I have read this article:




Especially where it says that three mass types of neutrinos arrive:

All the measured neutrinos from the 1987 supernova arrived on earth within about 10 seconds of one another. Think about an electron neutrino emitted from the supernova with an energy of 10 MeV (an MeV is a million eV [electron-volts], or 1/1000 of a GeV; read here for the definition of these terms). Well, that electron neutrino was a mixture of neutrino-1, neutrino-2 and neutrino-3, each of which traveled with a slightly different speed! Is this something we would have noticed? We don’t precisely know the masses of the neutrinos, but suppose that neutrino-2 has a mass-energy of 0.01 eV (electron-volts; see this article for the definition) and neutrino-1 has a mass-energy of 0.001 eV. Then their two velocities, remembering that they have the same energy, would differ from the speed of light and from each other by less than a part in a hundred thousand trillion

v1 – v2= c [ (m22 – m12) c4/ 2 E2 + … ] = 0.0000000000000000005 c

(all equations accurate and precise to a one percent or better.) That velocity difference would mean the neutrino-2 part and neutrino-1 part of the original electron-neutrino would both arrive at the earth within a millisecond of each other — a undetectable difference for a variety of technical reasons. (Keep in mind that OPERA claims a difference of neutrino speeds from the speed of light of one part in 100,000, a much, much larger effect, though the measurement involves neutrinos at an energy a few hundred times larger than those from the supernova.)

And about the superposition here:

Neutrino oscillation arises from a mixture between the flavor and mass eigenstates of neutrinos. That is, the three neutrino states that interact with the charged leptons in weak interactions are each a different superposition of the three neutrino states of definite mass. Neutrinos are created in weak processes in their flavor eigenstates[nb 1]. As a neutrino propagates through space, the quantum mechanical phases of the three mass states advance at slightly different rates due to the slight differences in the neutrino masses. This results in a changing mixture of mass states as the neutrino travels, but a different mixture of mass states corresponds to a different mixture of flavor states. So a neutrino born as, say, an electron neutrino will be some mixture of electron, mu, and tau neutrino after traveling some distance. Since the quantum mechanical phase advances in a periodic fashion, after some distance the state will nearly return to the original mixture, and the neutrino will be again mostly electron neutrino. The electron flavor content of the neutrino will then continue to oscillate as long as the quantum mechanical state maintains coherence. Since mass differences between neutrino flavors are small in comparison with long coherence length for neutrino oscillations this microscopic quantum effect becomes observable over macroscopic distances.

Now what I do not understand is,

  1. is it that physically (classically) three spatially separated neutrinos are in flight, and they arrive together (separated by only millisecond), we just can't seem to be able to differently observe them as they arrive

  2. Or is it that they are in a superposition (as per QM), so spatially they have a common wavefunction, and at the point of observation, the wavefunction collapses, and the neutrino mass and flavor type will be known (the one that dominates out of the three maybe)?

Based on the two articles, it is not defined, because one of them says that different mass types of neutrinos are in flight with different speeds, and they arrive within a millisecond and our apparatus can't distinguish between them, then the other article says that there is one single neutrino in flight with a single speed in a superposition of eigenstates. I assume those eigenstates cannot have different speeds. Or can a superposition of a particle also mean that it's speed and spatial coordinates, energy and mass are different at the same time for the different eigenstates? In this case the different eigenstates would differ not only by one characteristic (mass/flavor), but also by speed and spatial coordinate.


  1. is it 1. or 2., so three different neutrinos arrive with different speeds(and we don't have good enough apparatus to distinguish) or just a single neutrino in a superposition with one single speed?

2 Answers 2


I wrote a few words about timing resolution limits in neutrino detection in an earlier post, but in the case of supernova neutrinos the issue isn't a technical one at the detection end, it has to do with the generation of the neutrinos in the first place.

The problem with timing from supernova neutrinos is that the neutrinos are produced by the process $$ p \to n + e^+ + \nu_e $$ which occurs during the core collapse over the course of multiple milliseconds.

So even though you can know the finishing times for the race to sufficient precision, you don't know the relative starting time well enough to draw any conclusions.


The neutrinos start off not as “electrinos” (defined as the isospin partner of the electron) which is a coherent superposition of mass eigenstates. If the neutrinos released by beta decays all had exactly the same speed, then the phases of mass eigenstates would change in a predictable way, and the particles would arrive in a different but still coherent superposition. But since their speed varies, they actually arrive as a non-coherent mixture. It would be tough to tell the difference experimentally, even if you could detect low-energy muinos and tauinos by black magic.

  • $\begingroup$ can you please tell me more about the non-coherent mixture? what does it mean non-coherent mixture? $\endgroup$ Jun 10, 2018 at 4:43
  • $\begingroup$ The relative phases are either truly random or essentially unpredictable. $\endgroup$ Jun 10, 2018 at 10:17
  • $\begingroup$ For example, unpolarized light is a non-coherent mixture of any two orthogonal polarizations: horizontal and vertical, criss and cross, or LHC and RHC. They are all equivalent. $\endgroup$ Jun 10, 2018 at 10:30

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