A proof of operators in exponentials

[aop12]

Homework Statement

Assume C=[A,B]≠0 and [C,A]=[C,B]=0

Show
e^Ae^B=e^A+Be^{\frac{1}{2}[A,B]}

Homework Equations

All are given above.

The Attempt at a Solution

I recently did a similar problem (show e^ABe^-A = B + [A,B] + \frac{1}{2}[A,[A,b]]+...) by defining a function e^xABe^-xA and doing a taylor expansion, so I thought this might be done similarly, but I have gotten nowhere with this approach. I would like to figure this out myself, so I am really looking for guidance/hints if anyone has any. It would be much appreciated

 

[aop12]

So I think I figured it out, but I would appreciate input on whether it is right or not.
start with:

e^Ae^B = e^x

Then, looking for x

x = log(e^Ae^B)
Which can be found using the Baker–Campbell–Hausdorff formula

x = A + B + \frac{1}{2}[A,B]
x = A + B + \frac{1}{2}C

Thus,
e^Ae^B = e^{A+B + \frac{1}{2}}

since C and A, and C and B commute, C and A+B commute,

e^{A+B + \frac{1}{2}}=e^A+Be^\frac{1}{2}C

thus,
e^Ae^B=e^A+Be^{\frac{1}{2}[A,B]}

 

[Fredrik]

It looks correct (except for the fact that you forgot to type the C in a couple of places), but are you sure you're allowed to use the BCH formula? I would have guessed that the point of the exercise is to prove a special case of it. (I haven't thought about how to do that).

 

[Fredrik]

I found a solution in a book. They don't use the BCH formula. Instead, their strategy is to prove that the maps
\begin{align}&t\mapsto e^{tA}e^{tB}e^{\frac{t^2}{2}[A,B]}\\
&t\mapsto e^{t(A+B)}
\end{align} satisfy the same differential equation and the same initial condition.