A question involving the Uncertainty Principle

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Homework Statement:
A boy on top of a ladder of height (H) is dropping marbles of mass (m) to the floor
and trying to hit a crack in the floor. To aim, he uses equipment of the highest
possible precision. Show that despite his great care the marbles will miss the crack by
an average distance close to:
(h2pi/m)^1/2(H/g)^1/4
Where (g) is the acceleration due to gravity.
Relevant Equations:
Equations related to motion under gravity and the uncertainty principle
I have no idea how to approach. Hence would appreciate any help
 

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  • #2
PeroK
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Homework Statement:: A boy on top of a ladder of height (H) is dropping marbles of mass (m) to the floor
and trying to hit a crack in the floor. To aim, he uses equipment of the highest
possible precision. Show that despite his great care the marbles will miss the crack by
an average distance close to:
(h2pi/m)^1/2(H/g)^1/4
Where (g) is the acceleration due to gravity.
Relevant Equations:: Equations related to motion under gravity and the uncertainty principle

I have no idea how to approach. Hence would appreciate any help

You have to give this problem your best effort. Have you been studying Quantum Mechanics?

What's the width of the crack in the floor?

Personally, I don't see how QM stops you hitting something with a marble!

And:

:welcome:
 
  • #3
HallsofIvy
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You understand, I hope, that h, Plank's constant, is about [itex]10^{-33}[/itex] (Joule-seconds). This calculation is going to be much much smaller than anything I would call a crack (or even be able to see)!
 
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  • #4
PeroK
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... and much smaller than the width of a marble!
 
  • #5
haruspex
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What's the width of the crack in the floor?
I think the crack is to be taken as zero width. The marble always misses the crack, but the task is to find the average magnitude of the miss. (Or the questioner might want the RMS miss, but if so it is worded badly.)
 
  • #6
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This question is from a Quantum physics text book actually, chapter on the uncertainty principle !! Sorry, I should have mentioned this context.
Initially I didn't get it, and then I did. So you use equation of motion under gravity, H = 1/2 gt^2, and equate this to the the x component of velocity of the falling marble, viz Vx = Xt. And then apply the uncertainty relationship to substitute Vx. The point is to merely show that, given the 10^-34 value of h, QM need not have to applied to macroscopic systems.
Thanks anyway folks
 
  • #7
haruspex
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apply the uncertainty relationship
Which one?
I only know ones that set a lower bound to products like ΔpΔx. What relationship allows you to calculate an average?
 
  • #8
PeroK
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Which one?
I only know ones that set a lower bound to products like ΔpΔx. What relationship allows you to calculate an average?
In QM the deltas in the UP are standard deviations. The UP is a statistical law.
 
  • #9
PeroK
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This question is from a Quantum physics text book actually, chapter on the uncertainty principle !! Sorry, I should have mentioned this context.
Initially I didn't get it, and then I did. So you use equation of motion under gravity, H = 1/2 gt^2, and equate this to the the x component of velocity of the falling marble, viz Vx = Xt. And then apply the uncertainty relationship to substitute Vx. The point is to merely show that, given the 10^-34 value of h, QM need not have to applied to macroscopic systems.
Thanks anyway folks

Note that the UP relates position and momentum in the same direction:
$$\sigma_x \sigma_{p_x} \ge \frac \hbar 2$$
$$\sigma_y \sigma_{p_y} \ge \frac \hbar 2$$
The speed of the marble in the vertical direction has no relationship with the uncertaity in its position in the horizontal direction. Let's take the vertical to be ##x## and the horizontal perpendicular to the crack/line to be ##y##.
A proper analysis of this from a QM perspective might be: The particle is going to be dropped and will take an approximately well-defined time to reach the floor. The particle may be put in an initial state by the dropping apparatus and, in principle, the uncertainty in the ##y## direction may be arbitrarily small. But, the smaller the uncertainty in its y-position, the greater the uncertainty in its y-momentum. Moreover, the particle's wavefunction after being released will evolve and spread out over time, depending on its spread of y-momentum, leading to a spread of impact points - if the experiment is repeated.

The trick is to choose the initial state in such a way as to minimise the spread of x-positions at the time of impact with the floor. A full solution would involve the evolution of a minimum-uncertainty Gaussian wave packet. Which probably has the optimal balance of position and momentum uncertainty. Although, you could also consider the initial constraint to be a particle in an infinite square well. Then you want to choose the width of the well to mimimise the spread of the particle's wave function - again after a fixed time of evolution.

You could shortcut these with an estimate based only on the UP. This must, however, involve some uncertainty in initial y-position. The spread in y-momentum, for example, can be arbitraily small, but only if the uncertainty in the initial y-position is large.

In truth I'm not sure what calculation could be intended here. In my view, these dodgy half-classical, half-quantum hybrid questions do little to further your understanding of QM.
 
