Custodial Symmetry: Understanding the Transformation Law

  • Thread starter Thread starter TriTertButoxy
  • Start date Start date
  • Tags Tags
    Symmetry
TriTertButoxy
Messages
190
Reaction score
0
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!
 
Physics news on Phys.org
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:

H\rightarrow LHR^\dagger

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 \rho) 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:
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top