# Baker-Campbell-Hausdorff formula(e)

Given a linear operator  $A$, we define:

 $\exp{A}:=\sum_{k=0}^{\infty}\frac{1}{k!}A^{k}.$ (1)

It follows that

 $\frac{\partial}{\partial\tau}e^{\tau A}=Ae^{\tau A}=e^{\tau A}A.$ (2)

Consider another linear operator $B$. Let $B(\tau)=e^{\tau A}Be^{-\tau A}$. Then one can prove the following series representation for $B(\tau)$:

 $B(\tau)=\sum_{m=0}^{\infty}\frac{{\tau}^{m}}{m!}B_{m},$ (3)

where $B_{m}=[A,B]_{m}:=[A,[A,B]_{m-1}]$ and $B_{0}:=B$. A very important special case of eq. (3) is known as the Baker-Campbell-Hausdorff (BCH) formula. Namely, for $\tau=1$ we get:

 $e^{A}\;Be^{-A}=\sum_{m=0}^{\infty}\frac{1}{m!}B_{m}.$ (4)

Alternatively, this expression may be rewritten as

 $[B,e^{-A}]=e^{-A}\left([A,B]+\frac{1}{2}[A,[A,B]]+\cdots\right),$ (5)

or

 $[e^{A},B]=\left([A,B]+\frac{1}{2}[A,[A,B]]+\cdots\right)e^{A}.$ (6)

There is a descendent of the BCH formula, which often is also referred to as BCH formula. It provides us with the multiplication law for two exponentials  of linear operators: Suppose $[A,[A,B]]=[B,[B,A]]=0$. Then,

 $e^{A}e^{B}=e^{A+B}e^{\frac{1}{2}[A,B]}.$ (7)

Thus, if we want to commute two exponentials, we get an extra factor

 $e^{A}e^{B}=e^{B}e^{A}e^{[A,B]}.$ (8)
Title Baker-Campbell-Hausdorff formula(e) BakerCampbellHausdorffFormulae 2013-03-22 13:39:51 2013-03-22 13:39:51 Mathprof (13753) Mathprof (13753) 13 Mathprof (13753) Definition msc 47A05 BCH formula Baker-Campbell-Hausdorff formula Baker-Campbell-Hausdorff formulae