Consider infinite potential well i.e. Hilbert space $L^2 \bigl([0,1]\bigr)$. Next we consider subset $$D_\theta = \left\{ \psi \in L^2 \bigl([0,1]\bigr) | \; \psi \; \text{is absolutely continuos and } \psi (0) = e^{i \theta} \psi (1) \right\} $$
on which we define operator $p_\theta = i \frac{\partial}{\partial x}$.
Denote by $\psi_{n, \theta} = e^{i (2\pi n – \theta) x}, \; n \in \mathbb{Z}$ eigenfunctions of $p_\theta$ to the eigenvalues $\lambda_{n, \theta} = 2\pi – \theta$.
Now move on to commutator $[x,p_\theta]$. Typically it would be equal to $-i$, but one can write:
$$\langle \psi_n | [x, p] \psi_n \rangle = \langle \psi_n | (xp – p x) \psi_n \rangle = \lambda_n \langle \psi_n | (x-x) \psi_n \rangle = 0 \neq -i \langle \psi_n | \psi_n \rangle = -i$$
My question is: how one should cope with uncertainty principle in infinite potential well?
Best Answer
The problem in the evaluation is that the function $x \psi_n$ lies outside the domain $D_{\theta}$ since it does not satisfy the boundary condition.
The operator $p{_\theta}$ is not self adjoint on this class of functions. Therefore, the step of its evaluation on the ket is not correct.
The correct way to perform the computation is by integration by parts and in this case, the boundary term at $x = 1$ gives the correct answer.
Please see the following article by: F. Gieres: "Mathematical surprises and Dirac's formalism in quantum mechanics", where examples of similar errors are given. Please see especially, example 5 in page 6 and its solution in page 39, which is very similar to the problem at hand.