Quantum Mechanical Pictures

$\hat{H}=\frac{{\hat{P}}_x^2+{\hat{P}}_y^2+{\hat{P}}_z^2}{2m}$.

Then, $\frac{d}{dt}{P}_i=-\frac{i}{\hslash }[{\hat{P}}_i,\hat{H}]=0$.

And, $\frac{d}{dt}{R}_i=-\frac{i}{\hslash }[{\hat{R}}_i,\hat{H}]=\frac{P_i}{m}$.

Since, momentum is a constant of the motion ($P_i=C_i$), ${\hat{R}}_i(t)={\hat{R}}_i(0)+\frac{{\hat{P}}_i}{m}t$.

$\hat{H}=\frac{{\hat{P}}_x^2+{\hat{P}}_y^2+{\hat{P}}_z^2}{2m}+\hat{V}(R)$.

Then, $\frac{d}{dt}{P}_i=-\frac{i}{\hslash }[{\hat{P}}_i,\hat{H}]=-\frac{\partial }{\partial R_i}\hat{V}(R)$, by Virial Theorem.

And, $\frac{d}{dt}{R}_i=-\frac{i}{\hslash }[{\hat{R}}_i,\hat{H}]=\frac{P_i}{m}$.

$\frac{d^2}{d{t}^2}{R}_i=\frac{1}{m}\frac{d}{dt}{P}_i=\frac{1}{m}{\mathrm{\nabla }}_i\hat{V}(R)$.

Thus, we have Ehrenfest Theorem, $m\frac{d^2}{d{t}^2}\overrightarrow{R}=-\overrightarrow{\mathrm{\nabla }}\hat{V}(\overrightarrow{R})$.