I Open Questions about Neutrinos Today

[vanhees71]

That exact same argument suggests you can measuree Lx and Lz simultaneously. Youyr argument is with quantum mechanics, not particle physics.

The answer to this "dilemma" (surprised nobody called it a "paradox" yet) is in Message #9.

A. You need to use an apparatus large enough. The scale is usually measured in kilometers.
B. Even if you could do this, you would see three decays:

• X \rightarrow Y + e^- + \overline{\nu}_1
• X \rightarrow Y + e^- + \overline{\nu}_2
• X \rightarrow Y + e^- + \overline{\nu}_3

You would not somehow "find the mass of the flavor eigenstate".

I think what's measured as particles are indeed the "other" particles rather than neutrinos as "particles". You cannot measure neutrinos as a "particle", because there's no interaction that projects to the mass eigenstates but only such that project to flavor eigenstates which are not mass eigenstate

Take the "neutrino-mass" determination with, e.g., the strategy to measure the endpoint of the electron-energy spectrum very accurately (a la KATRIN) by measuring the electron energy spectrum in tritium decay, i.e., $\text{t} \rightarrow ^3\text{He}+\text{e}^-+\bar{\nu}_{\text{e}}$. Let's make it idealized considering fully accurate measurements of arbitrary precision. So what's measured? You take an ensemble of tritium nuclei at rest and measure precisely the energy of the electron from the $\beta$ decay of each triton. Of course you don't measure the (anti-)neutrino but only the energy of the electron. Since the decay is not into a mass eigenstate of the anti-neutrino but in the electron-flavor eigenstate (which is established by measuring definitely an electron in the $\beta$ decay). Now the energy of the electron depends on the mass of the neutrino. So when measuring the endpoint of the electrons' energy spectrum you don't get a definite value but a distribution of end-point energies due to the superposition of neutrino-mass eigenstates. So what you'll measure is an average of the neutrino masses weighted with the corresponding PMNS matrix elements (modulus squared).

 

[snorkack]

I think what's measured as particles are indeed the "other" particles rather than neutrinos as "particles". You cannot measure neutrinos as a "particle", because there's no interaction that projects to the mass eigenstates but only such that project to flavor eigenstates which are not mass eigenstate

Take the "neutrino-mass" determination with, e.g., the strategy to measure the endpoint of the electron-energy spectrum very accurately (a la KATRIN) by measuring the electron energy spectrum in tritium decay, i.e., $\text{t} \rightarrow ^3\text{He}+\text{e}^-+\bar{\nu}_{\text{e}}$. Let's make it idealized considering fully accurate measurements of arbitrary precision. So what's measured? You take an ensemble of tritium nuclei at rest and measure precisely the energy of the electron from the $\beta$ decay of each triton. Of course you don't measure the (anti-)neutrino but only the energy of the electron.

Does not follow.
If you don´ t measure the antineutrino, but are otherwise free to make measurements to arbitrary precision, you can also measure the recoil of He-3, and the angle between electron and He-3 recoil. Which combined would give you the specific momentum and energy of antineutrino.

Since the decay is not into a mass eigenstate of the anti-neutrino but in the electron-flavor eigenstate (which is established by measuring definitely an electron in the $\beta$ decay). Now the energy of the electron depends on the mass of the neutrino. So when measuring the endpoint of the electrons' energy spectrum you don't get a definite value but a distribution of end-point energies due to the superposition of neutrino-mass eigenstates.

But if you measure the momenta of both electron and the nucleus recoil, you get the energy, momentum and rest mass of antineutrino for each decay event - near endpoint (antineutrino energy is small) or far from endpoint.

So what you'll measure is an average of the neutrino masses weighted with the corresponding PMNS matrix elements (modulus squared).

But if you measure momenta of both electron and nucleus, what would you get for neutrino rest mass?
A spectrum where all neutrinos fit, within measurement precision, to one of the three mass eigenvalues, and the three decay paths have a specific branching factor?

