5
$\begingroup$

Does there exist a simultaneous eigenbasis of the energy operator $T=\hat{p}^2/2m$ and the momentum operator $\hat{p}=-i\hbar\, d/dx$, for a particle in a 1-dimensional box of unit length?

Actually, one would expect that there is one because the commutator $[\hat{T},\hat{p}]=0$. However, I'm not sure about that because the full Hamiltonian has also a contribution of the (infinite) potential well.

Moreover, the eigenfunctions of the energy do vanish at the boundary of box, but the eigenstates of the momentum operator can never vanish there. However, the eigenfunctions of the momentum operator are also eigenfunctions of the energy. But the converse is not true.

$\endgroup$
  • $\begingroup$ See my answer to this question - physics.stackexchange.com/q/268672 $\endgroup$ – BoundaryGraviton Oct 17 '16 at 19:06
  • $\begingroup$ Sean Lake's answer is great, but if you want a more mathematical explanation, the momentum operator is not self adjoint in this situation (assuming the usual homogeneous boundary conditions). Therefore it has no eigenvectors. $\endgroup$ – Javier Oct 17 '16 at 21:30
  • $\begingroup$ @Javier Would you mind spelling that out in an explicit answer? Is it because the wave functions have a corner in them at the boundary? I ask because I'm imagining implementing the adjoint explicitly, using integration by parts, and that's the only point of breakdown that stands out. I'd also be interested to know how it would be affected if the boundary had Neumann style boundary conditions (vanishing derivative at edge, specified mean value). $\endgroup$ – Sean E. Lake Oct 17 '16 at 21:44
  • 1
    $\begingroup$ @SeanLake: I messed up. The momentum operator is indeed self-adjoint; the problem is that it is unbounded. This means that it need not have any eigenvectors or eigenvalues at all. The spectral theorem is subtle in infinite dimensions. $\endgroup$ – Javier Oct 17 '16 at 23:25
8
$\begingroup$

No. The eigen-functions of momentum are of the form $\operatorname{e}^{ikx}$, and they fail the boundary conditions of the box. If you were working on a ring, with periodic boundary conditions, this would be possible. As it is, with a box you have to construct the standing wave functions from even and odd combinations of the momentum eigen-functions, and select those standing waves that obey the boundary conditions. The constructed functions will be eigen-functions of $\hat{p}^2$, but not $\hat{p}$.

If you want to think of it in terms of the commutator, you are correct that you need to consider the contribution of the boundary walls. You can handle them using step functions of finite height that will later be extended upward in a limit. So: $$\hat{H} = \frac{\hat{p}^2}{2m} + D \left[\Theta(-\hat{x}) + \Theta(\hat{x} - a)\right],$$ for a box that extends from $0$ to $a$, and $\Theta(x)$ the Heaviside unit step function. Computing the commutator with $\hat{p}$ will give you: $$[\hat{p}, \hat{H}]_{\operatorname{x-basis}} = D \left[\delta(-x) + \delta(x - a)\right],$$ which shows problems at the wall boundaries, as expected.

$\endgroup$
  • $\begingroup$ Nice answer. Another way to put it is that for given boundary conditions an unbounded operator, in this case the momentum, may well be hermitic, but not self-adjoint. Here the set of bc-s that make the momentum self-adjoint on a box $[0,a]$ is of the form $\psi(0) = \alpha \psi(a)$ for $\alpha \in {\mathbb C}$, $|\alpha|^2 =1$, and does not include Dirichlet bc-s. $\endgroup$ – udrv Oct 18 '16 at 2:22
  • $\begingroup$ Thanks for the answer (+1). Under this circumstances, what are then the (possible) measurable momenta of the particle in the physical reality of the box? I guess they are countable. $\endgroup$ – kaffeeauf Oct 18 '16 at 5:46
  • $\begingroup$ See the momentum section of the Wikipedia article on the the particle in a box. Although I have some reservations about the possibility of a measurement returning a result that corresponds to a state that isn't physically realizable due to the infinite energy of the state - by postulate after measuring an observable the state is in an eigenstate of the observable that corresponds to the eigenvalue/result. $\endgroup$ – Sean E. Lake Oct 18 '16 at 14:12
  • $\begingroup$ @kaffeeauf I guess my reservation is resolved by switching off the box walls and then measuring the momentum. $\endgroup$ – Sean E. Lake Oct 18 '16 at 14:31

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.