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Homework Help: A proof of operators in exponentials

  1. Nov 28, 2012 #1
    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[itex]\frac{1}{2}[/itex][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] + [itex]\frac{1}{2}[/itex][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. jcsd
  3. Nov 28, 2012 #2
    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 + [itex]\frac{1}{2}[/itex][A,B]
    x = A + B + [itex]\frac{1}{2}[/itex]C

    Thus,
    eAeB = eA+B + [itex]\frac{1}{2}[/itex]

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

    eA+B + [itex]\frac{1}{2}[/itex]=eA+Be[itex]\frac{1}{2}[/itex]C

    thus,
    eAeB=eA+Be[itex]\frac{1}{2}[/itex][A,B]
     
  4. Nov 29, 2012 #3

    Fredrik

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    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).
     
  5. Nov 29, 2012 #4

    Fredrik

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    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.
     
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