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# 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) |

## Mathematics Subject Classification

47A05*no label found*

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## Comments

## equation (7)

How do you arrive at equation (7) ?

## Re: equation (7)

could you please explain it in detail?

## Re: equation (7)

> How do you arrive at equation (7) ?

could you please explain it in detail?