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.