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Heisenberg uncertainty question

  1. Aug 10, 2015 #1
    I read this article by Hawking which includes this quote "
    In classical mechanics one can
    predict the results of measuring both the
    position and the velocity of a particle.
    In quantum mechanics the uncertainty
    principle says that only one of these
    measurements can be predicted; the ob
    server can predict the result of measur
    ing either the position or the velocity but
    not both. Alternatively he can predict
    the result of measuring one combination
    of position and velocity."

    Can someone elaborate on the last sentence "Alternatively he can predict
    the result of measuring one combination of position and velocity"

    How is that supposed to work exactly? If you can know only one or the other then how do you measure the combination to predict anything?
  2. jcsd
  3. Aug 10, 2015 #2


    Staff: Mentor

    I don't know what Hawking meant. I can however tell you what the correct statement of the principle is.

    Suppose you have a large number of similarly prepared systems ie all are in the same quantum state. Divide them into two equal lots. In the first lot measure position to a high degree of accuracy. QM places no limit on that accuracy - its a misunderstanding of the uncertainty principle thinking it does. The result you get will have a statistical spread. In the second lot measure momentum to a high degree of accuracy - again QM places no limit on that. It will also have a statistical spread. The variances of those spreads will be as per the Heisenberg Uncertainty principle.

  4. Aug 10, 2015 #3


    User Avatar
    Science Advisor
    Gold Member

    Yes, this scenario is actually an application of the HUP. It is the product of 2 non-commuting observables' standard deviations which cannot be less than a certain quantity (a constant). A measurement (say p) which is not intended to be more accurate than a certain value can be executed. Another (say q) which is also not intended to be more accurate than a certain value can then be executed. As long as those are properly executed, you will know a combination of p and q as Hawking says.

    This could, for example, be done on entangled particles: execute a measurement of p on Alice, execute a measurement of q on Bob. Done properly, you would not violate the HUP and you would still know a lot about Alice (and Bob). You would know p +/- and you would know q +/-.

    Again, you could not execute such measurements with more precision than the HUP allows and expect useful information. This is *not* the EPR example, because in that example non-commuting observables are measured to very tight precision. So the proper execution in my example means the margin of error was intentionally made larger.
  5. Aug 10, 2015 #4
    This is a disappointingly misleading quote from Hawking. There is nothing to stop you from measuring both the position and momentum of a particle, as Hawking seems to imply at first. The HUP states that you cannot simultaneously know both of them at an arbitrary level of accuracy.
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