# Method and Apparatus For Transmitting Signals Faster than Light? (Based on EPR/QM)

1. Jan 17, 2006

### Dr. RMC

Method and Apparatus For Transmitting Signals Faster than Light?? (Based on EPR/QM)

Method and Apparatus For Transmitting Signals Faster than Light?? (Based on EPR/QM)

Does the following apparatus make sense? Please discuss and let me know where the logic or science is falling short, if it is falling short.

45Revolver: Method and Apparatus for Transmitting Signals Faster than
the Speed of Light

Feel free to email me for the paper and the figures:
REVOLVER45@gmail.com

Consider an apparatus with a source for two interacting spin ½ photons
at the center, as shown in Figure 1. On either side of the source is a
double slit at distance d1, and beyond each double slit is a screen at
distance d2.

The two interacting spin ½ photons separate and travel in opposite
directions. Each passes on through a double slit-double-slit1 and
double-slit2, and an interference pattern is observed on each of the
two screens, screen1 and screen2.

Suppose then that a measurement device is then placed in front of the
upper slit in the double-slit to the left of the photon source, as
shown in Figure 2. Now photons must either go through the upper slit
or lower slit in double-slit1, and the interference pattern disappears
on screen1 to the left of the photon source.

Because momentum is conserved, whenever one photon passes through the
upper slit in double-slit1 (situated to the left of the source), the
other photon must pass through the lower slit in double-slit2 (situated
to the right of the source), as shown in Figure 2. And whenever one
photon passes through the lower slit in double-slit1, the other photon
will pass through the upper slit in double-slit2. Thus the
interference pattern will also disappear on screen2 to the right of the
source, whenever a measurement device is placed in front of the upper
slit in double slit1 to the left of the source.

Thus the placement of the detector in front of double slit1 to the
left of the photon source affects the interference pattern on screen2
to the right of the source.

Because d1 and d2 can be changed without affecting the basic physics,
a situation in which d1 is very, very large with d2 just a little bigger (d1 ~ d2) is conceivable. In such a scenario, by
selectively placing a measurement device in front of double slit1 and
removing it at a controlled interval, the interference pattern at
screen2 will be created and destroyed at the same controlled interval,
thereby sending a signal.

As d1 becomes greater and greater as
compared to d2, the signal can be sent faster and faster, transcending
the velocity of light and approaching instantaneous.

There could be many practical physical embodiments of such a device or apparatus for sending signals faster than the speed of light.

One preferred embodiment would be a central photon source, and fiber
optics run to two distant locations on either side of the photon
source. A double slit or other interference apparatus would be
employed at each of the two distant locations. A removable measurement
device would be located at one or both locations, and its use or
none-use in the setup could instantaneously affect the interference
pattern at the other location.

45Revolver: Method and Apparatus for Transmitting Signals Faster than
the Speed of Light

Feel free to email me for the paper and the figures:
REVOLVER45@gmail.com

Last edited: Jan 17, 2006
2. Jan 17, 2006

### JesseM

I'm not sure, but my guess would be that no interference pattern is ever seen in the total pattern of photons on the screen in such a setup, even if no measurement is made of which slit each one through; instead, the interference pattern would only be seen in the some sort of coincidence-counting, like if you only looked at the subset of photons hitting the screen on the right side that are paired to photons that hit the bottom portion of the left screen, or something along those lines. My guess here is based on an analogy with the delayed choice quantum eraser experiment which is discussed on this thread.

3. Jan 17, 2006

### Dr. RMC

Thanks Jesse,

It seems strange to me that no interference pattern would ever be seen, as if a photon passes through both slits, it must interfere with itself.

If we pictured millions or billions of photons an interference pattern would emerge.

Or then how could a photon pass through both slits and not interfere with itself?

