DParlevliet Posted November 16, 2013 Author Share Posted November 16, 2013 This detector absorbs the photon, or in other words, it stops the wave at the hole. Since the wave can pass only at 1, the fringes aren't poduced. I suppose this is a partial explanation? For every absorbed photon a new photon must be emitted in the detector, otherwise none will arrive at the main detector. If the photon is absorbed, there is no wave anymore, also not in slit 1. When the photon goes through slit 1 is more intriguing, because something happens with the wave in detector. Link to comment Share on other sites More sharing options...
Enthalpy Posted November 17, 2013 Share Posted November 17, 2013 Absorbed in the detector at one slit (not reemitted) or passing through the other slit. Link to comment Share on other sites More sharing options...
DParlevliet Posted November 17, 2013 Author Share Posted November 17, 2013 But the standard set-up is detectors at both slits, which gives a signal in the main detector without interference pattern. Those detectors must detect+pass each photon, otherwise there would be no signal in the main detector. Link to comment Share on other sites More sharing options...
DParlevliet Posted November 17, 2013 Author Share Posted November 17, 2013 There is another item which puzzles me here. The explanation above depends on the operation of the slit-detector. If a detector would be possible which emits a coherent photon, then there should be an interference pattern. But in Wikipedia (Complementarity) it is explained as fundamental. By detecting a particle it is not possible anymore to have wave-properties, like interference.I suppose that is not right. When the slit-detector emits a new photon, it is a wave again, with unknown position of a particle, so can interfere. The other way around in the above set-up with one detector: if a photon is going through slit 1 it is not detected, so remains a wave, but still does not give an interference pattern. So my conclusion would be that the slit-detector does something to the wave, making it non-coherent Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted November 18, 2013 Share Posted November 18, 2013 I tried to produce an artists illustration of what I thought was going on accoding to the various quantum explanations . attached DSCF3299.JPG The photon is the bit shooting out at the speed of light after the electron has jumped from higher to lower orbit . Hope of some help. just a visualisation ( maths boffs dont like these, but I need them to remain sane ! ) Mike WOW! that is a nice work of art Link to comment Share on other sites More sharing options...
Toffo Posted November 19, 2013 Share Posted November 19, 2013 But the standard set-up is detectors at both slits, which gives a signal in the main detector without interference pattern. Those detectors must detect+pass each photon, otherwise there would be no signal in the main detector. Well, it sounds to me those detectors are some kind of advanced and modern detectors. So, where can I read about that standard set-up? A link to a simple description of the set-up would be nice. Link to comment Share on other sites More sharing options...
DParlevliet Posted November 19, 2013 Author Share Posted November 19, 2013 No, with standard setup I mean what is normally discussed: the double slit with a detector on each slit. How detectors work I don't know, that was my question. I thought that Enthalpy explained that a detector absorbs the photon and does not re-emit it. Link to comment Share on other sites More sharing options...
DParlevliet Posted November 19, 2013 Author Share Posted November 19, 2013 Another interesting case is X-ray diffraction (my job) where I finally found the formula I was looking for. X-ray diffraction measures the reflection of X-ray photons in a crystal, to examine the molecule structure. In a crystal molecules are positioned in regular layers which result in interference. The intensity of a reflection is the sum of the type and position of the atoms in the molecule. Therefore the intensity of reflection must be known accurate. X-ray-diffraction is based on Thomson scattering (X-ray Crystallographic Technology, Andre Guinier) The above Thomson formula describes the scatter intensity of a particle with mass m. The first conclusion is that the intensity is inverse dependent on the mass square. So scattering is done by electrons, minor by the heavy nucleus of the atom. The second conclusion is that the radiation is not spherical, but has a half cosine.And I am even more convinced that a photon in reflection is not absorbed with inherent wave collapse, but that it remains a wave which interacts which the charge of a electron. Link to comment Share on other sites More sharing options...
DParlevliet Posted November 21, 2013 Author Share Posted November 21, 2013 Now a snack in between. The left setup F gives an interference pattern. So the right one too, isn't it? But the path lengths are quite different. So if you measure the travel time between incoming detector and main detector, you know which path the particle went. Where is the error in this argumenting? Link to comment Share on other sites More sharing options...
