Right now I’m on the 4th floor of the physics building. If I walk down the hall to the balcony overlooking the atrium, I could drop a bouncy ball and watch its trajectory. It’ll fall to the ground, bounce up to some fraction of its initial height, and repeat the process with a loss of energy each time until it finally comes to rest. In a perfect world with no dissipation of energy, the ball would always bounce exactly up to its initial height and repeat its trajectory forever.
In principle, I can drop the magic ball at a given time on the stopwatch and thenceforth predict exactly where it will be at any point in the future. On the other hand, I might just choose to tell you that I have in fact made the drop from the 4th floor without telling you when exactly that drop was. If you passed your physics classes this semester, you could still calculate things like the bounce rate, impact velocities, and what have you. But you couldn’t say exactly where the ball is right now, because I haven’t given you the time of drop.
On the other hand, you do know that the ball is moving slower when it’s near the top of its trajectory and faster when it’s near the bottom. Thus it spends more time near the top and less time near the ground. You could, if you wanted, calculate the probability that the ball is near some given height h at a randomly chosen instant. In some sense, this is what a quantum mechanical wavefunction is*. We calculated the quantum wavefunction for just this problem not too long ago, and we can compare it to the classical probability distribution for the classical situation I’ve described.
We’ll do the comparison later. For now we’ll just calculate the “classical wavefunction”. We know that the probability P(x) that the ball will be found near a height x is proportional to how long it spends around x. How long it spends around x is inversely proportional to its velocity at x, because if it’s going fast it’s not going to be around long. So P(x) ~ 1/v(x). We should find v(x). From conservation of energy, we know that the total energy (kinetic and potential) of the ball is a constant. Call it E.
Remember that when I dropped the ball, its initial velocity is 0. This means that the actual numerical value of E is mgh, where h is the height from which I dropped it. Anyway, we can solve the above equation for v:
Since the probability is proportional to 1/v, we have:
But this isn’t quite a probability density. For one thing the units are wrong, for another it doesn’t integrate to 1. We need to find the normalization constant C such that:
Hey, it’s exam time. Work that out yourselves for practice. Most of the constants will cancel, and we’ll use some theorist-style simplification and define φ = x/h while picking units such that h = 1. In that case, the probability distribution is:
We can graph this:
From the graph, it’s clear that you’re a lot more likely to find the ball near the top – just as we expected. In fact, there’s about a 71% chance that you’ll find it above the halfway mark, and you’re more than 4 times more likely to find it in the top 1/5th than you are in the bottom 1/5th.
But this doesn’t look a whole lot like the Airy function, which is the quantum mechanical result. We’ll take a closer look at the reason next time.
*In other very important senses, the two situations are very different. For instance, one’s a stationary state and one isn’t. The quantum analogue of the classical problem is really something more like a coherent state rather than a high-n eigenvalue.