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

Disproving Heisenburg's Uncertanty Principle

  1. Mar 8, 2012 #1
    So I've been thinking about this for a long time. I understand that HUP means that every attempt to measure quantum particles alters them. For example, if you want to 'look' at a particle by hitting it with a light photon, you alter its trajectory, speed, and position by an unknown amount, and we can't measure the changes because that would require another measurement which changes it by an unknown amount, ad infinitum.

    Question 1: can a single quantum particle be isolated completely for observation?
    Question 2: what would happen if you bombarded the particle with two equal energy photons simultaneously to cancel the change out?

    If my understanding is off feel free to correct me.
     
  2. jcsd
  3. Mar 8, 2012 #2
    First off, your formulation is an interpretation of the UP, albeit a popular one. Although I'm no big fan of that formulation, I think I can clear up your problem: to bombard it the way you want to, you would already need to know its position to set up the apparatus properly (how else would you make sure the collissions happen the way you want them to?).

    Also, not condescendingly, just informatively: it's Heisenberg.
     
  4. Mar 8, 2012 #3
    Hmmm, I've only just started learning about Uncertainty Principle, and I find the notion of a "thought experiment" to be "thought provoking" (see what I did there?) :tongue2:

    anyways...

    AQ1. I'm not sure you would 'observe' something this minute. I mean, nowadays everything is done in bulk. You can't simply get 1 electron and 1 proton and just smash them together (not to my knowledge). Also to keep something still at absolute zero you need to have; 2 particles, completely still and a universe apart... and that leave no space for any observation equipment or humans to examine it! :rofl:

    AQ2. I have absolutely no clue about this... but I'll throw my 2 cents in.

    It would only be near impossible to get 2 photons to hit an electron (or any other quantum particle) simultaneously and on opposite poles.

    I'm guessing a whole bunch of people will now correct my 'hypothesis' (and also my spelling/grammar) on this subjective subject.
     
  5. Mar 8, 2012 #4

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    Not trying to correct anyone, just want to clear up a few things about the HUP:

    1. The HUP applies to non-commuting observables only. Which should be a big tip-off that the observation is NOT disturbing the particle in the conventional sense (for if it were, everything would be affected). For example, you can know x-spin and x-position to unlimited precision (they commute), but you cannot know x-spin and y-spin to unlimited precision (they do not commute).

    2. The HUP applies to entangled particles which are distant from each other. This too should be a tip-off, in that the observation of one also constrains the other. As EPR/Bell shows us, you cannot obtain information (say position of one and momentum of the other) which exceeds the limits of the HUP.
     
  6. Mar 8, 2012 #5

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    By the way, welcome to PhysicsForums!

    :smile:
     
  7. Mar 8, 2012 #6
    To DrChinese's comment I would also like to add that you can mathematically prove the HUP through QM's basic postulates (systems satisfy the Schrodinger equation, observables correspond to linear operators). Therefore, debating about ways to circumvent it is just as futile as trying to trisect an angle with a straight edge and compass. Of course, this is assuming the truth of the postulates of QM.
     
  8. Mar 8, 2012 #7
    Do you really believe this yourself? Of course it's smart to try and find thought experiments trying to disprove an established result, cause when you fail you will be able to establish where exactly your attempt went wrong, eventually giving more insight into the principle.

    But when it comes to the HUP, it's more delicate, cause it's not at all clear that the HUP is indeed something that says "if you measure something, things bump in that way such that...". Thank you DrChinese for giving some clear arguments for why such statements seem rubbish. The only thing that still confuses me, and confuses me about all vague arguments in physics, is that it still seems to work out to give rough estimates. I suppose it's a self-selectioning process: stupid bad arguments disappear, cause they don't give correct results, so that the bad arguments that do give good results remain.
     
