Clarification about conserved currents in the Standard Model

[HomogenousCow]

I'm currently doing a course on the Standard Model and I think I've gotten confused on some of the symmetries and resulting currents. Here's my current understanding:

Before SSB:
Electroweak - U(1) symmetry on fermion spinors and Higgs complex doublet, 1 conserved current
Electroweak - SU(2) symmetry on Higgs complex doublet, quark and lepton spinor doublets (left-handed spinors), 3 conserved currents
Color - SU(3) symmetry on quark spinor triplets, 8 conserved currents

After SSB:
EM - U(1) residual symmetry on fermion spinors, 1 conserved current
Color - Same as before, 8 conserved currents

So my questions are follows:
1. Are the 3 currents from SU(2) completely gone after SSB?
2. Why must the Higgs field lose U(1) symmetry after SSB? It feels arbitrary that the Higgs field is chosen to be real after SSB.
3. Why do we talk about "color conservation" when there are in fact 8 conserved currents due to SU(3)?
4. I've read about a conserved "baryon number" due to U(1) symmetry, is this another distinct U(1) symmetry?
5. Why do the W-bosons carry electric charge?? Do they participate in the residual U(1) symmetry?

 

[Orodruin]

1. You can write down whatever currents you want. That does not mean that such a current will be conserved. It depends on what you mean by "completely gone".
2. What do you mean by "lose U(1) symmetry"? That the Higgs is not charged under the remaining U(1)? This is a direct consequence of how the symmetry is broken and does not have anything to do with choosing the vev to be real. The choice of a real vev is just an issue of a choice of basis. If you changed basis you would get a different generator of the remaining U(1) under which your new Higgs would still be uncharged.
3. I am sorry, but I do not understand the question. Color conservation would refer to all of those currents.
4. Yes it is a distinct symmetry, which is accidental in the SM, i.e., it is not imposed as a gauge symmetry but is a result of the fermion content of the model (the actual symmetry is B-L, baryon and lepton number individually are broken by non-perturbative effects in the SM).
5. This is a direct remnant from the SU(2) self-interactions before symmetry breaking and the fact that the photon has a non-zero component of the gauge boson $W_3$. Hence, the photon must couple to $W_1$ and $W_2$. Yes, $W^\pm$ interact electromagnetically.

 

[HomogenousCow]

1. You can write down whatever currents you want. That does not mean that such a current will be conserved. It depends on what you mean by "completely gone".

Yes what I meant to ask was, are the original 3 currents due to SU(3) still conserved after SSB?

3. I am sorry, but I do not understand the question. Color conservation would refer to all of those currents.

How are the 8 Noether currents related to the color charges (Red/Anti-Red etc.)?

 

[Orodruin]

How are the 8 Noether currents related to the color charges (Red/Anti-Red etc.)?

The assignment of red/green/blue is based on the fundamental representation of SU(3). The conserved Noether currents are based on the one-parameter continuous symmetries generated by each of the generators, i.e., the adjoint representation (i.e., the traceless part of ${\bf 3 \otimes \bar 3}$).

 

[HomogenousCow]

Okay, so the color assignments are not related to any conserved quantities? They're often compared to electric charge so I was under the impression that they were conserved in some way.

 

[king vitamin]

The way that "color" is often referred to in introductory material is misleading in my opinion. One would like to associate the three colors/anticolors with the three color indices (and conjugate indices for anticolors), but with this assignment the usual way people assign colors to hadrons doesn't make sense. For example, people often refer to mesons as being red*antired=white or blue*antiblue=white etc, but since it is really in the singlet representation (the trace of) of the tensor product \mathbf{3}\otimes\overline{\mathbf{3}}, a meson is really something like (red*antired + green*antigreen+blue*antiblue)/sqrt(3). It's rather important that there is only one combination of these which is a singlet ("white"), otherwise there would be more mesons than we actually observe.

This is especially true of baryons, which transform under the only singlet obtained from decomposing in the tensor product \mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3} into irreps. I don't even know how to write this out in the "color" language (and it wouldn't be very useful), but it's certainly not "red*green*blue".

 

[Orodruin]

This is especially true of baryons, which transform under the only singlet obtained from decomposing in the tensor product 3⊗3⊗33⊗3⊗3\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3} into irreps. I don't even know how to write this out in the "color" language (and it wouldn't be very useful), but it's certainly not "red*green*blue".

The singlet is the completely anti-symmetric combination. This is seen relatively easy if you use Young tableaux multiplications as the only way to create a singlet out of three boxes (fundamental representation) is to stack them on top of each other - ie, anti-symmetrisation.

This is why it is common to refer to it as rgb. Each part of the linear combination has one of each color:
(rgb + gbr + brg - rbg - bgr - grb)/sqrt(6)