Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 4 / 6B / Solution

Put $D=1+i\hslash t/m$ and write the Gaussian wave packet as $\psi ={\pi }^{-1/4}{D}^{-1/2}{e}^{-x^2/(2D)}$. Differentiation givesHence $\boxed{i\hslash {\psi }_t=-({\hslash }^2/2m){\psi }_{xx}}$, the free Time-dependent Schrödinger equation.

With $\tau =\hslash t/m$, its probability density isThe stated Gaussian integral shows its integral is one. Symmetry gives $⟨x⟩=0$; differentiating that Gaussian integral with respect to its coefficient gives $⟨x^2⟩=(1+{\tau }^2)/2$. Thus the position standard deviation isIt is not a stationary state: its probability density changes with time as the packet spreads. A stationary state would have only an overall time-dependent phase and a fixed position density.