1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A proof of operators in exponentials
  1. Operator proof (Replies: 3)

Loading...