  • #10
PeroK
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This question is from a Quantum physics text book actually, chapter on the uncertainty principle !! Sorry, I should have mentioned this context.
Initially I didn't get it, and then I did. So you use equation of motion under gravity, H = 1/2 gt^2, and equate this to the the x component of velocity of the falling marble, viz Vx = Xt. And then apply the uncertainty relationship to substitute Vx. The point is to merely show that, given the 10^-34 value of h, QM need not have to applied to macroscopic systems.
Thanks anyway folks

PS take a look at this lecture to see what I'm talking about. The key experiment (single slit diffraction) starts at 39 mins. This shows that you can squeeze a light beam into a smaller region on the screen by passing it through a slit, but after a certain critical point the smaller you make the slit, the more the light beam starts to spread out again.


Actually, the whole lecture is worth watching if you want to understand the UP. It also includes a much better analysis of why quantum effects are negligible in everyday mechanics.
 
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  • #11
haruspex
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In QM the deltas in the UP are standard deviations. The UP is a statistical law.
That's why I asked. Standard deviations will tell you about RMS error, not average modulus of error, which the question asks for.
 
  • #12
kuruman
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In truth I'm not sure what calculation could be intended here. In my view, these dodgy half-classical, half-quantum hybrid questions do little to further your understanding of QM.
I think the intended calculation is to consider projectile motion in the horizontal direction. The deviation from the crack upon landing on the floor is given by the SUVAT equation ##\delta=x_0+v_xt_f## with ##x_0=\Delta x##, ##v_x=\Delta p/m## and ##t_f=\sqrt{2H/g}##. The UP is used to eliminate ##\Delta p## from the equation and then ##\delta## can be minimized w.r.t. ##\Delta x##. Then one can go back to the SUVAT equaiton to find ##\delta_{min}.##

I agree that these hybrid questions do little to further one's understanding, but I think some of them go beyond that to being misleading. In this case, if one substitutes reasonable numbers ##m=20~\mathrm{g}## and ##H=3~\mathrm{m}##, the expression posted in #1 returns ##\delta=3\times 10^{-16}~\mathrm{m}##. This is smaller than an atomic nucleus. Considering that the marble's diameter must be much much larger than that, what does "missing the crack" mean? Also, if one is to assume that the marble is a "point mass" (a common assumption in classical mechanics) what does the very Uncertainty Principle that one is invoking here have to say about that?
 
  • #13
haruspex
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I think the intended calculation is to consider projectile motion in the horizontal direction. The deviation from the crack upon landing on the floor is given by the SUVAT equation ##\delta=x_0+v_xt_f## with ##x_0=\Delta x##, ##v_x=\Delta p/m## and ##t_f=\sqrt{2H/g}##. The UP is used to eliminate ##\Delta p## from the equation and then ##\delta## can be minimized w.r.t. ##\Delta x##. Then one can go back to the SUVAT equaiton to find ##\delta_{min}.##
That does give the book answer, but is not, it seems to me, an answer to the book question. E.g. it assumes Δx and Δp have the same sign.
Wouldn't the correct expression be along the line of ##\int\int|\phi(p)|^2|\psi(x)|^2|x+\alpha p|.dpdx##?
 
  • #14
kuruman
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I imagined a circle of hits centered at the target whose radius is determined by the optimum choice of Δx and Δp. For that I assumed that that these two quantities must have the same sign to account for the worst case scenario.

I do not understand your integral, specifically how the time evolution is taken into account. Even so, I don't think the author of the problem intended the use of wavefunctions.
 
  • #15
PeroK
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That does give the book answer, but is not, it seems to me, an answer to the book question. E.g. it assumes Δx and Δp have the same sign.
Wouldn't the correct expression be along the line of ##\int\int|\phi(p)|^2|\psi(x)|^2|x+\alpha p|.dpdx##?

Your integral looks right from a classical perspective: ##x_0## is distributed in some way; ##p## is distributed in some way (with some relation to the distribution of ##x_0##).

But, ultimately, in QM there will be an initial wavefunction (distribution of ##x##), which will evolve over time according to the Schroedinger equation (not based on SUVAT!).
 
  • #16
haruspex
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I imagined a circle of hits centered at the target whose radius is determined by the optimum choice of Δx and Δp. For that I assumed that that these two quantities must have the same sign to account for the worst case scenario.

I do not understand your integral, specifically how the time evolution is taken into account. Even so, I don't think the author of the problem intended the use of wavefunctions.
Based on your post #12, I thought you were taking the uncertainties in the initial x position and x velocity as governed by ΔxΔp = etc. (note equals, not greater than or equal to) and computing the resulting x displacement after the time to fall from that height. As I posted, that does give the book answer, provided we assume the two uncertainties have the same sign and, subject to that, they are tuned to minimise the resulting landing error. (Neither of which seem to be justified by the question being asked.)
The question does not mention a minimum error, so it does not seem to be about optimising Δx and Δp. Rather, it asks about the average distance from the target, and to get that we must have to integrate over the range of possibilities.

