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The backward direction is trivial and this one probably too, but I just can't find a convincing argument.

$A$, $B$ are Operators on a Hilbert Space (Ket Space).

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Jul 9 at 8:10
  • $\begingroup$ Maybe, but as physicists deal a lot with such commutators I asked here. Should I move the thread? $\endgroup$ – TheoreticalMinimum Jul 9 at 8:14
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    $\begingroup$ General tip: Never crosspost, but you might flag for migration. $\endgroup$ – Qmechanic Jul 9 at 8:23
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It is not true. Take for example $$ B = 2\pi \mathrm{i} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} , $$ then $\exp(B) = 1$ and $[A,\exp(B)] = 0$ for all $A$.

However, clearly, $[A,B] \neq 0$ for some $A$.

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  • $\begingroup$ This seems legit. Do you know where the mistake might be in my proof-scetch? $\endgroup$ – denklo Jul 9 at 8:25
  • $\begingroup$ You sure that exp(B)=1 ? I tried calculating exp(B) using Mathematica, and I'm getting exp(B) = {{1,1},{1,1}}, and not the value of {{1,0},{0,1}} (= Identity matrix). $\endgroup$ – Samuel Weir Jul 9 at 8:38
  • $\begingroup$ Let $B=2\pi i S$. Then $S^2=1$ and so $e^{2\pi i S}=\cos(2\pi)+i \sin(2\pi) S=1$. $\endgroup$ – Bob Knighton Jul 9 at 8:58
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    $\begingroup$ @SamuelWeir I also calculated it using Mathematica ;) remember to use MatrixExp and not Exp (which is elementwise exponentiation). $\endgroup$ – Noiralef Jul 9 at 9:25
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One may argue that $$ [ A,\exp(tB)] = 0 \\ \Rightarrow 0 = \partial_{t} [ A,\exp(tB)]\\ \Rightarrow 0 = \partial_{t|t=0} [ A,\exp(tB)]= [A,B\exp(tB)]_{t=0} = [A,B] $$ .

However, as Noiralef pointed out, one may not conclude $$ [ A,\exp(B)] = 0 \Rightarrow [ A,\exp(tB)] = 0 $$

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    $\begingroup$ From $[A, \exp(B)] = 0$ you can not conclude that $[A, \exp(tB)] = 0$ for all $t$ $\endgroup$ – Noiralef Jul 9 at 8:27
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    $\begingroup$ Ok, aggreed. I'll edit my answer and let it stand here. Maybe others can learn from it as well. $\endgroup$ – denklo Jul 9 at 8:30

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