Neutrino theory regarding rest masses

[Buzz Bloom]

Hi ChrisVer:

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

I think this must be a font problem. Thanks for interpreting it for me.

Can I ask you what your background is

I was trained in mathematics, mostly applied, and I also had some introductory physics cources. Befor retiring, I had a career in software development, especially concerning databases. As a lifetime hobby, I have tried to educate myself about a variety of scientic topics: mostly in (1) molecular biology (relating to the origin of life), and (2) in physics, especially GR and cosmology. Earlier this year I began to study some atmospheric physical-chemistry regarding global warming.

Very recently cosmology had lead me to issues about the neutrino, and also QM. I have found the theoretical and experimental physics about the neutrino to be a continuous fascinating mystery. I have formed the opinion that one cannot know just a small amount about the neutrino. Every time I thought I had learned something new about the neutrino, it soon became clear that what I had learned was not quite completely correct. It was just the tip of an iceberg which needed deeper study to clarify what I had thought I had just learned.

BTW, my first job after graduating college involved using a Marchant electro-mechanical calculator to find several eigen values and vectors of a 41x41 matrix. The method involved iteratively re-multiplying an arbitrary initial unit vector by the matrix until the process converged within the precision achievable with the calculator. After an eigen vector was found, a new initial unit vector was selected which was then made normal to all the previously found eigen vectors.

The physics forum has been extremely helpful.

Thank you for your patience and for all your help,
Buzz

 

[Buzz Bloom]

Hi ChrisVer:

even 400meV (their sum) is less than 45GeV

Another small but curious mystery. What makes the threshold value of 45GeV particulary special?

Thanks again,
Buzz

 

[Orodruin]

Hi ChrisVer:
Another small but curious mystery. What makes the threshold value of 45GeV particulary special?

Thanks again,
Buzz

It is half the mass of the Z boson and the constraints come from the Z boson hidden decays.

 

[Buzz Bloom]

Hi Orodruin:

It is half the mass of the Z boson and the constraints come from the Z boson hidden decays.

Thanks for your post,
Buzz

 

[Buzz Bloom]

Hi fzero:

Then this [(M)] is a real matrix with entries involving the eigenvalues and products of sines and cosines of the PMNS angles.

I have been pondering this for a while, trying to remember what I think I learned while an undergraduate. I am pretty sure that (M) can not have real components, since the components of it's eigenvectors are complex. I think that the eigenvectors of any real matrix must have real components. If you are not sure whether this is correct or not, I will start a thread in the math sub-forum.

Thanks for the discussion,
Buzz

 

[ChrisVer]

I am not really sure, but think about this... if $u$ is an eigenvector, isn't $iu$ an eigenvector too? Eigenvectors are defined from:
$A \cdot \textbf{v} = \lambda \textbf{v}$
So both $\textbf{v}$ or $i\textbf{v}$ satisfy the above.
But you may be right in figuring out that M doesn't have to be real - it should only be hermitian since the mass eigenvalues cannot be imaginary... I am saying "may be right" because I haven't performed the calculations and I am not going to do that either (since wolfram is really bad in helping me perform them computationally). Maybe someone with a better program at hand can input the PMNS matrix with the cos,sin and exp[i*delta] and find whether the result is purely real or not.

 

[fzero]

$M$ must be Hermitian (and you can prove that from the expression $UDU^\dagger$), but my comment was based on looking at the actual entries of $U$ and how the phases entered that product. I suspect that trig identities will make the terms proportional to $e^{\pm i \delta}$ vanish. The actual calculation is tedious, but not too difficult, so you're invited to check.

 

[ChrisVer]

Let's try my luck... Instead of taking all the products etc, I will only check a suspicious term (from the form of the PMNS matrix):]

Now I will take the $M_{12}$ that is the:
\begin{align}
M_{12} &= U_{1m} (DU^\dagger)_{m2} \notag \\
&=-c_{12} c_{13} s_{12}c_{23} d_{11}- c_{12}^2 s_{23} c_{13} s_{13} e^{-i\delta} d_{11}+ s_{12} c_{13} c_{12}c_{23} d_{22} - s_{12}^2 s_{13}c_{13} c_{23} d_{22} e^{-i\delta} + s_{13}c_{13}s_{23}e^{-i\delta} d_{33} \notag
\end{align}

I don't see how you could get rid of the $e^{i\delta}$'s...

