I have shown a proposition:
Suppose $A$ and $B$ are two linear operators on a (complex) Hilbert space, where the domains may not be the whole. Then, if either $A$ or $B$ is normal and has an eigenvector in the space, then the commutator $[A,B]=AB-BA$ cannot be a nonzero scalar multiple of the identity.
A proof follows:
Suppose that $A$ is normal and has an eigenvector $f$/eigenvalue $a$, and $[A,B]=k\in\mathbb{C}-\{0\}$. Then, $$0 = (a^*-a^*)\langle f,Bf \rangle = \langle A^*f,Bf \rangle - \langle f,BAf \rangle \\= \langle f,[A,B]f \rangle = \langle f,kf \rangle = k^*||f||^2 \ne 0.$$
Is it sound? If then, I want to generalize it further.
(1) Is the converse also true? I mean, if a (normal) operator $A$ has no eigenvector in the space, then can we say that there exists another operator $B$ such that the commutator $[A,B]$ is a nonzero scalar multiple of the identity?
(2) We can weaken the condition "normal"? Is there a general theorem on the eigenvectors of $A$ and $A^*$? For example, do there always exist pairs $Af=af$ and $A^*g=a^*g$ such that $\langle f,g \rangle \ne 0$ holds? If not, what condition do we need?