I am studying conformal field theory, and I have run into an problem with calculations involving the quantum-mechanical translation and dilatation operators. It actually boils down to an issue about ordinary differential operators.
Consider an analytic function $\phi\colon \mathbb{R} \to \mathbb{R}$. Define two differential operators $\mathcal{P} = \frac{d}{dx}$ and $\mathcal{D} = x \frac{d}{dx}$. The goal is to compute $e^{\varepsilon\mathcal{D}} \phi(a)$ for $a, \varepsilon \in \mathbb{R}$. I have two different methods which give different results.
Method 1: Like in this question, we write $\mathcal{D} = \frac{d}{d\ln x}$ and find $$ e^{\varepsilon\mathcal{D}} \phi(a) = e^{\varepsilon \frac{d}{d\ln x}} \phi(e^{\ln a}) = \underline{\phi(e^{\varepsilon} a)} $$ which follows from the Taylor expansion of $\phi(e^y)$ around $y = \ln a$.
Method 2: We note that $e^{a\mathcal{P}} \phi(0) = e^{a\frac{d}{dx}} \phi(0) = \phi(a)$, which follows from the Taylor expansion of $\phi(x)$ around $x = 0$. We can therefore write $$ e^{\varepsilon\mathcal{D}} \phi(a) = e^{\varepsilon\mathcal{D}} e^{a\mathcal{P}} \phi(0) = e^{\varepsilon\mathcal{D}} e^{a\mathcal{P}} e^{-\varepsilon\mathcal{D}} e^{\varepsilon\mathcal{D}} \phi(0). $$ The final factor $e^{\varepsilon\mathcal{D}} \phi(0)$ is equal to $\phi(0)$, by Method 1 or by noting that setting $x = 0$ kills all terms except the first in the expansion $e^{\varepsilon\mathcal{D}} \phi(0) = \sum_{n = 0} \frac{\varepsilon^n}{n!} \left(x\frac{d}{dx}\right)^n \phi(0)$.
We then write $$ e^{\varepsilon\mathcal{D}} e^{a\mathcal{P}} e^{-\varepsilon\mathcal{D}} = e^{a e^{\varepsilon\mathcal{D}} \mathcal{P} e^{-\varepsilon\mathcal{D}}} $$ (which follows termwise from the expansion of $e^{a\mathcal{P}}$) and use the nameless identity (it seems to be called Hadamard's lemma or the Hausdorff formula?) $$ e^A B e^{-A} = e^{[A,\cdot]} B = B + [A,B] + \frac{1}{2}[A,[A,B]] + \cdots $$ and calculate the commutator $[\mathcal{D}, \mathcal{P}] = -\mathcal{P}$ to evaluate $$ e^{\varepsilon\mathcal{D}} \mathcal{P} e^{-\varepsilon\mathcal{D}} = e^{\varepsilon[\mathcal{D}, \cdot]} \mathcal{P} = e^{-\varepsilon} \mathcal{P}. $$ Putting everything together, we have $$ e^{\varepsilon\mathcal{D}} \phi(a) = e^{a e^{-\varepsilon} \mathcal{P}} \phi(0) = \underline{\phi(e^{-\varepsilon} a)}. $$
Question: Why do these two methods give different results? There seems to be a sign error somewhere, but I have tried and failed to find it for several hours, so I am starting to suspect that there may be a deeper problem with my reasoning. Is there a flaw in my Method 2?
