= Solution
Put $s=1+it$. Logarithmic <differentiation> gives
$$
\frac{\partial_t\Psi}{\Psi}=-\frac{i}{2s}+\frac{imx^2}{2\hbar s^2},\qquad \frac{\partial_x^2\Psi}{\Psi}=-\frac{m}{\hbar s}+\frac{m^2x^2}{\hbar^2s^2}.
$$
Therefore
$$
i\hbar\partial_t\Psi=\left(\frac{\hbar}{2s}-\frac{mx^2}{2s^2}\right)\Psi=-\frac{\hbar^2}{2m}\partial_x^2\Psi,
$$
so the appropriate <potential energy> is $\boxed{V(x)=0}$. The <wavefunction> never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.
The squared modulus is
$$
\boxed{|\Psi(x,t)|^2=\frac1{\sqrt{1+t^2}}\exp\left(-\frac{mx^2}{\hbar(1+t^2)}\right).}
$$
It changes width and height with time, so \b[this is not a <stationary state>]. Its <integral> is the constant $\sqrt{\pi\hbar/m}$; multiplying the <wavefunction> by $(m/(\pi\hbar))^{1/4}$ normalizes it. The normalized <Gaussian wave packet> stays centred at zero but spreads, with <position> <variance> $\hbar(1+t^2)/(2m)$. It represents a localized <free particle> with a spread of momenta, rather than an <energy eigenstate>. <Free Gaussian wave packet evolution> broadens the packet even with zero mean <momentum>.
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