Faster-than-light communication

[Albert V]

Many seem to be skeptical whether faster-than-light communication is possible. Consider the experiments performed by Alain Aspect where the polarization of one of the entangeled photons force the other to collapse into the opposite axis.

http://en.wikipedia.org/wiki/Alain_Aspect

The Wikipedia article says that the experiment do not demonstrate faster than light communicaiton and I wonder why.

If one of the polarization filters is switched from axis to axis, and the other filter kept steady, it should according to my logic, be possible to transfer information. What am I missing?

 

[Bob_for_short]

Learning information is different from transmitting.

 

[Fredrik]

If one of the polarization filters is switched from axis to axis, and the other filter kept steady, it should according to my logic, be possible to transfer information.

How? Suppose e.g. that you flip a coin and want to send a 1-bit message (1="heads", 0="tails") to let someone at the other end know what the result was. How would you do this?

 

[Albert V]

The information is sent like this:

Entangeled photons are created with a frequency that is known to the receiver. By keeping his filter steady he would have a "click" in his detector when a photon passes his filter, and a delay when the photon is stopped by the filter. If the experiment works like this it is possible to send a binary string of information. However, the problem seems to be that if the filters are adjusted to the same axis it will be random whether the photon is stopped by your filter, or the filter at the receiver. And I think it will be the same in a situation were you are located much closer to the photon transmitter (anhilation point) than the receiver.

Am I correct?

 

[Albert V]

Ok, new idea:

By carrying out a Bell test experiment we will observe the following:

1) When the polarisers are oriented along opposite axis we will get clicks in both detectors

2) When the polarisers are oriented along the same axis we have 50% chance of getting a hit in the detectors at both sites.

Now we use 20 photons to transmit 1 bit of information. Data that gives us 100 % hits, versus data that gives us only 50 % hits.



Is this faster-than-light communication or can somebody tell me where I go wrong?

Best
Albert

 

[DrChinese]

Ok, new idea:

By carrying out a Bell test experiment we will observe the following:

1) When the polarisers are oriented along opposite axis we will get clicks in both detectors

2) When the polarisers are oriented along the same axis we have 50% chance of getting a hit in the detectors at both sites.

Now we use 20 photons to transmit 1 bit of information. Data that gives us 100 % hits, versus data that gives us only 50 % hits.



Is this faster-than-light communication or can somebody tell me where I go wrong?

Best
Albert

There are always clicks at the detectors, assuming you put a polarizing beam splitter in front of a pair of detectors. Each photon is randomly polarized. They wouldn't be polarization entangled otherwise. As previously mentioned, entangled photons cannot be used to transmit information.

 

[Count Iblis]

You could try to create two entangled observers. A large quantum computer implements an observer. We take two such quantum computers and intitialize them in some mutually entangled state of the form

Sum over_j A_j |j>|j>

where |j> denotes some bitstring in a |0>, |1> basis which corresponds to some classical brain state of the observer and state of the virtual world the observer finds himself in. The quantum algorithm is such that bistrings in the |0>, |1> basis evolve deterministically. So, we assume that a classical determinisic algorithm can represent the mind of a real person.

Suppose that the two entangled quantum computer pairs are separated by a large distance. Then if one observer makes what is in his mind a random choice for some numbers, he knows that his copy will have made exactly the same decision.

 

[Albert V]

There are always clicks at the detectors, assuming you put a polarizing beam splitter in front of a pair of detectors. Each photon is randomly polarized. They wouldn't be polarization entangled otherwise. As previously mentioned, entangled photons cannot be used to transmit information.

Yes, this is also how I understand the system. I have taken this fact into account. Therefore one and one photon can not be used to transmit information. But let us start from the beginning with all my assumptions.

1) A continuous source of photons emitted at an even frequence

2) When the polarisers are oriented along opposite axis we will get clicks in both detectors (the receiver will therefor obtain clicks according to the given frequence. Let us say that 10 photons is enough data for him to conclude that the filter at the transmitter is oriented along the opposite axis)

3) When the polarisers are oriented along the same axis we have 50% chance of getting a hit in the detectors at both sites. In this situation the receiver will have clicks but also delays regarding the agreed frequence. Statistically he will have 50% less clicks than in the former situation. 10 photons gives him enough data to conlude that the filter at the transmitter is oriented along the same axis.


