Paul Draw said:
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?
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 [itex]W^{1,2,3}_\mu, B_\mu[/itex] (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 [itex]W^{\pm}_\mu,Z_\mu[/itex].
The massless boson exists because you have the additional freedom of rotating again your field due to the residual [itex]U_{em}(1)[/itex]. 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.