The distance between person A and person B is, for example, 10 light minutes.
We have two sources between those two persons, each of the source releasing simultaneously pairs of two entangled photons, one pair every 4 seconds ( time period is of course debatable ), which polarization is at 90 degrees of difference between themselves, and at 45 degrees in regard to 0 degrees axis of polarization.
At person A and B places are two filters , each filter for each particular source.
Source 1 is releasing one pair of entangled photons in opposite direction, one photon toward filter 1 at A place and other photon toward filter 1 at B place.
Source 2 is also releasing one pair of entangled photons in opposite direction, one photon toward fitter 2 at A place and other photon toward filter 2 at B place.
Photons are arriving at filter 1 and 2 at A place a bit sooner than B photons are arriving at filter 1 and 2 at B place.
Now, A wants to communicate to B "yes".
A set up his filter 1 at 45 degrees. As I understand probability is that 50 % from two 45 degrees in regard to 0 axis polarized entangled photon would pass filter 1 set up at 45 degrees ( or set up at 315 degrees ). So, as soon as one of the photons passed filter 1 at 45 degrees at A place other one arriving at B place would pass through B filter 1 already set up at 315 degrees, and therefore as soon as A filter 1 passed through a photon A would know that B got an information "yes". If on the other hand A did not want to communicate "yes" to B , A set up his filter 1 46 degrees, so no photons are passing through 315 degrees filter 1 at B place.
If on the other hand A want to communicate to B "no" he would do the reverse with source and filters 2. A set up his filter 2 at 315 degrees, and B filter 2 is already set up at 45 degrees. As soon as photon passes A filter 2 at 315 degrees A knows that B filter 2 let through a photon at 45 degrees, and A knows that B got an information "no". If A did not want to communicate "no" to B then A set up his filter 2 at 314 degrees, so no photons passes through 45 degrees of filter B 2.
This is the basic idea. If I have not overlooked something obvious this process could be speed up, perhaps with four pairs of filters arranged as described and automated for some binary two ways faster than light transfer of information/communication.
Mistake is where?
Anton
Fivedoughnuts simple guide to hyperspacial communication ..... a sketch of 4 easy to follow steps.
Righty ho folks, listen good & proper....
Step 1: Collect positrons and electrons from gamma photon decays (lots are recommended)
Step 2: Entrap them cryogenically in pre-purged magnetic containment devices (ultra vacuum/particle density near zero) to minimize possibility of annihilation.
Step 3: Separate the electron and positron containment devices (at any distance you choose fit).
Step 4: finally mobilise (by affecting containment field) either the electrons or positrons ..... then by the magic of what used to be called quantum interconnectivity, the corresponding anti-particles will anti-wobble, thus producing an inductive effect on their containment coils which are simply amplified as a signal.
I'd like the 1st hyperspacial transmission to be the song "I've got a lovely bunch of coconuts".
Damn fine technology, same concept allows for hyperspacial processing too!
Righty ho folks, listen good & proper....
Step 1: Collect positrons and electrons from gamma photon decays (lots are recommended)
Step 2: Entrap them cryogenically in pre-purged magnetic containment devices (ultra vacuum/particle density near zero) to minimize possibility of annihilation.
Step 3: Separate the electron and positron containment devices (at any distance you choose fit).
Step 4: finally mobilise (by affecting containment field) either the electrons or positrons ..... then by the magic of what used to be called quantum interconnectivity, the corresponding anti-particles will anti-wobble, thus producing an inductive effect on their containment coils which are simply amplified as a signal.
I'd like the 1st hyperspacial transmission to be the song "I've got a lovely bunch of coconuts".
Damn fine technology, same concept allows for hyperspacial processing too!
QUOTE (fivedoughnut+Aug 12 2007, 04:12 AM)
Fivedoughnuts simple guide to hyperspacial communication ..... a sketch of 4 easy to follow steps.
Righty ho folks, listen good & proper....
Step 1: Collect positrons and electrons from gamma photon decays (lots are recommended)
Step 2: Entrap them cryogenically in pre-purged magnetic containment devices (ultra vacuum/particle density near zero) to minimize possibility of annihilation.
Step 3: Separate the electron and positron containment devices (at any distance you choose fit).
Step 4: finally mobilise (by affecting containment field) either the electrons or positrons ..... then by the magic of what used to be called quantum interconnectivity, the corresponding anti-particles will anti-wobble, thus producing an inductive effect on their containment coils which are simply amplified as a signal.
I'd like the 1st hyperspacial transmission to be the song "I've got a lovely bunch of coconuts".
Damn fine technology, same concept allows for hyperspacial processing too!
Great! Get teleportation set ups here! I mean..there!
Righty ho folks, listen good & proper....
Step 1: Collect positrons and electrons from gamma photon decays (lots are recommended)
Step 2: Entrap them cryogenically in pre-purged magnetic containment devices (ultra vacuum/particle density near zero) to minimize possibility of annihilation.
Step 3: Separate the electron and positron containment devices (at any distance you choose fit).
Step 4: finally mobilise (by affecting containment field) either the electrons or positrons ..... then by the magic of what used to be called quantum interconnectivity, the corresponding anti-particles will anti-wobble, thus producing an inductive effect on their containment coils which are simply amplified as a signal.
I'd like the 1st hyperspacial transmission to be the song "I've got a lovely bunch of coconuts".
Damn fine technology, same concept allows for hyperspacial processing too!
Great! Get teleportation set ups here! I mean..there!
QUOTE (Neil Farbstein+Aug 12 2007, 04:23 AM)
Great! Get teleportation set ups here! I mean..there!
Thanks Neil ..... you've got the cash ..... build it, and cut me in for a mere 1% of the profits .... probably would still make me a couple of trillion dollars.
Thanks Neil ..... you've got the cash ..... build it, and cut me in for a mere 1% of the profits .... probably would still make me a couple of trillion dollars.
You two comedians. 
That is really admirable entrepreneur spirit that you shown. You two gone out there where no man has gone before.
A correction from initial post...
If we want to communicate "yes" we set up both filters at A place at 45 degrees polarization.
At B place filter 1 is set up at 315 degrees and filter 2 is set up at 45 degrees of polarization.
If from two photons arriving at A place both photons passes filters 1 and 2 at A place, which are set up at 45 degrees , only photon at B place which would pass it's filter would be one which would pass through filter 1 set up at 315 degrees . While filter 2 set up at 45 degrees would block a photon which entangled other photon passed a filter 2 at A place also set up at 45 degrees.
Where is a mistake here? I do not say there is no mistake, I simply ask what is it.
Anton
PS. I assumed that if photon A 1 is blocked at A 1 filter, which is set up at 45 degrees of polarization, that that means that that photon assumed 315 degrees of polarization, and considering that that it's other entangled photon cannot pass filter B 1 which is set up at 315 degrees because that photon at B 1 filter assumed 45 degrees of polarization.
Perhaps here is some mistake?
That is really admirable entrepreneur spirit that you shown. You two gone out there where no man has gone before.
A correction from initial post...
If we want to communicate "yes" we set up both filters at A place at 45 degrees polarization.
At B place filter 1 is set up at 315 degrees and filter 2 is set up at 45 degrees of polarization.
If from two photons arriving at A place both photons passes filters 1 and 2 at A place, which are set up at 45 degrees , only photon at B place which would pass it's filter would be one which would pass through filter 1 set up at 315 degrees . While filter 2 set up at 45 degrees would block a photon which entangled other photon passed a filter 2 at A place also set up at 45 degrees.
Where is a mistake here? I do not say there is no mistake, I simply ask what is it.
Anton
PS. I assumed that if photon A 1 is blocked at A 1 filter, which is set up at 45 degrees of polarization, that that means that that photon assumed 315 degrees of polarization, and considering that that it's other entangled photon cannot pass filter B 1 which is set up at 315 degrees because that photon at B 1 filter assumed 45 degrees of polarization.
Perhaps here is some mistake?
Now I see where is a mistake.
If my assumption that if photon A 1 is blocked at A 1 filter which is set up at 45 degrees that then it's other entangled photon cannot pass filter B 1 which is set up at 315 degrees is correct, then if photon A 2 is blocked at 45 degrees then it's other photon B 2 would pass a filter B2 which is set up at 45 degrees.
