When is classical mechanics valid for describing motion of atoms?

Question

In Molecular Dynamics simulations, the Newton's equation of motion is used to calculate the time evolution of system. Once, I read in an introductory text that when the thermal de Broglie wavelength $$\Lambda=\frac{h}{\sqrt{2\pi mkT}}$$ is much smaller than the interparticle distance, using classical mechanics is justified and it can be used instead of quantum mechanics. Why? I mean I'd like to start from the Schrodinger equation or a theorem which is based on it (e.g. Ehrenfest's theorem) and using the above criterion obtain the Newton's equation of motion.

May you help me?

Answer 1

I'd start from Ehrenfest theorem, $$\frac{d\langle{p}\rangle}{dt}=-\left\langle{\frac{dV(x)}{dx}}\right\rangle$$ expanding the right side about $\langle{x}\rangle$, $$\frac{dV(x)}{dx}=\frac{dV(\langle{x}\rangle)}{d\langle{x}\rangle}+\frac{dV(\langle{x}\rangle)^2}{d\langle{x}\rangle^2}(x-\langle{x}\rangle)+\frac{1}{2}\frac{dV(\langle{x}\rangle)^3}{d\langle{x}\rangle^3}(x-\langle{x}\rangle)^2+\mathcal{O}\left(\langle{x}\rangle^4\right)$$ now, $\langle{x}-\langle{x}\rangle\rangle=0$, and $\langle(x-\langle{x}\rangle)^2\rangle=\sigma_x^2$, so that if $V$ varies slowly on $x$, we may just consider the first terms of the expansion, $$\frac{d\langle{p}\rangle}{dt}=-\frac{dV(\langle{x}\rangle)}{d\langle{x}\rangle}-\frac{1}{2}\sigma_x^2\frac{dV(\langle{x}\rangle)^3}{d\langle{x}\rangle^3}$$ and now for the variance $\sigma_x^2$ to be neglected, we may suppose that the size of the wave function is much smaller than the variation of the potential $V$. This way we get the desired result $$\frac{d\langle{p}\rangle}{dt}=-\frac{dV(\langle{x}\rangle)}{d\langle{x}\rangle}$$ We can interpret this as that the spatial extent of each wavefunction, practically meaning the deBroglie wavelength, need be much smaller than the separation of particles, that may represent the source of the potential $V$.