Double slit experiment - QM or wave?

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Cthugha

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Not true. This is what QM predicts, not wave theory. In wave theory, there are just as many photons hitting the dark fringes as there is hitting the bright fringes, only they're not setting off the detector because in these exact positions, they are interering destructively. Therefore, the detection count is significanlty lower compared to firing rate in wave theory.
No, this is exactly the wrong point here. I do not know of any theory, which would predict such strange behaviour, especially not wave theory. You just neglect constructive interference. Consider the case of an ideal detector with 100% quantum efficiency, which means that any photon will be detected. While your understanding of destructive interference would cause photons to go through undetected. Howwever at the places of constructive interference nothing would happen. Any photon would be detected anyway due to the 100% QE, so there are just two conclusions left:

a) The constructive interference will lead to an increase in photon detections, which cancels the decrease of the destructive interference area. Therefore no decrease in the total number of photon counts will occur.

b) There are no effects of constructive interference at all. The number of photons detected at the areas of wannabe constructive interference equals the number of photons detected with no double slit present. This is either a violation of conservation of energy or (if you think that the wave will just travel through the screen and be detected somewhere else - however this might happen) easily disproved by putting a second detector screen after the first one, which should then show bright spots, where there have been areas of destructive interference at the first screen. You can easily build your own double slit at home and show, that the second option is plain wrong.
 
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Why is the beam splitter not identical to the double slit? After all, it provides the superposition of path the SAME way as the 2-slit experiment.
From a QM point of view there's no difference I'm sure, there's only two outcomes - detection or no detection. But if you consider a wave, the process of beamsplitting is an imperfect one, there's not just a right or left choice. With low-intensity waves resulting intensities may be considerably affected by properties of the beamsplitter itself, microscopic irregularities or other discrepancies. And measuring the splitted waves would not be the same as measuring the actual superposition.

You still haven't mentioned how you are going to reconcile with the FACT that you get single DOTS on a screen when low-intensity light sources are used.
You get dots because there is a limitation to how low-intensity light we actually can detect. What you would consider a photon, is what I would define as the lowest intensity light we're able to detect. I believe there is a non-uniform intensity distribution covering the detectors constantly, but simply too low-intensity to detect. So a detection by the CCD would in this scenario represent a higher-intensity-anomaly in the intensity distribution, and the source is filtered to such an extent that these events are so rare as to be considered "individual photons".

I believe both Cthugha and muppet had clearly described this already. I don't quite understand what more you want.
There must be some data on this. Obviously we're not gonna convince each other of either interpretation, so there's no point debating. Surely someone has compared counts with and without the double slit? There's probably people here that can do it at work.

Edit: BTW, I've pointed out Marcella's paper several time already. It might be useful to review the QM derivation of the single, double, and multiple slit formalism without invoking ANY classical wave picture, and yet, one gets exactly the expected results that we know and love. So this is a reference one might want to keep:
I don't see relevance. The probability distribution is not what sets the interpretations apart.
 
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No, this is exactly the wrong point here. I do not know of any theory, which would predict such strange behaviour, especially not wave theory. You just neglect constructive interference.
How do I neglect constructive interference?

Consider the case of an ideal detector with 100% quantum efficiency, which means that any photon will be detected. While your understanding of destructive interference would cause photons to go through undetected. Howwever at the places of constructive interference nothing would happen.
define "nothing". And I'm not saying the photons will "go through" the detector. I'm not sure what happens with them, maybe they're reflected and scattered?

a) The constructive interference will lead to an increase in photon detections, which cancels the decrease of the destructive interference area. Therefore no decrease in the total number of photon counts will occur.
This is the QM interpretation, yes. Any change will only redistribute the photons, so the total count remains the same.

b) There are no effects of constructive interference at all. The number of photons detected at the areas of wannabe constructive interference equals the number of photons detected with no double slit present. This is either a violation of conservation of energy or (if you think that the wave will just travel through the screen and be detected somewhere else - however this might happen) easily disproved by putting a second detector screen after the first one, which should then show bright spots, where there have been areas of destructive interference at the first screen. You can easily build your own double slit at home and show, that the second option is plain wrong.
I don't believe they will go through the detector, no. But you could test it by splitting the detector like I illustrated in my initial setup C:
http://physics.soderholms.com/doubleslit_setup.jpg [Broken]
 
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You still haven't mentioned how you are going to reconcile with the FACT that you get single DOTS on a screen when low-intensity light sources are used.

