In The Dirac-delta function as an initial state for the quantum free particle, Emilio Pisanty
states that if an object's position wave function, $\Psi(x,0)$, is a delta function (at $t = 0$), then $$ \Psi(x,t)=\begin{cases}\delta(x) & t=0\\ \sqrt{\frac{m}{2\pi\hbar |t|}}e^{-i\mathrm{sgn}(t)\pi/4}\exp\left[i\frac{mx^2}{2\hbar t}\right]&t\neq 0\end{cases} $$ In that case, at $t>0$, the momentum wave function is$$ \Phi\left(k,\, t\right) = \frac{1}{\sqrt{{2\pi}}}\sqrt{\frac{m}{2\pi\hbar t}}e^{-i\mathrm{sgn}(t)\pi/4}\int_{-\infty}^{\infty}\exp\left[i\frac{mx^2}{2\hbar t}\right]e^{ikx}\mathrm{d}x $$
According to Engineering Tables/Fourier Transform Table 2, and if we set $\mathcal{x'}=-x$,$$ \begin{alignat}{7} \Phi\left(k, \, t\right) & ~=~ -\sqrt{\frac{m}{2\pi\hbar t}} \, e^{-i\mathrm{sgn}(t)\pi/4} \, \sqrt{\frac{\hbar t}{m}} \, \exp\left[{-i\left(\frac{\hbar t k^2}{2m} - \frac{\pi}{4}\right)}\right] \\ & ~=~-\frac{1}{\sqrt{{2\pi}}} \, e^{-i\mathrm{sgn}(t)\pi/4} \, \exp\left[{-i\left(\frac{\hbar t k^2}{2m} - \frac{\pi}{4}\right)}\right] \\ & ~=~-\frac{1}{\sqrt{{2\pi}}} \, \exp\left[{-i\frac{\hbar t k^2}{2m} }\right] \\ & ~=~ -\frac{1}{\sqrt{{2\pi}}} \, \exp\left[{-i\omega t }\right]\end{alignat}$$
This seems to indicate that, for $t>0$, both $\Psi^2$ and $\Phi^2$ are constant in space, though $\Psi^2$ diminishes in time. That means that both position and momentum are totally uncertain in space for $t>0$.
Did I get that right?
I know that a position eigenstate is not a solution of Schrodinger's equation and so this is all somewhat outside the scope of orthodox QM, but this does seem to violate the idea that the less certain is one variable, the more certain is it's conjugate variable.
