Built on Facts

Clearing the air on the Airy fuction

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:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

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):

i-ec8d2bc9698de2bf6063e486cb0cc968-g1.png

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:

i-30ec64ac881f15c96f03db209a447ba5-g2.png

So will E = 4.08795:

i-c250b5394d71554fa8bbd4a11856b1e1-g3.png

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:

i-3cc003759e870542a167e65b3e51980e-g4.png

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.

Comments

  1. #1 Omega Centauri
    November 10, 2010

    You got me thinking about the airy function. How could I compute such a beast. Its easy to create a power series about x=0, but since it
    contains Bi(x) as well as Ai(x), and we know the former function blows up strongly at x values that aren’t even that big. That implies for nontrivial X values that the cancellation of terms will be ferocious. It ain’t ganna be easy to not just lose all your digits of precision by contaimination with even a timy little bit of Bi(x) (which is impossible to totally eliminate numerically). The formal definition from the fourier domain doesn’t look promising either, an indefinite integral of a function which oscillates faster and faster for ever. Trying to find a way to numerically integrate it would be a project in itself. I’m guessing the easiest thing to do is find a value
    of ai(x) and d/dx(Ai(x)) at a largish value of X and maybe integrate towards zero, that way epsilion worth of Bi will be supressed rather than grow like heck….

    Anyway, I think computationally, it sounds like an interesting problem.

    Of course if we don’t put in a boundary condition but instead let
    the wave form extend to minus infinity, it isn’t square integral, but instead has an amplitude like the dirac delta of zero! So you need your left side BC. Add in a rightward B.C. and you can then allow some Bi(x) to creep in.

    ? Adding two diferent modes? Aren’t these waveforms complex, and multiplied by exp( i * omega * t ), where the omega values will differ for each mode? Or are PSI(x) real and time invariant (hey its been like almost 40years for me since I did this stuff)?

  2. #2 agm
    November 11, 2010

    OC, if I were dealing with such a problem, I’d definitely use a technique like integral transforms to move this into a space where

    a) boundary conditions are integrated in the interpretation and solution of the problem and/or

    b) the problem has easier mathematical properties.

    As the recent post on Laplace transforms showed, integral transforms often turn ugly calculus problems into much simpler algebraic problems. Need quadrature of a signal to makes various signal processing algorithms useful for seismic? Hilbert transform. Need to investigate poles in the parameter space associated with an electrical system’s description? Laplace transform. Want to remove certain types of coherent noise from a data set? Tau-p transform and mute the noise. Mr Fourier has many friends.

    Odds are, some of these spaces will resolve a lot of the computational issues you are thinking about.

  3. #3 Raskolnikov
    November 11, 2010

    Great job! I’m looking forward to the next post!

  4. #4 JR
    November 11, 2010

    The digital library of mathematical functions has a short discussion on how to compute them. http://dlmf.nist.gov/9.17

    Apparently one of the simplest methods is to solve the differential equation directly numerically.

  5. #5 Omega Centauri
    November 11, 2010

    Thanks to all, especially JR, that NIST publication looks to be of great value. I was only entertaining the idea of a real function of a real value, and had considered that integration towards the neg x direction assuming one can obtain an accurate seed Ai and d/dx of Ai(x) at some suitably largish X value. In any case I don’t actually need the function, but I like to know that if I needed one (or something similar) that I could figure out how to do it.

    I was frustrated, I had spotted my Abramowitz & S in some odd corner of the house recently but can’t locate it. It looks like that reference could be a massively improved replacement for it.

  6. #6 Eric Lund
    November 12, 2010

    Alternatively, if you already have code to compute Bessel functions of fractional order, you can use that code to compute Airy functions. The relations are given in section 9.6 of the NIST link above, or if you can find your copy of A&S look for equations 10.4.14 through 10.4.21 (page 447 in my edition).

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    February 8, 2012

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