Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The Uncertainty Principle Uncertainty

  1. Nov 1, 2014 #1
    My friend and I had this argument about whether or not the uncertainty principle is applicable to stationary particles. I maintain that it is, because the principle is really about predictability ( isn't it?) But he maintains that it doesn't. So I would just like to clear things up . Does it or doesnt it? Thanks in advance.
     
  2. jcsd
  3. Nov 1, 2014 #2

    phinds

    User Avatar
    Gold Member
    2016 Award

    "stationary" is not a meaningful term in this context. Motion is relative. There is no such thing as absolute motion in which something can be absolutely stationary. You have to say stationary RELATIVE TO WHAT.

    For example, you, yourself, right now as you read this, are moving at .9999999c and are massively time dilated. Does that have any impact on you?
     
  4. Nov 1, 2014 #3

    atyy

    User Avatar
    Science Advisor

    The uncertainty principle does apply to stationary particles (particles with zero average velocity).
     
  5. Nov 1, 2014 #4
    You are absoloutely right. I do have to specify relative to what. I meant stationary relative to the observer.That also answers my question , so thanks.
     
  6. Nov 1, 2014 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The uncertainty principle applies to all objects and for all observers.
     
  7. Nov 1, 2014 #6
    Right, thanks a lot
     
  8. Nov 1, 2014 #7

    jtbell

    User Avatar

    Staff: Mentor

    You can never say for sure that a particle's momentum is exactly zero. All you can ever do is say that ##p = 0 \pm \Delta p## for some value of ##\Delta p##.
     
  9. Nov 1, 2014 #8

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    In QM you cant know both position and momentum simultaneously ie there is no observation that gives both at the same time. Its is a misconception to believe QM forbids measuring either with as great an accuracy as you like, but what you cant do is measure them both at the same time.

    That being the case a particle at rest in QM is not possible because that would mean knowing it position and momentum simultaneously - unless of course you mean by at rest zero momentum but you don't know where it is - that is possible - but I don't think that's what's usually meant in saying an object is at rest.

    Thanks
    Bill
     
  10. Nov 1, 2014 #9

    DaveC426913

    User Avatar
    Gold Member

    In fact, HUP is so applicable to particles that it means there is no such thing as A "stationary" particle.

    Attempts have been made to bring atoms to a (relative) stop by reducing their temperature to 0K. Presumably, an atom at 0K will cease all motion, and thus its position and momentum (zero) could be precisely known.

    Atoms aren't havin' any of that.

    As atoms are reduced to 0K, and their motion stops, the atoms lose their individual identity and become "smeared out" into what's called a Bose-Einstein Condensate, nicely cooperating with HUP.
     
  11. Nov 2, 2014 #10
    Yeah thanks
     
  12. Nov 2, 2014 #11
    The uncertanty priciple is not only true for delta p and delta x thus postion and momentum, but it is far more general. The reason why we have h(bar)/2 is because the comutators of the X and P ,namely[X,P], are equivalent to -ih(bar). The general uncertanty principle is ΔAΔB>=<[A,B]>*1/2.
     
  13. Nov 2, 2014 #12
    How to describe UP of a photon? It is at location 5 m from us and its momentum is 10^-20 kg m/s.

    Because, the simplest explanation of UP is fourier transformation of a position wave function of a rest particle. This gives a momentum wave function. Square of absolute value of both of them are gaussian distributions and this gives thicknesses of these distributions and then product of thicknesses of both distributions.

    How to describe gaussian wave function of the photon.
     
    Last edited: Nov 2, 2014
  14. Nov 2, 2014 #13
    Yes,If the the observables do not comute. Else the uncertanty is zero in the product of the observables.
     
  15. Nov 2, 2014 #14

    Dale

    Staff: Mentor

    I thought that it was more that you could not prepare a state where both were well defined, but you could measure them as accurately as you like.
     
