Is the electric charge of bosons w1 w2 w3 well defined?

You're starting from a very bad place if you have fermions in eigenstates of the broken symmetry but are discussing the fields of the unbroken symmetry.

 

Disclaimer: I have not thought about this for a while so apologies if I screw it up :)

Ok, so you are talking about the electroweak lagrangian but just the bit with electron and electron neutrino say, not any other fermions? So, on the wikipedia page you linked to there is the lagrangian [itex]L_f[/itex] in the "before electroweak symmetry breaking" section. The last two terms in this lagrangian describes the interactions between the left handed SU(2) doublet (electron and neutrino) and gauge bosons, and the right handed SU(2) singlet electron and gauge bosons, respectively.

For the doublet, there are seperate terms for each of the w1 w2 w3 and B fields if you expand this term, because the covariant derivative is just a linear thing with a piece for each field. So it is these basic fields that are being exchanged. For the singlet, the w1 w2 w3 fields are not in there, just the B field, because being an SU(2) singlet it has no SU(2) gauge fields in it's covariant derivative.

Now, the confusing thing is that during symmetry breaking, it is the w3 and B fields that mix into the Z and A (photon). So the w1 and w2 fields are really the same fields as W+ and W-, which is why they use the same names for them both in the wikipedia article. The only trick is that they pick up a mass term from the higgs kinetic term (first term in [itex]L_h[/itex])

Well being eigenvectors of T3 and Y means their weak isospin and hypercharge are unchanged when acted on by W3 and B, which since Z and A are linear combinations of W3 and B means Z and A also do not change these things, thus Z and A are neutral current interactions. Or you could say that the electron and neutrino are also eigenstates of the newly created U(1)em, and they are eigenstates of the screwed up SU(2) that remains.

W1/W+ and W2/W- screw with the SU(2) eigenstate because they are raising and lowering operators really, which is why they turn electrons into neutrinos and vice versa. They move you around in the SU(2) space. But after symmetry breaking the electrons get a mass, so the electrons/neutrino doublet is not a proper SU(2) doublet anymore, since you will notice if you rotate this doublet in SU(2) space. Previously the electron bit looked just the same as the neutrino bit. I think. Not sure what happens with the electric charge, I must think about it.
Anyway, the W+ and W- will still move you in this approximate SU(2) space just the same, so you can still change electrons into neutrinos with them.

Ok that last paragraph is a bit of a mess, please someone who is better at group theory come along and explain properly what my mess is trying to get at :).

 

No - you have given the fermions electric charge. Before symmetry breaking, you don't have electric charge. My experience is that you really need to work in one or the other basis unless you really, really know what you are doing.

 

Before symmetry breaking you don't have a photon. The right representation is not the A potential.

 

The wiki page you link to alludes to this without explaining it terribly much:

"In the Standard Model, the W± and Z0 bosons, and the photon, are produced by the spontaneous symmetry breaking of the electroweak symmetry from SU(2) × U(1)Y to U(1)em, caused by the Higgs mechanism (see also Higgs boson).[3][4][5][6] U(1)Y and U(1)em are different copies of U(1); the generator of U(1)em is given by Q = Y/2 + I3, where Y is the generator of U(1)Y (called the weak hypercharge), and I3 is one of the SU(2) generators (a component of weak isospin)."

I.e. you do not have Q (electric charge) until you have U(1)em, which is only a useful thing to talk about after symmetry breaking happens. Your leptons do however have Y (hypercharge), which tells you things about how they interact with the B gauge bosons, and is basically an exact copy of how electric charge works, but it is in fact different.

 

You can define the electric charge before symmetry breaking if you really want to, the U(1)em subgroup still exists in the fundamental SU(2)_L x U(1)_Y group, however there are lots of other U(1) subgroups that you could have picked instead and nothing to distinguish them from each other. So the electric charge is meaningless.

I refer you again to the wikipedia page. Notice in the "after electroweak symmetry breaking" lagrangian there is a piece called the neutral current, L_N. The electromagnetic current part of that contains the electric charge, and is what defines how the fermions interact with photons, which is the whole point of electric charge. There is no such term in the lagrangian before symmetry breaking, so the electric charge, although you can "define" it just the same, doesn't appear in any term in the lagrangian so you may as well not talk about it.
There IS a totally analogous piece which defines the hypercharge current, so the interaction between the fermions and the B field, in which the hypercharge appears playing the same role as electric charge.

 

Well, OK, but if you're having trouble with something, and that's the root of it, the disbelief is unlikely to be much help.

 

the coupling constants of _L and _Y fix the U(1), because you want it to be a non-chiral force. I guess that you could forgot about symbreaking and just define the photon in this way.

Now, perhaps Vanadium could give more details about how then the complex representations of the SU(2)_LxU(1)_Y group combine to allow to extract a real representation of U(1)_EM. I'd try myself, as really I am interested on the details of this discussion, but I have 8 hours flight in a couple hours, plus jetlag...

I am particularly worried, once the representation issue is clarified, about the charges of massless winos.

 

This is just a semantics issue. He refers to the doublet as the charged lepton doublet because it is the doublet that ends up having a charged lepton in it. It remains the same object, but it's properties change with symmetry breaking.

Hmm, I can't remember how that works, must go look...

The leptons and quarks are in the complex representation of U(1)_EM I believe, isn't the real one the 2x2 representation?

I think this is just the same thing, i.e. they have no electric charge when they are massless because it is before symmetry breaking. In fact they can have a mass and still no electric charge because they have mass terms coming from the soft supersymmetry breaking part of the lagrangian as well as from the higgs sector. I assume you are talking about MSSM winos here.

 

Let me try this again.

You need to decide whether you are working in the broken or the unbroken symmetry basis. (Colloquially, "after" or "before" the break) This is exactly analogous in mechanics to picking your coordinate system. Mixing the basis is exactly as bad as mixing coordinates systems; you might get the right answer, but it would surely be accidental.

 

"So to summarize, yes, the electron is charged before the break and if you look in the right direction of weak isospin space, the left and right components have the same electrical charge."

Well, sure that is correct enough, but I think it is a bit misleading. The electromagnetic force has not separated from the weak force before symmetry breaking. Consider that he says "if you look in the right direction of weak isospin space". That's fine, but before symmetry breaking any direction in weak isospin space is as good as any other, so you really have no reason to pick out "the right direction". The bit that is electromagnetism is just a little piece of a larger group, and that larger group is the force you should be concerned about before symmetry breaking.

In additional, the left and right handed pieces of the electron are quite separate things before symmetry breaking since there is no mass term to mix them together.

 

My jetlag keeps ongoing, plus a flu I surely got in Dubai hub. But anyway I am interested on knowing why I am wrong; I thought that the electromagnetic field was perfectly defined by asking it to couple vector to the electron field and zero to the neutrino field. People keeps telling that there is "lots of other U(1)", could you please parametrize these lot?

 

My understanding is that the breaking of a gauge symmetry can never create charges but only destroy them. The charge superselection sectors form a group under the combination of charges which is dual to the (unbroken part of the) global gauge group (Dopplicher Haag Ruelle theory). Hence breaking the gauge group will result in a merging of charge states but cannot create new ones.