4. Jan 17, 2006

### JesseM

Well, look over the delayed choice quantum eraser experiment, which is also discussed on this thread. In this experiment, you have an entangled pair of photons, one labeled the "signal photon" which ends up at detector D0 (which can also be replaced with a screen as in the double-slit experiment), another labeled the "idler photon" which ends up at either D1, D2, D3, or D4. If it ends up at D3 or D4, then you know which location (analogous to the slits) the pair of photons came from, like in the double-slit experiment where you measure which slit the photons go through; if it ends up at D1 or D2, then the which-path information is lost, like in the double-slit experiment where you don't measure which slit it went through. If you look only at the subset of signal photons whose idlers end up at D3, you don't see an interference pattern, and likewise with D4. On the other hand, if you look at the subset of signal photons whose idlers end up at D1, you do see an interference pattern, and likewise with D2, so in a sense you could say the photon has "interfered with itself", as you say. But what's interesting is that the peaks of the D1 interference pattern line up with the valleys of the D2 interference pattern, so if you look at the sum of the D1 and D2 interference patterns (the subset of signal photons whose idlers ended up at either D1 or D2), you get a pattern that shows no interference. Again, look at that earlier thread I linked to above and the paper I linked to in my post there, which includes a diagram of the setup and a graph of the various signal photon patterns, showing how the peaks of the D0/D1 interference pattern line up with the valleys of the D0/D2 interference pattern.

So like I said, I would guess that something similar happens in the setup you describe--if you look at the proper subsets of photons going to the left screen which are paired with photons going to the right screen, you would get an interference pattern, but when you summed all the possible interference patterns based on such subsets, the peaks of some would line up with the valleys of others in such a way that the total pattern shows no interference. That's just a guess since I don't know how to explicitly calculate the correct QM prediction in this setup, but I can't see any other way that FTL signalling (or even backwards-in-time signalling!) could be avoided here.

Last edited: Jan 17, 2006
5. Jan 17, 2006

### Dr. RMC

Thanks Jesse,

I looked over the link and the paper.

If there is never any interference pattern upon the screens in the setup I describe, then the respective photons to the right and left of the source must not be passing through both slits in the double slits.

Hence, even without being measured, the photons are behaving like particles, with definitive positions and momenta.

But if the photons are behaving as waves, then an interference pattern must be produced.

Thanks.

6. Jan 17, 2006

### JesseM

But what if there is interference when considering various subsets of photons in your setup--like considering the subset of all photons hitting the right screen that are paired with photons hitting the bottom half of the left screen--but when all these patterns are added together, the result is a noninterference pattern?

Perhaps it would be better to talk about this issue in terms of the delayed choice quantum eraser experiment, where it's clearer what subsets we're considering. If we look at the total pattern of signal photons on the screen and see no interference, would you say this means the signal photon/idler photon pairs are not coming from both radioactive atoms A and B (which are analogous to the slits)? If that's what you would say, how do you explain the fact that when we consider only the subset of signal photons whose idlers end up at D1, we do see an interference pattern? Does this mean that the photons in this subset are coming from both atoms?

To make this problem sharper, suppose we replace the beam-splitters BSA and BSB in the fig. 1 of that paper with mirrors, so all the idlers end up at D1 and D2. Again, if you look at the subset of signal photons whose idlers ended up at D1, you see an interference pattern; and if you look at the subset of photons whose idlers ended up D2, you see an interference patter; and every photon falls into one subset or the other. So does this mean every photon did come from both radioactive atoms? But then if you look at the total pattern of signal photons (the sum of photons from the first subset with photons from the second), you see a non-interference pattern. Does this mean every photon did not come from both radioactive atoms?

It seems to me the problem is that your terminology is fuzzy--I don't really see what it means to say that a given photon did or didn't go through both slits/come from both atoms, when a single photon can be viewed both as a member of a subset that does show interference, and simultaneously as a member of a larger subset which does not.

7. Jan 17, 2006

### Dr. RMC

I guess the basic question is this:

Suppose we have a beam of two interacting spin 1/2 photons.

The beam of photons separates, with each of the two interacting photons traveling in opposite directions.

If in each direction there exists a double slit setup, will an interfence pattern be created on a screen beyond the double slit?

8. Jan 17, 2006

### JesseM

Will the total pattern of all photons hitting the screen create an interference pattern? I think the answer is no, but I also think that certain well-defined subsets of the photons hitting the screen will create an interference pattern.

Did you follow the details of the delayed choice quantum eraser setup well enough to be able to understand the questions I was asking in my previous post? If not, I can try to explain it further; if so, it would be helpful if you would address them.

