One thing I've never really had a good feeling for is what neutrino oscillations would look like for a neutrino at rest.

Would the neutrino be sitting there with a mass that was uncertain until measured? I suppose that since the mass of a particle without electric charge would be pretty difficult to measure, that this might be part of the issue.

Or would the mass be certain, but oscillating? How does this work with conservation of energy?

I'm not making a rhetorical point here, smarter people than me are on board with massive neutrino oscillations, I just don't know how this idea works out and I haven't seen an explanation in terms of concepts.


1 Answer 1


Well, yes, neutrino flavor oscillations will present very differently in the extreme non relativistic limit, instead of the ultra-relativistic limit we are used to from standard experiments in the neutrino industry today.

In the static limit, motion will be supplanted by Gaussian wave packet diffusion. But the neutrino flavors you would detect in this fantasy-land would also change in time.

For simplicity, take only two flavors, e and μ, and two massive states 1 (red) with mass m, and 2 (blue) with higher mass M, and so only one mixing angle, (like Cabbibo's for quarks), call it Pontecorvo's: $$ \nu_e= {\color {red}\nu_1 } \cos \theta +{\color {blue} \nu_2}\sin \theta \\ \nu_\mu= -{\color {red}\nu_1} \sin \theta + {\color {blue}\nu_2} \cos\theta . $$ Recall the red and blue states are "for real" and evolve in time with fixed masses, unlike the flavored states (unreal, convenience driven--a feature of detection techniques).

Suppose you produce a $\nu_\mu$ at rest by some magic, at the origin x = 0, so, then the superposition state of noninteracting red and blue modes, with wave function $$ \psi(\vec{r},t=0)= {\color {blue}\psi_2} \cos \theta -{\color {red} \psi_1} \sin \theta. $$

Take the initial profile to be Gaussian, with initial probability distribution proportional to $$ { \Large e^{-\frac{\vec{r}\cdot \vec{r}}{a}}} = (\cos^2 \theta +\sin^2 \theta)~{ \Large e^{-\frac{\vec{r}\cdot \vec{r}}{a}}}, $$ where a is the initial wave packet width-squared.

That is, the same noninteracting red and blue initial Gaussian profiles $\psi_1(\vec{r},0)$ and $\psi_2(\vec{r},0)$ will now spread freely as usual, with controlling masses m and M , respectively, in their (kinetic) energy. They will end up driving the uncertainty relation very far indeed from the t = 0 saturation ($\Delta x ~ \Delta p_x=\hbar/2$, etc) but the red faster than the blue, $$ {\color {red}\psi_1}(\vec{r},t) = \left({a \over a + i\hbar t/m}\right)^{3/2} { \Large e^{- {\vec{r}\cdot\vec{r}\over 2(a + i\hbar t/m)} }},\\ {\color {blue}\psi_2}(\vec{r},t) = \left({a \over a + i\hbar t/M}\right)^{3/2} { \Large e^{- {\vec{r}\cdot\vec{r}\over 2(a + i\hbar t/M)} }}. $$

Thus, upon spreading with time, the respective widths-squared, $$ \sqrt{a^2 + (\hbar t/m)^2 \over a} $$ for red (1), rapidly, and $$ \sqrt{a^2 + (\hbar t/M)^2 \over a} $$ for blue (2), more slowly; ultimately linearly in time. (The differing velocities are due to the light red mode's larger relative spread in the initial velocity distribution, for the same momentum spread controlled by a.)

Sitting far away at L with your neutrino flavor detector, you will be observing the red mode arriving at you diffusively $$ {\color {red}\nu_1}=\nu_e \cos \theta -\nu_\mu \sin \theta $$ much before the blue one, so you will find a transition $\nu_\mu\to \nu_e$ with amplitude $$ \langle \nu_e ,t | \nu_\mu, o\rangle\propto \sin\theta \cos\theta ~ \psi^*_1 (L,t) \psi_1(0,0), $$ versus persistence, $$ \langle \nu_\mu ,t | \nu_\mu, o\rangle\propto \sin^2\theta ~ \psi^*_1 (L,t) \psi_1(0,0), $$ so, for small mixing angle, you'll be detecting more $\nu_e$s first. The blue bump has not reached you yet, so there is no interference--you might as well have handled the relative probabilities, squares of the above.

I'm too lazy to work out the complete time-dependent oscillation formula...

At the origin, as the proportion of red to blue decreases steadily, you may also see that $\nu_e$ starts appearing there as well, even though you started with none, almost paradoxically, due to the disruption of the delicate alignment of the red and blue wave packets.

  • Your side questions: Energy is always conserved, and no energy uncertainty feature arose here. The masses are parameters of the theory, not some type of direct "observable", and would enter into and be fitted by the relative e/μ profiles measured in that highly unrealistic fantasy world.

  • Why highly unrealistic? Because it is time to confess the above notional static Gaussian is nothing like the "rest frame" of the neutrino detected in the lab: There is no such frame! The given energy lump your detector in Minnesota observes is actually two (3 in the real world) different "real" mass eigenstates (think red, blue, here) with the same energy, but, perforce, different momenta and hence velocities. You cannot Lorentz-transform to the rest frame of both, because there is no such frame. Your detector selects the suitable momentum (velocity) components of each to match the energy, and tallies the rates described by the probability estimated with the given parameter masses, and L. So, above, I just serviced the fantasy...


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