I'm having trouble understanding the interaction of radiation with matter in (elementary non-relativistic QED) in Coulomb gauge ($\nabla\cdot\boldsymbol{A}=0$). We saw how to quantize the free electromagnetic field in a cavity of finite dimension, by writing the hamiltonian in terms of creation and annihilation operators, i.e.: \begin{equation} \hat{H}_{\mathrm{em}}^{\mathrm{free}}=\sum_{\lambda=1}^{2}\sum_{\boldsymbol{k}}\hbar\omega_{\boldsymbol{k}}\left(\hat{a}^{\dagger}_{\lambda\boldsymbol{k}}\hat{a}_{\lambda\boldsymbol{k}}+\frac{1}{2}\right) \end{equation} where $\lambda$ runs on the two polarization of the field. Thus the hamiltonian for field-particle system in the non-relativistic approximation is: \begin{equation} \begin{split} \hat{H}&=\sum_{a}\left(\frac{\hat{\boldsymbol{p}}_{a}}{2m_{a}}+V_{a}\right)+\sum_{\lambda=1}^{2}\sum_{\boldsymbol{k}}\hbar\omega_{\boldsymbol{k}}\left(\hat{a}^{\dagger}_{\lambda\boldsymbol{k}}\hat{a}_{\lambda\boldsymbol{k}}+\frac{1}{2}\right)+\sum_{a}\left(-q_{a}\frac{\hat{\boldsymbol{A}}\cdot\hat{\vec{p}}_{a}}{m_{a}c}\right)=\\ &=\hat{H}_{\mathrm{particle}}+\hat{H}_{\mathrm{em}}^{\mathrm{free}}+\hat{H}_{\mathrm{int}} \end{split} \end{equation} The problem I have is that to compute the transition probability we used Fermi's Golden Rule: \begin{equation} P_{i\rightarrow f}=\frac{2\pi}{\hbar}\left\vert W_{fi}\right\vert^2\rho\left(E_f\right)\end{equation} (f and i being the final and initial state) while this formula (in all books I read about non relativistic quantum mechanics) is usally derived upon considering the perturbation hamiltonian time independent or a perturbation of the form: \begin{equation} \hat{H}_{\mathrm{int}}=\hat{T}_{+}e^{-i\omega t}+\hat{T}_{-}e^{i\omega t} \end{equation} (where $\hat{T}_{-}=\hat{T}_{+}^{\dagger}$), while our perturbation hamiltonian is \begin{equation} \hat{H}_{\mathrm{int}}\propto\sum_{\lambda\boldsymbol{k}}\left[\hat{\boldsymbol{p}}_{a}\cdot\boldsymbol{e}_{\lambda\boldsymbol{k}}\left(\hat{a}_{\lambda\vec{k}}e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+\hat{a}^{\dagger}_{\lambda\boldsymbol{k}}\right)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}} \right] \end{equation} ($\boldsymbol{e}_{\lambda\boldsymbol{k}}$ being the two polarizations vector), and the $\hat{a},\hat{a}^{\dagger}$'s are time dependent. I hope I made my point clear enough.
P.S.: What we did in order to quantize the EM field was to write a Fourier series expansion of the vector potential \begin{equation}\boldsymbol{A}\left(\boldsymbol{x},t\right)=\sum_{\boldsymbol{k}}\boldsymbol{a}_{\boldsymbol{k}}\left(t\right)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}\end{equation} and by plugging this expression in $\Box \boldsymbol{A}=0$ one find $\ddot{\boldsymbol{a}}_{\boldsymbol{k}}+\omega_{\boldsymbol{k}}\boldsymbol{a}_{\boldsymbol{k}}=0$ so that $\boldsymbol{a}_{\boldsymbol{k}}{}_{\pm}\propto e^{\pm i\omega_{\boldsymbol{k}}t}$