 

[vanhees71]

Does not follow.
If you don´ t measure the antineutrino, but are otherwise free to make measurements to arbitrary precision, you can also measure the recoil of He-3, and the angle between electron and He-3 recoil. Which combined would give you the specific momentum and energy of antineutrino.

But if you measure the momenta of both electron and the nucleus recoil, you get the energy, momentum and rest mass of antineutrino for each decay event - near endpoint (antineutrino energy is small) or far from endpoint.

But if you measure momenta of both electron and nucleus, what would you get for neutrino rest mass?
A spectrum where all neutrinos fit, within measurement precision, to one of the three mass eigenvalues, and the three decay paths have a specific branching factor?

Yes, and with this measurement you'd project to a corresponding neutrino-mass eigenstate, but the state you are measuring is not a neutrino-mass eigenstate. That's why you won't find the same neutrino mass in each measurement but with some probability, given by the corresponding modulus squared PMNS matrix elements, one of the three neutrino masses.

 

[Vanadium 50]

First, it is absolutely not true that neutrinos only interact with matter in flavor eigenstates. There are neutral current events: ν + X → ν + X.

Second, the term "flavor eigenstate" is confusing people. It may be helpful to think "flavor projection of the mass eigenstate" instead.

If you had mass resolution that was good enough, the decay X \rightarrow Y + e^+ + \nu_1 would occur with twice the rate as X \rightarrow Y + e^+ + \nu_2 (I am going to use positron emission as an example so I don't have a zillion overlines to include.)

If I had a beam of pure \nu_1, it would have twice the reaction cross-section on a target of Y as \nu_2. (The inverse process)

If I get a beam of neutrinos from X decay and use them to induce the inverse process on a target of Y, the rate is 5/9 in appropriate units.

Now, if I cannot tell whether I have a \nu_1 or \nu_2 in flight, the two states interfere (which some people call "oscillate") and the strength varies (with L/E) between 1 and 1/9.

 

[ohwilleke]

The Particle Data Group has five review articles on neutrinos.

 

[vanhees71]

First, it is absolutely not true that neutrinos only interact with matter in flavor eigenstates. There are neutral current events: ν + X → ν + X.

Second, the term "flavor eigenstate" is confusing people. It may be helpful to think "flavor projection of the mass eigenstate" instead.

If you had mass resolution that was good enough, the decay X \rightarrow Y + e^+ + \nu_1 would occur with twice the rate as X \rightarrow Y + e^+ + \nu_2 (I am going to use positron emission as an example so I don't have a zillion overlines to include.)

If I had a beam of pure \nu_1, it would have twice the reaction cross-section on a target of Y as \nu_2. (The inverse process)

If I get a beam of neutrinos from X decay and use them to induce the inverse process on a target of Y, the rate is 5/9 in appropriate units.

Now, if I cannot tell whether I have a \nu_1 or \nu_2 in flight, the two states interfere (which some people call "oscillate") and the strength varies (with L/E) between 1 and 1/9.

Sure, there are also the neutral-current reactions. The point is that it's with neutrinos as with any other (von Neumann) measurements in QT. You get some result of the measured quantities with probabilities given by Born's rule using the state the measured object is prepared in.

The phenomenon is called "oscillations" because all kinds of similar cases are called "oscillations" (e.g., Rabi oscillations). Of course, everything is described by Hilbert space vectors and operators as for any quantum system.

 

[ChrisVer]

I was wondering however, if the three "real" neutrinos (those that are propagated) have different masses, wouldn't they travel at different speeds? Wouldn't that disintegrate the flavour neutrinos (as e.g. the neutrinos of lower mass would arrive first and will be followed by the higher mass neutrinos)? Or is that another reason why we observe the "oscillation"?

 

[vanhees71]

That's why we observe the oscillation. The flavor eigenstates are superpositions of the mass eigenstates. Only the mass eigenstates have a well determined interpretation as particles with the usual "on-shell condition" for energy and momentum, and as such a well-determined velocity $\vec{v}=c \vec{p}/\sqrt{m^2 c^2+\vec{p}^2}$.

