To start, the macroscopic system is subject to the same laws as the microscopic one, although it is harder to isolate it from it's environment. In any case, your harmonic oscillator can be considered to be made out of many particles each with their individual Hamiltonian operator, along with interactions between each particle, so the total Hamiltonian is just the sum of these, and acts on the wavefunction for the entire system. Now, you can always choose different variables to describe the system, and a convenient variable change turns out to be $x_\mathrm{com}$, the center of mass of the system, and $x_i$, the position of the $i^\mathrm{th}$ particle relative to the center of mass. You will mostly find that the variable $x_\mathrm{com}$ doesn't enter any of the interaction terms, due to translation invariance of the problem, except where it enters the potential function $V(x_\mathrm{com})$ and the kinetic energy. Therefore you can write your wavefunction as
$\Psi(x_\mathrm{com}, x_1,x_2,\dots) = \Psi_\mathrm{com}(x_\mathrm{com}) \times \psi(x_1,x_2,\dots)$, or, at least, a superposition of such states. Once you have a solution that satisfies $$(T_\mathrm{com} + V_\mathrm{com}) \Psi_\mathrm{com}(x_\mathrm{com}) = E_\mathrm{com} \Psi_\mathrm{com}(x_\mathrm{com})$$
you can substitute that solution into the wavefunction for the entire system and solve for the motion of the other particles. In other words, the center of motion factors out and it's dynamics can be considered separately.

Having said that, the normal solutions you obtain for harmonic oscillators aren't much good for macroscopic systems, because they have such large uncertainty, and don't look at all like classical behaviour, so you would then consider coherent states instead.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user lionelbrits