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Natural Science Forum / Physics / Research / December 2007On the uncertainty principle for photons. An experimental counter example??
From QM follows that the position of a photon can not be determined better than L (de Broglie wavelength). Suppose one creates a wave in the meter range - say L=2m. Could the photons in this wave go through a tube of radius 1 cm? If one registers a click of a photon counter placed inside the tube - would this be a counterexample of QM - e.g. the position of the photon determeined better than de Broglie's wavelength??? > From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). [quoted text clipped - 3 lines] > would this be a counterexample of QM - e.g. the position of the photon > determeined better than de Broglie's wavelength??? "near field" microscopy 107,000 hits Subwavelength apertures are commonplace. Exit intensity rapidly decays with conduit length.  SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 > "near field" microscopy 107,000 hits > [quoted text clipped - 4 lines] > Uncle Alhttp://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/lajos.htm#a2 So it's a counterexample which takes common place??? Regards: Kevin > From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). If you think so, please show a derivation from first principles. (and please, wavelength will do. No need to involve De Broglie) > Suppose one creates a wave in the meter range - say L=2m. Could the > photons in this wave go through a tube of radius 1 cm? Yes. Try it. > If one registers a click of a photon counter placed inside the tube - > would this be a counterexample of QM - e.g. the position of the photon > determeined better than de Broglie's wavelength??? No, unless you supply the proof required. You can do better than that though. Suppose you throw a photon with sufficient energy at an atom, and detect the resultant photo-electron. If you see an electron you have fixed the position of the photon to an atomic diameter, much smaller than its wavelength. For further amusement you may note that the atom can by a hydrogen atom in a Rydberg state, which allows you to detect even radio-frquency waves. Best, Jan > From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). [quoted text clipped - 3 lines] > would this be a counterexample of QM - e.g. the position of the photon > determeined better than de Broglie's wavelength??? I think the misunderstanding is that QM does not say that the energy of a photon cannot be captured in a volume much smaller than its wavelength, but that the exact path that the photon travels between source and detection cannot be measured more precisely than this. As other replys have pointed out, the energy in a radio frequency photon can originate from a volume with dimensions orders of magnitude smaller than the wavelength, and when the photon is detected the location of detection can be determined with much greater precision as well. It is the location of the "photon" in between these two events that is uncertain. Rich L. >> From QM follows that the position of a photon can not be determined >> better than L (de Broglie wavelength). [quoted text clipped - 16 lines] > > Rich L. Sure, but what about the Delta x times Delta p >= hbar/2 problem? If the position is momentarily at the size of an atom (a few tenths of nm) and much smaller than wavelength, then the momentum uncertainty is more than the wavelength - but it can't be. There must be a good answer, but is it simple? > Sure, but what about the Delta x times Delta p >= hbar/2 problem? If the > position is momentarily at the size of an atom (a few tenths of nm) and much > smaller than wavelength, then the momentum uncertainty is more than the > wavelength - but it can't be. There must be a good answer, but is it simple? The position-momentum uncertainty as you wrote it applies only to Gaussian wave packets, you can't apply it to a plane wave (unless you think it as a limit for Delta x -> infinity). Anyway, if you light the atom with a radio frequency wave, and you get a photo-electron, you know that you've hit it. Now, how do you know where is the atom? The only information you have is that you've hit it, but you don't know where you did it better than the wavelength. If you want to pin down the position of the atom better than that, you can start to use more spatially- (or time-) compressed wave packets, and you will start to lose the energy information, just as the uncertainty principle says. For example, if you want to know the position within 0.1nm, you'll need a wave packet with an energy dispersion of at least 1keV. Regards, Guillermo > >> From QM follows that the position of a photon can not be determined > >> better than L (de Broglie wavelength). [quoted text clipped - 22 lines] > wavelength - but it can't be. There must be a good answer, but is it simple? >- Hide quoted text - That would be true for the photon, but once it is "detected" it doesn't exist anymore. The energy of the photon is now in an electron confined (possibly) to an atom. The uncertainty principle doesn't say anything about what happens when the wave function "collapses" like this. Rich L. > From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). [quoted text clipped - 3 lines] > would this be a counterexample of QM - e.g. the position of the photon > determeined better than de Broglie's wavelength??? de Broglie wave length L = h/p uncertainty (delta L) => h/(4pi(delta p)) There is a difference. Richard >> From QM follows that the position of a photon can not be determined >> better than L (de Broglie wavelength). [quoted text clipped - 11 lines] > > Richard The Heisenberg uncertainty principle requires that when the uncertainty in a photon's position is less than it's wavelength (lambda), the uncertainty in its momentum must be greater than hc/lambda (ignoring factors of 2 and pi). This is possible because the photon momentum is complex, with both a real part, with magnitude hc/lambda, and an imaginary part, which makes the total magnitude of the momentum greather than hc/lambda. The imaginary component of the momentum required for subwavelength resolution corresponds to an attenuation of the photon wave. Using a narrow tube to define the photon position comes at the cost of an exponential loss of intensity. J.J.Lodder wrote: kvblake <kvblake2...