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Interpretation of the uncertainty principle

  1. Oct 29, 2004 #1
    I have a fundamental question regarding the uncertainty principle (position and momentum). This principle states that it is impossible simultaneously to know both position and momentum with arbitrarily good precision - even in principle.

    Now, what does "know" mean? No one can prevent you from doing a simultaneous measurement on position and momentum, and the values you get must be the position and momentum at the time of measurement (within the accuracy of the apparatus, which - in principle - can be made as small as you like). The position and momentum becomes well-defined when measured.

    By my understanding the only thing that the uncertainty principle says is that if we perform identical experiments and we get a small spread in x-values we will get a large spread in p-values according to [tex] \Delta x \Delta p_x \geq \bar{h} [/tex]. [tex]\Delta x[/tex] should be interpreted as the standard deviation in the measured x-values and [tex]\Delta p_x[/tex] as the standard deviation in the measured p-values.
     
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  3. Oct 29, 2004 #2

    dextercioby

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    Excellent.

    That's the simplified version due to Werner Heisenberg.The uncertainty principle is much more general and applies to any non commuting self adjoint operators associated by quantization to classical (Hamiltonian) observables.

    Knowing means determining the observables with arbitrary (hence even maximum) precision.That is to say that for observables A and B,[tex] \Delta A [/tex] and [tex] \Delta B [/tex] can be 0,either both,or only one (irrelevant which).


    Now comes the beautiful part of QM.Since,for an observable A,u define [tex] \Delta A =: A-<A>[/tex],which in self adjoined operator context reads [tex] \Delta A\hat =: A\hat-<A\hat>_{I\psi>}[/tex],and the commutation relationship between the self adjoint operators [tex] x\hat \mbox{and} p_{x}\hat [/tex] tells u that their commutator is proportional to the unit operator,then the principle stated above steps in and tells u that whichever the quantum state [tex]{I\psi>}[/tex] pure or mixed,u determine the averages of the operators [tex] x\hat \mbox{and} p_{x}\hat [/tex] u have no chance of finding [tex] \Delta x \mbox{or/and} \Delta p_{x} [/tex] nil.
    So yes,u can make the accuracy of the apparatus as small as u like,but when measuring simultaneously position on an axis and momentum on the same axis always the standard deviations will obbey Heisenberg formula.

    Depends on what u mean by "identical" experiments.To me,there are no "identical" experiments,therefore the theory of measurement (in QM especially) is merely statistical.

    The theory of measurement in QM is very important and relies on the basic concepts of compatible and noncompatible observables.I would advise u to read upon these matters...


    DANIEL-CRISTIAN CIOBOTU.
     
  4. Oct 29, 2004 #3

    HallsofIvy

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    Nature certainly can! In order for you to measure position "as small as you like" you will have to use light, or electrons (as in an electron microscoe), or whatever, with a very small wavelength. The smaller the wavelength, the greater the energy contained in a "quantum"- and whatever you are measuring with changes the momentum of the object you are measuring. The point of the uncertainty principle is that there ARE pairs of quantities that CANNOT be measured simultaneously- position and momentum are one such pair.
    (In terms of quantum mechanics, only quantities whose corresponding operators COMMUTE can be measured simultaneously- and most operators do not commute.)
     
  5. Oct 29, 2004 #4
    What are [tex]\Delta x[/tex] and [tex]\Delta p_x[/tex] exactly? They are not deviations from the "true values" of [tex]x[/tex] and [tex]p_x[/tex] (like classical uncertainties) since position and momentum are in general not well-defined...?

    Let's say we have a stream of electrons accelerated from rest through a constant potential and directed through a single slit and recorded on a screen. If the slit is very narrow the x-positions of the electrons are well-defined when going through, but the x-momentum has a huge spread corresponding to a wide pattern on the screen. Is this not the uncertainty principle in action?

    Please check out this thread https://www.physicsforums.com/showthread.php?t=48579 (especially post #9), which I think contradicts some of the things you are saying.

    I'm not saying you're wrong (you probably aren't). I just want a better understanding of this.
     
  6. Oct 29, 2004 #5

    dextercioby

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    They are the deviations from the average values,as stated above:[tex] \Delta \hat{A} =: \hat{A} - <\hat{A}>_{|\psi>} \hat{1}[/tex].

    Yes it is,it's basically what Feynman says in his book.

    In what sense??

    The "classical" slit experiment is the perfect example of an experimental proof that Heisenberg's principle doesn't just hold,but it actually rules the bahavior of matter at microscopical level.
     
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