Double slit experiment - QM or wave?

[Hydr0matic]

I propose a setup to distinguish between the quantum interpretation and the wave interpretation of the double slit experiment.

Briefly, they are:

Quantum
The photon gun fires a photon that travels through either slit A or slit B, then hits a detector cell registering a spot in the detector pattern. All photons follow the same procedure, and with the help of quantum mechanics, the result is a nice interference pattern.

Wave
The photon gun fires a photon that travels through both slit A and slit B, then hits a detector cell with either destructive or constructive interference, depending on the location of the detector cell. If the superposition is "too destructive", no spot will register in the detector pattern. All photons interfere destructively or constructively and the result is a nice interference pattern.


So how do we distinguish between these interpretations? We do it by counting detections.

If the photon gun fires 1000 photons per minute, quantum mechanics predict that, during a minute, a 1000 detections will be registered (ideally). The wave theory on the other hand, predict that during a minute, only photons that are not interfering destructively will register at the detector.

So the setup is simple. First fire the photon gun without the double slit to determine the photon firing rate, then add the double slit and count the detections.

Since I adhere to the wave interpretation, I predict that there will be a significant difference between firing rate and detection rate. However, if this was the case, you could still argue that the non-detections "got lost" somewhere at the slits. But even if this was the case, you could still distinguish the theories from each other. Because, if the quantum interpretation is correct, the "lost photons" are lost, just random mishaps at the slits. On the other hand, in the wave interpretation all "lost photons" are destructive interference. This means that, if we alter the detector so that the destructive areas (dark bands) become constructive instead of destructive, there should be a significant increase in detections which cannot be explained by quantum mechanics.

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


Any thoughts?

 

[ZapperZ]

Hate to break this to you, but there's a more elegant experimental demonstration that has been done for undergraduate physics labs based on the Mach-Zehnder interferometer. See T.L. Dimitrova and A. Weis, Am. J. Phys. v.76, p.137 (2008).

One of the component of the experiment that was done by the students was this:

The demonstration, whose result is astonishing for students, is realized in the following way. First the fringe pattern is locked to a photodiode as explained in Sec. IV B, and the photomultiplier is moved to a fringe minimum, as characterized by a low photon count rate which can also be displayed acoustically. If now path A of beam 1 is blocked inside the interferometer, it is possible to hear (and see) a distinct increase of the click rate. This result demonstrates that if we give each photon the choice of taking either path A or path B, it has a low probability to appear at the detector. In contrast, if we force the photon to follow a specific path by blocking the other path, then the probability to arrive at the detector is much higher. The puzzling fact that a two-path alternative for each photon prevents it from reaching the detector, while blocking one of the paths leads to a revival of the clicks, is most intriguing for beginning students. This experiment is well suited for illustrating this remarkable quantum mechanical effect, which can be explained only if we assume that each photon simultaneously takes both paths A and B; that is, each photon, in the phrasing of Dirac, "interferes with itself."

You'll have a tough time convincing anyone that your "wave interpretation" is valid with such observation.

Zz.

 

[Hydr0matic]

You'll have a tough time convincing anyone that your "wave interpretation" is valid with such observation.

Oook? .. The quote you supplied completely proves my point. I don't get it.

The "revival of the clicks" cannot be interpreted as anything other than elimination of destructive interference.

I'm guessing this is more of a semantics issue now then. What would you say is the difference between a wave and a photon that "interferes with itself"?

 

[ZapperZ]

Oook? .. The quote you supplied completely proves my point. I don't get it.

The "revival of the clicks" cannot be interpreted as anything other than elimination of destructive interference.

I'm guessing this is more of a semantics issue now then. What would you say is the difference between a wave and a photon that "interferes with itself"?

The difference happened at the beam splitter. You need to read the entire article, and also the http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf" .

Zz.

 

[Hydr0matic]

The difference happened at the beam splitter. You need to read the entire article, and also the http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf" .

