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Experimental test of the Uncertainty principle

  1. Apr 28, 2010 #1
    Sorry if this question has already been asked many times, but I found no answer after a quick search. Have already made an experimental test of the Heisenberg's inequalities:
    [tex]\Delta p \Delta x \geq \hbar[/tex]

    I'm quite sure our instruments' imprecision are larger than Planck's constant, but I prefer be sure.
     
  2. jcsd
  3. Apr 28, 2010 #2
    you mean that if we have such experiments where the experimental uncertainty is smaller than the calculated HUP?
     
  4. Apr 28, 2010 #3

    Doc Al

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    Those deltas refer to distributions of measured values, not to the accuracy of any one particular measurement.
     
  5. Apr 28, 2010 #4
    yes, I forgot to point that out
     
  6. Apr 28, 2010 #5
    Sorry, my english is bad so it's hard for me to explain myself.

    Those deltas (Heisenberg's ones) refer to the spatial extension and the extension in Fourier Space (momentum space), like the width of a Gaussian, but what i am saying is, could we make a measurement of a electron, for example, where the measurement uncertainty of his position and his momentum are violating the Heisenberg's inequalities. Or let's say it in another way: Is actual instrument precision enough to dispute the uncertainty principle
     
  7. Apr 28, 2010 #6
    You can not speak of HUP for one measurment
     
  8. Apr 28, 2010 #7

    Doc Al

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    Can the "uncertainty" of a particular measurement be less than the delta in Heisenberg's inequality? Absolutely! Does that dispute the uncertainty principle? No.
     
  9. Apr 28, 2010 #8

    ZapperZ

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    You need to learn a little bit more of the thing that you want to test. If not, you are going to be testing the wrong thing.

    Look carefully at the definition of those "delta's". It is a SPREAD in measurements, i.e. MANY measurements.

    I've written in another thread the misconception of the HUP using the single slit as an illustration. Look for it.

    Zz.
     
  10. Apr 28, 2010 #9
    Last edited by a moderator: Apr 25, 2017
  11. Apr 28, 2010 #10
    Thank you for the reference Zapperz, I've found your note and read it. So now I understood why my example was absurd. But I've a new one which I think more relevant:

    Let's take your one slit montage and send this time a Gaussian wave packet in it. We fix aperture of the split, to fix the [tex] \Delta x[\tex]. We then make a series of measure of the momentum [tex] p_x[\tex] on the screen. We should have a Gaussian repartition of the momentum. But since it's a Gaussian repartition even really high values of [tex] p_x[\tex] will be measured (with really low probabilities ofc) and the HUP only deals with the width of a Gaussian at half amplitude (sorry I think I am not clear at all). So we could violate the HUP, except if the inequality is define by the width at middle amplitude for a Gaussian without excluding firmly an "exotic value" of [tex]p_x[\tex]; but then could it still be violated for an non-Gaussian wave packet?
     
  12. Apr 28, 2010 #11
    Last edited by a moderator: Apr 25, 2017
  13. Apr 28, 2010 #12

    SpectraCat

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    That paper looks highly suspect to me .. I have not read it carefully yet, but I am not sure the authors really understand the HUP. I found the following statement on the second page:

    This is very different from my own understanding of the HUP, which is that it is connected at a very deep level to the fact that quantum states are defined as vectors in a Hilbert space. It is then fairly simple to prove that an HUP relation emerges from simple vector algebra for any pair of non-commuting operators acting on kets in that space. Perhaps this is somehow equivalent to the author's definition above, but I cannot see how ... his definition looks much more restrictive.

    I will have to read the rest of the paper at another time, but I suspect that in the end, what they are presenting is not true violations of the HUP, but rather (at most) apparent violations of a more narrowly defined special case.
     
    Last edited by a moderator: Apr 25, 2017
  14. Apr 28, 2010 #13

    ZapperZ

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    Er.. what?

    If you send a gaussian wave packet, it means that you're sending a large amount of photons that are not monochromatic! The pattern you get on the screen would be VERY difficult to decipher, much less, to apply the HUP to it.

    The HUP does NOT deal with that gaussian wave packet. The HUP is a fundamental consequence of the non-commutation of observables, i.e. it is a fundamental property of QM!

    I still think you do not realize that [itex]\Delta A = \sqrt{<A>^2 - <A^2>}[/itex].

    Zz.
     
  15. Apr 28, 2010 #14
    By a Gaussian wave packet, I mean that we have prepared an electron in a way that his distribution of probablity is like a Gaussian, for example:
    [tex]\psi = Ae^{i\omega t}e^{-(x-x_0(t))^2/2\sigma^2}
    \psi \psi^* = AA^*e^{-(x-x_0)²/\sigma^2}
    [/tex]
     
  16. Apr 28, 2010 #15

    DrChinese

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    I think there are a couple of things that help to highlight the HUP:

    a) You can measure commuting observables without limit. The HUP only applies to non-commuting observables. This should be a good indicator that observational issues are not relevant.

    b) With a pair of entangled particles, you can do the following: Alice measures an observable to high precision. Bob measures the non-commuting complementary observable, also to high precision. You now know the values of a non-commuting pair to higher precision than the HUP allows. Or do you? No, the HUP still applies. A subsequent investigation will show that the particles no longer have the values you thought they did, but they will follow the HUP.

    In other words: ZapperZ is saying that measuring the values in and of itself tells you nothing more about the particle than the HUP allows. You can "think" you have beaten it, but you will not have.
     
  17. Apr 28, 2010 #16

    ZapperZ

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    This still doesn't tell you what you get AFTER you pass through the slit. The uncertainty of the position of the electron AT the slit is still the width of the slit! The uncertainty in the lateral momentum after the slit will still be equivalent to the transverse distance at the screen.

    BTW, where you have seen such a gaussian description of a single, free electron going through a slit?

    Zz.
     
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