# Proof of Baker-Campbell-Hausdorff involving Bernoulli numbers

1. Mar 8, 2016

### CAF123

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. Mar 10, 2016

### Twigg

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.