And if not, why? What is the difference to neutrinos oscillations?


1 Answer 1


In principle, the charged leptons may oscillate as well but it would need rather strange initial states that could be easily obtained in the ultrarelativistic limit only and the experimental arrangement would have to be very unusual, anyway.

The reason why the neutrinos oscillate is that the initial state isn't an energy eigenstate. It is a superposition of two energy eigenstates. For example, one may create a neutrino $\nu_e$ which is a general superposition of the energy eigenstates $\nu_1,\nu_2,\nu_3$. We may be sure that the initial state is $\nu_e$ because this is the $SU(2)$ partner of $e^-$ – an energy eigenstate – and that's why it's easy to "convert" $e^-$ to $\nu_e$.

This initial state $\nu_e$ may still be an eigenstate of energy, $p^0$, with a rather well-defined value. The energy eigenstates $\nu_1,\nu_2,\nu_3$ then imply slightly different values of the momentum. In practice, $p^0$ of the neutrinos is much larger than the neutrino rest masses which is why $|\vec p|\approx p^0$ and the deviations are very small.

If you try to reverse the role of the neutrinos and the charged leptons, everything works in principle but not in practice, due to the high charged leptons' masses. If you want a general superposition of the mass eigenstates $e^-$ and $\mu^-$ with a well-defined energy $p^0$ (and all realistically produced states are superpositions of "approximately" the same energy eigenstates), this energy must exceed the muon mass and the corresponding momentum $|\vec p|$ must be huge, nearly $p^0$, for the electron part of the wave function. If $p^0=m_\mu+\epsilon$, then the electron part is moving at a much higher speed than the muon part: they inevitable separate from each other rather quickly (and it becomes awkward to interpret it as "one" oscillating particle at a rather well-defined place). This is why the initial state is an unnatural superposition of two "macroscopically" different states which is hard to produce in practice.

It's also a "macroscopic", Schrödinger-cat-like superposition because we need to prepare the initial state e.g. as the $SU(2)$ partner of the $\nu_1$ neutrino mass eigenstate. But it's very hard to bring neutrinos to rest which may be needed to be sure – because of the initial measurement – that we have $\nu_1$ with no probability of having $\nu_2$, for example, as the initial state.

All those superpositions that inevitably lead to oscillations of some kind do exist in the Hilbert space. But in practice, we measure the mass eigenstates of the heavier charged leptons much more easily than the mass eigenstates of the neutrinos. That's why the mass eigenstates of charged leptons – $e^-,\mu^-,\tau^-$ – may easily be prepared as the initial state, and so can their $SU(2)$ (neutrino) partners which are not mass eigenstates, and therefore oscillate. It's not obvious what the procedure to prepare the mixture of the charged leptons mass eigenstates would be. But even if there were such a procedure, one will need these leptons to move almost at the speed of light (just like in the case of neutrinos) to prevent the components from spreading to different locations.

  • $\begingroup$ regardless of the difficulty to prepare states appropriate for observing charged lepton flavor oscillation, how does this explain the very low experimental upper limit for neutrinoless muon decay? $\endgroup$
    – Ali Moh
    Commented Apr 15, 2015 at 8:43
  • 2
    $\begingroup$ @Ali, wouldn't such a decay violate energy/momentum conservation? $\endgroup$ Commented Apr 15, 2015 at 11:38
  • $\begingroup$ $\mu\rightarrow e\gamma$ only violates lepton number conservation $\endgroup$
    – Ali Moh
    Commented Apr 15, 2015 at 16:29
  • $\begingroup$ @Ali Moh That's just it: given the huge $\Delta m_l^2$ compared to $\Delta m_\nu^2$, neutrino less muon decay naturally opts for the loop neutrino conversion channel, cf box, so $P\sim (\Delta m_\nu^2 /m_W^2)^2$. $\endgroup$ Commented Aug 13, 2020 at 13:45

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.