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?
Best Answer
If you look at your distribution in momentum space, you can see that it is an even function about $p=0$. For this reason, the average value of the momentum is zero, and so the center of the wave packet will remain stationary.
However, you can "imprint" a momentum on the original position-space wave function by multiplying by a plane wave. That is, if the initial state is given by $$ \psi(x,0)=(2\pi\sigma^2)^{-1/4}e^{ik_0x}e^{-x^2/4\sigma^2}\,, $$ then the center of the momentum-space wave function is at $p = \hbar k_0$, indicating that the average value of the momentum is $\hbar k_0$. Then, the center of the wave-packet in position space will be at $x=\frac{\hbar k_0}{m}t$, i.e., the wave-packet will move with a speed give by $p_0/m$.