# Gauge eigenstates vs. Mass eigenstates

1. Sep 27, 2013

### Kontilera

Hello fellow physicsforumists.
I am currently looking at the standard model and one of the key ingridients is to rotate the gauge eigenstates to the mass eigenstates by a transformation acting on their family index. The problem is that I can't really see what we are doing.

The mass eigenstates are such that the massterm coefficients matrices are diagonal. But how do we define gauge eigenstates to begin with?

Best Regards
Kontilera

2. Sep 27, 2013

### tiny-tim

3. Sep 28, 2013

### andrien

gauge eigenstates are unphysical.Each quark in standard model has right handed components because they are massive.Lagrangian is first written in terms of the doublet and singlet fields which contain these unphysical quarks which are termed as gauge quarks or sometimes gauge eigenstates.When you use yukawa coupling to give masses to quarks,a mass matrix is generated.these determine the masses and flavour mixing of quarks.The quark fields used before(SSB) are unphysical gauge eigenstes,you have to find the physical or mass eigenstates by transforming the quark mass matrices into diagonal form.

4. Sep 28, 2013

### dauto

The unphysical gauge eigenstates are defined based on the other fields that will be connected to it through the gauge interaction, so for instance there is a term like uwd' in the Lagrangian which represents the interaction of a u quark with a W boson and a d' unphysical quark. The d' along with the s' and d' can be related to the physical (mass eigenstates) fields d, s, and b through a "rotation".

5. Sep 29, 2013

### Kontilera

Thanks for the answers. I helped me some but I'm not sure I understand why they are refered to as gauge eigenstates.. is there an eigenvalue equation?

6. Sep 29, 2013

### andrien

They are more commonly referred as gauge quarks,gauge eigenstate is just misnomer.They are written like like that because they relate to physical mass eigenstates.

7. Sep 30, 2013

### kurros

I'm a little rusty on my group theory, but I don't think it's a misnomer. The relevant eigenvalue equations are the ones where the operator is a Cartan generator of one of the gauge groups right? Or some such thing. For instance if we consider SU(2), then before symmetry breaking the 3 (massless) gauge bosons are eigenvectors of the diagonal SU(2) generator, in the 3x3 (I think this is the adjoint?) representation. Similarly the left handed (massless) fermions are eigenvectors of the same generator but in the 2x2 (fundamental?) representation.

8. Sep 30, 2013

### kurros

Err actually maybe it is only the fermions that work like that; I am pretty sure they at least are all eigenvectors of some gauge-group related operator or another.