how will we know the wavefunction which is normalized for time t=0 is also normalized for all times?

Question

abb0t · Answer

There is an integral function that defines this. I am sure you can find this in your book:
$$\frac{ d }{ dt } \int\limits |\psi (x,t)|^2dx=\frac{ i \cal{h} }{ 2m }[\psi^*\frac{ \partial\psi }{ \partial x }-\frac{ \partial \psi^* }{ \partial x }\psi ]$$
Now the right hand side needs to be shown to be zero but this is only true if we make the assumption that $\psi$ is normalizable at all time, t. However, you can find functions which change from normalizable to non normalizable and still satisfy the schrodinger Equation.

abb0t · Answer

btw, that $h$ is the hamiltonian operator. I didn't know how to change it using LaTex.

abb0t · Answer

For example, if ur given a 1-D finite square well potential, $V_0$. The spectrum of the Hamiltonian, $h$ includes one or more (normalizable) bound states with $E < 0$, and a continuum of un-normalizable scattering states with $E > 0$. Suppose initially the system is in a bound state with energy $E_0 < 0$. Then, at $t=0$, you turn on an additional time-varying potential $V_1$ which oscillates at frequency $\omega$. And $E_0 + h\omega > 0$, so this gives the system enough energy to escape the well to infinity. In fact since $V_1$ includes a step function at $t = 0$, it contains a continuous range of frequencies, and as a result the system's energy will also have a continuous range. In terms of perturbation theory, $V_1$ is a perturbation connecting a normalizable state with an unormalizable one, and the system has evolved from one to the other.