17.1: Particle in a Well

We begin with the one-dimensional case of a particle oscillating about a local minimum of the potential energy $V(x)$. We’ll assume that near the minimum, call it $x_{0}$ the potential is well described by the leading second-order term, $V(x)=\dfrac{1}{2} V^{\prime \prime}\left(x_{0}\right)\left(x-x_{0}\right)^{2}$ so we’re taking the zero of potential at $x_{0}$, assuming that the second derivative $V^{\prime \prime}\left(x_{0}\right) \neq 0$, and (for now) neglecting higher order terms.

Figure $\PageIndex{1}$

To simplify the equations, we’ll also move the $x$ origin to $x_{0}$, so

$$\begin{equation} m \ddot{x}=-V^{\prime \prime}(0) x=-k x \end{equation}$$

replacing the second derivative with the standard “spring constant” expression.

This equation has solution

$$\begin{equation} x=A \cos (\omega t+\delta), \text { or } x=\operatorname{Re}\left(B e^{i \omega t}\right), \quad B=A e^{i \delta}, \quad \omega=\sqrt{k / m} \end{equation}$$

(This can, of course, also be derived from the Lagrangian, easily shown to be $L=\dfrac{1}{2} m \dot{x}^{2}-\dfrac{1}{2} m \omega^{2} x^{2}$.