Double Slit Experiment questions.

[Sam99]

Hi all.

Been reading on the DS experiment & some things still seem cloudy to me.

How & where exactly is the measurement made the determines which slit the electron goes through? I'm under the assumption that a detector gadget or material of some sort is placed in front of one slit & is there to register if it went through the slit. It cant be light or a camera watching it placed near one slit? That precise locating would make it fan out like a wave as it leaves the slit right?

Observed or not, im assuming after leaving the slits the particle instantly go back to behaving like a wave?

I read something like if information about which slit the particle came through is erased before the particle hits the detector the interference pattern is restored. But how is the info erased? Has this been done? If the experimenter found out which slit it goes through & keep it a secret. Then his co worker angrily killed him never knowing. Would this restore the pattern as the particle hits the detector way down the ling?


This bugs me, I must be missing something vital. But wouldn't the slit itself collapse the wave? As the whole wave washes apon the gap between the two slits & wouldn't the wave of the quanta bounce back like water waves?

[Shaula]

But how is the info erased? Has this been done? If the experimenter found out which slit it goes through & keep it a secret. Then his co worker angrily killed him never knowing. Would this restore the pattern as the particle hits the detector way down the ling?

No, it is rather more involved than that! The wiki page on Quantum Erasers gives some details on how it is done. It also gives some ideas about how you can tell which slit the object went through - imagine if you put polarisers over each slit and measured the polarisation state of each photon hitting the screen. You would know which slit it went through and not see a pattern.

Rather than look at the wavefunction collapsing I rather like the decoherence approach. That said it is just another interpretation of what is going on 'really'. It seems to all boil down to information at some level - if the slit doesn't give away any information about the interactions it has with the particle then no wavefunction collapse/decoherence happens.

[WayneFrancis]

Hi all.

Been reading on the DS experiment & some things still seem cloudy to me.

How & where exactly is the measurement made the determines which slit the electron goes through? I'm under the assumption that a detector gadget or material of some sort is placed in front of one slit & is there to register if it went through the slit. It cant be light or a camera watching it placed near one slit? That precise locating would make it fan out like a wave as it leaves the slit right?

Observed or not, im assuming after leaving the slits the particle instantly go back to behaving like a wave?

I read something like if information about which slit the particle came through is erased before the particle hits the detector the interference pattern is restored. But how is the info erased? Has this been done? If the experimenter found out which slit it goes through & keep it a secret. Then his co worker angrily killed him never knowing. Would this restore the pattern as the particle hits the detector way down the ling?


This bugs me, I must be missing something vital. But wouldn't the slit itself collapse the wave? As the whole wave washes apon the gap between the two slits & wouldn't the wave of the quanta bounce back like water waves?

You can't detect which slit it goes through and get a interference pattern. As soon as you do those that you've detected will not produce a interference pattern but will actually form a pattern like it was a more classical particle. So it isn't that the detection causes the interference pattern. It is the opposite. We like to think of things as one or the other, waves or particles, and many scientists call them "particle waves" but this is us just trying to get our heads around this peculiar effect.

For photons I know of no way to detect a photon mid route and this is the issue. All you can say is where the photon landed and maybe an idea of the possible route it took by some clever timing. IE if you have 2 paths and they differ in length then you can say it went one route and not the other but I'm not sure how you'd apply that to the double slit experiment.

note this is my understanding of the topic and I may be WAY off base.

[BadTrip]

From wiki:

"The delayed-choice experiment and the quantum eraser are sophisticated variations of the double-slit with particle detectors placed not at the slits but elsewhere in the apparatus. The first demonstrates that extracting "which path" information after a particle passes through the slits can seem to retroactively alter its previous behavior at the slits. The second demonstrates that wave behavior can be restored by erasing or otherwise making permanently unavailable the "which path" information."