 

[Orodruin]

I don't see how you could get rid of the $e^{i\delta}$'s...

You cannot. If you could, there would be no possibility for neutrino oscillations to violate CP.

Note that the M you are talking about here is actually $mm^\dagger$ if neutrinos are Dirac, where $m$ is the neutrino mass matrix (which is proportional to its Yukawa couplings). The matrix $m$ would be a completely general complex matrix and it would take a biunitary transformation to diagonalise it. Having the product between it and its conjugate results in a Hermitian matrix.

If neutrinos would be Majorana, then $m$ is a complex symmetric matrix, diagonalisable as $d = UmU^T$, where the phases of the entries in the diagonal matrix $d$ depend on $U$ (which is not unique). Again, the expression $mm^\dagger$ is Hermitian and has absolute values squared of the entries of $d$ as eigenvalues.

 

[Buzz Bloom]

Hi ChrisVer:

I think that the eigenvectors of any real matrix must have real components.

if uu is an eigenvector, isn't iuiu an eigenvector too? Eigenvectors are defined from:
A⋅v=λv A \cdot \textbf{v} = \lambda \textbf{v}
So both v\textbf{v} or ivi\textbf{v} satisfy the above.

You are correct. I should have said, "may have all real components." I was confused by my experience when I was only working with real normalized eigenvectors.

Thanks for your post,
Buzz

 

[Buzz Bloom]

Hi fzero, ChrisVer, and Orodruin:

MM must be Hermitian

I don't see how you could get rid of the eiδe^{i\delta}'s...

You cannot.

To summarize from the discussion: it is certain that:

(1) M must be Hermitian, that is, it's conjugate transpose is it's own inverse
(2) at least one of M's nine components is not real.​

Thanks for the discussion,
Buzz

 

[Orodruin]

(1) M must be Hermitian, that is, it's conjugate transpose is it's own inverse
(2) at least one of M's nine componets is not real.

1) No. M is hermitian, but you have given the description of a unitary matrix. A hermitian matrix is equal to its own hermitian conjugate.
2) No. We do not know this. This is still to be determined experimentally.

 

[ChrisVer]

This is still to be determined experimentally.

You mean like determining δ=0?

 

[Orodruin]

Yes, if $\delta = 0$ or $\pi$, neutrino oscillations are not violating CP.

 

[Buzz Bloom]

Hi Orodruin:

M must be Hermitian (and you can prove that from the expression UDU†UDU^\dagger)

Sorry about the appearance of the special characters in the quote, I think there is somthing flaky in my computer.

M is hermitian, but you have given the description of a unitary matrix.

Underlining in above quotes is mine.

Sorry about my confusion. The vocabulary for the variety of complex matrix types is not yet well re-embeded in my mind.

I also said:

at least one of M's nine componets is not real.​

You commented:

No. We do not know this. This is still to be determined experimentally.​

If my mind is now working OK, if M is Hermetian, then M's three diagonal components must all be real.
Also, Wikperida https://en.wikipedia.org/wiki/Hermitian_matrix defines a Hermetian matrix:

a square matrix with [at least some] complex entries that is equal to its own conjugate transpose (bracketed text my addition)​

Therefore M is Hermetian also implies that two or four or all six of it's non-diagonal element are complex.

Thanks for your post,
Buzz

 

[Orodruin]

Therefore M is Hermetian also implies that two or four or all six of it's non-diagonal element are complex.

No, your logic is failing here. The off diagonal terms can also be real, they do not need to be, but they may be. There may be zero non-real elements in the matrix.

 

[ChrisVer]

A real symmetric matrix is Hermitian... Hermitianity is the relation that $A^\dagger = A$... A real symmetric matrix is satisfying the hermitianity condition.
$A = \begin{pmatrix} a & b \\ b &c \end{pmatrix}$ with $a,b,c \in \mathbb{R}$ has $A^\dagger = A^T =\begin{pmatrix} a& b \\ b & c\end{pmatrix}= A$.

In this case again, it's as I asked Orodruin too, if $\delta =0$ (or $\pi$) then the $e^{\pm i \delta}$ doesn't stand for a complex number...it's $\pm 1$...and there are no possible complex elements in $M$ (of course the real numbers are just a subset of the complex numbers)...except for a Majorana case(?).

