$A$ and $B$ commute on a dense set but $e^{iA}$ and $e^{iB}$ do not

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?

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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 8 at 7:10
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. –  Argument Dec 13 at 8:14