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If the Hamiltonian of a system changes abruptly (over a very short time interval) from one form to another, we would expect the wave function not to change much, yet its expansion in terms of the eigenfunctions of the initial and final Hamiltonians may be different but the wave function in first is the same.

now my problem : consider the famous problem that we have for example a particle is initially in the ground state of an infinite, one-dimensional potential wall with walls at x=0 and x=a. first if the wall at x=a is suddenly moved to x=8a,we can expansion the wave function(in first moment it's not change)in new eigenfunctions and there is no problem!but if we the wall at x=a is suddenly moved to a/2 then the wave function in new situation is not normalized and we can not expand it in new eigenfunctions because it's not zero at x=a/2!so what's wrong with this?is this told us that we can not do the second scenario and it's impossible?why?

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    $\begingroup$ When you combine infinite potential walls with instantaneous changes in the Hamiltonian, you can get indeterminate results that involve $0\cdot\infty$, particularly if you use the wrong basis. That's not the whole issue in the problem you are discussing, but it is important to keep in mind. $\endgroup$ – Buzz Jun 8 at 0:33
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Leaving aside the a-physical approximation of a mathematically infinite potential, the validity of the sudden approximation relies on the alteration of the Hamiltonian imposing sufficiently small changes on the state of the system, and that really isn't the case when you push the boundary far into the original space in the well.

How do we know?

Because the time-evolution of the wave-function is given by the application of the Hamiltonian to the wave-function $$ \frac{\partial}{\partial t}\Psi(x,t) \propto \left[\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \right] \Psi(x,t) \;, $$ but from the conditions of the problem and the initial state we know that during the time the well is evolving $V(x)\Psi(x,t)$ is a large potential applied to a non-trivial wave-function.

So the time-evolution of the state will be non-trivial, so treating the system as un-perturbed by the action is a problem.


Contrast this with the "expanding the well" case where $\Psi(x,t)$ is small (exactly zero in the ideal case) in the spatial region where the large change in $V$ occurs.

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