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Physicist, PLEASE answer this puzzle

Discussion in 'Science and Technology' started by Unes, Jun 3, 2018.

  1. Unes

    Unes Active Member
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    For the mirror reflection if each photon of the incoming image on the surface of the mirror, it has to go through the uncertainty in the direction of the emission, and also it has to deal with the uncertainty in the time of the emission, as it is discovered by QED, then the reflected image from the surface of the mirror cannot possibly resemble to the incoming image, and the reflected image should get scrambled beyond any recognition. What happens at the surface of the mirrors that our well-tested uncertainty rule of QED for the reflected photons totally fails, and instead, we have the certainty of the Optic Law for the reflected photons?


    QED All-Path method and Wavefunction method are not supposed to implement the detection point which it is dictated by the Optic Law, because QED tells us “the possible detection points” for almost all the emitted photons are very different from “the detection point” that it is offered by the Optic Law.
     
  2. David T

    David T Well-Known Member
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    Well there you go. You have asked for an interpretation Of QED when it's major
    Contributor Richard feynman said", if anyone says they understand it they don't. It works"

    So factually no one here is a Richard Feynman I am sure there will be a whole host of interpretations.

    That said you are delving into pilot wave theory good luck.

    Pilot wave theory - Wikipedia
     
    #2 David T, Jun 3, 2018
    Last edited: Jun 3, 2018
  3. Sustainer

    Sustainer Well-Known Member

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    There is nothing complex about a reflection. A reflection is an invert of the inverted.
     
  4. exchemist

    exchemist Well-Known Member

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    I think it will be because the uncertainty introduced is far too small to be perceptible for a macroscopic image, that's all. Remember how small h/4π is. h ~ 6.6 x 10⁻³⁴ J-sec.
     
    #4 exchemist, Jun 3, 2018
    Last edited: Jun 3, 2018
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  5. LegionOnomaMoi

    LegionOnomaMoi Veteran Member
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    The Feynman path integral formulation of quantum mechanics and quantum field theory (including QED, which was the first successful quantum field theory and remains a fundamental component of the standard model) refers to possibilities such as possible "paths" that e.g., a photon or electron may be thought of as traversing after "emission" but prior to any measurement or detection.
    The closest thing to a wavefunction method in QED is the Dirac equation (or, if you are really desperate to get close to the wavefunction from standard QM, then the so-called Klein-Gordon equation) and neither are really adequate specifically because they don't handle the relativistic interactions of quantized electromagnetic fields.
    The uncertainty relations given in simplistic undergraduate and beginning graduate level treatments of QM are not those we find in QED or any QFT. In particular, the uncertainty relations of relativistic quantum mechanics or QFT combine with the energy-mass conversion from relativistic kinematics and dynamics to yield to vital differences found when comparing QM with a relativistic QFT such as QED:
    1) Particle number is not conserved and in the case of certain particles like photons isn't even well-defined. The uncertainty relations of QM translate into something wholly new when relativistic effects are accounted for in that the oscillations of quantum systems mean the creation and annihilation of particles.
    2) For QED specifically, even vacuum state systems do not prohibit the emission and absorption of photons or other virtual particles and in general any interaction over any time can result in the emission and absorption of photons (similar results are true for electrons and positrons, but the statistics differ here and the dynamics do as well). In particular, the uncertainties in QED are perhaps best expressed by the manner in which probability amplitudes are given to all possible diagrams for any particular process- there are always infinitely many but they are weighted differently according to likelihood in terms of how they contribute to the overall interaction and therefore how they are renormalized.

    In QED, as in QM, detection points are given as possibilities up until they are detected. Nothing in any theory tells us that uncertainty relations dictate we cannot "implement the detection point" given by quantum optics, although that phrasing doesn't have much meaning. In any event, quantum optics bypasses a good deal of the difficulties dealt with in a real treatment of photons by ignoring certain classes of interactions and processes that aren't relevant for the task at hand. Optics, however, whether classical or quantum, tells us fundamentally about how to work with particular kinds of instruments and while very important for research in fundamental theories and foundation physics, is not a framework within which we find any description of fundamental constituents of reality such that optic laws somehow supersede or negate QED or any fundamental theory seeking to describe reality (or at least if not reality then to describe operationally something that does not depend upon a particular choice of instruments and applications but the broadest possible approach to any way in which fundamental constituents can be treated, such as the way in which QED deals with photons or the relevant groups of the standard model deal with electroweak interactions including photons and electrons).
     
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  6. exchemist

    exchemist Well-Known Member

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    Er, how does this answer the poster's question?

    Or are you copy-pasting as a joke?
     