  9. Mar 9, 2012 #8

    mysearch

    User Avatar
    Gold Member

    My questions are in the spirit of the comment above. I am in the 'early' stages of trying to better understand some of maths of QM/QFT, starting with Dirac's notation and linear vector spaces. As such, I am currently watching Professor V Balakrishnan video series of 31 lectures, although I am only at lecture-6. However, before diving into the abstraction of linear vector space, lecture 2-5, he starts with HUP in lecture-1, so I had started to think about some of the surrounding issues. As a slight aside, I also found the following Mathpages reference very helpful in understanding Dirac notation in terms of matrix vectors, which underpin both Heisenberg’s and Dirac’s formulation of QM and presumably led Heisenberg to the HUP. At the end of the Mathpages article there is a brief mention of HUP from which the following is taken by way of reference:
    “From this it follows that quantum mechanics would entail no fundamental indeterminacy if it were possible to simultaneously diagonalize all the observables. In that case, all observables would commute, and there would be no “Heisenberg uncertainty”. However, we find that it is not possible to simultaneously diagonalize the operators corresponding to every pair of observables. Specifically, if we characterize a physical system in Hamiltonian terms by defining a set of configuration coordinates q1, q2, … and the corresponding momenta p1, p2, …, then we find the following commutation relations”​

    However, I am not sure how the mathematical premise of the non-commutative relationship between [q] and [p] is rationalised in physicality terms.

    [1] [itex]q_mp_n – p_nq_m = i \hbar[/itex]

    For example, wouldn’t [q] and [p] be real observables of a system. If so, why does the result have a complex component. How is [1] reconciled with the HUP in [2].

    [2] [itex] \Delta x \Delta p \geq \frac {\hbar}{2}[/itex]

    While I realise that HUP is often said to be a fundamental characteristic of the quantum realm and not a limitation of experimental measurement, I would like to try to clear up a couple of issues that have always confused me somewhat. It is sometimes stated that the description of the wave-particle duality (WPD) just reflects a limitation of classical language, when applied to the quantum realm. However, might it be argued that it is the classical concept a point-particle, which is the more conceptually difficult to reconcile with QM/QFT. For example, in QFT, is it not more ‘logical’ to assume some form of wave process associated with the field in order to explain the transport of scalar energy in spacetime. If you follow this train of thought, wouldn’t a wave description have some uncertainty in its exact location [Δx] associated with its wavelength, such that [2] above might also be interpreted as:

    [3] [itex]p= \hbar \kappa= \frac{h}{\lambda}[/itex]

    [4] [itex] \Delta p = \frac {h}{\Delta x}[/itex]

    The equating of [itex]\Delta x[/itex] with [itex]\lambda[/itex] when defining location is explained further below. However, we might first try to describe a photon in these terms, i.e.

    [5] [itex] p= \frac{h}{\lambda}= \frac{hf}{c}[/itex]

    Now there would appear to be no ambiguity in terms of the frequency of a given photon from which the wavelength can be unambiguously calculated in vacuum using [itex]c=f \lambda[/itex], as presumably can the momentum [p]. So, in this case, is it is the location of a photon that might be said to be totally ambiguous, not the momentum itself, which we appear to be able to calculate independently of any exact knowledge of location. However, if we now try to describe a deBroglie matter-wave in terms of some localised wave packet, we appear to have no certainty of a single frequency/wavelength within a superposition wave model, such that the relationship below would appear to also contain some degree of uncertainty.

    [6] phase velocity=[itex] \frac {\omega}{\kappa} = f \lambda[/itex]

    [7] group velocity=[itex] \frac {\partial \omega}{\partial \kappa} = \partial f* \partial \lambda[/itex]

    So in the case of an electron, for example, as opposed to a photon, it seems we cannot calculate the momentum with any absolute precision, as in [3], if both the exact frequency and wavelength are somewhat ambiguous within the description of a wave-packet, where the exact location of the particle would also appear to ambiguous, i.e. distributed within the length of the wave packet. Therefore, in this respect, [itex]\Delta x[/itex] might be said to be analogous to the definition of deBroglie’s wavelength, i.e. not exactly certain.

    While there may well be a flaw in this logic, it seems that HUP is primarily linked to the uncertainty of location. Therefore, I not sure why QM claims that [itex]\Delta x[/itex] can be zero under any circumstances. Given that theory has to be verified by empirical measurements, how do practical experiments overcome these issues. Thanks
     
  10. Mar 9, 2012 #9
    Re: Disproving Heisenberg's Uncertanty Principle

    Thanks for the spelling correct vodka.