Zz.

I would like to know more about this. could it be that the energy isnt absorbed directly by the atoms but rather by the field produced by the oscillating charges within the atom which are in turn produced by the absorption of light energy? if the process is unstable then one atom could end up with all the energy thereby starving the surrounding atoms.
 
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I would like to know more about this. could it be that the energy isnt absorbed directly by the atoms but rather by the field produced by the oscillating charges within the atom which are in turn produced by the absorption of light energy? if the process is unstable then one atom could end up with all the energy thereby starving the surrounding atoms.
After all, the emission of light is accomplished by the surrounding field. why not the absorption of light?
 

Cthugha

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How do I neglect constructive interference?
As I said before: imagine an ideal detector behind a double slit. According to your theory, there won't be any redistributions of photons from areas of destructive interference towards areas of constructive interference. So in the areas of (pseudo) constructive interference there will be at most the number of counts, which would be there without the double slit, but no additional counts as the detector is already ideal and there is no redistribution. So in your theory constructive interference is just not there and does not have any effect.

define "nothing". And I'm not saying the photons will "go through" the detector. I'm not sure what happens with them, maybe they're reflected and scattered?
What are you saying then? Reflected by what? Which wave theory describes such strange behaviour? Not even common classical wave theory does. What should happen to these photons?

There must be some data on this. Obviously we're not gonna convince each other of either interpretation, so there's no point debating. Surely someone has compared counts with and without the double slit? There's probably people here that can do it at work.
What is wrong with the Thorn paper?
However, you can try your idea at home. Building a double slit should not be much of a problem and some low cost photodiodes or a cheap CCD should not be too expensive as well.
 

ZapperZ

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From a QM point of view there's no difference I'm sure, there's only two outcomes - detection or no detection. But if you consider a wave, the process of beamsplitting is an imperfect one, there's not just a right or left choice. With low-intensity waves resulting intensities may be considerably affected by properties of the beamsplitter itself, microscopic irregularities or other discrepancies. And measuring the splitted waves would not be the same as measuring the actual superposition.
If you are so worried about that, then you need to worry about your "light detector" as well! What instrument do you think you will need to "count" these photons to verify your claim? What principle do these photodetectors use?

You get dots because there is a limitation to how low-intensity light we actually can detect. What you would consider a photon, is what I would define as the lowest intensity light we're able to detect. I believe there is a non-uniform intensity distribution covering the detectors constantly, but simply too low-intensity to detect. So a detection by the CCD would in this scenario represent a higher-intensity-anomaly in the intensity distribution, and the source is filtered to such an extent that these events are so rare as to be considered "individual photons".
That would make no sense that these non-uniformity would somehow conspire to produce JUST coincidentally a pattern that eventually matches the double slit interference pattern. I don't buy that, and this is true, then I could use the SAME argument against your setup, which makes any kind of detection of any kind suspect. I would bring out the CCD data from my photoemission experiment as proof that your assertion is wrong, and that each of these "dots" DO correspond to the location where such a photon.

Zz.
 
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If you are so worried about that, then you need to worry about your "light detector" as well! What instrument do you think you will need to "count" these photons to verify your claim? What principle do these photodetectors use?
I do worry. Both source and detector are imperfect, which is why I'd prefer no other manipulation between them.

That would make no sense that these non-uniformity would somehow conspire to produce JUST coincidentally a pattern that eventually matches the double slit interference pattern. I don't buy that, and this is true, then I could use the SAME argument against your setup, which makes any kind of detection of any kind suspect. I would bring out the CCD data from my photoemission experiment as proof that your assertion is wrong, and that each of these "dots" DO correspond to the location where such a photon.
You asked me how a singe dot could be produced, not a single dot in an interference pattern. Ofcourse, if you add a double slit the principle would be the same, only with interference.
 

ZapperZ

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You asked me how a singe dot could be produced, not a single dot in an interference pattern. Ofcourse, if you add a double slit the principle would be the same, only with interference.
Er.. no, I asked about a single dot on the detector produced by sending light of extremely low intensity through the double slit. So the double slit IS there.

BTW, based on Marcella's paper, where does it confirm your claim of what QM would produce using your scenario?

Zz.
 