Two questions:
1) regarding the delayed-choice version of the experiment; Can anyone provide a link or even further explanation of how it's interpreted that the extraction of "which path" information AFTER THE PARTICLE HAS PASSED THROUGH THE SLIT can seem to "RETROACTIVELY" alter it's previous behavior? I'm not arguing, I'm just asking for more info and a deeper understanding
2) So in the quantum erasor version of the experiment... is the issue, the main point, the crux ...that the collected information was never interpreted/reviewed? Again, looking for deeper understanding of this version of the experiment.

Thanks any and all!

EDIT: As I was looking a bit....
"The experiment has two stages: first the experimenter marks through which slit each photon went, without disturbing their movement, and demonstrates that the interference pattern is destroyed. This stage shows that it is the existence of the "which-path" information which causes the destruction of the interference pattern. The second stage goes by erasing the "which-path" information, and demonstrating that the interference pattern is recovered. It does not matter whether the erasure procedure is done before or after the detection of the photons."

................what?! How does one erase the information regarding which path the photon took.........before detecting the photon?

Again...not arguing.... asking for an education.

[StupendousMan]

This does not make sense when interpreted using our everyday reasoning. It's a quantum effect.

If you don't understand it, you're doing it right. If you think you do understand it, you're almost certainly wrong.

Welcome to the non-intuitive world of quantum effects.

[RobA]

Here's a diagram I drew to help describe a Delayed Choice Quantum Ersaser to some friends of mine (hey, with friends like me ..... ) It's based more on this experiment than the previous article, but hope you don't mind if I give a short summary as well (as much as anything, so the knowledgeable folks here can tell me if I've been telling my friends porkies over the years!).
DelayedChoiceQuantumEraser.png

Briefly, the secret's all in the entanglement. The incoming photon passes through the slits, and we then break it into two entangled photons - termed the Signal and the Idler. From the particle view, this means we have one pair of particles - either US and UI (for the Upper slit) or LS and LI.

We take the Signallers to the the normal double-slit detection. Now, since we've constructed this experiment, we know how long everything takes - so we know exactly how long the detector will take to record a result. I marked this with an arrow - so say it takes X ms for the signal to get to the arrow.

The Idlers go to another set of detectors. Again, we know how long these things take, so we can move the detectors so far away that the Idler photons take more than X ms to get to them (or alternatively, since you asked "How does one erase the information regarding which path the photon took.........before detecting the photon", we can arrange to erase before X ms by bringing the detectors closer).

Now, if D1 fires, we know that the original photon must have gone through the upper slit. If D2 fires, it must have gone through the lower. Mapping the output of D0 for these cases gives us no interference pattern.
If D3 or D4 fire, then the which-way information has been erased, and D0 shows the interference pattern.

But given that D1, D2, D3 and D4 are all positioned after D0 would have received and recorded the position of the incoming photon (ie. after X ms), then how can the D0 output be correlated in this fashion???? As StupendousMan says, welcome to the world of Quantum Effects.

[Strange]

Briefly, the secret's all in the entanglement.

That is a very nice summary. (By which I mean, I think I understand it!) Thanks.

I assume the entanglement is important because that maintains the correlation between where and how the photons are detected - even across separation in space (and time?). So if the photons are detected at D1 or D2, this "spookily" correlates with non-interference at D0; if they are detected at D3 or D4 then this correlates with the appearance of the interference pattern. At which point, we have to remember that correlation doe not imply causation.

Is it possible to split each photon into non-entangled pairs (but still use the same setup to determine the path)? If so, would we get different results?

[BadTrip]

"Stupendousman: If you don't understand it, you're doing it right. If you think you do understand it, you're almost certainly wrong."



heh... well I'm happy say that, based upon your advice and review, I'm absolutely doing it right! hehe

[RobA]

Is it possible to split each photon into non-entangled pairs (but still use the same setup to determine the path)? If so, would we get different results?

As soon as you do so (assuming such a thing is possible), D0 would never give an interference pattern. At first, my gut feeling was "well, if you left the original photon untouched, then maybe there'd be no change" - just shows how careful you have to be thinking about this.

