Neutrino theory regarding rest masses

[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.

 

[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.