Zz.

I don't have access to the article via AJP. The Thorn paper seems to be more of a bell test setup with beam splitting and coincidence counts, i.e. no interference. So how is it relevant?

Concerning duality, different experiments show different things, and one does not take precedence over another (even if the wave interpretation is more easily observed). My original point was that the double slit is one of those experiments that cannot be explained satisfactory with a quanta of light. But you obviously believe otherwise, so I ask you. Why is the double slit not explained with a wave model of light? Which interpretation would Occam prefer?

 

[ZapperZ]

I don't have access to the article via AJP. The Thorn paper seems to be more of a bell test setup with beam splitting and coincidence counts, i.e. no interference. So how is it relevant?

The point being that "interference" by itself isn't sufficient, because both scenario results in the same observation (unless you want to consider higher order interference). What is important is the "which-way" observation that can't be resolved via the wave scenario.

Concerning duality, different experiments show different things, and one does not take precedence over another (even if the wave interpretation is more easily observed). My original point was that the double slit is one of those experiments that cannot be explained satisfactory with a quanta of light. But you obviously believe otherwise, so I ask you. Why is the double slit not explained with a wave model of light? Which interpretation would Occam prefer?

But it can't. If you look at ALL the observation, your application of Occam will require the minimal sent of description to describe ALL, not just one, not just two, observations. I haven't even yet invoked other phenomena such as the antibunching observation, multiphoton photoemission, etc. that current has zero description via wave scenario. Do you think your Occam principle will allow you to switch gears of describing light one way for one set of experiments, and switching to another to describe another set? I can stick to just one for all of them.

Zz.

 

[Cthugha]

If the photon gun fires 1000 photons per minute, quantum mechanics predict that, during a minute, a 1000 detections will be registered (ideally). The wave theory on the other hand, predict that during a minute, only photons that are not interfering destructively will register at the detector.

So the setup is simple. First fire the photon gun without the double slit to determine the photon firing rate, then add the double slit and count the detections.

Since I adhere to the wave interpretation, I predict that there will be a significant difference between firing rate and detection rate.

The whole idea is completely flawed. Interference is a phenomenon arising due to superposition of possible paths/probability amplitudes. There is no such thing as photons just interfering destructively and vanishing completely. For every zone of destructive interference, there is also a zone of constructive interference, so the total photon number does not change anyway. Otherwise this would be a harsh violation of energy conservation.

 

[Hydr0matic]

The point being that "interference" by itself isn't sufficient, because both scenario results in the same observation (unless you want to consider higher order interference). What is important is the "which-way" observation that can't be resolved via the wave scenario.

Which scenarios? and what about the "which-way" observation can't be resolved?

Both my setup, the quote you supplied, and Cthugha's post tells you why interference is enough, and both interpretations do not result in equal observations. QM predicts no change in clicks / number of detections when blocking one path or changing the resulting interference pattern. Wave theory does.

But it can't. If you look at ALL the observation, your application of Occam will require the minimal sent of description to describe ALL, not just one, not just two, observations. I haven't even yet invoked other phenomena such as the antibunching observation, multiphoton photoemission, etc. that current has zero description via wave scenario. Do you think your Occam principle will allow you to switch gears of describing light one way for one set of experiments, and switching to another to describe another set? I can stick to just one for all of them.

I'm not sure you can. How do you fit longer wavelength radiation in your description? radio transmission, microwave ovens... and refraction, diffraction? .. Even if you could, would you consider your description to be the final say in the matter? I can certainly appreciate the benefits and practicality of having a single description for all phenomena, but that doesn't mean I'm ruling out all other possibilities.

So please, I'm interested... Just looking at the double slit experiment, what about it makes you consider it to be better explained with quantum mechanics than with a wave model? Because I've always wondered why physicists use it as an example of how special QM is, when the actual explanation involves quanta of light "interfering with itself". What does this mean, and how does it differ from a wave model?

