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Problem with non-commutative functions(quantum mechanics)

  1. Apr 8, 2012 #1
    1. I am working through E.Merzbacher quantum mechanics. The problem is;
    if G(λ) =eλAeλB for two operators A and B, show that
    dG/dλ=[A+B+λ[A,B]/1!+λ2[A,[A,B]]/2!+.............]G



    2. [A,B] is taken to mean AB-BA



    3. The only way I can think of proving this is by taylor expanding G(λ) and then differentiang each term in the expansion with respect to λ; this has not worked! Can anyone please help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 8, 2012 #2
    Just to expand on my attempt at a solution so far, which I think is barking up the wrong tree even though it seems close is;
    1) taylor expand G(λ) at f(0) I get

    G=1+λ[A+B]+λ2(A2+2AB+B2)/2!+.....
    =1+λ[A+B]+λ2(A+B)2/2!+λ3(A+B)3/3!
    and then what makes this tempting is that differentiating with respect to λ I get
    dG/dλ=[A+B]+λ(A+B)2/1!+λ2(A+B)3/2!

    Why I am sure this is the wrong approach is that;
    1. the whole expansion is supposed to be multiplied by G, which clearly it is not
    2. I have no idea if the (A+B)2, (A+B)3 can be expressed it the commutative notation [A,B], [A,[A,B]] etc. probably not so back to square one.
     
  4. Apr 10, 2012 #3
    be more careful when expanding quadratics since (A+B)(A+B) = A^2 +AB +BA +B^2 = A^2 +[A,B] +2BA +B^2
     
    Last edited: Apr 10, 2012
  5. Apr 11, 2012 #4
    Sorry perhaps I was unclear, I know that the quadratic expansions do not equal where I am (was) trying to get.
    Anyway I have since cracked it. The procedure is to use the prouct rule to get dG/dλ. You then multiply by eλbe-λb. You end up with
    dG/dλ=[B+eλbAe-λb]G
    The term eλbAe-λb] is an identity provided in the textbook. Substitution of this identity gives the result. Thanks.
     
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