Neutrino theory regarding rest masses

[Buzz Bloom]

Hi fero:

I think degenerate refers to the fact that, in the cosmological modeling, the active neutrinos are assumed to have the same mass in order to reduce the complexity of the computation.

This is very helpful.

Thanks for your post,
Buzz

 

[Buzz Bloom]

Hi Orodruin:

No, this is wrong. The flavour states are (quite well known) linear combination of the mass eigenstates.

This is also a misunderstanding. The mixing matrix is what it is and does not depend on time

This is what any high precision physics experiment of today will do.

"Degenerate" is typically used in the neutrino community to refer to a situation where the absolute neutrino mass scale is such that the neutrino masses are approximately equal.

I much appreciate you very helpful answers to my questions.

Thanks for your post,
Buzz

 

[snorkack]

I see what might be a problem with this approach. The problem involves knowing the vector direction of the neutrino's momentum vector realtive to the vectors of the other particles. I don't think this vector information is experimentally available. Is reliable theorectical knowledge about this vector available?

Just measure the sizes AND directions of the momenta of all three visible particles in the interaction. Then by conservation of momentum and vector addition, you get the direction and value of neutrino momentum.

 

[Buzz Bloom]

Hi snorkack:

Just measure the sizes AND directions of the momenta of all three visible particles in the interaction. Then by conservation of momentum and vector addition, you get the direction and value of neutrino momentum.

You are right - a mental lapse while trying to explain my thought. I was thinking about the momentum and kinetic energy of the nucleus to be hit by the neutrino before it is hit. The measurements you can make are with respect the stationary frame of all the detectors. Measuring the nucleus energy and momentum before the hit alters those values so they will differ at the hit. Perhaps (in theory) this effect on the nucleus (considering Heisenberg uncertainty) can be kept adequately small (for the precision purposes of the neutrino calculations) by allowing the time and the spatial "location" of the nucleus measurements to be sufficiently large - but maybe not.

Thanks for your post,
Buzz

 

[snorkack]

I was thinking about the momentum and kinetic energy of the nucleus to be hit by the neutrino before it is hit. The measurements you can make are with respect the stationary frame of all the detectors. Measuring the nucleus energy and momentum before the hit alters those values so they will differ at the hit. Perhaps (in theory) this effect on the nucleus (considering Heisenberg uncertainty) can be kept adequately small (for the precision purposes of the neutrino calculations) by allowing the time and the spatial "location" of the nucleus measurements to be sufficiently large - but maybe not.

Since the nucleus to be hit by neutrino is a long-lived state, its energy can be measured with arbitrary precision.

 

[Orodruin]

So, suppose that our E and p have been measured with enough precision to compute m so as to identify the mass eigenstate. Will every individual neutrino interaction show one of the 3 rest mass eigenstate values to the precision of measurements (rather than intermediate values)?

Yes, if you could determine the neutrino momentum to arbitrary precision (in reality, with enough precision to resolve the different mass eigenstates), then you would be able to identify which mass eigenstate was involved in each interaction. Such neutrinos would not oscillate but have fixed probabilities of interacting with a W and each charged lepton, just as all up type quarks have fixed probabilities of interacting via a W and each down type quark (the probability amplitudes in that case given by the CKM matrix instead of the PMNS). In a simplified picture, this would follow from the wave packets of different mass eigenstates decohering.

The practical difficulties in doing so would be immense. Not only the energy resolution of detectors would be an issue, but also things such as thermal motion of the involved target/decaying nucleus would generally be sufficient to spoil any such measurement.

 

[Vanadium 50]

The practical difficulties in doing so would be immense.

That's an understatement. For neutrino beams, you can't do this at all. The pion lifetime is too short - its mass uncertainty due to its decay width precludes measuring a neutrino mass to better than about 1 eV. You can avoid this problem by going to atomic nuclei decays (at a huge cost in flux) but you need to make sure your decay pipe is big - of order a kilometer in all directions. Otherwise the quantum mechanical effects of putting this particle in a box is enough to spoil the energy resolution you will need.

