= Einstein mean-square displacement
{c}
{title2=$\langle x^2\rangle=\hbar\coth(T_E/(2T))/(2m\omega_E)$}
For one Cartesian direction in an <Einstein solid>, $\langle x^2\rangle=\hbar[2\langle n\rangle+1]/(2m\omega_E)$. Substituting the thermal oscillator occupation proves the displayed result. At $T\gg T_E$, <equipartition> gives $\langle x^2\rangle=k_BT/(m\omega_E^2)=\hbar^2T/(mk_BT_E^2)$. The total displacement variance of an isotropic atom is three times the one-coordinate result, a distinction that matters in applying a melting threshold.
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