Quantum Mechanics Uncertainty problem

In summary, according to quantum mechanics, the closest an object of mass m can fall on average to the base of a tower of height d is (d/g)^(1/4) (h/m)^(1/2), determined using the Uncertainty Principle and the equations of motion. Despite a possible misprint, this solution provides a good understanding of the problem.
  • #1
CatWoman
30
0

Homework Statement



An object of mass m is dropped from a tower of height d. Show that according to
quantum mechanics the closest the object can fall on average to the base of the
tower is:
(d/g)^(1/4) (h/m)^(1/2)

Homework Equations


I don't know - that is the problem to know what to use.


The Attempt at a Solution


Using v^2=u^2+ 2as => velocity when mass reaches ground is v=√2gd

so momentum p=mv=m√2gd
using Uncertainty Principle ∆x∆p=h => ∆x=h/∆p=h/(m√2gd)

This is not the right answer so what have I done wrong?
 
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  • #2
You are confusing p with [itex]\Delta p[itex].
 
  • #3
I do not know how to find delta p. I have often got confused with the Uncertainty Principle and the difference between a value and it's uncertainty. I thought that maybe the value is uncertain until you measure or calculate it, then the measured or calculated value is the degree of uncertainty. But this is obviously wrong. I havn't been able to find a decent explanation. Then even if I find delta p, I am confused between what would then be meant by x or delta x.
 
  • #4
Start by assuming some initial uncertainty in horizontal position Δxi. What minimum spread in horizontal velocity does that imply? Use that to deduce the horizontal spread when object hits the ground.
 
  • #5
So before the mass falls delta x = h/m delta v ?
Then how is that used to calculate delta x after the fall?
Many thanks for your suggestions so far.
 
  • #6
Do this: Assume some initial spread in x (delta x). Now calculate the spread in velocity. Use that calculated delta v to figure out the new spread in x when it reaches ground level.
 
  • #7
I do not know how to start other than delta v = h / (m . delta x)
I then put delta v into v^2=u^2+ 2as substituting delta v in place of v, then converted back to delta x using delta x = h / (m . delta v)

I think I am barking completely up the wrong tree with this method, especially using the classical equations of motion, but can't see anything better to do!
 
  • #8
CatWoman said:
I do not know how to start other than delta v = h / (m . delta x)
Right: Δv = h/(mΔxi)
I then put delta v into v^2=u^2+ 2as substituting delta v in place of v, then converted back to delta x using delta x = h / (m . delta v)
Careful: Δv is the spread in horizontal velocity. Use this to figure out the associated spread in horizontal position at the bottom of the drop. (Hint: How long does it take the mass to fall? How far does it move sideways in that time?)
 
  • #9
I used s = ut + 1/2 at^2 to get t=(2d/g)^1/2
then delta x = t . delta v = (2d/g)^1/2 (h/m delta x)
then rearranging to get delta x = (h/m)^1/2 (2d/g)^1/4.

So the answer is almost right except I have a factor of 2 in the answer that I cannot get rid of as the question says I should get: delta x = (h/m)^1/2 (d/g)^1/4.

Many thanks for your help with this.
 
  • #10
CatWoman said:
I used s = ut + 1/2 at^2 to get t=(2d/g)^1/2
then delta x = t . delta v = (2d/g)^1/2 (h/m delta x)
then rearranging to get delta x = (h/m)^1/2 (2d/g)^1/4.
Looks good.

So the answer is almost right except I have a factor of 2 in the answer that I cannot get rid of as the question says I should get: delta x = (h/m)^1/2 (d/g)^1/4.
I don't see how they got rid of the factor of 2. What textbook is this from? (Assuming it's a textbook problem.)
 
  • #11
Hi, no it's not from a textbook, it's on a worksheet so it's probably a mis-print then. I really appreciate your guidance with this problem.
 

1. What is the quantum mechanics uncertainty problem?

The quantum mechanics uncertainty problem, also known as the Heisenberg uncertainty principle, is a fundamental principle in quantum mechanics that states that it is impossible to simultaneously measure certain physical properties, such as position and momentum, with perfect accuracy. This is due to the inherent uncertainty and randomness at the quantum level.

2. How does the uncertainty principle affect our understanding of the physical world?

The uncertainty principle challenges our classical understanding of the physical world, where it was believed that all physical properties could be measured with absolute precision. It shows that there are limitations to our ability to observe and understand the behavior of particles at the quantum level.

3. Can the uncertainty principle be overcome or bypassed?

No, the uncertainty principle is a fundamental aspect of quantum mechanics and cannot be overcome or bypassed. However, there are ways to work around it, such as using statistical methods to describe the behavior of particles or using quantum entanglement to make more accurate measurements.

4. How does the uncertainty principle relate to the wave-particle duality of particles?

The uncertainty principle is closely related to the wave-particle duality of particles, which is the concept that particles can exhibit both wave-like and particle-like behaviors. The uncertainty principle explains why we cannot have precise knowledge of both the position and momentum of a particle simultaneously, as this would violate the wave-like behavior of particles.

5. Is the uncertainty principle only applicable to subatomic particles?

No, the uncertainty principle applies to all particles, including macroscopic objects. However, the effects of uncertainty are typically only noticeable at the quantum level due to the small scale and sensitivity of particles. For larger objects, the effects of uncertainty are negligible and can be ignored in most cases.

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