There was some dissension in the comments of my post on solving the Schrodinger equation with a linear potential. What the post boiled down to was that the solution was Ai(u), where we found that u was:

The point of the post was to work through and get that coefficient that’s in front of the (Fx + E). The point wasn’t to actually solve a physics problem with that, so I glossed over the fact that in solving the physical problem we’ll need to deal with boundary conditions, which means paying attention to both types of Airy function Ai(u) and Bi(u), as well as taking into account that only certain values of E will work, and that any linear combination of those Airy functions with those working E values is itself a solution. Which was fine with me, since the point was to actually do the math involved in getting that coefficient since I have not seen that clearly done elsewhere online.

But leaving it there seems to have confused some people, or at least irritated some people who think others might be confused. Fair enough! Let’s use what we did in the last post to actually solve a physics problem.

Let’s say we have a problem which we might call the quantum bouncing ball. We have a region of linear potential for x > 0 (say, the gravitation field of the earth), and we have an infinite potential at x < 0, which just means the wavefunction will be 0 at and below x = 0. Quantum mechanics requires that wavefunctions be continuous, so as the wavefunction approaches x = 0 from the right, that function had better go to zero. For convenience, pick units such that u = (x - E) which is possible if you remember that F is negative. In general, the solution will be some additive combination of the two Airy functions Ai and Bi.

If E = 0, our functions Ai(u) and Bi(u) look like this (for clarity in this and all other plots the horizontal axis is an x axis, not a u axis):

Well, the Bi function is the one that goes shooting up to infinity on the right. Wavefunctions can’t do that, they have to be normalizable. So we can say with assurance right away that all of the Bi terms will have 0 as a coefficient. They won’t be involved. But as is, this Ai(u) doesn’t work either. It has to be zero at the origin. We have to pick an E value such that it is. By looking up the zeros of the Ai function in a table, we can find that E = 2.33811 will do the trick:

So will E = 4.08795:

And so will any other E that shifts the function to the right exactly far enough so that one of the nodes happens at x = 0. These are the eigenvalues for this problem, which correspond to the energy levels of the quantum bouncing ball.

Any linear combination of these Ai(u(E)) with the different allowed E’s will be a solution to the problem, with the particular combination given by the initial conditions and the normalization requirement. Remember that we’ve graphed the whole Airy function which is just equal to the wavefunction on the side of the graph with x > 0, the wavefunction itself will be 0 for the side of the graph with x < 0.

One possible example is an equally weighted combination of the eigenfunction for the first and sixth energy eigenvalues. Normalized, it looks like this:

In general, given a particular initial condition (say, a Gaussian wavepacket representing a just-released bouncing ball) you can construct the linear combination of Ai(u(E)) that satisfy the initial conditions by exploiting the orthogonality of the Ai function, just as you would with any other eigenfunctions of the Schrodinger equation.

Well, I’m not sure if this will clear things up or not. But it should at least provide something of a bridge between the raw material of “The Airy function solves the Schrodinger equation in a linear potential” and an actual solution to a physical problem.

We’ll revisit this shortly and compare this quantum solution to the purely classical Newtonian solution of a bouncing ball. There’s some interesting differences – and some interesting similarities.