Let $H$ be an infinite dimensional Hilbert space. Is it true that we can always find two (continuous) linear operators $A, B: H \to H$ such that $AB, BA$ are of trace class, or at least $[A, B] = AB - BA$ is, such that $\mathrm{Tr}([A, B]) \neq 0$?
This question originated from the observation $[\partial, x] =1$ and that can be used to show that $\mathbb{C}[x]$ is infinite dimensional as a vector space since $\mathrm{Tr}([A, B]) = 0$ for finite dimensional vector spaces (although we can show it directly). In some sense, I wonder about the converse of this statement. Maybe there’s nothing to do with Hilbert space, but we can restrict to that case anyway.
This answer provides an example of two operators $A,B$ such that $AB-BA$ is in the trace class and ${\rm Tr}\,(AB-BA)\neq 0.$
Let $\{e_n\}_{n=0}^\infty $ denote an orthonormal basis in $\mathcal{H}$. Let $Ae_n=e_{n+1}.$ Then $A^*e_0=0$ and $A^*e_n=e_{n-1}.$ We get $A^*A=I$ and $AA^*= I-P_0$, where $P_0$ is the orthogonal projection onto $\mathbb{C}e_0.$ Hence $A^*A-AA^*=P_0$ and ${\rm Tr}\,(A^*A-AA^*)=1.$
If $\mathcal{H}$ is not separable we can decompose the space as $\mathcal{H}= \mathcal{H}_1\oplus \mathcal{H}_2,$ where $\mathcal{H}_1$ has a countable orthonormal basis. Then we define $A$ acting on $\mathcal{H}_1$ as above and acting on $\mathcal{H}_2$ as the identity operator. Then ${\rm Tr}\,(A^*A-AA^*)=1.$