How do neutrinos oscillate to more massive states?

Question

Neutrino oscillations show that neutrinos change flavor as they propagate, if I understand correctly, and that this is done by allowing different mass and flavor eigenstates states to mix. Do they have definite mass states when they are measured (or at any time)? If so then how is it possible for a neutrino with a lower mass eigenstate to later have the potential to oscillate into a higher mass eigenstate (or whate makes this not possible)? Is mass and energy conserved by a sudden change in the neutrino's energy?

Answer 1

I didn't read the papers mentioned above. But I am trying to explain this by the very basics of quantum mechanics.

First, if we can somehow measure the mass eigenstate of neutrinos, then definitely this eigenstate will not oscillate anymore (without further interactions involved in the future). Suppose for a mass eigenstate we have $H|m_i\rangle=E_i|m_i\rangle$, then $$H e^{-iHt}|m_i\rangle=e^{-iHt}H|m_i\rangle=E_i(e^{-iHt}|m_i\rangle),$$
meaning that the state will keep to be at the $m_i$ eigenstate under the time evolution.

The oscillations happen only when the neutrinos are at the flavour eigenstates which are superpostions of mass eigenstates: $$|\nu_i\rangle=M_{ij}|m_j\rangle$$. Then under the time evolution we have $$e^{-iHt}|\nu_i\rangle=e^{-iE_jt}M_{ij}|m_j\rangle,$$ where three $j$ should be understood to be summed once. This new state then in general will be superpostions of the flavour eigenstates, i.e. $$e^{-iE_jt}M_{ij}|m_j\rangle=e^{-iE_jt}M_{ij}M^{-1}_{jk}|\nu_k\rangle.$$ When we have $E_i=E,\ i=1,2,3$, then we have $$e^{-iHt}|\nu_i\rangle=e^{-iE_jt}M_{ij}M^{-1}_{jk}|\nu_k\rangle=e^{-iEt}|\nu_i\rangle,$$ meaning that there is no oscillations and hence the oscillations depends on the mass difference rather than the mass itself. This is simply a fact that the operator $F$ that has flavour eigenvalues does not commute with the Hamiltonian $[F,H]\neq 0$, i.e. $$Fe^{-iHt}|\nu_i\rangle\neq e^{-iHt}F|\nu_i\rangle=v_i e^{-iHt}|\nu_i\rangle.$$

So there is no energy-momentum violation because at the superpositions only the average value has meanings. We can easily understand that for a position eigenstate $|x\rangle$, we have no definite momentum for it.

Answer 2

To carry on with Eld's comments, the mass matrix for neutrinos is not diagonal with respect to the Hamiltonian. The Hamiltonian basis $|E\rangle$ and the neutrino basis $|\nu\rangle$ are such that $\langle\nu|E\rangle~=~e^{iM}$, where $M$ is the CKM or PNMS matrix. The physics is similar to Feynman's discussion of the Kaon problem in the third volume of his $Lectures~on~Physics$.