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You still haven't mentioned how you are going to reconcile with the FACT that you get single DOTS on a screen when low-intensity light sources are used.
Ok, sorry. I should have assumed you meant through a double slit.
 
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BTW, based on Marcella's paper, where does it confirm your claim of what QM would produce using your scenario?
I'm guessing equations 24/25? Why do you ask?
 
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As I said before: imagine an ideal detector behind a double slit. According to your theory, there won't be any redistributions of photons from areas of destructive interference towards areas of constructive interference. So in the areas of (pseudo) constructive interference there will be at most the number of counts, which would be there without the double slit, but no additional counts as the detector is already ideal and there is no redistribution. So in your theory constructive interference is just not there and does not have any effect.
Ok, I see your point now. And you are right, for some reason I was assuming that the non-slit distribution would be the equivalent of a totally constructive area. So I guess my original statement about the predictions of wave theory were incorrect. Thank you Cthugha for pointing this out so clearly.

I guess the only way to seperate the theories now would be to test my setup C, and see if the number of detections in the constructive areas redistribute.


What are you saying then? Reflected by what? Which wave theory describes such strange behaviour? Not even common classical wave theory does. What should happen to these photons?
Like I said, I'm not sure. I don't know enough about light-material interactions from a wave point of view to make a qualified guess. But if we assume that there's no perfect destructive interference anywhere, I would guess the waves would interact with the detector in some way, just not enough to actually register a detection.

What is wrong with the Thorn paper?
This is my problem:
J.J. Thorn et al paper said:
Hence, if a single quantum of light is incident on the beamsplitter (BS), it should be detected at the transmission output or at the reflection output, but not both: there should be no coincident detections between the two outputs.
The Thorn paper is not about interference, it's about how a beamsplitter works. And the whole experiment is set up to refute the idea that wave theory predicts a perfect 50/50 split, no matter how low-intensity light you use. Which I find quite ridiculous, unless you have a perfect source, a perfect beamsplitter and a perfect detector.

But if you don't have perfect measuring devices, and you define the limit of what a detector can measure by 1 (i.e. a "photon"). And ..
1. We have a source that fires lightwaves with intensity 1 ± 0.5.
2. We have a beamsplitter that splits a wave 50/50 ± 25.
3. We have a detector that detects 1.25 ± 0.25.

In this scenario, there is NO WAY that both R and T detectors go off simultaneously. In fact, a prerequisite of simultaneous detection, is that the source fires lightwaves with at least double the intensity of what the detector can detect. But ofcourse, if both detectors were to go off simultaneously, experimenters would assume that their source sent off 2 photons, and that would invalidate the experimental condition of a single-photon source.

So in my opinion, the setup is designed to have a detection at either R or T, never at both. And it doesn't matter if you go with wave theory or QM.

However, you can try your idea at home. Building a double slit should not be much of a problem and some low cost photodiodes or a cheap CCD should not be too expensive as well.
Tried to google but I don't know what to look for. Any brands or models you can recommend that's used for physics? Most have other applications, like photography and video.
 

Cthugha

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This is my problem: The Thorn paper is not about interference, it's about how a beamsplitter works. And the whole experiment is set up to refute the idea that wave theory predicts a perfect 50/50 split, no matter how low-intensity light you use. Which I find quite ridiculous, unless you have a perfect source, a perfect beamsplitter and a perfect detector.

[...]

In this scenario, there is NO WAY that both R and T detectors go off simultaneously. In fact, a prerequisite of simultaneous detection, is that the source fires lightwaves with at least double the intensity of what the detector can detect. But ofcourse, if both detectors were to go off simultaneously, experimenters would assume that their source sent off 2 photons, and that would invalidate the experimental condition of a single-photon source.

So in my opinion, the setup is designed to have a detection at either R or T, never at both. And it doesn't matter if you go with wave theory or QM.
Well, it is at least a version of one of the standard experiments of quantum optics, but let me ask you a question on what you wrote also:

But if you don't have perfect measuring devices, and you define the limit of what a detector can measure by 1 (i.e. a "photon"). And ..
1. We have a source that fires lightwaves with intensity 1 ± 0.5.
2. We have a beamsplitter that splits a wave 50/50 ± 25.
3. We have a detector that detects 1.25 ± 0.25.
Why is there a limit on what a detector can measure? If this limit is defined only by the incident intensity, this is already the qm picture of quantized excitations of the em-field.

