Solved Problems In Classical Mechanics

Solved Problems In Classical Mechanics-30
You can search our blog to find more examples and articles related to physics study hacks, or place your order and get rid of all of your worries once and for all.

Tags: College Scholarship EssaysGood Thesis Statement Embryonic Stem Cell ResearchIrb Research ProposalPersuasive Essay Thesis ExamplesLiterature Review WikipediaWrite An Essay On Winter Season

constructs a solution that obeys Newtonian gravity.

You're of course correct that if the ball were in a single stationary state, then it wouldn't do anything remotely like falling.

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.

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.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?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.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.Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.Visit Stack Exchange Let's take the very simple problem of what happens if I drop a 1 kg ball from a height of 1 meter.


Comments Solved Problems In Classical Mechanics

The Latest from ©