Sigh.
Anton
If my assumption that if photon A 1 is blocked at A 1 filter which is set up at 45 degrees that then it's other entangled photon cannot pass filter B 1 which is set up at 315 degrees is correct, then if photon A 2 is blocked at 45 degrees then it's other photon B 2 would pass a filter B2 which is set up at 45 degrees.
Sigh.
Anton
What about this.
The source is between A and B, and it is releasing a pair of entangled photons.
Photon A arrives at A place sooner than photon B at B place.
At A place it is an ordinary detector.
At B place is a barrier with double slits and detector behind a barrier, so DS apparatus.
Now, we want to communicate "yes" from A place to B place, and we already agreed that interference pattern of hits on the detector at B place would mean "yes".
So, we move the detector at A place and we let a photon A to pass through unobserved, and we let a photon B to pass through slit/s in a state of superposition , and photon B hits the detector in, I assume, an interference pattern of hits.
Correct so far?
Now, we want to communicate "no" from A place to B place, and we already agreed that particle pattern of hits on the detector at B place would mean "no".
Now, photon A arrives at A place and it is blocked by detector. At that moment both entangled photons changed their state from state of superposition into after an observation state.
In that moment of detection of photon A, photon B is about to pass through slits.
If photon B would now pass through slit/s already in the after an observation state should not photon B hit a detector in particle pattern of hits?
If it would, is not this faster than light transfer of information/ communication?
Anton
The source is between A and B, and it is releasing a pair of entangled photons.
Photon A arrives at A place sooner than photon B at B place.
At A place it is an ordinary detector.
At B place is a barrier with double slits and detector behind a barrier, so DS apparatus.
Now, we want to communicate "yes" from A place to B place, and we already agreed that interference pattern of hits on the detector at B place would mean "yes".
So, we move the detector at A place and we let a photon A to pass through unobserved, and we let a photon B to pass through slit/s in a state of superposition , and photon B hits the detector in, I assume, an interference pattern of hits.
Correct so far?
Now, we want to communicate "no" from A place to B place, and we already agreed that particle pattern of hits on the detector at B place would mean "no".
Now, photon A arrives at A place and it is blocked by detector. At that moment both entangled photons changed their state from state of superposition into after an observation state.
In that moment of detection of photon A, photon B is about to pass through slits.
If photon B would now pass through slit/s already in the after an observation state should not photon B hit a detector in particle pattern of hits?
If it would, is not this faster than light transfer of information/ communication?
Anton
Today version.
For "yes" information everything stays the same. Photon A is let pass unobserved, undetected by detector at A place, so photon B goes in it's state of superposition through slit/s , and photon B hits the detector inside one of the areas where photons are hitting the detector in the interference pattern of hits ( which areas are experimentally determined before ).
Now, for "no" information.
Instead arranging the set up in the way that A photon is detected at A place just before photon B is about to pass through slit/, we change the set up slightly in the following manner. We arrange the set up so that photon A is detected at A place at the very moment when photon B is "inside" the slit/s.
If photon B changes from it's state of superposition into after the observation state when photon B is "inside" the slit/s , then photon B cannot be at both slits in the sense of Copenhagen interpretation of QM, and therefore photon B must "choose" inside which slit it is. Therefore photon B would surely hit the detector in the particle pattern of hits.
So, "yes" signal and signal "no" are transfered faster than light in this manner?
Anton
PS. If photon B would immediately hit inside some of the areas where photons never hit when they are hitting the detector in interference pattern, then we would immediately know that the information is "no".
But if that would not be the case then some number of photons, already agreed and experimentally established that they are certain to produce visible particle pattern of hits, would be used for each "no" signal.
The same for "yes" signal.
For "yes" information everything stays the same. Photon A is let pass unobserved, undetected by detector at A place, so photon B goes in it's state of superposition through slit/s , and photon B hits the detector inside one of the areas where photons are hitting the detector in the interference pattern of hits ( which areas are experimentally determined before ).
Now, for "no" information.
Instead arranging the set up in the way that A photon is detected at A place just before photon B is about to pass through slit/, we change the set up slightly in the following manner. We arrange the set up so that photon A is detected at A place at the very moment when photon B is "inside" the slit/s.
If photon B changes from it's state of superposition into after the observation state when photon B is "inside" the slit/s , then photon B cannot be at both slits in the sense of Copenhagen interpretation of QM, and therefore photon B must "choose" inside which slit it is. Therefore photon B would surely hit the detector in the particle pattern of hits.
So, "yes" signal and signal "no" are transfered faster than light in this manner?
Anton
PS. If photon B would immediately hit inside some of the areas where photons never hit when they are hitting the detector in interference pattern, then we would immediately know that the information is "no".
But if that would not be the case then some number of photons, already agreed and experimentally established that they are certain to produce visible particle pattern of hits, would be used for each "no" signal.
The same for "yes" signal.
QUOTE (Mate+Aug 11 2007, 11:06 PM)
So, as soon as one of the photons passed filter 1 at 45 degrees at A place other one arriving at B place would pass through B filter 1 already set up at 315 degrees, and therefore as soon as A filter 1 passed through a photon A would know that B got an information "yes". If on the other hand A did not want to communicate "yes" to B , A set up his filter 1 46 degrees, so no photons are passing through 315 degrees filter 1 at B place.
A will have a 50% chance of having the photon pass his filter, and likewise 50% chance of rejecting the filter.
B will get the same result as A, whichever that was.
How does that communicate anything?
Or, if A decides not to use the filter, he won't alter the wave function. B will have a 50% chance of it passing his filter. If A had used his filter, he would have gotten the same result as B. So?
How does that allow A to send a message to B?
At best, it allows them to synchronize their pre-arranged plans to a random event. Good for coordinating a flanking maneuver, perhaps, not not useful to communicate those plans to begin with.
A will have a 50% chance of having the photon pass his filter, and likewise 50% chance of rejecting the filter.
B will get the same result as A, whichever that was.
How does that communicate anything?
Or, if A decides not to use the filter, he won't alter the wave function. B will have a 50% chance of it passing his filter. If A had used his filter, he would have gotten the same result as B. So?
How does that allow A to send a message to B?
At best, it allows them to synchronize their pre-arranged plans to a random event. Good for coordinating a flanking maneuver, perhaps, not not useful to communicate those plans to begin with.
QUOTE (LearmSceince+Aug 13 2007, 06:51 PM)
A will have a 50% chance of having the photon pass his filter, and likewise 50% chance of rejecting the filter.
B will get the same result as A, whichever that was.
How does that communicate anything?
Or, if A decides not to use the filter, he won't alter the wave function. B will have a 50% chance of it passing his filter. If A had used his filter, he would have gotten the same result as B. So?
How does that allow A to send a message to B?
At best, it allows them to synchronize their pre-arranged plans to a random event. Good for coordinating a flanking maneuver, perhaps, not not useful to communicate those plans to begin with.
Yep, I realized where I made a mistake.
But what about the last proposal?
Anton
B will get the same result as A, whichever that was.
How does that communicate anything?
Or, if A decides not to use the filter, he won't alter the wave function. B will have a 50% chance of it passing his filter. If A had used his filter, he would have gotten the same result as B. So?
How does that allow A to send a message to B?
At best, it allows them to synchronize their pre-arranged plans to a random event. Good for coordinating a flanking maneuver, perhaps, not not useful to communicate those plans to begin with.
Yep, I realized where I made a mistake.
But what about the last proposal?
Anton
QUOTE (Mate+Aug 14 2007, 02:13 PM)
Yep, I realized where I made a mistake.
But what about the last proposal?
Anton
Same thing. Each observer will see a random result, until they compare notes.
I had a hard time following your text. If you want to post another one, you might try drawing a diagram.
But what about the last proposal?
Anton
Same thing. Each observer will see a random result, until they compare notes.
I had a hard time following your text. If you want to post another one, you might try drawing a diagram.
QUOTE (LearmSceince+Aug 16 2007, 06:41 AM)
Same thing. Each observer will see a random result, until they compare notes.