    My summary wasn't meant to be completely accurate, just a generalization of my perception on HUP. So I appreciate the info on HUP dr. Chinese, but how does hitting one with a photon affect the particles then?

    I don't mean this in a practical way. But my hypothesis is that if you were lucky enough (or had some new way to determine location) to strike a particle with two photons simultaneously, in theory, shouldn't the forces negate one another, thus allowing the accuracy of measurement without the changes that such observation causes? And, if you could do this enough times in a row, hypothetically, couldn't you use the observations to determine velocity with complete accuracy. As I understand it, the two are non-commuting observables, so my question still applies.

    I only ask this because even if the chances of success are 1:10^1000000, it's still greater than 0%. My question is, if someone could conduct this experiment successfully, would it break HUP? I ask because I only have a limited understanding of HUP.
     
  11. Mar 9, 2012 #10
    Absolutely. Note that I added:

     
  12. Mar 9, 2012 #11
    I don't understand your comment that it's futile. We don't yet understand why or how things like entanglement happen, do we? So HUP must not be all the answer. Even if we can't disprove HUP, I can't think of a better thing to try in order to stumble on the reasons behind the strange behavior of quantum particles.

    I never assume when I can theorize.

    By the way, I do have a theory which could explain the strange behavior, but my understanding of quantum physics limits me so I'm using this thought experiment to learn more and refine or disprove my idea without sounding stupid or uneducated.
     
  13. Mar 9, 2012 #12
    Hello usererror.

    My interpretation of the HUP is that we cannot answer both questions
    where? and when? with perfect accuracy.

    So we can pick a point in space but we cannot then say exactly when the particle will be there.

    Or we can pick a time but we then cannot say exactly where the particle will be at that time.

    Given those thoughts how would you arrange for the exact coincidence of your two photons with the particle?

    The uncertainty principle is another way of expressing the probabilistic nature of QM.

    go well
     
  14. Mar 9, 2012 #13

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    Re: Disproving Heisenberg's Uncertanty Principle

    As mentioned, no. That is because your premise is that it is the disturbance that leads to the HUP. It isn't, the HUP is inherent within (a consequence of) the description of the underlying particles.

    And for proof of that, just look at what happens when you observe entangled particle pairs (which are like mirror clones of each other) which are spacelike separated. The HUP is still followed, which would not be the case if your hypothesis were correct.

    So I would ask you to look at ALL of the evidence I and others present, not just a portion.
     
  15. Mar 9, 2012 #14

    Dale

    Staff: Mentor

    Last edited by a moderator: May 5, 2017
  16. Mar 9, 2012 #15

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    The full article, in the archives:

    http://arxiv.org/abs/1201.1833

    The uncertainty principle generally prohibits determination of certain pairs of quantum mechanical observables with arbitrary precision and forms the basis of indeterminacy in quantum mechanics. It was Heisenberg who used the famous gamma-ray microscope thought experiment to illustrate this indeterminacy. A lower bound was set for the product of the measurement error of an observable and the disturbance caused by the measurement. Later on, the uncertainty relation was reformulated in terms of standard deviations, which focuses solely on indeterminacy of predictions and neglects unavoidable recoil in measuring devices. A correct formulation of the error-disturbance relation, taking recoil into account, is essential for a deeper understanding of the uncertainty principle. However, the validity of Heisenberg's original error-disturbance uncertainty relation is justifed only under limited circumstances. Another error-disturbance relation, derived by rigorous and general theoretical treatments of quantum measurements, is supposed to be universally valid. Here, we report a neutron optical experiment that records the error of a spin-component measurement as well as the disturbance caused on another spin-component measurement. The results confirm that both error and disturbance completely obey the new, more general relation but violate the old one in a wide range of an experimental parameter.
     
    Last edited by a moderator: May 5, 2017
  17. Mar 10, 2012 #16
    Very interesting Dale, as well as DrChinese :)
    Would be nice to know the history behind how Earle Kennard came up with his formulation.
     
