# Transformation of electromagnetic potential under local U(1) transformation

Let $$\mathcal{L}=-(\partial _{\mu} \Phi^*)(\partial ^{\mu} \Phi)$$

With $$\Phi , \Phi^*$$ being complex fields.

When looking at local U(1) transformations in class, we saw that $$\mathcal{L}$$ is not invariant under the transformations of the form $$\Phi \rightarrow e^{i\theta(\mathbf{x})}\Phi$$.

So we introduced covariant derivatives $$\partial^{\mu} \rightarrow D^{\mu}=\partial^{\mu}-iA^{\mu}(\mathbf{x})$$

As an exercise we should determine how $$A^{\mu}$$ has to transform to make $$\mathcal{L}$$ invariant.

I got $$A^{\mu} \rightarrow A^{\mu}+\partial^\mu \theta$$ as a result. I also showed that $$\mathcal{L}$$ is indeed invariant under this transformation.

Now what I am wondering is, how can I interpret the transformation of $$A^{\mu}$$? I haven't had electrodynamic classes so I don't really know much about electrodynamic potentials, but I think it might have something to do with U(1) transformations being rotations (atleast I think, they are).

In a second semester electrodynamics course you would learn about the gauge freedom of electromagnetism. If you have electric and magnetic fields $$\textbf{E}$$ and $$\textbf{B}$$ these are determined by the 4-vector potential A. However there is no single A that determines $$\textbf{E}$$ and $$\textbf{B}$$ uniquely. This is merely a more complicated version of in Newtonian gravity how you can always add a constant to the potential without changing the dynamics.
Given the $$\textbf{E}$$ and $$\textbf{B}$$ fields are contained in an antisymmetric tensor F, $$F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$$ you can see for example that $$F^{00} = 0$$, $$F^{01} = \partial^0 A^1 - \partial^1 A^0 = E_x$$, etc. Now if you add a divergence term to your 4-vector potential, $$A'^{\mu} = A^{\mu} + \partial^{\mu}f$$ we can compute the new $$\textbf{E}$$ and $$\textbf{B}$$ fields, \begin{align} F'^{\mu\nu} &= \partial^\mu A'^\nu - \partial^\nu A'^\mu \\ F'^{\mu\nu} &= \partial^\mu (A^\nu +\partial^{\nu}f) - \partial^\nu (A^\mu + \partial^{\mu}f)\\ F'^{\mu\nu} &= F^{\mu\nu} + \partial^\mu \partial^{\nu}f - \partial^\nu \partial^{\mu}f\\ F'^{\mu\nu} &= F^{\mu\nu}. \end{align} So the vector potential is only well defined up to an overall divergence term. Now specifically towards your question, you found that you need the vector potential to transform by a divergence to impose local U(1) symmetry, $$A^{\mu} \rightarrow{} A^{\mu} + \partial^{\mu}\theta(x)$$ but this would leave the $$\textbf{E}$$ and $$\textbf{B}$$ (along with energy, momentum, interactions) unchanged.