 

[snorkack]

Not only the energy resolution of detectors would be an issue, but also things such as thermal motion of the involved target/decaying nucleus would generally be sufficient to spoil any such measurement.

How does zero point movement of the decaying nucleus affect the measurements?

 

[fzero]

How does zero point movement of the decaying nucleus affect the measurements?

$kT$ at 300 K is 0.0257 eV, but a decaying nucleus could easily recoil with an order or two of magnitude more than this, which can be more than the possible mass splittings.

 

[fzero]

Hi fzero and ohwilleke;

In your posts #16 and #20 you both mentioned and discussed "mass eigenvalues". I believe I have a good unerstanding about matrices, eigen vectors, and eigen values, but the term "mass eigenvalues" confuses me. I assume the must be some 3×3 matrix M and 3 vectors V_i each with a cooresponding eigenvalue a_i, and they are related by the equation:

M×V_i = a_i×V_i,​

where the a_i's are the mass eigenvalues. I also underatnd that the three vectors V_I are orthogonal to each other.

I have several questions:

1) What does the matrix M represent physically?
2) How are the elements of M measured or calculated?
3) What do the three corresponding eigenvectors V_i represent physically?​

I have also seen the term "neutrino flavor eigenstate", for example in Wikipedia:
https://en.wikipedia.org/wiki/Neutrino_oscillation , in the section "Theory". This article does not discuss the equation involving M above. SInce a flavor is not a numerical value, I can't see how it can possibly be an eigen value of some matrix like M. I am tending to conclude that the concept of eigenstates in discussing neutrinos may well be a metaphor, rather anything to do with matematical eigen values and vectors. Could this be true?

Thanks for your discussions,
Buzz

So, first we need to understand why we have a mass matrix for neutrinos at all. You could say that electrons, muons, etc. also have a mass matrix, but since it is diagonal for most of the known elementary particles, we don't usually bother thinking about their masses that way.

First we need to remember from QM that in order to define a useful basis for the states of a system, we first start by naming the observables that we might use to label the states. Typically we might start with the Hamiltonian and try to find the states $|E\rangle$ that are the eigenvectors of the Hamiltonian; their eigenvalue is the energy of the state. For a free particle, we might also use the momentum operator, since we can construct the energy from the momentum through $E^2 = p^2 + m^2$ (in units with $c=1$).

Very often in real systems, there are multiple states with the same energy eigenvalue (we call these degenerate states), so the energy is not enough to fully label the state. Therefore we might want to use one or more other observables $A,B,C,\ldots$ to further label the states until we have a basis where each state $|E,a,b,c,\ldots\rangle$ is uniquely determined by specifying the corresponding eigenvalues. As a linear algebra problem, for the eigenvectors of $H$ to also be eigenvectors of another operator $A$, then the commutator $[H,A]=0$. We say that $H$ and $A$ are "compatible observables" if the operators commute. Similarly if we have to introduce another operator $B$ to label the states, then $[H,B]=[A,B]=0$, and so on for any other operators.

For example, in the absence of a magnetic field, the spin up and down states of an electron are degenerate in energy. Since there is no magnetic field, the spin operator commutes with the Hamiltonian and we can use the spin to further distinguish our basis states.

We also use the language "conserved quantity" to refer to an observable associated to an operator which commutes with the Hamiltonian, since in QM, the rate of change of an operator is directly proportional to its commutator with the Hamiltonian. This is the physical reason why we want to use observables that are compatible with the Hamiltonian to label states: we don't have to worry about their eigenvalues changing with time.

Now when we study particle interactions and decays, we find that energy, momentum, and angular momentum are conserved quantities, so we can start labeling states by their momentum, mass (since the momentum and mass can be used to specify the energy), and spin. But there are additional conserved quantities that we find. Most obviously we find that electric charge is conserved, so that's another label. But there are less obvious ones like the flavor numbers.

