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

[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?

 

[naima]

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?

 

[Vanadium 50]

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.

 

[naima]

All i consider is before symmetry breaking. (fermion and jauge bosons)
Look at the lagrangian http://en.wikipedia.org/wiki/Electroweak_interaction#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?

 

[kurros]

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 separate 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 $L_h$)

All i consider is before symmetry breaking. (fermion and jauge bosons)
Look at the lagrangian http://en.wikipedia.org/wiki/Electroweak_interaction#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 :).

 

[Vanadium 50]

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.

 

[naima]

interesting answers.

It is the first time i read that the charge appeared with symmetry breaking.
Could you elaborate or give a link?

thanks

 

[Vanadium 50]

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

 

[kurros]

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.

 

[naima]

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 e_L and its neutrino nu_e are grouped into a column matrix L_e
This suggests the introduction of a group of leptonic isospin for which L_e is a doublet while the right component e_R 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

 

[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. 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.

 

[naima]

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

 

[naima]

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.

 

[Vanadium 50]

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.

 

[naima]

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.

 

[naima]

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)
.

 

[arivero]

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.

 

[kurros]

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

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.

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.

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

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...

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

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

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.

 

[naima]

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.

This was not semantics
I sent a mail to the french cowriter of my handbook. He was kind enoug to answer my question.

I translate the end:

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.

 

[Vanadium 50]

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.

 

[kurros]

This was not semantics
I sent a mail to the french cowriter of my handbook. He was kind enoug to answer my question.

I translate the end:

"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.

 

[naima]

Nobody answered arivero's questions (eg, he mentionned the charge of the winos..)
We'll wait for the end of his jetlag, He said he was interested on the details.
perhaps you could begin.

 

[naima]

the problem of massless electric charges was already discussed https://www.physicsforums.com/showthread.php?t=107673"
in this forum (2006)

 

[arivero]

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?

 

[DrDu]

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.

 

[arivero]

Hmm, ok, I am reviewing usual textbook materials and it is not obvious if/how you can extract the electromagnetic field.

The whole point is that it is not an arbitrary combination of weak isospin component $W^3_\mu$ and hypercharge $B_\mu$; it is just the combination that is proportional to $g_Y W^3_\mu + g_W B_\mu$ (note the interchange of the coupling constants). It should be possible to characterize such peculiar combination without invoking symmetry breaking.

I guess that part of the problem, or part of the solution, (and probably the origin of a heated debate between Distler and Woit, time ago) is that hypercharge is secretly chiral, because while it does not seem so in the Lagrangian, it has actually different charges for the left and right components of a spinor.

 

[naima]

remember that we start with charged particles (e is the value of the electric charge) and with 2 coupling constants g g'. you do not need breaking to get the angle:
e = g sin a = g'cos a

 

[arivero]

remember that we start with charged particles (e is the value of the electric charge) and with 2 coupling constants g g'. you do not need breaking to get the angle:
e = g sin a = g'cos a

No, we don't start with electromagnetically charged particles, they are an outcome. But of course, we should not even need the angle. We want just$$e = { g g' \over \sqrt {g^2 + g'^2}}$$. The question is how to argue for this outcome if we start from SU(2)xU(1) without looking to symmetry breaking.

 

[naima]

The weinberg angle is often found in the breaking paragraph but could you find
one handbook where you find something like
the electron has acquired a mass and a charge after the symmetry breaking?

 

[arivero]

The weinberg angle is often found in the breaking paragraph but could you find
one handbook where you find something like
the electron has acquired a mass and a charge after the symmetry breaking?

Different things, charge and mass. Mass is definitely, in the Standard Model, from the electron coupling to the higgs field and then it is adquired after the symmetry breaking, via the yukawa coupling. This is the thing you are going to find in any handbook.

Charge, is the thing we are discussing here in this thread. No doubt that hypercharge and weak isospin are defined independently of symmetry breaking. Some handbooks just postulate them, some others "deduce" them from the requirement of getting the correct electromagnetic charge.

 

[arivero]

It seems that we are mixing -ahem- two different questions.

a) about the pure theory, with SU(2)xU(1) and not fermion content at all. Then the question is if there is a sensible choosing for an U(1) "electromagnetic" subgroup or if any subgroup will give $W^+$ a charge of one unit.

b) about the theory with electrons and neutrinos. Then we can require the coupling of the U(1) to be equal for left and right electrons (and its value actually defines the electric charge) and zero for neutrinos, and the the point is how arbitrary the hypercharge assignments for the neutrino and left and right electron can still be.

Both questions are, it seems to me, well formulated before symmetry breaking. Just for peace of mind, you can change the sign in the quadratic term of the higgs field and voila, you are in the unbroken theory.

The mixing a+b allows for a third question (perhaps it can be already done just in (a), can it?): if the charge that $W^+$ "carries" is also the same charge that it presents when we compute the $WW\gamma$ vertex.

 

[naima]

I'd like to come back to my initial problem.
In local gauge theories we add fields to the derivative so as to get it covariant.(eg w1 w3 w3)
Actually the exchanged bosons are W+ W- and W3. (W+ = W1 - iW2 and so on)
What is the trick to get these "physical" bosons out of first ones in general ? (eg in SU(5), SO(10) ..)

 

[arivero]

I'd like to come back to my initial problem.
In local gauge theories we add fields to the derivative so as to get it covariant.(eg w1 w3 w3)
Actually the exchanged bosons are W+ W- and W3. (W+ = W1 - iW2 and so on)
What is the trick to get these "physical" bosons out of first ones in general ? (eg in SU(5), SO(10) ..)

Well, I thought that your initial problem was really my question (a) above, or the a+b, perhaps.

As for what are these bosons you call "physical", it is more of a question on group (or lie algebra) representation theory, isn't it? It seems that "ladder operators", who are used to produce the whole representation, are the important ones. SU(2) as a single ladder operator, W+.

 

[kurros]

b) about the theory with electrons and neutrinos. Then we can require the coupling of the U(1) to be equal for left and right electrons (and its value actually defines the electric charge) and zero for neutrinos, and the the point is how arbitrary the hypercharge assignments for the neutrino and left and right electron can still be.

I agree with you, but is there a reason we should care if the left and right electrons have equal coupling if there is no symmetry breaking? Since they are massless they propagate entirely separately from each other and they are in different group representations, so is not our effort to associate them with each other arbitrary?

 

[DrDu]

I'd like to come back to my initial problem.
In local gauge theories we add fields to the derivative so as to get it covariant.(eg w1 w3 w3)
Actually the exchanged bosons are W+ W- and W3. (W+ = W1 - iW2 and so on)
What is the trick to get these "physical" bosons out of first ones in general ? (eg in SU(5), SO(10) ..)

As long as your symmetry isn't broken, you could use instead of f and f' as basis states the combinations 1/√2 (f+f') and 1/√2 (f-f') or 1/√2(f+if') and 1/√2(f-if'), which would be interchanged by something like w2±iw3 and w1±iw3, respectively. So before symmetry breaking, there is no reason to single out a special combination of the w_i.