Exercises
[1MW] Difficulty:*.In the general case (when we do not know if \(A,B\) commute) we proceed as follows. Let’s define \([A,B]=AB-BA\).
Setting \(B_ 0=B\) and \(B_{n+1}=[A,B_ n]\) you have
\begin{eqnarray*} B_ n& =& A^ nB - n A^{n-1}BA + \frac{n(n-1)} 2 A^{n-2}BA^ 2 + \cdots + (-1)^ n\, BA^ n =\\ & =& ∑_{k=0}^ n (-1)^ k\binom {n}{k} A^{n-k} B A^{k}~ ~ ; \end{eqnarray*}let’s define now \(Z=Z(A,B)\)
\begin{equation} Z{\stackrel{.}{=}}∑_{n=0}^∞\frac{B_ n}{n!}~ ~ ,\label{eq:Z(A,B)} \end{equation}20(note that \(Z\) is linear in \(B\)): prove that the above series converges, and that
\begin{equation} \exp (A)B\exp (-A)=Z~ ~ ;\label{eq:exp_ A_ B_-A_ Z} \end{equation}21from this finally it is shown that
\[ \exp (A)\exp (B)\exp (-A)=\exp (Z)~ ~ . \]
(These formulas can be seen as consequences of the Baker–Campbell–Hausdorff formula [ 39 ] ).
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