Consider muon decay $\mu^- \rightarrow e^- + \bar{\nu_e} + \nu_\mu$. If we assign a number $L_\mu=+1$ to the muon and its neutrino and a number $L_e=+1$ to the electron and its neutrino (with $L_i=-1$ assigned to the corresponding antiparticles), then $L_e$ and $L_\mu$ are conserved separately. It turns out that this lepton flavor number conservation persists in all Standard Model interactions. We can formally define an operator for, say, $L_e$ that acts on a state by adding the number of electrons and electron neutrinos in the state, $n_{e,\nu_e}$ and subtracts the number of antiparticles $\bar{n}_{e,\nu_e}$ (at the risk of confusion, the bar here does not mean complex conjugate. Then $L_e = n_{e,\nu_e} - \bar{n}_{e,\nu_e}$, with analogous formulae for the $\mu$ and $\tau$ flavors.

This was the state of affairs before definitive evidence for neutrino oscillations was discovered. Neutrino oscillations violate lepton flavor number conservation and this is where the mass matrix comes in. An elegant way of explaining neutrino oscillations is to suppose, first, that the neutrinos have non-zero mass and, furthermore, that the mass matrix for neutrinos has off-diagonal elements in flavor space. Then we would say that the neutrino mass terms in the Hamiltonian (though in quantum field theory we'd be more likely to use a Lagrangian) do not commute with the lepton flavor number operators. We still don't know what form the mass matrix $M_{ff'}$ even takes, but we can make models like the one involving a sterile neutrino. Despite our ignorance, we can introduce a matrix $U_{fk}$ that would diagonalize the mass matrix, which is precisely the PMNS matrix that defines the flavor eigenstates in terms of the mass eigenstates. The mass eigenstates evolve simply in time, so we can write the time evolution of the flavor states as a superposition and hence describe the neutrino oscillations.

To finally get to your other questions, the eigenvalues of the mass matrix are the physical masses that one would measure. Masses of interesting particles are almost never measured directly. If you recall Thompson's experiment to measure e/m for the electron by measuring the curve of the orbit in a magnetic field, that's about as direct a measurement as you can get. In a modern high-energy experiment, the particles whose mass we might want to measure either don't live long enough or don't interact strongly enough to make such a measurement possible. So a modern detector is designed to measure the energy and momentum of the decay products of the particles that we want to study. One then has to reconstruct the event to trace the momenta back to the particle in question. If this can be done accurately then you can add the vectors appropriately and infer the energy-momentum of the particle.

Physically the mass eigenstates are the states that the particle state would "collapse to" if we could make a direct measurement of the mass. For the other elementary particles, their mass eigenstates are the same as their "flavor" eigenstates. We don't have electrons oscillating into muons or taus.

 

[Orodruin]

So, first we need to understand why we have a mass matrix for neutrinos at all. You could say that electrons, muons, etc. also have a mass matrix, but since it is diagonal for most of the known elementary particles, we don't usually bother thinking about their masses that way.

Rather thansaying that it is diagonal, it would be more precise to say that we chose to work in a basis where it is diagonal. The mismatch of the diagonalisations of the up and down type quarks is what leads to quark mixing and the CKM matrix.

As such, I would not say that the neutrino mass matrix is non-diagonal per se. It is non-diagonal in the basis where the charged lepton mass matrix is diagonal.

That's an understatement.

I am sorry, I am not a native English speaker and have difficulties finding words stronger than "immense" ... How does "unsurmountable" sound?

 

[Buzz Bloom]

Hi snorkack:

Since the nucleus to be hit by neutrino is a long-lived state, its energy can be measured with arbitrary precision.

I am unsure about your assumptions regarding the measuring environment.

I am thinking about the practicalities of measuring the energy and momentum for all of the inputs and outputs of an interaction between, for example,
a neutrino ν and a nucleus N*_in -> an electron e* and a modified nucleus N_out.
Such a measurement will require (1) a very large number of νs, and (2) either (a) a very large number of N_ins or (b) a single N*_in and an extremely long time to wait for an interaction event. You may be assuming that the practicalities are irrelevant because it is only a thought experiment of what is theoretically possible insofar as no laws of physics are violated.

