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Let $\hat A, \hat B$ be two operators on ${L}^2[-1,1]$ such that

$$ \hat A\psi(x) = x\psi(x)$$ $$ \hat B\psi(x) = -i\hbar \frac{d\psi(x)}{dx}$$

It's easy to see that $\hat A$ is bounded and self-adjoint, while $\hat B$ is unbounded. To define its adjoint, we limit $\hat B$ to the following domain $\{\psi \in \mathcal{C}^1[-1,1] \vert \psi(-1)=\psi(1)\}$. These two operators do not commute, therefore we'd expected Heisenberg uncertainty principle to hold. However, consider the family of functions $f_n(x)$ for $n\in \mathbb{Z}$

$$ f_n(x) = \frac{1}{\sqrt{2}}e^{i\pi nx}$$

which forms a basis of orthonormal eigenvectors of $\hat B$ and also belongs to Dom($\hat B$). Since they're eigenvectors, $\hat B$'s uncertainty in the state $f_n(x)$ is zero. Meanwhile, $\hat A$'s uncertainty is also well-defined and finite, perhaps we have no reason to believe it wouldn't. In other terms,

$$ \Delta_{f_{n}} \hat A \Delta_{f_{n}}\hat B = 0 $$

which is conflict with Heisenberg uncertainty principles. The reason such contradiction emerges is because $f_n(x)$ does not belong to Dom$(\hat A\hat B)\cap$Dom$(\hat B \hat A)$, as Heisenberg principle of uncertainty requires. In particular, $\hat A f_n(x)$ does not satisfy boundary conditions. However, this motivation doesn't really convince me, especially from a physical point of view. Is such a system even buildable in our universe? How would uncertainty relations really apply if we were to build it? Is it just mathematical gibberish or there's also a more concrete explanation?

Matteo Campagnoli
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    The notation looks pretty complicated, but it would seem that this is just a restatement of the questions Heisenberg's uncertainty principle derivation in a ring and How can I solve this quantum mechanical “paradox”?, for example. An interesting paper on this subject is Levy-Leblond's Who is afraid of nonhermitian operators. – Philip May 04 '21 at 20:16
  • Thanks for linking me those questions. I think they definitely deal with the same issues, such as the domain of unbounded operators and the "ring behavior" (the wavefunction has the same value at both extremes of the interval), which push us to be more cautious in quantum mechanics. However, if I'm not mistaken, there doesn't seem to be any reason to believe this so-called paradox may contradict Heisenberg uncertainty principle on a general basis. – Matteo Campagnoli May 05 '21 at 15:52
  • From what I understand, for the the uncertainty principle to hold between two operators, certain properties need to be satisfied (for the particle on the ring, the "angle" operator isn't singlevalued, for example) which aren't satisfied by one of your two operators. Levy-Leblond's paper shows how to construct certain operators that do satisfy the uncertainty principle: it's just that they are no longer Hermitian, and therefore don't correspond to physical observables. Take a look at the paper, it's really quite fun to read! – Philip May 05 '21 at 15:58

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