Classical Mechanics Solved Problems

Classical Mechanics Solved Problems-36
Classically, $F = mg$ and $g \approx 10 \frac$, so the ball feels a force of \, \mathrm$.

And it turns out that $$\mathrm \,\varrho_(t,x_0,\xi_0)\hat_= S(t,x_0,\xi_0) \mathrm(\hslash)\; ,$$ where $S(t,x_0,\xi_0)$ is the of the particle at time $t$, corresponding to initial condition $(x_0,\xi_0)$.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.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

And it turns out that $$\mathrm \,\varrho_(t,x_0,\xi_0)\hat_= S(t,x_0,\xi_0) \mathrm(\hslash)\; ,$$ where $S(t,x_0,\xi_0)$ is the of the particle at time $t$, corresponding to initial condition $(x_0,\xi_0)$.

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.

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.

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And it turns out that $$\mathrm \,\varrho_(t,x_0,\xi_0)\hat_= S(t,x_0,\xi_0) \mathrm(\hslash)\; ,$$ where $S(t,x_0,\xi_0)$ is the of the particle at time $t$, corresponding to initial condition $(x_0,\xi_0)$.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.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.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 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 $\hslash$.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.

/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.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 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 $\hslash$.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.

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