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The momentum operator in the position representation is $\hat{p}=-i\hslash d/dx$. Differentiating the given Gaussian wave packet givesThe probability density $|\Psi {|}^2$ is an even function, so the term proportional to the odd function $x$ has integral zero. Normalization supplies ${\int }_{-\infty }^{\infty }|\Psi {|}^{2}dx=1$, and hence

The one-dimensional Time-dependent Schrodinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.