Consider the Schrödinger equation $H\Psi=i\hbar\frac{\partial \Psi}{\partial t}({\bf r},t)$. The hamiltonian $H$ is: \begin{equation} H=-\frac{\hbar^2}{2m}\nabla^2+V({\bf r},t) \end{equation} And the equation is: \begin{equation} \Big[-\frac{\hbar^2}{2m}\nabla^2+V({\bf r},t) \Big]\Psi({\bf r},t)=i\hbar\frac{\partial \Psi}{\partial t}({\bf r},t) \end{equation}
If $V$ is time-independent, then $H$ too. In this case the wave function can be factored as $\Psi({\bf r},t)=\psi({\bf r})e^{-i\omega t}$, and then $H\Psi=E\Psi$, so the eigenvalues of $H$ are the energies $E$.
If $V$ is time-dependent, we can't separate the variables ${\bf r}$ and $t$ like before, but $H$ is still an hermitian operator and it must have certain eigenvalues.
My question is: What are the eigenvalues of $H$ when it depends on time? (and if they are the energies, why?)
EDIT: If I try to separate the variables, I have $\Psi({\bf r},t)=\psi({\bf r})\varphi(t)$, putting this in Schrödinger equation (this is the method of separation of variables): \begin{equation} \Big[-\frac{\hbar^2}{2m}\nabla^2+V({\bf r},t) \Big]\psi({\bf r})\varphi(t)=i\hbar\frac{\partial}{\partial t}[\psi({\bf r})\varphi(t)] \end{equation}
\begin{equation} -\varphi(t)\frac{\hbar^2}{2m}\nabla^2\psi({\bf r})+V({\bf r},t)\psi({\bf r})\varphi(t)=i\hbar\psi({\bf r})\frac{\partial \varphi}{\partial t}(t) \end{equation} Dividing by $\Psi({\bf r},t)=\psi({\bf r})\varphi(t)$:
\begin{equation} -\frac{\hbar^2}{2m}\frac{1}{\psi({\bf r})}\nabla^2\psi({\bf r})+V({\bf r},t)=i\hbar\frac{1}{\varphi(t)}\frac{\partial \varphi}{\partial t}(t) \end{equation} But I can't continue with the method, because the left side depends on $t$ (it doesn't depend on ${\bf r}$ only).
