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(I have restated the question after the comments to make it clearer)

Is there any way mathematically a Dirac neutrino can have different mass for LH and RH state?

The dirac mass term (obtained after Spontaneous symmetry breaking) is

$m\bar{\nu}\nu = m\bar{\nu_R}\nu_L + m\bar{\nu_L}\nu_R$

(This term can be applicable to other fermion fields as well)

Now, what I see is that for neutrino (particle), whether I take its LH state or RH state, it will possess same mass for both the states.

Am I right with the concept?

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    $\begingroup$ How would you know the RH neutrino with a different mass isn't just a new type of Dirac neutrino? $\endgroup$
    – Peter Shor
    Commented Dec 3, 2017 at 12:52
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    $\begingroup$ How would you hope to find a RH neutrino? $\endgroup$
    – safesphere
    Commented Dec 3, 2017 at 12:57
  • $\begingroup$ I am saying that suppose we assume that we have found a RH neutrino. Now, by defination of dirac mass term $-m_D\bar{\nu}\nu=-m_D\bar{\nu_R}\nu_L-m_D\bar{\nu_L}\nu_R$ We do not distinguish between the mass of LH or RH neutrino. Dirac terms treats both LH and RH states at same level. While if we consider a majorana neutrino, both its LH and RH states would have different masses as defined by Seasaw mechanism. Thus, my question is that if we find a RH neutrino with different mass than LH neutrino, can I conclude that it is a majorana particle? $\endgroup$
    – kbg
    Commented Dec 3, 2017 at 14:44
  • $\begingroup$ So all I am trying to state is that: Is there any way by which a Dirac mass term can give different mass to LH and RH state of a particle? $\endgroup$
    – kbg
    Commented Dec 3, 2017 at 14:50
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    $\begingroup$ I wasn't trying to say anything deep. As an example, even if neutrinos are Dirac, the LH muon neutrino will have a different mass that the RH electron neutrino. If you find a heavy RH neutrino, how do you know whether it's a completely new neutrino that nobody has ever seen before? So even if you see a heavy RH neutrino, it doesn't prove that neutrinos are Majorana particles. Now that you've clarified the question, I think my earlier comment is irrelevant. $\endgroup$
    – Peter Shor
    Commented Dec 5, 2017 at 17:56

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