I am working through a quantum mechanics problem involving the time evolution of a free particle (the particle is a proton) given that the initial state is a Gaussian wave packet of the form: $$ \psi(x,0)=(2\pi\sigma^2)^{-1/4}e^{-x^2/4\sigma^2}\,, $$ where $\sigma$ is the width of the Gaussian. I worked through it and got that the evolution, $\psi(x,t)$, is $$ \psi(x,t)=\frac{\sqrt{\frac{1}{\sigma^2}}(\sigma^2)^{3/4}}{2^{1/4}\pi^{3/4}}\int_{-\infty}^{\infty}e^{i(kx-\hbar k^2t/2m_p)-k^2\sigma^2}dk\,. $$ To derive this, I used the Fourier transform to expand $\psi(x,0)$ in terms of the eigenfunctions of a free particle, which are the plane waves. Then, I computed $\phi(k)$ using the inverse Fourier transform and substituted this back into the Fourier integral for $\psi(x,0)$. To compute $\psi(x,t)$, I realized that the component waves of the wave packet must propagate independently from one another, and the time evolution of a general plane wave is given by $\psi(\vec{r},t)=Ae^{i(\vec{k}\cdot\vec{r}-\omega t)}$. I multiplied this by the integrand of $\psi(x,0)$ to obtain $\psi(x,t)$, where I substituted $\hbar k^2/2m_p$ for $\omega$ (I used the Planck relation and the energy eigenvalues of a free proton).
Performing the integration and substituting the necessary constants, I plotted the probability density $|\psi(x,t)|^2$ and got a stationary Gaussian centered about $x=0$ that spreads out in time. By stationary, I mean that it does not move along the $x$-axis. It stays symmetric about the origin. I have also seen Gaussian wave packets that move along the $x$-axis as they spread. So, what is the difference? What makes a Gaussian stay centered while some appear to move?