Case (a) All N_ins in the collection will have to be measured frequently and tracked individiually so that it is known which single one of the many, N*_in, has been hit by a ν. Between the measurement of the N*_in and the hitting event, N*_in must not have had time to interact with any other N_ins, since such an interaction will change the values just measured. Also, e* must not interact with any other electrons between the time of the hit and the time it's energy and momentum is measured. Is satisfying all these constraints theoretically possible?

Case (b) The single N*_in needs to be confined to the space S in which it can be measured. It will also have to be repeatedly measured so that the during the time between it's last measurement and it's being hit by ν it's interaction with whatever confines it to S will not significantly change it's energy or momentum. This case seems to me to be more plausibly possible than (a), but it might take millenia to complete the experiment.

Thanks for your post,
Buzz

 

[ohwilleke]

Why would someone look at so light sterile neutrinos?

We know from weak force interactions, that there are exactly three neutrino flavors under 45 GeV. But, early neutrino oscillation measurements using neutrinos generated by nuclear reactors suggested that 4 kinds of neutrinos were necessary to make equations that best fit the pattern in the data, something called the "reactor anomaly". The best fit to that anomaly suggested a fourth neutrino on the order of 1 eV in mass. But, because a neutrino of this mass could not interact via the weak force, it had to be sterile.

Subsequent reactor data has cast doubt on the 3+1 model of neutrino oscillation, making it look more like a statistical fluke.

 

[Buzz Bloom]

Hi fzero:

We still don't know what form the mass matrix Mff′M_{ff'} even takes, but we can make models like the one involving a sterile neutrino. Despite our ignorance, we can introduce a matrix UfkU_{fk} that would diagonalize the mass matrix, which is precisely the PMNS matrix that defines the flavor eigenstates in terms of the mass eigenstates.

Your post #45 explains a lot of what I have been confused about concerning the mass eigenvalues a_i and the corresponding eigen vectors V_i. In the above quote in particular, you answer two of my three questions about the mass matrix M whose eigen values are the three possible values for a neutrino. The context is the equation: M×V_i = a_i×Vⁱ.

1) What does the matrix M represent physically?
2) How are the elements of M measured or calculated?
3) What do the three corresponding eigenvectors V_i represent physically?​

(1) is not answered. Presumaly M is a 3×3 matrix of numbers. Does theory tell us whether the numbers are real or complex? Do the numbers have a physical interpretation:

(a) unitless real numbers representing probabilities
(b) unitless complex numbers representing amplitudes
(c) real or complex numbers with the units of mass
(d) something else.​

(2) is partially answered:

We still don't know what form the mass matrix [M] even takes.​

I am confused by the notation Mff′M_{ff'}. I would much appreicate a post from you explaining this.
(3) is answered. The columns of the PMNS matrix U are the three vectors V_i. Therefore:

U^T*×M×U = D​

where D is a diagonal matrix with mass units whose diagonal components are the three mass eigenvalues. ("*" means conjugate, and ^T means transpose.)

Thank you very much for your post 45,
Buzz

 

[snorkack]

Case (a) All N_ins in the collection will have to be measured frequently and tracked individiually so that it is known which single one of the many, N*_in, has been hit by a ν.

How about - we have a large number of indistinguishable nuclei whose locations are not individually well known but whose momenta are all known to be equally zero?

 

[fzero]

Hi fzero:
Your post #45 explains a lot of what I have been confused about concerning the mass eigenvalues a_i and the corresponding eigen vectors V_i. In the above quote in particular, you answer two of my three questions about the mass matrix M whose eigen values are the three possible values for a neutrino. The context is the equation: M×V_i = a_i×Vⁱ.

1) What does the matrix M represent physically?
2) How are the elements of M measured or calculated?
3) What do the three corresponding eigenvectors V_i represent physically?​

(1) is not answered. Presumaly M is a 3×3 matrix of numbers. Does theory tell us whether the numbers are real or complex? Do the numbers have a physical interpretation:

(a) unitless real numbers representing probabilities
(b) unitless complex numbers representing amplitudes
(c) real or complex numbers with the units of mass
(d) something else.​

(2) is partially answered:

We still don't know what form the mass matrix [M] even takes.​

I am confused by the notation Mff′M_{ff'}. I would much appreicate a post from you explaining this.
(3) is answered. The columns of the PMNS matrix U are the three vectors V_i. Therefore:

U^T*×M×U = D​

where D is a diagonal matrix with mass units whose diagonal components are the three mass eigenvalues. ("*" means conjugate, and ^T means transpose.)

