Why are the tau neutrino and mu neutrino stable? For example, why can't we have this decay?

$$\nu_\tau \rightarrow \nu_e+\gamma$$

This doesn't seem to be forbidden by conservation of energy-momentum, charge, angular momentum, or lepton number.

[EDIT] WP has some relevant info here: https://en.wikipedia.org/wiki/Lepton_number#Violations_of_the_lepton_number_conservation_laws

Lepton flavor is only approximately conserved, and is notably not conserved in neutrino oscillation.[4] However, total lepton number is still conserved in the Standard Model.

This leaves me unclear on whether the mu and tau neutrinos are in fact unstable in the standard model, but just with a very long lifetime.

There is also some discussion here: Rothstein, "What Do We Know About the Tau Neutrino?," https://arxiv.org/abs/hep-ph/9506443 . In eq. (8) he discusses the process I asked about, along with $\nu_\tau\rightarrow3\nu_e$ and decay into an electron neutrino plus a Higgs. Not being a particle physicist, I'm not having much luck extracting the fundamentals from this paper. It sounds like perhaps the standard model allows these decays, but with a very long lifetime, while modifications of the standard model might allow them to go a lot faster...?

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    $\begingroup$ I thought the moral of neutrino oscillations was that the flavor eigenstates are not stable, but mix in flight. $\endgroup$
    – rob
    Commented Nov 5, 2019 at 23:22
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    $\begingroup$ I asked a colleague about this, and he informed me that we don't even know for sure that the electron neutrino is the lightest. You can have an "inverted hierarchy," and this is not yet ruled out observationally. $\endgroup$
    – user4552
    Commented Nov 7, 2019 at 16:59

2 Answers 2


It is forbidden at tree-level because the gamma doesn't couple to an entity with neither charge nor magnetic moment and because it violates lepton-flavor number (which is conserved by the electromagnetic interaction).

There are plausible loop diagrams which it, but we can guess that they would be supressed because it involves a minimum of two weak couplings and by the small phase space for the final state.

Now, experimental considerations.

  • Good news: The photon energy is set (in the CoM) by the mass difference, but for a beam experiment that can be ferociously boosted and should be experimentaly accessible.
  • Bad news: a low rate implies a very low linear phton density along the beam. Even if they are decaying by this process we're going to be very hard-pressed to observe the event.
  • $\begingroup$ Thanks. That seems like a nice explanation of why it's at least strongly suppressed. But is it completely forbidden? And if so, why? $\endgroup$
    – user4552
    Commented Nov 5, 2019 at 23:02
  • $\begingroup$ @Ben I'm now at a loss as well. I've got a diagram drawn on my whiteboard which looks acceptable to me (the neutrino vertex spawns a $Z^0$ which creates a fermion loop that spits out the photon, so it's not even a very complicated diagram). The phase space is going to be tiny on top of the coupling issues, but all the quantum numbers and coupling restrictions seem to be met. $\endgroup$ Commented Nov 5, 2019 at 23:13
  • $\begingroup$ Thanks, that's interesting. I made some edits to the question, including a reference that may be of interest. $\endgroup$
    – user4552
    Commented Nov 5, 2019 at 23:19
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    $\begingroup$ But... it is not $\nu_\tau, \nu_e$ that have masses... These are just "convenience production ad absorption linear combinations" and not massive propagating states... $\endgroup$ Commented Nov 5, 2019 at 23:27
  • $\begingroup$ @CosmasZachos Yes. That's a complication. This should be about $vu_i \to \nu_j + \gamma$ rather than about flavor states as such. I talked with a specialist on a related topic once and put some note in my answer to an earlier question. $\endgroup$ Commented Nov 5, 2019 at 23:33

Such a diagram does exist through a loop. One thing to keep in mind is that neutrinos decay in mass eigenstates not flavor eigenstates. So in principle you have have $\nu_e$ decaying into $\nu_\tau$ even if the mass ordering is normal (although it will be smaller than the other way around based on the know parameters of the PMNS matrix). I think that the dominant diagram is $\nu_3$ goes to $e^- + W^+$ in the loop, the $W^+$ radiates a photon and then the loop ends with $\nu_1$. There are PMNS matrix elements at the neutrino vertices.

The lifetime for this process is extraordinarily small and completely undetectable. In the rest frame of the neutrino you have $\sqrt{s}\sim0.1$ eV initially, and then something $\sim100$ GeV running in the loop. Plus, all neutrinos we are likely to have at hand or to detect are extremely relativistic adding many more orders of magnitude to their life time in the lab frame.

  • $\begingroup$ Pure speculation: there could be an upper limit to how much energy you can temporarily add in a quantum interaction before the universe says no, and 100GeV is too much for .1eV. $\endgroup$
    – Joshua
    Commented Jun 11, 2021 at 14:32

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