new physics (beyond the SM) is necessary to account for stronger CP violation effects and for a departure from thermal equilibrium.

A nice mechanism based on heavy sterile neutrinos involved in the seesaw model was proposed and it is now better known as Baryogenesis through Leptogenesis or more simply as Leptogenesis.

In this model the Yukawa couplings of such particles provide both the source of CP violation, and ensure out of equilibrium decay conditions. Morevoer, the lepton number violation is implemented in the Majorana masses of these neutrinos. Sphaleron processes are then responsible for converting the lepton asymmetry into a baryon asymmetry.

Let us consider the simplest implementation of seesaw mechanism, namely Type I models. In these theories there is a certain number n_s of sterile Majorana neutrinos $N_s \equiv \nu_{sR}+\nu_{sR}^C$

The most general Dirac/Majorana Lagrangian for the Yukawa coupling reads (Φ stands for a Higgs doublet)

${\cal L}_{Y}^R=- Y^{\nu}_{ \alpha s} \overline{\nu_{\alpha L}} N_{s} \overline{\phi}_0 + Y^{\nu}_{ \alpha s} \, \overline{l_{\alpha L}} \, N_{s}\, {\phi}_- -Y^{\nu *}_{\alpha s} \, \overline{N_{s}} \, \, \nu_{\alpha L} \, \phi_0+ Y^{\nu*}_{ \alpha s} \, \overline{N_{s}} \, \, l_{\alpha L} \, \phi_+- \frac 12 M_s \, \overline{N_{s}} \, N_{s}$

At tree level, N_s that is a Majorana particle can decay in the following channels

$N_{s} \rightarrow \phi_0 + \nu_{\alpha L}$ $N_{s} \rightarrow \phi_+ + l_{\alpha L}$ $N_{s} \rightarrow \overline{\phi_0} + \nu_{\alpha L}^C$ $N_{s} \rightarrow \phi_- + l_{\alpha L}^C$

Few elements of Leptogenesis

Hence the total width reads

$\Gamma_{D_{s}} = \sum_{\alpha} \biggl[ \Gamma(N_{s} \rightarrow \phi_0 + \nu_{\alpha L}) + \Gamma(N_{s} \rightarrow \phi_+ + l_{\alpha L})+ \Gamma(N_{s} \rightarrow \overline{\phi_0} + \nu_{\alpha L}^C)+ \Gamma(N_{s} \rightarrow \phi_- + l_{\alpha L}^C) \biggr]= \frac{M_{s}}{8\pi} ( Y^{\nu \dagger} \, Y^\nu)_{ss}$

Let us denote with N₁ be the lightest right–handed neutrinos and assume for simplicity that its interactions have washed out any lepton number asymmetry previously generated by others N_s at temperatures T >> M₁. Under these assumptions the final asymmetry depends on N₁ only. If out of equilibrium condition are satified

$\Gamma_{D_{1}} < H (T\sim M_{1})$

With the decay of N₁a lepton asymmetry is generated due to the CP asymmetry arising from the interference of the tree level and one-loop diagrams, as shown in Figure.

Decays of right-handed neutrinos

Few elements of Leptogenesis

An imaginary part develops in the one-loop contribution

$\varepsilon_1 \equiv \frac{\Delta_1}{\Gamma_{D_{1}}} \simeq \frac{1}{8\pi} \frac{1}{( Y^{\nu \dagger} \, Y^\nu)_{11}} \sum_{s \neq 1} \mbox{Im}\bigg\{( Y^{\nu \dagger} \, Y^\nu)_{1s}^{2}\bigg\} \biggl[ f\biggl( \frac{M_{s}^{2}}{M_{1}^{2}}\biggr) + g\biggl( \frac{M_{s}^{2}}{M_{1}^{2}}\biggr)\biggr]$

where

$\Delta_1 = \sum_{\alpha} \biggl[ \Gamma(N_{1} \rightarrow \phi_0 + \nu_{\alpha L}) + \Gamma(N_{1} \rightarrow \phi_+ + l_{\alpha L}) - \Gamma(N_{1} \rightarrow \overline{\phi_0} + \nu_{\alpha L}^C) -\Gamma(N_{1} \rightarrow \phi_- + l_{\alpha L}^C) \biggr]$

The two function f(x) and g(x) are defined as

$f(x) \equiv \sqrt{x} \biggl[ 1-(1+x) \ln \biggl(\frac{1+x}{x}\biggr) \biggr]$

and

$g(x) \equiv \frac{\sqrt{x}}{1-x}$

Few elements of Leptogenesis

For hierarchical neutrino masses, M₁ << M₂,…Mn_s the asymmetry assumes the simple form

$\varepsilon_{1} \simeq -\frac{3}{8\pi} \frac{1}{( Y^{\nu \dagger} \, Y^\nu)_{11}} \sum_{s \neq 1} \mbox{Im}\bigg\{( Y^{\nu \dagger} \, Y^\nu)_{1s}^{2}\bigg\} \frac{M_{1}}{M_{s}}$

In case in which the decay temperature is << M₁ one can neglect the inverse decays and L violating scatterings. This holds as long as

$\frac{\Gamma_{D_1}}{H(T\sim M_{1})}\simeq \frac{m_{Pl}}{(1.66)(8\pi)\sqrt{g_{*}}}\frac{( Y^{\nu \dagger} \, Y^\nu)_{11}}{M_{1}} < 1$

which yields the following bound

$m_{\nu_1} \equiv( Y^{\nu \dagger} \, Y^\nu)_{11} \frac{v^{2}}{2 \, M_{1}} \simeq 20 \sqrt{g_{\ast}} \frac{v^{2}}{m_{Pl}} \frac{\Gamma_{D_{1}}}{H} \bigg|_{T=M_{1}} \lesssim 10^{-3} \; \mbox{eV}$

For the electroweak SM we have g_* ≈ 100 , while in supersymmetric modeld this value is almost doubled. In general, one has a certain amount of wash out of lepton number asymmetry produced by the N₁decay, parametrized by a coefficient k

$L = \kappa \frac{\varepsilon_{1}}{g_{*}}$

For small k or strong wash out regime all L violating interactions are in equilibrium, while k=1 corresponds to T_D <<M₁.

Few elements of Leptogenesis

More precise predictions for the leptogenesis model would require the solution of a set of Boltzmann equations, whose description is away from the aim of the present course.

For strongly hierarchical right-handed neutrino masses, when the asymmetry ε₁depends on the decay of the lightest right-handed neutrino leptogenesis becomes very predictive on lepton asymmetry, provided that N₁ decays at temperature T > 10¹² GeV. In particular, if the ratio M₁/M₂<< 1,one gets an upper bound for ε₁ better known as the “Davidson-Ibarra” bound

$|\varepsilon_{1}| \le \frac{3}{16\pi} \frac{M_{1}(m_{\nu_3}-m_{\nu_2})}{v^{2}} \equiv \varepsilon_{1}^{DI}$

Since

$|m_{\nu_3} - m_{\nu_2}| \le \sqrt{\Delta m_{32}^{2}} \sim 0.05 \, {\mbox eV}$

one gets a lower bound for M₁

$M_{1} \ge 2 \times 10^{9} \; \mbox{GeV}$

In case of degenerate light neutrinos, the leading terms in the expansion of ε₁ in M₁/M₂ and M₁/M3 vanish. A computation next to the leading order yields a looser bound

$|\varepsilon_{1}| \lesssim \mbox{Max}\biggl( \varepsilon^{DI}, \frac{M_{3}^{3}}{M_{1}M_{2}^{2}}\biggr)$