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Consider the Hilbert space $H = E^{\otimes n}$ where $E=\mathbb{C}^2$.

On $E$ we have an observable $O$ (i.e. a Hermitian matrix) that is diagonalizable in the standard basis with eigenvalues $1$ and $-1$. By tensoring $O$ with the identity on $E^{\otimes n-1}$, and doing so for each of the $n$ possible positions for the factor $E$, we get a set of commuting observables $O_i$, $i=1\dots n$. An observable that commutes with all these observables must belong to the algebra generated by the $O_i$'s.

Now if I have an observable $M$ on $H$ such that the operator norm of $[M,O_i]$ is at most $1$ for all $i$, how far can $M$ be from the algebra generated by the $O_i$'s, in the operator norm? What are explicit examples that are far from the algebra?

Any pointer or relevant remark for related questions welcome.

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  • $\begingroup$ The $O_i$ are not a complete set of commuting observables --- there aren't enough. The dimension of $H$ is $2^n$ not $n$. $\endgroup$
    – Nik Weaver
    Commented May 22, 2020 at 16:58
  • $\begingroup$ (Unless you mean something different than I do by a "complete" set.) $\endgroup$
    – Nik Weaver
    Commented May 22, 2020 at 17:00
  • $\begingroup$ @NikWeaver that's correct sorry. I meant the ring they generate is a complete set of commuting observables $\endgroup$
    – alesia
    Commented May 22, 2020 at 17:10
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    $\begingroup$ @NikWeaver: I think the question is basically how far away $M$ is from the space of polynomials in the $O_i$. $\endgroup$
    – Abdelmalek Abdesselam
    Commented May 22, 2020 at 17:23
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    $\begingroup$ @AbdelmalekAbdesselam that's correct, sorry for the bad phrasing $\endgroup$
    – alesia
    Commented May 22, 2020 at 17:26

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Well, you have $$ [\frac{1}{idx},x]=1/i $$ although $\frac{1}{idx}$ is far from the Identity.

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  • $\begingroup$ Yes, are you suggesting that taking the analog of this commutation relation over the finite field F_2^n instead of the reals would give an interesting example? $\endgroup$
    – alesia
    Commented May 22, 2020 at 17:18
  • $\begingroup$ What does your fraction mean? Should it be $\frac1 i\frac{\mathrm d}{\mathrm dx}$, or is there really some meaning to $\frac1{\mathrm dx}$? $\endgroup$
    – LSpice
    Commented May 22, 2020 at 17:22
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    $\begingroup$ I think the question is really finite dimensional in nature and this example does not help much. $\endgroup$
    – Abdelmalek Abdesselam
    Commented May 22, 2020 at 17:25

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