# Classical Mechanics Solved Problems

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).The function $\mathrm \,\varrho_(t,x_0,\xi_0)\hat_$ is a function of time $t$, of position $x_0$, momentum $\xi_0$ through the initial quantum condition $\varrho_(x_0,\xi_0)$, and of $\hslash$.But qualitatively, this is clearly different from the classical result.

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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).

Classically, $F = mg$ and $g \approx 10 \frac$, so the ball feels a force of $10 \, \mathrm$.

From $S = ut 1/2 gt^2$, we get that the ball hits the ground in about $\sqrt \, \mathrm$. In principle, quantum mechanics should also be able to do this.

Or, if you don't like the collapse phrasing, the interaction of any stray photon with the ball will entangle the ball's position with the photon's state.

This decoheres the enormous position superposition you've constructed and gives the ball a somewhat definite position to the perspective of any external observer.

So we start with the 3-dimensional time-independent Schrodinger equation with a potential of $V = -GM/r$: $$-\frac\nabla^2 \psi - \frac\psi = E\psi \,.$$ This is effectively the same equation as for solving the Hydrogen atom: a

This decoheres the enormous position superposition you've constructed and gives the ball a somewhat definite position to the perspective of any external observer.

So we start with the 3-dimensional time-independent Schrodinger equation with a potential of $V = -GM/r$: $$-\frac\nabla^2 \psi - \frac\psi = E\psi \,.$$ This is effectively the same equation as for solving the Hydrogen atom: a $1/r$ potential but with different coefficients.

So the answer should take the same form, and we get all the energy levels, the n/l/m quantum numbers, etc.

Am I missing some way to get the classical result from the Schrodinger equation?

Related but not quite the same: Is it possible to recover Classical Mechanics from Schrödinger's equation?

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This decoheres the enormous position superposition you've constructed and gives the ball a somewhat definite position to the perspective of any external observer.So we start with the 3-dimensional time-independent Schrodinger equation with a potential of $V = -GM/r$: $$-\frac\nabla^2 \psi - \frac\psi = E\psi \,.$$ This is effectively the same equation as for solving the Hydrogen atom: a $1/r$ potential but with different coefficients.So the answer should take the same form, and we get all the energy levels, the n/l/m quantum numbers, etc.Am I missing some way to get the classical result from the Schrodinger equation?Related but not quite the same: Is it possible to recover Classical Mechanics from Schrödinger's equation?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.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 $\varrho_(t,x_0,\xi_0)=e^\varrho_(x_0,\xi_0)\;e^$.Your proposed stationary states would delocalize the ball over the entire Earth, over a scale of thousands of miles.But if you just at the ball, you can measure where it is, collapsing this superposition.

/r\$ potential but with different coefficients.

So the answer should take the same form, and we get all the energy levels, the n/l/m quantum numbers, etc.

Am I missing some way to get the classical result from the Schrodinger equation?

Related but not quite the same: Is it possible to recover Classical Mechanics from Schrödinger's equation?

## Comments Classical Mechanics Solved Problems

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