Ehrenfest theorem proof I'm using this resource along with Griffith's Introduction to Quantum Mechanics to try and reproduce the Ehrenfest theorem.
From equation $(176)$ in the link above, we have:
$$\frac{d\langle p\rangle}{dt}=\int_{-\infty}^{\infty}\left[\frac{-\hbar^2}{2m}\frac{\partial}{\partial x} \left( \frac{\partial \Psi^*}{\partial x} \frac{\partial \Psi}{\partial x} \right)
+V(x)\frac{\partial|\Psi^2|}{\partial x} \right] dx$$
I am able to get to here without issues, but next we have to show that:
$$\int_{-\infty}^{\infty}\left[\frac{-\hbar^2}{2m}\frac{\partial}{\partial x} \left( \frac{\partial \Psi^*}{\partial x} \frac{\partial \Psi}{\partial x} \right)
 \right] dx = 0$$
Which would only be true if:
$$\left. \frac{\partial \Psi}{\partial x} \right|^{x=\infty}_{x=-\infty} = \left. \frac{\partial \Psi^*}{\partial x} \right|^{x=\infty}_{x=-\infty} = 0$$
Is there a way to know this generally? It's obviously true in certain cases of the wave function (e.g. $\Psi(x)=\exp[-x^2]$). In general, I thought the only condition for normalization was that:
$$\left. \Psi \right|^{x=\infty}_{x=-\infty} = \left. \Psi^* \right|^{x=\infty}_{x=-\infty} = 0$$
 A: Fun fact; it's not true in general! For example, this answer lists an example of a function that is totally square-integrable and therefore viable as a wave-function but whose derivatives do not have a well-defined limit at infinity.
The real reason you can get away with doing this approximation is that we assume implicitly in quantum mechanics, perhaps with not enough forcefulness, that wave-functions have "compact support", i.e., the functions and their derivatives are only nonzero on a closed, bounded subset of space.
Some toy examples of wave-functions eschew this requirement, such as the quantum free particle with exact momentum, but this is not a true wave-function as it is not square-integrable.
A: Another question's answer notes that the expectation value of kinetic energy is proportional to
$$\int_{-\infty}^\infty dx\, \psi^* \left(-\frac{\partial^2\psi}{\partial x^2}\right) = \int_{-\infty}^\infty dx\, \left|\frac{\partial\psi}{\partial x}\right|^2 - \left.\psi^* \frac{\partial\psi}{\partial x}\right|_{-\infty}^\infty.$$
In the Ehrenfest derivation, you have already been willing to set boundary terms that include a factor of $\psi$ or $\psi^*$ (without derivatives) to zero. So the above reduces to $\int_{-\infty}^\infty dx\, (\partial\psi^*/\partial x)(\partial\psi/\partial x)$. Now, on physical grounds, this expectation value of kinetic energy should be finite. Thus, the integrand approaches zero as $x \to \pm\infty$, which suffices to show the vanishing of the boundary term you asked about.
