Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 18D (Quantum Mechanics)

For the proposed Gaussian wave packet, differentiation givesInserting these in the Time-dependent Schrödinger equation with Hamiltonian $(p^2-x^2)/2$, its right-hand side becomesEquating the constant and quadratic coefficients with $i\hslash {\partial }_t\Psi$ yieldsConversely these two equations make the substitution an identity, so they are sufficient for the Gaussian evolution in an inverted harmonic oscillator.

Set $z=\varphi -it$ with real constant $\varphi$. ThenThus $B=\tan (\varphi -it)/(2\hslash )$ solves the Riccati equation wherever it is finite. Integrating the amplitude equation also giveswith the square root chosen continuously from $t=0$. For the normalizable solutions used in part (c), $\sin (2\varphi )>0$; equivalently one can choose $0<\varphi <\pi /2$ modulo $\pi$. The unrestricted real parameter solves the differential equation locally, but not every value defines a normalizable wavefunction.

Write $B=b+ic$ with $b>0$. Normalization of the Gaussian wave packet gives $|A{|}^2\int e^{-2b{x}^2}dx=1$, hence $|A{|}^2=\sqrt{2b/\pi }$. The supplied Gaussian integral ratio then givesThe momentum operator obeys $p\Psi =2i\hslash Bx\Psi$, and direct second differentiation gives $p^2\Psi =(2{\hslash }^{2}B-4{\hslash }^2{B}^2{x}^2)\Psi$. Taking the expectation value and substituting the first result givessince $2b-B=\bar{B}$. These are the second moments of a complex Gaussian wave packet.

For the solution in part (b), the trigonometric identitygives, writing $D=\cosh (2t)+\cos (2\varphi )$,Thus normalizability requires $\sin (2\varphi )>0$, and the exact expectation values areSince $\cosh (2t)\sim e^{2t}/2$,Their difference is the constant $-\hslash \cot (2\varphi )$, so the mean energy $⟨H⟩=-\hslash \cot (2\varphi )/2$ stays constant despite exponential spreading in both observables.