 

[ChrisVer]

so for example in a Supernova, you'd probably have the creation of many electron neutrinos \nu_e (OK basically it'd be antineutrinos but nevermind)... however from those neutrinos, the mass eigenstates, \nu_i, would propagate in space and be detected on Earth.
Wouldn't that give us an indication about the mass hierarchy of the mass eigenstates? As for example the mixture of \nu_1 would be observed a few minutes prior to that of \nu_3 (or vice versa) because it is lighter and travels faster? Or the problem is that we don't observe that many neutrinos from such sources to be able and make such a distinction?

 

[Vanadium 50]

you'd probably have the creation of many electron neutrinos (OK basically it'd be antineutrinos but nevermind)

No, they are mostly neutrinos. In SN1987A we saw antineutrinos because all we had were antineutrino detectors operating.

As for example the mixture of would be observed a few minutes prior to that of (or vice versa) because it is lighter and travels faster?

Minutes? Have to be pretty heavy to get minutes. You should work out how heavy neutrinos need to be to get a change of minutes.

The idea is correct, but the numbers don't work out. SN1987A gave a limit of 12 eV. That's at least 100x higher than the actual masses (and much higher than the mass differences).

 

[Vanadium 50]

A fun problem- if you want to constrain neutrino masses with supernovae, do you want one close by? Or far away? Or is there an optimal distance?

 

[ChrisVer]

A fun problem- if you want to constrain neutrino masses with supernovae, do you want one close by? Or far away? Or is there an optimal distance?

Well yes. Would be fun to take a couple of minutes and try to figure out at which distance of a SN the time difference would be ~X (=60?) seconds for neutrinos in the range of \Delta m^2 \approx 0.0025\text{ eV}^2 or so. However with the distance the flux would also drop, I'm not sure if that would be an issue.

 

[mfb]

However with the distance the flux would also drop, I'm not sure if that would be an issue.

How could it not be an issue? If you have a 1% chance to detect any neutrino you are clearly too far away.

The separation should ideally be longer than the neutrino pulse at emission. We know how long that was for SN1987A.

 

[ChrisVer]

How could it not be an issue?

Well maybe you are right. I spoke like that because, out of my head, I don't have a "feeling" about the magnitudes, with the cross-section dropping as r^{-2} and the actual number of neutrinos that are produced in an SN explosion. If you had told me to take the solid angle surface area into account, I'd have predicted that we couldn't detect neutrinos from a few 100,000's ly afar (and yet we did with SN1987A).
Of course then increasing the distance 10-fold, would decrease the flux by a 100... At the same time it would only increase the time available for separation between the 1-2-3 neutrinos (or the sensitivity to the masses compared to the SN1987A) by only 1 order of magnitude.

 

[mfb]

It's expected that the Milky Way has ~2 supernovae per century. With ~100 billion similar galaxies in the observable universe that's ~100 supernovae per second, give or take three orders of magnitude. Clearly we don't measure that many.

 

[Vanadium 50]

t the same time it would only increase the time available for separation between the 1-2-3 neutrinos (or the sensitivity to the masses compared to the SN1987A) by only 1 order of magnitude.

That's the right way to think about the problem. Your measurement improves as 1/r. In your example, you are 100x as sensitive to time differences, but your signal is 10x smaller, so you get a factror of 10. So that means you want to go as close as you can.

@mfb points out the next factor that needs to be considered: the difference in time-of-flight needs to be longer than the emission time for you to measure it it. And that means you want to go as far as you can.

Since you have two effects in opposite directions, there is an optimal distance you can calculate. I have not done this.

For antineutrinos, we're already at the best we can do with 1987A. The pulse length is comparable to the emission time, so you need to go out farther. But we can barely see the signal now, so you can't.

Neutrinos have a much shorter emission time.

 

[ohwilleke]

A fun problem- if you want to constrain neutrino masses with supernovae, do you want one close by? Or far away? Or is there an optimal distance?

The Sun is definitely too close for comfort. A nice healthy distance would be preferable.

Distance also puts more gap between the neutrino and photon signals allowing for better resolution of any speed differences (with the problem that you don't know how far spaced the original signals are from each other).