Assume some THING generates a non-entangled duplicate photon of the incoming photon, then sends the original to D0 and the duplicate down the idler path. This means the duplicate is carrying which-way information with it (ie. which photon is actually present). The only way this which-way information could be present is for our THING to have performed a measurement of the incoming photon, so destroying the interference pattern.

even across separation in space (and time?)

Bold mine

You've hit the nail on the head with this one - isn't time a fascinating subject ! In my post, I blithely said "Idler photons take more than X ms" - but how much more? A few extra milliseconds? Clearly, any amount of time is possible (eg. just by putting the idlers in some kind of loop and only letting them out when we're ready), and the correlation would be retained. This raises the question: Is this a mechanism we can use to send a message back in time?

So, let's push this :
I want to buy or sell some shares, depending on tomorrow's price So I put the idlers in a loop for a day, resolving to force them to hit D1/D2 if I want to buy, or D3/D4 if I want to sell. Can I read the result of this forcing on my D0 NOW (so if get a bell-curve, I buy, or an interference pattern I sell)?

It's similar to the EPR paradox. The results prove the particles are correlated, but the correlation is only evident by comparing the state of one with the state of the other - ie. we con't use it for FTL message transmission.

Likewise with our buy/sell. All D0 will show today is a big bell curve, so we can't "read the message". You only get the interference patterns (assuming I'm selling) by splitting the D0 results into the 2 charts D0-for-D3 and D0-for-D4. (and yes, this shows that adding 2 small interference patterns results in one big bell curve).

Anyway, I think I'm getting a bit carried away here. If anyone's interested, my "description for friends" was actually a 3-doc Layman-written-crash-course-in-QM tome!! (surprisingly, I still have the friends!) I've dropped it in my dropbox here , and I explore this more in #3 . Be warned, it's a bit (OK, a LOT) gee-whizzy, but hey!

Warning: Even though it's my document and I promise it's clean, I'm just some Rob on the internet.
You should ALWAYS run a virus scan on office docs you download (after unzipping!). There ARE ms-word viruses out there.

[BadTrip]

Thank you RobA for a very nice diagram and a lot of great info to research and review.

First question I need to research... a single photon can be split into two photons?

[BadTrip]

As soon as you do so (assuming such a thing is possible), D0 would never give an interference pattern. At first, my gut feeling was "well, if you left the original photon untouched, then maybe there'd be no change" - just shows how careful you have to be thinking about this.

So the experiment is fundamentally flawed. ?

[Strange]

As soon as you do so (assuming such a thing is possible), D0 would never give an interference pattern.

That was my (uninformed) assumption as well. I assumed it is either impossible to generate non-entangled pairs from a photon or if it is it "breaks" the experimental setup.

Assume some THING generates a non-entangled duplicate photon of the incoming photon, then sends the original to D0 and the duplicate down the idler path. This means the duplicate is carrying which-way information with it (ie. which photon is actually present). The only way this which-way information could be present is for our THING to have performed a measurement of the incoming photon, so destroying the interference pattern.

That makes sense.

[Strange]

First question I need to research... a single photon can be split into two photons?

Yes, by means of "spontaneous parametric down conversion", which sounds like something from Star Trek.
http://en.wikipedia.org/wiki/Spontan...own_conversion

So the experiment is fundamentally flawed. ?

Not really; that comment was a response to my "non-entangled photon splitting" question. That creates a different experiment (which you could say is fundamentally flawed as it doesn't demonstrate the desired effect!)

[BadTrip]

Thanks for that link Strange. From that wiki link:
"A nonlinear crystal is used to split photons into pairs of photons that, in accordance with the law of conservation of energy, have combined energies and momenta equal to the energy and momentum of the original photon"

So, just in a general sense, we'd anticipate each photon to have ~1/2 the energy and momenta of the original?