@yahoo.com> wrote: > From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). If you think so, please show a derivation from first principles. (and please, wavelength will do. No need to involve De Broglie) > Suppose one creates a wave in the meter range - say L=2m. Could the > photons in this wave go through a tube of radius 1 cm? Yes. Try it. > If one registers a click of a photon counter placed inside the tube - > would this be a counterexample of QM - e.g. the position of the photon > determeined better than de Broglie's wavelength??? No, unless you supply the proof required. You can do better than that though. Suppose you throw a photon with sufficient energy at an atom, and detect the resultant photo-electron. If you see an electron you have fixed the position of the photon to an atomic diameter, much smaller than its wavelength. For further amusement you may note that the atom can by a hydrogen atom in a Rydberg state, which allows you to detect even radio-frquency waves. Best, Jan ========================================================= Sorry about the delay. A huge lack of Internet connection... I think that was a trivial fact. Nevertheless the prove I found seems a little strange to me too. It's from QED of Beresteckii, Lifshitz (the intro). It's like that: 1. Min uncertainty of x -> Dx for a massive particle in its rest frame is Dx=h/mc (under h I mean Dirac's h not Heisenberg's h->h/2PI but I dont know how to write it with this bar up.. - that's why ...... was confused with this 4PI - ---- the second 2 is from some preciser formulation of DxDp=h/2) 2. Min uncertainty in the frame where the particle is moving with energy E is then Dx=ch/E 3. For ultrarelativistic particles E almost = cp hence Dx>h/p 4. Photons are always relativistic so Dx>h/p=L (lambda as introduced by de Broglie). In this book also stands the following. The above is valid for experiments from every result of which x can be determined (I would call this experiments of class I) If one uses impacts without probability one for the time of experiment from each deflection of the probe particle one can judge about x. But if deflection is 0 one can not tell anything about x (class II). I think my experiment is from class I and yours is from class II. ================================ To Rich: Rich L. wrote: I think the misunderstanding is that QM does not say that the energy of a photon cannot be captured in a volume much smaller than its wavelength, but that the exact path that the photon travels between source and detection cannot be measured more precisely than this. As other replys have pointed out, the energy in a radio frequency photon can originate from a volume with dimensions orders of magnitude smaller than the wavelength, and when the photon is detected the location of detection can be determined with much greater precision as well. It is the location of the "photon" in between these two events that is uncertain. Rich L. ========================== If there is a source of radiowaves and I put in front of it a tube of lenght 10 m and r=2 cm with a detector at the end and the detector clicks I could be absolute sure of the path and Dp wouldn't also be great provided 'c' is a constant. .. cut text... > ================================ > To Rich: [quoted text clipped - 21 lines] > clicks I could be absolute sure of the path and Dp wouldn't also be > great provided 'c' is a constant. If you did this experiment, you would find that the probability of detecting the radio frequency photon was extremely small, assuming the walls of the tube are opaque to radio waves and the wavelength is much greater than 2cm. This is no more mysterious than detecting a photon in the evanescent region where the photon theoretically cannot exist at all. The QM wave function exists in such areas, but is attenuating rapidly with distance into it. QM makes experimentally confirmed accurate predictions about such situations. Rich L. > .. cut text... > [quoted text clipped - 36 lines] > > - - Thanks for consideration: I am sure that probability for detection would be extremely small. But as you say there would be such detections even with opaque walls. What about the path!!! of that photon (even the impulse) - if there is just one exception the Heisenberg principle is threatened??? I can even remove a part of the tube and know the path of that photon in between?! Regards: Kevin > > .. cut text... > [quoted text clipped - 49 lines] > > - Show quoted text - I don't think so. The Heisenberg principle concerns the product delta_P*delta_X or delta_E*delta_T. In the case of the tube you are confining in the X and Y directions to the diameter of the tube (assuming it extends along the Z direction). Delta X and Delta Y are thus very small. The Uncertainty principle says the uncertainty in the momentum in the X and Y directions will thus be very large, which is is due to the high k (wavenumber) in order to meet the boundary conditions on the walls of the tube (the math is easier if you use a square tube). Thus at the end of the tube you do indeed have a very precise location for the photon, but you have a very poor idea of it's momentum in the X and Y directions. In fact it is because the momentum uncertainty is so large that the amplitude is decaying exponentially down the tube. Does that help? Rich L. > > > .. cut text... > [quoted text clipped - 69 lines] > > - - Not much. More questions arise? 