Do free electrons exist? In order to construct a gauge invariant theory (say for the electron), we introduce a gauge field in the Lagrangian,
\begin{align}
\mathcal{L} = \bar{\psi}(\, i \gamma^\mu \partial_\mu  + i e \gamma^\mu A_\mu - m) \psi - \frac{1}{4} F_{\mu \nu } F^{\mu \nu}.
\end{align}
The term $\bar{\psi} A_{\mu} \psi$, seems to imply that there is always a coupling between the electron and photon, and therefore it seems that free electrons do not exist.  But in reality, we do have free electrons.
What's wrong with my deduction?
 A: Indeed, totally "free" electrons do not exist.
An electron moving in a cavity containing zero photons can still emit photons, so the interaction is always on, even if the electromagnetic field naively appears to be 'off'.
More importantly, when you use the phrase 'free electron', you're contrasting it with a 'bound electron'. In the former case, the electromagnetic field appears to have no significant effect on the motion of the electron, while in the latter it does. 
However, this is also false. The electron carries electromagnetic field energy along with it, and this changes the mass of the electron, both classical and in QFT. So even electrons sitting in empty space are affected by the electromagnetic field. (In fact, the changes to the mass are formally infinite, and must be cancelled out by a negative infinite "bare" mass, so the effect of the electromagnetic field is extremely severe! But that's a long story for another time.)
A: All you're saying is that electrons are coupled to photons in the Standard Model.  But logically every particle in the Standard Model must be coupled to some other particle - otherwise we'd have no way to detect it!  So according to your definition of "free," it's trivially true that no free particles of any kind exist - this is in no way particular to electrons or photons.
