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Natural Science Forum / Physics / Research / July 2008single photon through glass
[Moderator's note: Lines reformatted to make them shorter. Although 80 is a maximum to enable nice display on almost all terminals, it's better to keep your own, original, unquoted text in posts at 72 characters per line or less, to enable it to be quoted a couple of times and still keep things at less than 80 characters per line total. Also, don't use more than 2 characters (including space) for a quote symbol. -P.H.] I'm trying to get a description of how an individual photon goes through glass. It seems to require the full power of field theory. The setup would be like the two slit experiment where a single photon at at time is fired from an emitter and then recorded after passing through a lens. The assumption is that the glass could still be transparent to an individual photon. This is like in the two slit experiment where the phonomenon still occurs when the light is sent one photon at a time. If the passage is a sequence of absorptions and reemissions, then the reemissions ought to be in random directions and the glass is not transparent. It seems that the absorptions and reemissions must be virtual. All possible routes, including the absorptions and reemissions, get merged into a single field. The field then gives a probability of the combination source-emission and recorder-absorption events. This from Wikipedia: "In classical optics light travels over all allowed paths, and their interference results in Fermat's principle. Similarly, in QED, light (or any other particle like an electron or a proton) passes over every possible path allowed by apertures or lenses." The inclusion of "lenses" is what I'm interested in. In what way does a lens enable or remove possible paths? Is the photon at the detector considered the same photon as emitted (in the same sense as light traveling in a vacuum)? Does light travel through glass mostly because of a lack of absorptions? Classically, the glass molecules are supposed to "resonate" to the visible light. What does this mean with an individual photon? Note: That light travels slower in glass is supposed to be derived, so I want to avoid that easy shortcut. Also, the shortcut that this is in the realm of classical EM shouldn't really be used because the context is a single photon. The questions are directed toward a cleaner understanding (in QED terms) of why light travels slower through glass. Thanks in advance as always, Ned Phipps > I'm trying to get a description of how an individual photon goes through > glass. > Is the photon at the detector considered the same photon as emitted (in > the same sense as light traveling in a vacuum)? Same photon? Different but having the same energy? Perfect copy? I would argue that's a metaphysics question, not a physics question, and thus cannot be answered by physics. -- Fran >> I'm trying to get a description of how an individual photon goes through >> glass. [quoted text clipped - 6 lines] > I would argue that's a metaphysics question, not a physics question, and > thus cannot be answered by physics. Not at all metaphysical. This is my current feel about how QFT works. There are two QFT variations: 1) It can be a sequence of absorptions and reemissions in series 2) It can be a merging of all possible paths - the basis of QFT. The second variation is the vacuum variation. Consider it as one photon if it is a merging of all possible virtual absorptions and reemissions. Virtual and merged, or real and in sequence. The calculations are different. One automatic answer has been that of course it is absorbed and reemitted. Light travels at c *always*. To travel slower it must be absorbed and reemitted. Some time must take place while absorbed. But, if it is reemitted, why is the reemission not in a random direction? The "virtual and merged" variation maintains the direction effect. Does this mean that the merging includes the absorptions and reemissions and delays? --Ned > -- > Fran > I'm trying to get a description of how an individual photon goes through > glass. It seems to require the full power of field theory. The setup . > This from Wikipedia: "In classical optics light travels over all allowed > paths, and their interference results in Fermat's principle. Similarly, [quoted text clipped - 3 lines] > The inclusion of "lenses" is what I'm interested in. In what way does a > lens enable or remove possible paths? Hi, Read the small pocket-book "QED: The strange theory of light and matter" by Feynman for an in-depth intuitive description of lenses, mirrors, re-emissions and absorptions. There he derives the probabilities and in particular, goes in-depth about what happens in the sum-over-histories model for light passing through glass. Without resorting to "shortcuts" as you describe it below, through EM or classical optics. It's not always about removing or enabling paths, but also in modifying paths or pathlengths (a path has a complex amplitude in the sum which enables interference effects and which is affected by the virtual absorptions and re-emissions). A good illustration of this is not the glass-lens but the Fresnel zone- plate or an honest-to-god grating. You get a higher intensity of transmittance at a point, by occluding certain paths in the sum which would otherwise interfere destructively! http://en.wikipedia.org/wiki/Zone_plate Never mind the description about "diffraction" etc, this is also a top- level semi-classical result which is understood in terms of the sum-over- histories. As you observe below, one of the major problems with teaching QM and QED seems to be the abundance of levels of physics involved in describing the same phenomena, for example the mix-up between fundamental QED descriptions and classical optics and Maxwells EM, and all their related keywords. > Note: That light