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Derivative of operators/commutators

  1. Oct 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove:

    ## \frac{dG}{d\lambda} = G \Big( A +B + \frac{\lambda}{1!} [A,B] +\frac{\lambda^2}{2!} [A,[A,B]]... \Big)##

    2. Relevant equations

    ##G(\lambda) = e^{\lambda A}e^{\lambda B} ##

    3. The attempt at a solution

    I tried taking the derivative with respect to lambda:
    ## \frac{dG}{d\lambda} = Ae^{\lambda A}e^{\lambda B} +e^{\lambda A}e^{\lambda B}B##

    ## = AG + GB##

    Doesn't really seem to help me.

    I'm assuming from the answer that the identity needs to be used:

    ##e^{\lambda A}Be^{-\lambda A} = B + \frac{\lambda}{1!} [A,B] +\frac{\lambda^2}{2!} [A,[A,B]]...##

    But not sure how to piece it all together.

    I'm reviewing for a test, and I'm working through some exercises.
    (This is exercise 3.16 in chapter 4 of Merzbacher if anyone wants to reference it)

    I think I'm just missing something really simple....
     
    Last edited: Oct 29, 2013
  2. jcsd
  3. Oct 29, 2013 #2
    Ah, Okay

    ##dGdλ = AG+GB = G (A+B) = G (A + e^{A \lambda}Be^{-A \lambda})##

    Then just use the identity.
     
  4. Oct 30, 2013 #3

    Mark44

    Staff: Mentor

    Typo above. I think you mean dG/dλ. I see where you get AG + GB. Is it reasonable to assume commutivity of multiplication? I'm thinking in terms of matrix multiplication, which is not generally commutative.

    In any case, how did you get from G (A + B), which I assume means G(λ) * (A + B), rather than "G of A + B", to what you show on the right side of the equation above?

    Also, what do the notations [A, B] and [A, [A, B]] mean?
     
  5. Oct 30, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I can do the notation question. That's pretty standard. [A,B] is commutator of A and B. [A,B]=AB-BA.
     
  6. Oct 30, 2013 #5

    Mark44

    Staff: Mentor

    Thanks, Dick!
     
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