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Why is heisenberg uncertainty not a limit of technology?

  1. May 4, 2012 #1
    How do we know that the uncertainty principle is a property of an electron and not a limit of our measuring ability? I understand that photons striking an electron alter its momentum, but imagine an electron that is not being observed. Couldn't it have both a position and a momentum at a given point in time?
     
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  3. May 4, 2012 #2

    Pythagorean

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    Because it's not a problem of measurement, it's a problem of definition of time and space as we know it. It's a theoretical, mathematical problem that can be experimentally verified; it's not an experimental nuisance that we gave a name.
     
  4. May 4, 2012 #3

    jtbell

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    There is a very general uncertainty principle that applies to all kinds of "wave packets", and can be derived from the mathematics of Fourier analysis:

    $$\Delta x \Delta k \ge \frac{1}{2}$$

    It applies to electromagnetic waves, sound waves, electrical signals in wires, etc. The HUP is simply the application of this principle to the wavelike behavior of particles. It is no more a reflection of technological limitations on measurement, than is the case with sound waves, electrical signal pulses, etc.
     
  5. May 4, 2012 #4

    Demystifier

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    We don't.

    Yes, it could. In fact, the Bohmian interpretation of quantum mechanics proposes a very precise value of both position and momentum at a given time. See e.g.
    http://xxx.lanl.gov/abs/quant-ph/0611032
     
  6. May 4, 2012 #5

    DrChinese

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    Welcome to PhysicsForums, danphan323!

    Demystifier made some comments which are accurate in a certain sense. However, the best answer is NO, particles do not have simultaneously well-defined values for non-commuting properties.

    Notice that I said "non-commuting". Commuting properties CAN have simultaneously well-defined values. So for example: spin and momentum can both be known, but not position and momentum.

    In addition, a well known paper from 1935 referred to as EPR (Einstein is the E) tackled this issue from your perspective. A series of works over nearly 50 years answered the question in the negative. See EPR, Bell (1965), Aspect (1981) for more on this.
     
  7. May 4, 2012 #6
    True. But maybe one should point out here also that the price you pay for this objectivity in Bohm-theory is that it is explicitly non-local, i.e. particles have to move (much!) faster than light.

    We also know this is true in general from Bell's theorem which shows that any model insisting on particles "having" propertied even when they are not measured (i.e. objective theories) will have to be non-local (i.e. allow super-luminal signaling), otherwise that model will contradict experimental results in the setups made with entangled particle pairs.
     
  8. May 4, 2012 #7

    Fredrik

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    I was thinking that too. Demystifier's comments are correct, as far as I can tell, but only because to "have a position" can mean something different from to "be prepared in a state represented by a sharply peaked wavefunction".

    I started writing a much longer explanation, but it's taking too long. So this short comment will have to do, at least for now.
     
  9. May 4, 2012 #8

    DrChinese

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    I always love Demystifier's answers. Always sharp. Of course with that Bohmian edge as well. :smile:
     
  10. May 4, 2012 #9
    are time and space non-commuting?

    can all non-commuting properties can be broken down (reducible to or derived from) time-space?
     
    Last edited: May 4, 2012
  11. May 6, 2012 #10

    zonde

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    You have made similar posts in number of thread about Heisenberg uncertainty. But is there discussion where this explanation has been discussed in more details?

    Anyways I would like to understand to what extent this explanation works so let me ask some questions.
    You can't make [itex]f(x)[/itex] and [itex]\hat{f}(\xi)[/itex] peak sharply at the same time so it seems very elegant explanation for uncertainty principle in QM. Now the question I have is what would be physical interpretation of functions [itex]\hat{f}(\xi)[/itex]. It takes as an argument frequency and produces amplitude and phase for particular frequency.
    It seems like dimension of frequency can't span real space or time so it should be something more complex and indirect, right?
     
  12. May 6, 2012 #11

    Pythagorean

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    no (pretty sure time and space straight forward operators) and I don't know about all non-commuting properties. I know a lot of properties in classical physics can be reduced to time and space (but only if you forgive that mass is a ratio of distances).
     
  13. May 6, 2012 #12
    Time is not an operator in QM.
     
  14. May 6, 2012 #13

    Pythagorean

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    hrm, so then it's not an observable?
     
  15. May 6, 2012 #14

    Pythagorean

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    so then how do they commute time and energy for the uncertainty principle involving them? I thought commutation was for observable operators.
     
  16. May 6, 2012 #15
    No, time is not an observable and the time-energy uncertainty is a different principle than the position-momentum relation (it doesn't help that they look the same).

    Unfortunately I don't really know how it's derived or why it should be true (but it kinda feels right in the light of special relativity).
     
  17. May 6, 2012 #16

    Fredrik

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    It's explained here.
     
  18. May 6, 2012 #17

    Pythagorean

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    Apparently, that's exactly where it came from ("it kinda feels right")

    http://en.wikipedia.org/wiki/Uncertainty_principle#Energy.E2.80.93time_uncertainty_principle

    addendum:

    thanks Fredrik
     
  19. May 6, 2012 #18

    jtbell

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    I think most practical applications of Fourier analysis (e.g. signal processing) use time and frequency as the conjugate variables. However, the same mathematics applies when you use position and wavenumber ##k = 2 \pi / \lambda## as the conjugate variables:

    $$\psi(x) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} {A(k)e^{ikx} dk}$$
    $$A(k) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} {\psi(x)e^{-ikx} dx}$$

    Wavenumber and momentum are of course related by ##p = \hbar k##.
     
    Last edited: May 7, 2012
  20. May 6, 2012 #19
    I thought this is something we do know. My understanding is that the suggestion that the uncertainty principle is merely a limit on our ability to measure (a particle that does in fact have a fixed position and velocity) is known as the "hidden variable interpretation." And whenever I see mention of the hidden variable interpretation, it is always coupled with a statement along the lines of "convincing empirical evidence has now falsified the hidden variable interpretation." I have no real idea what evidence they are referring to, but if this is still an active controversy, it certainly isn't represented as such in anything I've read...

    Am I mistaken? I only have a lay understanding of QM...
     
  21. May 6, 2012 #20
    As far as I know, there aren't experiments that can falsify hidden variables. What the experiments have done is put extreme requirements on any hidden variable theory. I have no professors that buy into a hidden variable interpretation, but there are a number of very smart people out there that are still making good arguments. It isn't mainstream, but it is far from crackpottery.

    According to the standard interpretation (the one that every undergrad textbook that I know of and most grad textsbooks buy into), the HUP is definitely a fundamental principle. That being said, 150 years ago galilean relativity was a fundamental principle.

    So... you are pretty much right, but these are very complicated things and it is hard to prove that an interpretation of statistical data is wrong.
     
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