My view is the the person setting the question did not think it through.

Not sure what you mean about a circle. The crack is a line, so surely it is the distance from a line that concerns us. And if you are optimising Δx and Δp in one place, why take a worst case scenario in another?
Do you get the book answer with that approach?
 
  • #17
haruspex
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not based on SUVAT!
Yes, I suspected that was the case. Is there a way to answer the question sensibly, or is it dependent on an unknown initial wave function?
 
  • #18
PeroK
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Based on your post #12, I thought you were taking the uncertainties in the initial x position and x velocity as governed by ΔxΔp = etc. (note equals, not greater than or equal to) and computing the resulting x displacement after the time to fall from that height. As I posted, that does give the book answer, provided we assume the two uncertainties have the same sign and, subject to that, they are tuned to minimise the resulting landing error. (Neither of which seem to be justified by the question being asked.)
The question does not mention a minimum error, so it does not seem to be about optimising Δx and Δp. Rather, it asks about the average distance from the target, and to get that we must have to integrate over the range of possibilities.

My view is the the person setting the question did not think it through.

Not sure what you mean about a circle. The crack is a line, so surely it is the distance from a line that concerns us. And if you are optimising Δx and Δp in one place, why take a worst case scenario in another?
Do you get the book answer with that approach?

The required calculation is just a quick and dirty estimate. It doesn't consider the nature of the distributions at all. Your integral in a previous post looks right for the classical case where the initial position and initial momentum have known distributions.

In QM, however, the initial distribution of position (wavefunction) evolves according to the Schroedinger equation (not according to SUVAT!). The QM solution would involve an estimate of the time evolution of the wavefunction and hence the distribution of position at some later time. Technically, you would want to start with the ground state of the infinite square well as the initial wavefunction. And then let that evolve in free space. Which is easier said than done.
 
  • #19
haruspex
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The required calculation is just a quick and dirty estimate. It doesn't consider the nature of the distributions at all. Your integral in a previous post looks right for the classical case where the initial position and initial momentum have known distributions.

In QM, however, the initial distribution of position (wavefunction) evolves according to the Schroedinger equation (not according to SUVAT!). The QM solution would involve an estimate of the time evolution of the wavefunction and hence the distribution of position at some later time. Technically, you would want to start with the ground state of the infinite square well as the initial wavefunction. And then let that evolve in free space. Which is easier said than done.
Yes, I understand that the correct approach is through evolution of the wave function, but which method have you in mind for the quick and dirty estimate? Does it lead to the book answer, and if so is it by means of SUVAT as I outlined (and with all the mistaken assumptions I mentioned)?
 
  • #20
PeroK
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Yes, I understand that the correct approach is through evolution of the wave function, but which method have you in mind for the quick and dirty estimate? Does it lead to the book answer, and if so is it by means of SUVAT as I outlined (and with all the mistaken assumptions I mentioned)?

I assume that intended method is:

Take the initial uncertainty in ##x## as a starting offset, ##x_0##; calculate the uncertainty in ##p## from the uncertainty relation and obtain a definite velocity, ##v##, from this; calculate the offset after time ##t## as ##x_0 + vt## (assuming ##v## has the same sign as ##x_0##; minimise this wrt ##x_0##; take this to be a worst case and divide by ##2## to get the answer.

That said, the ##2\pi## in the answer looks like it's in the wrong place.

To be honest, I've never really learned this quantum-SUVAT hybrid, although I've seen a few homework problems on here that require this approach.
 
  • #21
PeroK
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PS taking a look at my notes for the Gaussian (minimum uncertainty) wave-packet, we can use this to get a proper estimate here. If we start with a Gaussian wavefunction:
$$\Psi(x, 0) = \big ( \frac{2a}{\pi} \big )^{\frac 1 4} e^{-ax^2}$$
Then at time ##t## we have:
$$\langle |x| \rangle(t) = \frac{b(t)}{\sqrt{2\pi}}$$
Where ##\langle |x| \rangle(t)## is the expected value of the magnitude of the x-position at time ##t##. And:
$$b(t) = \sqrt{\frac 1 a (1 + (\frac{2\hbar at}{m})^2)}$$
We can, therefore, mininise the uncertainty in position at time ##t## by minimising ##b(t)## wrt the parameter ##a##. This results in:
$$a = \frac{m}{2\hbar t}; \ \ b_{min} = 2 \sqrt{ \frac{\hbar t}{m}}; \ \ \langle |x| \rangle_{min} = \sqrt{\frac{2\hbar t}{\pi m}}$$
Looking at the original problem now, it seems to me that a better idea is to use a quick and dirty estimate to justify that the uncertainty is proportional to ##\sqrt{\hbar t/m}##.

Note also that for this Gaussian the uncertainty in momentum is constant over time:##\sigma_p = \hbar \sqrt a##.
 
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  • #22
haruspex
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Thanks, @PeroK . So the correct answer is 1/π times the given answer.
 

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