I had a hard time following your text. If you want to post another one, you might try drawing a diagram.
I am not absolutely sure but it seems to me from your comments that you have not read the whole page at all? Right?
Anyway, as clearest as I can, on English.
Point A and point B are , for example, 180 000 000 km distant from each other.
At the point X , which is exactly between point A and B, is the source which is releasing 20 pairs of entangled photons ( arbitrary value ) every 0.5 seconds one pair ( arbitrary value ), and than it makes a pause of 2 seconds ( arbitrary value ). So, 2o pairs, 20 photons "a" toward point A, and 20 photon"b" toward point B, than pause of 2 seconds, and then again 20 pairs, and so on..
At point A is the ordinary detector. It does not measure anything except that particle hit the detector. But it can be programed that it moves itself from trajectory of photons "a" arriving at it for any number of photons "a", and reversely it can be programed to stay at it's position and registering hits for any number of photons "a".
At point B is the barrier with two slits, alike barrier in classic Double Slit experiment. Behind the barrier is the detector alike detector in classic DS, which also does not measure anything except where particle hit the detector. But it has additional features that it can memorizes where exactly some number of consequent hits hit the detector. And it has a feature that we can set the detector to memorize any number of hits, and then resets itself for the next number of hits. And it recognizes automatically is the pattern of 20 hits ( let us say that it is 20, but it can be some other number, of course ) an interference pattern or particle pattern of hits.
Now, when photon "a1" , for example, is hitting the detector at A point, photon "b1" is "inside" the slit/s at point B ( or very very close to the slits, that close that if we would detect that photon in that moment we would get a particle pattern of hits on the detector ).
So, you are at point A and you want to transfer signal "yes" from point A to point B. You programed your detector to move away from one group of 20 "a" photons which therefore passed by point A unobserved. At point B 20 photons "b" have passed the slits in their usual state of superposition, and they made an interference pattern of hits , which pattern is recognized automatically by the detector, and people at point B know that your signal is "yes".
You want to transfer signal "no", so you programed your detector to register 20 hits from one group of photons "a". As each of photon "a" from that group is been registered at the detector at A point in that moment each of those photon "a" changes from the state of superposition in the after the observation state. As some photon "a" from that group changes that photon "a" influence it's photon "b" from that particular pair. So that photon "b" also changes from state of superposition into after the observation state. And that photon "b" changes into after the observation state exactly when it is either inside the slits or very very close to slits.
The crucial assumption for this to work.
If photon "b" is in the after the observation state when it is inside or very close to the slits then that photon "b" cannot be at both slits at the same time ( which could, if remained in the superposition state ) in the sense of Copenhagen interpretation of QM, and therefore that photon "b" has to "choose" at which slit it is or through which slit it would pass. And in that case it would hit the detector in the particle pattern oh hits. And 20 hits like that the detector automatically recognizes as particle pattern. and people at point B know that your signal is "no".
Now, this is from top of the head version. If the basic idea could work then this could be improved, I suppose.
So, the question is. Could this work as I imagined? At least in principle?
Anton
I had a hard time following your text. If you want to post another one, you might try drawing a diagram.
I am not absolutely sure but it seems to me from your comments that you have not read the whole page at all? Right?
Anyway, as clearest as I can, on English.
Point A and point B are , for example, 180 000 000 km distant from each other.
At the point X , which is exactly between point A and B, is the source which is releasing 20 pairs of entangled photons ( arbitrary value ) every 0.5 seconds one pair ( arbitrary value ), and than it makes a pause of 2 seconds ( arbitrary value ). So, 2o pairs, 20 photons "a" toward point A, and 20 photon"b" toward point B, than pause of 2 seconds, and then again 20 pairs, and so on..
At point A is the ordinary detector. It does not measure anything except that particle hit the detector. But it can be programed that it moves itself from trajectory of photons "a" arriving at it for any number of photons "a", and reversely it can be programed to stay at it's position and registering hits for any number of photons "a".
At point B is the barrier with two slits, alike barrier in classic Double Slit experiment. Behind the barrier is the detector alike detector in classic DS, which also does not measure anything except where particle hit the detector. But it has additional features that it can memorizes where exactly some number of consequent hits hit the detector. And it has a feature that we can set the detector to memorize any number of hits, and then resets itself for the next number of hits. And it recognizes automatically is the pattern of 20 hits ( let us say that it is 20, but it can be some other number, of course ) an interference pattern or particle pattern of hits.
Now, when photon "a1" , for example, is hitting the detector at A point, photon "b1" is "inside" the slit/s at point B ( or very very close to the slits, that close that if we would detect that photon in that moment we would get a particle pattern of hits on the detector ).
So, you are at point A and you want to transfer signal "yes" from point A to point B. You programed your detector to move away from one group of 20 "a" photons which therefore passed by point A unobserved. At point B 20 photons "b" have passed the slits in their usual state of superposition, and they made an interference pattern of hits , which pattern is recognized automatically by the detector, and people at point B know that your signal is "yes".
You want to transfer signal "no", so you programed your detector to register 20 hits from one group of photons "a". As each of photon "a" from that group is been registered at the detector at A point in that moment each of those photon "a" changes from the state of superposition in the after the observation state. As some photon "a" from that group changes that photon "a" influence it's photon "b" from that particular pair. So that photon "b" also changes from state of superposition into after the observation state. And that photon "b" changes into after the observation state exactly when it is either inside the slits or very very close to slits.
The crucial assumption for this to work.
If photon "b" is in the after the observation state when it is inside or very close to the slits then that photon "b" cannot be at both slits at the same time ( which could, if remained in the superposition state ) in the sense of Copenhagen interpretation of QM, and therefore that photon "b" has to "choose" at which slit it is or through which slit it would pass. And in that case it would hit the detector in the particle pattern oh hits. And 20 hits like that the detector automatically recognizes as particle pattern. and people at point B know that your signal is "no".
Now, this is from top of the head version. If the basic idea could work then this could be improved, I suppose.
So, the question is. Could this work as I imagined? At least in principle?
Anton
Hi Mate, Learmscience,
I've been watching and thinking for a bit.
I don't know whether this is true but if it is it might eliminate a lot of trial and error.
Proposition .. the act of measuring/setting the state of (say) the signal photon cannot be distinguished from the act of knowing the state at the time the photons are 'created'.
?
Best wishes, C2.
I've been watching and thinking for a bit.
I don't know whether this is true but if it is it might eliminate a lot of trial and error.
Proposition .. the act of measuring/setting the state of (say) the signal photon cannot be distinguished from the act of knowing the state at the time the photons are 'created'.
?
Best wishes, C2.
QUOTE (Confused2+Aug 16 2007, 01:19 PM)
Hi Mate, Learmscience,
I've been watching and thinking for a bit.
I don't know whether this is true but if it is it might eliminate a lot of trial and error.
Proposition .. the act of measuring/setting the state of (say) the signal photon cannot be distinguished from the act of knowing the state at the time the photons are 'created'.
?
Best wishes, C2.
C2,
I am not quite sure what you propose. Can you please clarify in other words, preferably with some example how would that work as you imagined?
Regarding trial and error.
I suppose that the detector at B place is needed to be calibrated just once for all possible distances of the source from the slits, if the slits are remaining at same distance from the detector and remaining the same themselves in the sense of width , and if the distance between them and thickness of the barrier would also be the same/standard.. In short , one "standard" DS apparatus.
Of course it would be advisable to chose the one DS set up which can distinguish an interference from particle pattern with as less hits as possible.
What would also be needed is to make a software which can recognized an interference from particle pattern from that number of hits.
If then the whole process can be speed up so the source is emitting pairs every fraction of the second instead every half of the second then , I guess, babbling faster than light would be a real life possibility.
No matter what Einstein would think about that. : )
Anton
I've been watching and thinking for a bit.
I don't know whether this is true but if it is it might eliminate a lot of trial and error.
Proposition .. the act of measuring/setting the state of (say) the signal photon cannot be distinguished from the act of knowing the state at the time the photons are 'created'.
?
Best wishes, C2.