The whole idea is completely flawed. Interference is a phenomenon arising due to superposition of possible paths/probability amplitudes. There is no such thing as photons just interfering destructively and vanishing completely. For every zone of destructive interference, there is also a zone of constructive interference, so the total photon number does not change anyway. Otherwise this would be a harsh violation of energy conservation.

Read the quote in ZapperZ first post. Your claim is not true locally in an experiment like this, the photon detection does change. At the position where there is destructive interference, the light doesn't interact with the matter (eg. the detector), so it continues it's propagation and eventually interferes differently and interacts with something else. Energy is conserved, but not confined to the experiment.

 

[RandallB]

QM predicts no change in clicks / number of detections when blocking one path or changing the resulting interference pattern. Wave theory does.

You have grossly miss read something somewhere. Half the clicks with no interferance pattern just a dispersion patttern just like wave theroy does is excactly what QM says.

Just google for double slit basics or reread whatever your looking at.

 

[ZapperZ]

Which scenarios? and what about the "which-way" observation can't be resolved?

In a beam splitter with wave picture, half of the way goes through while the other half gets reflected. In a photon picture, a photon has a probability of 50% to go through and 50% to get reflected. This means that if you put your detector over both branch of the interferometer, you'll only get one detection, not both, in the photon picture. This is what the J.J. Thorn et al. expt. detected.

Both my setup, the quote you supplied, and Cthugha's post tells you why interference is enough, and both interpretations do not result in equal observations. QM predicts no change in clicks / number of detections when blocking one path or changing the resulting interference pattern. Wave theory does.

I don't see how Cthugha's post contradicts mine. I described a generic interference experiment and that you can described it with both pictures.

I'm not sure you can. How do you fit longer wavelength radiation in your description? radio transmission, microwave ovens... and refraction, diffraction? .. Even if you could, would you consider your description to be the final say in the matter? I can certainly appreciate the benefits and practicality of having a single description for all phenomena, but that doesn't mean I'm ruling out all other possibilities.

Actually, yes you can since YOU were the one who wanted to invoke Occam. Refraction and diffraction requires interaction with the material, and in condensed matter physics, optical transport are often described using the photon picture. See for example, the microscopic explanation on the interaction of photons with the phonon structure of a material that can result in the explanation for why things are opaque, transparent, and why light moves "slower" in a dispersive material.

So please, I'm interested... Just looking at the double slit experiment, what about it makes you consider it to be better explained with quantum mechanics than with a wave model? Because I've always wondered why physicists use it as an example of how special QM is, when the actual explanation involves quanta of light "interfering with itself". What does this mean, and how does it differ from a wave model?

Shoot one photon at a time, i.e. light with extremely low intensity. Look at it over time. Now do the same with electrons, protons, etc. Tell me if what you observe is different. If yes, then light is different than the "physical particles" such as electron, protons, etc. and you may have a leg to stand on. If they look the same and have the same characteristics, then you have a lot of explaining to do.

Zz.

 

[Cthugha]

Read the quote in ZapperZ first post. Your claim is not true locally in an experiment like this, the photon detection does change. At the position where there is destructive interference, the light doesn't interact with the matter (eg. the detector), so it continues it's propagation and eventually interferes differently and interacts with something else. Energy is conserved, but not confined to the experiment.

Of course the photon detection rate is changed locally. This is what interference is about. So the total count rate of the detector will only change, if the detector is small compared to the total dimensions of your interference pattern. But if you use a small detector and place it inside a spot of destructive interference, qm also tells you, there will be a significant difference between the counting rate with and without double slit in this spot. So I do not see any difference.

The only possibility to have no change in the counting rates would be to use a deterministic photon gun, which fires single photons in completely defined directions at completely defined moments. This would not produce an interference pattern. However this case corresponds to the case of completely incoherent light, where the distance of the slits is far longer than the coherence length of your wave. So any wave model will not predict any changes in the counting rates either.

