Are Neutrinos Dirac or Majorana: What Are the Implications?

[Matterwave]

Hi, I recently attended several lectures on the topic of neutrino astrophysics. I wanted to verify some of the fact that I gleaned for them, specifically about the Dirac vs Majorana nature of neutrinos.

1) The most basic fact first. If a neutrino is Dirac in nature, then it has 3 flavors, and 2 spin states (helicity states), for a total of 6 possible internal states. In addition, the anti-neutrino has 6 possible internal states. In total, there are 12 possible states. If a neutrino is Majorana in nature then anti-neutrinos are simply the right handed helicity states and regular neutrinos are the left handed helicity states, and so there are only 6 total possible states instead of 12. Is this true?

2) If fact 1 is true, then this is the "deduced fact" that I have gleaned from the talks. Given that the statistical weights attributed to neutrinos are 6 and not 12 (in, for example, neutrino decoupling calculations), is it therefore true then that either neutrinos are Majorana in nature, or, if they are Dirac in nature, then the right handed Dirac states are sterile states (6 sterile states)? This would be due to the left handed nature of the weak interactions?

3) If both fact 1 and 2 are true, then I have a follow up question. If neutrinos are Dirac in nature, what does it mean, since they are massive, for the right handed states to be sterile? Since right handedness and left handedness, helicity, is only a Lorentz invariant for massless particles (Weyl fermions), then there are some particles (Lorentz frames) which will "see" a right handed neutrino as a left handed one. In that case, what does it mean for the right handed neutrino to be sterile?

4) An additional follow up question. If neutrinos are Majorana in nature, then the anti-neutrinos are right handed helicity states. In that case, if the right handed Dirac neutrinos are sterile, then why are not the Majorana anti-neutrinos sterile since they are right handed?

Thanks.

 

[Orodruin]

Hi, I recently attended several lectures on the topic of neutrino astrophysics. I wanted to verify some of the fact that I gleaned for them, specifically about the Dirac vs Majorana nature of neutrinos.

Just out of curiosity: Was this a summer school? If so, which and who was giving the lectures?

1) The most basic fact first. If a neutrino is Dirac in nature, then it has 3 flavors, and 2 spin states (helicity states), for a total of 6 possible internal states. In addition, the anti-neutrino has 6 possible internal states. In total, there are 12 possible states. If a neutrino is Majorana in nature then anti-neutrinos are simply the right handed helicity states and regular neutrinos are the left handed helicity states, and so there are only 6 total possible states instead of 12. Is this true?

Correct.

I would here like to stress that if you do add the right-handed neutrinos, they are SM singlets and nothing is stopping you from giving them a Majorana mass of their own. This is the basics behind the (type-I) seesaw mechanism, where you end up with three light active Majorana neutrinos and three heavy Majorana states which are mainly sterile.

2) If fact 1 is true, then this is the "deduced fact" that I have gleaned from the talks. Given that the statistical weights attributed to neutrinos are 6 and not 12 (in, for example, neutrino decoupling calculations), is it therefore true then that either neutrinos are Majorana in nature, or, if they are Dirac in nature, then the right handed Dirac states are sterile states (6 sterile states)? This would be due to the left handed nature of the weak interactions?

If you are considering the cosmological situation of decoupling, neutrino masses will typically be too small to have any effect (the helicity flip would occur on time-scales longer than the decoupling time). The right-handed states of Dirac neutrinos should therefore not enter into these since they are sterile. Indeed this is due to the left-handed nature of weak interactions (and the fact that the RH states also have no other interactions).

3) If both fact 1 and 2 are true, then I have a follow up question. If neutrinos are Dirac in nature, what does it mean, since they are massive, for the right handed states to be sterile? Since right handedness and left handedness, helicity, is only a Lorentz invariant for massless particles (Weyl fermions), then there are some particles (Lorentz frames) which will "see" a right handed neutrino as a left handed one. In that case, what does it mean for the right handed neutrino to be sterile?

Essentially that you would typically need a mass insertion to create the right-handed neutrino - and neutrino masses are very very small.

4) An additional follow up question. If neutrinos are Majorana in nature, then the anti-neutrinos are right handed helicity states. In that case, if the right handed Dirac neutrinos are sterile, then why are not the Majorana anti-neutrinos sterile since they are right handed?

Because electroweak interactions couple to left-handed particles and right-handed anti-particles. Or in other terms, the right-handed anti-neutrino is part of the SU(2) doublet
$$
L = \left(\begin{array}{c} \nu_L \\ \ell_L \end{array}\right),
$$
which is what couples to the W. The Majorana mass term is of the form
$$
\overline{\nu_L^c} \nu_L + {\rm h.c.}
$$
and thus relates active states with active states.