If you think, that the limit is just defined by some minimal transition energy of the detector, which needs to be overcome, you can easily test this by first using some single photon source in the IR range and afterwards some single photon source in the range of green or blue, where the energy of a single photon is roughly double the energy of an IR photon. If both setups show identical results, the limit is not defined by the minimum energy of the transition, but by the photon number.

Tried to google but I don't know what to look for. Any brands or models you can recommend that's used for physics? Most have other applications, like photography and video.
I usually use stuff from Hamamatsu or ID Quantique, but I suppose, this stuff is clearly too expensive for doing experiments at home. I am pretty sure Thorlabs has some simple photodiodes in the 10-50€ range, but I suppose you will get just the photodiode and will have to worry about how to measure the photocurrent yourself. Are there any simple electronics markets near where you live? Most of them should at least have some basic diodes available.
 
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Why is there a limit on what a detector can measure? If this limit is defined only by the incident intensity, this is already the qm picture of quantized excitations of the em-field.
There is a limitation on both intensity and wavelength.

If you think, that the limit is just defined by some minimal transition energy of the detector, which needs to be overcome, you can easily test this by first using some single photon source in the IR range and afterwards some single photon source in the range of green or blue, where the energy of a single photon is roughly double the energy of an IR photon. If both setups show identical results, the limit is not defined by the minimum energy of the transition, but by the photon number.
Since blue light is closer to the electron wavelength it would require less intensity than the IR to set off the detector. So if both setups show identical results, that tells me that the blue source emits lightwaves with lower intensity than the IR source.

I usually use stuff from Hamamatsu or ID Quantique, but I suppose, this stuff is clearly too expensive for doing experiments at home. I am pretty sure Thorlabs has some simple photodiodes in the 10-50€ range, but I suppose you will get just the photodiode and will have to worry about how to measure the photocurrent yourself. Are there any simple electronics markets near where you live? Most of them should at least have some basic diodes available.
Thanks!
 

DrChinese

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1. And measuring the splitted waves would not be the same as measuring the actual superposition.

You get dots because there is a limitation to how low-intensity light we actually can detect. What you would consider a photon, is what I would define as the lowest intensity light we're able to detect. I believe there is a non-uniform intensity distribution covering the detectors constantly, but simply too low-intensity to detect. So a detection by the CCD would in this scenario represent a higher-intensity-anomaly in the intensity distribution, and the source is filtered to such an extent that these events are so rare as to be considered "individual photons".

...

2. This is my problem: The Thorn paper is not about interference, it's about how a beamsplitter works. And the whole experiment is set up to refute the idea that wave theory predicts a perfect 50/50 split, no matter how low-intensity light you use. Which I find quite ridiculous, unless you have a perfect source, a perfect beamsplitter and a perfect detector.
1. What are you measuring with a split wave then? *Something* shows up at the detector.

Again, your point is DIRECTLY answered in the Thorn paper. If the above scenario were true - the exact point they considered - there WOULD be higher order effects (i.e. fluctuations above or below the threshold for detection). So when a pair of photons emerge from a PDC crystal: one might be too low intensity to detect, the other might not. Most importantly, sometimes there might be 2 on one side and only 1 on the other. Instead, they are always detected in (within experimental limits). From the paper:

(7) g(2)(0)>=1 [predicted by wave theory you propose]
(11) g(2)(0)=0 [quantum mechanical prediction]

This is a clear line in the sand, no possibility of overlapping results! The experimental result was g(2)(0)=0.0177 by the way. To repeat ZapperZ's answer: the reason this experiment is not performed using a double slit setup is because it is answered perfectly by this one.

Your objection about "imperfections/irregularities/impurities" in a beamsplitter is inaccurate as such "impurities" would be easily detected in the Thorn experiment.

The OP question is sorta like asking why linear accelerators are used to prove theories about particle physics rather than using a double slit setup. The answer is because the ones being used are the most effective - and convincing - way to test a specific scientific hypothesis.


2. This represents a misunderstanding on your part, IMHO. The wave picture does not predict a *perfect* 50-50 split when low intensity light is used, simply that there is a transmitted T portion and a reflected R portion of the wave. It might be 60% T and 40% R for all we know. Let's assume that happened. Well, it turns out that only the 60% T one is detected and never the 40% R one. But how does it happen that our detectors magically only see the 60% one and not the 40% one? Is it because the 40% one is too low for our detector but the 60% one isn't? That fails too! We could add another beamsplitter in series - to further reduce the 60% intensity (which might now be 36% and 24% or something) - and see immediately that there is NO such lower threshold. 1 and only 1 photon is detected.

