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.

  • 2
    $\begingroup$ I did it for you this time, but note that you should use '\langle' and '\rangle' instead of '<' and '>'. The difference is $$\langle\psi|H|\psi\rangle$$ and $$<\psi|H|\psi>$$ $\endgroup$ Commented Apr 6, 2019 at 3:51
  • 2
    $\begingroup$ @AaronStevens Thank you, I didn't know that! $\endgroup$
    – Shamaz
    Commented Apr 6, 2019 at 3:55
  • 3
    $\begingroup$ In number 3) you evolve an approximate eigenket of the the full hamiltonian according to the time independent hamiltonian treatment where every ket gets a phase factor. However, because the Hamiltonian is time dependent, the Schrodinger equation doesn't have the nice looking $e^{iHt}$ evolution anymore, and what you did is invalid. Time-dependent perturbation theory is the only way to go. $\endgroup$ Commented Apr 6, 2019 at 4:04
  • $\begingroup$ @DinosaurEgg Couldn't it so be valid because the Hamiltonian is constant on $0<t<T$? $\endgroup$ Commented Apr 6, 2019 at 12:24
  • $\begingroup$ @DinosaurEgg I understand what you are saying, but couldn't I write then that, to the first order, $\Psi^1_1(x,t) = \Psi^1_1(x) e^{-iE^1_1 t/\hbar}$, where $E^1_1$ is the first order correction to the energy of the ground state? In that case every $\Psi^0_k(x)$ in my series get mutliplied by the same exponential. But in any case, this does not change the final result. Ho and why is the Hamiltonian time-dependant? $\endgroup$
    – Shamaz
    Commented Apr 6, 2019 at 15:47


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.