transition probabilites of atomic systems prone to some time-varying electromagnetic field are very often calculated using perturbation theory leading to expressions including the vector potential $\mathbf{A}$. This approach seems to be very complicated and I am wondering:
Is using the electrostatic potential $U$ to calculate "electric transitions" (approximately) sufficient?
This was the short version of my question; below you will find some further notes that (hopefully) contain all information needed to clearify it.
In relativistic quantum mechanics, the action of an an electrodynamic field on an electron can be realized by minimal coupling using the replacement $\partial_\mu \rightarrow \partial_\mu - \mathrm{i}eA_\mu$. Thus, in a certain sense, (quantum) electrodynamics can be derived from the Dirac equation using the invariance of the free-electron action with respect to a phase change.
Electromagnetic Fields and the Schrödinger Equation
Nevertheless, often one must ignore the beauty of gauge theory to be able to practically calculate something under certain simplifications using the non-relativistic Schrödinger-equation.
There, one can incorporate an external electromagnetic field applying the replacements $$H\left(\mathbf{p},\mathbf{x}\right) \rightarrow H\left(\mathbf{p}-e\mathbf{A}\left(t,\mathbf{x}\right),\mathbf{x}\right)+eU\left(t,\mathbf{x}\right)$$ to the Hamiltonian of the system where $\mathbf{A}$ is the known vector potential and $U$ is the potential corresponding to the electric field.
In the Coulomb gauge $\mathrm{div}\mathbf{A}=0$ one can always achieve $U\equiv 0$ if no sources are present and the overall Hamiltonian will look like $$H = H_0 -\frac{e}{m}\mathbf{A}\left(t,\mathbf{x}\right)\mathbf{p}$$ where $H_0 = \frac{\mathbf{p}^2}{2m} + V(\mathbf{x})$ and the term in $\mathbf{A}^2$, the ponderomotive potential has been neglected.
Time-Dependent Perturbation Theory
The last expression for $H$ has the form $$H = H_0(\mathbf{p},\mathbf{x}) + H_t(\mathbf{p},\mathbf{x},t)\ .$$ Using the Dyson series, one can formally write down the time-dependent solution in the interaction picture as $$|\psi_I(t)> = T\exp\left[\frac{1}{\mathrm{i}\hbar}\int_{t_{0}}^{t}H_{t,I}\left(t^{\prime}\right)dt^{\prime}\right]|\psi_I(t_0)> .$$
Then, one takes only the first two terms into account to find that transition amplitudes are approximately given by $$A_{n\rightarrow m}\approx\frac{1}{\mathrm{i}\hbar}\int_{t_{0}}^{t}\left\langle m\right|H_{t,I}\left(t^{\prime}\right)\left|n\right\rangle dt^{\prime}\ \mathrm{with}$$ $$H_{t,I}\left(t\right) = e^{\mathrm{i}\left(t-t_{0}\right)H_{0}/\hbar}H_t\left(t\right)e^{-\mathrm{i}\left(t-t_{0}\right)H_{0}/\hbar}\ .$$
Relation to the Question
In the given case we have $$H_t = -\frac{e}{m}\mathbf{A}\left(t,\mathbf{x}\right)\mathbf{p}$$ and assuming that $$\mathbf{A}\left(\mathbf{x},t\right) = \mathbf{A}\left(\mathbf{x}\right) T(t)\ ,$$ and further using $$\mathbf{p}=\frac{\mathrm{i}m}{\hbar}\left[H_{0},\mathbf{x}\right]\ ,$$ one will have to solve $$A_{n\rightarrow m}\propto <m|\mathbf{A}\left(\mathbf{x}\right)\cdot \mathbf{x}|n>\int_{t_0}^t e^{-\mathrm{i}\omega_{mn}\tau}T(\tau) d\tau\ +\ \mathrm{h.c.}$$ with $\hbar \omega_{mn} = E_m - E_n$, where the $E_i$ are energy levels of the unperturbed $H_0$.
Using here the vector potential $\mathbf{A}$ is the correct way to incorporate the field, of course. For me, it seems extremely complicated to do so since to calculate the transition probabilities for an arbitrary field distribution one will have to expand $\mathbf{A}$ into vector spherical harmonics.
So, the use of a scalar potential like $U$ would greatly simplify computations and my question can now be re-asked as
Why can't we just simply use $U(\mathbf{x})\cdot T(t)$ to calculate the transition amplitudes $A_{n\rightarrow m}$ for "electric excitations"?
Thank you very much for insights, comments and corrections.