The experimental conclusion is: there is no such thing as a low-intensity beam of light that is "too low" to be detected - unless it is of zero intensity. So it never can be just a little too low on one side, and just high enough on the other. This matches QM predicted view, but is at odds with the view you propose. In a double slit setup, constructive/destructive interference changes detection probabilities, but there is still just one photon to detect.

You said reject the beamsplitter experiment because you don't agree with the "interpretation" of how a wave traverses it; but it comes across as a person who rejects facts that do not agree with your interpretation. So I ask: how do you explain Thorn? That should serve as a litmus test of your interpretation.
 
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The OP question is sorta like asking why linear accelerators are used to prove theories about particle physics rather than using a double slit setup. The answer is because the ones being used are the most effective - and convincing - way to test a specific scientific hypothesis.
The OP question was not a general one about wave-particle duality. I simply stated that QM could not fully explain the double slit with light, nothing else.

2. This represents a misunderstanding on your part, IMHO. The wave picture does not predict a *perfect* 50-50 split when low intensity light is used, simply that there is a transmitted T portion and a reflected R portion of the wave. It might be 60% T and 40% R for all we know.
That's what I was saying.

Let's assume that happened. Well, it turns out that only the 60% T one is detected and never the 40% R one. But how does it happen that our detectors magically only see the 60% one and not the 40% one? Is it because the 40% one is too low for our detector but the 60% one isn't? That fails too! We could add another beamsplitter in series - to further reduce the 60% intensity (which might now be 36% and 24% or something) - and see immediately that there is NO such lower threshold. 1 and only 1 photon is detected.
Well THAT's interesting. Why don't you show me a paper with that setup. Add 4-5 levels of beamsplitters to the thorn setup and see if there's still a detection at some end. With only one beamsplitter, what does Thorn prove?

You said reject the beamsplitter experiment because you don't agree with the "interpretation" of how a wave traverses it; but it comes across as a person who rejects facts that do not agree with your interpretation. So I ask: how do you explain Thorn? That should serve as a litmus test of your interpretation.
You just explained yourself how the Thorn paper is interpreted with waves, so which facts am I rejecting? And then you say that you would get the same result with more beamsplitters. Am I gonna take your word on that?

I'm not a narrowminded person. If you show me that I'm wrong - like Cthugha did - I will recognise that. If there indeed is a Thorn setup with many levels of beamsplitters and still a detection at some end, I would have a very hard time explaining that.
 

DrChinese

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An illustration of the setup:
http://physics.soderholms.com/doubleslit_setup.jpg [Broken]


Any thoughts?
OK, let's examine your hypothesis per the diagram. I believe you are saying that you want to have a detection mechanism in which a portion can alternate between a position of destructive interference (we'll call that the D position) and a position of constructive interference (the C position). You propose that the detections will increase as a results of the C position versus the D position.

Here is the detail about that... QM predicts that this increase will occur too - at that particular spot at least! And where do those "new" photons come from? They end up as reductions in the other spots they would otherwise have arrived at. The total detections, per QM, will remain constant. You cannot move from D to C and get more total detections. (Conservation of energy and all that.)

So imagine, if you will, a single small detector which can move from a nearly perfect D position (0 detections) to a very C position near the center (so it gets a pretty high count %, maybe 10% of the total). Then, behind that, we position a large detector that picks up everything that doesn't hit the movable small detector. It should be obvious (per QM) that the total clicks will be constant regardless of whether the small detector is in the D or C position. On the other hand, you are predicting that the total count will be different by about 10%

This test should be pretty easy to do, you can use a small mirror in the D/C positions to bounce the light to one detector (out of the way, on a different plane perhaps), and have another detector behind it to catch everything else. So a double slit and 2 detectors, plus a suitable low intensity source.

Now here is the nice part: normally PDC photons cannot be used in a situation like this; but they can as long as there is no possibility of knowing which path they take through the double slit (some quantum erasers do this, for example). So it is possible to use a PDC source! This means you can get a very low intensity beam, and even coincidence count on both sides (comparing Alice's gate photon count to Bob's detected photon count). You would need the PDC crystal, plus another detector.

