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Uncertainty principle

  1. Jul 12, 2010 #1
    There is something (among many things of course) that confuses me with the uncertainty principle as it pertains to position and momentum of a photon. If one shoots a photon of red light, for example, at a screen, one can see where exactly the photon hits the screen so position is known. Also, because the speed of light in vaccuum is uniform at c, one can calculate, using simple algebra, the location of the photon in every step of the way. Where is the uncertainty here? I am not measuring the location of the photon inbetween but why should I, if c is a universal constant then c is a universal constant. Obviously, my reasoning is wrong, but I can't decide where my mistake is.
    1. Is it something to do with measuring the exact time of the impact? But why should I care, again c tells me how long it took for the light to hit the screen.
    2. Or maybe I cannot measure the exact distance to the screen? But then I should never be able to measure that distance, regardless of whether I am performing experiments with light or not.
     
  2. jcsd
  3. Jul 12, 2010 #2
    Hellow stone1,

    This is a quite common misunderstanding in Quantum mechanics. I didn't really understand it until I read Griffith's book.

    The uncertainty principle does not correspond to a single instance/experiment measurement, but it is a statistical results which states that if you repeat this experiment over and over again, you'll find that the results are scattered, not always the same value, with the product of the standard deviation of the measured position and momentum of IDENTICAL copies of the the experiment greater than or equal to h bar.
    i.e.
    [tex]
    \sigma_{position} \cdot \sigma_{momentum} \geq \hbar
    [/tex]

    It has nothing to do with the apparatus used in the measurement, and for sure, your measurement of a single experiment is fully deterministic, it is just if you repeat it over and over again you would not get the same result
     
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