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Heisenberg's Uncertainty Principle

  1. May 26, 2014 #1
    Hello,

    i'm solving some quite simple problems using the uncertainty principle, but I don't have access to the solutions and I really don't have a feel for what a 'sensible' answer is... When finding the minimum uncertainty in velocity, I end up with things greater than the speed of light, so I want to check i'm not making an error somewhere.

    1. The problem statement, all variables and given/known data

    Suppose that an electron is trapped in a small region and the uncertainty in it's position is [itex]10^{-15}[/itex]m. What is the minimum uncertainty in the particles momentum? What uncertainty in the electrons velocity does this correspond to?

    2. Relevant equations



    3. The attempt at a solution

    [itex](\Delta y)(\Delta p_{y}) \geq \frac{h}{4 \pi}[/itex]

    Since we're dealing the minimum uncertainty, we can equate the two.

    [itex](\Delta p_{y}) = \frac{h}{4 \pi (\Delta y)}[/itex] = 5.27 x 10-20 kgms-1

    [itex]p = mv[/itex]

    [itex]\Delta p_{y} = m (\Delta v_{y})[/itex]

    [itex]\Delta v_{y} = \frac{\Delta p_{y}}{m}[/itex] = 5.79 x 1010 ms-1

    This answer just seems absurd, maybe it's a consequence of the accuracy we know the electrons position to, but like I said I have no way of checking what a sensible answer is.

    So, have I gone wrong somewhere?

    Thanks for any help you can give.
     
  2. jcsd
  3. May 26, 2014 #2
    Your mistake is to use the formula p = mv which is the non-relativistic formula. Try using p = γmv, where γ = (1 - v2/c2)-1/2.
     
  4. May 26, 2014 #3
    Aha! Thanks, I was looking at an old example from my notes but evidently it was intended to show why the heisenberg uncertainty principle is negligible at the macroscopic level of a ping pong ball.
     
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