Is it possible to take square root of the operator $e^{a\frac{d}{dx}}$?

Question

Is it possible to take square root of the operator $e^{a\frac{d}{dx}}$ where $a$ is a real or complex constant? Actually in physics one can take the square root of $(\Box +m^2)$ operator associated with the Klein-Gordon equation and one obtains the Dirac equation. I was wondering whether it is possible to take the square root of translation operator in quantum mechanics which has the form $e^{-ia\hbar\frac{d}{dx}}$.

Answer 1

Let $$\begin{align} D&=\exp\left(a\frac{\mathrm{d}}{\mathrm{d}x}\right)\\ &=\sum_{n \in \mathbb{N}} \frac{1}{n!}\left(a\frac{\mathrm{d}}{\mathrm{d}x}\right)^n \end{align}$$ And we want to calculate $\sqrt{D}$. It will be $$\sqrt{D}=\exp\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)$$ because $$\begin{align} \sqrt{D}\sqrt{D}&=\exp\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)\exp\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)\\ &=\sum_{n \in \mathbb{N}} \frac{1}{n!}\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)^n\sum_{k \in \mathbb{N}} \frac{1}{k!}\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)^k\\ &=\sum_{i \in \mathbb{N}}\sum_{n=0}^{i} \frac{1}{n!}\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)^n \frac{1}{(i-n)!}\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)^{i-n}\\ &=\sum_{i \in \mathbb{N}}\sum_{n=0}^{i} \frac{1}{n!(i-n)!}\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)^{i}\\ &=\sum_{i \in \mathbb{N}}\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)^{i}\sum_{n=0}^{i}\frac{1}{n!(i-n)!}\\ &=\sum_{i \in \mathbb{N}}\left(\frac{1}{2}a\frac{\mathrm{d}}{\mathrm{d}x}\right)^{i}\frac{2^i}{i!}\\ &=\sum_{i \in \mathbb{N}} \frac{1}{i!}\left(a\frac{\mathrm{d}}{\mathrm{d}x}\right)^i\\ &=\exp\left(a\frac{\mathrm{d}}{\mathrm{d}x}\right)\\ &=D \end{align}$$ as it should be.