Sorry that I have to be spoonfed, but I need to know how these systems behave.

Best
Albert

 

[DrChinese]

Yes, this is also how I understand the system. I have taken this fact into account. Therefore one and one photon can not be used to transmit information. But let us start from the beginning with all my assumptions.

1) A continuous source of photons emitted at an even frequence

2) When the polarisers are oriented along opposite axis we will get clicks in both detectors (the receiver will therefor obtain clicks according to the given frequence. Let us say that 10 photons is enough data for him to conclude that the filter at the transmitter is oriented along the opposite axis)

3) When the polarisers are oriented along the same axis we have 50% chance of getting a hit in the detectors at both sites. In this situation the receiver will have clicks but also delays regarding the agreed frequence. Statistically he will have 50% less clicks than in the former situation. 10 photons gives him enough data to conlude that the filter at the transmitter is oriented along the same axis.


Sorry that I have to be spoonfed, but I need to know how these systems behave.

Best
Albert

I think you may have a slight misunderstanding of what the actual setup produces in the way of results.

There are 2 observers, Alice and Bob. They get a stream of photons arriving as you say, perhaps 10. They EACH have a polarizing beamsplitter and 2 detectors. When Alice gets a dectector hit on the first detector, she marks it a +. The other detector registers as a -. Ditto for Bob.

Alice can vary her beamsplitter orientation so she can "send" a message. (Not really of course as we will find out.) Bob holds his beamsplitter fixed and steady so he can receive a "message". The beamsplitter separates the photons to one of the 2 detectors as mentioned, and he records a + or a - according to which of his 2 detectors fires.

Now what does Bob see when he records his results? He see a random series of + and - results, something like this:

++-+-+---+

Since the "message" is random, no information is received. It doesn't matter what Alice does, Bob's message is totally random by any measure. That is because the source of the entangled photons is random itself. The pattern that indicates there is entanglement is ONLY evident when Alice AND Bob's results are correlated.

 

[Albert V]

Thanks Dr. Chinese for spending time one these questions.

I am a little confused by your beamsplitters because I am not familiar with how they work. I had a single channel Bell test in mind. And I use polarisers since they are more easy to understand.

After anihilation of a positron and an electron, two photons are generated with opposite directions and planes. The planes are not determined, but will be after one of the photons passes a polariser. Opposite orientation of the polarisers will always allow BOTH the photons to pass and same orientation will only allow one of the photons to pass.

Am I correct this far?

 

[DrChinese]

Thanks Dr. Chinese for spending time one these questions.

I am a little confused by your beamsplitters because I am not familiar with how they work. I had a single channel Bell test in mind. And I use polarisers since they are more easy to understand.

After anihilation of a positron and an electron, two photons are generated with opposite directions and planes. The planes are not determined, but will be after one of the photons passes a polariser. Opposite orientation of the polarisers will always allow BOTH the photons to pass and same orientation will only allow one of the photons to pass.

Am I correct this far?

Yes, for this setup, and notice that coincidence matching (CM) is necessary. That does not allow an opportunity for Alice to send a message to Bob. Alice must first send her results to be matched with Bob's. That occurs at light speed. So no FTL possible.

 

[Albert V]

Yes, for this setup, --

Hence, when the polarisers are oriented along opposite axis, Bob gets a message like this; ++++++++++, but when the polarisers are oriented at the same axis Bob will get a message like this; +delay++delay+delaydelay++ (where + signifies clicks at the preagreed frequence of photons)?

Am I also correct this far?

 

[DrChinese]

Yes, for this setup, --

Hence, when the polarisers are oriented along opposite axis, Bob gets a message like this; ++++++++++, but when the polarisers are oriented at the same axis Bob will get a message like this; +delay++delay+delaydelay++ (where + signifies clicks at the preagreed frequence of photons)?

