How Does Quantum Mechanics Affect the Landing of Marbles?

  • Thread starter Thread starter genius2687
  • Start date Start date
  • Tags Tags
    Mechanical Quantum
genius2687
Messages
11
Reaction score
0
This is a problem to a practice qualifying exam for graduate students.

A series of marbles, each with mass m, is dropped from a height H directly above a line on the ground. Although a high precision dropping device is used, each marble does not land on the line. Show that the typical distance from the line where a marble lands is

delta_x=(h_bar/m)^(1/2)*(2H/g)^(1/4)


I'm thinking of the uncertainty principle delta_x*delta_p>=h_bar/2 but other than that, I don't know how to solve this problem.

Trying to use the schrodinger equation to come up with <x>^2 and <x^2>
in order to get (delta_x)^2 = <x^2> - <x>^2 seems pretty much impossible since the potential used here is gravitational and written in the form V = mg(H-x), so therefore to find the wavefunction, you have a differential equation with variable coefficients (which requires a complicated power series), and will take a lot of time.

There should be an easy way to solve this since this is an exam problem.
 
Physics news on Phys.org
Hi genius2687,

This is a tricky one, but you can reason as follows. You can imagine that the marble is released from a definite horizontal position at t = 0 and measured at a definite horizontal position at t = T (when the marble hits the table). So what you really want to know is how much an initially localized wave function spreads in the time it takes for the marble to hit the floor. To determine this spread, you can make use of some heuristic reasoning using uncertainty principles. The marble falls for a time T = \sqrt{2 H/ g}, and since states of definite of position are not states of definite kinetic energy, the possible values of energy have some spread which can be characterized by \Delta E \sim \hbar/T. The momentum also has some spread which is set by the energy spread. With these hints you should be able to make it to the final answer. Of course, you can also just solve the Schrodinger equation for the free particle with a delta function initial condition. This doesn't take very long, and even without the full solution, you can estimate the size of the spread just on dimensional grounds.

Hope this helps!
 
Last edited:
Thanks for the advice. I got the answer except for a factor of 1/2. Though I used delta_x*delta_p=h_bar/2 (That's the relation in my QM textbook. I've seen the 1/2 factor left out, so I think I got the right idea.)
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top