 

[muppet]

Cthugha beat me to it. I'm afraid it looks like your conceptual understanding of the two-slit experiment is wrong. The experiment has been performed many times before, and the QM description wins.
Suppose your "wave" interpretation

The photon gun fires a photon that travels through both slit A and slit B, then hits a detector cell with either destructive or constructive interference, depending on the location of the detector cell. If the superposition is "too destructive", no spot will register in the detector pattern. All photons interfere destructively or constructively and the result is a nice interference pattern.

were correct. Then a single photon, traveling through the slits, would produce dots at the location of every constructive interference fringe. This simply isn't what you see when you do the experiment.
Also, it's technically wrong to say that in the quantum description

The photon gun fires a photon that travels through either slit A or slit B, then hits a detector cell registering a spot in the detector pattern.

The WHOLE POINT of QM is that this is NOT what we say happens. What we say happens in QM is that there is a wavefunction, which goes through both slits, and interferes with itself constructively and destructively at the interferences fringes along the screen. The square of the amplitude of this wavefuction at a point describes the probability that an electron will be detected at that point if we choose to make a measurement there. So the interference pattern is slowly manifested on the screen as increasing numbers of photons are detected in the places earmarked as 'probable' by the wavefunction. When a photon is detected in one place, its wavefunction is said to "collapse", and there's no longer any chance of finding it anywhere else.

 

[Marty]

So please, I'm interested... Just looking at the double slit experiment, what about it makes you consider it to be better explained with quantum mechanics than with a wave model? Because I've always wondered why physicists use it as an example of how special QM is, when the actual explanation involves quanta of light "interfering with itself". What does this mean, and how does it differ from a wave model?

I think you're right about the double slit experiment. Feynman once said that all the essential mysteries of QM are included in the double slit experiment, but he was talking about
a thought experiment using electrons, not light. He meant that the essential mystery is that particles (like electrons) show interference. He never said anything about the double slit experiment with regard to the wave vs particle theories of light.

People also invoke the image being built up one click at a time, or one silver halide crystal at a time as evidence of the particle nature of light. This argument of course has nothing to do with double slits or triple slits or any kind of interference. It seems convincing at first glance but is actually quite hard to nail down. The problem is that it's hard to know when you have exactly one photon at a time: and both the wave theory and the particle theory allow for a diminishing probability of coincidence counts as you make the source weaker and weaker.

 

[muppet]

The evidence that light has some particulate nature is nothing to do with the two-slit experiment. It's to do with the photoelectric effect.

 

[lightarrow]

The evidence that light has some particulate nature is nothing to do with the two-slit experiment. It's to do with the photoelectric effect.

I'm not disagreeing with you, just would like to add that, however, the two slits experiment, in my hopinion, has to do with the fact that a corpuscle of light has or not a definite individuality from the source to the detectors. What I mean is this:

If a corpuscle with such individuality exists, it must go through one or the other or both (splitting itself) the slits. It's well known that if it goes through one or the other, you don't get an interference pattern; it's probably less known that if it splits in two and goes through both...nothing changes because you have two particles passing through the slits and the same you don't have an interference pattern (ah yes, you could say it interferes with itself, but this is what a wave does, not a corpuscle).

The only possibility, in my opinion, is that the photon has not a definite individuality between the source and the detectors, the field constantly creates and destroyes it, so it's the field that passes through both slits (exactly how you wrote), interfering with itself and then creating the single photon which hits the detectors.

Just another crazy idea, of course.

 

[Hydr0matic]

Hey all. I appreciate you joining the discussion, but I really would like to narrow this thread down. Wave-particle duality is a huge subject that can be debated forever, so I would appreciate if we could focus on the original statement. Which was...

• QM and wave theory predicts different outcomes in the double slit experiment. Wave theory predicts a significant decrease in detection rate compared to firing rate, QM does not.

Is there any data on this subject?

