What would be the benefits to adding time of flight measurements to a double slit experiment? I was looking at this diagram and noticed the light source has an off-axis angle, meaning the path length to travel through either slit is different:
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If an accurate time of flight measurement was available for the light traveling, ie. maybe 100ps or faster photodiode detector array for the screen, and a fast pulsed "single photon" emitter for the light source, this could be used to help infer which slit the light is traveling through instead of directly measuring and interfering with the light as it travels. Of course it is probably common sense that the light is a wave and travels through both slits, but still ideas like the Copenhagen interpretation may think of the photons as particles that really do travel through one slit only, so in that case it should be useful to measure time of flight.
cheers, Jamie
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Any given photon will pass through one slit or the other, and the TOF
will correspond to which it selects. You'll see two distinct clusters
of flight times.
Ditto a beam splitter interferometer. A photon hits a half-silvered
mirror and half of the photons pass through, the other half are
reflected. If the photons finally arrive at a detector by
different-length paths, there will be two distinct transit times.
But in both cases, you can still observe interferance, as if each
photon actually took both paths.
Aside: I don't know of any way to emit just one photon per trigger, or
to make a uniformly spaced stream of photons, but I vaguely recall
seeing something like that.
John
Ok so if a photon truly travels through only one of the two slits, then it should be possible to do an experiment where you only see one cluster of flight times. For example, if there is a 50% probability of traveling through one slit or the other, in an experiment with a sequence of 10 photons there will be a 1 / (2^10) chance that all 10 photons will travel through the same slit, so if the same experiment is run ~1000 times, statistically one of the experiments results would have only a single measured flight time result. However if the light is traveling through both slits as a wave, I don't know if an experiment would ever have only a single measured flight time as a result.
cheers, Jamie
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A given photon travels through only one slit, or takes one of two possible paths through an interferometer. Another photon might take the other slit or path. If the path lengths are different, a histogram of flight times will have two distinct peaks.
If you use two photodetectors, one behind each slit or one in each arm of the interferometer, you'll see each photon hit one detector or the other, just like you'd expect particles to behave. The geometry of the setup determines the relative probability of the two paths. If a mirror is 90% silvered, the ratio of the photons will be 9:1.
For example, if there is a 50% probability of
Sure. That will be a rare event but it will happen once in a while.
As I recently said, folks who think of electromagnetic problems as photons bouncing around always get the wrong answer.
You have to apply Maxwell's equations, and then do the photon thing at the detector. Otherwise you wind up with nonsense.
Cheers
Phil Hobbs
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Thats what happened with the double slit experiment, the two incorrect assumptions of being able to fire particles of light or electrons, and fire them one at a time, leads to the paradox of wave/particle duality.
If you assume that you are "firing" waves, with an analog intensity, I think it makes more sense..
cheers, Jamie
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On a sunny day (Sat, 28 May 2011 01:10:39 -0400) it happened Phil Hobbs wrote in :
Exactly!
To make it a bit more down to earth, 'photon' is the amount of enery released when an electron is knocked out of orbit in the DETECTOR. Since our detectors consist of atoms with electrons bound to those, that is about the sensitivity or grain of the detector(1).
(1 inline why not) That is also the origin of Planck's constant, and Planck himself had a very different opinion on that, see wikipedia article on him.
It says absolutely NOTHING about what happens in-between. Waves. Of course waves can be waves in something much finer than the atoms and electrons.
The word 'ether' is forbidden, but they call it 'space filled with virtual particles that pop in and out of existence' these days. To give you some idea what I am talking about consider this:
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There is also my postings 'Explaining the photo electric effect from the wave perspective' long time ago in sci.physics.
In the 2-detector case, if I applied Maxwells equations, in both the double-slit and the beam-splitter cases, photons would be split in half, and both detectors would see a fraction of each photon. Which doesn't happen.
Photons behave like waves until you detect them, in which case they behave like particles.
Probably not, but it sounded a bit like you were agreeing with Jamie in the paragraph above.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net
Q switched laser and a very dense optical filter so that the probability of just one photon getting through is less than unity.
Either you detect something at the screen inside the timegated window or you don't. Crucially you don't want more than one photon in the apparatus if you are going to play this game. It doesn't matter if there are none - you just have to wait even longer to see the pattern.
