Built on Facts

Classical Wavefunctions

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.

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

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:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

Since the probability is proportional to 1/v, we have:

i-d18063683dcc0d42b9be45451a84d1e3-3.png

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:

i-80f9713f6d11461837a9f9b540684e36-6.png

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:

i-bb5c2c6b0452df43a61e3974bd9b473f-5.png

We can graph this:

i-7b7142f1efa09b47c3b8242f89bcb9f4-graph.png

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.

Comments

  1. #1 Simon
    December 14, 2010

    I’m interested in seeing how far you go with this calculation.

    I actually calculated the (approximate) coherent states for the quantum bouncer during my 4th year at uni. It’s fairly special in that like the harmonic oscillator, the phase space trajectory of (,

    ) match the classical system quite well. This does not happen in other systems such as the infinite square well or the Coulomb problem.

    I’ve planning on writing up a quick demonstrations.wolfram for it, when I get some time….

  2. #2 Simon
    December 14, 2010

    Damn html… it should be the expectation value of x and p inside those parentheses.

  3. #3 djlactin
    December 14, 2010

    um… amusing spelling fail in title.

  4. #4 Patrick LeClair
    December 14, 2010

    I assigned the probability distribution for a classical pendulum in my modern physics class this semester, and in looking for lecture material came across this interesting paper:

    Am. J. Phys. 63 (9), 823, 1995: R.W. Robinett “Quantum and classical probability distributions for position and momentum”

    Covers the harmonic oscillator, infinite well, and linear confining potential, as well as a couple of unbound systems.

  5. #5 Paul Murray
    December 14, 2010

    If the probability is inversely related to the velocity, doesn’t that mean that the probability of it being at the top of its arc is infinite?

  6. #6 J
    December 14, 2010

    Re: Paul

    The pdf will approach infinity as the ball reaches the top of its arc, but the probability that it is exactly at the top at any given point in time is 0 since the distribution is continuous.

  7. #7 Annonymous
    December 15, 2010

    That’s actually the density function not the distribution function.

    A more direct derivation is as follows -

    The random variable is f(t) = h -(1/2) g t^2

    So for 0<=x<=h

    P(x) = Prob({f(t) < x }) = Prob ({sqrt(2(h-x)/g < t })
    = 1 – sqrt(2(h-x)/g)/sqrt(2h/g) = 1 – sqrt(1 – x/h)

    So the distribution function is

    0 x<=0
    1 – sqrt(1-x/h) 0<=x<=h
    1 h<=x

    Then the density function is

    0 x<=0
    1/(2 h sqrt(1-x/h)) 0<=x<=h
    0 h<=x

  8. #8 Annonymous
    December 15, 2010

    My comment got garbled. The inequality signs show as
    equal signs. The distribution function is 1 – sqrt(1-x/h)
    for x between 0 and h, 0 for x less than 0 and 1 for x
    greater than h.

    The density function is 1/(2 h sqrt(1-x/h) ) for x between
    0 and h and otherwise 0

  9. #9 Annonymous
    December 15, 2010

    Some of the derivation got lost. The point is
    that the distribution function for the random
    variable h – (1/2) g t^2 comes immediately from
    the definition of the distribution function using
    just algebra. Then the density function is obtained by
    differentiation.

  10. #10 dmabu
    December 16, 2010

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    image.spreadshirt.com/image-server/image/composition/4006595/view/1/producttypecolor/1/type/png/width/378/height/378/e-mc2_design.png

    now we are going to bury you…

    And the lesson from all of this? DOUBLE!
    ____________________________

    What do you want, you little ****ers?

    more of these idiots

    youtube.com/watch?v=q4C5yzFmC80

    en.wikipedia.org/wiki/List_of_prizes_for_evidence_of_the_paranormal

    HOW N WON ALL THE PARANORMAL PRIZES!

    en.wikipedia.org/wiki/Nostradamus

    THE HIGH PRICE OF REVOLUTION

    youtube.com/user/xviolatex?feature=mhum

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