Mechanism for FTL information exchange.

[superpaul3000]

First, let me point out that this is probably wrong. Anytime I hear faster than light travel I am automatically skeptical. I just haven’t figured out why it’s wrong yet. Any ideas?


1. Two devices situated in between Bob and Alice produce two streams of entangled pairs of qubits. Each pair travels in opposite directions, two streams reach Bob and two reach Alice.

2. The production of one of these pairs is a bit closer to Bob and the production of the other pair is closer to Alice. From the prospective of either of the observers, the device closer to them produces the transmitting channel and the device further away produces the receiving channel.

3. Each observer possesses a transmitting instrument and a receiving instrument. The transmitting instrument simply measures a transmission channel qubit to form the classical bit 1 or does not measure the qubit to form the classical bit 0. The receiving instrument, perhaps an interferometer of some sort, records whether the incoming qubit from the receiving channel is in a superposition state implying the classical bit 0 or in a definite state implying the classical bit 1.

4. Bob can perform measurements on his transmission stream in such a way that a sequence of 1’s and 0’s is produced containing a message. This instantaneously makes Alice’s receiving channel assume definite states corresponding to Bob’s 1’s just before it reaches Alice. Then Alice can use her interferometer. If interference is seen, the particle is still in a superposition state and Alice records a 0. If no interference is seen, the particle is in a definite state and Alice records a 1.

5. Alice can likewise send Bob information with her transmission channel.

 

[San K]

First, let me point out that this is probably wrong. Anytime I hear faster than light travel I am automatically skeptical. I just haven’t figured out why it’s wrong yet. Any ideas?1. Two devices situated in between Bob and Alice produce two streams of entangled pairs of qubits. Each pair travels in opposite directions, two streams reach Bob and two reach Alice.

2. The production of one of these pairs is a bit closer to Bob and the production of the other pair is closer to Alice. From the prospective of either of the observers, the device closer to them produces the transmitting channel and the device further away produces the receiving channel.

3. Each observer possesses a transmitting instrument and a receiving instrument. The transmitting instrument simply measures a transmission channel qubit to form the classical bit 1 or does not measure the qubit to form the classical bit 0. The receiving instrument, perhaps an interferometer of some sort, records whether the incoming qubit from the receiving channel is in a superposition state implying the classical bit 0 or in a definite state implying the classical bit 1.

4. Bob can perform measurements on his transmission stream in such a way that a sequence of 1’s and 0’s is produced containing a message. This instantaneously makes Alice’s receiving channel assume definite states corresponding to Bob’s 1’s just before it reaches Alice. Then Alice can use her interferometer. If interference is seen, the particle is still in a superposition state and Alice records a 0. If no interference is seen, the particle is in a definite state and Alice records a 1.

5. Alice can likewise send Bob information with her transmission channel.

Both the entangled pairs need to be compared (with their entangled twins) by an instrument (known as co-incidence counter) before any information can be extracted/got.

Both the photon and its entangled twin (or atleast their state) have to be physically bought together in a coincidence counter (though they don't have to necessarily arrive at the same time)

No matter where the co-incidence counter lies, the shortest distance is still the straight line distance between Alice and Bob.

And the fastest way to get the photons together is the speed of light.

 

[SpectraCat]

First, let me point out that this is probably wrong. Anytime I hear faster than light travel I am automatically skeptical. I just haven’t figured out why it’s wrong yet. Any ideas?


1. Two devices situated in between Bob and Alice produce two streams of entangled pairs of qubits. Each pair travels in opposite directions, two streams reach Bob and two reach Alice.

That statement is fine

2. The production of one of these pairs is a bit closer to Bob and the production of the other pair is closer to Alice. From the prospective of either of the observers, the device closer to them produces the transmitting channel and the device further away produces the receiving channel.

That statement is also fine, except the distinction between "transmitting" vs "receiving channels is contrived .. the entangled pairs yield the same results irrespective of the order of measurement.

3. Each observer possesses a transmitting instrument and a receiving instrument. The transmitting instrument simply measures a transmission channel qubit to form the classical bit 1 or does not measure the qubit to form the classical bit 0. The receiving instrument, perhaps an interferometer of some sort, records whether the incoming qubit from the receiving channel is in a superposition state implying the classical bit 0 or in a definite state implying the classical bit 1.

The problem is with the vagueness of the details for the receiving instrument, which obfuscate the fact that in order to obtain a result from a quantum system, you need to perform a measurement .. there is no such thing as passive detection. So, you need to specify clearly how the receiver can tell if their measurement reflects a result pre-determined by the transmitter's measurement, or if the transmitter left the particle unmeasured, so it is the receiver who is actually breaking the entanglement. This is where all of these models break down, because once you understand the details, you will understand that it is impossible to design an experiment where the actions of the transmitter change the results at the receiver's end in any way that is detectable to the receiver.