1. The uncertianty of impulse means there should be photons of different (even very very high) energies / impulses/ ariving at the end of the tube. Where could this energies come from? [In the tought experiment of Heisenberg this is very clear - an electron gets an impulse/energy from the photon which is intended to locate it.] 2. The wave associated with the photon was originally in the radio wavelength range. If now the energy is high - the wavelength may be already in the visible range. So there should be a change in the wavelength of the particle. I never heard about such phenomenon - change of the wavelength of a particle even if it goes through obstacles. Maybe I just don't know? 3. What about that sectioin where there is no tube. The particle (photon) is not limited in X,Y - the wave diffracts there and many of the photons which had passed part 1 of the tube should spread; but not this one which goes into part 2 of the tube. It is not confined by the walls and nevertheless I can't imagine it has been travelling some other path but straight line into part 2. I can't imagine one of the spread photons can turn and come into part 2 and travel then in straight line. 4. For a moment I thought the Heisenberg principle is for ensembles (not for individual particles) but no it is an absolute prohibition for 1 particle! 5. My original question was not at all concerning impulse - for the massless particles it has been shown (as I cited in this thread before) in Beresteckii, Lifshitz that Dx>L (h/p - wavelength of de Broglie; h->h/2.PI) - maybe because Dp couldn't be bigger than p itself (I cited before the whole prove). That's why I thought of that counter example with the tube. Regards: Kevin .. > Not much. More questions arise? > > 1. The uncertianty of impulse means there should be photons of > different (even very very high) energies / impulses/ ariving at the > end of the tube. Where could this energies come from? The energy is proportional to the frequency. Confining the wave to the tube does not change the frequency. What is affected is the relationship between the momentum (aka wavelength) and energy. The wavelength across the radius of the tube is very short, which normally would correspond to a high frequency photon. However the "wavelength" in the direction of the tube is imaginary, and this cancels out the effect of the very short wavelength across the tube. It all works out. > [In the tought experiment of Heisenberg this is very clear - an > electron gets an impulse/energy from the photon which is intended to > locate it.] I don't see how that applies to a photon propagating down a tube? > 2. The wave associated with the photon was originally in the radio > wavelength range. If now the energy is high - the wavelength may be > already in the visible range. So there should be a change in the > wavelength of the particle. I never heard about such phenomenon - > change of the wavelength of a particle even if it goes through > obstacles. Maybe I just don't know? As stated above, there is no change in photon frequency and thus energy. > 3. What about that sectioin where there is no tube. The particle > (photon) is not limited in X,Y - the wave diffracts there and many of [quoted text clipped - 4 lines] > spread photons can turn and come into part 2 and travel then in > straight line. Once the photon is in free space you can no longer say with certainty where it is, or was. Just because it came out of a small hole in one place and entered a small hole in another does not mean the photon had to traverse a direct straight line between those two points. There are many experiments in diffraction that demonstrate this. > 4. For a moment I thought the Heisenberg principle is for ensembles > (not for individual particles) but no it is an absolute prohibition > for 1 particle! This is correct. > 5. My original question was not at all concerning impulse - for the > massless particles it has been shown (as I cited in this thread [quoted text clipped - 4 lines] > > That's why I thought of that counter example with the tube. I don't follow this last bit. What you are wondering about here seems to be just diffraction. For something that is much more difficult to make sense of, and fundamental to Quantum Mechanics, study the EPR paradox, Bells Inequality, and the Aspect experiments based on them. Then look up Quantum Eraser Experiment. These all have decent articles on Wikipedia. Rich L. > .. > [quoted text clipped - 12 lines] > effect of the very short wavelength across the tube. It all works > out. Sorry I've never read anywhere about imaginery wavelength. I would highly appreciate if you can cite a textbook or a website where this is analysed. In a past article in this thread Eliot Spech wrote about imaginary part of the momentum. I also dont know nothing about this no have ever seen in a textbook on QM. > > 3. What about that sectioin where there is no tube. The particle > > (photon) is not limited in X,Y - the wave diffracts there and many of [quoted text clipped - 10 lines] > to traverse a direct straight line between those two points. There > are many experiments in diffraction that demonstrate this. OK. Let the photons fly all those crooked paths of Feynmann. I hope the speed is still constant. Then if one puts a time 'filter' and gets one click on the detector he would know the path better than h/p = lambda? > For something that is much more difficult to > make sense of, and fundamental to Quantum Mechanics, study the EPR > paradox, Bells Inequality, and the Aspect experiments based on them. > Then look up Quantum Eraser Experiment. These all have decent > articles on Wikipedia. The most weird for me is the behaviour of a single photon in Mach- Zehnder interferometer. Regards:Kevin .. > Sorry I've never read anywhere about imaginery wavelength. I would > highly appreciate if you can cite a textbook or a website where this [quoted text clipped - 3 lines] > I also dont know nothing about this