Am I also correct this far?

No, Bob ALWAYS gets a "message" (in your terms) like:

+delay++delay+delaydelay++

Meanwhile, Alice gets a somewhat similar "message" depending on her polarizer orientation. At some orientations it will be the same message, at some it will be opposite, and at others a mixture.

 

[DrChinese]

Keep in mind that Alice is located one place, and Bob is located at another. Each sees one side (one half) of the stream of entangled photon pairs. Assuming they were equidistant from the source, they would receive photons going into their respective polarizer simultaneously (within a narrow window). Since those photon pairs have random (but opposite) polarization, both Alice and Bob see random streams. When the streams are compared at a later time, and you know the respective polarizer orientations, you can detect that the source pairs were (or were not) entangled.

 

[jtbell]

Two key points affect the results of an attempt to send a signal via entangled photons:

1. The setting on Alice's polarizer does not affect the result that Bob gets, for any individual photon. And vice versa, swapping Alice and Bob.

2. Also, neither Alice nor Bob can control the sequence of states in which the photons are produced. They can only control their polarizers.

First suppose that Alice and Bob both keep their polarizers horizontal. Each column indicates one photon pair. H and V indicate the polarizer settings, + and - indicate whether a photon got through that polarizer. They might get something like this. I generated Alice's sequence of + and - by flipping a coin thirty times.

Alice: HHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
       ++-++++-+-++-+-+-----++-+---++
Bob:   HHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
       ++-++++-+-++-+-+-----++-+---++

Now suppose that instead, Alice had kept her polarizer vertical, and Bob had kept his horizontal. For the same photons they would have gotten the following results

Alice: VVVVVVVVVVVVVVVVVVVVVVVVVVVVVV
       --+----+-+--+-+-+++++--+-+++--
Bob:   HHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
       ++-++++-+-++-+-+-----++-+---++

If Alice had alternated between horizontal and vertical polarizations, in "chunks" of five photons at a time, with Bob keeping his horizontal the whole time:

Alice: HHHHHVVVVVHHHHHVVVVVHHHHHVVVVV
       --+----+-+--+-+-+++++--+-+++--
Bob:   HHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
       ++-++++-+-++-+-+-----++-+---++

How does Bob extract Alice's "signal" (HHHHHVVVVVHHHHHVVVVVHHHHHVVVVV) from his data, when they cannot actually repeat the experiment with the same sequence of photons? Each time they repeat the experiment, they get a different sequence of photon polarization states, at random, like me flipping my coin another thirty times.

 

[StandardsGuy]

DrChinese & jtbell,

Isn't this thread an example of what you just told me in another thread was prohibited? It clearly is opposed to mainsteam theory. From your guidlines:

"It is against our Posting Guidelines to discuss, in most of the PF forums or in blogs, new or non-mainstream theories or ideas that have not been published in professional peer-reviewed journals or are not part of current professional mainstream scientific discussion."

And yet you made no objections and participated in it. Why?

 

[Albert V]

Thank you both for being patient with me

This is the point where I don't follow


I said:
Opposite orientation of the polarisers will always allow BOTH photons to pass and same orientation will only allow one of the photons to pass.

DrChinese said:
Yes

If Bob keeps his filter horizontal and Alice keeps her filter vertical, why does not the frequency of photons passing the polarizer go up compared to the situation where Bob and Alice both keep their polarizers at a, let's say, vertical position?

Albert

 

[Albert V]

I see now that these systems did not behave as I had understood. I thought that the wave-function collapsed into a certain plane when the photons hit a polarizer, well perhaps it does, but the plane can still be different from the orientation of the polarizer.

The measurer is unable to feed information into the system.

But what is the probability of passing photons. There are many more planes than horisontal and vertical.

I don't see why Einstein thought this was spooky. The planes of the photons are opposite, and that's it.

Thanks for helping me

 

[jtbell]

There are many more planes than horisontal and vertical.

