# Does $e^{X} e^{Y} = e^{X + Y}$ imply that the operators $X$ and $Y$ commute?

I am considering two self-adjoint, but unbounded operators $X$ and $Y$ on a Hilbert space. By the Baker-Campbell-Hausdorff formula, we know that $$[X, Y] = 0 \Rightarrow e^{X} e^{Y} = e^{X + Y}.$$ But is the converse also true? If I have the property $e^{X} e^{Y} = e^{X + Y}$, can I somehow prove that the operators must commute?

• The very similar question: math.stackexchange.com/questions/2276597/… May 19 '17 at 14:40
• Thanks for pointing out the similar question, but the answers there are not entirely helpful and my setting is also different, so I'm hoping for some helpful answers here!
– Luke
May 19 '17 at 14:43
• Are you assuming that $X+Y$ is essentially self-adjoint? May 19 '17 at 22:04
• I'm not quite sure if I am assuming it, but in the case that lead me to this general question, $X$ and $Y$ are essentially self-adjoint on the same dense domain.
– Luke
May 20 '17 at 20:57
• If you do not assume that $X+Y$ is (at least) essentially self-adjoint, then I do not think it is very clear what $e^{X+Y}$ means. May 24 '17 at 18:24