Understanding Weak Isospin in SU(2) Gauge Theory

[nikkkom]

In QCD, quark is in fundamental representation of SU(3) and thus it has to have 3 charges (what we came to call "colors"). Gauge bosons are in adjoint representation and there are 8 of them. The choice how to assign color charges to them is not unique, one popular choice is based on Gell-Mann matrices. Wiki has a good layman explanation: https://en.wikipedia.org/wiki/Gluon

Naively, I would expect that in SU(2) gauge theory, fermions are in its fundamental representation and thus have to have 2 charges. And it is "sort of" true - in SM, the "up" and "down" labels for quarks are basically those charges (and for leptons too: charged leptons carry "down" charge, neutrinos are "ups").

There are 3 gauge bosons in (unbroken) SU(2): W1, W2 and W3.

Is it possible to assign these charges, call them "flavor charges", to W1/2/3 bosons (say, based on Pauli matrices, analogously to gluons' colors)? Like this:

$(u\bar{d}+d\bar{u})/\sqrt{2}$
$-i(u\bar{d}-d\bar{u})/\sqrt{2}$
$(u\bar{u}-d\bar{d})/\sqrt{2}$

But when SU(2) gauge symmetry is mentioned in SM, this is never written like this. Somehow it simplifies to a single charge, weak isospin. This part I don't understand. How does it work?

 

[Orodruin]

Unlike the SU(3) colour group, the SU(2) of the SM is a spontaneously broken symmetry.

We use weak isospin because it is convenient after symmetry breaking. If you base a Yang-Mills theory on SU(2) without symmetry breaking you will have a generator of the SU(2) representation a field transforms according to in its interaction vertex with the gauge boson. The gauge bosons themselves transform according to the adjoint representation.

 

[nikkkom]

Unlike the SU(3) colour group, the SU(2) of the SM is a spontaneously broken symmetry

Sure, I know that. So, if you would need to describe an unbroken SU(2), you would need two "flavor charges", right?

We use weak isospin because it is convenient after symmetry breaking.

I suspect so. Basically, I'm asking for a more detailed explanation.

If you base a Yang-Mills theory on SU(2) without symmetry breaking you will have a generator of the SU(2) representation a field transforms according to in its interaction vertex with the gauge boson. The gauge bosons themselves transform according to the adjoint representation.

I think that's what I said in my introductory explanation.

 

[ChrisVer]

you would need two "flavor charges", right?

aren't you already getting it when you introduce tha I₃?

 

[nikkkom]

aren't you already getting it when you introduce tha I₃?

Doesn't look like it to me.
As I understand it, the charge which is just a number (can be positive or negative) is a feature of gauge groups with only one generator. Electromagnetism has electric charge. U(1) subgroup of electroweak group has weak hypercharge.

QCD has more than one generator and has three charges associated with it, not one.

SU(2) subgroup of electroweak group has more than one generator. Shouldn't it have more than one charge?

Of course, since charges are conserved (in unbroken symmetry), any linear combination of them is conserved too. Thus, if I3 is defined as I3 = 1/2 (u - d), it will also be a conserved quantity. (For example, any up quark carries one +1up charge, thus its I3 = +1/2. W+ boson carries +1up and -1down charges, thus its I3 is +1).

But this seems to be arbitrary. I can construct any number of such quantities. Say Z = r + 2g - 3b would be conserved in QCD (since r,g,b are individually conserved). It does not make Z useful.

I'm thinking there is an explanation for why I3? Maybe try starting from what "third component of" weak isospin means, and what happened to first and second conponents?

 

[Orodruin]

QCD has more than one generator and has three charges associated with it, not one.

No it doesnt. It has one charge, color charge. That charge is a unit vector in a three dimensional complex space.

It the same way an SU(2) charge is a unit vector in a two-dimensional complex plane.

What makes I3 so important is its role in SSB as it is associated with the neutral gauge boson that mixes with the hypercharge boson to become the photon and Z fields. As a consequence, electric charge is a linear combination of I3 and hypercharge.

 

[ChrisVer]

as for why the 3rd and not the 1st or the 2nd, it's just a matter of convention on which component's base you write stuff... in complete analogy to why we are using the 3rd component of the spin and not the other 2 (which can be expressed as a linear combination of the 3rd's due to completeness)