This answer explains how the Coulomb-potential can be calculated as the energy shift of the (photon) ground state for 2 charges fixed in place. This calculation has been done for a covariant "quantization" of the electromagnetic field, where one has 2 transverse photon degrees of freedom, and one longitudinal and one temporal one. The book by Cohen-Tanoudji ("Atom-Photon Interaction") explains this quantization in detail, and also points out what the coulomb potential amounts to in this way of describing things.
Now I wonder - Can the same calculation also be carried out when on has quantized the electromagnetic field in the Coulomb-gauge, in the way it is usualy done in lectures on quantum optics?
I tried calculating: \begin{align} \hat{H}_{int} = 2 \hat{\vec{p}}_1 \hat{\vec{A}}(\hat{\vec{x}}_1) + \hat{\vec{p}}_2\hat{\vec{A}}(\hat{\vec{x}}_2) \Delta E = \langle 0 | \hat{H}_{int} | 0 \rangle + \sum_n \frac{ \langle 0 | \hat{H}_{int} | n \rangle \langle n | \hat{H}_{int} | n \rangle }{-E_n} \end{align} With $\hat{\vec{A}}$ being replaced by the expansion in creation and annihilation operators, but because the $\hat{\vec{p}}_i$ become $k_i$ in the nominator instead of the denominator,the result differs from the linked answer.
I suspect that the answer is no, because when quantizing the electromagnetic field in the Coulomb-gauge, one implicitly "quantizes" only the transverse degrees of freedom, and set's the $A_0$ component of the 4-potential to 0.
What is the best way to see this?
