A proof of operators in exponentials

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Homework Statement


Assume C=[A,B]≠0 and [C,A]=[C,B]=0

Show
eAeB=eA+Be\frac{1}{2}[A,B]


Homework Equations


All are given above.


The Attempt at a Solution


I recently did a similar problem (show eABe-A = B + [A,B] + \frac{1}{2}[A,[A,b]]+...) by defining a function exABe-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
 
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So I think I figured it out, but I would appreciate input on whether it is right or not.
start with:

eAeB = ex

Then, looking for x

x = log(eAeB)
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,
eAeB = eA+B + \frac{1}{2}

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

eA+B + \frac{1}{2}=eA+Be\frac{1}{2}C

thus,
eAeB=eA+Be\frac{1}{2}[A,B]
 
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).
 
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.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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