[Strange]

So, just in a general sense, we'd anticipate each photon to have ~1/2 the energy and momenta of the original?

Yes, my understanding is that each photon has half the frequency (i.e. energy) and the sum of their momentum equals the original (that's vector sum, obviously, as they are heading off in different directions).

[BadTrip]

Yes, my understanding is that each photon has half the frequency (i.e. energy) and the sum of their momentum equals the original (that's vector sum, obviously, as they are heading off in different directions).

Thank you Strange.....

So these are junior photons then.

I'm sorry but if the intent is to not interact with the subject of the experiment, it would seem that we've failed before the experiment has even been initiated.

[a1call]

Some key information seems to be overlooked in this thread.
From Strange's link:

SPDC is stimulated by random vacuum fluctuations, and hence the photon pairs are created at random times. The conversion efficiency is very low, on the order of 1 pair per every 10^12 incoming photons[2].

What this means is that only a very small percentage of the split photons are entangled and that the majority are not.

What this indicates is that the detectors in the diagram have to be electronic detectors rather than photographic films in order to filter out the un-entangled portion of photon triggers.

What this means is that the setup can not be used to do FTL information transmission, because signals from all detectors will have to be processed in a central processor to perform the entangled-photons filtering. If all photons were entangled after passing through the crystal, then no filtering (in a central unit) would have been required and the interference patterns could have been instantly created/destroyed light years away at D0 by simply switching on/off D1-D2 / D3-D4 pairs.

[RobA]

I'm sorry but if the intent is to not interact with the subject of the experiment, it would seem that we've failed before the experiment has even been initiated.

Not really. The aim of the double-slit experiments is to investigate the behaviour of the superposition. By getting a couple of particles to exist in that one superposition means that we're still preserving the thing that we want to test.

Going back to some of the questions from your OP:

How & where exactly is the measurement made the determines which slit the electron goes through? ... It cant be light or a camera watching it placed near one slit?

Sure it can be light. You can imagine the light bouncing off the particle into a camera, that sounds a big horn when it spots the particle coming its way- that's one big measurement ! On the other hand, a single photon would probably not constitute a measurement - it would join the superposition as we discussed above. So how much light (or how many photons) would constitute a measurement? That's the big question about at where's the boundary between our classical world and the quantum effects.

Observed or not, im assuming after leaving the slits the particle instantly go back to behaving like a wave?

Yes.

If the experimenter found out which slit it goes through & keep it a secret. Then his co worker angrily killed him never knowing. Would this restore the pattern as the particle hits the detector way down the ling?

My guess is no, the original researcher made a measurement, but really it comes down to your views of Schrodinger's cat, and Einstein's question "Does the moon exist if nobody is looking at it?".

[RobA]

What this means is that only a very small percentage of the split photons are entangled and that the majority are not.

Yes, that's what the coincidence counters (that I dodn't put on my diagram) are for, so we only look at cases where the entanglement was successful (ie. we get both the Signal and the Idler photons).

What this means is that the setup can not be used to do FTL information transmission, because signals from all detectors will have to be processed in a central processor to perform the entangled-photons filtering. If all photons were entangled after passing through the crystal, then no filtering (in a central unit) would have been required and the interference patterns could have been instantly created/destroyed light years away at D0 by simply switching on/off D1-D2 / D3-D4 pairs.

No, the problem with FTL transmission is not due to the noise factor. Reading a message FTL would be impossible even if we had a 100% successful method of generating entangled particles. Reread my post #9 - the scenario I describe there is the same as you're suggesting. Basically, any FTL entanglement effects - this, or EPR - only become evident when bringing together the information from the pairs of particles involved.

It's sweet how nature arranges this - very frustrating, but sweet !

[Strange]

What this means is that only a very small percentage of the split photons are entangled and that the majority are not.

I read that as meaning that a small proportion are split (but that all split photons are entangled).

What this indicates is that the detectors in the diagram have to be electronic detectors rather than photographic films in order to filter out the un-entangled portion of photon triggers

And because we are dealing with single photon events that need to be correlated between detectors.