 

[Buzz Bloom]

Hi ChrisVer:

Hermitianity is the relation that A^†=A. ... A real symmetric matrix is satisfying the hermitianity condition.

(I edited the garbled quote that my computer put above to try to make it look like the original. How did you enter the dagger? is it a TeX command? I think my computer flakiness is related to TeX.)

I will try to change the definition at Wikipedia to make it clear that a Hermitian matrix may have complex components.

Thanks for your post,
Buzz

 

[ChrisVer]

The dagger I use in latex is " ^\dagger ".. ^ is for the powering.
Well again I'm saying that the real numbers case is just a special case of the complex numbers [where the imaginary part vanishes] so there is nothing wrong in saying it is complex and it happening to be real...it's just that the extra operation of complex conjugation * is trivial.

 

[Buzz Bloom]

Hi ChrisVer:

I made a correction at https://en.wikipedia.org/wiki/Hermitian_matrix . I hope it sticks.

Regards,
Buzz

 

[Orodruin]

Hi ChrisVer:

I made a correction at https://en.wikipedia.org/wiki/Hermitian_matrix . I hope it sticks.

Regards,
Buzz

The statement on Wikipedia was not wrong and should be reverted. Real numbers are a subset of complex numbers as Chris pointed out. What we argued against was your assertion that there had to be elements which were not real numbers. Do not edit Wikipedia unless you are 100% sure of what you are doing and have expertise in the field.

 

[Buzz Bloom]

Hi Orodruin:

The statement on Wikipedia was not wrong and should be reverted.

I agree with you completelty about the math. I found the original phrasing ambiguous and unnecessarily confusing, although correct mathematically. It seemed to suggest that the defintion of Hermitian implied at least one non-real component.

The discussion says clearly, "The diagonal elements must be real," and "Hence, a matrix that has only real entries is Hermitian if and only if it is a symmetric matrix, i.e., if it is symmetric with respect to the main diagonal. A real and symmetric matrix is simply a special case of a Hermitian matrix."

However, I think that many who did not already know what a Hermitian matrix was, and who read only what appears to be the definition in the first sentence would make the wrong interpretation. I think my revision avoids this ambiguity. I think it is unreasonable to require someone reading about a mathematical term to read an entire article to understand the definition of the term when one sentence can be sufficient.

Thanks for your post,
Buzz

 

[Orodruin]

However, I think that many who did not already know what a Hermitian matrix was, and who read only what appears to be the definition in the first sentence would make the wrong interpretation. I think my revision avoids this ambiguity. I think it is unreasonable to require someone reading about a mathematical term to read an entire article to understand the definition of the term when one sentence can be sufficient.

I strongly disagree. The definition was not the least bit unclear. A hermitian matrix is a matrix with complex entries which is its own hermitian conjugate. There is nothing ambiguous about that. The definition makes it perfectly clear that any matrix which satisfies this is hermitian. Real numbers are a subset of complex numbers and you should expect anyone who reads about hermitian matrices to know this. Therefore, a real and symmetric matrix is going to be hermitian. Saying that the elements "may be" complex is only adding confusion. In my opinion, you have destroyed a perfectly fine opening to a Wikipedia article.

Information on Wikipedia should be accurate and precise, which the original was. You should not edit it while learning a subject just because you think it would be more pedagogical in a different way. In general, people with significantly more experience in communicating the subject are going to have written the entries in the first place.

 

[Buzz Bloom]

Hi Orodruin:

You should not edit it while learning a subject just because you think it would be more pedagogical in a different way.

When I made the change I also added my reasons for the change to the talk page. If the more experienced people maintinaing Wikipedia articles agree with you, they will undo my change. I think they might possibly agree with me that pedagogical considerations are very important, and yet disgree with me that the article would benefit from the pedagolically oriented change I made, or that one was necessary in the original text -- or maybe not with respect to any combination of these possibilities.

I appreciate your sharing your thoughts about this with me.

Thanks for your post,
Buzz

 

[ChrisVer]

it is indeed a very bad mistake to use "may be" in a definition... it raises the ambiguity, when definitions should be fair and square...someone can say "then there might be the case that it is not be a complex number=>what is it?"... and strictly speaking "not a complex number" would also rule out the real numbers . The distinction between real and something else, is with real vs imaginary, and not real vs complex.

