# Throwing a stone with quantum mechanics

The "throwing a stone" problem is one of the most elementary classical mechanics problem a student encounters when starts learning physics.

Given an initial position $\mathbf{x}_0$ and an initial velocity $\mathbf{v}_0$, find the trajectory of an object, given a force field $\mathbf{F}(\mathbf{x},\mathbf{v},t)$. For example, $\mathbf{F}=\mathbf{g}=(0,0,-g)$, assuming a homogenous newtonian gravitational field.

I wish to understand the classical limit of QM, so I am curious as to how one may solve such a problem in QM. I am specifically interested in the case when the "stone" is considered of course, point-like, but as a macroscoping object, whose mass is close to that of an actual stone.

Basically, my line of thought is, even if such a problem is difficult to solve directly, since, based on our current understanding of the world, it is fully quantum-theoretic in nature, classical problems in theory, should be soluble in quantum mechanical ways.

Of course, I am aware of Schrodinger's and Heisenberg's equations, I am more interested in specific solutions, since the main thing that's unclear to me here is the initial-value problem.

The classical equation can be solved uniquely given an initial position and an initial velocity.

However, if we switch to QM, this method breaks down. For a stone to be thrown, from point $\mathbf{x}_0$, with velocity $\mathbf{v}_0$, one would have to assume that an inital state $|\psi_0\rangle$ for the stone is given, which is simultaneously a position and momentum eigenstate: $$\hat{x}_i|\psi_0\rangle=x_i|\psi_0\rangle \\ \hat{p}_i|\psi_0\rangle=p_i|\psi_0\rangle,$$ but this is, of course, impossible.

I guess the classical limit arises from the fact that $\hbar$ is very small compared to any action/angular momentum dimension quantity that appears in this problem, and as such, the $\hat{x}$ and $\hat{p}$ operators "nearly commute", hence there are states which are "nearly position eigenstates" and "nearly momentum eigenstates" simulataneously.

I, however, have absolutely no idea how to formulate these "nearlies" mathematically.

Question: How to treat mathematically the quantum mechanical problem of throwing a macroscopic stone? My main motivation is to show through a specific example that the behaviour quantum mechanics ascribes to a macroscopic object is effectively the same as classical behaviour, however I have no idea how to treat the initial value problem. How can we determine the initial state of the stone?

Note: I am aware that the classical-quantum mechanical correspondance can be shown the most easy way using path integrals, but I have always gotten the impression that it was a quite general procedure, not related to any specific problem. I am not interested in answers in that direction. I am only interested in finding a suitable initial state for this stone.

• see my answer to a similar question here physics.stackexchange.com/q/330144 – anna v May 1 '17 at 6:56
• @annav For the purposes of this question, I am ignoring the fact that a stone is a composite. You can take the stone to be an electron with $m=1 kg$ or whatever and no EM interactions present. I meant "macroscopic" only in the sense that all possible $xp$ products are far greater than $\hbar$. – Bence Racskó May 1 '17 at 7:04
• You then do not have a quantum mechanical problem. What is the potential that will define the quantum mechanical wavefunction? the gravitational is very very weak . what is throwing the stone? – anna v May 1 '17 at 7:24
• @annav Is it not possible to encode the initial conditions fully in the state vector at $t=0$? – Bence Racskó May 1 '17 at 7:37
• throwing a stone involves the classical gravitational potential, which controls the trajectory. A quantum mechanical trajectory is a probability distribution from a wavefunction solution in some potential problem. For example an electron scattering off an electron . then the probability can be calculated given the electromagnetic potential between them. hyperphysics.phy-astr.gsu.edu/hbase/Particles/imgpar/feynm2.gif there are initial momenta and directions but also a potential – anna v May 1 '17 at 7:40

You consider your initial state to be a wavepacket with some uncertainty in both, momentum and position, and as you mentioned, because of the size of the mass, you can make both uncertainties pretty small from a macroscopic point of view. For a 1kg mass you can have uncertainties of $10^{-17}m$ and $10^{-17}m/s$ For position and speed.