# A proof of operators in exponentials

1. Nov 28, 2012

### aop12

1. The problem statement, all variables and given/known data
Assume C=[A,B]≠0 and [C,A]=[C,B]=0

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

2. Relevant equations
All are given above.

3. 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

2. Nov 28, 2012

### aop12

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

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]

3. Nov 29, 2012

### Fredrik

Staff Emeritus
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).

4. Nov 29, 2012

### Fredrik

Staff Emeritus
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.