So consider the one-dimensional time-independent Schrodinger equation:

In some ways it’s not really an equation as such, because you have to plug in some function V(x) that describes the potential in the problem you’re solving. When you first learn quantum mechanics you’ll learn the big ones: V(x) = 0, V(x) = V > E, V(x) = 1/2 kx^2, V(x) = k/x, etc. One that’s conspicuously absent is usually one of the simplest looking potentials: V(x) = -Fx. That’s just the potential of a constant force, say a particle in a uniform gravitational field. Rarely do textbook authors bother to work out a solution – they’ll just say “This can be solved with the Airy function”, and maybe give the solution as an accomplished fact.

Which isn’t too bad – in most cases if you’re the kind of person who knows what an Airy function is you can probably work it out yourself. But there’s something to be said for a walkthrough for those who’re just getting to the point of understanding how to solve differential equations in terms of special functions. So let’s do it:

For the Schrodinger equation with a linear potential, we have to solve:

Rearrange and cancel the negative signs:

This looks kinda sorta familiar. The Airy function – by defintion – is the function solving the differential equation:

Our equation looks kinda sorta like that. What we’d like to do is to finagle our Schrodinger equation into that form, and then we’d know that the Airy function of some argument would be our solution. We might suspect the substitution u = Fx + E would be a good start. But that can’t quite be it, because we’d still have that hbar/2m term floating out in front of the derivative. There may be some coefficient that can simultaneously get rid of the stuff in front of the derivative and turn Fx + E into just u. Let’s call that coefficient (if it exists) A. We then have:

By the chain rule,

If the u-x relationship weren’t linear we’d have to be a little careful here, but it is and so it’s clear that:

So we need the coefficient in front of the u to be -1, so go ahead and multiply through by -A:

Plug that sucker in and we have:

The solution to which is, by definition:

Well, there’s also the Airy function of the second kind Bi(x), but that diverges for positive arguments and usually isn’t relevant. In occasional circumstances it is, and you should be aware of it.

I hereby declare us done. Now in practice you have to apply some boundary conditions. For instance, in solving the problem of a quantum ball bouncing on a hard surface we have the boundary condition psi(x = 0) = 0, and only particular values of E will satisfy those conditions. Those are the eigenvalues that make up the energy spectrum. That’s just details though.

Details which you’ll have to know if you’re studying for exams – which are rapidly approaching if you’re a student. Good luck! It’s also interesting to compare this quantum result to the classical one, which we’ll do shortly.