= Gaussian evolution in an inverted harmonic oscillator
{c}
{title2=$B(t)=\frac{\tan(\phi-it)}{2\hbar}$}
For the <Hamiltonian (quantum mechanics)> $H=(p^2-x^2)/2$, a <Gaussian wave packet> $Ae^{-Bx^2}$ satisfies $\dot A=-i\hbar AB$ and $\dot B=-i/(2\hbar)-2i\hbar B^2$. The displayed solution is normalizable for real $\phi$ exactly when $\sin(2\phi)>0$; its <second moments of a complex Gaussian wave packet> are
$$
\langle x^2\rangle=\frac{\hbar[\cosh(2t)+\cos(2\phi)]}{2\sin(2\phi)},\qquad
\langle p^2\rangle=\frac{\hbar[\cosh(2t)-\cos(2\phi)]}{2\sin(2\phi)}.
$$
Both grow as $\hbar e^{2t}/[4\sin(2\phi)]$, while their difference and hence the mean <energy> remain constant.
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