When working with the wave function of a particle ($\Psi(\vec{r}, t)$), my professor always makes the assumption that not only does said wave function fulfill $lim_{r\to\infty} \Psi = 0$ but in fact he assumes that this function goes to $0$ faster than its derivatives. This gives rise to two questions:
Under what circumstances can we consider $\Psi$ to be an element of a Schwartz space? Since it is a rapidly decreasing function, I'm guessing it generally does belong to said space, but to be honest I don't know enough math to back up this statement. I'm more interested, however, in knowing when I can assume $\Psi$ tends to zero faster than its derivatives than in the formal mathematical aspect of the wave function
Given that $lim_{x\to\infty} \Psi(x,t) = 0$, then does this mean that the first derivative is also $0$ when approaching infinity? That is to say, if $\Psi$ tends to $0$ when approaching infinity, then $\frac{\partial \Psi(x,t)}{\partial x} \to 0$ when $x\to \infty$? Intuitively, since $\Psi$ is getting all flat, the derivative should go to zero as well. The reason I'm asking this is because, after integrating by parts, we often get a term such as $\left. \Psi\cdot \frac{\partial^n \Psi}{\partial x^n}\right |_{-\infty}^{\infty}=0$ and my professor argues this happens because $\Psi$ tends to zero faster than its subsequent derivatives. However, if all of its derivatives approach zero when $r\to \infty$ then I don't understand why he doesn't simply rule the expression out as $0\cdot 0=0$. The only reason why he'd do that is, I presume, because some derivatives actually go to infinity, which doesn't make sense at all.
