# Is $[A,\exp{B}]=0 \Rightarrow [A,B]=0$ true?

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).

• Would Mathematics be a better home for this question? – Qmechanic Jul 9 at 8:10
• Maybe, but as physicists deal a lot with such commutators I asked here. Should I move the thread? – TheoreticalMinimum Jul 9 at 8:14
• General tip: Never crosspost, but you might flag for migration. – Qmechanic Jul 9 at 8:23

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 $$A$$.

• This seems legit. Do you know where the mistake might be in my proof-scetch? – denklo Jul 9 at 8:25
• 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). – Samuel Weir Jul 9 at 8:38
• Let $B=2\pi i S$. Then $S^2=1$ and so $e^{2\pi i S}=\cos(2\pi)+i \sin(2\pi) S=1$. – Bob Knighton Jul 9 at 8:58
• @SamuelWeir I also calculated it using Mathematica ;) remember to use MatrixExp and not Exp (which is elementwise exponentiation). – Noiralef Jul 9 at 9:25

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 $$[ A,\exp(B)] = 0 \Rightarrow [ A,\exp(tB)] = 0$$

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