Consider a quantum mechanical system with hamiltonian $\hat{H}$, and eigenstates $\psi_n$ with energies $E_n$. I want to `force' this system into a specific eigenstate $\psi_0$ that has energy $E_0$. My idea is to add a periodic perturbation $\epsilon \hat{V}(t)$ that is periodic in time with period $\frac{\hbar}{E_0}$; here $\epsilon$ is a small real number. The new system has Hamiltonian $$\hat{H}+\epsilon\hat{V}(t).$$ My intuition says that, regardless of the initial condition at $t=0$, the periodic perturbation would cause the whole system to be periodic with period $\frac{\hbar}{E_0}$ in the limit $t\rightarrow\infty$; and in the unperturbed system, the phase of eigenstate $\psi_0$ also oscillates with this period. So when $\epsilon$ is very small, the system would eventually move into a state close to $\psi_0$. However, I have not managed to prove this. Is this true in general?
Own attempts
Here are my attempts using variation of constants and Floquet's theorem. Firstly, write $\psi(t)=\sum_n c_n(t)\psi_n$ with all $c_n(t)\in\mathbb C$. Now we get from the Schrödinger equation: $$\frac{\partial c_n(t)}{\partial t} = \frac{1}{i\hbar}\left(E_nc_n(t)+\sum_m \langle \psi_n\mid\epsilon\hat{V}(t)\mid\psi_m\rangle c_m(t)\right).$$ This is a linear system of differential equations with time-periodic coefficients. So we can apply Floquet's theorem. This gives us \begin{equation} \begin{pmatrix} c_1(t)\\ c_2(t)\\ \vdots\\ c_N(t) \end{pmatrix} = P(t)e^{tB}\mathbf{v} \end{equation} where $P(t)$ is a time-periodic $N\times N$-matrix with period $\frac{\hbar}{E_0}$, $B$ is a constant $N\times N$-matrix, and $\mathbf{v}$ is a $N\times 1$-vector. We know that $|c_n(t)|^2\leq1$ for all $n,t$, so it follows that all eigenvalues of $B$ have real part at most $0$.
At this point I get confused. How can I prove that when $t\rightarrow\infty$, the system will converge to a state that is close to $\psi_0$? Or is that not true in general? With `close' I mean that when $\epsilon\downarrow0$, the limiting state will converge to $\psi_0$.