Indeed, and the cases where the polarizers are not at right angles to each other, are the really "spooky" ones! In these cases you cannot explain the experimentally observed probabilities of photons passing through the polarizers (more precisely the correlations between pairs of entangled photons) by using a "natural" simple model that explains the case where the polarizers are at right angles.

Dr. Chinese's Web site has a lot of useful information and links to original papers related to entanglement and Bell's Theorem. For simplified examples that illustrate the "spooky" correlations see particularly

Bell's Theorem with Easy Math (Dr. Chinese's own example)

http://www.iafe.uba.ar/e2e/phys230/history/moon.pdf from Physics Today magazine (April 1985).

 

[DrChinese]

I don't see why Einstein thought this was spooky. The planes of the photons are opposite, and that's it.

Thanks for helping me

Yes, it is true that when you talk about opposite or parallel orientations, then there does not seem to be anything weird happening. Einstein essentially said that in such cases, the results must in fact be predetermined. This is often referred to as the Bertlmann's socks anaology (you can look it up if you like).

But that is not the tricky situation. Bell discovered that at other specific angles, the quantum statistics were incompatible with the idea of predetermination. Experiments have since shown that quantum statistics are correct, and the idea of predetermination cannot be.

So in the end, and to use Einstein's words: either there is spooky action at a distance or the moon is not there when no one is looking.

 

[DrChinese]

Dr. Chinese's Web site has a lot of useful information and links to original papers related to entanglement and Bell's Theorem...

Thanks! And I have recently been expanding some of the references so that more advanced & related areas - such as GHZ, Leggett and Hardy - are also covered. For those not otherwise familiar with them: these theorems are separate but similar in conclusion to Bell. I.e. basically that realistic or local realistic theorems cannot hold.

 

[Albert V]

Hi folks, I'm back with another idea. This would not be of any practical use, but it is interesting to discuss whether FTL is possible in principle.

We use a similar experimental setup:View attachment FTL.pdf

The principle for sending information will be the two states superposition (0) and polarized (1). Alice is the sender of the information and Bob function as a receiver.

In superposition it is possible to see a single photon interference on the screen (I don't know exactly how many photons that must hit the screen in order to produce an interference pattern, but it should be a limited amount - we assume that Alice and Bob know exactly how many in order for Bob to see a pattern or not. After this preeagreed number Bob resets the screen).

If Alice introduces a polarizer Bob will not see the interference pattern since his photon will be in a polarized stage and will not be able to interfere with itself, but choses the left or the right side of the wire (depending upon vertical or horisontal polarization) .

I am not sure what happens if Alice does not get a click in her detector. I assume that the photon still has taken on a value, and that the entangled pair is not longer in its superposition? But still, the presence of a polarizer should in the long run give another pattern on the screen than when not present.

By introducing and removing the polarizer, Alice should be able to send Bob a message.

 

[DrChinese]

Hi folks, I'm back with another idea. This would not be of any practical use, but it is interesting to discuss whether FTL is possible in principle.

We use a similar experimental setup:View attachment 19808

The principle for sending information will be the two states superposition (0) and polarized (1). Alice is the sender of the information and Bob function as a receiver.

In superposition it is possible to see a single photon interference on the screen (I don't know exactly how many photons that must hit the screen in order to produce an interference pattern, but it should be a limited amount - we assume that Alice and Bob know exactly how many in order for Bob to see a pattern or not. After this preeagreed number Bob resets the screen).

If Alice introduces a polarizer Bob will not see the interference pattern since his photon will be in a polarized stage and will not be able to interfere with itself, but choses the left or the right side of the wire (depending upon vertical or horisontal polarization) .

I am not sure what happens if Alice does not get a click in her detector. I assume that the photon still has taken on a value, and that the entangled pair is not longer in its superposition? But still, the presence of a polarizer should in the long run give another pattern on the screen than when not present.

By introducing and removing the polarizer, Alice should be able to send Bob a message.

Great idea. (I came up with that one a few years ago myself before someone here shot me down.) The problem is that you don't get an interference pattern on Bob's side... ever. Entangled photons do not exhibit the kind of interference pattern you describe. This is a specific physical and detectable difference between entangled photons and "normal" photons.