travels slower in glass is supposed to be derived, so I > want to avoid that easy shortcut. Also, the shortcut that this is in [quoted text clipped - 4 lines] > Thanks in advance as always, > Ned Phipps Regards, Bjorn >> I'm trying to get a description of how an individual photon goes through >> glass. It seems to require the full power of field theory. The setup [quoted text clipped - 37 lines] > descriptions and classical optics and Maxwells EM, and all their related > keywords. Thanks. Forgot about that Feynman book. I have it and just re-read it. The photon takes all paths through the glass. Some of the paths include one or more scattering. Each scattering results in a 90 degree phase shift. When the paths are merged with the phase shifts included, the effect is equivalent to a slower velocity. It was the inclusion of the scatterings in the merging process that is confusing. I thought that event type things were supposed to make the field collapse and the process start over. The only lingering questions have to do with the phase shift. Why is it 90 degrees? (or what determines the delay of the re-emission)? Perhaps it is naturally nearly 90 degrees, but there is a mechanism that pushes it to act like it is exactly 90 degrees? Feynman mentioned that materials that scatter with <90 degrees phase shift are opaque. No mention of >90 degrees. My guess is that the phase shift is actually variable and the path merging makes it equivalent to a single 90 degrees value. --Ned >> Note: That light travels slower in glass is supposed to be derived, so I >> want to avoid that easy shortcut. Also, the shortcut that this is in [quoted text clipped - 7 lines] > Regards, > Bjorn > Thanks. Forgot about that Feynman book. I have it and just re-read it. > [quoted text clipped - 9 lines] > I thought that event type things were supposed to make the field collapse > and the process start over. So the photon interacting with the glass atoms does not constitute a measurement that would localise the photon?  SignatureDirk http://www.transcendence.me.uk/ - Transcendence UK Remote Viewing classes in London >> Thanks. Forgot about that Feynman book. I have it and just re-read it. >> [quoted text clipped - 13 lines] > So the photon interacting with the glass atoms does not constitute a > measurement that would localise the photon? That's a core point for looking at this. I'm still not sure. Feynman' setup uses a monochromatic source. His other examples were almost always single photons. Not the case for the glass reflection/transmission example. There was a reflection analysis (grating) that required the originating time at the source for the photon to vary. This implies that multiple source photons went into the calculation for a single destination photon. The transmission analysis is getting more confusing each time I review it. I'm wanting to build the correct model of QED equations. I feel I'm almost ready to get started. It feels like a photon scattering in glass is: 1) an absorption by a glass molecule 2) a glass molecule in an excited state for a time 3) a re-emission at the same energy The glass molecule holds the excited state for a quarter cycle. I'm wondering what determines the duration of an excited state. It seems it must be associated with the ability to complete another interaction. The interaction is virtual in the sense that it gets merged with all variations. The only measured interaction is the initial emission to final absoption. I would say no. The photon is never localised within the glass. --Ned > http://www.transcendence.me.uk/ - Transcendence UK > Remote Viewing classes in London > That's a core point for looking at this. I'm still not sure. Feynman' > setup uses a monochromatic source. His other examples were almost [quoted text clipped - 5 lines] > source photons went into the calculation for a single destination > photon. Actually, I posted a related question about this here a couple of weeks ago but there were no takers :) In case people don't know the passage I'm mentioning here, the story is that the photon propagator according to the first 100 pages of the Feynmans short QED book is described as having a phase component that varies with the pathlength. All kinds of experiments involving interference and sum-over- histories are described and explained from this principle. "The little arrow turns as the particle moves". Then at around p 100, he in passing mentions that, oh by the way, the propagator does not modify the phase. It is really the light-source (emission) which has a *time-dependent* phase-rotation, thus combined with non-phase- modifying propagation it gives the *effect* of a path-length-dependent photon propagation phase. Is this correct btw ? I have never calculated with the real QED propagators, is the photon propagator phase-conserving ? Without knowing what went on, I assume he did it this way to avoid having to describe a space-time diagram directly at page 1 in the book (which would be necessary to describe interference-effects arising not only from the spatial configuration but also from the time-dependence of the emissions etc). So yes, according to that, multiple virtual emissions at different times have to be combined and interfered for the correct classical result to emerge.. /Bjorn >> That's a core point for looking at this. I'm still not sure. Feynman' >> setup uses a monochromatic source. His other examples were almost [quoted text clipped - 33 lines] > propagators, > is the photon propagator phase-conserving ? He used a coherent source to help with the reflection variation. The reflection off deeper layers has longer path lengths so I think that multiple source photons might be part of the reflection variation. Don't really know. The transmission variation looks at all paths straight through. These all have the same length, but there is one absoption and re-emission on each path (except