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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?

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    $\begingroup$ If the domains are not the whole space, what does it mean for $[A,B]$ to be a scalar multiple of the identity? It might not be defined anywhere except $0$. $\endgroup$ – Robert Israel Aug 15 '14 at 7:07
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    $\begingroup$ Hmm.. clearly, the case you've mentioned is too trivial to deal with. It might seem to be confusing. I've thought of the case of the canonical commutation relation, $[x,p]=ih/2\pi$. I think if $\text{Im}(B)\subset \text{Dom}(A)$ and $\text{Im}(A)\subset \text{Dom}(B)$, it makes sense on $\text{Dom}(A)\cap\text{Dom}(B)$ as a domain-restricted identity and it can surely be a nontrivial large set. $\endgroup$ – user169867 Aug 15 '14 at 7:16
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    $\begingroup$ $\text{Dom}(A) \cap \text{Dom}(B) = \{0\}$ is a very real possibility; in some situations it can even be "generic". See e.g. math.ubc.ca/~israel/papers/is.ps $\endgroup$ – Robert Israel Aug 15 '14 at 7:20
  • $\begingroup$ Oh, thanks. I've overlooked that point. Um.. then, what if I suppose that both $A$ and $B$ are densely defined? $\endgroup$ – user169867 Aug 15 '14 at 7:26
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    $\begingroup$ You might also look at en.wikipedia.org/wiki/Canonical_commutation_relation and en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem $\endgroup$ – Robert Israel Aug 15 '14 at 15:12

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