What does $\exp\left( ax\frac{d}{dx} \right)$ do on $\psi(x)$? I'm trying to find out
$$\exp\left(ax\frac{d}{dx}\right)\psi(x)= \  \ ? $$
I tried spending the exponential and then operating the derivatives one by one but I found no pattern. Besides, it gets messier and messier.
It looks to me more like:
$$\exp\left( -\phi_0\frac{\partial }{\partial \phi}\right)\psi(\rho,\phi)=\psi(\rho,\phi-\phi_0)$$
But $\phi_0$ is suppose to be constant. Can any help me through this?
 A: A short proof is to note that
$$
\frac{\partial}{\partial \alpha} \left(\exp\left \{\alpha x \frac d{dx} \right\}\psi(x)\right)  = x  \frac d{dx} \left(\exp\left \{\alpha x \frac d{dx} \right\}\psi(x)\right)
$$
and
$$
\frac{\partial}{\partial \alpha} \psi(e^\alpha x) =  x \frac d{dx} \psi(e^\alpha x)
$$
so $\psi(e^\alpha x)$ satisfies the same 1st order ODE in $\alpha$ as does  $\exp\left \{\alpha x \frac d{dx} \right\}\psi(x)$ with the    same  initial condition.
A: I had a professor who called this the "Dilation Operator", though I haven't seen it referred by that name anywhere else. You can compute it using a neat coordinate transformation. To start, you should know that:
$$\exp\left({a\frac{\text{d}}{\text{d}y}}\right)g(y) = \sum_{n=0}^\infty \frac{a^n}{n!}\frac{\text{d}^n}{\text{d}y^n}g(y) = g(y+a),$$ which follows quite simply from the definition of the Taylor Series.
The trick to evaluate the operator you have is to perform a coordinate substitution from $x$ to some $y$, such  that $$x \frac{\text{d}}{\text{d} x} \longrightarrow \frac{\text{d}}{\text{d}y}$$
You should be able to see that if you choose $y = \ln x$, this would be true, so $$a \left(x\frac{\text{d}}{\text{d}x}\right) f(x) = a \left(\frac{\text{d}}{\text{d}y} \right) f(e^y),$$
and from this it's quite easy to show that $$\exp\left({a x\frac{\text{d}}{\text{d}x}}\right) f(x) = \exp\left({a \frac{\text{d}}{\text{d}y}}\right) f(e^y) = f(e^{y+a}) = f(e^a x),$$
and voila, there you have it!

Edit: Here's another (simpler?) way to show it, using the properties of the Euler Operator $\left(x \frac{\text{d}}{\text{d}x}\right)$, whose eigenfunctions are the monomials $x^n$, and eigenvalues $n$.
$$\left(x \frac{\text{d}}{\text{d}x}\right) x^n = n\,\, x^n \quad \quad \implies \quad \quad \left(x \frac{\text{d}}{\text{d}x}\right)^m x^n = n^m x^n$$
If you can write $\psi(x) = \sum_n c_n x^n$, and expand the exponential operator using its power series definition, then:
\begin{aligned}
\exp\left(a \frac{\text{d}}{\text{d}x}\right) \psi(x) &= \sum_m \frac{a^m}{m!}\left(x \frac{\text{d}}{\text{d}x}\right)^m \sum_n c_n x^n\\
&= \sum_n c_n \underbrace{\sum_m \frac{a^m}{m!}n^m}_{\exp{an}} \,\, x^n\\
&= \sum_n c_n \left(e^a x\right)^n\\
&= \psi(e^a x)
\end{aligned}
