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Quantum Mechanics Uncertainty problem

  • Thread starter CatWoman
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  • #1
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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?
 

Answers and Replies

  • #2
HallsofIvy
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You are confusing p with [itex]\Delta p[itex].
 
  • #3
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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
Doc Al
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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
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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
Doc Al
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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
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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
Doc Al
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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
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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
Doc Al
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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
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Hi, no it's not from a text book, it's on a worksheet so it's probably a mis-print then. I really appreciate your guidance with this problem.
 

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