no have ever seen in a textbook on > QM. Look up "evanescent waves" in wikipedia, or in most optics text books (I don't have one with me at the moment to cite). You encounter the same mathematics in the theory of waveguides, when the width of the waveguide is smaller than the wavelength of the wave. In all these cases, the wave number "k" becomes imaginary in the direction down the waveguide, (or away from the surface in the case of evanescent waves). The result of "k" being imaginary is that the exponential is decaying, whereas the real "k" results in an oscillitory wave. > > > 3. What about that sectioin where there is no tube. The particle > > > (photon) is not limited in X,Y - the wave diffracts there and many of [quoted text clipped - 15 lines] > Then if one puts a time 'filter' and gets one click on the detector > he would know the path better than h/p = lambda? It is not practical to set a time "filter" (more commonly called a time "gate") so tight that you could perfectly determine the path of the photon. In certain cases you can limit the range of the photon, but never enough to violate the uncertainty principle. A very interesting and accessible paper on some real experiments on this subject are at http://arxiv.org/abs/quant-ph/9501016v1. > > For something that is much more difficult to > > make sense of, and fundamental to Quantum Mechanics, study the EPR [quoted text clipped - 8 lines] > > - Show quoted text - Rich L. > .. > [quoted text clipped - 58 lines] > > - - Thank you very much for directions. One question: Did Heisenberg knew that? I dont think I understand the movement of photons or other particles no I think anywhere is a clear explanation. Just let see a whole new example- very simple - a slot and a screen. Before the slot one could not tell where the particle could go (the momentum is undefined in parallel to the slot directions). Heisenberg principle. But if I register a particle on the screen - I know it originated from the slot. So all the possible 'paths' would lie in a cone with a top in the registration point and base the slot. Very near to the region of absorption (registration) the uncertainty would tend to zero thus breaking Heisenberg principle. Or else the particle must makes jumps with infinite velocity. So I think the Heisenberg principle applies only before registartion of a particle. The particles must have (if are points) paths and we know where a particle was going through space when we register it. The photon do not have parts which travel different paths (as it seems to be in Mach-Zehnder). Regards:Kevin > Thank you very much for directions. > One question: Did Heisenberg knew that? [quoted text clipped - 24 lines] > > - Show quoted text - Kevin, The Heisenberg principle does not say anything about what is known after the photon is absorbed. Somehow all the energy in the photon ends up in a rather confined space (typically an electron orbiting an atom with a diameter on the order of 10^-10 meter). Instead it refers to the ability to MEASURE AT THE SAME TIME two complementary properties of a particle. In the case of the photon you can determine that it passed through the slit (thus is localized to a delta_x equal to the width of the slit), but now if you try to measure its momenum transverse to the slit, you will measure a large delta_p. That is, after going through a very narrow slit you have no idea what direction the photon is going. Even if before going through the slit you prepared the photon beam to be highly collimated. Conversely if you measure the transverse momentum (get a small delta_p) then you necessarily get a large delta_x. In optics it is well known that to get a beam that will propagate a long distance without spreading out that you must have a large diameter beam. This is why when they do laser ranging to the moon they direct the laser into a telescope to expand the beam to a meter or two in diameter. If they didn't do this, the beam would expand to many times the size of the moon before it got there. Again, by preparing a beam with small delta_p you get a large delta_x. If you take that large beam (small delta_p) and send it through a narrow slit, the beam that passes through has a small delta_x (the width of the slit) but now it has a large delta_p (it spreads out rapidly in the direction of the narrow dimension of the slit. You might get a better feel for this by studying Gaussian Beams. If you look this up on Wikipedia, notice the relationship between the waist diameter of the beam, w0, and the divergence of the beam THETA. This is what Heisenberg is talking about. Rich L. > You can do better than that though. > Suppose you throw a photon with sufficient energy [quoted text clipped - 7 lines] > > Jan There are certain rules to this type of thing E = p*c (a variation of work energy theorem) p = h/lambda (h^2/2me) (1/(2*a))^2 = h*c/lambda h = planck's constant = 6.58E-27 erg sec c = speed of light = 3E10 cm/sec me = electron mass = 9.11E-28 g a = Bohr radius = 5.3E-9 cm lambda = wave length cm then lambda = 9E-7 cm (ultraviolet) "hydrogen atom in a Rydberg state" would indicate a larger 'a' but this would have to very large (low energy) (perhaps existing in the quiet of outer space) to 'detect' radio-frequency waves. The two states have to match such that momentum is conserved. Richard Saam > > You can do better than that though. > > Suppose you throw a photon with sufficient energy [quoted text clipped - 32 lines] > > Richard Saam Maybe I make a simple error but I can't find explanation. I obviously receive contradiction to Heisenberg's principle. This is in an another form what I wanted to stress at the beginning of the thread and also formulated in the last post to Rich. I agree with what he has said in his answer but that's not about what I am concerned. So I try to put it in another equivant form. Let's have a flux (z-axis) of free photons towards a slit d. The wavelength is L=h/2.PI.pz =a/pz. pz is the impulse in z. Px is 0. After the slit there is diffraction and the Heis law says Dpx.d>a or there must!!! be photons of px>a/d. Now the greatest possible value for px is to get equal to pz (if the photon doesn't take energy/impulse from the walls of the slit - in which I dont believe) - conservation of impulse law. But we can surely make the slit as narrow as we like d->0 (or its enough d<L). Then there must be px>pz which is impossible (as I has shown). I hope I made the problem a little more clear and hope someone is in ability to show what is wrong in exactly this situation.