Gauge eigenstates vs. Mass eigenstates


by Kontilera
Tags: eigenstates, gauge, mass
Kontilera
Kontilera is offline
#1
Sep27-13, 04:01 AM
P: 101
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?

Please, if you have the energy to write some word about this I would be thankful.

Best Regards
Kontilera
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tiny-tim
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#2
Sep27-13, 04:19 AM
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Hello Kontilera!
Quote Quote by Kontilera View Post
The mass eigenstates are such that the mass term coefficients matrices are diagonal. But how do we define gauge eigenstates to begin with?
For flavour eigenstates, see eg http://en.wikipedia.org/wiki/Neutrin...3Sakata_matrix
andrien
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#3
Sep28-13, 09:18 AM
P: 977
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.

dauto
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#4
Sep28-13, 10:16 AM
P: 1,244

Gauge eigenstates vs. Mass eigenstates


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".
Kontilera
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#5
Sep29-13, 03:28 AM
P: 101
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?
andrien
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#6
Sep29-13, 04:55 AM
P: 977
Quote Quote by Kontilera View Post
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?
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.
kurros
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#7
Sep30-13, 04:18 AM
P: 313
Quote Quote by Kontilera View Post
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?
Quote Quote by andrien View Post
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.
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.
kurros
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#8
Sep30-13, 05:57 PM
P: 313
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.


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