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Particle in a box and Heisenberg Uncertainty principle paradox?

  1. Dec 10, 2012 #1
    Say you have a particle in a one-dimensional box of length L.
    The particle can only have momentum values of the form
    [itex] p_{n} = \frac{nh}{2L} [/itex] according to the De Broglie standing wave condition.

    Now say I don't measure the position of the particle, but I know for certain that it is in the box. Then the uncertainty in the position is [itex] Δx = \frac{L}{2} [/itex].
    Therefore, the uncertainty of the momentum is [itex] Δp ≥ \frac{h}{\pi L} [/itex] (where I used Heisenberg's Uncertainty Principle that [itex] Δx Δp ≥ \frac{h}{2π} [/itex].

    Now, suppose my instruments are good enough so that their only limit is that set by Heisenberg. Then [itex] Δp = \frac{h}{\pi L} [/itex]. But then, suppose I measured a momentum of [itex]\frac{2h}{2L} [/itex]. Then the uncertainty of my momentum measurement is less than the distance to the next integer-multiple momentum value. Therefore I know, for sure, that the momentum is exactly 2h/(2L) since assuming otherwise would mean accepting values in-between, contradicting quantization. But then the uncertainty is 0, contradicting Heisenberg.

    I reach a contradiction. I assume Quantum mechanics are correct, so there must be a mistake in my reasoning. But I don't see it. Can you show me?

    My only guess, for now, is that one of the formulas I used are not exactly correct, and a more advanced course in QM will explain it better. (By the way, I'm only in a college-level course so I don't know much about QM formalism)
    Last edited: Dec 10, 2012
  2. jcsd
  3. Dec 10, 2012 #2


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  4. Dec 10, 2012 #3

    Thank you for the article. I read the abstract and introduction and it does seem to answer the question, but it is way too advanced for me (I have no idea what a "self adjoint operator" is).
  5. Dec 11, 2012 #4


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    There is a simpler, less mathematical way to answer your question. As long as the particle is confined within the box, the point is that you simply can NOT measure the momentum with perfect accuracy. To measure the momentum with such a good accuracy you would need an apparatus bigger than the box, which would destroy confinement of the particle within the box. As a result, the uncertainty of the position would become bigger than the size of the box.
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