The breaking of the flavour permutational symmetry

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In summary, the article discusses how to generate mass for the first two generations of quarks by breaking the flavour permutation symmetry S3. The top of the page 5 of the pdf link provides some additional information about the symmetric group's representations.
  • #1
btphysics
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Hello,I read an article about fo generating mass for the first two generations of quarks by breaking the flavour permutation symmetry S3. In the top of the page 5 of the following article: http://arxiv.org/pdf/hep-ph/9807214v2.pdf ,
they mention that "Mq1 transforms as the mixed symmetry term of the doublet complex
4tensorial representation of the S(3)d diagonal subgroup of SL(3) ⊗ SR(3)". What does this means? What transformations on the left and on the right leave the following matrix invariant:

[tex] \begin{pmatrix} A1 & iA2 &-A1-iA2\\ -iA2& -A1 &A1+iA2\\ -A1+iA2&A1-iA2&0 \end{pmatrix}[/tex]

Best regards.
 
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  • #3
Thanks for the the answer. In that article they give the same explanation, and the matrix is almost equal. I continua to don´t see any clues about what means that expression and moreover under what transformations that matrix is invariant.
 
  • #4
Please please guys never link to the PDF
 
  • #5
I preferred that paper because it gives some extra information on the symmetric group's representations.
 
  • #6
ChrisVer said:
I preferred that paper because it gives some extra information on the symmetric group's representations.
The paper might be fine, but please link to the abstract, not the pdf. There is no point in loading the whole pdf just to read the abstract (=what most readers will do).
 
  • #7
mfb said:
The paper might be fine, but please link to the abstract, not the pdf. There is no point in loading the whole pdf just to read the abstract (=what most readers will do).

:uhh: well I didn't send the paper to the OP to read the abstract...but OK for the rest readers you are right...
 
  • #8
Returning to the OP, that should be an easy problem to do if one has some computer-algebra software like Mathematica. I did it in a couple minutes with Mma. Should I give my solution here?
 
  • #9
lpetrich said:
Returning to the OP, that should be an easy problem to do if one has some computer-algebra software like Mathematica. I did it in a couple minutes with Mma. Should I give my solution here?
Perhaps even the Mathematica code, better than the solution :)
 
  • #10
Links to the abstracts of the papers:
[1304.6644] Quark sector of S3 models: classification and comparison with experimental data
[hep-ph/9807214] The breaking of the flavour permutational symmetry: Mass textures and the CKM matrix

GIven matrix A in the OP, I think that btphysics wants a matrix T such that T.A.T^(-1) = A. Multiply on the right by T and one gets T.A = A.T, and that is a bunch of linear equations in the components of T.

My Mathematica code:
Code:
(* The OP's matrix -- it is Hermitian *)
amat = {{A1,I*A2,-A1-I*A2},{-I*A2,-A1,A1+I*A2},{-A1+I*A2,A1-I*A2,0}};
avars = {A1,A2};
(* Decomposition assuming that A1 and A2 vary independently *)
amlist = D[amat,#]& /@ avars;
(* Matrix that will leave it invariant *)
trmat = Array[t,{3,3}];
(* This ought to be zero *)
conds = (trmat.# - #.trmat)& /@ amlist;
(* Find the transformation matrix *)
trmsol = trmat /. ToRules[Reduce[Thread[Flatten[conds]==0]]];
SparseArray[trmsol - ((t[3,3]-t[3,2])*IdentityMatrix[3]+t[3,2])]
So trmat has form a*I + b.

The original matrix has form
$$
\{A1, A2\} \cdot
\left\{\left(
\begin{array}{ccc}
1 & 0 & -1 \\
0 & -1 & 1 \\
-1 & 1 & 0 \\
\end{array}
\right),\left(
\begin{array}{ccc}
0 & i & -i \\
-i & 0 & i \\
i & -i & 0 \\
\end{array}
\right)\right\} $$
(Output courtesy of Mathematica TeXForm)

I found its eigenvalues with Eigenvalues[amat]
$$ \left\{0,-\sqrt{3} \sqrt{\text{A1}^2+\text{A2}^2},\sqrt{3}
\sqrt{\text{A1}^2+\text{A2}^2}\right\} $$
 
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  • #11
Sorry for only answer now, but last weeks i had some personal issues. I am much grateful for your replies, and i have already understand the problem leptrich, you understood my doughts. With this type of matrix next one can generate mass for the 1st generation and even sufficient CP violation.
 

What is the flavour permutational symmetry?

The flavour permutational symmetry refers to the principle that the fundamental interactions of particles do not depend on the specific flavour of the particles involved. This means that, for example, the interactions between a down quark and an up quark are the same as the interactions between an up quark and a down quark.

How does the flavour permutational symmetry play a role in particle physics?

The flavour permutational symmetry is an important concept in particle physics because it helps us understand the fundamental interactions between particles. It allows us to simplify complex interactions and make predictions about particle behavior.

What is "the breaking" of the flavour permutational symmetry?

"The breaking" of the flavour permutational symmetry refers to situations where the symmetry is not strictly observed. This can happen due to various factors such as mass differences between particles or the effects of the Higgs field.

Why is the breaking of the flavour permutational symmetry important in particle physics?

The breaking of the flavour permutational symmetry is important because it allows us to explain some observed phenomena that cannot be explained by a strict adherence to the symmetry. For example, it can help explain the different masses of particles and the origin of CP violation.

How do scientists study the breaking of the flavour permutational symmetry?

Scientists study the breaking of the flavour permutational symmetry by conducting experiments, analyzing data, and developing theoretical models. They also use tools such as particle colliders and detectors to observe and measure the behavior of particles and their interactions.

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