Of course the photon detection rate is changed locally. This is what interference is about. So the total count rate of the detector will only change, if the detector is small compared to the total dimensions of your interference pattern. But if you use a small detector and place it inside a spot of destructive interference, qm also tells you, there will be a significant difference between the counting rate with and without double slit in this spot. So I do not see any difference.

What I'm talking about, is using a detector that will cover the entire area behind the slits, so that all photons passing the slits can be detected. In that case, QM predicts that... when a photon is fired, it's probabillity wave will pass the slits, interfere, and collapse at the detector determining where the photon will strike. This means that all (ideally) photons will strike the detector. This is NOT what wave theory predicts. In wave theory, the dark fringes in the pattern is a result of the fact that this area was hit by a photon (wave) that was interfering destructively, and therefore didn't set off the detector. This means that wave theory predicts a significant decrease in detection rate compared to firing rate. Unlike QM.



Off Topic:

See for example, the microscopic explanation on the interaction of photons with the phonon structure of a material that can result in the explanation for why things are opaque, transparent, and why light moves "slower" in a dispersive material.

And why would a quanta bend it's path when interacting with the material?

Shoot one photon at a time, i.e. light with extremely low intensity. Look at it over time. Now do the same with electrons, protons, etc. Tell me if what you observe is different. If yes, then light is different than the "physical particles" such as electron, protons, etc. and you may have a leg to stand on. If they look the same and have the same characteristics, then you have a lot of explaining to do.

You know as well as me that these experiments are on a completely different scale when shooting particles. Based on what I've been able to find on the web the setup is very different, but results are kind of similar, yes. They both display fringes anyway. I have some questions though. Have a look at the video on this site: http://www.hqrd.hitachi.co.jp/em/doubleslit.cfm
Here's a quick analysis of the brightness of the fringes: http://physics.soderholms.com/electron_fringes.jpg
Which fringe would you say is the center one? A typical double slit pattern has an intensity distribution looking something like this: http://img.sparknotes.com/figures/C/c33e2bffc162212e1d9aa769ad3ae54f/envelope.gif
It looks to me like, either there's something shifty going on with the intensities, or the center fringe is a minimum. So I'm not convinced it's actually an interference pattern we're looking at. In the setup (http://physics.soderholms.com/electron_biprism.jpg ) there's a few (to say the least) things in the way of the detector. And the paper (American Journal of Physics 57 117-120) doesn't say anything about what it looks like without the biprism, or what happens if you change the distance to the detector, or block one of the paths. It's basically an experiment set up and performed only to confirm quantum mechanics. They might as well just put the conclusion in there.

Suppose your "wave" interpretation were correct. Then a single photon, traveling through the slits, would produce dots at the location of every constructive interference fringe. This simply isn't what you see when you do the experiment.

What are you saying, that light can't be wave because we can shine a dot on the wall? That light must be corpuscular because it travels in straight lines? That's a bit 18th century of you.

Anyway. Too much off topic now.

 

[DrChinese]

Wave-particle duality is a huge subject that can be debated forever, so I would appreciate if we could focus on the original statement. Which was...

• QM and wave theory predicts different outcomes in the double slit experiment. Wave theory predicts a significant decrease in detection rate compared to firing rate, QM does not.

Is there any data on this subject?

ZapperZ is correct and you are closing your eyes to an experiment that answers directly the question you are asking: whether photons are discrete or not. The Thorn experiment (which is simply an undergrad version of experiments previously performed by others) asks: are photons simply waves of probability? The answer is a big NO. There are definite predictive differences - as you imagine - between various possibilities. The one you suggest is clearly ruled out, by 377 standard deviations if I recall. The QM predictions are confirmed in the process. *Read* the paper. I might suggest you quit insisting that the experiment is not relevant when it obviously is. How can anyone take your suggestion seriously when you ignore facts you don't like?