(e.g the wavelength L bigger than the width opf the slit d). Kevin > From QM follows that the position of a photon can not be determined > better than L (de Broglie wavelength). >From relativity, it follows that the photon has no position at all, because there is no rest frame to which it might be referred. The photon is all momentum. It is not a ball, it does not get "exchanged" like spies passing pregnant newspapers around. It is, simply, a quantum of the electromagnetic field, and nothing more. Who is teaching our students relativity these days? -drl On 29 =EE=CF=C5=CD, 00:58, DRLunsford <antimatte...@yahoo.com> wrote: > On Oct 17, 12:41 pm, kvblake <kvblake2...@yahoo.com> wrote: > [quoted text clipped - 11 lines] > > -drl Do you mean that when a photon is created there emerges a EM field at once in the whole universe or the existing EM field changes at a constant value immidiately everywhere? I read about localized photons in Mandel's quantum optics. Regards: Kevin > Do you mean that when a photon is created there emerges a EM field at > once in the whole universe or the existing EM field changes at a > constant value immidiately everywhere? > I read about localized photons in Mandel's quantum optics. > > Regards: Kevin No, I mean exactly what relativity implies - things that go at C cannot be localized in any way. The electromagnetic field already exists everywhere in the universe, and can absorb and release energy- momentum anywhere and any-when. The photon as a thing, a localized object, does not exist, and can't possibly exist. In fact the entire idea of propagation is what is at issue here. In relativity, the world is 4-d, and propagation is a primitive fact, like distance in Euclidean geometry, and can't be reduced. So the EM field absorbs energy here, and releases it there, with the difference of here and there being an interval lying on the light cone. Photons are the units of absorption and release. The 4-d-ness of the world really has to be taken at a face value. The conflict between relativity and QM is stark enough without making it worse. Neither ceases to be true because of the other. -drl > > Do you mean that when a photon is created there emerges a EM field at > > once in the whole universe or the existing EM field changes at a [quoted text clipped - 21 lines] > > -drl QED works with "point particles" and the photon is a point particle like any other. Feynman is clear on this subject and he is correct. There are a number of ways of seeing this. (a) Experimentally, we are unable to distinguish between a photon with a small (uh, very very small) mass and a massless photon. Theoretically, you can assume that a photon has a small mass, make a calculation, and then let the mass go to zero. The limit will be well defined and you will get the right answer as if you had left the photon massless from the beginning. Therefore, our theory and experiment cannot tell you what the mass of the photon really is. If you are a gambling man, you can place your bet anywhere along the real line from - 1.0e-56 kg to +1.0e-56 kg and your number for the photon mass will be compatible with experiment, and therefore, ipso-facto, equally likely as the value "0" that it is assumed to be. An oft quoted fact about nature is that anything that is not forbidden is mandatory. A massive photon is not forbidden. And the set of photon masses where the photon is truly a massless particle, as compared against the experimentally allowed photon masses, is a set of measure zero, a mighty thin assumption to stand upon. (2) The definition of "point particle" in a QFT is quite clear and the photon is one of these. Some might confuse that discussion with the question of whether or not a photon can be "localized". You might as well ask if an electron can be localized. Both are on the same basis with respect to quantum mechanics as each other. Yes, it is true that it is easy to find good QM references that say the photon does not have a position-space wave function, but these references are out of date. Margaret Hawton solved the problem about a decade ago. For a recent article of hers, see http://arxiv.org/abs/0705.3196 and references. The above is not well known, despite being published in Phys Rev A. For example, see: http://arxiv.org/abs/quant-ph/0508202 CarlB <carl@brannenworks.com> writes >QED works with "point particles" and the photon is a point particle like any >other. Feynman is clear on this subject and he is correct. The problem I have with a 'point particle' is the word 'point'. I'm perfectly happy to accept it as a precise mathematical fiction with limited physical validity whose use results in superbly accurate predictions when compared to experiment. As an example the probability of finding a particle at 'a point' must be zero even if finding one 'in a small volume' may be very high. Once one introduces infinitesimal physical things, one should rather worry about nasty infinities rearing their heads where you least expect it. >(2) The definition of "point particle" in a QFT is quite clear and the >photon is one of these. Some might confuse that discussion with the question >of whether or not a photon can be "localized". You might as well ask if an >electron can be localized. Both are on the same basis with respect to >quantum mechanics as each