Thank you very much for your post 45,
Buzz

What I meant by $M_{ff'}$ was the mass matrix in flavor space, so $f,f' = e,\mu,\tau$ as opposed to the mass eigenstates that are usually labeled by $i,k=1,2,3$. I had thought that you'd used the notation in an earlier post, so didn't clarify.

As for the nature of the mass matrix, we can make some comments based on the measured parameters of the PMNS matrix. We can take $D=\text{diag}(m_1,m_2,m_3)$ and compute

$$(M)_{ff'} = U_{fi} (D)_{ij} (U^\dagger)_{jf'}.$$

I haven't gone through the algebra explicitly, but I think the complex phases drop out of the final expression. Then this is a real matrix with entries involving the eigenvalues and products of sines and cosines of the PMNS angles.

As for interpretation, it's hard to give a direct one, since only the eigenvalues are measured "directly" (quotes because as I've mentioned even the mass measurements are not truly direct). The PMNS angles appear in the expressions for the amplitudes that we would compute for interactions involving neutrinos, so they can be deduced by carefully determining processes that depend on them most strongly. But neither they or the elements of the mass matrix are themselves probabilities or amplitudes (Edit Except for the earlier discussed role the elements of $U$ play in the probability to measure a particular mass eigenvalue in a flavor eigenstate). The mass eigenvalues and PMNS parameters should be thought of as additional parameters for the extended Standard Model.

Now, I should probably explain the comment I made about not knowing the form that the mass matrix takes. You could argue that the expression above is a pretty clear description. But what I meant was that in quantum field theory, what we mean by the mass matrix is usually the expression that appears directly in the Lagrangian and to write that we need more information. The whole reason people were satisfied with thinking that neutrinos were massless was that you couldn't write mass terms down for them within the Standard Model. Technically this has to do with the absence of a right-handed neutrino, which could be used to write down so-called Dirac mass terms, as is done for the electron.

Now that we know that neutrinos have a non-zero mass, we have to ask what is the new ingredient that let's us write mass terms. This is where the sterile neutrino proposal comes in. If we add at least one right-handed sterile neutrino, then we can write Dirac mass terms that couple the sterile neutrino to the active ones. But we're not sure that this is the correct explanation.

 

[Buzz Bloom]

Hi snorkack:

How about - we have a large number of indistinguishable nuclei whose locations are not individually well known but whose momenta are all known to be equally zero?

You are assuming a solid lattice at T=0 (or extrremely close to 0) made of identical atomic isotopes. The nuclei must have at least 2 protons, since if the proton of any form of H is hit by the neutrino, the resulting nucleus will be all neutrons, and the N_out particle(s) resulting energy and momentum would be very hard to measure.

I made a small search to find a possible isotope that might work. Here is a quote from http://www.physics.udel.edu/~glyde/Solid_H13.pdf .

Since helium is light, it's thermal wavelength, λ_T, is long, e.g., at T = 1.0 K, λ_T ≈ 1.0 Å for ⁴He.
Helium is therefore difficult to localize. Attempts to localize
it lead to a high kinetic or zero point energy.​

Lithium is also a bad choice. Here is a quote from https://en.wikipedia.org/wiki/Lithium#Isotopes :

Both natural isotopes have anomalously low nuclear binding energy per nucleon compared to the next lighter and heavier elements, helium and beryllium, which means that alone among stable light elements, lithium can produce net energy through nuclear fission.​

⁹Be may possibly work OK. ⁸Be and ¹⁰Be are radioactive.
(See https://en.wikipedia.org/wiki/Beryllium#Isotopes_and_nucleosynthesis .)
The problem with Be (and worse atoms with higher atomic numbers) is the number of electrons (4 for Be) in the latice environment along the path the output electron will follow while moving through the lattice. The longer the path, the more likely this electron will interact with another electron. I suppose that the lattice might by in the form of somewhat thin sheet, but that reduces the likelihood of a hit by a neutrino. I dont't have the knowledge to make a calculation of the optimum lattice thickness, but if this approach is not impossible I would expect it to be extrremely difficult.

