How to derive the dipole interaction term in coulomb gauge QED from minimal coupling? In quantum optics, one usually "quantizes" in the coulomb gauge, with amounts to the assumption that there are no charges present $\rho = 0$. I will introduce my question by roughly scetching the procedure along the lines of Brien Hatfield (QFT of pointparticles and strings, p.81 ongoing). The classic coulomb law $\vec{\nabla} \vec{E} = 0$, which is not an equation of motion, but instead only a constraint to one timeslice, is then satisfied by the then field-momentum operators $\vec{\nabla} \hat{\vec{E}}= 0$ and the fields $\vec{\nabla} \hat{\vec{A}}= 0$.
The ccr read
\begin{align}
[A^i(\vec{x}), E^j(\vec{y}] = \delta_{ij}^{\text{trans}}(\vec{x}-\vec{y})
\end{align}
and everyone is happy, because everything is consistent. We effectively only have 2 degrees of freedom.
However, we put in the information that there are no charges, and set the scalar potential $\phi$ to zero, so $E^j = - \partial_0 A^j$.
Now my question is - what happens if we introduce charges, that means in this context if we have the EM quantum field interacts with another field, or another system? For this question I ask about the specific 2nd system, which is the nonrelativistic point particle electron. Via minimal coupling, the electron is coupled to the field by
\begin{align}
\hat{H} = (\hat{\vec{p}}- \frac{q}{c}\vec{A})^2 - q \phi(\vec{x})
\end{align}
This term also resembles the usual $j_\mu A^\mu$ term from classical mechanics or the $\Psi^{\dagger} \gamma_\mu A^\mu \Psi$ term from QED. We usually use the minimal coupling hamiltonian to derive an interaction of the form $\hat{\vec{x}} \hat{E(t, \vec{x_0})}$ (with $\vec{x}_0$ being the position of the atom) for atomic transitions. But how can we do that when we already have set $\phi$ to zero in the beginning of the quantization? I don't see a way to derive (with suitable approximations) the dipole interaction term, only using the (quantized) vector potential $\hat{A^i}$. How would one proceed here?
The standard treatment to arrive at the
\begin{align}
q \hat{\vec{x}}\vec{E}(\vec{x_0},t)
\end{align}
from the
\begin{align}
q \phi(\hat{\vec{x}})
\end{align}
is to make the approximation that $\nabla \phi$ is essentially constant around $\vec{x_0}$, so that it can be written as $q \vec{x} \hat{E}(\vec{x_0})$, but we can't do that here. So how can we still arrive at an interaction that involves the electric field (operator) and the dipole moment of the electron, but without looking at the $0-potential operator (which we have set to zero in the beginning of the EM-Field quantization)?
 A: This is a very interesting question and to some extent still an open question in the current literature. As such, I will not attempt to fully answer it, but simply give a few hints on how things are usually done.

But how can we do that when we already have set  to zero in the beginning of the quantization?

If we are being strict, you cannot. What is usually done in the Coulomb gauge is that the electromagnetic field given by the nucleus of an atom is treated as a classical potential which is added to the given quantized Hamiltonian for free space. For quantum optics, this procedure is sufficient most of the time. If you want to go deeper, one can add QED correction terms via a perturbation methodology, which is often employed in precision physics (e.g. to calculate Lamb shifts, $g$-factors etc.).

In quantum optics, one usually "quantizes" in the coulomb gauge, with amounts to the assumption that there are no charges present =0. [...] So how can we still arrive at an interaction that involves the electric field (operator) and the dipole moment of the electron, but without looking at the $0-potential operator (which we have set to zero in the beginning of the EM-Field quantization)?

The first sentence here is somewhat true as a starting point, but not really the whole story. In quantum optics, one usually moves away from the Coulomb gauge in a second step using the Power-Zienau-Woolley transformation (see e.g. this textbook). The resulting Hamiltonian then allows to obtain the dipole interaction Hamiltonian or its multipole generalisation in the long wavelength limit.
There are some nice sources to go deeper into this topic. As a freely accessible one, I would like to recommend these lecture notes: https://atomoptics.uoregon.edu/~dsteck/teaching/quantum-optics/