other. This is absolutely so. We cannot precisely localise either an electron or a photon. One can do it only to the precision of the (invariably a quantised system) detector. Over many years, and many switches between all the common paradigms, I have come to the conclusion that everything is in fact completely wavelike. This has no infinities or zero things to confound it and seems to allows one to explain all the QM things I am able to properly comprehend, in at least a handwavy sort of way. It has the rather serious downside that there are no convenient mathematical methods for obtaining the precision offered by the pointlike methodology. However that doesn't make it wrong from a physical perspective. SignatureOz This post is worth absolutely nothing and is probably fallacious. Thus spake Oz <Oz@farmeroz.port995.com> >CarlB <carl@brannenworks.com> writes > [quoted text clipped - 9 lines] >physical things, one should rather worry about nasty infinities rearing >their heads where you least expect it. Imv this is the wrong definition of the word point. I think what qed is telling us is that a point particle does not have to be at a point in a spacetime manifold. Indeed, if as is empirically the case, the points of the manifold only refer to relative positions of matter with respect to reference matter, then we have no way to talk about infinitesimal volumes, only finite volumes as dictated by the resolution of measurement. >>(2) The definition of "point particle" in a QFT is quite clear and the >>photon is one of these. Some might confuse that discussion with the question [quoted text clipped - 13 lines] >methodology. However that doesn't make it wrong from a physical >perspective. Remember that the apparent wave-like structure only occurs as a means of calculating probabilities for the results of measurement with respect to reference matter. Regards SignatureCharles Francis moderator sci.physics.foundations. charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and braces) Oh No <NotI@charlesfrancis.wanadoo.co.uk> writes >Thus spake Oz <Oz@farmeroz.port995.com> >>CarlB <carl@brannenworks.com> writes [quoted text clipped - 18 lines] >volumes, only finite volumes as dictated by the resolution of >measurement. Yes, but that resolution of measurement carries an energy 'penalty'. IMHO somehow this needs to be an integral part of the measurement. We also note that there must be some finite minimum resolution which will be closely related to the size of a black hole of the measurement size==energy. >>This is absolutely so. We cannot precisely localise either an electron >>or a photon. One can do it only to the precision of the (invariably a [quoted text clipped - 11 lines] >calculating probabilities for the results of measurement with respect to >reference matter. I would beg to differ. IMHO the wave is the particle by any reasonable definition of "is". SignatureOz This post is worth absolutely nothing and is probably fallacious. > > Do you mean that when a photon is created there emerges a EM field at > > once in the whole universe or the existing EM field changes at a [quoted text clipped - 8 lines] > momentum anywhere and any-when. The photon as a thing, a localized > object, does not exist, and can't possibly exist. You refute the idea of Einstein about photons (explanation of photo effect) by his own special relativity theory. I've never heard that Einstein had claimed anywhere the photons dont exist because they are bound to move at c. I have read some textbooks about relativity though am not a specialist. I havent seen anywhere in some textbook of SRT to stand refutation of photons. > In fact the entire idea of propagation is what is at issue here. In > relativity, the world is 4-d, and propagation is a primitive fact, [quoted text clipped - 3 lines] > are the units of absorption and release. The 4-d-ness of the world > really has to be taken at a face value. I dont think the field just releases pure momentum any-where and any- then. If you deny propagation how would you explain 1. the light doesnt go though object (cancellation of propagation) 2. final speed > The conflict between relativity and QM is stark enough without making > it worse. Neither ceases to be true because of the other. > > -drl Action through distance is not better. Sincerely Kevin DRLunsford wrote {bd0e5aef-0731-4e1b-b005-7004acc80633@s8g2000prg.googlegroups.com} on Tue, 11 Dec 2007 06:29:26 +0000: > The photon as a thing, a localized object, does not exist, and can't > possibly exist. Feynman thought that the EM fields of classical electrodynamics are not physical objects but ficticious entities arising from a fundamental direct particle action. See his classical paper with Wheeler "Classical Electrodynamics in Terms of Direct Interparticle Action" [1]. In basis to that previous work, Feynmann developed his approach to QED [2]. There photons are the quanta of the EM field, but like in the classical limit, the photons are product from direct electron-electron action. Here photons on a collection of charges are collective phenomena somewhat like phonons in solids. Those views are generalized in [3] where classical and quantum electrodynamics are developed without the introduction of real fields or photons. The theory developed in [3] is more economical, consistent, and general than classical and quantum electrodynamics based in fields and photons. [1] Classical Electrodynamics in Terms of Direct Interparticle Action. Rev. Mod. Phys. 1949, 21, 425. Wheeler J. A; Feynman R. P. [2] Space-Time Approach to Quantum Electrodynamics. Phys. Rev. 1949, 76, 769. Feynman, R. P. [3] Cosmology and action-at-a-distance electrodynamics. Review of Modern Physics 1995, 67(1), 113. Hoyle F; Narlikar J. V. SignatureI follow http://canonicalscience.com/guidelines.txt On 11 =E4=C5=CB, 08:29, DRLunsford <antimatte...@yahoo.com> wrote: > On Dec 4, 3:08 pm, kvblake <kvblake2...