9. Jan 17, 2006

### Dr. RMC

Yes--I followed your argument and will address it in a bit.

But first I would like to remain focused on my apparatus.

Consider a beam of photons on either side of the apparatus I describe.

If the beam is incident upon a double slit, how could it not create an interference pattern upon a screen just beyond the double slit?

What could possibly cause a beam of photons to not create an interference pattern?

Thanks!

10. Jan 17, 2006

### JesseM

If I'm right that there would be no interference in the total pattern of photons, presumably the "cause" would be that photons which are members of entangled pairs do not behave in quite the same way as photons which are not, in the context of the double-slit experiment. After all, there is some possibility that by performing the right measurement on the other member of the pair, you can determine which slit the first photon went through, even without performing measurements at either of the two slits the first photon could have passed through. If this is correct, then you should be able to show that this is the case using the rules of quantum mechanics, so the ultimate "cause" is just that that's how the laws of quantum mechanics say the photon should behave, trying to translate these mathematical laws into some kind of intuitive verbal explanation may not always be possible.

11. Jan 18, 2006

### Dr. RMC

Would this then mean that photons which are members of entangled pairs show no interference pattern? And then would this mean that if a photon was once a member of an entagled pair, it could never again behave live a wave and interfere with itself?

Suppose we consistently measured one of the entangled photons, thereby liberating the other photon from the entangled state. Woul the liberated photons be capable of behaving like photons and creating an interference pattern after passing hrough a double slit? Because this is another manner in which signals may be sent faster than the speed of light.

12. Jan 18, 2006

### ZapperZ

Staff Emeritus
Er... suppose we DON'T have a beam of interacting spin 1/2 photons!

Zz.

13. Jan 18, 2006

### ZapperZ

Staff Emeritus
Entangled photons do interfere, but not within the entangled unit. It has been shown that entangled photons beat the diffraction limit[1,2], as it should if the entangled glob cannot be considered as separable (mathematically) and thus have a higher energy (freq) as an entity.

Zz.

[1] P. Walther et al., Nature v.429, p.158 (2004).
[2] M.W. Mitchell et al., Nature v.429, p.161 (2004).

14. Jan 18, 2006

### Dr. RMC

Suppose we consistently measured one of the entangled photons, thereby liberating the other photon from the entangled state. Woudl the liberated photons be capable of behaving like photons and creating an interference pattern after passing hrough a double slit? If so, then signals may be sent faster than the speed of light.

If one can intentionally alter an interference pattern from afar at velocities faster than light, then an apparatus may be designed for faster-than-light communication.

15. Jan 18, 2006

### JesseM

That's not how it works. The other photon will continue to behave in an entangled way until it is measured; if you compare the first measurement of photon A with the first measurement of photon B you should find the correlations that are characteristic of entanglement, even if photon A was measured long before photon B. Subsequent measurements of particles that were previously entangled would no longer show entanglement though, it's only the first measurement of each that counts.

Last edited: Jan 18, 2006
16. Jan 24, 2006

### Sherlock

The interference pattern disappears on screen1 because you've, in effect, closed one of the openings in double-slit1.
The side that has both slits open will show an interference pattern. The side that has one slit closed will not show an interference pattern.
No. Closing one of double slit1's openings will not affect the raw data interference pattern produced by double slit2. Double slit2 produces interference banding (after about 60 to 70k detections) because both its slits are unobstructed. Double slit1, with an obstructed opening, will not.

17. Jan 24, 2006

### JesseM

Are you sure about that? By conservation of momentum, if you know which slit one photon went through, doesn't that mean you know which slit its entangled twin went through? The pattern you describe isn't how it works in the delayed choice quantum eraser--there instead of two slits a particle can go through you have two atoms which periodically emit pairs of photons, one labeled the "signal" photon which goes towards a screen similar to the double-slit experiment, and the other labeled the "idler" photon which goes through a system of beam-splitters and mirrors arranged so that, depending on which detector the idler ends up at, you either recover the "which-path" information about which atom it came from (and thus which atom the corresponding signal photon came from), or the which-path information is "erased". But in neither case do you see any interference in the total pattern of signal photons on the screen, despite the fact that you had no measuring device to detect which of the two atoms a given signal photon came from--you only see interference when you do coincidence-counting between the subset of idler photons that ended up at a detector where there which-path information was erased and the corresponding subset of signal photons. So, I guessed that something similar would be true in the experiment described in this thread, where you would never see any interference pattern in the total pattern of photons on either screen, interference would only be visible if you did coincidence-counting between various subsets (like maybe the subset of all photons on one screen whose twin ended up in a certain narrow region on the other screen).

