Derivative of wavefunction of a quantum system

Question

If $\psi(x)$ represents the wavefunction of a 1D quantum system, it satisfies the Schrodinger equation, has a unit norm, and $\lim_{x\rightarrow \infty }\psi(x)=0.$ Then is it true that $\lim_{x\rightarrow \infty }\dfrac{d\psi(x)}{dx}$ also zero or another constant?

Answer 1

In a formal mathematical sense, no. Consider

$$ \psi(x)=\sin(x^2)/x $$ This wavefunction is smooth and square integrable, and goes to zero at infinity. However, it's derivative does not go to zero at infinity, as you can verify.

However, for convenience we often assume all wavefunction have vanishing derivative at infinity. This is required to make $P^2$ Hermitian.