From: Aaron Bergman Subject: Re: Baker-Campbell-Hausdorff Date: Tue, 15 Aug 2000 18:42:38 -0400 Newsgroups: sci.math Summary: [missing] In article , Matthew Nobes wrote: > Hello all, > > I've forgotton my math tables at the office and am in dire need of the 4th > term in the BCH series. I'm sorry. > Can anybody out there help me? C = log(e^A e^B) = A + B + (1/2)[A,B] + (1/12)[A,[A,B]] + (1/12)[B,[B,A]] + ... The general expression is C = B + \int_0^1 g(e^{t Ad[A]} e^{Ad[B]})(A) dt where g(z) = log z / (z - 1) = \sum_{j=0}^\infty \frac{(1-z)^j}{j+1}. Ad[A] is an operator such that Ad[A](B) = [A,B]. But hopefully you won't have to do much with the general formula. Working out the next term was an assignment in my math methods class. Aaron -- Aaron Bergman