# On commutator of bounded operators

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.)

• 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. Commented May 6 at 11:50
• 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. Commented May 6 at 14:14

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.