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Dirac vs Majorana Neutrinos 
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#1
Aug114, 02:28 AM

Sci Advisor
P: 2,951

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 antineutrino has 6 possible internal states. In total, there are 12 possible states. If a neutrino is Majorana in nature then antineutrinos 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 antineutrinos are right handed helicity states. In that case, if the right handed Dirac neutrinos are sterile, then why are not the Majorana antineutrinos sterile since they are right handed? Thanks. 


#2
Aug114, 05:58 AM

P: 675

I would here like to stress that if you do add the righthanded 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 (typeI) seesaw mechanism, where you end up with three light active Majorana neutrinos and three heavy Majorana states which are mainly sterile. $$ L = \left(\begin{array}{c} \nu_L \\ \ell_L \end{array}\right), $$ 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. 


#3
Aug114, 06:45 AM

P: 93

r
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. 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) 


#4
Aug114, 07:39 AM

P: 675

Dirac vs Majorana Neutrinos



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