 

[Orodruin]

Just because you misunderstood the definition does not mean that someone else will. In my mind, anyone who is well versed in the terminology of complex numbers should get the definition correctly. Add on top of that the reasons given by Chris and you should realize that the change is a very very bad idea. This is the key part:

The distinction between real and something else, is with real vs imaginary, and not real vs complex.

 

[Buzz Bloom]

Hi ChrisVer and Oradruin:

You have convinced me that my pedagogical change can be impoved. I have added a word as follows:

In mathematics, a Hermitian matrix (or self-adjoint matrix) is a square matrix that may have non-real complex entries, and that is equal to its own conjugate transpose​

I do not see in what way this new definition is still ambiguous. I am hopeful that the Wikipedia people who look at my current change will agree that compared with the original text some pedagogical change would be helpful. If they see any ambiguity in my text, I hope they will improve it.

Thanks for your discussion,
Buzz

 

[Orodruin]

Please just reverse it to what it was before. Even textbooks will give the definition that was there before. This new version of yours is even making it worse. It is completely unintuitive what "may have non-real complex components" means and the word "may have" has nothing to do in a definition as remarked by Chris. And next time you consider making a change, check the exact text with someone who is experienced on the subject before making the change.

 

[Vanadium 50]

Buzz, yesterday you didn't know what a Hermitian matrix was (and I am not certain you do even today). I don't understand why people feel compelled to edit Wikipedia on subjects that they are new to, but all you have done is left a mess on Wikipedia for someone else to clean up. You should revert it - the original definition was fine, and your new one is work.

 

[Orodruin]

In addition, you are quoting this very thread in the talk page as the reason for your edit. This could seriously damage the reputation of Physics Forums. Since you seem unwilling to revert the edit yourself, I am going to do it. Please stop doing things like this.

 

[vanhees71]

I don't understand the problem. A finite-dimensional complex matrix $\hat{M}$ is called Hermitean iff $\hat{M}^{\dagger}=\hat{M}$. In finite-dimensional unitary spaces a linear operator is self-adjoint iff it is Hermitean. That's it.

It's more complicated when it comes to hermitizity and self-adjointness in infinite-dimensional Hilbert spaces. There you have to distinguish between the weaker Hermitizity from the stronger self-adjointness.

 

[Orodruin]

I don't understand the problem.

The problem is that Buzz is equating "complex" with "not real" (i.e., having imaginary part not equal to zero).

 

[Buzz Bloom]

Hi @Vanadium 50 and @Orodruin:

I appologize for offending you. I am glad Orodruin restored the original text. You are probably not interested in my reasons for not doing it myself, and they are not particularly important anyway.

I very much enjoy, and learn from, the discussions in the PF. It never occurred to me that my referring on the Talk page about my being corrected in the PF might harm the PF. If you think it important to do so, I will edit that page and remove that reference.

Regretfully,
Buzz

 

[ChrisVer]

I didn't feel offended...(I hope nobody did- "speaking for others" )
What everyone has been trying to tell you (even after your 2nd correction) was that the original definition was better (and more correct) , and for that reason it should remain unchanged.
As for PF, as a result of dropping the quality of the definition (and for that referring the PF), you are also dropping the quality of PF...And nobody in here said that the original definition was wrong...so we didn't "correct" the definition, but we corrected your misinterpretation of it...

and as a final note: keep your thread on topic (this has gone a bit astray)...it's yours and you should protect it. The conversation about the hermitian conjugation and all that followed, is not helping in that...

 

[Buzz Bloom]

New Question

Quote from
https://en.wikipedia.org/wiki/Pontecorvo–Maki–Nakagawa–Sakata_matrix#Parameterization .

The PMNS matrix is not necessarily unitary and additional parameters are necessary to describe all possible neutrino mixing parameters, in other models of neutrino oscillation and mass generation, such as the see-saw model, and in general, in the case of neutrinos that have Majorana mass rather than https://en.wikipedia.org/w/index.php?title=Irac_fermion&action=edit&redlink=1 .​

Does the quote mean that the PMNS mass mixing matrix includes values for probabilites that take into account both of two currently undecided possibilites regarding the nature of a neutrino's mass type: Majorana or Dirac?