18. Jan 24, 2006

### Sherlock

I'm not sure about anything yet. But, afaict, Dr. RMC wasn't describing a delayed choice quantum eraser setup. In the OP setup there was just doubleslit1 with a detector obstructing one of the slits, and doubleslit2 with both slits open. The raw data on the detecting screen1 will show no interference because the light going through the slit where the detector is placed next to doubleslit1 is absorbed by the detector and can't interfere with the light transmitted by the other, open, slit. On the other side, neither slit of doubleslit2 is obstructed, so you eventually get an interference pattern on screen2.

If you pair the photons via coincidence detection and then select out the portion of side 2's data corresponding to what is detected at side 1's screen slit detector, then there will be no interference pattern on side 2. I don't know if this can be done in real time or not. If not, and if the data output is irreversible (say, a photographic plate as opposed to a video monitor), then what you will see is an interference pattern on side 2, but what you will statistically calculate will be a clumping corresponding to side 1's output.

Or, you can erase the which way info on side 1 (in effect, remove the detector obstruction from the closed slit --- recreating, in effect, the original situation with both slits open in doubleslit1) by coincidence matching with side 2 and reintroducing the missing statistical info to side 1 to get an interference pattern (even though the raw data output might not actually show an interference pattern on screen1).

It's not which way info, per se, that determines whether you get an interference pattern or not --- it's that in order to generate which way info you have to, in effect, close one of the slits in the double slit device. And, to erase which way info you have to, in effect, reopen the closed slit.

I've noted the thread on Scully's paper you referenced. It's on my list of things to read. Also, I found this:
http://www.bottomlayer.com/bottom/kim-scully/kim-scully-web.htm

... which looks like it might be a good explanation of it.

Any corrections wrt what I've written are appreciated.

Last edited: Jan 24, 2006
19. Jan 24, 2006

### JesseM

I know, but it seemed to me like there was some kind of analogy there--in both setups, on one side you have a photon coming from one of two locations with no measuring device to tell you which one it came from (the photon going to the right in Dr. RMC's setup vs. the signal photon coming from one of two atoms in the DCQE), but on the other side you have its twin where, depending on what kind of measurement you make on the twin, you may be able to tell where both photons came from (the photon going to the left in Dr. RMC's setup, where you can put in a detector to tell which slit it went through and then deduce which slit the other photon went through by conservation of momentum vs. the idler photon whose which-path information can be discovered if it goes to certain detectors in the DCQE). In the DCQE there is no interference in the total pattern of signal photons no matter what measurement is made of the idler, so I thought something similar would probably be true of the setup Dr. RMC describes. After all, suppose you did see interference in the total pattern of photons on the right side, but then you had a detector on the left side so you always knew which slit each photon on the left went through; by conservation of momentum you should then know which slit each photon on the right went through, so if you see interference in the photons on the right doesn't that contradict complementarity?
But that's what I'm saying, I don't think you can assume that you will get interference when both slits are open but the photons are members of entangled pairs. Again, if you know which slit each photon on screen1 went through, that would seem to tell you which slit each photon on screen2 went through (despite the fact that both slits to screen2 were open with no detectors near them), since conservation of momentum should hold for members of entangled pairs, as I understand it. Again, there's the analogy with the DCQE where the signal photon can come from either of the two atoms (which are analogous to the slits) and so looking at the signal photons alone it might seem like you should get interference in their total pattern, but in fact you don't, you only get it in coincidence-counting between idlers whose which-path information was erased and the corresponding signal photons.
But what if the detector on side 1 was on all the time, so you know which slit every photon on that side went through? This should also tell you which slit every photon on side 2 went through, but whether you see interference in the total pattern of photons on side 2 can't depend on whether the detector is on or off on side 1, or this would lead to the possibility of FTL or backwards-in-time signalling just as Dr. RMC suggested.
But isn't that part of the basic idea of complementarity, that an experiment can reveal particle-like or wavelike behavior but not both at the same time? If there could be a double-slit type experiment where you see an interference pattern on the screen but you also know which slit each particle went through, this would seem to be a violation of that idea. But I only have an intuitive understanding of what "complementarity" means so I'm not sure about this (I'm not even sure if 'complementarity' is even an idea that's ever been formalized, I've seen some debates between professional physicists about issues involving complementarity where neither ever referred to any formal definition of the principle to settle the argument).
You don't necessarily have to close it--if you do the double-slit experiment with electrons or some other particle that can be detected without being absorbed (I don't think this is possible for photons although I'm not sure), then you can just have a detector at one of the slits that detects by scattering some other particle like a photon off the electron, still allowing the electron to go through that slit.

