Let $\Psi(x, t)$ be the wavefunction for a particle of mass $m$ moving in one dimension in a potential $U(x)$. Show that, with suitable boundary conditions as $x \rightarrow \pm \infty$,

$$\frac{d}{d t} \int_{-\infty}^{\infty}|\Psi(x, t)|^{2} d x=0$$

Why is this important for the interpretation of quantum mechanics?

Verify the result above by first calculating $|\Psi(x, t)|^{2}$ for the free particle solution

$$\Psi(x, t)=C f(t)^{1 / 2} \exp \left(-\frac{1}{2} f(t) x^{2}\right) \quad \text { with } \quad f(t)=\left(\alpha+\frac{i \hbar}{m} t\right)^{-1}$$

where $C$ and $\alpha>0$ are real constants, and then considering the resulting integral.