@yahoo.com> wrote: > [quoted text clipped - 23 lines] > > -drl Some of my knowledge on photon comes from Feynmann's QED-strange theory of light. He explicitly states that the photon is a pointlike particle and we must be glad that Newton confinced himself that the light is made up of particles. Only the paths are unknown - e.g. how they move. Is he wrong? - (Feynmann) Are they particles only when being released or absorbpt - which can happen anytime but only near an atom excited properly? Ilian On Dec 14, 2:00 am, il...@abv.bg wrote: > Some of my knowledge on photon comes from Feynmann's QED-strange > theory of light. He explicitly states that the photon is a pointlike > particle .. I don't remember that, but something doesn't become true by dint of authority. The photon is not a point particle. QED is a good book, but you must understand that corners are cut in attempting to communicate the essence of the subject without assuming detailed knowledge of either math or physics. -drl Thus spake DRLunsford <antimatter33@yahoo.com> >On Dec 14, 2:00 am, il...@abv.bg wrote: > [quoted text clipped - 5 lines] >>Only the paths are unknown - e.g. how they move. >>Is he wrong? I don't think he is wrong, but Feynman's view, to which I adhere, is currently unfashionable. >I don't remember that, but something doesn't become true by dint of >authority. The photon is not a point particle. Nor does it become untrue by dint of authority. You should remember that there is no complete mathematical structure for qed, and there is no accepted interpretation of quantum theory, let alone for qed. To paraphrase Newton, if one wants to see far one should stand on tall shoulders. Feynman may be out of step with the "modern view", but in my view his shoulders are a lot taller than those who see a different picture, and the view he expresses is much more coherent. >QED is a good book, but you must understand that corners are cut in >attempting to communicate the essence of the subject without assuming >detailed knowledge of either math or physics. That is something else you have forgotten about the book. Feynman expresses an aim to correctly express the concepts without delving into the maths. He actually does a very good job of describing what the formulae say, but without using the formulae themselves. Regards SignatureCharles Francis moderator sci.physics.foundations. charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and braces) > I don't think he is wrong, but Feynman's view, to which I adhere, is > currently unfashionable. It is? > Nor does it become untrue by dint of authority. You should remember that > there is no complete mathematical structure for qed, and there is no Sure there is. That is how it produces measurable numbers. It's not legitimate, but it works. > accepted interpretation of quantum theory, let alone for qed. To > paraphrase Newton, if one wants to see far one should stand on tall > shoulders. Feynman may be out of step with the "modern view", but in my > view his shoulders are a lot taller than those who see a different > picture, and the view he expresses is much more coherent. The photon is not a point particle, no matter whose shoulders are being illuminated. > That is something else you have forgotten about the book. Feynman > expresses an aim to correctly express the concepts without delving into > the maths. He actually does a very good job of describing what the > formulae say, but without using the formulae themselves. It's a false aim then. You should try to get people to really understand something and leave it up to them if they want to do the math. -drl I've been reading this discussion with interest, and I'd like to add my two cents. Long ago I tried to think of the photon as a particle of some sort similar to a massive particle (e.g. an electron) but massless and moving at the speed of light. After trying for many years to make sense of such phenomena as diffraction, lasers, Bell's inequality and quantum mechanical experiments such as the Aspect experiments, and considerations of Relativity, I've ended up concluding trying to think of a photon as a particle is probably futile. All we really observe is interactions. It is clear that electromagnetic interactions are quantized as described by QM. Between these interactions we really don't know anything about the "photon", and the QM experiments suggest that we really can't know anything until the interactions are known, i.e. have "happened" already in our frame. Even then there are severe limits to what you can say about the "path" of a photon between where it is emitted and where it is absorbed. It seems to be to be a misguided attempt to apply what appears to be happening on our macroscopic scale to what is happening on the quantum mechanical scale. On the macroscopic scale we really don't see particles between interactions either. We see a baseball flying through the air because of light constantly being reflected off of it. It follows a well defined and localized trajectory because its mass, energy and the uncertainty principle give very small deviations from the mean of the quantum mechanically probable trajectories. To apply intuition learned at this scale to elementary particles is to miss what QM is really telling us about physics. QED does not make any assumptions about how a photon gets from point A to B. All possible paths are summed to get the net probability. Every path has an effect, although most are canceled by other complementary paths. I'm thinking that this picture is probably closer to reality than attempts to understand exactly what path the photon traveled. My own thinking is that the "photon" really is only an attempt to understand the discrete nature of light absorbtion and emission, but that the connection between the emission and absorbtion is not nearly as tight as we are inclined to think. I have great difficulty reconciling the experiments and the apparent