First, here's the problem statement.
Suppose you have an infinite square well of length $a$, where the box extend from $x=0$ to $x=a$. At $t=0$, you add a perturbation $H'$ of the form: \begin{array}{ll} H' =V_0 \text{ if } 0 \leq x \leq a/2 \\ H' =0 \text{ otherwise} \end{array}
If the initial state is the ground state of the unperturbed well, $\Psi^0_1(x)$, and the pertubertation is turned off at $t=T$, what is the probability to find the particle in the first excited state, $\Psi^0_2(x)$, at $t =T$.
I first solved this problem using time-independant perturbation theory. To avoid confusion, note that the subscripts correspond to the energy quantum numbers and the superscripts correspond to the order of correction. Here's what I did:
1) Find the first-order correction to the ground state using $$\Psi^1_1(x) = \Psi^0_1(x) + \sum_{k\neq1} \Psi^0_k(x) \frac{\langle\Psi^0_k|H'|\Psi^0_1\rangle}{E^0_1 - E^0_k}$$
2)Normalize $\Psi^1_1(x)$. This part was tricky as it involves an infinite sum, but I manage to get an exact solution with mathematica. We'll let $A$ be the normalization constant.
3) Evolve $\Psi^1_1(x)$ with time.
$$\Psi^1_1(x,t) = \frac{1}{A}\left(\Psi^0_1(x)e^{-iE^0_1 t/\hbar} + \sum_{k\neq1} \Psi^0_k(x) \frac{\langle\Psi^0_k|H'|\Psi^0_1\rangle}{E^0_1 - E^0_k}e^{-iE^0_k t/\hbar}\right)$$
4) The probability at time T to find the particle in the first excited state is then $$|\langle\Psi^0_2|\Psi^1_1(T)\rangle|^2 = \left|\frac{1}{A}\frac{\langle\Psi^0_2|H'|\Psi^0_1\rangle}{E^0_1 - E^0_2}e^{-iE^0_2 T/\hbar}\right|^2$$
But now comes the important realization. If you compute $\langle\Psi^0_2|H'|\Psi^0_1\rangle$, you get a real answer (The wavefunctions are all sines and $H'$ is simply a constant). This implies that $$|\langle\Psi^0_2|\Psi^1_1(T)\rangle|^2 = \left|\frac{1}{A}\frac{\langle\Psi^0_2|H'|\Psi^0_1\rangle}{E^0_1 - E^0_2}\right|^2$$ i.e you get a time independant answer.
This is not really suprising as the perturbation was itself time-independant. The thing is, if I instead use time-dependant perturbation theory, the first-order correction is:
$$\langle\Psi^0_2|\Psi^1_1(T) \rangle = -\frac{i}{\hbar}\int_0^T e^{i(E^0_2-E^0_1)t'/\hbar}\langle\Psi^0_2|H'(t')|\Psi^0_1\rangle dt' = -\frac{i}{\hbar}\langle\Psi^0_2|H'|\Psi^0_1\rangle\int_0^T e^{i(E^0_2-E^0_1)t'/\hbar}dt' = -\frac{\langle\Psi^0_2|H'|\Psi^0_1\rangle}{E^0_2-E^0_1}(e^{i(E^0_2-E^0_1)T/\hbar}-1)$$
Hence $$|\langle\Psi^0_2|\Psi^1_1(T)\rangle|^2 = \left|\frac{\langle\Psi^0_2|H'|\Psi^0_1\rangle}{E^0_2-E^0_1}(e^{i(E^0_2-E^0_1)T/\hbar}-1)\right|^2$$ obviously has some time dependance.
So my questions are, why don't I get the same answer for both techniques? More importantly, how can one have time dependance and the other one does not have time dependance? Also, which technique yields the best approximation? I suspect time-independant perturbation is probably better, as if time-dependant was better, then there would really be no point in learning time-independant perturbation.
