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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    – AlphaLife
    Apr 27, 2021 at 14:50

1 Answer 1


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.

  • $\begingroup$ 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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