Quantum Bouncing Ball

May 19th, 2008 · 2 Comments

In honor of the NBA playoffs, how about some quantum mechanics? Ok, maybe it’s a stretch! But let’s say we have a particle bouncing on a hard surface. We can model this by defining a floor at x = 0. For x > 0, the potential V(x) = mgx. For x < 0, the potential is V(x) = ∞. Solving this explicitly is not easy, though we will do it in a future post. But if we just want to find the ground state, we have a good trick up our sleeve for finding that energy. The key is simply to use Heisenberg’s uncertainty principle.

This is an approximation, of course. In actuality, the equality should be a “greater than”, and the Dirac constant should be divided by 2. But the h/2 is only achieved in the harmonic oscillator, and just using h as an approximation isn’t awful in most systems in the ground state. The ground state proviso is important: the minimum uncertainty occurs where the position and momentum are both in some sense maximally confined, which is of course in the ground state. As the uncertainty principle is a lower limit, there’s of course no real possibility of usefully applying it to higher energy states. But for the ground state, we can use it to our advantage by assuming that Δx = x and Δp = p. In other words, the expected position and momentum are approximately the same as their uncertainty.

Using E to represent energy now, we know that E = mgx, implying x = E/mg. We also can relate this to momentum with the famous relationship

Substitute these into the uncertainty relationship

And simply solve for E.

So we have a factor which contains the relevant units and variables, and a factor of 2^(-1/3) which gives the actual energy. 2^(-1/3) ≈ 0.7937, by the way. Is this accurate? It turns out the stuff under the radical is exactly right, as indeed it pretty much had to be in order to get the right units. But the error in the prefactor is significant. In reality, the answer is about 2.338 instead of 0.7937. The uncertainty is considerably greater than our estimate of h. It turns out to be more like 5h. But for an order of magnitude estimate it’s pretty good.

Tags: Undergraduate Physics Major · Worked Problems

2 responses so far ↓

1 meichenl // May 25, 2008 at 12:39 pm

I was confused reading through this, because you managed to eliminate the variable x in some step I thought I had glossed over. It took a minute before I realized you set

E = p^2/2m
AND
E = mgx

But this ought to be

E= p^2/2m + mgx

I don’t really understand why delta(x) = x and delta(p) = p, even as approximation. Could you elaborate a bit?

With that assumption, I’d then say

x = h/p

then plug this into the total energy

E = p^2/2m + m*g*h/p

then minimize this expression to find an estimate of ground state energy

doing this, I get

E = 1.5 * (m*g^2*h^2)^1/2

Matt replies: ~~Your answer doesn’t have units of energy, so I think there’s a glitch somewhere~~. I think your method is on the right track though - a variational attack should give a more precise answer than my estimate, though I wanted to avoid calculus since it’s aimed at beginning college students. UPDATE: Actually I think it was just a typo. Doing the math myself I think you should have a (1/3) instead of (1/2) as the power. This gives exactly what you need. It’s a better estimate too, as variational calculations usually are. But again I wanted to avoid calculus for the time being. Thanks for commenting, as your method will help those students who want to go the extra mile and get a more precise answer!

For the delta(x) = x and delta(p) = p approximation, think of it macroscopically. A bouncing ball will have an average potential energy roughly on the order of mgx, where x is the “average” height. But that average will be closer to the maximum height since the ball is traveling slower there. You can make a similar argument for the momentum, so then you’ve got both x and p in terms of some “average” E. But quantum mechanically, there’s a limit on how precisely you can know position and momentum. So since the ground state will in some sense have a maximally localized wavefunction (in both x and p space), the average x won’t be much higher than delta(x) because then it’s no longer maximally localized. Ditto for p space. Of course this is only even a loose approximation for the ground state, and quickly falls apart for higher states.
2 meichenl // May 25, 2008 at 2:04 pm

thanks for clarifying the approximation with uncertainty and expectations. it’s a nice trick.

there was a typo in my last line for the expression of the energy. the units actually come out exactly the same as yours, but with a coefficient of 1.5 rather than .7

Matt replies: I guessed that while you were typing this comment. Thanks again, and I hope you’ll be back! It’s great to have interested and qualified people help me make sense. :)

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