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Proof of Baker-Campbell-Hausdorff involving Bernoulli numbers

  1. Mar 8, 2016 #1

    CAF123

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    Gold Member

    1. The problem statement, all variables and given/known data
    The BCH formula states that the product of two exponentials of non commuting operators can be combined into a single exponential involving commutators of these operators. One may write that ##\ln(e^A e^B) = \sum_{n \geq 1} c_n(A,B)## where $$c_{n+1} = \frac{1}{n+1} \left( -\frac{1}{2} [c_n, A-B] + \sum_{m=0}^{\lfloor{n/2}\rfloor} \frac{B_{2m}}{(2m)!} \sum_{k_1, \dots, k_{2m} \geq 1,\,\, k_1 + \dots + k_{2m} = n} [c_{k_1}, [ \dots, [c_{k_{2m}}, A+B] \dots]]\right)$$ and ##B_n## are the Bernoulli numbers.

    Using this I want to construct the relation $$\ln(e^A e^B) = A+B + \frac{1}{2}[A,B] + \frac{1}{12} \left( [A,[A,B]] - [B,[A,B]]\right) + \dots$$

    2. Relevant equations
    Bernoulli numbers are defined through $$\frac{x}{e^x-1} = \sum_{n=0}^{\infty} \frac{B_n}{n!}x^n$$

    3. The attempt at a solution
    So the terms in the expansion are the ##c_n(A,B)## - since ##n \geq 1## using this formula I am not sure if I can directly compute ##c_o## - is it perhaps done recursively? And for the case ##n=1## the first sum there only includes one term corresponding to the case ##m=0## but then what does the sum delimiters mean? In particular, the term in the second sum with ##m=0## involves ##k_0## but what is ##k_0##?

    Thanks for any tips!
     
  2. jcsd
  3. Mar 10, 2016 #2

    Twigg

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    When in doubt try the easiest thing first: just suppose ##c_{1}= 1## ( I don't think ##c_{0}## is needed at all).

    See what happens to ##[c_{n},A - B]##, it goes to 0 because anything commutes with 1. Since n=1, floor(n/2) = 0, so there is no ##c_{k_{2m}}## fancy business going on for the nasty looking sum, like you're taking "0" commutators on A + B, so it just spits A + B back out again. And that's the first term in the BCH formula you're looking for. You still have to check that the scalars for the n=1 term come out to 1*(A+B).

    Try picking up at n=2 and see if you can't extend it.
     
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