 Solution

For one Cartesian displacement, the harmonic potential is $m_{\mathrm{Ar}}{\omega }_E^2{x}^2/2$. At $T\gg T_E$, the equipartition theorem assigns $k_{B}T/2$ to this quadratic potential term. HenceThis is a one-coordinate mean squared displacement, not the total three-dimensional displacement variance, which is three times larger in the isotropic model.

For comparison, the exact Einstein mean-square displacement isUsing $\coth y\simeq 1/y$ recovers the requested high-temperature result. Its nonzero zero-temperature limit is the displacement contribution of the zero-point energy.