C2,
I am not quite sure what you propose. Can you please clarify in other words, preferably with some example how would that work as you imagined?
Regarding trial and error.
I suppose that the detector at B place is needed to be calibrated just once for all possible distances of the source from the slits, if the slits are remaining at same distance from the detector and remaining the same themselves in the sense of width , and if the distance between them and thickness of the barrier would also be the same/standard.. In short , one "standard" DS apparatus.
Of course it would be advisable to chose the one DS set up which can distinguish an interference from particle pattern with as less hits as possible.
What would also be needed is to make a software which can recognized an interference from particle pattern from that number of hits.
If then the whole process can be speed up so the source is emitting pairs every fraction of the second instead every half of the second then , I guess, babbling faster than light would be a real life possibility.
No matter what Einstein would think about that. : )
Anton
QUOTE (Mate+Aug 16 2007, 10:27 AM)
You want to transfer signal "no", so you programed your detector to register 20 hits from one group of photons "a". As each of photon "a" from that group is been registered at the detector at A point in that moment each of those photon "a" changes from the state of superposition in the after the observation state. As some photon "a" from that group changes that photon "a" influence it's photon "b" from that particular pair. So that photon "b" also changes from state of superposition into after the observation state. And that photon "b" changes into after the observation state exactly when it is either inside the slits or very very close to slits.
The above paragraph I had difficulty understanding. I think you mean that observing "a" will also observe "b", destroying the interference pattern?
The critical assumption is wrong.
The mechanism at "b" will always see the interference pattern, regardless of what is happening at "a".
You are confusing entanglement and superposition to some degree.
But in any case, measuring a definite position or momentum of one photon will have no effect on the "which path" behavior of its twin which was originally entangled to have opposite position and momentum. Meanwhile, just going through the slits will observe the position/momentum state of b and change it. That just makes it more confusing. If you were still entangling spin states, the same would still hold: observing one particle does not cause a wave-function collapse of the other.
Hi Mate,Learmscience,
I think we may have an implicit reference to something like this :-
http://grad.physics.sunysb.edu/~amarch/
My own feeling is that the conspiracy will run pretty deep. For example the act of preparing photons in the same state will leave Alice in doubt as to whether her photon didn't pass through a polarizing filter because it didn't or whether it simply wasn't sent. I could be wrong but I suspect the opposite state to 'known' is 'not known' which might make the water seem rather muddy.
Maybe also .. the absence of a coincidence detector in a long range 'delayed choice eraser' experiment may open up new possibilities for 'the system' to find loopholes.
Best wishes - C2.
I think we may have an implicit reference to something like this :-
http://grad.physics.sunysb.edu/~amarch/
My own feeling is that the conspiracy will run pretty deep. For example the act of preparing photons in the same state will leave Alice in doubt as to whether her photon didn't pass through a polarizing filter because it didn't or whether it simply wasn't sent. I could be wrong but I suspect the opposite state to 'known' is 'not known' which might make the water seem rather muddy.
Maybe also .. the absence of a coincidence detector in a long range 'delayed choice eraser' experiment may open up new possibilities for 'the system' to find loopholes.
Best wishes - C2.
QUOTE (LearmSceince+Aug 20 2007, 08:01 PM)
...observing one particle does not cause a wave-function collapse of the other.
LearmSceince,
if that what you said would be the case why one entangled photon, which usually can assume with 50% of probability some spin up or down on a certain axis, assumes the opposite spin in regard to other photon from some pair of entangled photons, if that other photon has been already observed/measured?
Anton
LearmSceince,
if that what you said would be the case why one entangled photon, which usually can assume with 50% of probability some spin up or down on a certain axis, assumes the opposite spin in regard to other photon from some pair of entangled photons, if that other photon has been already observed/measured?
Anton
QUOTE (Confused2+Aug 26 2007, 10:25 PM)
Hi Mate,Learmscience,
I think we may have an implicit reference to something like this :-
http://grad.physics.sunysb.edu/~amarch/
My own feeling is that the conspiracy will run pretty deep. For example the act of preparing photons in the same state will leave Alice in doubt as to whether her photon didn't pass through a polarizing filter because it didn't or whether it simply wasn't sent. I could be wrong but I suspect the opposite state to 'known' is 'not known' which might make the water seem rather muddy.
Maybe also .. the absence of a coincidence detector in a long range 'delayed choice eraser' experiment may open up new possibilities for 'the system' to find loopholes.
Best wishes - C2.
C2,
I have been told by one college professor of physics that this proposal of mine is already known as so called double-double-slit experiment.
And that the reason why this proposal cannot be used for faster than light communication is that it is impossible to have the source and the points A and B to "rest" at exact positions/locations because of the uncertainty principle.
Now, I think that I found in the meantime the way how to solve that problem. Waiting for him to evaluate the idea.
Anton
I think we may have an implicit reference to something like this :-
http://grad.physics.sunysb.edu/~amarch/
My own feeling is that the conspiracy will run pretty deep. For example the act of preparing photons in the same state will leave Alice in doubt as to whether her photon didn't pass through a polarizing filter because it didn't or whether it simply wasn't sent. I could be wrong but I suspect the opposite state to 'known' is 'not known' which might make the water seem rather muddy.
Maybe also .. the absence of a coincidence detector in a long range 'delayed choice eraser' experiment may open up new possibilities for 'the system' to find loopholes.
Best wishes - C2.
C2,
I have been told by one college professor of physics that this proposal of mine is already known as so called double-double-slit experiment.
And that the reason why this proposal cannot be used for faster than light communication is that it is impossible to have the source and the points A and B to "rest" at exact positions/locations because of the uncertainty principle.
Now, I think that I found in the meantime the way how to solve that problem. Waiting for him to evaluate the idea.
Anton
I was trying to find someone who can answer the following question with relevant argumentation but no luck so far.
The question.
Unobserved/undetected photons are in the usual state of superposition. That state "allows" those photons in accordance to Copenhagen interpretation of quantum mechanics to pass/able-to-be at both slits simultaneously, then to interfere with themselves, with the result in the interference pattern of hits on the detector in DS experiment.
Now, if we have a pairs of entangled photons as I understand both of those entangled photons are also in the state of superposition, until one of them from particular pair is observed/detected, which would influence the other one from particular pair, So, if one entangled photon from some particular pair of entangled photons is observed/detected, that means that that photon is no longer in the state of superposition. Does that also mean that other photons from that pair is also no longer in the state of superposition?
If yes, and if that other photon would then approach the slits in DS experiment would that photon still be able to interfere with itself and hit the detector in an interference pattern of hits or would that photon hit the detector is particle pattern of hits?
Thanks.
The question.
Unobserved/undetected photons are in the usual state of superposition. That state "allows" those photons in accordance to Copenhagen interpretation of quantum mechanics to pass/able-to-be at both slits simultaneously, then to interfere with themselves, with the result in the interference pattern of hits on the detector in DS experiment.
Now, if we have a pairs of entangled photons as I understand both of those entangled photons are also in the state of superposition, until one of them from particular pair is observed/detected, which would influence the other one from particular pair, So, if one entangled photon from some particular pair of entangled photons is observed/detected, that means that that photon is no longer in the state of superposition. Does that also mean that other photons from that pair is also no longer in the state of superposition?
If yes, and if that other photon would then approach the slits in DS experiment would that photon still be able to interfere with itself and hit the detector in an interference pattern of hits or would that photon hit the detector is particle pattern of hits?
Thanks.
Hi Anton,
Sorry this doesn't immediately answer the question .. but it might help.
The wiki entry for Decoherence didn't bite me but ..
See yquantum's post here :- http://forum.physorg.com/index.php?showtop...ndpost&p=260330
Of the long list the best I've found so far is :-
"Can the Decoherent Histories Description of Reality be Considered Satisfactory?"
http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9811/9811050v4.pdf
My interpretation (could well be wrong) is that the wavefunction becomes a representation of the family of possible outcomes .. it doesn't 'collapse' on detection .. you just get one outcome from all the possibilities. If the 'whatever' is entangled then the selection of (say) the local outcome is as much a part of the remote wavefunction as it is of the local wavefunction. There is no 'forcing' .. only selecting.
