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If I have three operators such that $[A,B] = C$, and I know that $A$ and $C$ are Hermitian, does it follow that $B$ is anti-hermitian? If $A$ and $B$ were Hermitian, $C$ would be anti-Hermitian, so $B$ clearly isn't Hermitian (unless its trivial); but I'm having trouble coming up with a counter example or a proof. (The specific problem I'm interested in has $C$ equal to a nonzero constant, if that helps.)

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It is not true that for every $B$ with $[A,B]=C$, $B$ is anti-hermitian. If it were, you can always add $A$ to $B$ without changing the commutator making the resulting $B$ clearly not anti-hermitian. I suspect that you can always find a $B$ such that $B$ is antihermitian and the commutator relation is fulfilled. –  Fabian May 15 '11 at 20:18
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