 

[lightarrow]

• QM and wave theory predicts different outcomes in the double slit experiment. Wave theory predicts a significant decrease in detection rate compared to firing rate, QM does not.

Sorry but I don't understand it. Can you elaborate? Also, where exactly is the difference between firing rate and detection rate?

What I'm talking about, is using a detector that will cover the entire area behind the slits, so that all photons passing the slits can be detected. In that case, QM predicts that... when a photon is fired, it's probabillity wave will pass the slits, interfere, and collapse at the detector determining where the photon will strike. This means that all (ideally) photons will strike the detector. This is NOT what wave theory predicts. In wave theory, the dark fringes in the pattern is a result of the fact that this area was hit by a photon (wave) that was interfering destructively, and therefore didn't set off the detector. This means that wave theory predicts a significant decrease in detection rate compared to firing rate. Unlike QM.

If 10 photons are sent in a time interval of one second, ten photons will be detected in a time interval of one second, wave theory or QM.

Also, when you say that:

Quantum
The photon gun fires a photon that travels through either slit A or slit B, then hits a detector cell registering a spot in the detector pattern. All photons follow the same procedure, and with the help of quantum mechanics, the result is a nice interference pattern.

Wave
The photon gun fires a photon that travels through both slit A and slit B, then hits a detector cell with either destructive or constructive interference, depending on the location of the detector cell. If the superposition is "too destructive", no spot will register in the detector pattern. All photons interfere destructively or constructively and the result is a nice interference pattern

1. Your description of the "Quantum interpretation" is wrong, as others have said, because "a photon that travels through either slit A or slit B" is exactly a photon of which you have the "which way" information, and that means no interference pattern. Example: you detect the photon near one of the slits putting there a little detector; clearly this is not the two slits experiment we are talking about.

2. About your wave interpretation. As I've written in my previos post, if you are certain that the particle splits in two and passes through both slits, nothing changes, it's as if you had two photons passing simultaneously through both slits; it could seem strange, but you have the "which way" information the same, and so no interference pattern. Think to the fact that to say "the particles splits in two and both passes through the slits" means that you have that specific information, and to have it you should put two little detectors near every slit and detect the two particles (the two "half photon") near the slits; so, no interference pattern. Unless you invoke De Broglie Bohm interpretation, the only possibility is that no particles passes through one or the other or both slits. (In my opinion).

 

[muppet]

What are you saying, that light can't be wave because we can shine a dot on the wall? That light must be corpuscular because it travels in straight lines? That's a bit 18th century of you.

I'm not saying that light can't be a wave because we can shine a dot on the wall, and I'm certainly not saying that it's corpuscular because it travels in straight lines.
What I'm saying is that if light were a classical wave, then an arbitrarily small "amount" of light would show evidence of an interference pattern when diffracted through one or more slits. It's perfectly possible to "shine a dot on the wall" with a laser pointer (say), but if you take a laser pointer (emitting innumerable photons) and shine it through a diffraction grating you see an interference pattern, clear as day. If light were a pure, classical wave, then it wouldn't be possible to shine any amount of light through a two-slit setup in such a way that you didn't see interference fringes immediately.

Lightarrow: Schroedinger originally suggested that there were no particles, only localised wave packets. Lorentz (I think!) demonstrated that such "matter waves" would spread out over time due to dispersion, and lose their localisation, which of course conflicts with the existence extremely well localised particles in the world today. This is the reason why people don't think of the wavefunction as describing a particle as being physically spread out. I suppose I can't honestly say, however, that your idea is any crazier than the idea that a particle pops into existence when we decide to look for it

 

[Marty]

ZapperZ is correct and you are closing your eyes to an experiment that answers directly the question you are asking: whether photons are discrete or not.

