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I am wondering if someone has ever studied the following operator in the context of quantum mechanics:

$$Q : L^2(\mathbb{R}) \to L^2(\mathbb{R})$$

$$(Q f)(x) := s(x) f(x),$$

where $s(x)$ is the integer part of $x$. The operator $Q$ may also be thought of as $s(q)$, where $q$ is the usual position operator.

For example it is easy to see that it possess as eigenvectors all the function with support on a interval $(n,n+1)$, and the operator $P$ defined as $U^\dagger Q U$, where $U$ is the Fourier transform, is the same as $s(p)$, where $p$ is the usual momentum operator.

Can you suggest me some reference for such operator? For example I'd like to know the commutator $[Q,P]$.

Thanks

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  • $\begingroup$ How is this "smooth"?? $\endgroup$
    – Nik Weaver
    Commented Mar 20, 2014 at 20:52
  • $\begingroup$ Changed "smooth" to "coarse", you made your point. $\endgroup$
    – giulio bullsaver
    Commented Mar 20, 2014 at 21:00
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    $\begingroup$ There's no general definition of the commutator of two unbounded operators, and in this case I really don't see any sensible way to interpret your $[Q,P]$. $\endgroup$
    – Nik Weaver
    Commented Mar 20, 2014 at 23:53
  • $\begingroup$ what are you using to define U, the unitary operator? it doesn't seem like your function would necessarily have a closed form expression. it would be nice though. it seems as if, in the limit, your functional would return the value of the function at the position, however, it wouldn't be very well behaved. $\endgroup$
    – physicalmaths
    Commented Mar 21, 2014 at 13:22
  • $\begingroup$ @physicalmaths: it's the Fourier transform, suitably normalized to make it unitary. I think the formula for $P$ is a typo, it should just be $P = U^* Q U = s(p)$. $\endgroup$
    – Nik Weaver
    Commented Mar 21, 2014 at 18:36

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