  7. Milton Platt

    Milton Platt Well-Known Member
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    Is there an English translation of this available? LOL
     
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  8. LegionOnomaMoi

    LegionOnomaMoi Veteran Member
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    The problem with any answer to the OP's question beyond that of "this question takes a bunch of differing concepts and mashes them together in a nonsensical way" is that such an answer would have to deal with the problematic presentation of said concepts.
    Also, this is not the first time this kind of problematic question has been asked:
    I tried to answer, in some detail, the problems with the presentation of QED and related physical theories and concepts in that thread. Here, I did something similar but in less detail.
    Basically, the problem here is again (as was true in the previous thread) that some mostly outdated theory, framework, or approach from physics or engineering such as classical optics (in this case) or radio waves (in the previous case) is used to "uncover" some problem with the nature of quantum electrodynamics.
    The first issue is that one does not try to critique a theory like QED or even classical electromagnetism by using out-dated "laws" and descriptions from optics (or radio waves), however useful they may be in application today as approximations.
    The second, deeper issue is that QED is not simply quantum mechanics applied to photons and light. It is drastically different from QM in structure, mathematical nature, sophistication, etc. One cannot simply take for granted popular presentations of e.g., the uncertainty principle as it is to be found in texts on QM and assume that either the conceptual structure or the theoretical structure of the uncertainty principle remains in QED. It doesn't.
    Basically, this question:
    doesn't make sense to begin with because there is nothing like what is described above that was "discovered" or that exists in QED. That's without the problem of dealing with the fact that in quantum optics and other areas of physics in which semiclassical, quantum, and other non-relativistic approaches to photons in which it makes sense to ask questions about single photons, this isn't a sensical question.
    The uncertainty principle in QM is not something that anything goes "through"; rather, it is due to the non-vanishing of the commutator of observables in the operator algebra(s) of QM. It is thoroughly quantitative.
    If you wish to understand what would happen in a given scenario QM, QED, optics, etc., you can in general find the answer (or at least whether the question makes any sense) by actually writing down the theory or theories in question and then putting in the appropriate terms for the situation. This is a major advantage in physics: theories are not merely frameworks as they are in the sciences in general but tend to be mathematical equations or sets of equations. You can write down QED (and indeed the standard model) rather easily and literally point to the relevant portion of the Lagrangian (or, if you want to get really simplistic, you can eschew equations entirely and draw the relevant diagrams using the appropriate Feynman rules).
    It is difficult, in this case, to do so not because QED fails, but because the question starts and continues in a manner that should make any attempt to actually formulate the question in a sensible manner immediately render clear its futility. The question continues in a similar manner:

    There is no such uncertainty rule in QED and there is no optic law that can simultaneously tell us what happens to a photon in a physical situation that also includes the interaction of the photon with a macroscopic instrument (which, even in the simplest quantum mechanical situation, means that special kind of entanglement we normally just call "measurement" or even "collapse of the wave-function" if we wish to exclude the macroscopic device we cause to interact with the quantum system). The entire question is posed in such a way that a naive treatment of photons given in popular accounts of QM is said to hold for QED when it doesn't and them this treatment is supposed to somehow be put into question because of some "optic law" which would fail to account for photons in the first place, let alone how or if we can make sense of a single photon interacting with a mirror in the manner described in the question. Then we are told:

    Here the problem changes somewhat. Now we are some what beyond simply the issues involved in trying to answer a question which confuses a number of concepts and theories and is fundamentally ill-posed. We are now given problematic descriptions of these concepts and how "the Optic Law" should give us pause here. I honestly am not quite sure what the first assertion is intended to refer to when it conflates Feynman integrals and wavefunctions into some method that has some implementation rule concerning detection points and the Optic Law. I DO know that there is no method in physics as described and the OP is confused about something when it comes to detection points and this "optic law" notion.
     
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  9. exchemist

    exchemist Well-Known Member

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    Would you agree the OP asks, in effect, why the uncertainty principle does not prevent the formation of a reflected image?

    If so, how would you answer that question, in a paragraph?
     
