# What does the EM gauge freedom have to do with U(1)

1. Jul 16, 2015

### hideelo

I know that we have a free parameter in the relativistic potential for electromagnetism. I also know that we can introduce this as a scalar field ψ which gives this free parameter. I understand that this can be related to U(1) by mapping ψ: ---> e^iψ, which is the U(1) group. It just seems a little forced. I mean, sure you can map every scalar field into U(1) but why do it. It seems to me that the symmetry here is more analagous to some one dimensional translational symmetry since at each point in spacetime I can choose any point on the real line with no change n the corresponding equation of motion. Am I missing something?

2. Jul 17, 2015

### fzero

In classical electrodynamics, we couple the EM potential to a 4-current via $-A_\mu J^\mu$ and gauge invariance of the Lagrangian implies that $\partial_\mu J^\mu =0$, i.e., the current is conserved. We can simply take the gauge group to be $(\mathbb{R},+)$ with no ill consequences.

In quantum electrodynamics, we must introduce a separate quantum field for the matter degrees of freedom, e.g., the electron, with kinetic term $\bar{e} \gamma^\mu \partial_\mu e$ ($e$ is a spinor, $\gamma^\mu$ are Dirac matrices). The gauge transformation on the electron field is $e\rightarrow U e$ and the kinetic term transforms as

$$\bar{e} \gamma^\mu \partial_\mu e \rightarrow \bar{e} U^\dagger U \gamma^\mu \partial_\mu e + \bar{e} U^\dagger \gamma^\mu (\partial_\mu U) e.$$

The second term is canceled by including the coupling of the electron current to the gauge field $- \bar{e} \gamma^\mu A_\mu e$. From the first term, the Lagrangian will only be invariant if $U^\dagger U = 1$, which means that $U$ is a unitary transformation. So our gauge group must be $U(1)$ rather than $(\mathbb{R},+)$.

3. Jul 17, 2015

### hideelo

First of all, thanks for answering. Secondly, I wont say I understand everything you are talking about, but at least I know what it is that I am missing.