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Neutrions are produced and detected as flavor eigenstates $\nu_{\alpha}$ with $\alpha=e, \mu, \tau$. These states have no fixed mass, but are the combinations of three mass eigenstates $\nu_{k}$ with $k=1, 2, 3$, with mass $m_1$, $m_2$ and $m_3$, respectively. My questions are:

a) do neutrinos travel from source to the detector as flavor eigenstates or mass eigenstates?

b) is it possible to know which mass eigenstate the neutrino is in?

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    $\begingroup$ The Hamiltonian for free propagation has mass as its eigenvalue. $\endgroup$ – ZeroTheHero Nov 21 '17 at 15:15
  • $\begingroup$ So as you say, the neutrino indeed are produced and detected as flavour eigenstates; but these are nothing more than mixes or superpositions of the mass eigenstate. It is the mass eigenstates which evolve according to Schrodinger's equation, and hence what we see is just their superposition. Now, it turns out, that if all the mass eigenstates had the same mass, then they would always stay in the same flavour estate as they travel. But if their masses differ (as in experiment), the mass eigenstates decohere/go out of phase. This means the flavour eigenstate is allowed to change as it travels. $\endgroup$ – MKF Nov 21 '17 at 15:23
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    $\begingroup$ First, due diligence. Also. $\endgroup$ – Cosmas Zachos Nov 21 '17 at 15:27
  • $\begingroup$ Is every superposition of mass eigenstate a flavor eigenstate? I mean, for example, can A|nu_1> + B |nu_2> + C|nu_3> be always a flavor eigenstate for arbitrary values of A, B and C? $\endgroup$ – Seeker Nov 21 '17 at 17:14
  • $\begingroup$ @MKF I would leave the words 'nothing more than' out of "but these are nothing more than mixes or superpositions of the mass eigenstate." as the mass states don't have any kind of universal claim on being more special than any other basis. It would be no more or less correct to say that the mass states are "nothing more than" superpositions of flavor states. Mass states are the proper basis for evolving the free states because they are the eigenfunctions of the free Hamiltonian. Flavor states are the proper basis for interaction because they are the eigenfunctions of the weak Hamiltonian. $\endgroup$ – dmckee Nov 21 '17 at 17:48
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(a) They start as a flavor eigenstate, which is a super position of mass eigenstates. The mass eigenstates have different time evolution, hence the state is, in general, a mixed state in either basis.

(b) No. As an analogy, consider polarized photons and Faraday rotation--it may start out + polarized, rotate to a mixture of + & - (with coefficients a & b) and then at your + detector you see it $a^2$ fraction of the time, and $b^2$ you don't. In either case, you can't say which state a particular photon was in.

(b') Can we detect a $\nu_e$ and know it's mass? Can it have the mass of a $\nu_{\tau}$? The $\nu_e$ doesn't have "a" mass, it has 3:

$|\nu_e\rangle=0.82|\nu_1\rangle+0.54|\nu_2\rangle-0.15|\nu_3\rangle$

while a tau-neutrino:

$|\nu_{\tau}\rangle=0.44|\nu_1\rangle-0.45|\nu_2\rangle-0.77|\nu_3\rangle$

So, "yes", if we measure it's mass, then it will have a mass that a tau neutrino mass measurement could yield.

In theory: it's not a sensible question to ask, since flavor eigenstates aren't mass eigenstates.

In practice: we do not know the masses of the mass eigenstates, and their differences are much less than an eV--so how are you going the measure that?

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  • $\begingroup$ Sorry for the ignorance, but do specific particles not have a corresponding specific mass on detection? For example, can we detect an electron-neutrino with the mass of tau-neutrino? I guess not? Then what is the physical meaning of your statement that "a flavor eigenstate [...] is a super position of mass eigenstates"? $\endgroup$ – safesphere Nov 21 '17 at 16:44
  • $\begingroup$ @safesphere . It is a quantum mechanical framework, and it means that if you measure the four vector of a neutrino (invariant mass) you will have a corresponding probability of its being one of the three, even though it started as a nu_e . How will you measure it? by making it reveal its flavor in an interaction in the detector and measuring electrons and taus and muons. Quantum mechanics is not algebra. $\endgroup$ – anna v Nov 23 '17 at 6:20
  • $\begingroup$ @annav I get this part. My confusion is, why are we talking about mass and flavor as independent concepts? If we can't detect an electron-neutrino with the mass of tau-neutrino, then what's the meaning of "a flavor eigenstate [...] is a super position of mass eigenstates" and similar statements? $\endgroup$ – safesphere Nov 23 '17 at 6:27
  • $\begingroup$ @safesphere But in a sense we do, if we start with muon antineutrinos in an accelerator beam, we find taus generated from the probability distribution of the muon antineutrino to oscillate into a tau neutrino. $\endgroup$ – anna v Nov 23 '17 at 7:23
  • $\begingroup$ @annav I think this may be a bit different from what I mean. I don't doubt neutrinos oscillate, as this is rather a well known fact. And I do understand that measuring the mass of neutrinos is a tough task. However, conceptually, if we had the technology to measure the mass, then I believe, on every actual detection of the tau-neutrino we would measure it with the tau-neutrino mass. So I guess the point I am still missing is why we consider their masses and flavors to be independent superpositions, as they are not independent on detection. This is a deep topic though, so no worries :) Thanks! $\endgroup$ – safesphere Nov 23 '17 at 7:45

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