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Hesienburg Uncertainity Principal

  1. Apr 2, 2014 #1
    As per Heinsburg uncertainity principle, Δx*Δp≈h. It means we cannot actually measure simentenously both position and momentum of the matter. They reason out this because of wave-particle duality of matter. So my question is wheather we dont't know the exact x and p of matter under consideration or the matter itself deosnt know it what is its position or momentum. To make my question more clear, does uncertainity means our knowledge in limited but nature is absolute or nature itself doesnt know in which state it is?
    Also can Δx or Δp be ever zero. I'm asking this because if this is not the case, then my interpretation that position of say electron is then not localised but spread, so does this give rise to wave like nature of electron (because wave are also spread). Correct me if I am wrong.
    Thanks:
     
  2. jcsd
  3. Apr 2, 2014 #2

    DrChinese

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    1. As far as can be determined scientifically today, it appears that quantum objects do not possess simultaneously well defined values for non-commuting operators. So X and P are not both defined, and it is not an issue of knowledge.

    2. There appears to be no lower limit to uncertainty for the measurement of a single observable. So Δx or Δp can be zero.
     
  4. Apr 2, 2014 #3
    I would have thought that the uncertainty of Δx or Δp can't be zero, as it would violate the principle.
     
  5. Apr 2, 2014 #4

    Borek

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    This thread sets new standards when it comes to spelling uncertainty. Please, my eyes are bleeding.
     
  6. Apr 5, 2014 #5

    DevilsAvocado

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    LOL :rofl: Thank god it wasn't Principal Hindenburg ... ouch my stomach huirts


    * updating signature *
     
  7. Apr 5, 2014 #6

    DevilsAvocado

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    Maybe your missing or? Afaik, both Δx and Δp can't be zero, as this would violate quantum vacuum zero-point energy (i.e. the energy of its ground state).

    600px-Harmoszi_nullpunkt.png

    The energy of the quantum ground state is [itex]E_0 = \hbar \omega / 2[/itex]

    Where [itex]\omega[/itex] is the angular frequency at which the system oscillates and [itex]\hbar[/itex] is Planck's constant [itex]h[/itex] dived by [itex]2\pi[/itex], which is an extremely tiny number ≈ 1.05457×10−34 J s.
     
  8. Apr 5, 2014 #7

    kith

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    As long as the HUP is respected, Δx or Δp can be arbitrarily close to zero but not exactly zero.

    This is because the corresponding states (delta distributions and plane waves) are not normalizable which means that we cannot interpret |ψ|² as a probability (density) as needed for QM.
     
  9. Apr 5, 2014 #8

    bhobba

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    Indeed such states are not physically realizable, and are not square integrable which they must be to be physical.

    But they are introduced for mathematical convenience being part of the Rigged Hilbert space formalism QM really depends on. The test functions of the space are the physically realizable states, things like the Dirac delta function are linear functions defined on those states to make the math easier eg so things like position operators have eigenvectors.

    Thanks
    Bill
     
  10. Apr 6, 2014 #9
    This is the greatest thread ever
     
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