To carry out a non-perturbative calculation from QED it is usual to use the Foldy-Wouthuysen transformation. This is necessary to ensure that the time evolution of states matches the time evolution of field operators, without which constraint the phase differences corrupt the definition of the momentum operator. It is possible to simplify the Foldy-Wouthuysen transformation (which incorporates spin) and define the field picture
$$|f_F(t)\rangle = e^{-iH_It} |f(t)\rangle = e^{-iH_0t} |f(0)\rangle $$
In the field picture, kets evolve as in the Schrödinger picture for non-interacting
particles. The momentum operator in the field picture is
$$P_F^a= e^{-iH_It}i\partial^ae^{iH_It} $$
In the semi-classical correspondence, evolution may be treated for small $t$ as a
perturbation to the evolution of a non-interacting particle, by replacing the interaction
Hamiltonian with its expectation (in effect summing diagrams for the non-perturbative case). For a classical particle with
position $x$ and velocity $\dot x$, the classical current is
$$J=-e\dot x$$
The expectation of the interaction Hamiltonian is
$$\langle H_I\rangle=J \cdot\langle A \rangle = -e \dot x \cdot\langle A \rangle $$
Replacing the interaction Hamiltonian with its expectation gives a semiclassical
model in which the electron is quantum but the field is classical. In this
semi-classical model, the momentum operator in the field picture is
$$ P_F^a = e^{ie \dot x \cdot\langle A \rangle}i\partial^a e^{-ie \dot x \cdot\langle A \rangle} = i\partial^a-e\langle A^a\rangle $$
Thus, the expectation, $\langle A^a\rangle$, of the operator which creates and annihilates
photons acts in the manner of a classical vector field, modifying energy and
momentum. This is the standard formula for generalised momentum in the
presence of a classical field, often assumed on phenomenological grounds, but here seen from the emission and absorption of photons in interaction. Replacing momentum in the Dirac equation with generalised momentum gives the interacting Dirac equation (covered in many textbooks).
Again working in the field picture we have, from Ehrenfest’s theorem,
$$ {d \over dt}\langle P^a_F\rangle= \langle {d \over dt} P^a_F\rangle + i\langle[H,P^a_F]\rangle $$
Replacing the interaction in the Hamiltonian with the expectation as before
$$H=H_0 + H_I \approx H_0 + \langle H_I\rangle =H_0 -e\dot x\cdot\langle A\rangle $$
Substituting, using generalised momentum, and dropping the subscript F (since expectations are the same in any picture)
$$ {d \over dt}\langle P^a\rangle= e {d \over dt}\langle A^a\rangle +i\langle [ H_0 -e\dot x\cdot\langle A\rangle, i\partial^a-e\langle A^a\rangle]\rangle $$
$$ {d \over dt}\langle P^a\rangle= e {d \over dt}\langle A^a\rangle -e\partial^a \dot x\cdot\langle A\rangle $$
To intepret this, write it in the rest frame of the particle (so that we have proper time)
$$ \partial^0 \langle P^a\rangle= e \partial^0\langle A^a\rangle -e\partial^a \langle A^0\rangle $$
Then we only have to do a Lorentz transformation to find the Lorentz force law in terms of the Faraday tensor.
The derivation of Maxwell's equations is more straightforward, working from the Gupta-Bleuler gauge condition which yields Lorenz gauge, because it is not necessary to use the Field picture. I have given a full treatment in A Construction of Full QED Using Finite Dimensional Hilbert Space and in The Mathematics of Gravity and Quanta