4. Bob can perform measurements on his transmission stream in such a way that a sequence of 1’s and 0’s is produced containing a message. This instantaneously makes Alice’s receiving channel assume definite states corresponding to Bob’s 1’s just before it reaches Alice. Then Alice can use her interferometer. If interference is seen, the particle is still in a superposition state and Alice records a 0. If no interference is seen, the particle is in a definite state and Alice records a 1.

5. Alice can likewise send Bob information with her transmission channel.

That is not how interferometry works ... single photons do not produce interference patterns. Even if you are talking about ensembles of photons (i.e. in your example, Bob sends a "1" by leaving the transmitter on for 1000 photons, and a "0" by leaving 1000 photons unmeasured), there will be no difference in the pattern observed at Alice's end. After all, in both cases she is performing measurements on single photons ... why would you expect the *single-photon* results to be different if the photon is in a superposition state or not? What changes based on entanglement are the *coincidence results* generated considering measurements at *both* ends of the system. Only the coincidence results contain any information about the effects of choices made by Alice at Bob's detector, and vice-versa.

Hope that clears things up.

 

[DrChinese]

Just to add to SpectaCat's fine analysis: Entangled photons have a quirk. They do not produce interference patterns. Can you see why?

http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf

See Fig. 2.

 

[superpaul3000]

Thanks for pointing out the vagueness SpectraCat. I think I need to add more detail to this picture. By interferometer I do not mean something like a Fabry-Perot interferometer but something like a double slit experiment. DrChinese thanks for that link. This is precisely what I am talking about. Look at figure 3 in that paper. As you said, there is no interference pattern produced by the second entangled photon if the detector reveals the path of the second photon going through the double slit by measuring the first photon. However, an interference pattern will occur if no information about either photon’s state is being measured. The experiment in figure 3 is similar to my original idea. The following picture is a modified version of that experiment and a more detailed representation of my idea. The following quote is from the caption of figure 3. Now that my idea is a little clearer, can you pinpoint where the problem is?

“FIG. 3. Two photons and one double slit. A pair of momentum-entangled photons is created by type-I parametric down conversion. Photon 2 enters a double-slit assembly and photon 1 is registered by the Heisenberg detector arranged behind the Heisenberg lens. If the Heisenberg detector is placed in the focal plane of the lens, it projects the state of the second photon into a momentum eigenstate which cannot reveal any position information and hence no information about slit passage. Therefore, in coincidence with a registration of photon 1 in the focal plane, photon 2 exhibits an interference pattern. On the other hand, if the Heisenberg detector is placed in the imaging plane at 2 f, it can reveal the path the second photon takes through the slit assembly which therefore connot show the interference pattern (Dopfer, 1998).”

 

[DrChinese]

I think you will quickly find that you cannot put photons in the D2=0 position at will. Only a few will end up like that (because they are registered randomly). Not really sure of the percentages and numbers though. Maybe someone else can provide that. Keep in mind that the unmatched photons (i.e. only one within a time window) are all over the place. There may be 10-100 times more of those. In the end, nothing looks different at Alice based on Bob's actions, until you coincidence count.

 

[superpaul3000]

I think you will quickly find that you cannot put photons in the D2=0 position at will. Only a few will end up like that (because they are registered randomly). Not really sure of the percentages and numbers though. Maybe someone else can provide that. Keep in mind that the unmatched photons (i.e. only one within a time window) are all over the place. There may be 10-100 times more of those. In the end, nothing looks different at Alice based on Bob's actions, until you coincidence count.

So in any real experiment the non-entangled photons would vastly outnumber the entangled photons. Alice would see 1 for when Bob sends a 1 and 1 99.9% of the time Bob sends 0 making communication impossible. You say the use of a coincidence counter will filter out the entangled particles but that of course is limited by the speed of light, making this is a big waste of time.

The speed of light seems safe for now but one thing I find interesting is if you could somehow find an EPR source that generates 99.9% entangled photons, this FTL quantum telephone should in theory work. Right?

 

[San K]

So in any real experiment the non-entangled photons would vastly outnumber the entangled photons. Alice would see 1 for when Bob sends a 1 and 1 99.9% of the time Bob sends 0 making communication impossible. You say the use of a coincidence counter will filter out the entangled particles but that of course is limited by the speed of light, making this is a big waste of time.

