I want to solve the problem indicated in the title with $C$ a normalization constant and $d$ a scaling factor and $k_0$ the expected value of the wave number. $\widetilde\psi(k) = Ce^{-(k-k_0)^2d^2}$ describes a gaussian wave packet. Here's what I've got so far:
1) $$\psi(x,t) = \frac{1}{\sqrt{2\pi}} \int \widetilde\psi(k) e^{i(kx-\frac{\hbar k^2}{2m}t)} dk $$ = something quite complicated (using substitution, completing square and integration of the standard gaussian) yielding
2) $$|\psi(x,t)|^2 = \frac{|C|^2}{2\sqrt{d^4+(\frac{\hbar t}{2md})^2}}e^{\frac{-(x-\hbar k_0t)^2}{2(d^2+\frac{\hbar t}{2md}^2)}}$$
Now I'd say $\sigma_x = \sqrt{(d^2+(\frac{\hbar t}{2md})^2)}$ by comparison with a gaussian probability distribution. Also I think that using $p = \hbar k$ you can get $\sigma_p =\frac{\hbar}{\sqrt{2}d}$ by rewriting $\widetilde\psi(k) = Ce^{-(k-k_0)^2d^2}$ as $\widetilde\psi(k) = Ce^{-\frac{(p-p_0)^2d^2}{\hbar^2}} = Ce^{-\frac{(p-p_0)^2}{2\frac{\hbar^2}{2d^2}}} $ and comparing with the gaussian probability density. However, the correct answer is supposed to be $\sigma_p =\frac{\hbar}{2d}$.
What am I missing here?