## Problem

Let $A$ and $B$ be linear operators, which commute with the commutator of $A$ and $B$, i.e.

$\bigl[A,[A,B]\bigr]= \bigl[B,[A,B]\bigr]=0\,.$

Show that the following formulae hold:

$e^A e^B = e^B e^A e^{[A,B]}\,,\\ e^{A+B} = e^A e^B e^{-[A,B]/2}\,,$

which are known as the Baker-Campbell-Hausdorff identities. They play an important rule, e.g. in quantum mechanics, whenever one deals with exponentiation of the position and the momentum operator.

## Solution

We will start by proving the first equation. To this end we introduce an auxiliary operator $O_1(\lambda)$, which depends on a real parameter $\lambda$:

$O_1(\lambda) = e^{\lambda B}e^Ae^{[A,\lambda B]}e^{-\lambda B}e^{-A}\,.$

We will show, that $O_1(\lambda)= 1$, which will prove the claim. Clearly this statement is true for $\lambda=0$. The derivative of $O_1(\lambda)$ with respect to $\lambda$ is

$\frac{d}{d\lambda}O_1(\lambda) = e^{\lambda B}\Bigl(Be^A e^{[A,\lambda B]}e^{-\lambda B}+e^A[A,B]e^{[A,\lambda,B]}e^{-\lambda B}-e^Ae^{[A,\lambda B]}Be^{-\lambda B}\Bigr)e^{-A}\notag\\ \phantom{\frac{d}{d\lambda}O_1(\lambda)} = BO_1(\lambda)+[A,B]O_1(\lambda)-e^{\lambda B}e^Ae^{[A,\lambda B]}Be^{-\lambda B}e^{-A}\notag\\ \phantom{\frac{d}{d\lambda}O_1(\lambda)} =[A,B]O_1(\lambda)-e^{\lambda B}[e^A,B]e^{[A,\lambda B]}e^{-\lambda B}e^{-A}$

It is easy to see by induction, that under the conditions imposed on $A$ and $B$,

$[A^n,B]= n[A,B]A^{n-1}\,,$
which implies
$[e^A,B]= [A,B]e^A$

and thus $\frac{d}{d\lambda}O_1(\lambda)=0$ and consequently validates the first identity.

In order to prove the second equation, we will do a similar trick and define

$O_2(\lambda)=e^{-\lambda(A+B)}e^{\lambda A}e^{\lambda B}e^{-\lambda^2[A,B]/2}\,.$

Obviously, $O_2(0)= 1$, and the derivative

$\frac{d}{d\lambda}O_2(\lambda)= -(A+B)O_2(\lambda)+e^{-\lambda(A+B)}Ae^{\lambda A}e^{\lambda B}e^{-\lambda^2[A,B]/2}\notag\\ \phantom{\frac{d}{d\lambda}O_2(\lambda)=}+e^{-\lambda(A+B)}e^{\lambda A}Be^{\lambda B}e^{-\lambda^2[A,B]/2}-\lambda[A,B]O_2(\lambda)\notag\\ \phantom{\frac{d}{d\lambda}O_2(\lambda)}=-BO_2(\lambda)+[e^{-\lambda(A+B)},A]e^{\lambda A}e^{\lambda B}e^{-\lambda^2[A,B]/2}+e^{-\lambda(A+B)}Be^{\lambda A}e^{\lambda B}e^{-\lambda^2[A,B]/2}\notag\\ \phantom{\frac{d}{d\lambda}O_2(\lambda)}=0\,.$

In the second step I have interchanged $B$ with $e^{\lambda A}$ in the third term. The commutation rest cancels the fourth term.

This completes the proof.

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## About goobypl5

pizza baker, autodidact, particle physicist
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### 1 Response to Let gooby do teh hoemwerk: Baker-Campbell-Hausdorff identities

1. I remember I had to prove this theorem in a “Quantum Mechanics II” course homework exercise a few years ago, and it took forever to find clues for the solution online, until I found out some page where this was proven, but it was called “Glauber’s formula” there.