# Baker-Campbell-Hausdorff formula question

1. Feb 17, 2009

### Karliski

2. Feb 17, 2009

### Ben Niehoff

The Hadamard formula is easy to show. The full BCH formula is a ***** (I spent several hours yesterday trying to do it, but I didn't understand enough about Lie groups to get there). Anyway, start with this function:

$$f(s) = e^{sA} B e^{-sA}$$

Then differentiate it a few times with respect to s:

$$f'(s) = e^{sA} A B e^{-sA} - e^{sA} B A e^{-sA} = e^{sA} [A,B] e^{-sA}$$

$$f''(s) = e^{sA} A [A,B] e^{-sA} - e^{sA} [A,B] A e^{-sA} = e^{sA} [A, [A,B]] e^{-sA}$$

$$f'''(s) = e^{sA} [A, [A, [A,B]]] e^{-sA}$$

etc.

Now construct the Taylor series for f(s):

$$f(s) = f(0) + s f'(0) + \frac12 s^2 f''(0) + \frac1{3!} s^3 f'''(0) + ...$$

$$e^{sA} B e^{-sA} = B + [A,B] s + \frac12 [A, [A, B]] s^2 + \frac1{3!} [A, [A, [A, B]]] s^3 + ...$$

Finally, evaluate the above at s=1 to get the result.

3. Feb 17, 2009

### Karliski

Hi,

Thanks, yes that is what I also did. The "parametric induction" term threw me off.