I am confused as to how time-independent and time-dependent perturbation theories in QM give consistent results at the instant the perturbation is switched on. Suppose I have a two-level system which obeys the time-independent Schrödinger equation,
$\hat{h}_0 \psi_{0,1}=\epsilon_{0,1}\psi_{0,1}$
which is initially in the ground state $\psi_0$. Then suppose I switch on a time-dependent perturbation $\delta v(t)$ at some time $t_0$. The system will then be described by a mixed state $\psi(t)$, with
$\psi(t)=c_0(t)\psi_0 + c_1(t)\psi_1$.
Up to 1st order in $\delta v(t)$, the coefficients $c_{1,2}(t)$ are given by
$c_0^{(1)}(t)=1$
$c_1^{(1)}(t)=-i\int_{t_0}^t h'_{1,0}(t')e^{i\omega_0 t'} dt'$, with
$h'_{1,0}(t)=\langle \psi_1 | h_0 + \delta v(t) | \psi_0 \rangle$ and $\omega_0=\epsilon_1-\epsilon_0$.
(These expressions are taken from David Griffith's book on quantum mechanics and I've ignored $\hbar$'s). This implies that the system remains (up to first order) completely in its ground state at time $t_0$ when the perturbation is turned on, since the integral expression for $c^{(1)}_1(t)$ vanishes at $t_0$.
However, if we consider the static Schrödinger equation at $t_0$, then could we not use time-independent perturbation theory (with the perturbation given by $\delta v(t_0)$) to determine the state $\psi(t_0)$? In which case, $\psi(t_0)$ would be given by
$\psi(t_0) = \psi_0 + \frac{h'_{1,0}(t_0)}{\epsilon_0-\epsilon_1}\psi_1$.
So it appears the state is already a mixed state of both $\psi_0$ and $\psi_1$. In my mind, this seems to be in contradiction with the result from time-dependent perturbation theory, where $\psi(t_0)=\psi_0$.
I'm guessing there is some flaw in using time-independent perturbation theory at the initial time, but can someone explain to me why this is the case or what the problem is if it's not that?