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What is a neutrino state if not a particle?

Physics Asked on March 14, 2021

When reading about the 2015 Nobel prize and how this led to the possibility of the existence of sterile neutrinos I am told that:

"(…) three active neutrinos $nu_e$, $nu_mu$, $nu_tau$, are superpositions of three massive neutrinos $nu_1 $, $nu_2 $ ,$nu_3 $ with respective masses $m_1$, $m_2$, $m_3$"(…)

I know that there are only 3 neutrinos in the standard model, but the way this is written it makes it sound like there are 6 neutrinos, and 3 of them are consequences of the interaction of the other 3. I know this must be incorrect, but I don’t know why.

I have further read about neutrino oscillations and superposition of flavour states and neutrino masses but I don’t understand this quantum superposition of mass states of neutrinos and how these do not represent the existence of extra physical particles. I understand that it works like the superposition of waves, but I don’t see how this works with particles (and I also get confused due to the particle-wave duality).

One Answer

From what I understand, the two groups are linear combinations of states from the other group. $$|nu_1rangle = c_{11} |m_1rangle +c_{12} |m_2rangle+c_{13} |m_3rangle |nu_2rangle = c_{21} |m_1rangle +c_{22} |m_2rangle+c_{23} |m_3rangle |nu_3rangle = c_{31} |m_1rangle +c_{32} |m_2rangle+c_{33} |m_3rangle $$ And the above relation can be inverted to express mass states in terms of the flavour states.

So you can see that the space is spanned by three orthogonal vectors so the neutrinos must be of three kinds only. All the craziness comes from the fact that mass and flavour states are not the same, $langle m_i|nu_jranglene0$ for any $i,j$.

Answered by Superfast Jellyfish on March 14, 2021

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