# Cyclicity of the trace for operators

I know that if I have two operators $$A$$ and $$B$$ and one is bounded and the other is trace class, then $$\mathrm{Tr}(AB) = \mathrm{Tr}(BA).$$ Another case where this works is when $$A$$ and $$B$$ are both Hilbert-Schmidt operators.

But I heard that it is actually sufficient to have $$\mathrm{Tr}(|AB|)<\infty$$ and $$\mathrm{Tr}(|BA|)<\infty$$. Has anyone a reference about that? Are there other cases where the cyclicity of the trace still works, or at least where it works "in a certain sense"? In particular I am interested in the case where $$A$$ is a really nice operator and $$B$$ is unbounded.

For example, say $$B=x$$ is the unbounded operator of multiplication by $$x\in\Bbb R$$ and $$A$$ is a compact positive operator acting on $$L^2$$ functions $$\varphi$$ through the formula $$A\varphi(x) = \sum_j \lambda_j\, \psi_j(x) \int_{\Bbb R} \psi_j\,\varphi$$ with $$\sum_j \lambda_j\int_{\Bbb R} |\psi_j(x)|^2\,(1+|x|)\,\mathrm d x< C$$. Then $$\mathrm{Tr}(AB) = \mathrm{Tr}(BA) = \sum_j \lambda_j \int_{\Bbb R} |\psi_j(x)|^2\,x\,\mathrm d x$$

Remark: Another quite borderline case where I know how to do the proof is if $$A$$ and $$B$$ are positive operators and $$\sqrt A\sqrt B$$ is an Hilbert-Schmidt operator and one defines $$\mathrm{Tr}(AB) := \mathrm{Tr}(\sqrt B\,A\,\sqrt B) = \|\sqrt A\sqrt B\|_2^2$$. Then by invariance of the Hilbert-Schmidt norm by taking the adjoint, $$\mathrm{Tr}(AB) = \|\sqrt A\sqrt B\|_2^2 = \|\sqrt B\sqrt A\|_2^2 = \mathrm{Tr}(BA).$$

## 1 Answer

The case of $$\mathrm{Tr}(\vert AB\vert)<\infty$$ and $$\mathrm{Tr}(\vert BA\vert)<\infty$$ for bounded $$A,B$$ is addressed in Simon's book Trace Ideals and Their Applications, Corollary 3.8.

The idea is to show that $$AB$$ and $$BA$$ have the same nonzero eigenvalues including multiplicity and then applying Lidskii's theorem. The former is done in Deift's paper Applications of a commutation formula (this is referenced in Simon); it also treats a very specific case with unbounded operators, namely when $$B=A^*$$ (but in this case $$A^*\!A$$ looks to be unbounded, hence not trace class).