DParlevliet Posted November 26, 2013 Author Share Posted November 26, 2013 No comment. So everyone agrees that you can have both interference and know the path of the photon?In Wikipedia "Complementarity" the double slit is used to show that complementarity means that you cannot have both wave properties (interference) and particle properties (which path). I was not sure about that. The uncertainty of position is something in the micro/planck world, not the macro world of slit distances. But more important: there is indeed no interference at the moment the particle is detected, but when is transmitted it is a wave again. The same above. Both paths are waves, with inherent uncertainty of particle position. Only afterwards the human intellect can conclude which path it followed (average, not accurate) and the photon has no property which knows what goes around in our head.. Link to comment Share on other sites More sharing options...
Mike Smith Cosmos Posted November 26, 2013 Share Posted November 26, 2013 (edited) Where did you get your drawing graphics, I love them, I need them, for my own ! Mike Edited November 26, 2013 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted November 27, 2013 Share Posted November 27, 2013 ya its right according to quantum physics that electron emits energy as photon when jumps from higher to lower orbit but can you suggest that how much energy or how much photon it can release at the time of orbit change??? and how much time it continue the jumping process??? I don't mean to jump in ask this, but I cannot help to.. Does this mean that the electron has both mass and " no mass"? In otherwords here.. Does the electron, " transform into a photon?? Also if you don't mind please, I am reading that the " shape" of the electron is " not fully understood by scientist?? Meaning that scientist don't "really" know the shape of the electron? What about electron microscopes? Link to comment Share on other sites More sharing options...
DParlevliet Posted November 27, 2013 Author Share Posted November 27, 2013 I don't mean to jump in ask this, but I cannot help to.. Please don't go off-topic. Make a new one. Where did you get your drawing graphics, I love them, I need them, for my own ! I made them myself. You can download them and use as you wish. Link to comment Share on other sites More sharing options...
Mike Smith Cosmos Posted November 27, 2013 Share Posted November 27, 2013 Please don't go off-topic. Make a new one. I made them myself. You can download them and use as you wish. Great ! Where do I find the download ? Mike Link to comment Share on other sites More sharing options...
DParlevliet Posted November 27, 2013 Author Share Posted November 27, 2013 http://parlevl0.home.xs4all.nl/forum or (in my SeaMonkey) double-click on the image and choose Save Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted November 28, 2013 Share Posted November 28, 2013 (edited) Please don't go off-topic. Make a new one. I made them myself. You can download them and use as you wish. Shape of the wave of a single photon Is the title of the OP However, I think I have read enough already, "the shape is not understood, and is not known" "" period--->." Edited November 28, 2013 by Iwonderaboutthings Link to comment Share on other sites More sharing options...
DParlevliet Posted November 28, 2013 Author Share Posted November 28, 2013 (edited) Shape of the wave of a single photon Is the title of the OP However, I think I have read enough already, "the shape is not understood, and is not known" "" period--->." Yes, of the photon. Not of the electron or how electrons generate photons. The wave is perhaps not understood, but a lot of properties has shown here. At the end of the discussion I will make a summary. Edited November 28, 2013 by DParlevliet Link to comment Share on other sites More sharing options...
Toffo Posted November 28, 2013 Share Posted November 28, 2013 Now a snack in between. The left setup F gives an interference pattern. So the right one too, isn't it? But the path lengths are quite different. So if you measure the travel time between incoming detector and main detector, you know which path the particle went. Where is the error in this argumenting? Again there seems to be some advanced detector: The incoming detector. The incoming detector may detect a quite exact time, or quite uncertain time. I quess we have here such detector that it detects a quite exact time. So then, that makes the photon short. I mean the photon that the detecrtor emits. Or the photon that the detector observes passing by, I don't know what this detector exactly does. If incoming detector shortens long photons, it causes a loss of interference. If incoming detector elongates short photons, it causes an interference to pop up when otherwise there might not have been interference. Link to comment Share on other sites More sharing options...
DParlevliet Posted November 28, 2013 Author Share Posted November 28, 2013 I don't know what you mean with long or short photon. If you mean the wave length, that does not matter in this set-up. The detector can be regarded as a new photon source. A triggered photon source (if possible) could also be used. All input photons of the 50% mirror have the same wave length, short or long. Link to comment Share on other sites More sharing options...
