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

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

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?

• I think there is a couple of mistakes in your last expression: $p$ and $x$ should not be operators, and $\langle p |x \rangle$ should be outside the brackets (?) – caverac Nov 6 '16 at 1:25
• @caverac you are absolutely right – Gvxfjørt Nov 6 '16 at 1:31
• The $O(\epsilon^2)$ term is just the second order (and higher order) term(s) of the Taylor expansion for the exponential (the first order term is just 1, which gives you the $\langle p|x\rangle$ part). – OperaticDreamland Nov 6 '16 at 1:43