Transitions in constant perturbations using time independent perturbation theory
The perturbed Hamiltonian is given as.
$$H=\begin{cases} H^{(0)}\;for\;t\leq 0 \ H^{(0)}+V(x)\;for\;t>0\end{cases}.$$
Here $V(x)$ does not depend explicitly on time but it can depend on coordinates If the initial state of system at $t<0$ is $|i\rangle$, then the probability of the transition from $|i\rangle$ to $|f\rangle$ for some $t_o>0$ is (calculated through time dependent perturbabtion theory)
$$P_{i\to f}(t_o)=|V_{fi}|^2\Bigg(\frac{4\sin^2(\frac{\omega_{fi}t_o}{2})}{(E_f-E_i)^2}\Bigg)\tag{1}. $$
If $V^{(0)}$ is constant, then the state does not change but only the energy eigenvalues gets scaled.
So, for $t>0$,
$(H^{(0)}+V^{(0)})\psi(x)=E\psi(x)$
$H^0\psi(x)+V^{(0)}(x)\psi(x)=E\psi(x)$
$\implies H^0\psi(x)=\epsilon\psi(x)$ where $\epsilon=E-V^{(0)}$
So, the eigenstates of the perturbed Hamiltonian does not change only the energy gets scaled.
So, in this case transition probability from $|i\rangle\to|f\rangle$ for $i\neq f$ is $0$ if the system is in $|i\rangle$ for $t\leq 0$.
But suppose now $V=V(x)$ has position dependence. 669
So we can see this $H(x)=H^{(0)}+V(x)$.
Now for $t>0$ we can use time independent perturbation theory.
If all the states are non-degenerate, then first order correction in $|i\rangle$ is $$|i^{(1)}\rangle=-\sum_{n\neq i}\frac{V_{fi}}{E_n-E_i}|n\rangle.$$
We know that $|i^{(0)}\rangle$ (zeroth order correction i.e., $|i\rangle$) is orthogonal to $|i^{(1)}\rangle$
So, transition probability from $|i\rangle$ to $|f\rangle$ for $i\neq f$ is
$$P_{i\to f}(t_o)=\Big(\frac{|V_{fi}|^2}{(E_f-E_i)^2}\Big)\tag{2}$$
We can see that $(1)$ and $(2)$ are not same. There is no sine term in $(2)$.
We get both $(1)$ and $(2)$ using the first order correction in the unperturbed wavefunctions but we are getting different results.
Why are we getting different results, I was expecting that both time dependent and time independent perturbation theory will give the same result for such perturbations.
Addendum
I am improving my question as I think I was not able to properly frame it.
For some $t_o>0$, we have perturbed Hamiltonian. Now at that instant we can look the problem through the perspective of time independent perturbation theory because perturbed Hamiltonian for $t>0$ does not depend explicitly on time.
Now I get $(2)$ as my result and it does not depend at which instant we are looking the system for $t>0$.
Now I can make $t$ sufficiently small (greater than 0) my solution through time independent perturbation theory won't change.
Now we can see that before $t<0$ we have state $|i>$ and after $t>0$ we have a perturbed wave function (upto first order) which is the superposition of basis states of unperturbed wave function $(\psi^{(1)}=\sum_nc_n|n>)$.
Now we can see that from $(2)$, time dependence in the perturbed wave function can come only in its phase, not in observables or probability density. 669