  16. Nov 2, 2014 #15

    phinds

    User Avatar
    Gold Member
    2016 Award

    Yes, that's my understanding as well, but there are strong proponents here on this forum for both points of view --- (1) you can measure both simultaneously to any degree of accuracy but the next measurement from the same setup won't give the same values, and (2) setup doesn't matter since you can't measure them simultaneously to arbitrary accuracy anyway. I tried once to get it pinned down but the thread I started didn't achieve that, just made it clear that there ARE two points of view here on this forum, so I've been unable to tell from this forum which it is.
     
  17. Nov 2, 2014 #16

    atyy

    User Avatar
    Science Advisor

    ZapperZ and I had a discussion about this, and I believe there is only one point of view (I can't find the post now) - non-commuting observables cannot be simultaneously and accurately measured on an arbitrary unknown state. People often read ZapperZ's blog post as meaning that simultaneously accurate measurement of position and momentum is possible, but that is not what he means. What he means is that in a double slit experiment, if one puts the screen far away, and measures position accurately at the screen, then one can use that position result to accurately calculate (the "simple" method he uses gives the right answer, but one can also obtain it by a correct quantum mechanical calculation using the Fraunhofer limit), and thus to accurately measure the transverse momentum of the particle immediately after the slit. Thus a later accurate position measurement can be used to accurately measure an earlier, non-simultaneous, momentum. As the slit is narrowed, the initial position does become more certain, but the narrowing of the slit changes the wave function immediately after the slit, so it is not a more accurate measurement of "the same" wave function.

    So to summarize:

    (1) The textbook uncertainty relations do not refer to simultaneous measurement. They say that a state cannot have simultaneously well-defined position and momentum. This means that in accurate measurements of position on an ensemble of particles in a state and separate, non-simultaneous accurate measurements of momentum on a different ensemble of particles in the same state will yield position and momentum spreads that satisfy the textbook uncertainty relation.

    (2) The textbook uncertainty relations are derived from non-commuting observables. It is not possible to simultaneously and accurately measure non-commuting observables on an arbitrary unknown state. There are exceptions if the state is not arbitrary and we know something about it. For example, if the state is known completely, we can simply write down simultaneous, accurate "measurement results". Also, if we know (for discrete variables) that the state is one of several eigenstates of an observable, then a measurement of that observable will leave the state undisturbed, so that the same state is still available for a non-commuting observable to be measured. Another famous special state is the EPR state that allows simultaneous measurement of position and momentum because of entanglement. (There's a "controversy" in the literature, but it's semantics about how one should define the "measurement error". Let's just say that for the various choices of "measurement error", I believe these papers are correct.)

    http://arxiv.org/abs/1306.1565
    Proof of Heisenberg's error-disturbance relation
    Paul Busch, Pekka Lahti, Reinhard F. Werner

    http://arxiv.org/abs/1304.2071
    How well can one jointly measure two incompatible observables on a given quantum state?
    Cyril Branciard

    http://arxiv.org/abs/1212.2815
    Correlations between detectors allow violation of the Heisenberg noise-disturbance principle for position and momentum measurements
    Antonio Di Lorenzo


    There are a couple of famous, earlier papers in the literature. Park and Margenau's 1968 paper is correct, because it shows the simultaneous measurement of conjugate position and momentum for a restricted, non-arbitary set of states. Ballentine's 1970 review is wrong, because position and momentum are not conjugate in his example.
     
    Last edited: Nov 2, 2014
  18. Nov 2, 2014 #17

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    Moriheru directly answered the original question.

    Just to be more specific, I would point out the example that energy-time as a form of uncertainty that has nothing to do with particle motion.
     
  19. Nov 2, 2014 #18
    Right, its just that position- momentum uncertainty is a specific example.
     
  20. Nov 2, 2014 #19
    Yeah, so, which is it, then? Just wondering, of the two viewpoints which one is correct?
     
  21. Nov 2, 2014 #20

    phinds

    User Avatar
    Gold Member
    2016 Award

    If you are asking me, the I have to assume you did not understand the last sentence of the post you just quoted. If you're asking the rest of the world. read the posts in this thread. We continue to have both views, although atvy has put forth citations for one side of the argument.
     
    Last edited: Nov 2, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The Uncertainty Principle Uncertainty
Loading...