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Mass falling to ground (uncertainty principle)

  1. Sep 16, 2009 #1
    This is a 2d problem.

    If I drop a ball with mass m from a height L over ground. How large is the interval that the ball can possibly be on if m = 1 g and L = 2m. Use the uncertainty principle [tex]\Delta[/tex]x[tex]\Delta[/tex]p ~h to determine the interval.

    Am I supposed to think of this as a classical mechanics problem and just use the uncertainty principle? I'm not sure what [tex]\Delta[/tex]x or [tex]\Delta[/tex]p would be? Or am I supposed to construct a solution to the Schrodinger equation and work with that?
  2. jcsd
  3. Sep 16, 2009 #2
    I guess the idea is this: you're trying to drop the ball exactly on a given point. So you try to precisely position the ball right over the point, and let it go with an initial momentum of 0. However there will be some error [tex]\Delta x[/tex] in your initial position and some error [tex]\Delta p[/tex] in your initial horizontal momentum, both of which will result in the ball landing off target. You could make [tex]\Delta x[/tex] very small but then [tex]\Delta p[/tex] grows large and the ball flies off very far from the target. You could make [tex]\Delta p[/tex] small but then [tex]\Delta x[/tex] grows large and you end up dropping the ball from an initial position far from the target. So you're supposed to calculate (classically) the total error in the final location of the ball resulting from an initial position error of [tex]\Delta x[/tex] and an initial momentum error of [tex]\Delta p[/tex] and then minimize the error in the final position, under the constraint that [tex]\Delta x \Delta p = \frac{\hbar}{2}[/tex].
  4. Sep 16, 2009 #3
    I have that the velocity is [tex]\sqrt{2gL}[/tex] at landing and momentum at landing is [tex]m\sqrt{2gL}[/tex]. How do I connect this to my original [tex]\Delta p[/tex] and [tex]\Delta x[/tex]
  5. Sep 16, 2009 #4
    The vertical velocity is unimportant, I think. However if you have an uncertainty in your initial horizontal momentum of [tex]\Delta p[/tex], and a fall time of [tex]\sqrt{2L/g}[/tex] then this translates into an uncertainty in the landing position of [tex](\Delta p / m) \sqrt{2L/g}[/tex], which adds to the uncertainty in initial horizontal position, [tex]\Delta x[/tex]. So you need to minimize [tex]\Delta x + (\Delta p / m) \sqrt{2L/g}[/tex]
  6. Sep 17, 2009 #5
    How do you get that uncertainty from the fall time??

    Also, how am I to minimize that function. Is it a derivative that is set to zero? It can´t be that, if it would be then derivative with respect to what?
  7. Sep 17, 2009 #6
    Suppose you have an object with horizontal velocity [tex]v[/tex], which falls for time [tex]t[/tex]. It will cover a horizontal distance [tex]vt[/tex] during that time. Or equivalently, if you have an object with uncertainty in velocity [tex]\Delta v[/tex], after time [tex]t[/tex] this translates to an uncertainly in position of [tex](\Delta v) t[/tex]. Velocity is related to momentum by [tex]v = p/m[/tex], so [tex]\Delta v = (\Delta p) / m[/tex]. So the final uncertainty in position arising from the initial uncertainty in velocity is [tex](\Delta v) t = t (\Delta p) / m = (\Delta p / m) \sqrt{2L/g}[/tex], which you can just add on to the initial uncertainty in position [tex]\Delta x[/tex].

    So you get uncertainty as a function of two variables, [tex]\Delta x[/tex] and [tex]\Delta p[/tex]. But you can eliminate [tex]\Delta p[/tex] in favor of [tex]\Delta x[/tex], using the uncertainty principle, so you get uncertainty as a function of one variable, [tex]\Delta x[/tex]. Then, as you say, you can find the minimum as you do any function, by taking the derivative with respect to [tex]\Delta x[/tex] and setting it equal to zero.
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