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Post by robinpike on Nov 30, 2011 15:58:15 GMT 1
I'm still a bit confused, however. If, by interacting with the environment, a particle in a number of probabilistic locations at the same time 'collapses' to just one specific position how come that when the barrier is met (not the slits) the photon does not collapse but carries on as a wave through the slits to reach the plate beyond the double slits? Surely a collision with the solid part of the double slit screen will constitute a 'measurement.' What have I misunderstood? That is why QM as explanation is weak... perform the experiment with a barrier that absorbs light (such as carbon black), and if light were a wave, or a particle with uncertaintly as to its position, or a particle that can be in several places at the same time, etc, then far less of the light would be expected to get throught the slit? |
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Post by striker16 on Nov 30, 2011 16:49:05 GMT 1
I'm still a bit confused, however. If, by interacting with the environment, a particle in a number of probabilistic locations at the same time 'collapses' to just one specific position how come that when the barrier is met (not the slits) the photon does not collapse but carries on as a wave through the slits to reach the plate beyond the double slits? Surely a collision with the solid part of the double slit screen will constitute a 'measurement.' What have I misunderstood? That is why QM as explanation is weak... perform the experiment with a barrier that absorbs light (such as carbon black), and if light were a wave, or a particle with uncertaintly as to its position, or a particle that can be in several places at the same time, etc, then far less of the light would be expected to get throught the slit? I seem to remember Speaker saying that a photon isn't really a particle, nor is it a wave; it depends on how we measure it! Not very helpful. I suppose we are trying to define a photon as something we can relate to in the 'real' world, i.e. a particle or wave each time we make a measurement. Truth is it appears it is neither. |
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Post by eamonnshute on Nov 30, 2011 16:57:33 GMT 1
If photons are sent through one at a time then some of them will hit the screen with the slits instead of passing through. These will not reach the detector, so these will not produce a measurement.
That will not work. The only way to detect a photon is to absorb it, so you cannot say which slit the photon went through on its way to the detector.
You could do the analogous experiment with electrons as you can detect them without stopping them, but you would deflect them in the process, which screws up the experiment. So there is no way to determine which of the two slits the particle went through, and the experiment shows that it actually goes through both.
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Post by eamonnshute on Nov 30, 2011 17:09:30 GMT 1
[ I suppose we are trying to define a photon as something we can relate to in the 'real' world, i.e. a particle or wave each time we make a measurement. Truth is it appears it is neither. That is what Feynman says in the video earlier in the thread (yesterday at 5:52pm). You have to stop trying to think of quantum phenomena in terms of familiar macro objects, because they simply aren't like that! |
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Post by striker16 on Nov 30, 2011 17:25:42 GMT 1
[ I suppose we are trying to define a photon as something we can relate to in the 'real' world, i.e. a particle or wave each time we make a measurement. Truth is it appears it is neither. That is what Feynman says in the video earlier in the thread (yesterday at 5:52pm). You have to stop trying to think of quantum phenomena in terms of familiar macro objects, because they simply aren't like that! What kind of explanation do most physicists think might explain this strange state of affairs. Or don't they think anything? |
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Post by eamonnshute on Nov 30, 2011 17:35:29 GMT 1
What kind of explanation do most physicists think might explain this strange state of affairs. Or don't they think anything? Feynman said that nobody understands QM. I think this is the sort of thing he was talking about! |
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Post by robinpike on Nov 30, 2011 18:18:24 GMT 1
That will not work. The only way to detect a photon is to absorb it, so you cannot say which slit the photon went through on its way to the detector. You could do the analogous experiment with electrons as you can detect them without stopping them, but you would deflect them in the process, which screws up the experiment. So there is no way to determine which of the two slits the particle went through, and the experiment shows that it actually goes through both. True, but what about considering a different experiment, such as the partial reflection of light from the front surface of a block of glass? It is found that the amount of light reflected varies between 0% and 16%, depending on the thickness of the glass. Having measured what the percentage of reflected light is for a particular block of glass, if a millionth of an inch is then shaved off the back surface of the glass, the percentage of light reflected from the front surface of the glass is different. If individual photons are used and counted, how does QM explain the results of this experiment? |
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Post by eamonnshute on Nov 30, 2011 18:53:20 GMT 1
True, but what about considering a different experiment, such as the partial reflection of light from the front surface of a block of glass? It is found that the amount of light reflected varies between 0% and 16%, depending on the thickness of the glass. Having measured what the percentage of reflected light is for a particular block of glass, if a millionth of an inch is then shaved off the back surface of the glass, the percentage of light reflected from the front surface of the glass is different. If individual photons are used and counted, how does QM explain the results of this experiment? It is pretty much the same as for the double slit experiment. In both cases you calculate the wave amplitude at each point, and that gives the probability that a photon will arrive there. |
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Post by robinpike on Dec 1, 2011 9:58:14 GMT 1
It is pretty much the same as for the double slit experiment. In both cases you calculate the wave amplitude at each point, and that gives the probability that a photon will arrive there. I appreciate that the calculation works, but how does QM explain the dependence of the reflection of a single photon on the thickness of the glass? |
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Post by eamonnshute on Dec 1, 2011 10:41:21 GMT 1
I appreciate that the calculation works, but how does QM explain the dependence of the reflection of a single photon on the thickness of the glass? Back to Feynman again! This "collapse of the wave function" is still a mystery. en.wikipedia.org/wiki/Wave_function_collapse |
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Post by robinpike on Dec 1, 2011 11:04:36 GMT 1
The thing about this particular experiment and its result, i.e. that the amount of reflected light is dependent on the thickness of the glass, is that something besides QM needs to be involved.
Unlike the two slit experiment, where it could be argued that a single photon is able to pass thru 'both slits at the same time and interfere with its own path', a single photon cannot 'wait' at the front surface of the glass while also passing thru the glass, get reflected by the back surface and meet up with itself at the front surface of the glass and 'interfere with its own path'.
And if something else is involved in this particular experiment, then perhaps it is also involved in the two slit experiment as well. In which case, is QM real at all?
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