one to one connection beteen an emission at one place and an absorbtion at another, and at the same time being consistent with QM. I suspect that even QED is using a misleading picture by even calculating these paths. I suspect that this represents a limitation of our imaginations to conceive how these interactions really work. All that we really observe is that energy and momentum are somehow transported from point A to B. Our constructions of the E and M fields, and QM wave functions are mathematical tricks that allow us to calculate some aspects of these transport events. QED explains electromagnetic "forces" as interations between charged particles and virtual photons associated with the fields. I wonder to what degree this is a mathematical model that simulates what is happening, but is quite different. To make an analogy with a previous paradyme, epicycles on epicycles were a reasonably good calculation model for elliptical orbits, especially if they were sun based epicycles instead of earth based. Kepler and Newtons elliptical orbits were simpler for a given level of precision, although still not perfect (you have to take into account interactions from other planets and, in the case of small satellites solar wind to get current levels of precision). Another analogy is the mechanical models of the aether that were attempted in the late 1800's. I think we are getting closer to the "truth" (although I hesitate to use that word here), but we don't see it yet. My thoughts here tie in with another current discussion on the SPR and SPF groups. Physics has gotten terribly mathematical and has lost much of it's original basis in physical explainations for the theory. This concerns me because theory and mathematics are human constructions and are capable of describing anything at all. Some have accused String Theory as being a theory of anything. As long as physics remains constrained by experimental fact tying physics to our real world we might be OK. The difficulty with coming up with physical bases for the current theories is that the physical bases are too strange to us. The QM world (and the relativistic world) are too foreign to us to provide useful intuition. Or to even provide an easy to grasp conceptual foundation for the theories. Hence many follow the "shut up and calculate" operating principle. I personally feel it is important that we continue to search for such a conceptual foundation, but when (if) we find it I expect it will be something that appears quite alien to us. If it is too alien, if it requires too much explaination and teaching for a student to "get it", this might reduce its usefulness in guiding further physics. Rich L. Rich L. <ralivingston@sbcglobal.net> writes >QED does not make any assumptions about how a photon gets from point A to B. >All possible paths are summed to get the net probability. Every path has an [quoted text clipped - 7 lines] >connection beteen an emission at one place and an absorbtion at another, and >at the same time being consistent with QM. I am inclined to agree. However there is the problem, in a bent spacetime, of the particle having a probability not just 'here' but 'here and then'. Even more generally the light cone can (and typically does) intersect with the limit of the observable universe from several observers. Such fun! SignatureOz This post is worth absolutely nothing and is probably fallacious. On Dec 15, 6:12 pm, CarlB <c...@brannenworks.com> wrote: > (a) Experimentally, we are unable to distinguish between a > photon with a small (uh, very very small) mass and a massless > photon. Theoretically, you can assume that a photon has a Sure you can. If the mass is anything not zero then it has an undefined helicity. > Therefore, our theory and experiment cannot tell you what > the mass of the photon really is. If you are a gambling man, [quoted text clipped - 8 lines] > experimentally allowed photon masses, is a set of > measure zero, a mighty thin assumption to stand upon. What you are talking about is not the photon of QED. It is a hypothetical massive particle with completely different properties (Proca particle). It does not matter how small the mass is, once it's non-zero it is no longer a photon. > Yes, it is true that it is easy to find good QM references > that say the photon does not have a position-space wave > function, but these references are out of date. .and therefore any idea of localization of the photon is moot. The references are not "out of date". I suppose one can argue anything, but eventually it's "pointless" :) -drl On Dec 17, 11:27 am, il...@abv.bg wrote: > You refute the idea of Einstein about photons (explanation of photo > effect) by his own special relativity theory. I've never heard that Not at all. All the photo effect shows is that EM energy is quantized. > Einstein had claimed anywhere the photons dont exist because they are > bound to move at c. > I have read some textbooks about relativity though am not a > specialist. I havent seen anywhere in some textbook of SRT to stand > refutation of photons. I know, it is too easy to conclude that a photon is a fast bullet and once this image is implanted it's hard to shake. It's a great source of misunderstanding and a hindrance to learning QED. > I dont think the field just releases pure momentum any-where and any- > then. > If you deny propagation how would you explain > 1. the light doesnt go though object (cancellation of propagation) > 2. final speed I'm not denying propagation, I'm saying it's built into the geometry of the world (the isotropic points of the geometry are the light cone). This really has to be grasped. There is no need for a medium of propagation because it is a primitive fact in the geometry. > Action through distance is not better. It's not action at a distance because the distance is zero for events on the same light cone. -drl |
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