Can the operator $\operatorname{id}_V-(AB-BA)$ be nilpotent in the infinite-dimensional case.

Question

Let $V$ be a infinite-dimensional vector space over the field of characteristic $0$ and $A,B$ be linear operators of $V$. Let $\operatorname{id}_V$ be an identical operator.

Using trace function it is not easy to show that the operator $$ \phi = \operatorname{id}_V-(AB-BA) $$ can not be nilpotent in the case, where $V$ is finite-dimensional.

My question. Can the above operator $\phi$ be nilpotent in the case, where $V$ is infinite-dimensional?

Answer 1

Another example (inspired by quantum mechanics) is given by the operators $$A:f(x)\longmapsto xf(x)\qquad\text{and}\qquad B:f(x)\longmapsto\frac{d}{dx}f(x)$$ on $V=C^\infty(\mathbb{R})$. Again, their commutator is the identity, so that you obtain $\phi=0$.

Answer 2

Yes. Be $V$ spanned by the basis $\{e_i:i\in\mathbb N\}$. Be $Ae_i = \sqrt{i}e_{i-1}$ for $i>0$, $Ae_0=0$. Be $Be_i=\sqrt{i+1}e_{i+1}$. Then you can easily check that $(AB-BA)$ is the identity function, and therefore $\phi=0$ which of course is nilpotent.