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Let $A$ and $B$ be unbounded, symmetric operators on a Hilbert space $H$ with a common domain $D$. If $AB = BA$ on $D$, is it necessarily that case that $e^{iA}$ and $e^{iB}$ also commute? If $A$ and $B$ are bounded, then I know that this must be the case. However, I am not sure if the same must be true for unbounded operators. Does anyone have a proof or a counterexample?

Adam Zalcman
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user15464
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    No, it's not necessarily the case. Nelson produced some counterexamples which are discussed in Reed and Simon, vol 1, section VIII.5. There's an nlab-page outlining the idea (the discussion seems to follow Reed and Simon quite closely). – Martin Jan 08 '13 at 07:10
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    One would have to require $ A $ and $ B $ to be self-adjoint and not just symmetric in order for $ e^{i A} $ and $ e^{i B} $ to be defined. – Berrick Caleb Fillmore Dec 13 '13 at 08:14
  • Without additional conditions ensuring strong commutativity, we cannot conclude that $e^{iA}$ and $e^{iB}$ will commute. – Furdzik Zbignew Oct 23 '24 at 15:54

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