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B Position of a harmonic oscillator

  1. May 16, 2016 #1

    A. Neumaier

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    Here is the new thread. Please justify your criticism of my statement.
     
    Last edited: May 16, 2016
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  3. May 16, 2016 #2

    bhobba

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    I was replying to the statement:
    'the variance simply tells one something about the theoretical limit accuracy in trying to measure the oscillator position'

    The uncertainty relations place no limits on the accuracy you can measure position or momentum. You can measure them as accurately as you like. QM places no restriction at all.

    What the HUP says is if you take a large ensemble of similarly prepared systems; divide them into two lots, and measure position in one lot and momentum in the other, the variance of the results will be as per the HUP relations. Each measurement can be as accurate as you like.

    Thanks
    Bill
     
  4. May 16, 2016 #3

    A. Neumaier

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    What should it mean for a measurement to be accurate to more than the uncertainty (given by the standard deviation)? It cannot be checked, hence is not operationally defined. But measurements that have no operational definition cannot be carried out.

    Note that just reading a number off from a meter doesn't mean that this number is the measurement result, since any measurement of a continuous variable has an uncertainty associated with it. This is the official consensus of practicing physicists; see http://physics.nist.gov/cgi-bin/cuu/Info/Uncertainty/international1.html:
     
  5. May 16, 2016 #4

    bhobba

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    The HUP is a relation between position and momentum. Even in the loose interpretation of it you can measure position to any variance in accuracy you like if you are willing to have a large variance in momentum.

    For example in an electron going through a slit just behind the slit its position is known to the accuracy of the width of the slit - but that width can in principle be as narrow as desired. I am speaking of matters of principle - not with current technology exactly how narrow a slit can be made.

    Thanks
    Bill
     
  6. May 16, 2016 #5

    vanhees71

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    That's a very common misconception, bhobba rightly points out here. It's simply the opposite! To ensure that a probability distribution of some observable has (with a certain significance) some claimed standard deviation you have to measure the quantity with a significantly higher accuracy than the claimed standard deviation.

    In other words: there is not only no restriction to the accuracy of measuring the position of the particle in the harmonic-oscillator potential in a given preparation of this particle but to verify the very claim of the particle having a given "position uncertainty" you must be able to measure the position with much higher accuracy than that given by this standard deviation.

    Of course, the true art of experimental science is to evaluate both the statistical and systematic errors of measurements. If it only were so easy that some "world's highest authority in the field of measurement science" could solve this by decree. And as expected the CIPM only gives a recommendation how to communicate and notate the corresponding uncertainties in general. In detail, any good experimental paper takes quite a long part of its prose to precisely describe the systematical-error estimates and statistical methods to evaluate the statistical ones, but all this has not too much to do with the standard deviations described by the Heisenberg-Robertson uncertainty principle.
     
  7. May 16, 2016 #6

    A. Neumaier

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    But if the momentum in the direction of flight is known accurately (as it is in typical beams) one cannot measure the position in the direction of flight arbitrarily accurately - by the uncertainty relation. Thus it depends on the prepared state how well the position can be measured. My statement takes account of this. In any state you cannot in principle measure an observable to more than its standard deviation computed from the state.
     
  8. May 16, 2016 #7

    A. Neumaier

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    No. As long as the measurement errors are random rather than systematic, it is sufficient to perform the experiment often enough - even when each measurement is of the same low accuracy. This gives quite a good estimate of the uncertainty.

    And if one knows the accuracy of a measurement device then a single measurement is adequate, as one can get the uncertainty from the previous calibration.

    Nothing you (or bhoppa) said contradicted my statement
    One doesn't need the Heisenberg uncertainty relation to make this statement.
     
  9. May 16, 2016 #8

    bhobba

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    I would like an experimentalist to comment on that. For example I suspect if you shoot another beam at right angles and you get a collision you know its position. But its not my area and I would like someone that knows more to give an opinion.

