# Quantising the magnetic monopoles the make Maxwell symmetric

I don't believe this has already been asked, but I might be wrong; sorry. One can add a magnetic charge density/magnetic monopoles to Maxwell's equations to make the theory symmetric between the electric and magnetic fields. What happens when we quantise this theory?

Since QED is a $U(1)$ gauge theory, would we expect this theory to be a $U(1) \times U(1)$ gauge theory? This seems a bit naive, but it makes some sense since you have two charges instead of one, acting in the same way (possibly different coupling constants), hence two gauge groups instead of one. This gauge theory is still abelian, so it seems similar enough to QED.

I wonder about how the electron and monopole interact in this theory, and if that needs to be added in some way, and if perhaps we have too much additional symmetry, since QED already contains magnetism - it just lacks sources.

This is a great question --- indeed, we can add another matter field to QED and couple it to a second photon and get a $$U(1) \times U(1)$$ gauge theory. But the charges of these theories are independent; there isn't any duality, so the second $$U(1)$$ charges are not magnetic monopoles.
If you naively try and introduce a relation between the two potentials and charges to give duality you actually end up reducing the degrees of freedom such that you get one-charge electrodynamics. It is possible to develop a two-potential quantum field theory of electrodynamics with monopoles, but you need a modification of the definitions of the field strength tensor: if we denote the two potentials $$A_\mu$$ and $$B_\mu$$ we end up with
$$F_{\mu \nu} = n^\alpha \left[n_\mu (\partial_\alpha A_\nu - \partial_\nu A_\alpha - \tfrac{1}{2} \varepsilon_{\mu \nu \rho \sigma} n^\rho (\partial_\alpha B^\sigma - \partial^\sigma B_\alpha) \right]$$
and a very similar expression for the field tensor of $$B_\mu$$. Here $$n^\alpha$$ is an arbitrary fixed unit 4-vector. This prescription was originally due to Zwanziger.