quantum mechanics - Product of exponential of operators

in the context of non-relativistic quantum mechanics I want to show that, for any $A$ and $B$ operators

$$e^{A}e^{B}=e^{A+B}$$

if and only if

$$[A,B]=0$$

I remember my professor told use about looking for a differential equation but I don't remember the details, and I want to be able to prove it. Brute force the series doesn't seem to be a good idea.

Any hint will be appreciated thanks.

Answer

There is no 'only if' because it is not true: $$\begin{align} e^{A+B} = e^A e^B \end{align}$$ does not necessarily imply $[A,B] = 0$.

One can easily find an example of this using matrices. Here's one: $$\begin{align} A= \begin{pmatrix} 0 & 0 \\ 0 & 2\pi i \end{pmatrix}, B=\begin{pmatrix} 0 & 1 \\ 0 & 2 \pi i \end{pmatrix}. \end{align}$$ $[A,B] \neq 0$ but $e^{A+ B} = e^A e^B = I$.

Edit: Let me help with the if part, using a differential equation as OP desires. Compute $$\begin{align} \frac{d}{dt}(e^{t(A+B)}e^{-tA}e^{-tB}), \end{align}$$ and show that it is $0$ if $[A,B] = 0$.

That implies that $e^{t(A+B)}e^{-tA}e^{-tB}$ is independent of $t$. In particular, plugging in $t = 0$ gives $e^{t(A+B)}e^{-tA}e^{-tB} = I$ for all $t$. Then plug in $t = 1$ to get $e^{(A+B)}e^{-A}e^{-B} = I$.

QED.