    Thanks
    Bill
     
  10. May 16, 2016 #9

    A. Neumaier

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    You get decay products from which you can reconstruct a collision event, and to some limited accuracy the place where the event happened. But one even knows that position before doing the experiment, as it will be at the intersection of the two beams. This makes it clear that the position is not measured in the arrangement.

    To see the same in another way: In detecting a single collision, one loses most of the time information about the event. Thus you have measured the position at a very uncertain time - which makes the position at a given time still very indeterminate.
     
  11. May 16, 2016 #10

    bhobba

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    But like the slit I don't think there is any limitation in principle how thin that right angle beam can be. If you get a collision you know its position to the accuracy of the thinness of the beam. But I really want an experimentalist to comment - its beyond my background.

    Thanks
    Bill
     
  12. May 16, 2016 #11

    A. Neumaier

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    If you have infinitely thin beams carefully calibrated so that they meet and can scatter you know the exact position of their intersection without any measurement at all. And whatever you measure can at best confirm this position. No experimentalist is needed to see this.

    Thus the measurement reveals nothing about the position. It only measures that a particle has been present in the beam and passed at some uncertain time - something one could have known already by the preparation.
     
  13. May 16, 2016 #12

    bhobba

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    I don't get it. There will be no scattering unless it encountered something. Again I want someone with a background in this to comment. Its just something that occurred to me off the top of my head - there may be some other way of doing it.

    Thanks
    Bill
     
  14. May 16, 2016 #13

    A. Neumaier

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    Yes, that's why it is a measurement of the presence of a particle in the beam, but not one of position, since the position is known already before the measurement. The thinner the beam the lower the probability of scattering, so most particles will be missed. Hence to detect a particular particle to high resolution is impossible.
     
  15. May 16, 2016 #14

    rubi

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    If the particle beams are contrained to a small region of space, then their momentum distribution can't be sharply peaked due to the HUP.
     
  16. May 16, 2016 #15

    bhobba

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    That's right.

    Maybe go back to the good old screen but this time with a pinhole - you know the particle must have been at the position of the pinhole if its detected the other side.

    Thanks
    Bill
     
  17. May 16, 2016 #16

    vanhees71

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    No! The accuracy of a measurement is due to the measurement apparatus. You can have an electron with a very undetermined position, but you can measure its position at any accuracy you have available with your equipment. You can measure its position in direction of the beam quite accurately by simply putting a usual CCD with as high a resolution you have available into that beam and measure the position of an electron hitting this screen, although the positions of the electrons was not determined very well at all in preparing the beam of electrons. That's even true for photons for which a position doesn't even exist in the strict sense. Still you can put a screen detecting them into the screen and accurately measure the "position", although to be very strict, one has measured the location of an interaction between the electromagnetic field and the charges within the screen, but that's semantics and implied by the statement that one measures the position of some quantum.

    One can read the statement that the state of the object determines how accurately you can measure its observables quite often since Heisenberg, but that doesn't make Bohr's correction of this statement obsolete! Bohr was right in pointing out this misunderstanding of Heisenberg's own uncertainty relation!
     
  18. May 16, 2016 #17

    bhobba

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    Not only Bohr - but Ballentine - page 225 to 227.

    Thanks
    Bill
     
  19. May 16, 2016 #18

    dextercioby

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    The formulation advocated by Bhobba (as the interpretation of the HUP) is according to the ideas of Leslie Ballentince - the so-called „ensemble interpretation of QM” - usually thrashed here by atyy.
    This formulation is famous for rejecting the strict (projection operator (measure)) von Neumann projection postulate and is - to my mind - the closest formulation to a 100% interpretation of QM as a statistical theory.
     
  20. May 16, 2016 #19

    bhobba

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    Decoherent histories might make that claim as well where QM is the stochastic theory of histories. Interestingly a history is a sequence of projection operators so enshrines projections.

    I wouldn't say it rejects the projection postulate but rather gets rid of the collapse interpretive part of it.

    Thanks
    Bill
     
  21. May 17, 2016 #20

    strangerep

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    Yeah, like this:
    ? :-p

    [SCNR -- I wonder how long before this post gets pulled.]
     
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