Toffo Posted November 29, 2013 Share Posted November 29, 2013 I don't know what you mean with long or short photon. If you mean the wave length, that does not matter in this set-up. The detector can be regarded as a new photon source. A triggered photon source (if possible) could also be used. All input photons of the 50% mirror have the same wave length, short or long. I think this has been covered already. Photon sorce launches photons, remembers the launch times --> photon waves will be short ones. if we want to produce long photon waves, a photon container with a small hole is good for this purpose: no triggering --> no definite launch time --> long photons, I mean photon waves. Link to comment Share on other sites More sharing options...
DParlevliet Posted November 29, 2013 Author Share Posted November 29, 2013 I think this has been covered already. Photon sorce launches photons, remembers the launch times --> photon waves will be short ones. Not in this topic. When you look back here you will see that experiments (and Feynman) show that a single photon has a spherical radiating wave with intensity 1/r2 until infinity, independent how it is generated. Link to comment Share on other sites More sharing options...
Toffo Posted November 29, 2013 Share Posted November 29, 2013 With this kind of photon container we can produce photon wave bars of any desired length: With this device any photon wave bars produced by the previous device can be converted to any desired length: With this photon container we can do interference experiment with photon waves of any length: Link to comment Share on other sites More sharing options...
Sensei Posted November 30, 2013 Share Posted November 30, 2013 Not in this topic. When you look back here you will see that experiments (and Feynman) show that a single photon has a spherical radiating wave with intensity 1/r2 until infinity, independent how it is generated. ??? Do you know inverse square law? Intensity is quantity of photons per area. Very large quantity. The further from source emitting photons (or other particles), particles are spread on larger area. If you have 100 W light bulb, and it's emitting light at 525 nm wavelength, each photon has E=h*c/525nm =3.786e-19 J 100 W = 100 J/s (let's forget about loses) 100 J/3.786e-19 J=2.641*10^20 photons in the all directions, per second. 1 m from light bulb quantity of photons is 2.641*10^20 / 4*PI*1^2 = 2.1*10^19 per second per 1m^2 2 m from light bulb quantity of photons is 2.641*10^20 / 4*PI*2^2 = 5.254*10^18 3 m from light bulb quantity of photons is 2.641*10^20 / 4*PI*3^2 = 2.33*10^18 1000 m from light bulb quantity of photons is 2.641*10^20 / 4*PI*1000^2 = 2.1*10^13 We can go further and there will be distance with 1 photon per 1 m^2 per second And further there will be gaps without any photons in unit of area. Quantization of energy is essential to not have absurds/paradox like infinite energy. 1 Link to comment Share on other sites More sharing options...
DParlevliet Posted December 3, 2013 Author Share Posted December 3, 2013 Brian Greene in The fabric of the Cosmos is also on the track of complementary and refers to a more complicated setup in a paper of Kim et al. A simplified graphic from wikipedia: A laser beam goes through a double slit to a BBO crystal which causes a photon conversed in two photons of half energy, one going up, one going down. Thereafter a Glan-Thomson prism (for reason I don't know). The photons going up are focussed by a lens on the main detector D0. The photons going down passes several mirrors with detectors D1-D4. Only those photons in D0 are counted which coincidence with D1-D4. If you look to photons which coincidence with D3, which are sure blue photons, there is no interference pattern. If you look to photons which coincidence with D1, which can be both red or blue, there is an interference pattern.Especially strange is that the distance from BBO to D1-D4 is 2,5 m longer then to D0, which would indicate that during detection in D0 it must knows that 8 ns later its sister-photon will be detected by D1 (interference) of D3 (no interference). So a kind of looking in the future.But I don't understand why there is interference at all. Photons are emitted in red or blue area on the BBO. If a red photon is emitted there is no blue photon, so with what should it interfere? Because the lens is focused in infinity and not on the BBO, the read and blue area of the BBO will be out of focus on D0. All angles are fixed, so where is the interference?The article of Kim et all has also an explanation in QM-math, which I am not familiar with. Link to comment Share on other sites More sharing options...
Sensei Posted December 4, 2013 Share Posted December 4, 2013 Thereafter a Glan-Thomson prism (for reason I don't know). http://en.wikipedia.org/wiki/Glan%E2%80%93Thompson_prism "The prism can therefore be used as a polarizing beam splitter." Link to comment Share on other sites More sharing options...
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