20. Jan 26, 2006

### Sherlock

Afaik, with both slits open, there's no way to know if a single-photon detection was produced by light that went through both slits or only one. What's known is only that both slits were open and an interference pattern eventually emerged.
If you put a detector behind one of the slits (between the double-slit device and the screen), then you could think of this as both slits still being open, in which case, after a single-photon detection at the slit or at the screen there's still no way to know if all of the light incident on the double-slit device went through both slits or only one.
In the setup Dr. RMC describes, without pairing the photons via coincidence processing, and then doing some additional data processing, then what would be expected is that screen1 will not show an interference pattern and screen2 will show an interference pattern.
If, after coincidence matching, a logic pulse is sent from the slit detector on the left side via circuitry that then blocks the transmission of its twin to the the right side screen2 detector, then no interference pattern will emerge on screen2.

I have to read up on complementarity (I forget exactly how Bohm presented it in his textbook) --- but isn't it just part of Bohr's qualitative (Copenhagen) interpretation? If so, then violating it wouldn't diminish the validity of the mathematical formulation of QM --- which, afaik, predicts an interference pattern for screen2 of Dr. RMC's setup and no interference pattern for screen1.
The fact that paired photons are entangled just allows some processing that wouldn't otherwise be allowed. Without any additional processing, and just transmitting through an unobstructed double-slit device, then you'll get an interference pattern on the screen. Afaik, there's no reason to assume that the components of an entangled pair are in any way affecting each other as they move apart.
The photon isn't what went through one slit, or the other, or both. Afaik, nobody knows what's happening at that level. The photon is the detection attribute. If the single-photon slit detector on side 1 registers a detection, then nothing further is known about the qualitative behavior of the source-emitted light or the slit(s)-transmitted light (did something go through only one slit or both?) corresponding to that detection other than that the single-photon slit detector on side 1 registered a detection.

So, the results on side 1 don't really give qualitative which-way info. The only thing that can be said is that if the slit detector registers then for sure something went through the slit where the detector is, and if screen1 registers then for sure something went through the unobstructed slit. On side two, where the interference pattern emerges, everything that's known about light suggests that it's reasonable to assume that the light incident on the double-slit device corresponding to a single-photon detection at screen2 went through both slits. But what is actually happening at that level is still a mystery.
It doesn't. Unless you do some signal or data processing, what's done on one side of the setup won't affect what happens on the other side.
With both slits of the double-slit device open you get a wave-like distribution on the screen that's characteristic of two waves transmitted by and spreading omnidirectionally out from the two parallel openings and interfering.

With only one slit of the double-slit device open you get a wave-like distribution on the screen that's characteristic of one wave transmitted by and spreading omnidirectionally out from a single opening.

In both cases, the wave-like detection patterns are composed of particle-like single photon detections.