No it doesn't. You are trying to put a far too classical billiard balls interpretation on something which is decidedly non-classical. You can say nothing at all about how the light got from its source to the point where you detected it on the screen beyond the probability wavefunction and that is determined by the shape of the aperture in the dividing screen by the Fourier transform relationship.
If you could construct an apparatus that provides sufficiently asymmetric paths and stay inside the coherence lengths then I expect what would happen is that if you were to time gate the measurement so that only the photons going through the shorter path could get detected you would observe a diffraction pattern consistent with exactly that.
I strongly suspect without having done the detailed mathematics that the diffraction pattern occurs only once there has been enough time elapsed for the wavefunction to have propogated along both paths to reach the screen. This is the same problem that however you try to determine which slit a given photon went through you necessarily disturb the experiment and extinguish the two slit interference pattern.
It is horribly misleading to think of the photon as having passed through either slit in a diffraction experiment. You never observed it doing that and if you had tried to do so you would of necessity break the coherence that leads to an interference pattern.
The OP's question is however a good one. It might now be possible with a suitably fast Q switched laser to construct a time gated device where the behaviour of interference patterns or fringes could be tested in a single photon interference system where there are two unequal paths available to the light beams but where time gating at the detector does not permit sufficient time for the longer path to be fully explored and still reach the detector. My money would be on the fringes vanishing in the same way that closing off one slit would do.
Sure, but once in a while you'll get a pileup, and get two or three. To keep the pileup probability down, you's have to allow maybe 0.1 or less photons through per shot.
OK, let's fire single photons at some measured start times. Run them through a beam-splitter, then through two different path lengths, so the delays from the source to the detector are T1 and T2 measured in nanoseconds to the photodetector. Measure the times of flight and histogram. Assume monochromatic photons, so coherence length is large. If "You can say nothing at all about how the light got from its source to the point where you detected it" what does the time histogram look like?
If you have two photodetectors, one behind each slit, that's exactly what you'll see. A photon goes through one slit or it goes through the other.
I believe that's been done. As soon as you do anything to determine (or constrain) that the photons are in one leg or the other, the interferance vanishes.
A photon hitting a beam splitter will take both paths, only being localized upon measurement. Google "quantum non demolition measurement" and "delayed choice experiment"
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Dirk
http://www.neopax.com/technomage/ - My new book - Magick and Technology
Suppose a photon entered an array of beam splitters and mirrors such that it had, say, 20 possible paths through the maze, and each of the
20 paths had a unique path length. All paths end at a single photodetector. Assume we can launch a photon into the maze and measure the time until it hits the detector.
Any measured time will correspond to one of the 20 possible paths, in agreement with the path length. So if you know the arrival time, you know the exact path it took.
I think of it this way and it seems to work out for me:
A photon is NOT like a ball, it is instead a wave, a wave that is quantized in amplitude, and 1 photon is simply the smallest quanta of the wave. Since it is a wave, you CAN get an interference pattern with just one photon.
Think of a signal generator with only a step attenuator to control the output level.
On a sunny day (Sat, 28 May 2011 15:10:58 -0700 (PDT)) it happened George Herold wrote in :
My view (tm) is that it is just a detector thing. The 'light' generated by your bulb is just a wave. A wave of some specific strength. The chance of it knocking lose a photon in one or the other detector (slit) is close to 50 % (so of being detected) if the setup is right. So if you make the light intensity low enough sometimes one detector will see something behind one slit, and the other one sometimes something behind the other slit. What should be verified is that sometimes BOTH detectors (I mean both slits) see something, The argument is then that more than one photon was 'emitted'. That is not correct. All you are doing is having 2 wave detectors with threshold of Planck's constant x frequency. In WHATEVER FORM those detectors come, need not be PMTs, it is all a DETECTION issue. No mystery. This is because we detect with matter, our detectors are made of atoms and electrons bound to those. I can however think of a better detector :-)
Yes, but until you it actually hits the detector it takes all paths, with a different probability for each. The measurement collapses the probability wave to a single outcome.
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Dirk
http://www.neopax.com/technomage/ - My new book - Magick and Technology
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