# About Abelian U(1) Higgs model

1. Aug 17, 2015

### Paul Draw

1.In Abelian higgs model,we can come to the results that the gauge bosons acquire mass, implying the photons also acquire mass in this model；however,photons are massless in reality.How to explain this dilemma?
2.What does U(1) mean in this model?Does it correspond to some coservation laws,or selection rules?

2. Aug 17, 2015

### ChrisVer

1. So obviously what undergoes the Higgs mechanism is not U(1).... it's rather SU_L(2)xU_Y(1) that is spontaneously broken. However maybe that's useful in applications where the photons get an effective mass (like in superconducting matterial?)
2. U(1) is the symmetry of your Lagrangian (?), it's a choice which allows you to add generally a phase change to your fields.

3. Aug 18, 2015

### Paul Draw

Sorry,I am new to this topic,and I still confuse about some details.
Here is what I realize: U(1) symmetry correspond to conservation of electric charge,so the physics meaning of "U(1)" in this model is to make sure that the charge of the complex scalar field is conservative.
Am I right?

4. Aug 18, 2015

### Mordred

Last edited: Aug 18, 2015
5. Aug 19, 2015

### ChrisVer

Somewhat, but the thing is that it doesn't have to correspond to the electric charge. It can be any kind of "charge" of an abelian group. In the Standard Model SU(3)xSU(2)xU(1), the U(1) is the hypercharge and has almost nothing to do with electromagnetism before the symmetry breaking (the gauge bosons are the $W^{1,2,3}_\mu, B_\mu$ (not to be confused with W bosons or photon)... after the spontaneous symmetry breaking of SU(2)xU(1) you get a remaining/residual U(1) which will be for the electromagnetism with the massless photon and the heavy gauge bosons $W^{\pm}_\mu,Z_\mu$.

The massless boson exists because you have the additional freedom of rotating again your field due to the residual $U_{em}(1)$. If you break it, you will indeed get a mass for the photon.

A U(1) in general is the symmetry that allows you to redefine your fields by some change in their phase (global if it doesn't depend on the position, or local if it does). This corresponds to a conserved current (and charge) via Noether's theorem.

When you deal with a U(1) and they tell you that "the vector field is the photon", they are actually simplifying the process... it is a photon-like field, that's for sure, but it's not the photon.

Last edited: Aug 19, 2015
6. Aug 23, 2015