Operators – Commutator of Exponential Derivative and Position Operators

Question

What is the commutator of the exponential derivative operator and the exponential position operator?
\begin{align}
\left[\exp(\partial_x),\exp(x)\right]
=\exp(\partial_x)\exp(x)
-\exp(x)\exp(\partial_x)=~?
\end{align}

I first wrote down the exponentials into there power series sum
\begin{align}
\left[\exp(\partial_x),\exp(x)\right]
=
\sum_{n=0}^{\infty}
\dfrac{\partial_x^n}{n!}
\sum_{m=0}^{\infty}
\dfrac{\partial_x^m}{m!}
–
\sum_{n=0}^{\infty}
\dfrac{x^n}{n!}
\sum_{m=0}^{\infty}
\dfrac{\partial_x^m}{m!},
\end{align}
and then grouped like terms
\begin{align}
\left[\exp(\partial_x),\exp(x)\right]
=
\sum_{n=0}^{\infty}
\sum_{m=0}^{\infty}
\dfrac{
\left[
\partial_x^n
x^m
–
x^n
\partial_x^m
\right]
}{n!m!},
\end{align}
and then evaluate the derivatives
\begin{align}
\left[\exp(\partial_x),\exp(x)\right]
=
\sum_{n=0}^{\infty}
\sum_{m=0}^{\infty}
\dfrac{
\left[
\partial_x^n
(x^m)
+
x^m
\partial_x^n
–
x^n
\partial_x^m
\right]
}{n!m!}.
\end{align}
I am not sure if$$
\partial_x^n\left(x^m\right)~=~\frac{m!}{\left(m-n+1\right)!} \, x^{m-n}
\,,$$and then I got stuck.

Answer 1

Suresh is right, the Baker-Campbell-Hausdorff formula will save you. If $\left[A,B\right]$ commutes with $A$ and $B$ you have,

$$ \exp\left(A\right)\exp\left(B\right) = \exp\left(A+B+\frac{1}{2}\left[A,B\right]\right) \, .$$

Then the answer to your question is

$$ \exp\left(\partial_x\right)\exp\left(x\right) = 2 \exp\left(x+\partial_x\right) \sinh(1/2) \, .$$