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CAF123
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Homework Statement
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 $$\ln(e^A e^B) = A+B + \frac{1}{2}[A,B] + \frac{1}{12} \left( [A,[A,B]] - [B,[A,B]]\right) + \dots$$
Homework Equations
Bernoulli numbers are defined through $$\frac{x}{e^x-1} = \sum_{n=0}^{\infty} \frac{B_n}{n!}x^n$$
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!