B Commutators of functions of operators

[LagrangeEuler]

I would like to ask whether if operators $A$ and $B$ commute also operators $e^A$ and $e^B$ commute? Also I have a question is it possible that
$e^A$ is matrix where all elements are $\infty$ so that $e^A \cdot e^B-e^B\cdot e^A$ has all elements that are $\infty$?

 

[PeroK]

I would like to ask whether if operators $A$ and $B$ commute also operators $e^A$ and $e^B$ commute?

Have you tried to prove that?

Also I have a question is it possible that
$e^A$ is matrix where all elements are $\infty$ so that $e^A \cdot e^B-e^B\cdot e^A$ has all elements that are $\infty$?

That makes no sense. Matrices cannot have infinite elements.

 

[hilbert2]

I would like to ask whether if operators $A$ and $B$ commute also operators $e^A$ and $e^B$ commute?

I would try to prove that by showing with mathematical induction that any partial sum MacLaurin expansions $\displaystyle\sum\limits_{k=0}^{N}\frac{1}{k!}A^k$ and $\displaystyle\sum\limits_{k=0}^{N}\frac{1}{k!}B^k$ with a finite $N$ commute when $[A,B]=0$. It results just from the fact that $A^m B^n = B^n A^m$ for any $m,n\in\mathbb{N}$ when $[A,B]=0$, so the order of operators doesn't make any difference when forming that polynomial of two operator variables.