The speed of light seems safe for now but one thing I find interesting is if you could somehow find an EPR source that generates 99.9% entangled photons, this FTL quantum telephone should in theory work. Right?

i think yes and had posted similar, namely:

if there is zero (or close to zero) noise from non-entangled/random photons...i.e. the only photons that are allowed in the experiment are entangled ones...FTL should in theory work, it seems

or in other words, if we can find a way (for example entanglement) to transfer/transmit information without use of mass-energy, we could have FTL

Fundamentally this means:

either

FTL is possible ....note: a bunch of photons (instead of a single photon) would be required though
or
noise is not the reason for FTL being impossible

 

[superpaul3000]

i think yes and had posted similar, namely:

if there is zero (or close to zero) noise from non-entangled/random photons...i.e. the only photons that are allowed in the experiment are entangled ones...FTL should in theory work, it seems

or in other words, if we can find a way (for example entanglement) to transfer/transmit information with use of mass-energy, we could have FTL

Fundamentally this means:

either

FTL is possible ....note: a bunch of photons (instead of a single photon) would be required though
or
noise is not the reason for FTL being impossible

Well either way there isn't a huge problem here because causality is preserved using this type of FTL communication. For instance, picture the SR twins using this FTL phone. When they are both on Earth with v=0, they can communicate FTL. When one twin starts moving towards a star 10 lyrs away at near the speed of light, the EPR source must travel at half that speed in the same direction. The fact that it is moving away from you makes the rate of photons decrease and your twin appears to talk slower just he appears to be moving slower in the sky. From your twin's perspective, space contracts and the photons come in at a faster rate so you appear to him to talk faster and move faster when he looks at Earth. When he arrives at the star, you have aged 10 yrs and he hasn't aged. You can once again communicate with him FTL. However, no matter how you try to arrange this you cannot communicate with the past or future and thus causality is preserved. Your twin thinks he traveled into the future but really he has just been running a slow clock for the past 10 yrs.

 

[San K]

Well either way there isn't a huge problem here because causality is preserved using this type of FTL communication. For instance, picture the SR twins using this FTL phone. When they are both on Earth with v=0, they can communicate FTL. When one twin starts moving towards a star 10 lyrs away at near the speed of light, the EPR source must travel at half that speed in the same direction. The fact that it is moving away from you makes the rate of photons decrease and your twin appears to talk slower just he appears to be moving slower in the sky. From your twin's perspective, space contracts and the photons come in at a faster rate so you appear to him to talk faster and move faster when he looks at Earth. When he arrives at the star, you have aged 10 yrs and he hasn't aged. You can once again communicate with him FTL. However, no matter how you try to arrange this you cannot communicate with the past or future and thus causality is preserved. Your twin thinks he traveled into the future but really he has just been running a slow clock for the past 10 yrs.

yes, agreed, causality is not violated by FTL, (almost) no one disagrees with that...:)

or in other words: causality is only violated when mass-energy is used for transmission of information

in entanglement we are not using mass-energy ...

 

[DrChinese]

So in any real experiment the non-entangled photons would vastly outnumber the entangled photons. Alice would see 1 for when Bob sends a 1 and 1 99.9% of the time Bob sends 0 making communication impossible. You say the use of a coincidence counter will filter out the entangled particles but that of course is limited by the speed of light, making this is a big waste of time.

The speed of light seems safe for now but one thing I find interesting is if you could somehow find an EPR source that generates 99.9% entangled photons, this FTL quantum telephone should in theory work. Right?

Ah, no. You would still have the issue that the pattern never changes. The changing pattern is an artifact of the coincidence counter. The fringe and anti-fringe patterns add to make the pattern you actually see (Walborn). In the Dopfer setup, I must admit I don't fully understand what patterns appear and if they ever change.

 

[San K]

Even when we get fringes on Bob side (via no-which-way) we won't get them on Alice side ...no matter what setup/instruments we put and even if we have zero noise.

Is the above statement correct?

I am trying to find out the reason why Alice won't get interference pattern. Maybe it has to do with inherent randomness.

 

[San K]

Even if you are talking about ensembles of photons (i.e. in your example, Bob sends a "1" by leaving the transmitter on for 1000 photons, and a "0" by leaving 1000 photons unmeasured), there will be no difference in the pattern observed at Alice's end. After all, in both cases she is performing measurements on single photons ... why would you expect the *single-photon* results to be different if the photon is in a superposition state or not?

This is the part I don't get yet.

If Bob gets an interference pattern, why would Alice not get it (assuming zero noise, which might not be possible but just for theory/understanding sake...)?

 

[San K]

Even when we get fringes on Bob side (via no-which-way) we won't get them on Alice side ...no matter what setup/instruments we put and even if we have zero noise.

Is the above statement correct?

I am trying to find out the reason why Alice won't get interference pattern. Maybe it has to do with inherent randomness.

I got the answer (thanks to earlier post by Cthuga):

The simple/solo double slit is NOT (exactly) the same as the double slit in DCQE.

Single photon and two-photon (bi-photon, entangled) are different things and don't have exactly similar properties...for example coherence within themselves...

The logic for the single photon is not, all, the same for bi-photon.

you get interference pattern in simple double slit but for the "DCQE double slit" the story (and its a long one) is different...

See posts 12 (and onwards) on this page----> https://www.physicsforums.com/showthread.php?t=263066

So FTL is not possible even via entanglement.