TITLE: matrix expression for $e^{iA}Be^{iA}$ in terms of anticommutators?
QUESTION [6 upvotes]: I'm familiar with the expression 
$$e^{-iA}Be^{iA} = \sum_{n=0}^{\infty} \frac{i^n}{n!}[..[B,A],\dots A]_{n \; \rm times}$$
for square matrices $A$ and $B$ and was wondering if equivalently
$$e^{iA}Be^{iA} = \sum_{n=0}^{\infty} \frac{i^n}{n!}\{..\{B,A\},\dots A\}_{n \; \rm times},$$
or something similar?

REPLY [2 votes]: Nice question and nice answers. There is a generalization that admits an even simpler proof that I think it's worth mentioning. 
For (square) matrices $X,Y,B$,  let me define
$$
B(t):=e^{tX} B e^{tY}.
$$
Differentiating with respect to $t$ we see that $B(t)$ satisfies the following differential equation:
\begin{align}
\frac{dB}{dt}(t) &= XB(t) + B(t) Y \\
 &= \mathcal{K}_{X,Y}(B)& = XZ+ZY, \tag{1}
\end{align}
having defined the linear (super) operator
$$
\mathcal{K}_{X,Y}(Z) = XZ+ZY
$$
Now the solution of $(1)$ with initial condition $B(0)=B$ is  $
B(t) =  e^{t \mathcal{K}_{X,Y}} B$. We have then
$$
B(t) = e^{tX} B e^{tY} = e^{t \mathcal{K}_{X,Y}} B =\sum_{n=0}^\infty \frac{(t\mathcal{K}_{X,Y})^n}{n!} B .
$$
Set $Y=\pm X$ to obtain your formulas with commutators/anticommutators.