A question on Baker-Campbell-Hausdorff (BCH) Formulae (a special case) Let $D=\frac{d}{dx}$ be the derivative operator. Is it possible to find an operator (or a function) $A(x)$ such that:
\begin{equation}
e^{A(x)} D^n e^{-A(x)}=D^n+x 
\end{equation}
For instance, for $n=1$, it is straightforward to show that
\begin{equation}
e^{-\frac{x^2}{2}} D e^{\frac{x^2}{2}}=D+x 
\end{equation}
I have tried to use the following formula (a variation of BCH formulae):
\begin{equation}
e^{A(x)} D^n e^{-A(x)}=D^n+[A(x),D^n]+\frac{1}{2}[A(x),[A(x),D^n]]+...
\end{equation}
I understand that, to obtain the right hand side of the first equation, $[A(x),D^n]=x$ and $[A(x),[A(x),D^n]]=0$. However, I failed to proceed further.
Edit: It is proven below that $A(x)$ cannot be a pure function. Therefore, the question is now concerned about the possibility of an operator. I am particularly interested in an operator of the form of $A(x)=a(x)D$.
 A: I am not super versed in operator theory but for an elementary argument on why there is no solution for $n = 2$:
Suppose $e^{-A}D^2e^{A} = D^2 + x$. Then for any twice differentiable $f$, we would have
$$\begin{split} xf + f''
& = (D^2 + x)f \\
& = (e^{-A}D^2e^{A})f \\
& = e^{-A}D(A'e^Af + e^Af') \\
& = e^{-A}((A''+(A')^2)e^Af + 2A'e^Af' + e^Af'') \\
& = (A''+(A')^2)f + 2A'f' + f''
\end{split}$$
so taking $f(x) = 1$ means that $A''(x) + A'(x)^2 = x$. But if this were true then taking $f(x) = x$ we would also have that $A'(x) = 0$, so $A$ is constant, a contradiction.
It's not hard to see from this strategy that if we want to solve $e^{-A}D^2e^{A} = D^2 + gD + h$ for some functions $g,h$ of $x$, then it is a requirement that $4h = 2g' + g^2$, and if this is so then $A$ is the integral of $g/2$. So for example if you wanted $h$ to be zero, you would need $g' = -g^2/2 \implies g(x) = \frac{2}{C + x}$ for some $C$. Thus $e^{-A}D^2e^{A} = D^2 + \frac{2}{1+x}D$ has the solution $A(x) = \ln(1+x)$. And if you want $g$ to be zero like in your problem then $h$ must be zero as well, so $e^{-A}D^2e^{A} = D^2 + h$ has no solution for any nonzero function $h$.
Edit: Again I don't know an argument for general operators, but operators of the form $A(x) = a(x)D$ definitely do not have a solution since they are always composition operators of the form $e^Af(x) = f(g(x))$ for some function $g$ (see here). So any operator of this form maps the constant $f(x) \equiv 1$ to itself, which violates $e^{-A}D^2e^{A} = D^2 + x$ since then the left hand side would take $f$ to $0$ but the right hand side would take $f$ to $x$.
A: This is not possible if $n = 2$.
You want $A(x)$ such that
$$
e^{A(x)} \left(e^{-A(x)}f(x) \right)'' = f''(x) + x f(x) 
$$
for all $f$. Now take $f(x) = e^{A(x)}$. Since $f''(x) = e^{A(x)} \left(A''(x) + \left(A'(x) \right)^2\right)$, this implies
$$
0 = A''(x) + \left(A'(x) \right)^2 + x \, .
$$
Next, take $f(x) = x e^{A(x)}$. Since then $f''(x) = e^{A(x)} \left(2 A'(x) +  xA''(x) + x\left(A'(x) \right)^2\right)$, this implies now
$$
0 = 2 A'(x) + x\left(A''(x) + \left(A'(x) \right)^2 + x\right) = 2 A'(x)
$$
that is, $A(x)$ would have to be a constant. But of course that does not work.
Presumably this fails for all $ n > 1$.
