Neutral intermediate boson to neutrinos are flavor preserving

[physciencer]

Hey guys!

How can this sentence be explained and what does it mean?

If the couplings $$Z\nu_{\alpha}\bar{\nu}_{\beta}$$ of the neutral intermediate boson to neutrinos are flavor preserving, they are also diagonal when expressed in terms of mass eigenstates $$Z\nu_{i}\bar{\nu}_{j}.$$

 

[Simon Bridge]

Welcome to PF;
where did you see this sentence?

But it pretty much means what it says. What's the problem.

 

[physciencer]

Thank you Mr Simon! It is in https://www.amazon.com/dp/0201328321/?tag=pfamazon01-20 this book. But I can't see it mathematically speaking. Mind explaining it in mathematical terms, like if there is a proof that I can read to picture this better? In other words, how does it come about?

 

[Simon Bridge]

Note: the context has to be accessible to me. Presumably you don't expect me to buy the book and skim through 300+ pages just to understand your question?
There's a second motivation in getting you to spell out the context - the act of trying to explain a problem to someone else so they understand it often leads to an insight that solves the problem.

Do you not understand some of the words?
What level of education does the explanation need to be pitched at?

 

[physciencer]

I can't find available pdf online. You said that it says what it says, so I assumed that this is clearly written. So if I want to restate my question, why is it that if the couplings of the neutral intermediate boson to neutrinos are flavor preserving, then they are also diagonal when expressed in terms of mass eigenstates?

Edit: Explanation of whatever you feel comfortable at. Graduate level explanation perhaps?

 

[Simon Bridge]

The sentence, by itself, describes an interaction - it does not say that one of the properties leads to the other one, just that the properties are properties of the interaction that will also be described in the resulting lagrangian.

Basically it means that you get the neutrinos out that you put in, and that the interaction provides a basis in which the M (in $\phi^\dagger M \phi$) is diagonal.

If it means anything else, that will be in the context.
You have the text, you can tel me what the chapter is about (i.e. neutrino oscillations) and you can reproduce the paragraph that it appears in just by typing it out ... whatever you think best describes the context that the author has in mind. Otherwise it is impossible to go into detail.

 

[physciencer]

So this was written at the end of the chapter and was part of questions the author thought was intuitive to answer because he didn't refer it to any paragraph or section in the book. In other words, nothing was attached to this question. So the Chapter was Electroweak Interactions of Leptons and I guess the author is thinking that one can answer this according to total understanding of the chapter (perhaps). I will list the subtitles of chapter 6 hoping they would clear what is the chapter handling.
An Effective Lagrangian for the weak interactions
Intermediate vector bosons: A first look
The standard electroweak theory of leptons
neutral-current interactions among leptons
The higgs Boson: A first Look
The Higgs boson, Asymptotic behavior, and the 1 TeV scale
Neutrino Mixing and Neutrino Mass
Renormalizability of the theory

I hope this will help in clearing things up.

 

[Orodruin]

This is a trivial matter of a basis change. If the interaction is diagonal with the same strength for all flavours, then it is proportional to the unit matrix. Changing basis will not affect the unit matrix. If the strengths are different, this is no longer true.

 

[physciencer]

May you please explain this using mathematical terms and how mass eigenstates practically enter? Why is it if they are flavor preserving they are also diagonal when expressed in terms of mass eigenstates? @Orodruin

Edit: Is there a passage in some books that explains what is going on here because as I mentioned the book I am using does not really explain this or where did it pop up from. If you could suggest a read for me to understand this better?

 

[Orodruin]

There really is not much more to understand than the fact that the identity operator is left invariant under a unitary change of basis. This should be in any basic textbook on (complex) linear algebra.

 

[physciencer]

How does this relate to mass eigenstates or mass operator?It is probably this that I can't link? @Orodruin

 

[ChrisVer]

Because you can make a change in your basis and go from flavor eigenstates to mass eigenstates.

 

[physciencer]

Then why is that when flavors are of the same strength then the matrix should be proportional to identity?

 

[Orodruin]

Then why is that when flavors are of the same strength then the matrix should be proportional to identity?

Because the interaction is based on electroweak gauge theory, which is where the coupling strength comes from.

 

[physciencer]

So electroweak theory says if flavors are of the same strength then the representative matrix is identity? If so, why does it say so?

 

[Orodruin]

Basically because all of the neutrinos are in the same type of SU(2) multiplet and the coupling strength comes from the covariant derivative.

