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In the classical problem it's important that the ball was dropped from rest, otherwise the time taken to hit the ground will vary.But the quantum solution doesn't involve the initial conditions at all. My first thought was that I need the time-dependent Schrodinger equation instead of the time-independent one, but that doesn't lead anywhere - it just means the ball oscillates between solutions.The average position of the particle is then given by $$\mathrm \,\varrho_(t,x_0,\xi)\,\hat_\; ,$$ where $\hat_$ is the position operator (I have put the $\hslash$-dependence on the position operator because in general quantum operators depend on $\hslash$, however in the standard QM representation of the canonical commutation relations the position operator is independent of $\hslash$, and all the dependence is on the momentum operator; one could change this by means of a unitary transformation).
For example, for the free particle, a constant wavefunction would give a totally stationary particle, while $e^$ would give a moving particle.
Since the ball is heavy, this property is stable under interaction with the environment, like the position.
It would occupy some standing wave about the Earth and not evolve in time.
The reason we never see macroscopic objects in such states is because they are unstable in the same sense as Schrodinger's cat.
Am I missing some way to get the classical result from the Schrodinger equation?
Steps To Solve Problems - Classical Mechanics Solved Problems
Related but not quite the same: Is it possible to recover Classical Mechanics from Schrödinger's equation?But qualitatively, this is clearly different from the classical result.It means the ball has many quasi-stable orbitals for example, and it doesn't give a precise prediction that the ball will hit the ground in $\sqrt \, \mathrm$.That is to say, after a generic interaction with the environment the ball will quickly end up with its position peaked about a narrow value.(Not necessarily so narrow that the uncertainty principle comes into play, but effectively zero for all macroscopic purposes.) Such states exist in the Hilbert space, as complicated linear combinations of the energy eigenstates.The exact classical result is recovered only in the limit $\hslash\to 0$.If one does consider $\hslash$ with its real value, one would get corrections to the classical result, in term of powers of $\hslash$ (such corrections for an object of mass 1kg are extremely small).Nonetheless, let us take $H_$ to be the quantum Hamiltonian of the system (the one you wrote for example), and $\varrho_\hslash(x_0,\xi_0)$ to be the density matrix associated to a so-called , a state having minimal quantum uncertainty (this is done to be in a case that corresponds to a classical point of the phase space $(x_0,\xi_0)$ (delta distribution); one could take a different quantum state, but then one should also take into account a different classical description of the system, corresponding to an initial classical probability distribution in phase space that may not be a delta).The evolved state of the quantum system is $\varrho_(t,x_0,\xi_0)=e^\varrho_(x_0,\xi_0)\;e^$.To see that the peak in the wavefunction obeys Newton's laws, you can appeal to Ehrenfest's theorem, $$m \frac = - \left\langle \frac \right\rangle$$ which immediately gives that result.You may still be troubled, because in classical mechanics we need to specify an initial position and initial velocity, while in quantum mechanics it seems we only need to specify the analogue of position. The "velocity" of a particle is encoded by how fast the phase winds around in position.