Your/other thoughts welcome.
Best wishes - C2.
Sorry this doesn't immediately answer the question .. but it might help.
The wiki entry for Decoherence didn't bite me but ..
See yquantum's post here :- http://forum.physorg.com/index.php?showtop...ndpost&p=260330
Of the long list the best I've found so far is :-
"Can the Decoherent Histories Description of Reality be Considered Satisfactory?"
http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9811/9811050v4.pdf
My interpretation (could well be wrong) is that the wavefunction becomes a representation of the family of possible outcomes .. it doesn't 'collapse' on detection .. you just get one outcome from all the possibilities. If the 'whatever' is entangled then the selection of (say) the local outcome is as much a part of the remote wavefunction as it is of the local wavefunction. There is no 'forcing' .. only selecting.
Your/other thoughts welcome.
Best wishes - C2.
QUOTE (Confused2+Sep 13 2007, 01:06 PM)
Hi Anton,
Sorry this doesn't immediately answer the question .. but it might help.
The wiki entry for Decoherence didn't bite me but ..
See yquantum's post here :- http://forum.physorg.com/index.php?showtop...ndpost&p=260330
Of the long list the best I've found so far is :-
"Can the Decoherent Histories Description of Reality be Considered Satisfactory?"
http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9811/9811050v4.pdf
My interpretation (could well be wrong) is that the wavefunction becomes a representation of the family of possible outcomes .. it doesn't 'collapse' on detection .. you just get one outcome from all the possibilities. If the 'whatever' is entangled then the selection of (say) the local outcome is as much a part of the remote wavefunction as it is of the local wavefunction. There is no 'forcing' .. only selecting.
Your/other thoughts welcome.
Best wishes - C2.
C2,
if you are right, and considering that all of those possibilities are binary, yes or no, up or down, etc, what is happening in the moment when one entangled photon is detected and absorbed ? Has that photon "selected" half of all possibilities from that combined wave function in the moment of absorption? Is other entangled photon still in the mixed state ?
Anton
Sorry this doesn't immediately answer the question .. but it might help.
The wiki entry for Decoherence didn't bite me but ..
See yquantum's post here :- http://forum.physorg.com/index.php?showtop...ndpost&p=260330
Of the long list the best I've found so far is :-
"Can the Decoherent Histories Description of Reality be Considered Satisfactory?"
http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9811/9811050v4.pdf
My interpretation (could well be wrong) is that the wavefunction becomes a representation of the family of possible outcomes .. it doesn't 'collapse' on detection .. you just get one outcome from all the possibilities. If the 'whatever' is entangled then the selection of (say) the local outcome is as much a part of the remote wavefunction as it is of the local wavefunction. There is no 'forcing' .. only selecting.
Your/other thoughts welcome.
Best wishes - C2.
C2,
if you are right, and considering that all of those possibilities are binary, yes or no, up or down, etc, what is happening in the moment when one entangled photon is detected and absorbed ? Has that photon "selected" half of all possibilities from that combined wave function in the moment of absorption? Is other entangled photon still in the mixed state ?
Anton
We have two points in space, point A and point B, which are distant from each other, for example, 180 000 000 km.
At point A is just an ordinary detector which just detects and absorbs photons.
At point B is a barrier with two slits, like a barrier with two slits in double slits experiment. Behind a barrier with two slits is a detector, like a detector in double slit experiment.
Between point A and point B is a source which is releasing pairs of entangled photons at very very high frequency, as highest frequency technologically possible in order that we can use as many entangled pairs of photons as possible for each of the "yes" or "no" signals, then source makes small break between signals in order to discern between signals, than again the same for the next signal. So series of entangled photons goes into direction of the point A, and series of photons from theirs pair goes into direction of the point B for one signal, then small break, than again for the next signal.
So...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector
Let us assume that intensity of gravity at the point A, B and the source is the same or almost the same , and let us assume that positions of the points A, B and the source are stationary in regard to each other as it is practically possible.
Now, in the, so called, Illustration above the source is at the equal distance in regard to the points A and B.
But we would place the barrier with slits and detector at point B closer to the source, alike this...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbb>>slits/B>>detector
The distance for which the barrier and detector at point B would be moved closer to the source is the one , "minimum one", which would guarantee with certainty that entangled photons "b" are hitting the slits before their entangled "a" photons from their pairs are hitting the detector at point A. That "minimum distance" must be big enough to compensate eventual movement of the points A, B and the source in regard to each other because the system cannot practically be at the absolute rest and the same distance from each other.
Now, instead having the classical DS barrier with slits and the detector at fixed place at point B, we instead that have the barrier with slits and the detector which are moving for each of the signal ( with certain speed, I do not know what would that speed be considering the whole context of the proposal ) in direction away from the source for each of the signal. For each signal for 2 "minimum distances". Then barrier with detector goes back at their starting position, and the detector resets itself for next signal.
So, for each signal this is starting position, and following movements of the barrier with slits and the detector...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (starting position )
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (1 "minimum distance" )
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (2 "minimum distances")
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector ( reseting, the starting position for next signal ).
Now, we want to send signal "yes" from point A to point B. We already agreed that pure interference pattern of hits on the detector at point B would be "yes" signal. For signal "yes" the detector at point A is removed so photons "a" stay unobserved by passing by point A undetected/unobserved for that "yes" signal. At point B photons "b" are hitting the slits in the mixed state, and consequently the detector behind would register a pure interference pattern of hits no matter that barrier is moving away from photons "b" because photons "b" would hit the silts with the same speed of light regardless are splits with detector moving away or not.
Now, we want to send signal "no" from point A to point B. For signal "no" the detector at point A stays at it's place. At the beginning of the signal "no" because the slits at point B are closer to the source than detector at point A, a photons "b" are hitting the slits in mixed state, which is making an interference pattern of hits on the detector behind the splits.
However, as the barrier and the slits and detector are moving away from the source at one moment we would have the situation when entangled photons "a" would hitting the detector at point A just before or in the exact moment when entangled photons "b" from their pairs would be at slits.
Then the barrier would move a bit more from the source and photons "a" would hitting the detector at point A before photons "b" are near the slits.
Then the systems resets and goes back on the starting position for the next signal.
Here is the crucial assumption. In the described situation when entangled photons "a" are hitting the detector at point A just before or in the exact moment when entangled photons "b" would be at slits, detector/absorption of the photons "a" would influence their photons "b" from their pairs to switch from mixed state onto after an observation state. If photons "b" are not at the slits ( or very close to the slits ) in the mixed state then those photons "b" cannot be at both slits at the same time in the sense of Copenhagen interpretation of quantum mechanics. And the consequence is that those "b" photons cannot pass both slits simultaneously , to interfere with itself, to hit the detector at point B in an interference pattern of hits. Those photons "b" would hit the detector at point B in a particle pattern of hits.
Now, if I am right so far the only thing to discern "yes" from "no" signal is to calibrate experimentally the detector at point B so that detector register and analyze ( by software for the sake of speeding the system ) and immediately recognizes hits on the detector where photons never hit while hitting in a pure interference pattern. Which means signal "no".
What do you think? Does this concept allows faster than light communication? Of course in improved version, if there is the room for that.
Anton
At point A is just an ordinary detector which just detects and absorbs photons.
At point B is a barrier with two slits, like a barrier with two slits in double slits experiment. Behind a barrier with two slits is a detector, like a detector in double slit experiment.
Between point A and point B is a source which is releasing pairs of entangled photons at very very high frequency, as highest frequency technologically possible in order that we can use as many entangled pairs of photons as possible for each of the "yes" or "no" signals, then source makes small break between signals in order to discern between signals, than again the same for the next signal. So series of entangled photons goes into direction of the point A, and series of photons from theirs pair goes into direction of the point B for one signal, then small break, than again for the next signal.
So...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector
Let us assume that intensity of gravity at the point A, B and the source is the same or almost the same , and let us assume that positions of the points A, B and the source are stationary in regard to each other as it is practically possible.
Now, in the, so called, Illustration above the source is at the equal distance in regard to the points A and B.
