Oscillator uncertainty and allocation of classical energy ($\Delta x_n^2=(n+1/2)\hslash /(m\omega )$)

For a one-dimensional oscillator eigenstate, parity gives $⟨x⟩=0$ and its coordinate variance is the displayed expression. A classical oscillator with that coordinate energy has turning amplitude $A=\sqrt{2E_x/(m{\omega }^2)}$, so $\Delta x=A/\sqrt{2}$. In several dimensions the total energy does not specify a given coordinate's classical amplitude until its energy allocation is fixed. zero-point energies in the other quantum coordinates must not silently be assigned to that one classical mode.