*The point is that, quantum mechanically, the initial conditions cannot be a position $x_0$ and momentum $\xi_0$ at a fixed time.In classical (statistical) mechanics, the initial condition is a probability distribution in the phase space (in the case we are considering, it is a delta distribution centered in the initial condition $(x_0,\xi_0)$).We know that sometimes being a student and trying to keep a descent GPA can be really hard.*

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 $\sqrt \, \mathrm$.

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?

It would occupy some standing wave about the Earth and not evolve in time.

## Writing Persuasive Essays - Solved Problems In Classical Mechanics

The reason we never see macroscopic objects in such states is because they are unstable in the same sense as Schrodinger's cat.

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.

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.

## Comments Solved Problems In Classical Mechanics

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