You can see this in an enlightening article by Anton Zeilinger, p. 290, Figure 2.

Experiment and the foundations of quantum physics

"FIG. 2. A source emits pairs of particles with total zero momentum.
Particle 1 is either emitted into beams a or a' and
particle 2 into beams b or b' with perfect correlations between
a and b and a' and b', respectively. The beams of particle 1
then pass a double-slit assembly. Because of the perfect correlation
between the two particles, particle 2 can serve to find
out which slit particle 1 passed and therefore no interference
pattern arises."

Now, don't confuse the fact that you can get interference in SUBSETs of photons on Bob's side with being able to send a message. There are so-called "quantum eraser" experiments in which it superficially appears to show interference on Bob's side. And it does, but only in correlated subsets. I.e. you must send data from Alice's side to Bob so the subset can be correlated. Since that information must be sent no faster than light speed, you are back where you started.

 

[Albert V]

I think that Zeilinger is not clear on this point:

When does our knowledge of the system interact?

Is it that the knowledge is "out there" in principle, but not yet obtained - or is it that we actually obtain the knowledge by performing a certain experimental setup?

 

[Demystifier]

Great idea. (I came up with that one a few years ago myself before someone here shot me down.) The problem is that you don't get an interference pattern on Bob's side... ever. Entangled photons do not exhibit the kind of interference pattern you describe. This is a specific physical and detectable difference between entangled photons and "normal" photons.

You can see this in an enlightening article by Anton Zeilinger, p. 290, Figure 2.

Experiment and the foundations of quantum physics

"FIG. 2. A source emits pairs of particles with total zero momentum.
Particle 1 is either emitted into beams a or a' and
particle 2 into beams b or b' with perfect correlations between
a and b and a' and b', respectively. The beams of particle 1
then pass a double-slit assembly. Because of the perfect correlation
between the two particles, particle 2 can serve to find
out which slit particle 1 passed and therefore no interference
pattern arises."

Now, don't confuse the fact that you can get interference in SUBSETs of photons on Bob's side with being able to send a message. There are so-called "quantum eraser" experiments in which it superficially appears to show interference on Bob's side. And it does, but only in correlated subsets. I.e. you must send data from Alice's side to Bob so the subset can be correlated. Since that information must be sent no faster than light speed, you are back where you started.

It's remarkable that I also had the same idea some time ago, and that I was inspired by the same Zeilinger paper above (which, as Albert also noted, is somewhat vague at some parts.) I needed few days to find out the mistake.

 

[Albert V]

Let us try again with an eraser. Now I have no which path information. Do you mean, DrChinese, that the system still refuses to produce fringes on the screen?

View attachment FTL2.pdf

 

[DrChinese]

Quantum eraser version

Let us try again with an eraser. Now I have no which path information. Do you mean, DrChinese, that the system still refuses to produce fringes on the screen?

View attachment 19819

OK: we agree that IF we choose to erase or not erase, and that conditions what we see at Bob (interference pattern or not), THEN we could send an FTL message.

So you have to believe as there is a Nobel for the person who figures that out, and Zeilinger would like one of those, so probably this setup won't serve for FTL. He has probably tried it once or twice just to be sure.

So now the question is: why not?

The answer is that when there is erasure of which path, there is an interference pattern... but that is a subset you cannot control or otherwise influence. How does Bob know which subset to use? He asks Alice. Of course, that doesn't get you very far because that communication must be performed at light speed or less.

Bob's pattern consists of 2 random subsets: one in which there is erasure and interference, and another in which there is no erasure and "anti-interference". The sum of the 2 subsets is a non-interference pattern. You can look at this paper for more information:

S. P. Walborn, M. O. Terra Cunha, S. Padua, C. H. Monken (2001): A double slit quantum eraser

It takes some time to get it, so don't be surprised if you can back and forth on it a while. You have to match the captions to the graphs, this version may be easier to see:

http://grad.physics.sunysb.edu/~amarch/Walborn.pdf

or this:

http://grad.physics.sunysb.edu/~amarch/

Good luck!

