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In a hypothetical scenario where the coupling of the fermions to the gauge fields were extremely small, would it be possible to observe quark and charged leptons oscillate in the same manner as we observe neutrino oscillations today?

i.e. would it be possible for the flavour eigenstate of a quark/lepton produced in a weak interaction to be detected as a different flavour eigenstate at a distant detector due to the states propagating according to their mass eigenstates?

$$ \left| {\nu_{\alpha}(T)} \right> = \sum_iU^*_{\alpha i} \; e^{ip\cdot x} \left| \nu_i \right> $$

dmckee --- ex-moderator kitten
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Eweler
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1 Answers1

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The short answer is no! This is because, for the charged leptons (and the quarks), the flavor eigenstates are the same as the mass eigenstates. The neutrino flavors are defined in terms of the charged lepton mass eigenstates.

It all has to do with how we "observe" neutrinos as compared to the charged leptons. Think of an electron. How do we "see" it? We observe its "track" directly through a magnetic field and calculate e/m directly.

Now think about how we detect neutrinos (Reines and Cowan showed us how!). We keep the earthly matter in the (probable) path of the neutrinos and hope some of them will interact with that matter. When some neutrinos interact, each of them produces a particular "flavor" of charged lepton (electron, muon, etc.), which we "observe." Therefore, we "tag" the neutrinos as per the leptonic flavor they produce. Unfortunately, this is the only way we have learned to detect neutrinos, and this is why it is said that we observe neutrinos through their flavor eigenstates, which may not coincide with their mass eigenstates. This is what makes the neutrinos oscillate. In other words, if we were clever enough to "observe" the neutrino mass eigenstates directly, they wouldn't oscillate! I guess the KATRIN experiment will do that by precisely measuring the tail of the beta decay curve, but I am not entirely sure.

ddphy
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