21. Jan 26, 2006

### JesseM

My understanding is that conservation of momentum would mean that if you know its twin went through the bottom slit on the left, this tells you that the photon on the right must have gone through the top slit. This is analogous to how, in the DCQE setup, if the idler is detected in such a way that its which-path information is preserved, then this means you also know which path the signal photon came from. Is your understanding different than this?
I don't understand--are you saying that even if you put a detector on one of the two slits on the left side, then this still doesn't tell you which slit photons on the left side went through? If it's really a valid "detector", then by definition doesn't that mean that if you put it next to one of the slits then you can always tell whether the photon went through that slit or not?
So I take it you disagree that conservation of momentum means that if we know which slit photon1 went through, that means we know which slit photon2 went through? Or are you saying that even though we know which slit photon2 went through, there can still be an interference pattern on screen2?
Well, take a look at the recent thread Does a single beam of entangled photons create interference?--it seems like the answers from vanesch and others are that you don't see interference in the total pattern of photons when they are members of entangled pairs, regardless of whether you know which slit they went through. When you say that there should be interference "as far as you know", is that based on having seen the math done on this problem or seeing it discussed somewhere, or is it just based on an analogy with the nonentangled version of the double-slit experiment, or maybe on your own understanding of complementarity (ie the idea that if you don't have knowledge of which slit a particle went through you should see wavelike behaviour)?

And I don't think complementarity has anything specifically to do with the Copenhagen interpretation, as I understand it it's basically just the principle that when you don't know which of multiple a particle took, the different possible paths all contribute to the final outcome in a wavelike way. This is basically just a matter of the probability distributions, whether they show wavelike interference or not, and since all interpretations predict the same probability distributions this idea of complementarity shouldn't be specifically tied to any one of them.

In any case, I'm not sure if you're saying complementarity actually is violated here or not--are you saying that we could see interference on screen2 even if we can use conservation of momentum to deduce which slit photon2 went through (which as I understand it would violate complementarity), or are you saying that knowing which slit photon1 went through + conservation of momentum is not enough to tell us which slit photon2 went through?
I don't think they are affecting each other "as they move apart" in the sense of any FTL connection between them, but I think a photon created in an entangled state will behave differently in the double-slit experiment then a photon which wasn't created in an entangled state, so it's as though the photon has a "memory" of how it was created. Again, the answers on that other thread seem to confirm this.
It's only when we don't measure which slit the photon went through that nobody knows what the photon was "really" doing (this would depend on which interpretation of QM you choose). But when you have some sort of detector that tells you which slit a particle went through, I don't think physicists would object to saying "this particle went through that slit" in a straightforward way. That's what "which-path information" means as I understand it, definite knowledge that a particle took one path or another.
But would you say the same thing about the DCQE? Scully (the main author of the DCQE paper linked to earlier) and other physicists I've seen talking about it seem to have no problem saying that registration at certain detectors can give you "which-path information" for the idler, and for the entangled signal photon.

22. Jan 26, 2006

### Sherlock

I agree that if you actually had the which-path info of one photon, then the conservation law would allow you to deduce the which-path info of its entangled twin.

The question I have is whether you actually ever have which-path info in the sense that it excludes the possibility that any light whatsoever could have gone through the other slit.
Sure, but if you put a detector in front of or behind one of the slits and leave the other slit open, then how do you know that some of the emitted light pulse that produced the slit-detected photon didn't also go through the other slit? You do know that enough light was absorbed by the slit detector to register a photon, but if you then want to say that the photon produced on screen2 (the side where both slits are unobstructed) came through only one slit, how do you know it didn't go through both slits (even though it certainly went through the slit that corresponds to the conservation law deduction).

Without something coming from both slits for each single-photon detection, then how can an interference pattern emerge? But if something is indeed going through both slits and interfering, then why does only one location on the screen register a detection?
I'm saying that if you leave both slits open, then the unprocessed raw data will (after enough single-photon detections have registered) show an interference pattern.
I'll check it out. My guess is that some data or signal processing is going on --- which makes all the difference.
I've done the math for non-entangled particles transmitted through a double-slit device I don't know that it should be any different if that particle is a member of an entangled pair as long as no signal or data processing is involved.

The results at the two ends (of a setup that's measuring entangled particle pairs) are random and unrelated if no other structure is imposed on them. If double-slit devices are placed on both sides between the emitter and a screen detector, then, afaik, you should see an interference pattern emerge on both sides. If you then obstruct one of the openings of the double-slit device on one side, then, afaik, you should not see an interference pattern on that side, but you'll still see an interference pattern on the other side.
I don't know. I was just assuming that complementarity (being Bohr's creation) was part of the CI. David Bohm, in his 1950 textbook Quantum Theory, has several pages on it --- and I'm not sure I understand what he's talking about.

