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## is the electric charge of bosons w1 w2 w3 well defined?

Hi

In GSW theory we start with two fermions f f' (charge 0 -1) and 3 bosons w1 w2 w3.
the charge of the bosons in only introduced when one mix w1 and w2 giving w+ and w-.
Does it mean that before symmetry breaking f and f' were exchanging w+/w- but not
the basic w1 w2 w3?
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 Recognitions: Gold Member I know gluons have no electric charge. In Gut Theories using other groups are there other electric charged bosons? Must we then use the same trick (W1 - iw2 style) to define them as electric charged?
 Mentor 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.

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## is the electric charge of bosons w1 w2 w3 well defined?

All i consider is before symmetry breaking. (fermion and jauge bosons)
Look at the lagrangian http://Before_Electroweak_Symmetry_Breaking
the massless neutrino and electron are eigenvectors of T3 and Y
But what can we say about the W1 W2 W3? ie for T3 and Y?

 Quote by naima Hi In GSW theory we start with two fermions f f' (charge 0 -1) and 3 bosons w1 w2 w3. the charge of the bosons in only introduced when one mix w1 and w2 giving w+ and w-. Does it mean that before symmetry breaking f and f' were exchanging w+/w- but not the basic w1 w2 w3?
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 $L_f$ 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 $L_h$)

 Quote by naima All i consider is before symmetry breaking. (fermion and jauge bosons) Look at the lagrangian http://Before_Electroweak_Symmetry_Breaking the massless neutrino and electron are eigenvectors of T3 and Y But what can we say about the W1 W2 W3? ie for T3 and Y?
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 :).

Mentor
 Quote by naima All i consider is before symmetry breaking.
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.
 Recognitions: Gold Member interesting answers. It is the first time i read that the charge appeared with symmetry breaking. Could you elaborate or give a link? thanks
 Mentor 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.

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I disagree with that (the charge of the leptons before Symmetry Breaking).

I'll quote not wiki but the handbook "Quantum Field Theory" by Itzykson Zuber p620:

THE WEIBERG SALAM MODEL
 The left helicity component of the charged lepton eL and its neutrino nue are grouped into a column matrix Le This suggests the introduction of a group of leptonic isospin for which Le is a doublet while the right component eR is a singlet. A leptonic hypercharge is also assigned to each of these fields in such a way that the analog of the Gell-Mann and Nishijima rule is satisfied: Q = T3 +Y/2 The left doublet has Y = -1 and the right singlet Y = -2. T and Y commute. Therefore the transformation group is SU(2) x U(1) We then construct a gauge theory with this invariance group, involving a triplet of gauge fields A for SU(2) with a charge g and a field B for U(1). The U(1) coupling constant wil be denoted g'/2. Since we want a single gauge field (the photon) to remain massless after the spontaneous breaking, we introduce a doublet of complex of scalar fields phi+ and phi0
This is not ambiguous except for the 3 bosons.
the fermions charge are defined (0 and -1) and a coupling constant is just defined for the bosons.
The title of the post is linked to this point
 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.
 Recognitions: Gold Member You do not comment Itzykson handbook. He says that the left charged leptons are in a doublet (T3 = -1/2 , 1/2) and that a Y = -1/2 hypercharge is assigned to them. This give Q = 0 , -1 the "official" charge after symmetry breaking. perhaps this is meaningless, perhaps it is not
 Recognitions: Gold Member According to you, what were the conserved "things" (for leptons and their jauge bosons) , when leptons interacted before symmetry breaking? In fact this is what i try to understand.

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 Quote by naima I disagree with that (the charge of the leptons before Symmetry Breaking).
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.
 Recognitions: Gold Member This is not a question of disbelief. I am studying handbooks and some ones introduce charges and mixing angles before spontaneous breaking. that's all. Please try to answer the question about the physics before SB. thanks a lot.
 Recognitions: Gold Member I found http://en.wikipedia.org/wiki/Electroweak_epoch. You can see that before Symmetry Breaking, cosmology says there were long range W, Z (the Z use the theta mixing angle) .

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 Quote by kurros 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.
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.