one). (should be complex cascade of variations). I'm not sure if this is supposed to be described as a time delay or a phase shift. A time delay allows for multiple cycles. Is there a core field theory concept here for only considering the phase? A time delay immediately gives us slower travel. But the phase shift seems to be the usual technique. As a time delay, the 90 degree time equivalent would vary 2X from red to blue. --Ned > Without knowing what went on, I assume he did it this way to avoid > having to [quoted text clipped - 10 lines] > > /Bjorn > He used a coherent source to help with the reflection variation. > The reflection off deeper layers has longer path lengths so I think that > multiple source photons might be part of the reflection variation. > Don't really know. Yes I looked at his real lectures and got a better understanding on this topic. I posted the link to the site which has video-recordings of these lectures at the bottom of this post, very interesting. Its the lectures that were the foundation of the book. The photon propagates without adding a phase-variation to the amplitude, there are no "blue" or "red" photons, thats yet another high-level phenomena which is the result of the oscillation of the emission amplitude... it depends on what you call a photon. A line in a feynman-diagram, or a physical click in a photomultiplier (the latter has a wavelength of course). So yes, you have to really consider superpositions of a multitude of photons even when doing the simple examples with reflection. I am betting on that Feynman had to do this to avoid getting way too complicated too early in these lectures. But it confused us who were trying to dig deeper into it :) > The transmission variation looks at all paths straight through. These > all have the same length, but there is one absoption and re-emission > on each path (except one). (should be complex cascade of variations). > I'm not sure if this is supposed to be described as a time delay > or a phase shift. A time delay allows for multiple cycles. Is there a core > field theory concept here for only considering the phase? I dug deeper into this as well and fetched a paper, "Tutorial on fundamentals of radiation physics: interactions of photons with matter" by R.H. Pratt. Many kinds of scattering can happen between a photon and an atom, but the dominant scattering at low photon energies (wavelength >> atom size) is Rayleigh-scattering which is what we've talked about - an atom absorbs and re-emits a photon without changing its state. Its called coherent scattering as well. When the photon energy increases, the photon starts to scatter incoherently with wavelength/momentum change (Raman-scattering) and scatters off free electrons (Compton-scattering). It can also be completely absorbed by the atom (excitation), ionize an electron or be involved in pair- production at even higher energy. Above this, you get second order effects etc. Overall it seemed as if there weren't that many experiments made to actually compare atom-scattering theory with reality, I guess the calculations for getting the scattering amplitude in glass from first principles for a macroscopic property like refraction is a more difficult problem than you might imagine :) > A time delay immediately gives us slower travel. But the phase shift > seems to be the usual technique. As a time delay, the 90 degree time > equivalent would vary 2X from red to blue. There are all kinds of "non-linear" optics and materials which "slow down light" in the labs.. All the effects have to be due to various superpositions of the scattering effects mentioned above for all angles, energies and atomic states.. Here is the link to the recorded lectures: http://www.vega.org.uk/video/subseries/8 regards, Bjorn W >>> Thanks. Forgot about that Feynman book. I have it and just re-read it. >>> [quoted text clipped - 41 lines] > > I would say no. The photon is never localised within the glass. That would appear to be obvious, but *why*? And how many atoms might one expect the photon to interact with (whatever that means)? Think fibre optic. SignatureDirk http://www.transcendence.me.uk/ - Transcendence UK Remote Viewing classes in London >>>> Thanks. Forgot about that Feynman book. I have it and just re-read it. >>>> [quoted text clipped - 41 lines] > And how many atoms might one expect the photon to interact with (whatever > that means)? Think fibre optic. I would say the default is that an event is not localized. We can localize the initial emission and the final detection only because the devices have locations. Everything inbetween... Feynman uses the number 0.2 as the amplitude for an interaction. 0.2^2 as the probability. I think that Feynman left off all the cascading variations to keep it simple. I think every path gets two variations with every molecule it encounters - a scattering and a no scattering - with each scattering variation actually a multitude of variations. One variation for each scattering direction. To keep it "simple", can we assume that all the side variations cancel out and we can calculate from only the straight forward and straight backward variations? They usually are assumed to cancel, but yet again there is another "why?". --Ned >>> So the photon interacting with the glass atoms does not constitute a >>> measurement that would localise the photon? [quoted text clipped - 3 lines] > > That would appear to be obvious, but *why*? Because you do not know WHICH glass atom/molecule the photon interacted with. Or didn't interact with. So you need to sum amplitudes over them all, and that sum of amplitudes over all possible paths is what makes for the peculiar properties of QM. > And how many atoms might one expect the photon to interact with > (whatever that means)? Think fibre optic. All of them! Ajiko tried to worry about "how long" each glass atom/molecule "held" the photon. This is