Can we think of spontaneous emission of a photon from an excited atom as a driven harmonic oscillator problem? This is a kind of strange question, but I'm wondering, in the context of a fully quantum field theoretic treatment of spontaneous emission, if there is any model or way of calculating the process that treats it simply as an excited electron in the atom driving the field through its own time evolution?
I'm envisioning something like this: Since we treat the EM field as simply a set of quantum harmonic oscillators, can we take the unitary time evolution of an excited electron to be simply like a driving force to the field? So we would take the wave function of the field $\psi_{field}(\mathbf{r},t)$ and couple it (in quantum mechanics, just multiplying it) with the wave function of the electron. Let's take the field to be in its vacuum state and define this as the zero point energy. The excited electron has energy $\hbar\omega_0$, so the full coupled state of field/particle is:
\begin{equation}
\psi_{total}(\mathbf{r},t)=\psi_{electron}(\mathbf{r})\psi_{field}(\mathbf{r})e^{i\omega_0{t}}\tag{1}
\end{equation}
Since the coupled field-electron state has the same time-dependent phase factor as the excited electron, then if $\omega_0$ is some allowed mode of the field, can't we treat this as the field oscillating as though it is in a mode corresponding to that frequency? And so we give some amplitude for the time evolution to be "in" the field in the form of a photon and some amplitude for it to be "in" the electron? We know that in the end we should end up with the Rabi state:
\begin{equation}
|\psi\rangle=\cos{\omega_0{t}}|e,0\rangle+\sin{\omega_0{t}}|g,1\rangle\tag{2}
\end{equation}
And it seems to me that there should be a way of deriving the amplitudes in equation (2) from the time evolution picture in equation (1). However, I don't see how we get something like the coupling Hamiltonian $\hbar\omega_0(\hat{\sigma}^{+}\hat{a}+\hat{\sigma}^{-}\hat{a}^{\dagger})$ directly from the time evolution picture.
I'm curious if there is a way of considering the time evolution prima facie as I propose and if anyone has worked out an explicit model that goes this way?
Edit: I'm going to ask this question in a slightly different way that gets at what I really care about a bit better. Let's say we have a bound particle in some finite potential. We are not told anything about its wave function or its energy eigen-decomposition. All we are told is that it has some ground state energy $E_1$ and is not in the eigenstate $|E_1\rangle$,and that during some non-zero interval of time, it acquires a time-dependent phase $e^{i\omega_0{t}}$. We couple it to the vacuum state of a cavity and we demand that the cavity has an allowed mode at frequency $|(E_1/\hbar)-\omega_0|$. Is this enough information for us to perform some calculation of the interaction Hamiltonian and show that the interaction Hamiltonian of the particle with the vacuum has some non-zero component like $\hat{a}^{\dagger}_{\omega_0}\hat{\sigma}^{-}_{\omega_0}$?
In other words, we know that if the interaction Hamiltonian contains the term $\hat{a}^{\dagger}_{\omega_0}\hat{\sigma}^{-}_{\omega_0}$, that there is some time-evolution factor $e^{i\omega_0{t}}$ in the wave function, so having this factor is a necessary condition. But can we go the other way and say that this factor is a sufficient condition such that, not knowing anything about the actual energy levels of the original wave function, we can get a term in the interaction Hamiltonian directly from this phase factor?
 A: 
However, I don't see how we get something like the coupling Hamiltonian $\hbar\omega_0(\hat{\sigma}^{+}\hat{a}+\hat{\sigma}^{-}\hat{a}^{\dagger})$ directly from the time evolution picture.

We typically work in the "interaction picture." In this picture the operators are granted time-dependence via some "unperturbed" Hamiltonian $H_0$, and the rest of the Hamiltonian is called the interaction part.
The interaction part of the Hamiltonian does looks a bit like what you wrote above. In general the exact written form depends on the gauge and on some other common approximations (e.g., do we describe the electron with a Dirac equation or a schrodinger equation). But, under certain approximations, the interaction looks like:
$$
H_{\text {int}} = \frac{e}{mc}\vec p\cdot\vec{A}(\vec r)\;,
$$
where $\vec p$ is the momentum operator of the electron and the photon is described via vector potential $A$, which looks like:
$$
\vec A(\vec r) = \frac{2\pi\hbar c}{kV}\sum_{\vec k}\sum_{i=1,2} \vec \epsilon_{k,i} a_{k,i} e^{i\vec k \cdot \vec r} + \epsilon_{k,i}^* a_{k,i}^\dagger e^{-i\vec k\cdot \vec r}
$$

If you want to calculate scattering amplitudes, you need some further approximations, but roughly, you just plug into Fermi's Golden rule at the lowest order where you actually get a result. E.g., for scattering electrons off of electrons we need two factors of the $H_{int}$ we wrote above.
