Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S3 / 2

The equation is $i\hslash {\partial }_t\psi =[-{\hslash }^2{\partial }_x^2/(2m)+V(x)]\psi$. The generator of time evolution is the sum of kinetic and potential energy operators. For a sharp stationary energy, insert $\psi (x,t)=\varphi (x)e^{-iEt/\hslash }$ to obtain $[-{\hslash }^2{\varphi }^{\prime \prime }/(2m)+V\varphi ]=E\varphi$. The probability in $dx$ is $|\psi {|}^{2}dx$, normalized to one; hence a normalized one-dimensional wavefunction has dimensions length${}^{-1/2}$. Its overall phase is unobservable, but relative phase affects interference.

Integrating $F=-kx$ gives $V=k{x}^2/2$ up to a constant. Thus $\boxed{-{\hslash }^2{\varphi }^{\prime \prime }/(2m)+m{\omega }^2{x}^2\varphi /2=E\varphi }$, with $\omega =\sqrt{k/m}$. The potential is a symmetric upward parabola and tends to infinity in both directions; normalizable states decay at large $|x|$.

For the Gaussian, ${\varphi }_0^{\prime \prime }=(x^2/{\sigma }^4-1/{\sigma }^2){\varphi }_0$. Using $\hslash /(m\omega )={\sigma }^2$, the $x^2$ terms cancel and leave $E_0=\hslash \omega /2$. For ${\varphi }_1\propto x{\varphi }_0$, differentiation gives ${\varphi }_1^{\prime \prime }=(x^2/{\sigma }^4-3/{\sigma }^2){\varphi }_1$, and the same cancellation gives $\boxed{E_1=3\hslash \omega /2}$. The normalization constants do not affect these eigenvalue equations.

Squaring the states gives $\boxed{p_0=e^{-x^2/{\sigma }^2}/(\sqrt{\pi }\sigma ),p_1=2x^2{e}^{-x^2/{\sigma }^2}/(\sqrt{\pi }{\sigma }^3)}$. The first peaks at zero; the second has a node there and peaks at $x=\pm \sigma$. For a centered interval of width $\delta x\ll \sigma$, $P_0\simeq \delta x/(\sqrt{\pi }\sigma )$, whereas $P_1\simeq (\delta x{)}^3/(6\sqrt{\pi }{\sigma }^3)$. Its leading linear term vanishes, but the probability in a finite interval is not exactly zero.

Both probability densities are even, so $⟨x⟩=0$. The supplied Gaussian moments give $⟨x^2{⟩}_0={\sigma }^2/2$ and $⟨x^2{⟩}_1=3{\sigma }^2/2$. Therefore $\boxed{\Delta x_0=\sigma /\sqrt{2},\Delta x_1=\sqrt{3/2}\sigma }$. The original plots mark these two symmetric rms intervals around the common mean.

Time evolution gives $e^{-i{E}_{0}t/\hslash }[{\varphi }_0+e^{-i\omega t}{\varphi }_1]/\sqrt{2}$. It becomes the opposite relative-sign superposition when $e^{-i\omega t}=-1$, first at $\boxed{t_{\min }=\pi /\omega }$. The overlap with the initial state is proportional to $1+e^{-i\omega t}$ and vanishes then; global phase is immaterial.