Seesaw mech, CP violation, leptogenesis, lept/bary asymmetry

1. Oct 23, 2015

Anchovy

I'm trying to get my story straight on explaining the universe's matter-antimatter imbalance.

So far I understand CP violation in neutrino oscillations (PMNS $\delta$ parameter). And I think I'm right in saying that then a 'leptogenesis' process generates an excess of leptons over antileptons? The amount of which is related to the size of $\delta$? And then some 'sphaleron' process converts the extra leptons to baryons, with a corresponding excess over antibaryons?

So first of all I'd like to know if what I've wrote above is roughly correct. Then the next thing I'm wondering is, I'm under the impression that the seesaw mechanism is involved somewhere. I understand how that mechanism itself works but I don't quite get where it fits in. Is it just to do with introducing a right-handed neutrino? And that $\nu_{R}$ slots into this picture somewhere?

2. Oct 24, 2015

Orodruin

Staff Emeritus
The amount of lepton asymmetry produced in leptogenesis does not depend on the PMNS phase in the canonical seesaw. Instead, it depends on CP-phases available only at higher energies (this can be shown by the introduction of the Casas-Ibarra parametrisation).

The seesaw model is central in leptogenesis as the lepton asymmetry is generated by out-of-equilibrium decays of heavy states, the right-handed neutrino in the case of the type-I seesaw.

3. Oct 24, 2015

Anchovy

Ok, so in the seesaw you have $N = \nu_{R} + \nu_{R}^{C}$ and apparently we get leptogenesis through $N \rightarrow L_{\alpha} \hspace{1 mm} H$ or $N \rightarrow \overline{L}_{\alpha} \hspace{1 mm} \overline{H}$? So we need a right-handed neutrino mass at least big enough to create a Higgs and a lepton mass. Which the seesaw happens to need also. Fair enough.

And also there'll be a lepton-antilepton asymmetry if there is any difference between decay rates $\Gamma(N_{i} \rightarrow L_{\alpha} \hspace{1 mm} H), \hspace{2 mm}\Gamma(N_{i} \rightarrow \overline{L}_{\alpha} \hspace{1mm} \overline{H})$, given for each flavour $\alpha$ by a non-zero $\epsilon_{\alpha}$:
$$\epsilon_{\alpha} = \frac{\Gamma(N_{i} \rightarrow L_{\alpha} \hspace{1 mm} H) - \Gamma(N_{i} \rightarrow \overline{L}_{\alpha} \hspace{1mm} \overline{H})}{\Gamma(N_{i} \rightarrow L_{\alpha} \hspace{1 mm} H) + \Gamma(N_{i} \rightarrow \overline{L}_{\alpha} \hspace{1mm} \overline{H})}$$

And this can supposedly be written in terms of PMNS matrix Uas follows:
$$\epsilon_{\alpha} = \frac{-3 M_{1}}{16 \pi v^{2}} \frac{Im(\sum_{\beta \rho} m_{\beta}^{1/2} m_{\rho}^{3/2} U_{\alpha \beta}^{*} U_{\alpha \rho} U_{1 \beta} U_{1 \rho}) }{ \sum_{\beta} m_{\beta} |R_{1 \beta}|^{2}}.$$

But you say this imaginary part of $U_{\alpha \beta}^{*} U_{\alpha \rho} U_{1 \beta} U_{1 \rho}$ is not anything to do with $\delta$ but rather some other phase? At this point I have to wonder what the $\delta$ that can be measured in neutrino oscillation experiments has to do with all this?

I must be misunderstanding you given that I've seen this plot of the ($\delta$-dependent l believe?) Jarlskog parameter vs. the baryon asymmetry:

Last edited: Oct 24, 2015
4. Oct 24, 2015

Orodruin

Staff Emeritus
The matrix that appears in the expression for the asymmetry is not the PMNS matrix, it is the neutrino Yukawa couplings.

Edit: Note that this is true in the canonical seesaw. There are some variants of the seesaw and leptogenesis where you might be able to influence it through the low energy CP phase.

5. Oct 24, 2015

Anchovy

Ahhhh right. It's a different matrix. OK. I've often read measuring $\delta$ is the ultimate goal of oscillation studies to understand matter/antimatter asymmetry. So I need to find out about these 'variants'. I've only heard of a type-1 and type-2 seesaw, and I think I briefly saw that the type-2 seesaw involves Higgs triplets... which only appear in beyond-Standard Model theories? More digging required it seems...

6. Oct 24, 2015

Orodruin

Staff Emeritus
Also right-handed neutrinos do not appear in the standard model, so whatever seesaw you select, it is BSM physics. The point here is that within these seesaws there are some alternatives which give you things like flavoured leptogenesis, which could change the behaviour of leptogenesis.

The target of oscillation experiments to measure $\delta$ is mainly to find CP-violation also in the lepton sector. Finding it would also establish that the lepton sector does violate CP and therefore to some extent give fuel to the idea of leptogenesis even if the phase is not directly related.

The third type of seesaw (seesaw type-III) uses fermion SU(2) triplets in place of the right-handed neutrinos. The neutral component of the triplets would essentially work as the right-handed neutrino, but you will have additional constraints coming from the charged components.

7. Oct 24, 2015

fzero

But the PMNS matrix is the matrix that diagonalizes the mass matrix, which depends explicitly on the Yukawas. So in principle, one could determine the combination of Yukawas (and Majorana masses) that maps to $\delta$. There is an expression in the Fukugita-Yanagida paper where they express the net lepton production, in a suitable limit, as directly proportional to $\delta$. In a general case, the relationship between the PMNS matrix and Yukawas would probably be too complicated to be illuminating, but they are not independent.

8. Oct 24, 2015

Anchovy

Ahh yes, very true.

OK. I assumed there would have to be some direct relationship. I'm hoping to find out a bit about how they might possibly be directly related though. I've found some slides that contain the following:

so I'll be satisfied with finding something that gets me to the point of that blue box.

9. Oct 24, 2015

Orodruin

Staff Emeritus
In the canonical seesaw limit, the combination of the Yukawas which combine to $\delta$ is independent of the combination which produces the lepton asymmetry. Seen from the Casas-Ibarra parametrisation, $vY = i U \sqrt{m} O \sqrt{M}$, where $U$ is the PMNS matrix, $m$ the diagonalised light neutrino mass matrix, $O$ a complex orthogonal matrix, and $M$ the diagonalised right-handed neutrino matrix, the combination of Yukawas entering the lepton asymmetry is $Y^\dagger Y$, implying that $U$ is irrelevant and only the high-energy parameters remain. For more discussion, see eg http://inspirehep.net/search?p=find+eprint+0902.2469

Edit: Note that I have used the opposite convention of the Yukawas compared to the slides Anchovy found.

10. Oct 24, 2015

fzero

Thanks for the explanation and reference; they make the picture very clear.