But we would place the barrier with slits and detector at point B closer to the source, alike this...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbb>>slits/B>>detector
The distance for which the barrier and detector at point B would be moved closer to the source is the one , "minimum one", which would guarantee with certainty that entangled photons "b" are hitting the slits before their entangled "a" photons from their pairs are hitting the detector at point A. That "minimum distance" must be big enough to compensate eventual movement of the points A, B and the source in regard to each other because the system cannot practically be at the absolute rest and the same distance from each other.
Now, instead having the classical DS barrier with slits and the detector at fixed place at point B, we instead that have the barrier with slits and the detector which are moving for each of the signal ( with certain speed, I do not know what would that speed be considering the whole context of the proposal ) in direction away from the source for each of the signal. For each signal for 2 "minimum distances". Then barrier with detector goes back at their starting position, and the detector resets itself for next signal.
So, for each signal this is starting position, and following movements of the barrier with slits and the detector...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (starting position )
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (1 "minimum distance" )
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (2 "minimum distances")
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector ( reseting, the starting position for next signal ).
Now, we want to send signal "yes" from point A to point B. We already agreed that pure interference pattern of hits on the detector at point B would be "yes" signal. For signal "yes" the detector at point A is removed so photons "a" stay unobserved by passing by point A undetected/unobserved for that "yes" signal. At point B photons "b" are hitting the slits in the mixed state, and consequently the detector behind would register a pure interference pattern of hits no matter that barrier is moving away from photons "b" because photons "b" would hit the silts with the same speed of light regardless are splits with detector moving away or not.
Now, we want to send signal "no" from point A to point B. For signal "no" the detector at point A stays at it's place. At the beginning of the signal "no" because the slits at point B are closer to the source than detector at point A, a photons "b" are hitting the slits in mixed state, which is making an interference pattern of hits on the detector behind the splits.
However, as the barrier and the slits and detector are moving away from the source at one moment we would have the situation when entangled photons "a" would hitting the detector at point A just before or in the exact moment when entangled photons "b" from their pairs would be at slits.
Then the barrier would move a bit more from the source and photons "a" would hitting the detector at point A before photons "b" are near the slits.
Then the systems resets and goes back on the starting position for the next signal.
Here is the crucial assumption. In the described situation when entangled photons "a" are hitting the detector at point A just before or in the exact moment when entangled photons "b" would be at slits, detector/absorption of the photons "a" would influence their photons "b" from their pairs to switch from mixed state onto after an observation state. If photons "b" are not at the slits ( or very close to the slits ) in the mixed state then those photons "b" cannot be at both slits at the same time in the sense of Copenhagen interpretation of quantum mechanics. And the consequence is that those "b" photons cannot pass both slits simultaneously , to interfere with itself, to hit the detector at point B in an interference pattern of hits. Those photons "b" would hit the detector at point B in a particle pattern of hits.
Now, if I am right so far the only thing to discern "yes" from "no" signal is to calibrate experimentally the detector at point B so that detector register and analyze ( by software for the sake of speeding the system ) and immediately recognizes hits on the detector where photons never hit while hitting in a pure interference pattern. Which means signal "no".
What do you think? Does this concept allows faster than light communication? Of course in improved version, if there is the room for that.
Anton
So, after more than 200 views since my last post there is still noone here who can say or perhaps there is still noone here who wants to bother to say is this proposal of mine above valid or not, and why? Is there some other place on the net where I could ask for an opinion about the proposal? Thanks.
Now the new proposal...
They are two polarization filters at two points, point A and point B, which are distant from each other, for example, 90 000 000 km.
Roughly between those two points is the source which is releasing, with highest frequency technologically possible, pairs of entangled photons in the manner that one entangled photon from each of the pairs is propagating in the direction of the filter A and other entangled photon is propagating in the direction of the filter B, for some time period X for one signal, then the source makes a small break, then it is releasing again for the same period X for one signal, then again makes a small break, and so on...
Those entangled photons are polarized at 45 degrees in regard to 0 degrees axis.
Polarization filter A is closer to the source so entangled photons "a" which are either passing through the filter A or being blocked by the filter A are determining the polarization of the entangled photons "b" from the each of the particular entangled pair/s.
If the filter A is set up at 0 degrees axis for "yes" signal then photons "b" are coming at filter B with:
1) horizontal polarization if "photon "a" passed the filter A.
2) vertical polarization if photon "a" has been blocked by the filter A.
So, for "yes" signal photons "b" which are passing the filter B are approaching the filter B as vertically polarized.
Now we want to send signal "no" from point A to point B and for that purpose we set up the filter A at 315 degrees of polarization, so photons "b" are coming at filter B with...
1)45 degrees of polarization if "315" degrees photon "a" passed the filter A.
2) 315 degrees of polarization if "45" degrees photon "a" has been blocked by the filter A.
So, for "no" signal photons "b" are polarized at 45 degrees in regard to vertical axis, and half of those photons "b" are passing the filter B.
Now, we know exactly the duration of each signal and pause between the signals, so we know exactly when photons "b" are supposed to hitting the filter B for each signal, and we know exactly duration of the pause between signals, and we also know that source is releasing, for example, 8 000 of entangled pairs of photons for each signal
(how much is currently possible I have no idea, but the principle is as much photons as it is technologically possible).
Now, again, if signal "yes" has been sent from point A to the point B then the filter A is set as it has been described and photons "b" would hitting/passing the filter B on the vertical axis, an axis alike this one
I
( it should be a pure vertical line/s but letter "I" and letter "X" will play the role of vertical axis and 315/45 degrees axis), and if signal "no" has been sent from point A to the point B then the filter A is set up as it has been described and photons "b" would hitting/passing the filter B on the 315 and 45 degrees axis alike these,
X
Now, I think it is possible to distinguish between "yes" and "no" signals in the following manner...
The filter B is not an ordinary filter. The filter B is rotating one whole circle starting from 0 degrees at the very beginning of the photons "b" arriving for the beginning of the particular signal, and ending rotation at the very end of that particular signal at 359 degrees. So, filter B makes 360 degrees rotation during duration of each of the signals. It can move with constant speed or it can move like a second hand on a watch, staying the exact period of time at each of 360 degrees.
Let us say that filter B is rotating alike second hand of a watch.
Let us also say that for each of the 360 degrees the number of photons "b" is counted, that is, how many photons "b" passed the filter B at each of the 360 degrees.
This is what I think it would happen...
If signal is "yes" then as filter B is rotating by moving at 1 degree, 2 degrees to the 90 degrees I think that number of photons "b" passing the filter B would gradually decrease until the number of photons "b" passing filter B comes at 0 in the moment when filter B is at 90 degrees.
Then the number would start to increase as the filter B is moving toward 180 degrees and it will achieve the maximum of photons "b" passing the filter B when the filter B is at 180 degrees.
Then the number would start to decrease until 0 photons "b" passing the filter B in the moment when filter B is at 270 degrees.
And then the number would start to increase until reaches the maximum when the filter B comes at 360/0 degrees.
So, when we see distribution/graph we would see that filter B had two maximums of letting photons "b" through, at 0 and at 180 degrees, and we would see that filter B had two minimums at 90 and 270 degrees when it was blocking all photons "b".
And by look at that distribution/graph we would know that signal was "yes".
Now, if signal is "no" then either would distribution of photons "b" passing the filter B be uniformed or the maximums would be at 45-225/135-315 and minimums would be at 90-270/180-0 degrees axis. So, I am not sure would for signal "no" be uniformed distribution or distribution for signal "no" would be with stated maximums and minimums.
However, I am sure that by looking at distributions/graphs it would be clear was some particular signal "yes" or "no". Therefore, I think it would be possible to achieve faster than light communication as it proposed.
Am I right?
Anton
Now the new proposal...
They are two polarization filters at two points, point A and point B, which are distant from each other, for example, 90 000 000 km.
Roughly between those two points is the source which is releasing, with highest frequency technologically possible, pairs of entangled photons in the manner that one entangled photon from each of the pairs is propagating in the direction of the filter A and other entangled photon is propagating in the direction of the filter B, for some time period X for one signal, then the source makes a small break, then it is releasing again for the same period X for one signal, then again makes a small break, and so on...
