Imagine that one could theoretically trap a single electron in a small box, with walls that somehow prevent the electron from passing through and out of the box. Now, the box begins to move in on itself, gradually cornering and trapping the electron. The box continues trapping the electron until it perfectly encases it, disallowing the electron to move anywhere. I know that this scenario is seemingly impossible, but intriguing to think about. It has raised two questions in my mind:

  1. Would we then be able to know with exact certainty the electrons velocity and location, thereby negating the Uncertainty Principle?

  2. What would happen if the box kept on caving in even after perfectly encasing the electron? Could you actually crush an electron?


What you're describing is the particle in a box system, and for a 3D box the energy levels of this system are given by:

$$ E_{ijk} = \frac{\hbar^2\pi^2}{2mL^2}(n_i^2 + n_j^2 + n_k^2) \tag{1} $$

where $L$ is the size of the box. The numbers $n$ label the energy levels with $(1, 1, 1)$ being the lowest level and larger values of $n_i$, $n_j$ and $n_k$ giving higher energy levels.

The interesting thing about this system is that even in the lowest energy state, $(1, 1, 1)$, the energy is not zero. This is the phenomenon known as zero point energy, and a quick look at equation (1) shows that the magnitude of the zero point energy is inversely proportional to the size of the box squared:

$$ E_{111} = \frac{3\hbar^2\pi^2}{2m}\frac{1}{L^2} $$

So to you shrink the size of the box you have to put energy in by doing work on it. An electron is (as far as we know) a point particle and has no volume so to perfectly confine the electron would mean taking $L$ down to zero, and that would mean putting in an infinite amount of energy i.e. it can't be done.

The uncertainty principle comes in because the uncertainty in the electron's momentum scales roughly as the energy, so as $L \rightarrow 0$ the momentum uncertainty $\Delta p \rightarrow \infty$.

In fact even in principle we can't get $L$ smaller than about the Planck length, because at that point the energy density is so high that the system would form a black hole.

  • $\begingroup$ "Classic particle in a box" sounds almost like classical particle in a box, which is not what you meant :-) Maybe some rewording would help avoid confusion for some novice readers. $\endgroup$ – firtree Aug 25 '14 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.