From: Alex Selby <a.p.selby@REMOVE.pobox.com>
Subject: Re: Baker-Campbell-Hausdorff formula
Date: Wed, 03 Oct 2001 02:31:54 +0100
Newsgroups: sci.math
Summary: Exponentiating a sum of two (noncommuting) operators

Ray wrote:

> Let A,B be two operators such that [A,B] = some complex number lambda.
> Then e^(A+B) = e^A e^B e^(-lambda/2)
> Gzbluh?
> Why is this true?
> I get sum_N=0 to infinity (sum_k=0 to N (A^(N-k)*B^k /(k!(N-k)!))) *
> sum_r=0 to infinity (sum_t=0 to r (
> ((AB)^r-lambda^s*(AB)^(r-s))/(s!(r-s)!))) for the right side, since AB
> commutes with BA, but I don't know what to do with that.
>
> Thanks,
> Ray

Here is one way.

Let X(t)=e^(tA)*B*e^((1-t)A).
Then dX/dt = e^(tA)*[A,B]*e^((1-t)A)
           = lambda * e^(tA)*e^((1-t)A)
           = lambda * e^A
So [e^A,B]=X(1)-X(0)=lambda*e^A    (equation 1)

Now let Y(t)=e^(-tA)*e^(t(A+B))*e^(-tB).
Then dY/dt = e^(-tA)*(-A)*e^(t(A+B))*e^(-tB) +
             e^(-tA)*(t(A+B))*e^(t(A+B))*e^(-tB) +
             e^(-tA)*e^(t(A+B))*(-B)*e^(-tB)
           = e^(-tA)[B,e^(t(A+B))]e^(-tB)
           = -t*lambda*Y(t)
(substituting t(A+B) for A and t*lambda for lambda in equation 1).
This differential equation has the unique solution (with Y(0)=I)
     Y(t) = exp(-t^2/2*lambda)I
and t=1 gives the formula you quoted.

Alex