How was the $O(\epsilon ^2)$ term obtained?

Question

$$\newcommand{\brak}[1]{\left\langle #1 \right\rangle}$$

I'm trying to study Path Integral approach to Quantum Mechanics on my own, during the readings, I came across one part that I'm not certain how it was exactly derived, could use any possible hint and advice.

Consider the quantity: $\brak{x'\mid\exp(-iH\epsilon /\hslash)\mid x} =\int dp \brak{x'|p}\brak{p|\exp(-iH\epsilon/\hslash)|x}$

if we stick to the simple case $H =\frac{\hat{p}^2}{2m} + V(\hat{x})$, then should be able to reach

$\brak{p\mid\exp\left(-i\frac{\epsilon}{\hslash}\left[\frac{\hat{p}^2}{2m} + V(\hat{x})\right]\right)| x} = \exp\left(-i\frac{\epsilon}{\hslash}\left[\frac{\hat{p}^2}{2m} + V(\hat{x})\right]\right)\brak{p|x} + O(\epsilon^2)$

I'm wondering how this $O(\epsilon^2)$ term was obtained? If possible, can someone show me the derivation?

Answer 1

Use the Baker–Campbell–Hausdorff formula

\begin{eqnarray*} &&\langle p | \exp\left( -\frac{i\epsilon}{\hbar}\left(\frac{\hat{p}^2}{2m} + V(\hat{x})\right)\right)| x\rangle \\ &=& \langle p | \exp\left(-\frac{i\epsilon}{\hbar}\frac{\hat{p}^2}{2m}\right)\exp\left(-\frac{i\epsilon}{\hbar}V(\hat{x})\right)\exp\left(-\frac{\epsilon^2}{h^2}\left[\frac{\hat{p}^2}{2m},V(\hat{x})\right]\right)\cdots|x\rangle \\ &=& \langle p | \exp\left(-\frac{i\epsilon}{\hbar}\frac{\hat{p}^2}{2m}\right)\exp\left(-\frac{i\epsilon}{\hbar}V(\hat{x})\right)\left(1 + O(\epsilon^2)\right)|x\rangle \\ &=& \langle p | \exp\left(-\frac{i\epsilon}{\hbar}\frac{p^2}{2m}\right)\exp\left(-\frac{i\epsilon}{\hbar}V(x)\right)|x\rangle + O(\epsilon^2) \\ &=& \exp\left(-\frac{i\epsilon}{\hbar}\left(\frac{p^2}{2m}+V(x)\right)\right)\langle p | x \rangle + O(\epsilon^2) \end{eqnarray*}