  18. Mar 10, 2012 #17

    You are missing the definition of uncertainty. The uncertainty in position is the root-mean-square deviation (the standard deviation in ordinary statistics): [tex]\Delta x = \sqrt {\left\langle {\hat x^2 } \right\rangle - \left\langle {\hat x} \right\rangle ^2 } [/tex]. Likewise, the momentum uncertainty is [tex]\Delta p = \sqrt {\left\langle {\hat p^2 } \right\rangle - \left\langle {\hat p} \right\rangle ^2 } [/tex]. With these definitions you can obtain [tex]\Delta x\Delta p \ge \hbar /2[/tex] from [tex]\hat x\hat p - \hat p\hat x = i\hbar .[/tex] Most textbooks on quantum mechanics do the math for you.
    Best wishes
     
  19. Mar 11, 2012 #18

    mysearch

    User Avatar
    Gold Member

    Many thanks for the useful insight. I was generally aware that that final form of HUP was based on a statistical average, but have only started to look into the mathematical details of the derivation, which also seem to be linked to the wider issue of the maths used in quantum theory. For me, this will be quite a long term ‘project’, hence the following links for later reference.
    However, at this point, it is still unclear to me how the canonical commutation relationship between [q] and [p] ends up being a complex form of [h]. I was assuming [q] and [p] to be ‘real’ quantities, at least after measurement.
    While this may be true, the last link above suggests that ‘most textbooks’ may not go into all the details and it is unclear whether they may be primarily focused on the maths without necessarily questioning the physics of the system. For example, Heisenberg's microscope ‘experiment’ seems to highlight the practical problem of measuring the location [x] of a quantum particle using X-ray photons, i.e. short wavelength allows greater positional accuracy, but high frequency causes greater disturbance of the system. While this experiment shows a relationship between location [x] and momentum [p], it is unclear how momentum [p] is a fundamental variable of the particle, if it is a composite quantity proportional to a change in position [dx] with [dt], i.e. it cannot be unambiguously related to frequency/wavelength as per a photon. In part, this is why I also raised the issue of the ‘duality’ of a particle as a wave packet in post #8; for it seems to question how the exact location of the particle/wave packet can ever be exactly determined, i.e. it is [x] that is the root problem, but one that is understandable within the wave packet description. Please accept this is not an assertion, simply a question for clarification.

    Finally, I read somewhere that the term ‘uncertainty principle’ is an inaccurate translation of Heisenberg description, as he actually described this issue in terms of ‘indeterminacy’ rather than ‘uncertainty’. In this context, might it be interpreted that Heisenberg was simply implying that system observables, at the quantum level, have no definite value and, as such, there is no uncertainty, just an inability to assign an exact value of position [x], and therefore momentum [p], as a composite quantity, when approaching the quantum scale. Of course, I accept that such things may have been subject to much clarification over the last 80 years or so. Sorry to belabour what may be my own misunderstandings. Thanks
     
    Last edited: Mar 11, 2012
  20. Mar 11, 2012 #19
    You can obtain the commutator relation between position and momentum by doing (again) the math. Here, the operators are [tex]\hat x = x[/tex] and [tex]\hat p = - i\hbar \frac{\partial }{{\partial x}}[/tex], so that [tex]\left[ {\hat x,\hat p} \right]\psi (x) = \left[ {x, - i\hbar \frac{\partial }{{\partial x}}} \right]\psi (x) = x\left( { - i\hbar \frac{{\partial \psi }}{{\partial x}}} \right) - ( - i\hbar )\frac{{\partial \left( {x\psi (x)} \right)}}{{\partial x}} = i\hbar \psi (x)[/tex]. The imaginary unit comes from the momentum operator.

    The original “explanation” given by Heisenberg was in terms of classical-like interactions. But, the current understanding is that the uncertainty principle and complimentarity, among other effects, are purely quantum mechanical that reflect the statistical nature of quantum events. Uncertainty is not “caused” by an interaction or by classical ignorance.