Its an interesting idea you proposed. I hope other readers will comment.

Thanks for the post,
Buzz

 

[Buzz Bloom]

Hi fzero:

We can take D=diag(m1,m2,m3)D=\text{diag}(m_1,m_2,m_3) and compute

I would like to copy the equation in your post #51 imediately following the above quote, and then paste it into a new post. If I do this either by copy and paste, or by using the QUOTE feature, it dosen't come out looking the same. I assume there is some TeX way to do it.

Please help,
Buzz

 

[Buzz Bloom]

Hi fzero:

The more I think about that equation (the one I can't copy) the more questions come to mind. I am thinking that it might be better to start another thread to focus on discussing the PMNS matrix and the mass matrix and their relationship rather than discuss these topics here. What do you thnk?

Regards,
Buzz

 

[Orodruin]

Hi fzero:

The more I think about that equation (the one I can't copy) the more questions come to mind. I am thinking that it might be better to start another thread to focus on discussing the PMNS matrix and the mass matrix and their relationship rather than discuss these topics here. What do you thnk?

Regards,
Buzz

Before looking at the PMNS matrix, I suggest you familiarise yourself with the CKM matrix and quark mixing. You should be able to find relevant information in any textbook on particle physics.

The physics of lepton mixing is exactly equivalent to that of quark mixing, with the additional complication that neutrinos are very light, which leads to the possibility of creating mass eigenstates which do not decohere quickly, and that they may be Majorana fermions, which adds some slight complications in the definition of the PMNS matrix, which were not present in the case of the CKM.

 

[ChrisVer]

You can have a look in this thesis:
http://www2.physik.uni-bielefeld.de/fileadmin/user_upload/theory_e6/Diploma_Theses/dipl_kruppke.pdf
the parts from and after equation 1.13

 

[snorkack]

Hi snorkack:
You are assuming a solid lattice at T=0 (or extrremely close to 0) made of identical atomic isotopes. The nuclei must have at least 2 protons, since if the proton of any form of H is hit by the neutrino, the resulting nucleus will be all neutrons, and the N_out particle(s) resulting energy and momentum would be very hard to measure.

I made a small search to find a possible isotope that might work. Here is a quote from http://www.physics.udel.edu/~glyde/Solid_H13.pdf .

Since helium is light, it's thermal wavelength, λ_T, is long, e.g., at T = 1.0 K, λ_T ≈ 1.0 Å for ⁴He.
Helium is therefore difficult to localize. Attempts to localize
it lead to a high kinetic or zero point energy.​

Why solid lattice? Why attempt to localize? What we want is the momentum of neutrino, not its location.

 

[Buzz Bloom]

Hi snorkack:

How about - we have a large number of indistinguishable nuclei whose locations are not individually well known but whose momenta are all known to be equally zero?

Why solid lattice? Why attempt to localize? What we want is the momentum of neutrino, not its location.

I am confused by your two questions here, especially the second.

Why solid lattice?​

For each the nucleon to have a zero momentum, the temerpature T must be 0. I think this requires that the collection must be a solid, since a liquid of gas will have a non-zero T. A solid made up of identical nucleons I think must form a crystal lattice.

Why attempt to localize?​

I do not understand "localize" in this context. I did not specify that the solid lattice is limited in size, but I did suggest that a thickness be determined to optimize the tradeoff between a (1) large frequency of interation events and (2) a low frequency of disturbing the energy and momentum of the produced electron e and/or nucleon N_out. I did suggest the lattice might be a thin sheet, but it could also be a thin spherical shell with the neutrino generator at it's center.

