I am interested in the question of, if a particle is initially localised at some position $x_0$ what it will evolve to at a later time assuming a free Hamiltonian $H = p^2 /2m$. Long story short, I find that I am roughly able to get a result but what I get is nonsense.
In momentum-space, the TDSE is simply $$i \frac{\partial}{\partial t}\psi(p,t) = \frac{p^2}{2m} \psi(p,t)$$ This has the solution $\psi(p,t) = f(p)\exp(-i\omega_p t)$ where $\omega_p = p^2 /2m$. If the particle is initially localised in space at $x_0$, then $f(p) = \exp(-ipx_0)/2\pi$ (i.e. a delta function in position space). Thus we have the solution in momentum space $$\psi(p,t) = \frac{1}{2\pi}\exp(-ipx_0 - i\omega_p t)$$
However if I transform this back into position space I get (after completing the square in the exponent) $$\psi(x,t) = \frac{1}{2\pi} \sqrt{\frac{2m}{t}}e^{im(x-x_0)^2/2t} \int_{-\infty}^\infty e^{-is^2} ds$$ Aside from the fact that the integral over $s$ doesn't converge to anything, we have the additional problem that $|\psi(x,t)|^2 \propto 1/t$ doesn't depend on $x$. In other words, for all times the particle is completely delocalised in position-space. Have I made a mistake or is there a deeper problem with what I'm doing?
