 Solution

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.