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The breaking of the flavour permutational symmetry

  1. Sep 16, 2014 #1
    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.
     
  2. jcsd
  3. Sep 17, 2014 #2

    ChrisVer

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  4. Sep 17, 2014 #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.
     
  5. Sep 17, 2014 #4

    arivero

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    Please please guys never link to the PDF
     
  6. Sep 17, 2014 #5

    ChrisVer

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    I prefered that paper because it gives some extra information on the symmetric group's representations.
     
  7. Sep 17, 2014 #6

    mfb

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    Staff: Mentor

    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).
     
  8. Sep 17, 2014 #7

    ChrisVer

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    :uhh: well I didn't send the paper to the OP to read the abstract...but OK for the rest readers you are right...
     
  9. Sep 21, 2014 #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?
     
  10. Sep 21, 2014 #9

    arivero

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    Perhaps even the Mathematica code, better than the solution :)
     
  11. Sep 21, 2014 #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 (Text):

    (* 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]
    $$ \left\{0,-\sqrt{3} \sqrt{\text{A1}^2+\text{A2}^2},\sqrt{3}
    \sqrt{\text{A1}^2+\text{A2}^2}\right\} $$
     
  12. Sep 26, 2014 #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.
     
  13. Sep 26, 2014 #12

    arivero

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