That's not the question he was asking. Hydromatic is wrong on certain points of physics but he was quite clear on this point: he asks why people choose to use the double-slit experiment to support the particle theory over the wave theory. I wouldn't normally butt in on one of ZapperZ's discussions because of his tendency to use his power as a moderator to silence his opponents, but since you found it necessary to weigh in on his behalf, I thought it was only fair for me to point out that ZapperZ has not answered this question. In fact, the extensive quote he provides in his first answer shows that the wave theory rather than the particle theory is the obvious explanation for the double slit experiment. Hydromatic rightly pointed this out in his response.

 

[lightarrow]

Lightarrow: Schroedinger originally suggested that there were no particles, only localised wave packets. Lorentz (I think!) demonstrated that such "matter waves" would spread out over time due to dispersion, and lose their localisation, which of course conflicts with the existence extremely well localised particles in the world today. This is the reason why people don't think of the wavefunction as describing a particle as being physically spread out. I suppose I can't honestly say, however, that your idea is any crazier than the idea that a particle pops into existence when we decide to look for it

Ok, however I haven't actually said that there are no particles. What I say, I think, it's a little bit complex (but I can be totally wrong): the particles are present but they don't have a precise individuality, in the sense that we cannot say: the photon N. 109 has started from the source, now it's near the first slit, it has passed the first slit, now it has hit the detector N. 13,115. What I'm asserting, as personal idea (I'm sorry, sometimes I can't help having personal ideas, it's too appealing for me!) is that the photons are continuously created and destroyed by the field during the journey between source and detectors, so that it's not possible to say that a specific photon has passed through a specific slit and it's exactly for this reason, in my opinion, that we cannot have the "which way information" and have the interference pattern. Remember that the number of the photons, even if doesn't vary on average, it's not well determined: there is for example an indeterminacy relation between the number of photons and the phase of the electromagnetic field. The more one is determined, the less is the other, and you know that in an interference experiment the phase must be quite well defined (or you don't see it).

 

[ZapperZ]

That's not the question he was asking. Hydromatic is wrong on certain points of physics but he was quite clear on this point: he asks why people choose to use the double-slit experiment to support the particle theory over the wave theory. I wouldn't normally butt in on one of ZapperZ's discussions because of his tendency to use his power as a moderator to silence his opponents, but since you found it necessary to weigh in on his behalf, I thought it was only fair for me to point out that ZapperZ has not answered this question. In fact, the extensive quote he provides in his first answer shows that the wave theory rather than the particle theory is the obvious explanation for the double slit experiment. Hydromatic rightly pointed this out in his response.

Actually, I thought I HAD answered the question completely, and that I've stopped responding because I would simply be saying the SAME thing. If someone simply didn't care to accept what I've cited, then how many times does one have to say the same thing? And note that if I want to use my power as a moderator to silence my opponents, why didn't I shut down this thread already? The fact that it continued to go on clearly falsified your claim.

There are two separate issues here:

1. The OP's scenario is wrong in the first place. As has been mentioned many times by several people, the QM scenario of what would happen is simply not correct.

2. A double slit experiment can distinguish between "wave" and "particle" picture. Again, there's a huge amount of confusion here because the TYPE of such experiment. The Mach-Zenhder experiment is a perfectly valid "2-slit" experiment and allows for even more testing being done. And when this is done for low-intensity sources AND incorporating a which-way criteria, only then can one actually distinguish one from the other. again, refer to the J.J. Thorn et al. paper. In fact this is an experiment that can actually be done at the undergraduate level, so it isn't that difficult to do! One can easily test the OP's scenario using such set up!

So who's rushing to do this and get nominated for a Nobel Prize?

Zz.

 

[Hydr0matic]

There are two separate issues here:

1. The OP's scenario is wrong in the first place. As has been mentioned many times by several people, the QM scenario of what would happen is simply not correct.

2. A double slit experiment can distinguish between "wave" and "particle" picture. Again, there's a huge amount of confusion here because the TYPE of such experiment. The Mach-Zenhder experiment is a perfectly valid "2-slit" experiment and allows for even more testing being done. And when this is done for low-intensity sources AND incorporating a which-way criteria, only then can one actually distinguish one from the other. again, refer to the J.J. Thorn et al. paper. In fact this is an experiment that can actually be done at the undergraduate level, so it isn't that difficult to do! One can easily test the OP's scenario using such set up!