The probability distributions associated with transmission through a single-slit and a double-slit are both wave-like distributions. The double-slit distribution shows an interference pattern. What is complementary (ie., mutually exclusive, yet related to the overall picture) are single detections and multiple detections.

It's not enough to tell us that the light incident on the double-slit2 device only went through one slit. Is it? I don't know.
Why should it? A photon is a photon. Afaik, the results on side 1 and side 2 are unrelated unless you do some additional processing. This processing is something that physicists do --- not the particles by themselves.
Particles that have interacted, or have been acted upon similarly, or have a common emitter are related wrt various detection parameters. This is the conjectured physical basis of entanglement at the level of an underlying quantum world. But there's no qualitative apprehension of this at any level other than the instrumental level.
A slit-detector registration allows you to say that the detection was produced by something coming through that slit. It doesn't allow you to say that nothing went through the other slit. It just isn't known. Suppose you put single-particle detectors right next to each of the slits. Then these detectors never register simultaneously. Apparently whatever was incident on the slits only went through one. But remove them and replace them with a screen detector a bit further away and after 60k or so single-particle detections a distinct interference pattern emerges. Apparently whatever was incident on the slits, associated with each single-particle detection, went through both slits (otherwise, what sort of physical phenomenon could produce the interference pattern that eventually emerges), but only produces a single-particle detection on the screen. It's not just a statistical phenomenon --- after all, the raw data that you're looking at shows an interference pattern. This is a deep mystery, isn't it?
The which-path interpretation, the complementarity interpretation ... well, that's the point ... they're interpretations. Maybe they're useful, maybe not. I don't know enough yet to say with any high degree of certainty. If you noticed, I sort of changed my way of thinking/talking about this stuff during the course of our exchanges. I have no doubt that my thinking on this will change some more.
Your points and questions seem well considered. Maybe after reading the other thread(s) and Scully's paper I'll understand your insistence that entangled particles interact with double-slit devices differently than non-entangled particles.