 

[Albert V]

OK: we agree that IF we choose to erase or not erase, and that conditions what we see at Bob (interference pattern or not), THEN we could send an FTL message.

Well this is actually not my suggestion. I say we keep the eraser and introduce our polarizer in front of each detector again. The message is still sent by introducing and removing the polarizers. We just had to introduce the eraser to get rid of the which path information in order to get our inteference pattern back - but again I don't know whether the system will produce a pattern or not.

I guess that removing the eraser is equivalent.

And maybe Zeilinger has tried it, but doesn't want to tell us

 

[Demystifier]

Let us try again with an eraser. Now I have no which path information. Do you mean, DrChinese, that the system still refuses to produce fringes on the screen?

View attachment 19819

That is true. Particles on one side do not produce interference fringes, no matter what you do with particles on the other side. The interference fringes that can be seen are not produced by particles on one side. Instead, they are a property of the correlations between particles on BOTH sides. But you cannot observe these correlations without conventional subluminal communication between the two sides.

 

[Albert V]

I agree - it is actually easier to understand when you look at the math. It is the beamsplitter that actually filter out half of the photons by rotating them 90 degrees.

http://www.flownet.com/ron/QM.pdf

 

[IMP]

There are 2 observers, Alice and Bob. They get a stream of photons arriving as you say, perhaps 10. They EACH have a polarizing beamsplitter and 2 detectors. When Alice gets a dectector hit on the first detector, she marks it a +. The other detector registers as a -. Ditto for Bob.

Alice can vary her beamsplitter orientation so she can "send" a message. (Not really of course as we will find out.) Bob holds his beamsplitter fixed and steady so he can receive a "message". The beamsplitter separates the photons to one of the 2 detectors as mentioned, and he records a + or a - according to which of his 2 detectors fires.

Now what does Bob see when he records his results? He see a random series of + and - results, something like this:

++-+-+---+

Since the "message" is random, no information is received. It doesn't matter what Alice does, Bob's message is totally random by any measure. That is because the source of the entangled photons is random itself. The pattern that indicates there is entanglement is ONLY evident when Alice AND Bob's results are correlated.

But the results Bob gets could be thought of as 1101010001 (++-+-+---+ as you stated). Just because Bob cannot confirm the string until he and Alice get together does not mean the string is random. Correct?

 

[DrChinese]

But the results Bob gets could be thought of as 1101010001 (++-+-+---+ as you stated). Just because Bob cannot confirm the string until he and Alice get together does not mean the string is random. Correct?

Recall that polarization entangled photons always have a random orientation. Therefore the string Bob sees is ALWAYS random. Ditto for Alice. But when you correlate the results, you can estimate the angle between their polarizer settings by using Malus.

 

[Albert V]

A question just for fun:

It seems that it is the beamsplitter in the quantum eraser that ruins our FTL communication and Nobel. What if we throw out the beamsplitter and insert a black box containing a device that inserts a mirror by random. We still have no which path information and no polarizer that interacts with the system, but will we have the interference fringes?

View attachment Black box eraser.pdf

 

[DrChinese]

A question just for fun:

It seems that it is the beamsplitter in the quantum eraser that ruins our FTL communication and Nobel. What if we throw out the beamsplitter and insert a black box containing a device that inserts a mirror by random. We still have no which path information and no polarizer that interacts with the system, but will we have the interference fringes?

View attachment 19840

Recall that an uncorrelated group of entangled photons do not display the interference pattern ever.

Although is it amazing that the entire set does not produce interference, but Alice-Bob correlated subsets clearly display the interference fringes and anti-fringes at all times (i.e. when considered as 2 groups). So actually, some kind of interference is occurring behind the scenes. But the summation is NO interference!

[fringe] + [anti-fringe] = [no interference pattern]

 

[Descartz2000]

Refuting the existence of free choice experiments is a way to explain the results without turning to FTL.