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If a large impulse $F(t)= A \delta (t)$ is applied on a system say particle in a box, how to find the final state just after the impulse? This I can get. But I want to get an expression for probability to be found in the old hamiltonian eigen state.

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  • $\begingroup$ Hello and welcome to the physics stack exchange! Since this is not a typical help site, it would be great to ask questions such that it benefits the wider audience. You could improve your post by elaborating your question, explaining where you encounter the problem and what your progress is so far. Have a great time! $\endgroup$
    – AlphaLife
    Apr 27, 2021 at 14:50

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If you're trying to solve the time-dependent 1-D Schrödinger equation, $$ - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V(x,t) \psi(x,t) = i \hbar \frac{\partial \psi}{\partial t}, $$ you could treat the impulse as though it was due to a time-varying potential. Specifically, if $F(t) = A \delta(t)$, this could be treated as though it arose from a time-varying potential $V(x,t) = - A x \delta(t)$. (Note that classically $F = - \partial V/\partial x$.)

Now, in the region around $t = 0$, the spatial derivatives will be negligible compared to the time derivative and the potential, and so we can approximate the equation as $$ \frac{A x}{i \hbar} \delta(t) \psi(x,t) + \frac{\partial \psi}{\partial t} \approx 0. $$ This is a differential equation of the form $u' + P u = 0$, which has as its formal solution $u(t) = C e^{-\int P \, dt}.$ In our case, this implies that in the region of time around $t = 0$, we have $$ \psi(x,t) = f(x) e^{i A x \Theta(t)/\hbar} $$ where $\Theta(t)$ is the Heaviside theta-function (i.e., the antiderivative of the Dirac delta-function.)

In other words, the wavefunction will be shifted by a position-dependent phase. If you know what your wavefunction is at times preceding $t = 0$, it will simply be multiplied by $e^{i A x/\hbar}$ afterwards. Note that this effectively shifts $p \to p + A$ in momentum space, which is exactly what we would expect this impulse to do.

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  • $\begingroup$ I think that doesn't make sense because $\psi$ could be changing significantly during the small time interval, so you can't just integrate both sides treating it as a constant. What really should be happening is that if locally $\psi(x) = A(x) e^{i \theta(x)}$ then $d\theta/dx$ jumps. $\endgroup$
    – knzhou
    Apr 27, 2021 at 20:57
  • $\begingroup$ @knzhou: Hmm, fair point. My underlying idea was to treat this like the standard treatment of the delta-function potential $V(x) \propto - \delta(x)$ (for the time-independent Schrödinger equation.) In that case, you get a discontinuity in the first derivative of $\psi$ with respect to $x$. But in this case, the derivative I'm integrating is first-order in $t$, leading to a discontinuity in $\psi(x,t)$ itself at $t = 0$. And since $\psi(x,t)$ is discontinuous, it's not clear what $\psi(x,0)$ should mean. $\endgroup$ Apr 27, 2021 at 21:26
  • $\begingroup$ @knzhou: Updated with something that seems more plausible. $\endgroup$ Apr 27, 2021 at 21:44
  • $\begingroup$ ,but when i try to get the probability to be found in old hamiltonian eigen state i am not getting a probability equal to 1 is it correct $\endgroup$
    – Raghu
    Apr 28, 2021 at 13:23
  • $\begingroup$ @Raghu: I don't know for sure, but I wouldn't expect the system to stay in the same state except in the limit of $A \to 0$ (i.e., no impulse.) $\endgroup$ Apr 28, 2021 at 14:15

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