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Violation of uncertainty principle?

  1. Dec 4, 2011 #1
    Suppose I have an electron gun that shoots electron with momentum p (no uncertainty) at t=0, then at t=t0 I can calculate the exact location of my electron, with no uncertainty.

    Violation of uncertainty principle?
  2. jcsd
  3. Dec 5, 2011 #2


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    If there is no uncertainty in the momentum of the electron, then your gun will need to have an infinitely wide barrel.
  4. Dec 5, 2011 #3
    A single measurement of the position tells you nothing about its uncertainty. Only if you make many measurements of position, and always get the same location, can you say there is no uncertainty in position. Quantum uncertainty is about predictability, not about an exact measurement. What we call the "uncertainty" in quantum mechanics is called the "standard deviation" in ordinary statistics:
    The uncertainty in position is [tex]\Delta x = \sqrt {\left\langle {\left( {x - \left\langle x \right\rangle } \right)^2 } \right\rangle } [/tex]

    If you know the momentum with no uncertainty (meaning that when you repeatedly measure momentum, you always get the same momentum value), then you cannot "calculate the exact location of my electron". Your assumption that you can do so is incorrect. In quantum mechanics, we can only calculate the probability of finding the electron at a particular location when we measure its position. If the electron is known to be in a momentum eigenstate, as in your example, the calculated uncertainty in position is infinite, not zero as you suggest! Quantum mechanics predicts an equal probability of finding the electron anywhere. If we do your experiment many times, each position measurement will most likely find the electron at a different location. There is no violation of the uncertainty principle.
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