But, you need a distance close enough to get a resolvable two messenger signal.

 

[exponent137]

A new important measurement about the neutrino mass bound with KATRIN:


We have been waiting for this since Sep. 2019 https://arxiv.org/abs/1909.06048
Or when we first hear this old result?

$m^2$ of the old result was

And the upper bound was $1.1$ eV.
Now $m^2$ is positive.

 

[ChrisVer]

A new important measurement about neutrino mass bound with KATRIN:


We have been waiting for this since sep. 2019 https://arxiv.org/abs/1909.06048
Or when we first hear this old result?

$m^2$ of the old result was

View attachment 281719

Now it is positive. And the upper bound was $1.1$ eV.

What does a negative mass-squared value mean as a "best-fit" result (which is unphysical as they also claim in the same paper for the upper limit calculation)?

 

[mfb]

Measure the length of a stick, then measure the same length of the stick plus a sheet of paper and subtract. It's quite possible that the difference is negative, but it should be consistent with the positive sheet of paper within uncertainties.

They subtract two very large similar values from each other, the transition energy and the highest observed electron energies. It's a bit more complicated because they are also sensitive to the shape of the spectrum and the fit works with the squared mass, but the overall concept is the same.

 

[Vanadium 50]

I'm not sure I would consider KATRIN "new and important". The limit has moved down from 1.1 eV to 0.9 eV, or about 20%. But it's still an order of magnitude below the cosmological limit - that's that the sum of the neutrino masses is below 0.26 eV. (Which means a 0.09 eV upper limit on any of them)

 

[exponent137]

I'm not sure I would consider KATRIN "new and important". The limit has moved down from 1.1 eV to 0.9 eV, or about 20%. But it's still an order of magnitude below the cosmological limit - that's that the sum of the neutrino masses is below 0.26 eV. (Which means a 0.09 eV upper limit on any of them)

I admit that the analyses of DESI, EUCLID, and of other cosmological measurements will be really important because they will give the lower bound of the neutrino masses. (The results of DESI will even happen in 5 years.) Project 8 and Holmes will give better results than KATRIN, and conditionally some beta decays will give the mass of the neutrino.

But, when we wait, KATRIN will give an independent confirmation and every few percents are important. This measurement is from the first principle, where cosmological measurements have some additional unknown parameters.

I read that the cosmological upper bound for the sum of neutrino masses is 0.12 eV. So, do you think that 0.26 eV is a more conservative bound?

 

[ohwilleke]

I'm not sure I would consider KATRIN "new and important". The limit has moved down from 1.1 eV to 0.9 eV, or about 20%. But it's still an order of magnitude below the cosmological limit - that's that the sum of the neutrino masses is below 0.26 eV. (Which means a 0.09 eV upper limit on any of them)

The credible cosmology based limits are as low as 110 meV (i.e. 0.11 eV) for the sum of the three masses, and assuming that the differences between the three neutrino masses from oscillation data are correct to within their margins of error, absolute neutrino mass boundaries can be quite tightly bounded, with the best fit value being less than that.

The difference between the first and second neutrino mass eigenstate is roughly 8.66 +/- 0.12 meV, and the difference between the second and third neutrino mass eigenstate is roughly 49.5 +/- 0.5 meV, See, e.g., the Particle Data Group global averages.

This implies that the sum of the three neutrino mass eigenstates cannot be less than about 65.34 meV with 95% confidence, in addition to being not more than 110 meV.

Assuming the 0.11 eV sum of neutrino mass constraints, the neutrino mass differences from oscillation data, and a normal hierarchy (which almost all observational data favors, although not necessarily decisively), implies the following bounds on absolute neutrino mass, most of the uncertainty in which is driven by the uncertainty in the lightest neutrino mass which is shared in all three absolute mass estimates:

v1: 0 meV to 12 meV
v2: 8.42 meV to 21.9 meV
v3: 56.92 meV to 72.4 meV

By comparison, Katrin bounds v1 to less than 900 meV, which is 75 times the bound derived from cosmology and neutrino oscillations and an assumption of a normal mass hierarchy.

