# Does Leptogenesis require lepton number violating vertices?

One of the Sakharov condition for Baryogenesis requires Baryon number violating interactions in the theory. However, it is also possible to generate a baryon asymmetry via a lepton asymmetry or Leptogenesis. Therefore, for Baryogenesis through Leprtogenesis, shouldn't the theory necessarily have lepton number violating (interaction) vertices?

However, there is a contradiction. This is not the case for "Dirac Leptogenesis" where the relevant decay vertices leading to lepton number asymmetry are lepton number conserving!! Does it mean that lepton number violating vertices have nothing to do with Leptogenesis?

## Short answer (we're hung up on terminology):

If you use the word "Leptogenesis" exclusively for theories with lepton violating vertices, then "Dirac Leptogenesis" is a misnomer.

If you use the word "Leptogenesis" more broadly, to describe theories where leptonic interactions are important for Baryogenesis, then calling it "Dirac Leptogenesis" is OK.

"Dirac Leptogenesis" works differently from usual leptogenesis. It leads to a baryon asymmetry without lepton violating vertices (perturbative interactions). Both a baryon and lepton asymmetry are generated in this scenario from non-perturbative effects. Let me be more precise...

In order to allow for the dynamical generation of a Baryon Asymmetry in the Universe (BAU), it is sufficient to fulfill the Sakharov conditions (reminder):

• Baryon number violated
• C and CP symmetries violated
• Departure from thermal equilibrium occurs

Even though these conditions are not necessary conditions, we will stick to them in what follows (evading them often requires nasty assumptions, such as CPT violation).

''Leptogenesis'' comprises a class of theories where the baryon number ($B$) violation comes from a lepton number ($L$) violation. In the simplest realisations, there are lepton number violating interactions in the Lagrangian (as you expect) which give rise to the lepton asymmetry in the early Universe. After generating some $L$ one is then interested in converting it to $B$. This conversion can happen through non-perturbative processes called electroweak sphalerons, which are there already in the Standard Model (SM). Sphaleron processes are active only when the Universe is very hot, and they change (violate) $B$ and $L$ simultaneously while keeping the difference $B-L$ constant.

''Dirac Leptogenesis'' corresponds to a different class of theories, and in fact the name ''Neutrinogenesis'', given in the original paper, might be better suited. Even though these theories rely on interactions with leptons to generate the BAU, the key point is that for this class of models to work no lepton number violating interactions are necessary (aside from the above-mentioned non-perturbative sphalerons). Of course, then, generating a nonzero $B$ will require a more involved mechanism than usual.

So here is how ''Dirac Leptogenesis'' works (assume we start from $B = L = 0$):

• Some CP-violating but $L$-conserving process produces a nonzero lepton number for left-handed particles $L_\ell$, and an equal and opposite nonzero lepton number for right-handed particles, $L_r = - L_\ell$. Notice that total lepton number $L = L_\ell + L_r = 0$ is unchanged.
• SM interactions between left and right chirality particles (such as the electron Yukawa interactions, $y_e \overline{l_\ell} \phi e_R$) dilute the asymmetries in each chirality, driving them back to zero: $L_\ell, L_r \rightarrow 0$.
• This left-right equilibration happens quickly for all leptons except Dirac neutrinos, whose Yukawa interaction is assumed to be very small (connected to the smallness of neutrino masses). Thus, with the right ingredients we can keep $L_r = - L_\ell \neq 0$ stored as neutrinos.
• The sphalerons -- which are associated with nontrivial $\textrm{SU}(2)_\ell$ vacuum structure -- only touch left-handed particles. So they convert $L_\ell$ to $B$ (keeping $B - L_\ell$ constant) while $L_r$ is left undisturbed.
• As the universe cools down, sphalerons eventually become ineffective and the $B$ we got from them doesn't go anywhere. The remainder of $L_\ell$ will slowly equilibrate with some of the $L_r$, leaving behind some total $L=B$.

To conclude, there is no contradiction. It is just that ''Dirac Leptogenesis''/''Neutrinogenesis'' relies on a different mechanism than usual ''Leptogenesis'' to produce a net baryon number $B$ from leptonic interactions.