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.