2. I'm not satisfied with a reference to this experiment because it has other components that we don't share the same interpretation about. So in order for me to argue the detection rate issue, I would first have to argue the beam splitting issue for example. And I would prefer not to because it's not really relevant to the double slit experiment, and the key issue is much simpler. All you need is a photon gun, a double slit, a detector and two values - one count rate when there is no double slit, and one when there is.

1. So, if I'm wrong, what would QM predict with regards to the detection rate vs firing rate?

 

[Hydr0matic]

Sorry but I don't understand it. Can you elaborate? Also, where exactly is the difference between firing rate and detection rate?

Firing rate is the actual number of photons fired. The number you would detect if there is nothing in between the photon gun and the detector. The detection rate is the detectors count.

If 10 photons are sent in a time interval of one second, ten photons will be detected in a time interval of one second, wave theory or QM.

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.

1. Your description of the "Quantum interpretation" is wrong, as others have said, because "a photon that travels through either slit A or slit B" is exactly a photon of which you have the "which way" information, and that means no interference pattern. Example: you detect the photon near one of the slits putting there a little detector; clearly this is not the two slits experiment we are talking about.

This is just me expressing myself poorly as usual. By saying "either slit A or slit B", I'm not implying that we have any which-way information, just that - after the collapse of the wave function - the quanta travels through either slit A or B. But we don't know which. But we don't need to consider any which-way information to determine the two count rates.

2. About your wave interpretation. As I've written in my previos post, if you are certain that the particle splits in two and passes through both slits, nothing changes, it's as if you had two photons passing simultaneously through both slits; it could seem strange, but you have the "which way" information the same, and so no interference pattern. Think to the fact that to say "the particles splits in two and both passes through the slits" means that you have that specific information, and to have it you should put two little detectors near every slit and detect the two particles (the two "half photon") near the slits; so, no interference pattern. Unless you invoke De Broglie Bohm interpretation, the only possibility is that no particles passes through one or the other or both slits. (In my opinion).

I have read a lot about which-way tests but haven't seen one that wouldn't alter the conditions for interference in the first place. But all this is irrelevant, we don't need any which-way information to determine count rates. We don't need the count rate of slit A and B seperately, only the total number that was detected and the total number that was fired.

 

[ZapperZ]

2. I'm not satisfied with a reference to this experiment because it has other components that we don't share the same interpretation about. So in order for me to argue the detection rate issue, I would first have to argue the beam splitting issue for example. And I would prefer not to because it's not really relevant to the double slit experiment, and the key issue is much simpler. All you need is a photon gun, a double slit, a detector and two values - one count rate when there is no double slit, and one when there is.

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.

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.

1. So, if I'm wrong, what would QM predict with regards to the detection rate vs firing rate?

I believe both Cthugha and muppet had clearly described this already. I don't quite understand what more you want.

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:

T.V. Marcella Eur. J. Phys. v.23, p.615 (2002).

The preprint can be obtained here:

http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

Zz.

 

[Cthugha]

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.

 

[Hydr0matic]

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 going to 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.

 

[Hydr0matic]

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

 

[granpa]

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 isn't 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.

 

[granpa]

I would like to know more about this. could it be that the energy isn't 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]

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 going to 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]

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.

 

[Hydr0matic]

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]

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.

 

[Hydr0matic]

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.

 

[Hydr0matic]

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?

 

[Hydr0matic]

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 separate 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:

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]

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.

 

[Hydr0matic]

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]

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 sort of 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.

 

[Hydr0matic]

The OP question is sort of 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 going to 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]

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


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?

 

[DrChinese]

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.

 

[Hydr0matic]

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?

 

[Hydr0matic]

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]

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.)