Does single electron interact with its own electromagnetic field? The QED equations are given by
$\square A^\mu=e\bar{\psi}\gamma^\mu\psi$
$i\gamma^\mu\partial_\mu\psi=(m+e\gamma^\mu A_\mu)\psi$
These equations suggest, that electrons interact with field created by themselves, which seems absurd, because if a point particle would interact with its own field (that is equal to infinity in the particle's position). That creates a problem.
If electron does interact with its own field, how does one overcome the problem described above?
 A: This is known as the self-energy problem in quantum electrodynamics. Computing higher order corrections to the two-point function of the electron field (the electron propagator) generates divergent terms that have to be "renormalized". In short, the mass parameter $m_0$ (and also the charge parameter $e_0$) in the tree Lagrangian you are starting with, do not correspond to the physical (observable) mass $m_{\rm phys}$ and the physical charge $e_{\rm phys}$ the experimentalists are measuring. The unphysical mass parameter $m_0$ has to be tuned in such a way that it absorbes the divergent mass shift $\delta m$ obtained from the two point function, such that $m_{\rm phys}= m_0 + \delta m$ remains finite (order by order in the loop expansion).
Likewise, the three-point function ($ee \gamma$ vertex) receives divergent higher order contributions generating a divergent shift $\delta e$, requiring a suitable tuning of the unphysical charge parameter $e_0$, such that the observed charge $e_{\rm  phys} = e_0+ \delta e$ becomes finite.
For details, see the standard text books on quantum electrodynamics and quantum field theory, respectively.
