While writing an answer to this question, I started doubting about the interpretation of the uncertainty principle for the particle in a box.

In the 1-dimensional particle in a box problem, explicit solutions for the energy eigenstates exist, and are essentially of the form $\sin nx$ outside the support of the potential, and 0 where the potential is infinite. These somehow feel as if they should have definite momentum (up to direction), since they are free where the potential is 0, so that their energy is kinetic energy, and they cannot be in the region where they are not free.

On the other hand, position is strictly restricted to some interval. Consequently, any state must have bounded support in positional representation, and one would say there can be no states with bounded support in momentum space (by Paley-Wiener if you want). Indeed, the energy eigenstates involve an infinite number of momentum eigenstates (as they are zero outside an interval).

What is the way out? I think it must be that part of the energy in the energy eigenstates is potential energy. To see that, we can think of the idealized infinite potential as a limit of some sequence of finite potentials, which also is necessary in order to be able to give a meaning to the second derivative appearing in the Hamiltonian. This second derivative in the boundaries of the well tends to a multiple of a Dirac mass which has to be compensated in a very essential way by the term $V(x)\psi(x)$ in the Schrӧdinger equation $\psi''(x) + V(x)\psi(x) = \psi(x)$ (up to constants), so that in any approximation to the infinite potential will have a non-negligible contribution from the potential energy.

Is this a correct interpretation?


The eigenstates for a particle in a box are neither states of definite momentum nor superpositions of just two momentum eigenstates of equal magnitude and opposite sign. For a box of length $L$ centered at the origin, the energy eigenstates are

$$ \psi_n(x) = \begin{cases} \sqrt{\frac{2}{L}} \cos \left( \frac{n \pi x}{L} \right) \qquad &\text{for $n$ odd} \\ \sqrt{\frac{2}{L}} \sin \left( \frac{n \pi x}{L} \right) \qquad &\text{for $n$ even} \end{cases}$$

inside the box and zero elsewhere. Fourier transforming via $\psi_n(k) := \int_{-L/2}^{L/2} e^{-i k x} \psi_n(x)\, dx$ gives

$$ \psi_n(k) = \begin{cases} (-1)^{(n-1)/2} 2 \pi n \sqrt{2 L} \frac{\cos(k L/2)}{(n \pi)^2 - (k L)^2} \qquad &\text{for $n$ odd} \\ (-1)^{n/2} 2 \pi i n \sqrt{2 L} \frac{\sin(k L/2)}{(n \pi)^2 - (k L)^2} \qquad &\text{for $n$ even} \end{cases}$$ and as we'd expect from the Paley-Weiner theorem, the wavefunctions have unbounded support in momentum space.

The reason that the eigenstates don't have definite momentum magnitude (as we'd expect) is that momentum is only a good quantum number in translationally invariant systems, and the hard walls in the Hamiltonian break translational invariance. If we were to impose periodic rather than hard-wall boundary conditions, then translational invariance would be restored and the energy eigenstates $\psi_n(x) = \frac{1}{\sqrt{L}} e^{i n \pi x/L} $ would indeed be states of definite momentum $k_n = n \pi / L$. But in this case, the spatial manifold is compact and periodic and so is topologically equivalent to the circle $S^1$ rather than the real line $\mathbb{R}$, so we have Fourier series that are only defined at discrete values in Fourier space instead of Fourier transforms, and the Paley-Weiner theorem doesn't apply.


But the momentum is not definite.

The standing wave state is a mixture of waves with wave number $k$ and $-k$ (here I am assuming a square-well for simplicity of notation). In more than one dimension we have to write those as vectors, but the same principle applies: there is a mixture of multiple states with different momenta.

And the smaller the box the larger the ground state energy, which also implies a larger magnitude for $k$.

  • $\begingroup$ Thanks, good point. However, the state is not a superposition of the states of definite wave numbers $\pm k$ either, but only the restriction of these wave functions to the box. Also the problem would remain: the support of the momentum is bounded (even if not a single point), so that the position would have to be unbounded. $\endgroup$ – doetoe Jun 13 '14 at 0:44
  • $\begingroup$ @doetoe: you do have an uncertainty in momentum, because the sign of the wave number is undefined. For the uncertainty principle we are interested in the momentum, not the magnitude of it. So the momentum is uncertain by something like $\hbar k$ and the position by $\frac 1k$ $\endgroup$ – Ross Millikan Jun 13 '14 at 2:15
  • $\begingroup$ @RossMillikan i realize that (at least after dmckee's answer). It may not be the uncertainty principle that would be violated but another property of the Fourier transform, namely that not both position and momentum can have bounded support $\endgroup$ – doetoe Jun 13 '14 at 2:33
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    $\begingroup$ The momentum does not have bounded support. We tend to ignore the fact that there are much higher momentum states involved, but it is true (though they have small coefficients). Since the position sine wave has a derivative discontinuity at the boundaries, the transform will have positive coefficients going to infinity $\endgroup$ – Ross Millikan Jun 13 '14 at 2:40
  • $\begingroup$ I edited the question a bit to make it clearer and to correct the erroneous assertion that the momentum feels definite (its magnitude does) $\endgroup$ – doetoe Jun 13 '14 at 2:48

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