Those entangled photons are polarized at 45 degrees in regard to 0 degrees axis.
Polarization filter A is closer to the source so entangled photons "a" which are either passing through the filter A or being blocked by the filter A are determining the polarization of the entangled photons "b" from the each of the particular entangled pair/s.
If the filter A is set up at 0 degrees axis for "yes" signal then photons "b" are coming at filter B with:
1) horizontal polarization if "photon "a" passed the filter A.
2) vertical polarization if photon "a" has been blocked by the filter A.
So, for "yes" signal photons "b" which are passing the filter B are approaching the filter B as vertically polarized.
Now we want to send signal "no" from point A to point B and for that purpose we set up the filter A at 315 degrees of polarization, so photons "b" are coming at filter B with...
1)45 degrees of polarization if "315" degrees photon "a" passed the filter A.
2) 315 degrees of polarization if "45" degrees photon "a" has been blocked by the filter A.
So, for "no" signal photons "b" are polarized at 45 degrees in regard to vertical axis, and half of those photons "b" are passing the filter B.
Now, we know exactly the duration of each signal and pause between the signals, so we know exactly when photons "b" are supposed to hitting the filter B for each signal, and we know exactly duration of the pause between signals, and we also know that source is releasing, for example, 8 000 of entangled pairs of photons for each signal
(how much is currently possible I have no idea, but the principle is as much photons as it is technologically possible).
Now, again, if signal "yes" has been sent from point A to the point B then the filter A is set as it has been described and photons "b" would hitting/passing the filter B on the vertical axis, an axis alike this one
I
( it should be a pure vertical line/s but letter "I" and letter "X" will play the role of vertical axis and 315/45 degrees axis), and if signal "no" has been sent from point A to the point B then the filter A is set up as it has been described and photons "b" would hitting/passing the filter B on the 315 and 45 degrees axis alike these,
X
Now, I think it is possible to distinguish between "yes" and "no" signals in the following manner...
The filter B is not an ordinary filter. The filter B is rotating one whole circle starting from 0 degrees at the very beginning of the photons "b" arriving for the beginning of the particular signal, and ending rotation at the very end of that particular signal at 359 degrees. So, filter B makes 360 degrees rotation during duration of each of the signals. It can move with constant speed or it can move like a second hand on a watch, staying the exact period of time at each of 360 degrees.
Let us say that filter B is rotating alike second hand of a watch.
Let us also say that for each of the 360 degrees the number of photons "b" is counted, that is, how many photons "b" passed the filter B at each of the 360 degrees.
This is what I think it would happen...
If signal is "yes" then as filter B is rotating by moving at 1 degree, 2 degrees to the 90 degrees I think that number of photons "b" passing the filter B would gradually decrease until the number of photons "b" passing filter B comes at 0 in the moment when filter B is at 90 degrees.
Then the number would start to increase as the filter B is moving toward 180 degrees and it will achieve the maximum of photons "b" passing the filter B when the filter B is at 180 degrees.
Then the number would start to decrease until 0 photons "b" passing the filter B in the moment when filter B is at 270 degrees.
And then the number would start to increase until reaches the maximum when the filter B comes at 360/0 degrees.
So, when we see distribution/graph we would see that filter B had two maximums of letting photons "b" through, at 0 and at 180 degrees, and we would see that filter B had two minimums at 90 and 270 degrees when it was blocking all photons "b".
And by look at that distribution/graph we would know that signal was "yes".
Now, if signal is "no" then either would distribution of photons "b" passing the filter B be uniformed or the maximums would be at 45-225/135-315 and minimums would be at 90-270/180-0 degrees axis. So, I am not sure would for signal "no" be uniformed distribution or distribution for signal "no" would be with stated maximums and minimums.
However, I am sure that by looking at distributions/graphs it would be clear was some particular signal "yes" or "no". Therefore, I think it would be possible to achieve faster than light communication as it proposed.
Am I right?
Anton
Opps, I forgot one little detail.
If signal is "yes" than half of the photons "b" are approaching the filter B with vertical polarization and other half with horizontal polarization.
So if the filter B would rotate as I imagined distribution would actually be uniformed because as the filter is moving away from the vertical axis it is simultaneously moving closer to the horizontal axis. I assume that in that situation a loss of vertically polarized photons "b" would be compensate with increasing number of horizontally polarized photons "b". Which, I suppose, would result in uniformed distribution around 360 degrees.
Anton
If signal is "yes" than half of the photons "b" are approaching the filter B with vertical polarization and other half with horizontal polarization.
So if the filter B would rotate as I imagined distribution would actually be uniformed because as the filter is moving away from the vertical axis it is simultaneously moving closer to the horizontal axis. I assume that in that situation a loss of vertically polarized photons "b" would be compensate with increasing number of horizontally polarized photons "b". Which, I suppose, would result in uniformed distribution around 360 degrees.
Anton
yes it would need to be in a resonant state.
QUOTE (Zarabtul+Sep 26 2007, 05:31 AM)
yes it would need to be in a resonant state.
Zarabtul,
yep, I will try to find some other approach.
Since you commented. Can you also give me your opinion about the proposal before this last one? Would that proposal work or not ? And why?
Thanks.
Anton
Zarabtul,
yep, I will try to find some other approach.
Since you commented. Can you also give me your opinion about the proposal before this last one? Would that proposal work or not ? And why?
Thanks.
Anton
Hi, (I) posted in another forum (some time back) how the Dual photon emitters that are used in the "Strange actions at a distance" experiment can be utilized as exactly what Mate is attempting.
Given that the Dual photon emitter sends out a pair of equal and opposite (in direction) photons, and that it has been demonstrated that, by using a crystal to deflect-deviate the flight path of one of the photons, that causes the other photon to deflect-deviate equally, and opposite in direction to it's - Now - far off twin. Detecting that occurrence (the secondary deflection) gives us the fundamentals of a Binary ability, simply by having the crystal that is the cause of the deflection-deviation to waggle slightly, side to side, as to indicate either a one or a zero.
Long range, Faster then Light speed, communications is available to us now, IF we can set it up properly - in space - as you need to have a central emitter sighted along a line vast spatiality as to be able to hit Both a detector at one end and a deflecting-deviational Crystal at the other end.
This method is Instantaneous across whatever distance of space that you could-can line up (Not easy) the proper emitters and detectors across.
Given that the Dual photon emitter sends out a pair of equal and opposite (in direction) photons, and that it has been demonstrated that, by using a crystal to deflect-deviate the flight path of one of the photons, that causes the other photon to deflect-deviate equally, and opposite in direction to it's - Now - far off twin. Detecting that occurrence (the secondary deflection) gives us the fundamentals of a Binary ability, simply by having the crystal that is the cause of the deflection-deviation to waggle slightly, side to side, as to indicate either a one or a zero.
Long range, Faster then Light speed, communications is available to us now, IF we can set it up properly - in space - as you need to have a central emitter sighted along a line vast spatiality as to be able to hit Both a detector at one end and a deflecting-deviational Crystal at the other end.
This method is Instantaneous across whatever distance of space that you could-can line up (Not easy) the proper emitters and detectors across.
QUOTE (Mr. Robin Parsons+Sep 26 2007, 01:18 PM)
Hi, (I) posted in another forum (some time back) how the Dual photon emitters that are used in the "Strange actions at a distance" experiment can be utilized as exactly what Mate is attempting.
Given that the Dual photon emitter sends out a pair of equal and opposite (in direction) photons, and that it has been demonstrated that, by using a crystal to deflect-deviate the flight path of one of the photons, that causes the other photon to deflect-deviate equally, and opposite in direction to it's - Now - far off twin. Detecting that occurrence (the secondary deflection) gives us the fundamentals of a Binary ability, simply by having the crystal that is the cause of the deflection-deviation to waggle slightly, side to side, as to indicate either a one or a zero.
Long range, Faster then Light speed, communications is available to us now, IF we can set it up properly - in space - as you need to have a central emitter sighted along a line vast spatiality as to be able to hit Both a detector at one end and a deflecting-deviational Crystal at the other end.