23. Jan 26, 2006

### JesseM

Which-path information doesn't mean you exclude the possibility light went through the other slit, it just means that for every photon you detect on the screen, you know which of the two slits it went through.
If you assume an idealized perfectly efficient detector, then any particle that hits the screen that was not detected at one slit must have gone through the other, no?
It's all about the different probability distributions, and how you predict them according to the rules of quantum mechanics. In terms of this language of "going through both slits" vs. "going through one slit", maybe it would be easiest to think of this in terms of the path integral approach to QM, where you do a quantum-mechanical sum over all paths that are compatible with your detection events, with different paths interfering with each other. In this approach, the particle "going through both slits" can be thought of as a colloquial way of describing the idea that you are supposed to do a quantum-mechanical sum over paths going through both of the two slits, which leads to a certain probability distribution for where you'll detect the photon on the screen. On the other hand, if you measured the particle as having gone through a particular slit, you would only sum over paths going through that particular slit, which would lead to a different probability distribution. These two probability distributions represent different predictions about what results you'll actually get when you perform the experiment many times, so the question of whether it "went through both slits" or not is not just a matter of interpretation, if you understand the phrase to have this meaning in terms of the path integral approach. And I'm pretty sure that the rules of how to apply the path integral approach to the double-slit experiment tell you that if you have a perfectly efficient detector at one of the slits, you are not supposed to do a quantum-mechanical sum over paths through both slits at once with all the different paths interfering with each other, and you will not see the interference pattern that is characteristic of that sum.
Well, it sounds like you're talking about "interpretational issues" (ie the question of what's 'really' going on between measurements), but my understanding is that when physicists colloquially talk about the photon "going through both slits" this isn't really meant to commit to any particular interpretation, it's more just a shorthand for the rules for calculating what the probability distribution on the screen will be when both slits are open and you have no which-path information. So again, it's not a matter of interpretation, you can check experimentally when you see the probability distribution that's a result of letting paths through different slits interfere with each other in the path integral, and when you see the non-interfering probability distribution.
Do you think that's true even when we have a detector there?
I'm pretty sure they're just talking about the total pattern of photons on the screen, when all the photons are members of entangled pairs.
But that's exactly the question we're debating, isn't it? I think it should be different. In terms of the path integral approach, my guess is that conservation of momentum would mean you would only sum over paths where the two entangled photons went through opposite slits, which means if you had a detector to see which path photon1 went through, then you won't be summing paths through both slits for photon2, meaning you won't get interference. And if that's true, then presumably you still won't get interference even if you don't know which path photon1 went through, because if photon2's pattern depended on whether or not you put a detector on one of the two slits photon1 went through, that would be a recipe for an FTL communication device (presumably there is some less ad hoc way of showing why photon2 would not show interference if it's a member of an entangled pair, but that would involve actually going through the math for such an entangled pair).
You say "afaik", but it sounds like you aren't basing this on having done the calculations for this case or on having seen some physicist comment on the issue, you're just assuming that an entangled photon in the double-slit experiment would behave the same as a non-entangled one. Again, I just don't think that's true.
I think it would be, not in an "interpretational" sense but in the sense that I think it would be incorrect in this case to calculate the probability distribution for photon2 by doing a quantum-mechanical sum over paths that went through both slits.
Yes, but in quantum mechanics it's really the idea of a system's state that's fundamental, and the entangled state is different from the single-photon state.
Well, they are unrelated, in the sense that you can calculate the probabilities for what you will see on one side based only on information in the past light cone of your measurement on that side. The creation of the particle in an entangled state is part of that past light cone, although the results of any measurements on the other member of the entangled pair after they separated may not be.
What do you mean "processing"? Are you just talking about coincidence-counting? If you just look at the subset of measurements on one particle that correspond to the twin particle being measured at a particular detector, and you see some interesting pattern in this subset, I'm not sure I'd agree that this is purely something that the physicists "do" as opposed to something the particles themselves "do", I'm not even sure what you mean by this distinction.
Again, I think you're confusing interpretational issues with purely empirical ones. "Entanglement" doesn't refer to any particular hypothesis about a deep connection between separated particles or anything like that, it just refers to the fact that the outcome of a measurement on one particle will be correlated with the outcome of a measurement on the other in certain predictable ways.
See above, you're still talking in "interpretational" terms, but if you measure a particle at one slit I think it does mean that you should ignore paths through the other slit when using the path-integral approach to calculate the probability distribution on the screen. It probably wasn't clear before that I wasn't talking about such issues of interpretation, but hopefully this point about the practical procedure for constructing a path integral will clarify what I'm getting at here.

24. Jan 27, 2006

### Sherlock

JesseM, thanks for your comments. I think I understand the source of my confusion. I had temporarily lapsed into conceptual (or interpretational) thinking. Entangled states, as with any quantum phenomenon, don't exist in nature independent of the experimental setups that produce them.

The entangled results are produced by 1) a source known to produce entangled particle pairs, 2) detectors measuring the same general property (eg., position, momentum, etc.) at each end of the setup, and 3) pairing the individual results at each end via coincidence matching.

In Dr. RMC's original scenario, with unobstructed double-slit devices on both sides, you will get interference banding because the distributions on each side are both associated with both slits of the double-slit device --- that is, the terms representing emissions from each slit of the double-slit device are combined.

Then, if you add the slit-detector to the double-slit1 device you will destroy the interference pattern at screen1 unless you include the detections at the slit-detector. On the other side, the results at screen2 are entangled with the results at both screen1 and the slit-detector, and you will still get an interference pattern on screen2 unless you filter the results by removing either the results entangled with screen1 or the results entangled with the slit-detector --- thereby removing one of the emission terms relating to the slits of the double-slit2 device from the calculation of the distribution at screen2.

I haven't learned the path integral method yet.

25. Jan 27, 2006

### Sherlock

JesseM, I just looked at the thread where they're talking about SPDC produced entangled photons. I don't understand it yet. Apparently, SPDC photons, or maybe any entangled photons, when directed through an unobstructed double-slit device don't produce the interference banding. Another mystery. :-)