not the best way to think of this. It is the interference among ALL of the atoms/molecules that causes the delay, because the path lengths for them are all different and the interference is constructive only for a speed of advance of c/n (n is related to the density of charges and their interactions with the light). Tom Roberts > >>> So the photon interacting with the glass atoms does not constitute a > >>> measurement that would localise the photon? [quoted text clipped - 8 lines] > all, and that sum of amplitudes over all possible paths is what makes > for the peculiar properties of QM. Hi, I think what was bothering the OP was the nagging feeling that it would be reasonable if some kind of information of the photons interaction with so many atoms should "leak". If it leaks, the superposition is lost and the photon would be "localized" in the lingo above even if you as an experimenter cannot figure out which atom really took the recoil or whatever happened. OTOH, QM is not reasonable per default ;) /Bjorn >>>> So the photon interacting with the glass atoms does not constitute a >>>> measurement that would localise the photon? [quoted text clipped - 20 lines] > is constructive only for a speed of advance of c/n (n is related to the > density of charges and their interactions with the light). So why are not fibre optics completely transparent? What does end up localising the photon? SignatureDirk http://www.transcendence.me.uk/ - Transcendence UK Remote Viewing classes in London > So why are not fibre optics completely transparent? > What does end up localising the photon? The photons will be localised within the field modes of the fibre. Indeed, mostly a quantum optician would use those modes as a basis upon which to quantise the field, so that photons are explicitly constucted to inhabit those modes. I assume all the other discussion in this thread refers to some other notion of localisation? The various posts seem rather vague on what they mean by "localisation". Signature---------------------------------+--------------------------------- Dr. Paul Kinsler Blackett Laboratory (PHOT) (ph) +44-20-759-47734 (fax) 47714 Imperial College London, Dr.Paul.Kinsler@physics.org SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/ On Jul 9, 8:15 am, Dirk Bruere at NeoPax <dirk.bru...@gmail.com> wrote: > > Ajiko tried to worry about "how long" each glass atom/molecule "held" > > the photon. This is not the best way to think of this. It is the [quoted text clipped - 5 lines] > So why are not fibre optics completely transparent? > What does end up localising the photon? Anyone of the inelastic scattering modes (including old-fashioned complete absorption) *can* dissipate the information in the material (to a very large degree in a normal material, I assume), and these occur eventually for some photons obviously. Unless you use a photon wavelength which has practically zero coupling to any energy-levels in your material, but then you wouldn't be able to bend the photons with the fiber in the first place :) Incidentally, this is what you exploit and try to control when you're building ion-trap quantum-computing qubits. Its an ion contained in a magnetic field which is so isolated it can be kept in a superposition after it is hit by a photon or photon sequence. There are other examples of photon- controlled qubits which are engineered molecules with the right set of energy- levels which can live long enough in a superposition without dissipating the information ("decohere"). /Bjorn >>>> So the photon interacting with the glass atoms does not constitute a >>>> measurement that would localise the photon? [quoted text clipped - 22 lines] > > Tom Roberts The best description I've found so far describing light through glass is from Feynman's "Strange Theory of Light and Matter". There he described the scattering as a 90 degree phase shift. This seems to be a simplification (for the book) and I am still looking for the phase description of a single scatter. I am guessing that the phase is random. All variations merged. Resulting in an effective 90 degree phase shift. A single scatter is shown in a diagram with an electron absorbing a photon, then interacting a number of times with the nucleus, then emitting a new photon. This as one variation to be merged with all others. How does that 90 degree phase shift really happen? --Ned Phipps > I'm trying to get a description of how an individual photon goes through > glass. It seems to require the full power of field theory. The setup [quoted text clipped - 3 lines] > individual photon. This is like in the two slit experiment where the > phonomenon still occurs when the light is sent one photon at a time. There really is no need to worry about individual photons. The whole treatment is independent of the number of photons that pass through, Jan >> I'm trying to get a description of how an individual photon goes through >> glass. It seems to require the full power of field theory. The setup [quoted text clipped - 9 lines] > > Jan Wouldn't that make the single photon treatment the cleanest? It is mainly an exercise in field theory. I'm wanting to be able to describe light through glass according to QED and get an analysis that at the end says the time between emission and detection is <X> giving the apparent speed of light in the medium which we can describe as 1/n. I'm hoping to get a clear understanding of what is the property of glass that allows this to happen. Various ideas: 1) Glass reemits the photon very quickly, whereas other substances hang on to it for a while. 2) Glass reemits the photon with an absolutely random delay and the phase shifts cancel when all paths get merged. (Why wouldn't everything be transparent?) Feynman seems to indicate the reemission in glass always has a 90 degree phase shift from the absorption. Why? --Ned |
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