This spherical configuration would make the measuring devices for the energy an momentum of N_out and e a complicated engineering problem.
BUT
if (a) an optimum thickness could be calculated,
and (b) it could result in an adequate number of events in which
(c1) e's and N_out's energy and momentum were not significantly disturbed,
or (c2) any such disturbances could be indentified so that those events could be ignored,
THEN
I would agree that this might be a "practical" way to make the desired measurements so that the energy and momentum of the neutrino can be calculated.

I still think my Case (b) alternative is more likely to result in success, even if the experiment might have to be continued for millenia.

If your intention is that we are discussing a thought experiment, I think either Case (a) or (b) might possibly qualify as possible, but Case (b) is a simpler to describe scenario. However, that may just be an aesthetic choice: simplification is in the mind of the thinker.

Thanks for the discussion,
Buzz

 

[Buzz Bloom]

Hi ChrisVer:

You can have a look in this thesis:
http://www2.physik.uni-bielefeld.de/fileadmin/user_upload/theory_e6/Diploma_Theses/dipl_kruppke.pdf
the parts from and after equation 1.13

I scanned through the thesis, and there is a lot more there that I think I want to know, at least for the present. On the other hand, Chapter 7 "The State Vectors for Flavour Neutrinos" looks particularly interesting. However, I saw immediately that I would have some difficulties with the notation used. In equation 7.1

1) what does ":– " mean?
2) what does "*" mean?

Thans for your help,
Buzz

 

[ChrisVer]

For some reason it appears weirdly in your PC...

1) it is a := and is the notation of a "definition"...some other times it can be a = with a ^ from above.
2) * is the complex conjugate... $a=x+iy$ then $a^* = x -iy$ with a a complex number and x,y its real and imaginary parts.

Can I ask you what your background is like and why exactly are you interested in neutrino oscillations? I mean, how are you supposed to understand a quantum mechanical phenomenon if you lack knowledge on quantum mechanics?
I quoted a certain part in the thesis where you can find how the PMNS matrix appears by changing the flavor interactions part of the Lagrangian when diagonalizing the mass matrix term.

 

[snorkack]

I am confused by your two questions here, especially the second.

Why solid lattice?​

For each the nucleon to have a zero momentum, the temerpature T must be 0. I think this requires that the collection must be a solid, since a liquid of gas will have a non-zero T.

Um. Both isotopes of He are liquid at absolute zero.
Say you have a pool of liquid He-3 at absolute zero, so no vapour pressure and vacuum above the surface.
And then you are operating it as electron antineutrino detector. By reaction
He-3+nuebar->t+e+
Can you measure the energy of the positron emitted?

A solid made up of identical nucleons I think must form a crystal lattice.

Why attempt to localize?​

I do not understand "localize" in this context. I did not specify that the solid lattice is limited in size, but I did suggest that a thickness be determined to optimize the tradeoff between a (1) large frequency of interation events and (2) a low frequency of disturbing the energy and momentum of the produced electron e and/or nucleon N_out. I did suggest the lattice might be a thin sheet, but it could also be a thin spherical shell with the neutrino generator at it's center.

So how are electron energies measured with a great precision, like in these tritium decay experiments?

 

[Buzz Bloom]

Hi Orodruin:

I suggest you familiarise yourself with the CKM matrix and quark mixing.

Before tracking down a textbook, I decided to look at the Wikipedia article
https://en.wikipedia.org/wiki/Cabibbo–Kobayashi–Maskawa_matrix .
Here is a quote I would like to ask about:

The constraints of unitarity of the CKM-matrix​

This seems to be saying that the CKM-matrix is unitary. To refresh my memory from a linear algebra course I took as an undergraduate in the 1950s, I found the definition at
https://en.wikipedia.org/wiki/Unitary_matrix :

In mathematics, a complex https://www.physicsforums.com/javascript:void(0) https://www.physicsforums.com/javascript:void(0) U is unitary if its conjugate transpose U* is also its inverse.​

Since you suggested I might learn about the PMNS matrix by studying the CKM-matrix,
does the first Wkipedia quote imply that the PMNS matrix is also unitary?