Why hasn't this been done before? Probably has many times, but like many experiments... it adds little to our bank of knowledge so it might not get published. On the other hand, maybe someone can find a reference that puts this to bed. (Other than the two ZapperZ already provided... :)

Assuming I was clear about the experimental setup, do you agree? Or do you still insist that the total photon count rate goes up?
 
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DrChinese

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Well THAT's interesting. Why don't you show me a paper with that setup. Add 4-5 levels of beamsplitters to the thorn setup and see if there's still a detection at some end. With only one beamsplitter, what does Thorn prove?
Well I guess some folks think the 377 standard deviations of the result put an end to the question. Keep in mind that they already split the beam in half and saw no significant change in the count rate. Most scientists would call that compelling.

I have chatted with one of the authors of the Thorn paper previously, so I will ask if he has ever done that version of the experiment. It would be a nice way to put the icing on the cake, considering that this can be done in an undergrad setting. In effect, they could set a lower limit on the intensity of a probability wave at which it still acts as a discrete photon.
 
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Here is the detail about that... QM predicts that this increase will occur too - at that particular spot at least! And where do those "new" photons come from? They end up as reductions in the other spots they would otherwise have arrived at. The total detections, per QM, will remain constant. You cannot move from D to C and get more total detections. (Conservation of energy and all that.)
My interpretation does not violate conservation of energy. There are no "new photons" as you call it. ALL light hits the detector, but if the light is interfering destructively at some point, it will not set off the detector. Instead it will continue it's propagation somewhere (not clear) and interact with something else.

Assuming I was clear about the experimental setup, do you agree? Or do you still insist that the total photon count rate goes up?
Well, there's nothing in your post that I didn't already know, so yes. I believe the detection rate will go up. I also believe that this test can be done in a much simpler way. All I need is a laser, a double slit, some cardboard and some standard light intensity meter. I place the light meter in the center constructive area at all time. I then change the destructive areas at the sides to constructive by moving them back. If QM is correct, I should see a decrease in intensity on my meter, as the photons are redistributed. It shouldn't matter if I shoot 1 photon at a time or a billion, the redistribution should still occur. Correct?
 
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Well I guess some folks think the 377 standard deviations of the result put an end to the question. Keep in mind that they already split the beam in half and saw no significant change in the count rate. Most scientists would call that compelling.
All this tells me is that the detection limit is close to a 50% split, so that there is a small chance of a "non-detection" (ie exact 50/50 split). So if you remove the beamsplitter you would only get an insignificant increase in detections.

I have chatted with one of the authors of the Thorn paper previously, so I will ask if he has ever done that version of the experiment. It would be a nice way to put the icing on the cake, considering that this can be done in an undergrad setting. In effect, they could set a lower limit on the intensity of a probability wave at which it still acts as a discrete photon.
Please do so, would appreciate that very much.
 

DrChinese

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Well, there's nothing in your post that I didn't already know, so yes. I believe the detection rate will go up. I also believe that this test can be done in a much simpler way. All I need is a laser, a double slit, some cardboard and some standard light intensity meter. I place the light meter in the center constructive area at all time. I then change the destructive areas at the sides to constructive by moving them back. If QM is correct, I should see a decrease in intensity on my meter, as the photons are redistributed. It shouldn't matter if I shoot 1 photon at a time or a billion, the redistribution should still occur. Correct?
Sure. But you need a second detector to demonstrate that the area that moves (from D to C) is now seeing hits. Not all of the hits at C will come at the expense of the center area. If you are not detecting the hits for all of the area, then you will need to estimate what the effect would be for just the area you are measuring. As you describe it, there may not be enough of a change to clearly indicate one way or the other. But in a rigorous test, yes, the redistribution will occur.

By the way, you are missing a better way to push your interpretation (at least on the surface): A beam of light incident on a mirror does not take a single path to a detector either. There is constructive and destructive interference from a variety of possible paths beside the "normal" one. If you etch the mirror very exactly in the areas where there is destructive interference, then the intensity of the light detected will actually increase. Where does that added light come from? You would argue that by eliminating areas of cancellation, more light is bound to come through. And of course that is mostly correct. (But there is no change in the total photon count from that, as some of the photons went elsewhere previously - i.e. were not detected as part of the original beam. So there is no conflict with QM.)
 

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