[a1call]

I read that as meaning that a small proportion are split (but that all split photons are entangled).

It looks like you read right:
http://www.davidjarvis.ca/entangleme...nglement.shtml

From what I understand the crystal will direct each photon as a whole in one of the two directions with one of the two polarization states similar to a double refraction. But on rare occasions a photon will split into two entangled photons with oposing polarizations and sent in two different directions.

This would still require filtering out the noise aka un split photons.
Corrections are welcome.

The same site has a neat eraser section:

http://www.davidjarvis.ca/entanglement/spookiness.shtml

[a1call]

Yes, that's what the coincidence counters (that I dodn't put on my diagram) are for, so we only look at cases where the entanglement was successful (ie. we get both the Signal and the Idler photons).


No, the problem with FTL transmission is not due to the noise factor. Reading a message FTL would be impossible even if we had a 100% successful method of generating entangled particles. Reread my post #9 - the scenario I describe there is the same as you're suggesting. Basically, any FTL entanglement effects - this, or EPR - only become evident when bringing together the information from the pairs of particles involved.

It's sweet how nature arranges this - very frustrating, but sweet !

I read post #9 but not the zipped docs. I do not understand. If all the photons are entangled, then there is no noise. Then there is no need for a coencidence counter. Then there is no need for comparing the pairs. Then the interference pattern could be light years away from the D1-D4 detectors. Then if the pattern would be dependent on which-way info obtained at the D1-D4 detectors, then what will prevent us from altering the state of the pattern at D0, instantly and light years away?

ETA:
As always corrections are appreciated.
The way I understand it, the purpose of he coencidence counter is to filter out the un-entangled photons by only registering hits if photons are present at both D0 and D1-D4 sets. This is why hits at both sets need to be compared. The necessity of the coencidence counter is the only obstacle I can see preventing us from having FTL communication. If there are any other obstacles a clarification is greatly appreciated.

[RobA]

If all the photons are entangled, then there is no noise. Then there is no need for a coencidence counter.

That's right, that's all the coincidence counters do - filter out the noise from the unsplit photons. So we can ignore all that when discussing the experiment, and just concentrate on the cases where we got entanglement.

Then if the pattern would be dependent on which-way info obtained at the D1-D4 detectors, then what will prevent us from altering the state of the pattern at D0, instantly and light years away?

Nature is very subtle here, as with the EPR. Let's say we're monitoring D0, and our partner wants to send us a message by firing 100 photons (with 100% success in entanglement), and forcing into D1/D2 or D3/D4. Notice that he can only send them into a pair of detectors; he can't control into an individual one.

So, from 100 photons, around 50 will go into one detector, the other 50 into the other. When subsequently correlated with D0, the photons going into D1 form a 50-photon bell curve, and the photons from D3 form a 50-photon interference pattern (and likewise for D2 and D4). Now, the sum of two complementary 50-photon bell curves is as 100 photon bell curve. It turns out that the sum of two complementary 50-photon interference patterns is not a 100-photon interference pattern, but a 100 photon bell curve as well !

So when we're monitoring D0, we get the sum of the patterns from the two detectors. Whichever way our partner organised it, we get a 100-photon bell curve ! It's only when our partner sends us a message saying "Photons 1, 3, 4, 5, 7, ... went into one detector, the rest went into the other" that we can split our bell curve into the two charts and see the interference pattern (if D3/D4).

Neat, eh!

