# Do we know whether the “electron” neutrino is lighter than the “muon” neutrino?

The labeling of the neutrino mass eigenstates is arbitrary [PDG Neutrino Review]. Regularly it is assumed for convenience that $$\Delta m^2_{21} = m_2^2 - m_1^2 > 0.$$ Then, values for the mixing angles are given that lead to the PMNS matrix $V$ with the $V_{1,1}$ entry being the largest in both its row and column [i.e. Wikipedia]. This implies that the lighter mass eigenstate is mostly of electron flavor (even in an inverted hierarchy setting, between $\nu_e$ and $\nu_\mu$, it is always $\nu_e$ that is shown to be lighter).

The vacuum oscillation observables only depend on the cosine of $\Delta m^2$, or the squared sine, so its acutal sign is irrelevant here. In matter, there is a CP violation induced that is proportional to the $\Delta m^2$, but since we have three flavor mixing and $\Delta m_{31}^2 \gg \Delta m_{21}^2$, these effects will be used to determine the sign of $\Delta m^2_{31}$.

What is the status of the sign of $\Delta m^2_{21}$?

What implications would a negative sign of $\Delta m^2_{21}$ have?

• I thought $\Delta m^{2}_{21}$ was positive by definition, rather than it being a convenient assumption. – dukwon Dec 19 '16 at 13:09
• @dukwon Well, you can use it do define a numbering scheme, but the question then becomes what I wrote in the second paragraph: How do we know the lighter of the two states is mostly of electron flavor? – Neuneck Dec 19 '16 at 13:51