In an inverted hierarchy of neutrino masses, the minimum sum of the three neutrino masses given current neutrino oscillation data is around 98 +/- 1 meV (which leaves only about 4 meV of shared uncertainty in each of the three neutrino masses if indeed the sum of the three is not more than 110 meV), which is again, still far less than Katrin bound.

 

[mfb]

Mixing only gives you differences between squared masses (excluding smaller effects they are not sensitive to yet). If you take the square root of these you get a maximal mass difference, not a mass difference. Mixing alone doesn't set relevant upper limits on the neutrino masses.
The 0.11 eV cosmological constraint is not without criticism, direct measurements are useful for verification.

 

[exponent137]

At the end of this article, I saw two additional future projects of the sky surveys, CSST and PFS, beside of DESI, EUCLID, and SDSS-V.
https://phys.org/news/2021-04-scientists-dark-energy.html
Are there any other future surveys in progress?

I suppose that all of these surveys give also the upper bound of the neutrino masses? And, it is predicted, that they will give also the lower bound.

The most probably disagreement of two sorts of measurements of the Hubble constant does not influence the neutrino masses?

 

[exponent137]

exponent137 said:

We have been waiting for this since Sep. 2019 https://arxiv.org/abs/1909.06048
Or when we first hear this old result?

$m^2$ of the old result was

View attachment 281719
And the upper bound was $1.1$ eV.
Now $m^2$ is positive.

A combination of the Runs 1 and 2 is additionally a little better, the upper bound is 0.8 eV.
https://www.sciencenews.org/article/neutrino-max-possible-mass-tiny-new-estimate-particle-physics

Some interesting headings of presentations are here. Can someone view them?
http://meetings.aps.org/Meeting/APR21/Session/Q14

 

[exponent137]

By comparison, Katrin bounds v1 to less than 900 meV, which is 75 times the bound derived from cosmology and neutrino oscillations and an assumption of a normal mass hierarchy.

In an inverted hierarchy of neutrino masses, the minimum sum of the three neutrino masses given current neutrino oscillation data is around 98 +/- 1 meV (which leaves only about 4 meV of shared uncertainty in each of the three neutrino masses if indeed the sum of the three is not more than 110 meV), which is again, still far less than Katrin bound.

I am adding here another aspect: The goal of KATRIN is "<200 meV". This new result "<800 meV" gives a promise that this goal will be achieved. But this wished bound will be so 17 times the bound derived from cosmology, what will not be too big, and what will help to confirm the results from cosmology.According to posts #52 and #55, I am adding here the projects WFIRST and LSST as another two future cosmological projects for neutrino masses. And the following analysis will help at all of them:
https://phys.org/news/2021-05-supernovae-twins-possibilities-precision-cosmology.html

 

[exponent137]

And here is the KATRIN preprint of "<800 (900) meV":
https://arxiv.org/abs/2105.08533

Besides, the main phase of DESI is beginning.
https://newscenter.lbl.gov/2021/05/17/start-of-dark-energy-survey/
Maybe it will be the first one that will give the lower bound of the neutrino masses.

 

[ohwilleke]

And here is the KATRIN preprint of "<800 (900) meV":
https://arxiv.org/abs/2105.08533

Besides, the main phase of DESI is beginning.
https://newscenter.lbl.gov/2021/05/17/start-of-dark-energy-survey/
Maybe it will be the first one that will give the lower bound of the neutrino masses.

Thanks for the links. Mixing gives a pretty meaningful low bound on the sum of the neutrino masses, but a lower bound on the lightest neutrino mass, or even a confirmation that it was non-zero, would be serious progress.

 

[exponent137]

Thanks for the links. Mixing gives a pretty meaningful low bound on the sum of the neutrino masses, but a lower bound on the lightest neutrino mass, or even a confirmation that it was non-zero, would be serious progress.

Yes, I was not precise enough, they try to measure the sum of masses of the three neutrinos, now they measured only their upper bound.
https://arxiv.org/abs/2006.09395
It will not be perfect, but it will be the next level.