Now it turns out that if we use the information that the particle is approximately at a particular location, we must give up the idea that the particle has a definite energy. The resulting probability density then does depend on time: the middle of the probability density (which in the above we connect with the idea of "a" position) is seen to move much as classical mechanics says its mythical point particles would move.
Let's start by seeing how this would all work classically: we say the particle starts approximately at x0 but with some uncertainty: x0. We release the particle on average with zero initial velocity, but because of Heisenberg uncertainty relation, there must be some spread of velocity v0, which is sometimes to the right and sometime to the left. If we knew exactly where the particle was starting from and its speed, we'd know exactly where it was:
(Why is =2 in our dimensionless variables?)
But because of the initial uncertainty in the velocity, there is varying uncertainty in position:
x(t)= [(x0cos(2t))2+ (½v0sin(2t))2]½
Thus the packet of probability expands and contracts as the central value of the probability follows the usual classical relation:
Our particular application will have x0=10, x0=0.31, v0=3.8.
Here is the behavior of the classical probability distributions:
Here are the quantum mechanical results for t=0,.2,.4,...,1.6
Here is the return motion. Here is QuickTime movie showing these positions.
The last plot shows the average position of the wavefunction vs time. The red curve is the classical result. Note that the "broadening" of the wavefunction is simply the result of our uncertainty in the initial velocity. Since the motion is exactly periodic, one period (T=) later the wavefunction is exactly like that at t=0.
Here is data on the location of the particle plotted above.