Electric field associated with a stationary electron

As in any quantum theory, observables in QFT are represented by operators on a Hilbert space. In QFT, observables are constructed from field operators. Field operators, in turn, satisfy nice equations of motion that look like classical equations of motion — except that the field operators don't commute with each other.

In the simplest version of QED, we have two fields: the electron/positron field $psi$, and the EM field. These are non-commuting operators, but they still satisfy familiar-looking equations of motion (warning — I didn't check the signs carefully):

• The Dirac equation: $(gamma^mu(ipartial_mu-e A_mu)+m)psi=0$

• Maxwell's equations: $partial_mu F^{munu}= eoverlinepsigamma^nupsi$ and $F_{munu}=partial_mu A_nu-partial_nu A_mu$.

The $nu=0$ component of the first Maxwell's equation is the Gauss-law constraint $nablacdotmathbf{E}= rho$ where $rho equiv eoverlinepsigamma^0psi$ is the charge-density operator (an observable) and $mathbf{E}$ is the electric field operator (also an observable). This equation says that the observables $rho$ and $nablacdotmathbf{E}$ are equal to each other: they are the same operator. Thus any state with a non-zero charge automatically has the associated Coulomb field.

For example, writing $langlecdotsrangle$ for the expectation value in any state, we always have $langlenablacdotmathbf{E}rangle=langle rhorangle$. This holds for all states, no matter how many or how few electrons/positrons they happen to contain. In particular, it holds for single-electron states.

Whenever we write down equations of motion for the field operators like the ones I wrote above, we're using the Heisenberg picture. In the Heisenberg picture, the field operators at all times are just combinations of the field operators at any single time (say $t=0$), even though the interpretations depend on which time we're considering. The Gauss-law constraint is similar, except that time is not involved: it says that some combinations of field operators can be written in terms of others, in particular that $nablacdotmathbf{E}$ can be written in terms of $rho$, even though their interpretations may be different. This, of course, is the whole point of the equations of motion: to tell us how different observables are related to each other.

I've glossed over some tricky technicalities related to the fact that the field operators don't commute with each other. In particular, constructing a representation of these operators as operators on a Hilbert space is tricky because of gauge invariance. However, it can be done, and then the Gauss-law constraint $nablacdotmathbf{E}= rho$ can be viewed as the condition that physical states must be gauge invariant (at least up to a physically irrelevant overall constant phase factor).

In any case, here's the bottom line: any state that has a charge (as defined by the observable $rho$) automatically also has the associated Coulomb field field (as defined by the observable $mathbf{E}$), because the operators representing those two different observables are related to each other by the Gauss-law constraint. This is true for any (gauge-invariant) state, including single-electron states.

This is a comment on this part of your question:

that a stationary free electron

Free particles cannot be modeled by the plane waves fields of QFT, one has to go to the wave packet solution.

The mathematics of getting the electric field from the wave packet is beyond me.