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?

  • 1
    $\begingroup$ This is a subtle issue. You're right to point out that this violates energy-momentum conservation. The resolution is that you need to consider the full quantum state, which includes entanglement with the other products of the weak decay that produced the neutrinos. More details are in this paper. $\endgroup$
    – knzhou
    Aug 9, 2016 at 5:38
  • 1
    $\begingroup$ This is a good question; it has been addressed here arxiv.org/abs/hep-ph/9901399 by Lipkin, and it is a full paper that I cannot summarize $\endgroup$
    – anna v
    Aug 9, 2016 at 6:29
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/21351/2451 , physics.stackexchange.com/q/92008/2451 and links therein. $\endgroup$
    – Qmechanic
    Aug 9, 2016 at 7:17
  • $\begingroup$ The paper by Lipkin is amazing. It explains the physics and the right way to interpret the experiments, and the right picture (wave or particle or semi classical) to use and why. Recommend it. Eld's answer is a real nice QM derivation using basic QM of why there is no energy momentum violation, and how the flavor not commuting with the Hamiltonian explains the effects. $\endgroup$
    – Bob Bee
    Aug 10, 2016 at 4:02

2 Answers 2


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.

  • $\begingroup$ Regarding your time evolved equation of a flavor eigenstate in a superposition of mass eigenstates, how do you know when the mass eigenstates have evolved sufficiently to change the neutrino flavor (i.e. change the flavor index $i$)? $\endgroup$ Aug 11, 2016 at 13:46
  • $\begingroup$ @curiousStudent It is not necessary for a flavour eigenstate to evolve into another flavour eigenstate. But in general, due to the mass difference a flavour eigenstate will evolve into a superpositon of different flavour eigenstates so that you have chance to meausre the oscillation. $\endgroup$
    – Wein Eld
    Aug 11, 2016 at 20:06

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 $Lectures~on~Physics$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.