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Custodial Symmetry

  1. Mar 21, 2009 #1
    I have never really understood the approximate 'Custodial symmetry' in the Standard Model. I've seen it being described in many texts, but I can't seem to be able to put my finger on it.

    Would somebody please write down the transformation law for the Higgs fields under a 'custodial SU(2) transformation?' It would really help if I got that!
  2. jcsd
  3. Mar 22, 2009 #2


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    custodial isospin in the electroweak interaction is defined many ways, depending on what you want to do with it (they're all effectively the same up to field redefinitions). but a good starting place is the following: posit a GLOBAL symmetry:

    SU(2)_L x SU(2)_R

    Now gauge the SU(2)_L and identify it with the standard model gauge group, but only gauge the U(1)_R subgroup of the SU(2)_R part, and identify that with hypercharge. The special treatment of the U(1)_R explicitly breaks the R part of the symmetry, but if we turn off that special treatment (let g'=0) then the R symmetry is restored (up to the Yukawa couplings). So one can do a spurion analysis with g' and the fermion yukawa couplings.

    To the extent that g' is small, this describes the standard model. The SU(2)_R is the "custodial isospin" symmetry (sometimes it is the SU(2)_D, but I usually use the former in my research; as I said, they're the same up to field redefinitions).

    If you want to make the custodial isospin symmetry manifest, you can let the Higgs transform as a bifundamental of the L-R symmetry:

    [tex]H\rightarrow LHR^\dagger[/tex]

    This symmetry is realized in the standard model if you group the right-handed fermions into doublets of the SU(2)_R symmetry and let g'=0. If you have g' nonzero, this symmetry is only realized if R=1.

    Custodial isospin symmetry is very important for EW precision measurements - it ensures that the W-Z mass ratio (called [itex]\rho[/itex]) cannot get too large, for example. Its corrections must be proportional to g' and yukawas, especially the top quark yukawa coupling. This is a well known result.

    The extent to which custodial isospin symmetry is broken is a very powerful test for physics beyond the standard model.

    Hope that helps.
    Last edited: Mar 22, 2009
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