# Charge of the W bosons in the Higgs Mechanism

1. May 1, 2015

### nigelscott

I am confused about how the gauge boson W+ and W- get their charge under spontaneous symmetry breaking in the Higgs mechanism. Here's what I have so far:

The covariant derivative for a SU(2)⊗U(1) is

DμΦ = (∂μ + igWμiσi/2 + ig'Bμ)Φ where g and g' are coupling constants.

SU(2) is associated with weak isospin and the W0, W1 and W2 gauge fields.
U(1) is associated with weak hypercharge and the B gauge field.
The Higgs field, Φ, is a doublet and has a weak hypercharge of 1.

SU(2)⊗U(1) -> U(1) yields W0, W1 and W2 bosons and a B Boson.

W1 and W2 combine to give W+, W-, W0 and B combine to give a photon and a Z boson.

W+ somehow gets a charge of 1
W- somehow gets a charge of -1

Can somebody explain how all these things play with each other to produce the end result. I am looking for a qualitative explanation rather than a mass of equations. One of my key sticking points is the difference between a B boson and a photon since they both are associated with U(1).

2. May 1, 2015

### The_Duck

The photon field $A_\mu$ is a linear combination of $B_\mu$ and $W^0_\mu$. The idea is that the Higgs VEV changes under a SU(2) transformation or a U(1) transformation, but there is a particular combination of an SU(2) rotation and a U(1) rotation that leaves the Higgs VEV invariant. Therefore that symmetry is unbroken and the corresponding field $A_\mu$ remains massless.

The fact that the W is charged just means that it interacts with the photon field. We know that the W should interact with the photon because the original SU(2) group is nonabelian, so it has interactions between the three SU(2) gauge bosons. Part of the field $A_\mu$ comes from the SU(2) gauge field $W^0_\mu$, which interacts with the $W^+_\mu$ and $W^-_\mu$ fields via the original SU(2) self-couplings.

It really is helpful to slog through the equations here, but that's the high-level view.

3. May 1, 2015

### Orodruin

Staff Emeritus
These are different U(1) groups. Before EWSB, the electroweak sector has an unbroken SU(2)xU(1) gauge group. The Higgs vev breaks both the SU(2) and the U(1) symmetries, but leaves a particular U(1) subgroup unbroken, this subgroup is not the original hypercharge U(1).

4. May 1, 2015

### nigelscott

OK. This might be an iterative process for me. Are you saying SU(2)U(1)γ -> U(1)em but there is a subgroup of U(1)em that is still has unbroken symmetry (the photon field)? But I still don't get where the charges for the W0, W1, W2 boson comes from. In the case of fermions, the Higgs field swaps its hypercharge to conserve the particle hypercharge as its chirality changes. Is this mechanism in any way involved with the bosons? Sorry if I am somewhat ignorant about this. I am a retired engineer with a background in QM not Group Theory.

5. May 1, 2015

### Orodruin

Staff Emeritus
No, the U(1)em is the unbroken subgroup of SU(2)xU(1)Y. The corresponding generator is the photon.

The W0 does not become charged, it becomes part of the photon and Z boson. The W+ and W- charges arise as a remnant of the W1 and W2 interactions with the W0, which is due to the SU(2) being non-abelian, and the photon being part W0.

6. May 2, 2015

### vanhees71

The generator of the symmetry groups are the charges, not the gauge bosons. The latter couple to the conserved charges (usually via minimal coupling). This charge conservation is a necessary (but not sufficient) condition for invariance under the local gauge transformations.

7. May 2, 2015

### ChrisVer

Why don't you try to act with the charge operator on W+,W- ?