  10. LegionOnomaMoi

    LegionOnomaMoi Veteran Member
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    No, I wouldn't agree. And if that were the question, then the answer would be first that when we refer to the formation of any images, then we are asking about how macroscopic systems (like mirrors or photosensitive screens or whatever) interact with photons by causing them to have a definite position. In terms of the uncertainty principle, we can know the position arbitrarily well, but we will know nothing about any observable that doesn't commute with the position operator, such as the primary one referred to most frequently: the momentum.
    Basically, the uncertainty principle in its most common form says that the more precisely we know the position of a quantum "particle", the less exactly we know its momentum. When we form images in any way we are detecting photons, which means we are forcing them to register as being in such-and-such a place at a particular time. The uncertainty principle tells us that doing this will prevent us from knowing anything about where the photons head off to after detecting them (i.e., we can know positions at a particular time perfectly well but this entails that we will lose any information about momentum).
    Also, a reflected image can be formed from photons scattering off of a mirror only after we have lost all information about the actual photons that interacted with the mirror to form the reflected image. The only way we can obtain any information about the reflected photons would be to have the mirror double as a detection device, but here again whatever information we obtained about the incoming photons would be lost after they scattered off of the mirror to form the reflected image.
    Trying to fix this more exactly by considering only a single photon doesn't work. Even if we could ensure that a single photon interacted with the mirror, the reflection would send the photon off in a random direction (uncertainty principle matters here) and the reflected image (i.e., the detection of a photon that we are imagining we could say had been scattered off of the mirror) would again give us information only about the position, not about the momentum. In a more accurate analysis, we couldn't ever say that there was any single photon ever (there is no way to ever tell this, as photons with the same quantum numbers can occupy the same space at the same time, so what we detect as a single photon could be thousands or millions), we can't say anything about the path traversed from the mirror, and we can't say more about a photon reflected other than that it was detected at a particular place at a particular time which again is consistent with the uncertainty principle.
     
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  11. exchemist

    exchemist Well-Known Member

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    So are you saying that, according to QM, an image cannot be formed?
     
  12. LegionOnomaMoi

    LegionOnomaMoi Veteran Member
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    Having said that I don't agree that was the question asked (if it were, we wouldn't be talking about the uncertainty principle and QED like this), but granting that I can give you the answer in a sentence: it doesn't violate the uncertainty principle because the required commutation relations hold- we never know both the position and momentum of any photon at the same time such that the uncertainty principle would be violated.
     
  13. exchemist

    exchemist Well-Known Member

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    Er, quite.
     
  14. LegionOnomaMoi

    LegionOnomaMoi Veteran Member
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    No, I'm not. Images can be formed in all kinds of physical theories, and QM isn't a particularly appropriate one here, but there is nothing in the example of images forming that conflicts with the uncertainty principle in QM. Images are formed by the interaction of light with some detector. This can give us information about position. It tells us nothing about momentum.
     
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  15. exchemist

    exchemist Well-Known Member

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    All that is saying is that the uncertainty principle is not violated, which is a statement of the obvious.

    So, do you then agree with my answer, which is that an image is formed because the magnitude of the uncertainty is too tiny to be relevant in practice?
     
  16. exchemist

    exchemist Well-Known Member

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    But you can analyse reflected light with a spectrometer without destroying the image.
     
  17. LegionOnomaMoi

    LegionOnomaMoi Veteran Member
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    Not to the poster.
    No. You seem to be asserting something to the effect that there is some generic property of uncertainty in QM and that somehow in this case its magnitude is small. This is wholly false. The uncertainty principle in QM has to do with the algebra of observables: operators that don't commute, such as position and momentum, cannot be simultaneously known because the operators representing the observables don't commute/vanish!
    The magnitude of the uncertainty is irrelevant (and doesn't really relate to the uncertainty principle). What matters is what information about a quantum system we seek at a particular time. To obtain information about a quantum system we must be able to encode the information sought in an operator for the relevant observable. The uncertainty principle comes into play when we seek to obtain information about a quantum system using an operator such as position and another such as momentum simultaneously. This we can't do, but this wasn't a problem in the example, so the uncertainty principle is irrelevant (even without getting into the issues of the references to QED and so forth in the OP).
     
  18. LegionOnomaMoi

    LegionOnomaMoi Veteran Member
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    Again, the uncertainty principle concerns non-commuting observables. images always involve the detection of the positions of photons and tell us nothing about momentum of these photons. Thus, the uncertainty principle is irrelevant.
     
  19. LegionOnomaMoi

    LegionOnomaMoi Veteran Member
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    The above confuses the issue of classical-to-quantum transition or how the "classical" realm is supposed to be able to be recovered from the quantum world by taking the limit where Planck's constant goes to 0. This simplistic account (given in many textbooks and widely repeated!) of how we recover classical results is misleading at best and is actually closer to just plain wrong, but for simplicity lets just say that it is correct: quantum effects disappear and the classical world can be recovered when we can ignore Planck's constant or when it is can be treated as being 0 without changing anything relevant.
    This is not the uncertainty principle. It's true that the commutator of the position and momentum operators yield something that involve Planck's constant, but the uncertainty principle and when Planck's constant can be treated as negligible enough to ignore are wholly different. The issue of how small h/4π is concerns what we can treat as negligible in practice. The uncertainty principle tells us what we can measure or know even in principle.
     
  20. exchemist

    exchemist Well-Known Member

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    But we are talking about practice here, are we not, namely how a macroscopic image can be formed by reflected light? Or do you think the OP (who has not responded) meant to discuss only atomic scale blurring of the image?
     
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