BTW, I confess I tend to stay away from trying to use a textbook as a reference source. If it is about a topic I know little about, I find it very difficult to learn anything specific I want to understand. Most textbooks I've looked at recently require reading thoroughly from the beginning, and remembering what is read, since any later disccusion does not refer back to definitions of technical terms or notation, and usually there is no glossary or suitable index. In other words, they are terrible reference sources unless you have previously taken a course using the particular textbook, and you still retain a reasonably good memory.

Thanks for your suggestion,
Buzz

 

[Orodruin]

Since you suggested I might learn about the PMNS matrix by studying the CKM-matrix,
does the first Wkipedia quote imply that the PMNS matrix is also unitary?

The unitarity of the CKM matrix is a prediction from the Standard Model and has to be tested experimentally (and it has been). The PMNS matrix is also generally assumed to be unitary under some conditions, but there are some theoretical ideas which would make it almost unitary, but with small corrections.

 

[Buzz Bloom]

Hi Orodruin:

The PMNS matrix is also generally assumed to be unitary under some conditions, but there are some theoretical ideas which would make it almost unitary, but with small corrections.

Can you post citations of articles that explain
"assumed to be unitary under some conditions",
and
"some theoretical ideas which would make it almost unitary, but with small corrections"?

Also re

The unitarity of the CKM matrix is a prediction from the Standard Model and has to be tested experimentally (and it has been).

Can you post citations of articles about the experiments that confirmed the unitarity of the CKM matrix?

Thanks for your discussion,
Buzz

 

[Orodruin]

Can you post citations of articles about the experiments that confirmed the unitarity of the CKM matrix?

See http://ckmfitter.in2p3.fr and references from there.

Can you post citations of articles that explain
"assumed to be unitary under some conditions",
and
"some theoretical ideas which would make it almost unitary, but with small corrections"?

See hep-ph/0607020 and references therein.

 

[Buzz Bloom]

Hi Orodruin:

Thanks for the citations. I am sure it will take me quite a while to digest them.

Regards,
Buzz

 

[Buzz Bloom]

Hi Orodruin:

I scanned the article hep-ph/0607020 you cited regarding my question:

Can you post citations of articles that explain
"assumed to be unitary under some conditions",
and
"some theoretical ideas which would make it almost unitary, but with small corrections"?​

As whole, it is clearly way over my head, but it what I asked for. Here is a quote that I think I almost understand.

Without attaching ourselves to any particular model, we have studied a minimal
scheme of unitarity violation -MUV-, considering only three light neutrino species and
with the usual unitary matrix U_PMNS replaced by the most general non-unitary one.​

I underlined the phrase I would particularly like to undestand. I found several articles on the internet that discussed "light neutrino species", but none defined it. Could you do that for me please.

Thanks for your help,
Buzz

 

[ChrisVer]

Light neutrino species is that you have 3 light (rest mass less than 45 GeV) neutrinos , such as $\nu_e, \nu_\mu, \nu_\tau$.

 

[Buzz Bloom]

Hi ChrisVer:

Light neutrino species is that you have 3 light (rest mass less than 45 GeV) neutrinos , such as νe,νμ,ντ\nu_e, \nu_\mu, \nu_\tau.

The article was from 2007. I understand that it is now generally accepted, as very likely to be so, that the sum of these 3 masses is about 430 meV. This is about 11 orders of magnitude less than this 45 GeV threshold for being a "light neutrino". Can you summarize the likely implications regarding the conclusions of this paper about the possible non-unitarity of the neutrino mixing matrix in the light of this enormous difference between the concept then of a light neutrino and the reality of today's understanding about these masses?

Thanks for you post,
Buzz

 

[ChrisVer]

The experiment in LEP showed that there are three active (=subject to weak interactions) light (of mass less than 45GeV) neutrino species...that's what fitted the experimental data best... this doesn't seem such an enormous difference, at least not to me... even 400meV (their sum) is less than 45GeV... it's just that there are no other light neutrinos in the inbetween spectrum.
Plus I don't know about PMNS non-unitariness.