[a1call]

Hi RobA,

I appreciate your patience.
With EPR, the FTL communication fails because the outcome of Alice's measurements are random and thus Bob can not determine if she has performed a measurement or not by performing a measurement of his own (which are also going to be random either way) :

It turns out that the usual rules for combining quantum mechanical and classical descriptions violate the principle of locality without violating causality. Causality is preserved because there is no way for Alice to transmit messages (i.e. information) to Bob by manipulating her measurement axis. Whichever axis she uses, she has a 50% probability of obtaining "+" and 50% probability of obtaining "-", completely at random; according to quantum mechanics, it is fundamentally impossible for her to influence what result she gets. Furthermore, Bob is only able to perform his measurement once: there is a fundamental property of quantum mechanics, known as the "no cloning theorem", which makes it impossible for him to make a million copies of the electron he receives, perform a spin measurement on each, and look at the statistical distribution of the results. Therefore, in the one measurement he is allowed to make, there is a 50% probability of getting "+" and 50% of getting "-", regardless of whether or not his axis is aligned with Alice's.

However here (assuming 100% entanglment) if Alice turns Off D3/D4 leaving D1/D2 On (by replacing the 1st set of beam splitters by mirrors) then she will know exactly which slit each of the 100 photons has traveled through. This will destroy the fringe pattern at D0 where Bob is sitting in observance. He will know for sure that Alice has left only D1/D2 on. Likewise if Alice turns Off D1/D2 leaving D3/D4 On (by removing the 1st set of beam splitters) then she will only know that a photon has arrived but she will not know which slit each photon has travelled through. This will preserve the wave function and the entangled photons will form fringes at D0. Bob upon observing the fringe lines will know D1/D2 are off. The only obstacle in reality preventing this from being applied light years apart is the necessity of the coincidence counter which would require the filtering out of the noise at a central point thus requiring below-light-speed/light-speed information transmission to-and-from such central point for interpretation.

[Jeff Root]

For the original poster, Sam99, regarding the original
question:

It isn't clear to me whether you understand this key fact:
When a photon is detected, it is absorbed. A photon can
be detected at the screen where the interference pattern
shows up. That is, a photon detector can be used as the
screen. Placing the detector there means that you can't
tell which slit the photons went through. If a detector is
placed closer to the slits, it will absorb photons there so
that they can't reach the screen, and you won't get an
interference pattern.

-- Jeff, in Minneapolis

[RobA]

Hi RobA, I appreciate your patience.

My pleasure - I love exploring all this

With EPR, the important part of the quote is "there is no way for Alice to transmit messages (i.e. information) to Bob by manipulating her measurement axis. ". Basically, we only get "read-only" access to the particles. Alice can't set her particle to "+" and so force Bob get "-". The same is true here. Yes, Alice can force the particle into one pair of detectors, but she can't force it into a specific one. This means while Bob is monitoring D0, he'll only see the SUM of the results of the pair of detectors that Alice used.

So, let's set Bob a challenge. Alica and Charlie decide which pair of detectors they'll use, and Bob (monitoring D0) has to decide who is using which.

On Monday, Alice goes first. She's picked D1/D2 (which-way is known), and fires her 100 particles. Bob receives a 100-particle bell curve on D0.
On Tuesday, Charlie fires 100 particles at D3/D4 (which-way is not known). Bob still receives a 100-particle bell curve ! (Remember I mentioned earlier that "It turns out that the sum of two complementary 50-photon interference patterns is not a 100-photon interference pattern, but a 100 photon bell curve as well !")

ETA: This is also expected, since looking at my diagram in #6, when the D0 measurement is made, the Idlers are still intransit - ie. which way information is still potentially available since it has not yet been erased. Therefore, D0 cannot be displaying an intereference pattern at this time.

So, Bob has now received 100 particles from each - two identical 100-particle bell curves. He cannot tell which sender used which detectors, so this mechanism cannot be used for FTL transmission of information .

So where do the intererence fringes come in? They're the output from the individual detectors, not the sum of the pair. That's the next step of the story.

On Wednesday, Charlie sends his log "Photon 1 triggered D3, 2 triggered D3, 3 triggered D4", etc. Note that, as with EPR, this is a "read-only" measurement, that cannot be forced or decided beforehand. When Bob generates the plots of the D0 positions for each detector, he'll get two interference patterns. If Alice sends her log, Bob's two plots (for D1 and D2) will show two bell curves.