    If we had perfect measuring instruments, there would still be an uncertainty principle. It has nothing to do with the accuracy of a measuring device. There is always an uncertainty principle for non-commuting observables, even with a perfect apparatus! We must remember that the things we measure, like momentum and position, are operators, unlike anything in classical physics. Classical physics is determinant while quantum physics is not. Many, including Bohr, called it the indeterminacy principle. When we make repeated measurements of position, say, we generate the entire eigenvalue spectrum of the position operator and the position is said to be uncertain, or indeterminant. If we always get the same position value in repeated measurements, then the position is certain, or determinant. I hope this helps.

    Best wishes
     
  21. Mar 12, 2012 #20

    mysearch

    User Avatar
    Gold Member

    Eaglelake, while I really appreciate your help, please try to understand that what may have become obvious to you can represent a steep learning curve of a very broad spectrum of mathematical concepts, which do not appear to exist in the form of a 1-page tutorial. Equally, I think that many people need to, at least, try to anchor some of these abstracted concepts back into some form of physical interpretation. While I realise that your patience may have run out, it would be helpful to clarify a few points from post #19.

    [itex] \psi = e^{i \left( \kappa x - \omega t \right)} [/itex]

    [itex] \frac { \partial \psi}{\partial x} = i \kappa e^{i \left( \kappa x - \omega t \right)} = i \kappa \psi[/itex]

    [itex] \frac { \partial}{\partial x} = i \kappa = i \frac { \hat p}{\hbar} [/itex]

    [itex] \hat p = \frac {1}{i} \hbar \frac { \partial}{\partial x} = i \hbar \frac { \partial}{\partial x} [/itex]

    First, the equations above are simply provided as my understanding of the source of the momentum operator. In terms of a matter wave, the wave function then appears to be subject to a Fourier distribution around the wave number [k] due to dispersion of each superposition wave within the particle-wave packet. This seems to be the basis of the following paper: ‘Derivation of the HUP’ previously referenced , although it is interesting that this author does the derivation using trigonometric functions without the apparent introduction of the complex form.

    [itex]\left[ {\hat x,\hat p} \right]\psi (x) = \left[ {x, - i\hbar \frac{\partial }{{\partial x}}} \right]\psi (x) = x\left( { - i\hbar \frac{{\partial \psi }}{{\partial x}}} \right) - ( - i\hbar )\frac{{\partial \left( {x\psi (x)} \right)}}{{\partial x}} = i\hbar \psi (x)[/itex]

    Unfortunately, I am not sure that I follow all the steps in the equation above. I am assuming it is based on the non-commutative relationship between [x] and [p], i.e.

    [itex]\left[ {\hat x,\hat p}\right]\psi (x) = \left[ \hat x \hat p - \hat p \hat x \right]\psi (x) [/itex]

    If so, we can substitute for the momentum operator above, but while the first terms of the expansion seems to align to [xp], I am not sure about the expansion of [px] where [x] seems to be taken inside the partial differential operator, i.e.

    [itex]( - i\hbar )\frac{{\partial \left( {x\psi (x)} \right)}}{{\partial x}}[/itex]

    Equally, I am not sure how this then collapses to:

    [itex] i\hbar \psi (x)[/itex]

    Clearly, it would be very useful if anybody could provide a link that describes this form of algebraic manipulation in a little more detail or the specific example in question.
    I need to really think some more about non-commutative observables, especially in terms of [x] and [p]. Starting from a classical perception, which I assume many people do, it is difficult to understand momentum [p] as anything other than a composite quantity derived from 2 positions in time. As such, any uncertainty in [x] would imply an uncertainty in [p]. Equally, any attempt to associate a quantum description to some sort of superposition matter-wave seems to also lead to some form of uncertainty within the spread of the wavelengths involved. Again, I am not arguing this point, only explaining where my head is currently stuck. I will continue working on this log-jam.
    I agree that these are key arguments that I am simply trying to better understand. However, I cannot help feeling that we have a mathematical model that yields up statistical probabilities that align to experiments, while at the same time, it appears so vague in its description of any underlying physical process, e.g. dispersive superposition waveforms, that a myriad of interpretations are possible. As such, the following thread might be a good example of the spread of opinion. While post_51 and post_44 seem to be a reasonable rationalisation of the present status quo. Again, have appreciated your help. Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Disproving Heisenburg's Uncertanty Principle
Loading...