- #1
arkobose
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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!
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!
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