# A tractable Baker-Campbell-Hausdorff formula

1. Aug 24, 2011

### arkobose

1. Let A and B be two matrices, and $\lambda$ be a continuous parameter.

2. Now, define a function $f(\lambda) \equiv e^{\lambda A}e^{\lambda B}$. We need to show that $\frac{df}{d\lambda} = \left\{A + B + \frac{\lambda}{1!}[A, B] + \frac{\lambda^2}{2!}[A, [A, B]] + ... \right \}f$

Once this is shown, setting $\lambda = 1$, and $[A, [A, B]] = [B, [A, B]] = 0$ gives us a Baker-Campbell-Hausdorff formula.

3. I had shown this result quite a while ago, but now I have forgotten completely what I had done. This time, I tried differentiating $f(\lambda)$ w.r.t the argument, and then using the commutation was able to get the first two terms on the R.H.S., but thereafter I got stuck. The very minimal hint would be all that I need.

Thank you!

Last edited: Aug 24, 2011
2. Aug 28, 2011

### diazona

Without seeing your exact steps I can't say much, but you may need to expand out an exponential or two and work out some commutators term-by-term.

3. Aug 28, 2011

### arkobose

I solved it. Thanks anyway!