 Thermal expansion of a weakly anharmonic bond

For a bond free of applied tension, use a local potential energy $V(a+x)=V(a)+\kappa x^2/2+\gamma x^3/6+\cdots$ with $\kappa >0$. Classical equipartition gives $⟨x^2{⟩}_0=k_{B}T/\kappa$; vanishing mean force then yields $⟨x⟩=-\gamma k_{B}T/(2{\kappa }^2)$ to first order in anharmonicity. Thus the linear thermal expansion coefficient is $\alpha =-\gamma k_B/(2a{\kappa }^2)$. For a Morse potential, $\gamma =-3\kappa /\xi$, giving positive expansion $3k_B/(2a\kappa \xi )$. This is a local bound-state approximation; the isolated cubic truncation is not a globally stable equilibrium potential.