How to find the transition probability in impulse approximation? 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.
 A: 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.
