 Solution

Direct differentiation verifies the free-particle Schrodinger equation, so $V=0$. At $t=0$, normalisation of $A{e}^{-x^2/2}$ gives $A={\pi }^{-1/4}$. Symmetry gives $⟨x⟩=0$, while Gaussian integration givesIts growth shows wave-packet spreading and therefore a nonstationary state.