So, as with EPR, we find Nature operates FTL, but keeps the mechanism hidden so we can't use it!

[a1call]

On Tuesday, Charlie fires 100 particles at D3/D4 (which-way is not known). Bob still receives a 100-particle bell curve ! (Remember I mentioned earlier that "It turns out that the sum of two complementary 50-photon interference patterns is not a 100-photon interference pattern, but a 100 photon bell curve as well !")

It all boils down to this part.
I was not aware of this. I still don't see why this would happen.
Regardless we could eliminate all detectors on Alice/Charlie end altogether. Now interference patterns will have to form at D0, because that's what light does passing through double slits (entanglement withstanding or not). The only way to destroy the pattern at D0 is to utilize D1/D2 using mirrors instead beam splitters(or to simplify just put the sensors directly where the beam splitters are right now in the diagram.
This way by not having any detectors on Alice/Charlie end versus having detectors which definitively provide which-way info we can manipulate the formation of interference fringes. Assuming 100% entanglement and thus no need for the coencidence counter will provide means of FTL commiunications.

[Bob Angstrom]

Alice can't set her particle to "+" and so force Bob get "-".

Anton Zeilinger “Mr. Beam” (as in beam me up) has demonstrated a way around this. Alice has an entangled particle in a box that can contain two particles if, and only if, they are in opposite quantum states + and -. Alice then takes a – particle and puts it into the box forcing her entangled particle into a + state without ever observing it. This collapses the entanglement forcing Bob's particle to a – state. Alice puts a – particle in and Bob gets a – particle out “quantum teleportation.“ There is a transmission glitch here in that Alice's observation of the particle and putting it in the box are simultaneous. She can't observe the state of a particle and then decide whether or not to put it in the box.

I agree that we can't use entanglement for instant communication because entanglement necessarily extends both forward and backward in time if it is to “instantly” connect two remote particles. If Alice and Bob are 300,000 km apart and Alice sends a message to Bob, the message arrives instantly which is one second forward in time relative to Alice. If Bob instantly sends a message in return, the message travels the opposite direction which is one second forward in time relative to Bob. So the return message arrives at Alice's location two seconds forward in time. She then has to wait two seconds to read an “instant” message and that defeats the purpose of “instant” communication. An outside observer can only tell if a communication is instant via entanglement or by classical means if he can observe the photon in flight.

[RobA]

I was not aware of this. I still don't see why this would happen.
Regardless we could eliminate all detectors on Alice/Charlie end altogether. Now interference patterns will have to form at D0, because that's what light does passing through double slits (entanglement withstanding or not).

Here's an image of how it happens - note the peaks of one align with the troughs of the other. You must remember that this isn't adding two random interference patterns, but explicitly just for two derived from complementary sources.
SumInterference.png

It is inevitable that Bob gets the bell curve, since when the D0 measurements are made, the Idler photons are stil in flight. A superposition is a system that is comprised of every particle involved. So, if which-way information is even potentially available (ie. measurements can still be made on any particle in the superposition to determine which-way info), then that is enough to destroy the system's interference pattern. Once which-way information has been created, it must be explicitly erased to get the interference pattern back. So Entanglement does withstand!

[RobA]

Anton Zeilinger “Mr. Beam” (as in beam me up) has demonstrated a way around this.

That's really interesting. Offhand, I can't see how this would work - I would have imagined that the attempt to put the - particle in the box would have force a measurement of the particle already in the box, in order to decide whether to allow the new particle in or not. Would you have a link, or remember the name of the experiment?

I agree that we can't use entanglement for instant communication because entanglement necessarily extends both forward and backward in time if it is to “instantly” connect two remote particles.

Yes, and that was another big reason Einstein didn't like QM. We all get the quotes about his views of the probabilities and "God doesn't play dice", but IMHO the apparent conflict between Entanglement and FTL would have cut much deeper.