Feynman’s path integral and an electron in a Penning trap

ajw1

Last night the BBC repeated Brian Cox, A Night With the Stars (https://www.youtube.com/watch?v=HTU4CUhSmuw). At some point a calculation is done using a simplified version of Feynman’s path integral, where the mean time is estimated for a diamond to be found outside a small box:

t>[itex]\frac{x Δx m}{h}[/itex]

The box was not expected to be found empty then after several hundred billion times the age of the universe.
But yet in 1973 it was found to be possible to capture a single electron in a Penning trap, and keep it there for over a week (http://www.iap.uni-bonn.de/lehre/ss0...Wineland73.pdf). So when I apply the formula above to the electron, I find that I only have to wait for about a second to have a reasonable chance to find it one meter outside of that trap.
So why doesn’t this experiment invalidate Feynman’s formula?

mfb

He used a simplified formula - in particular, the potential height is not part of the final result, but it has to for the electron.

ajw1

It is always dangerous to use simplified formulas, as you show here.

So is it right to conclude that a free electron will swap its position randomly in space by roughly 1 meter every second (I take Δx = 1 mm and x=1m, and arrive at t=1.4sec)?

The_Duck

The interpretation of the formula in the original post is this: suppose I have an object of mass ##m## in free space, and I know that the particle is somewhere in a region of size ##\Delta x##. Because of the uncertainty principle, the object cannot have exactly zero momentum. Therefore eventually it will move out of these region of space. The formula gives the time ##t## at which the object can be expected to have moved a distance of about ##x## as a result of this nonzero momentum.

In particular, that formula only applies to an object in free space. If we confine the object somehow, such as electron in a trap, that formula has nothing to say about this situation. As an example, if the only object in the universe were a hydrogen atom, consisting of one electron bound to one proton, the electron would *never* escape the proton. (The uncertainty principle still applies, but its consequences are more subtle).

Applying the formula to a diamond in a box is thus misleading, because the box has walls that will keep the diamond inside the box for vastly longer than the formula suggests. The diamond can in principle eventually escape, but only through quantum tunneling, which, for the case of a diamond in a box, would take an almost infinite amount of time.