 

[ChrisVer]

Also if flavor is conserved, can't you think of the interaction Hamiltonian commutes with the mass matrix? so they both can be simultaneously diagonalized?

 

[physciencer]

Please elaborate, was that a question or an answer? If an answer, why would the interaction Hamiltonian commute with the mass matrix if the flavor is conserved. I can't relate.

 

[Orodruin]

Also if flavor is conserved, can't you think of the interaction Hamiltonian commutes with the mass matrix?

Not necessarily, it would not be the case if flavour was conserved but different flavours had different coupling strengths. If they have the same coupling strength, then yes, everything commutes with the identity operator.

 

[physciencer]

But you were already talking about strength rather than flavors. I am confused now.. What happened? Is it that conserved flavors make the matrix proportional to identity matrix or is it not? :(

 

[TEFLing]

Maybe it would be worthwhile to plod through and define every word...

Neutrinos have mass eigenstates with well defined masses...

But the neutrinos that come out of a Weak Force interaction are NOT in mass eigenstates ... They are in FLAVOR eigenstates ( electron muon tauon ) depending upon the other leptons in said interaction

The flavor eigenstates are superpositions of the mass eigenstates

Note I think the opposite is true as well, if the neutral boson interaction is not diagonal in flavor then it is also not diagonal in mass ?

 

[Orodruin]

No, conserved flavours does not imply proportional to unity, conserved flavour along with equal coupling does. My last post was a response to Chris' question.

 

[Orodruin]

Note I think the opposite is true as well, if the neutral boson interaction is not diagonal in flavor then it is also not diagonal in mass ?

No, this is not true. You could perfectly well have an interaction which is diagonal in one basis but not in another, just not the normal electroweak gauge interaction, which is proportional to the unit matrix in both bases.

 

[physciencer]

But the problem doesn't talk about equal couplings?

 

[physciencer]

This is a trivial matter of a basis change. If the interaction is diagonal with the same strength for all flavours, then it is proportional to the unit matrix. Changing basis will not affect the unit matrix. If the strengths are different, this is no longer true.

The question was: Why is it if the couplings of the neutral intermediate boson to neutrinos are flavor preserving then ...
They did not say any thing about the "same strength" you mentioned in what I quoted. Did you assume that to make the sentence hold?

 

[Hepth]

If you write a Lagrangian for the couplings of the Flavor-States of the neutrinos to the Z it would be something like :

$L = Z \nu_{i} V^{ij} \bar{\nu}_{j}$

where V is the coupling-strength constants in matrix form.

if there is NO FLAVOR VIOLATION, meaning the interaction is FLAVOR PRESERVING, then there are NO OFF-DIAGONAL elements of $V^{ij}$.

If all 3 are the same "Strength" then you can write it as the identity matrix times a constant.
If NOT, then you have a matrix that is like {{a,0,0},{0,b,0},{0,0,c}}

I was going to respond that its trivial to show that under a unitary transformation this leaves the couplings diagonal, but now I am not so sure...

Anyone have a solution?

I really don't think it implies that the mass-eigenstates are diagonal in coupling constants at all. A unitary transformation does not conserve this, and you get flavor-state violation between the mass eigenstates where the off-diagonal elements are proportional to the difference between the strengths of the couplings for each flavor state. Right? (everyone else)

 

[Orodruin]

I really don't think it implies that the mass-eigenstates are diagonal in coupling constants at all. A unitary transformation does not conserve this, and you get flavor-state violation between the mass eigenstates where the off-diagonal elements are proportional to the difference between the strengths of the couplings for each flavor state. Right?

Correct. This is also seen directly for the vacuum Hamiltonian of neutrino oscillations. You then have a Hamiltonian which is diagonal in mass basis, but with off diagonal elements proportional to the differences in flavour basis.

They did not say any thing about the "same strength" you mentioned in what I quoted. Did you assume that to make the sentence hold?

If they did not, then they should have or they made that assumption implicitly, since this is a prediction from electroweak theory (at least at tree level).

 

[ChrisVer]

if you make a unitary transformation to the neutrino : $\nu \rightarrow U \nu$ and $\bar{\nu} \rightarrow \bar{\nu} U^\dagger$

So

$Z \bar{\nu} V \nu \rightarrow Z \bar{\nu} (U^\dagger V U ) \nu$

The thing is that $U^\dagger V U$ is still diagonal if V is diagonal...?