Let $\mathbb H$ be a Hilbert space and let $\mathcal B(\mathbb H)$ be the bounded operators on $\mathbb H$. Let $J,K\in \mathcal B(\mathbb H)$ such that $ J=J^*, K=-K^*. $ Then the commutator $[J,K]$ is selfadjoint, equal to $JK+(JK)^*$.

Claim. If $[J,K]\ge 0$, then $[J,K]=0$.

Question. Is it true, is it well-known? What I know is that it is impossible to have $[J,K]=I$, showing that Quantum Mechanics needs unbounded operators to get the Uncertainty Principle, based upon the non-commutation of operators such as $D_x=\frac1{i}\frac{d}{dx}=J$ and $ix=K$. Here, we have a stronger statement, claiming that it is not even possible that the spectrum of the self-adjoint bounded operator $[J,K]$ is included in the non-negative half-line. (I guess that the above Claim also prevents the spectrum to be included in the non-positive half-line.)

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    $\begingroup$ If one of the operators is trace class, it follows immediately from the fact that the trace of $[J,K]$ (or any commutator) vanishes. It shouldn't be hard to get the general case by approximation. $\endgroup$ Commented May 6 at 11:50
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    $\begingroup$ More generally, the claim is easy if one of $J$ or $K$ has a basis of eigenvectors. Indeed, if $\xi$ is an eigenvector for $J$ or $K$, then $\langle [J,K] \xi,\xi\rangle=0$. The approximation argument does not seem straighforward to me, though. $\endgroup$ Commented May 6 at 14:14

1 Answer 1


If $H$ is finite dimensional there is a one-line solution: ${\rm tr}(JK - KJ) = 0$, so $JK - KJ$ cannot have positive spectrum.

But it is false in general! Let $V$ be a partial isometry from $H$ onto a proper subspace $K$, so that $V^*V = I$ and $VV^* = P \neq I$ is the projection onto $K$. Then let $J = V + V^*$ and $K = V - V^*$; we have $[J,K] = 2(V^*V - VV^*) = 2(I - P) \geq 0$, but it is not zero.

  • $\begingroup$ Nice example, Nik. $\endgroup$ Commented May 6 at 18:45
  • $\begingroup$ @MikaeldelaSalle thank you! $\endgroup$
    – Nik Weaver
    Commented May 6 at 20:02
  • $\begingroup$ Nice! The shift operator has struck again... $\endgroup$ Commented May 6 at 20:32
  • $\begingroup$ Sharp and natural: a great example. Thx. $\endgroup$
    – Bazin
    Commented May 6 at 21:37

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