 Solution

Let $a$ be a stable equilibrium separation: $V^{\prime }(a)=0$, $K=V^{\prime \prime }(a)>0$. With bond extension $\delta =x-a$, the Taylor expansion isThe harmonic approximation applies when the typical thermal and zero-point bond fluctuations satisfy $|\delta |\ll a$ and the higher terms are negligible, for example $|V^{\prime \prime \prime }(a)\delta |\ll 3K$ and $|V^{\prime \prime \prime \prime }(a){\delta }^2|\ll 12K$. The relevant small quantity is the relative displacement of neighbouring atoms. Small bond excursions about a stable minimum give an amplitude-independent spring constant $K=V^{\prime \prime }(a)$. Large thermal excursions, bond breaking or appreciable anharmonicity invalidate the approximation.