No bounded operators satisfy the canonical commutation relation with a nonzero commutator

[The hypothetical operators are nonzero] Suppose $A,B\in \mathcal{B}(H)$ satisfy $AB-BA=i\mathrm{\hslash }I$. Since $H\mathrm{\ne }\{0\}$, $I\mathrm{\ne }0$ and $\mathrm{\parallel }I\mathrm{\parallel }=1$ by [F1]. Because $\mathrm{\hslash }>0$, the commutator is nonzero; hence neither $A$ nor $B$ is zero.

[Commutators with powers] For every integer $n\ge 1$, [step 1.1, algebra] $[A,B^n]=ni\mathrm{\hslash }{B}^{n-1}\mathrm{.}$ The case $n=1$ is the assumed relation. If it holds for $n$, then $[A,B^{n+1}]=[A,B^n]B+B^n[A,B]=(n+1)i\mathrm{\hslash }{B}^n$, proving it by induction. No power $B^n$ can be zero: if one were zero, this identity would force $B^{n-1}=0$, and iteration would give $I=B^0=0$, contradicting [F1].

[Factorial lower bound] By step 2.1 and [F1]-[F2], for every $n\ge 1$, [step 1.1, step 2.1, F1, F2, algebra] $n\mathrm{\hslash }\mathrm{\parallel }{B}^{n-1}\mathrm{\parallel }=\mathrm{\parallel }[A,B^n]\mathrm{\parallel }\le \mathrm{\parallel }A{B}^n\mathrm{\parallel }+\mathrm{\parallel }{B}^{n}A\mathrm{\parallel }\le 2\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }{B}^n\mathrm{\parallel }\mathrm{.}$ Put $c:=\mathrm{\hslash }\mathrm{/}(2\mathrm{\parallel }A\mathrm{\parallel })>0$. Iterating the resulting inequalities $\mathrm{\parallel }{B}^n\mathrm{\parallel }\ge nc\mathrm{\parallel }{B}^{n-1}\mathrm{\parallel }$ from $\mathrm{\parallel }{B}^0\mathrm{\parallel }=\mathrm{\parallel }I\mathrm{\parallel }=1$ gives $\mathrm{\parallel }{B}^n\mathrm{\parallel }\ge c^{n}n!$ for every $n\ge 1$.

[Contradiction] By [F1], $\mathrm{\parallel }{B}^n\mathrm{\parallel }\le \mathrm{\parallel }B{\mathrm{\parallel }}^n$. Let $K:=\mathrm{\parallel }B\mathrm{\parallel }\mathrm{/}c$ and choose an integer $M>K$. For $n\ge M$, $n!\ge M!M^{n-M}$, so [step 3.1, F1, algebra] $\frac{c^{n}n!}{\mathrm{\parallel }B{\mathrm{\parallel }}^n}\ge \frac{M!}{M^M}{\left(\frac{M}{K}\right)}^n,$ which tends to $+\mathrm{\infty }$ because $M\mathrm{/}K>1$. This contradicts the simultaneous lower and upper bounds on $\mathrm{\parallel }{B}^n\mathrm{\parallel }$. Thus no bounded pair $A,B$ satisfies the commutator relation, and the displayed existential claim is refuted.