Note: The Majorana mass term as it stands breaks SU(2) invariance so it should not just be introduced as it is but as the result of some high energy completion of the SM. The lowest order effective operator with SM fields is then the Weinberg operator, which after EW symmetry breaking gives exactly a Majorana mass term of the form above.

 

[ofirg]

r

The most basic fact first. If a neutrino is Dirac in nature, then it has 3 flavors, and 2 spin states (helicity states), for a total of 6 possible internal states. In addition, the anti-neutrino has 6 possible internal states. In total, there are 12 possible states. If a neutrino is Majorana in nature then anti-neutrinos are simply the right handed helicity states and regular neutrinos are the left handed helicity states, and so there are only 6 total possible states instead of 12. Is this true?

The Number of degrees of freedom (D.O.F) doesn't depend on whether the neutrinos are Majorana or Dirac. It depends on if you include/consider right handed neutrinos. Left handed neutrinos(which are the neutrinos we have observed) have 3*2=6 D.O.F and with right handed neutrinos is goes up to 12.

From now on I'm including right handed neutrinos.
If neutrinos are Majorana, then you will have 6 neutrinos, each one with 2 spin states. 3 of them will be the light neutrinos that we know while the other 3 are presumably much heavier.

If fact 1 is true, then this is the "deduced fact" that I have gleaned from the talks. Given that the statistical weights attributed to neutrinos are 6 and not 12

In cosmological calculation, only the 3 neutrinos we have observed put in, so 6 D.O.F.
The presumed right handed are expected to be much heavier and with very small coupling to other matter. (They might have never been in thermal equilibrium)

If both fact 1 and 2 are true, then I have a follow up question. If neutrinos are Dirac in nature, what does it mean, since they are massive, for the right handed states to be sterile? Since right handedness and left handedness, helicity, is only a Lorentz invariant for massless particles (Weyl fermions), then there are some particles (Lorentz frames) which will "see" a right handed neutrino as a left handed one. In that case, what does it mean for the right handed neutrino to be sterile?

Here there is a distinction between chirality and helicity. Only the left handed chirality will have a coupling to the gauge bosons ( Z and W). Since the neutrino is massive sometimes a weak process will produce a neutrino which has right handed helicity in the lab frame. But as long as the neutrino is very relativistic, the fraction of these events will be suppressed.

An additional follow up question. If neutrinos are Majorana in nature, then the anti-neutrinos are right handed helicity states

If neutrinos are Majorana, they have no distinct antiparticles. simply 6 neutrinos, each one with two helicity states, three approximately active, three approximately sterile ( yet to be found)

 

[Orodruin]

From now on I'm including right handed neutrinos.
If neutrinos are Majorana, then you will have 6 neutrinos, each one with 2 spin states. 3 of them will be the light neutrinos that we know while the other 3 are presumably much heavier.

This is an assumption which is common but not at all necessary. The right-handed Majorana mass can be essentially anything (there are some experimental constraints ruling out the range where this mass is similar to the Dirac mass from the Yukawas, but it may be above or below this range). With a very low or zero RH Majorana mass we get (pseudo) Dirac neutrinos, with a heavier RH Majorana mass we can get anything from canonical type-I seesaw to light sterile neutrinos. The phenomenology is very rich and just assuming that the RH mass is very large is not highly motivated.

In cosmological calculation, only the 3 neutrinos we have observed put in, so 6 D.O.F.
The presumed right handed are expected to be much heavier and with very small coupling to other matter. (They might have never been in thermal equilibrium)

Sterile neutrinos are actively being considered as a source of additional radiation degrees of freedom in the early Universe. If the RH mass is very large, then the Yukawas are typically also large in order to reproduce the light neutrino masses through the seesaw, which means the heavy neutrinos will be produced through these, which in most cases will put them in thermal equilibrium unless the reheating temperature is lower than their mass. At some point the temperature will fall and the heavy neutrinos may decay out of equilibrium - this is the basic mechanism behind leptogenesis.

If neutrinos are Majorana, they have no distinct antiparticles. simply 6 neutrinos, each one with two helicity states, three approximately active, three approximately sterile ( yet to be found)

It should be mentioned that the additional sterile states do not need to exist to provide Majorana masses. Majorana masses may also be produced by any other high energy completion that provides the Weinberg operator when integrating out the heavy degrees of freedom. The prime example of this is the type-II seesaw mechanism, which introduces an additional SU(2) triplet scalar instead of singlet fermions.

 

[Matterwave]

Thanks for the answers. Cleared this up for me. :)