This method is Instantaneous across whatever distance of space that you could-can line up (Not easy) the proper emitters and detectors across.
Mr Robin Parsons,
I am not familiar with what you are saying. Perhaps a further clarification or/and some material/article which presents the idea of yours more throughly?
Anton
Given that the Dual photon emitter sends out a pair of equal and opposite (in direction) photons, and that it has been demonstrated that, by using a crystal to deflect-deviate the flight path of one of the photons, that causes the other photon to deflect-deviate equally, and opposite in direction to it's - Now - far off twin. Detecting that occurrence (the secondary deflection) gives us the fundamentals of a Binary ability, simply by having the crystal that is the cause of the deflection-deviation to waggle slightly, side to side, as to indicate either a one or a zero.
Long range, Faster then Light speed, communications is available to us now, IF we can set it up properly - in space - as you need to have a central emitter sighted along a line vast spatiality as to be able to hit Both a detector at one end and a deflecting-deviational Crystal at the other end.
This method is Instantaneous across whatever distance of space that you could-can line up (Not easy) the proper emitters and detectors across.
Mr Robin Parsons,
I am not familiar with what you are saying. Perhaps a further clarification or/and some material/article which presents the idea of yours more throughly?
Anton
Maybe, if you are nice, Mr. Dustbunny will feeel your need, and present with proof that he has a Cup, and a very nearly empty Bottle.
But remember to be nice.
But remember to be nice.
QUOTE (Mate+Sep 26 2007, 12:32 AM)
Opps, I forgot one little detail.
If signal is "yes" than half of the photons "b" are approaching the filter B with vertical polarization and other half with horizontal polarization.
So if the filter B would rotate as I imagined distribution would actually be uniformed because as the filter is moving away from the vertical axis it is simultaneously moving closer to the horizontal axis. I assume that in that situation a loss of vertically polarized photons "b" would be compensate with increasing number of horizontally polarized photons "b". Which, I suppose, would result in uniformed distribution around 360 degrees.
Anton
So I overlooked the fact that "yes" photons"b" are hitting the filter B both vertically and horizontally. Therefore a new approach about the problem of distinguishing vertically and horizontally polarized photons "b" of signal "yes" from photons "b" of signal "no" ,which are polarized at 45 degrees in regard to vertical axis.
The crucial assumption for this proposal is that signal is "yes" if there is no evidence to the contrary.
The filter B does not rotate a whole circle alike in the initial proposal but it can move/be at 45/225 and 315/135 degrees.
The filter B changes it's position from 45 to 315 and from 315 to 45 degrees position in regular sequence/manner. What exactly would have to be that sequence so that it yields the most reliable results it is matter of discussion. Let assume just for the sake of presenting the basic idea that sequence is,
45-315-45-315-45-315-45 and so on.
The filter is changing it's position for each of the "b" photon arriving at it. Considering that we know the frequency of the source and we can also calibrate experimentally the filter B, it would be possible to set up the filter B to change it's position between each of the photons arriving at it.
Now, if signal is "yes" then each of the photons "b" during signal "yes" is arriving at filter B with either vertical or horizontal polarization Considering that filter is always at 45/225 or 315/135 position that means that filter is always at 45 degrees in regard to vertical and horizontal axis , which would consequently result that one half of photons "b" during "yes" signal would pass the filter B.
Now, if signal is "no" then each of the photons "b" during signal "no" is arriving at filter B with either 45/225 or 315/135 degrees of polarization.
If particular photon"b" is of 45 degrees polarization and if it is hitting the filter B
which is in 45 degrees position that particular photon would pass the filter B, but if filter B is at 315 degrees position that particular photon would be blocked by the filter B. The same for 315 degrees photon "b", it passes the filter if filter is at 315 degrees position and it would be blocked if filter is at 45 degrees position.
So, how to distinguish "yes" from "no" signals?
Photons "b" are arriving in random sequence in regard to their polarization. In the case of "yes" signal they can pass either vertical or horizontal axis because no matter is particular "yes" photon "b" of vertical or horizontal polarization the filter is always at 45 degrees in regard to both vertical or horizontal polarization , being in the 45/225 degrees position or at 3157135 degrees position. So half of the photons "b" during "yes" signal are going to pass the filter B.
However, photons "b" during signal "no " are also hitting the filter B in random sequence. Considering that filter is moving from 45/225 to 3315/135 degrees of position in regular sequence then that can result in the situation when filter B would be more than 50 % times at 45/225 degrees position when photons "B" hitting the filter would be of 315/135 degrees of polarization, or vice versa.
In short. Even the number of 45/225 and 315/135 photons "b" would hit the filter B in equal number of times during "no" signal it may very well happen that either more than half or less than half of those photons would pass the filter B.
Therefore signal "no".
Anton
If signal is "yes" than half of the photons "b" are approaching the filter B with vertical polarization and other half with horizontal polarization.
So if the filter B would rotate as I imagined distribution would actually be uniformed because as the filter is moving away from the vertical axis it is simultaneously moving closer to the horizontal axis. I assume that in that situation a loss of vertically polarized photons "b" would be compensate with increasing number of horizontally polarized photons "b". Which, I suppose, would result in uniformed distribution around 360 degrees.
Anton
So I overlooked the fact that "yes" photons"b" are hitting the filter B both vertically and horizontally. Therefore a new approach about the problem of distinguishing vertically and horizontally polarized photons "b" of signal "yes" from photons "b" of signal "no" ,which are polarized at 45 degrees in regard to vertical axis.
The crucial assumption for this proposal is that signal is "yes" if there is no evidence to the contrary.
The filter B does not rotate a whole circle alike in the initial proposal but it can move/be at 45/225 and 315/135 degrees.
The filter B changes it's position from 45 to 315 and from 315 to 45 degrees position in regular sequence/manner. What exactly would have to be that sequence so that it yields the most reliable results it is matter of discussion. Let assume just for the sake of presenting the basic idea that sequence is,
45-315-45-315-45-315-45 and so on.
The filter is changing it's position for each of the "b" photon arriving at it. Considering that we know the frequency of the source and we can also calibrate experimentally the filter B, it would be possible to set up the filter B to change it's position between each of the photons arriving at it.
Now, if signal is "yes" then each of the photons "b" during signal "yes" is arriving at filter B with either vertical or horizontal polarization Considering that filter is always at 45/225 or 315/135 position that means that filter is always at 45 degrees in regard to vertical and horizontal axis , which would consequently result that one half of photons "b" during "yes" signal would pass the filter B.
Now, if signal is "no" then each of the photons "b" during signal "no" is arriving at filter B with either 45/225 or 315/135 degrees of polarization.
If particular photon"b" is of 45 degrees polarization and if it is hitting the filter B
which is in 45 degrees position that particular photon would pass the filter B, but if filter B is at 315 degrees position that particular photon would be blocked by the filter B. The same for 315 degrees photon "b", it passes the filter if filter is at 315 degrees position and it would be blocked if filter is at 45 degrees position.
So, how to distinguish "yes" from "no" signals?
Photons "b" are arriving in random sequence in regard to their polarization. In the case of "yes" signal they can pass either vertical or horizontal axis because no matter is particular "yes" photon "b" of vertical or horizontal polarization the filter is always at 45 degrees in regard to both vertical or horizontal polarization , being in the 45/225 degrees position or at 3157135 degrees position. So half of the photons "b" during "yes" signal are going to pass the filter B.
However, photons "b" during signal "no " are also hitting the filter B in random sequence. Considering that filter is moving from 45/225 to 3315/135 degrees of position in regular sequence then that can result in the situation when filter B would be more than 50 % times at 45/225 degrees position when photons "B" hitting the filter would be of 315/135 degrees of polarization, or vice versa.
In short. Even the number of 45/225 and 315/135 photons "b" would hit the filter B in equal number of times during "no" signal it may very well happen that either more than half or less than half of those photons would pass the filter B.
Therefore signal "no".
Anton
Try here a Page from; Thomas Jefferson National Accelerator Facility
QUOTE (Mr. Robin Parsons+Sep 27 2007, 12:17 PM)
Try here a Page from; Thomas Jefferson National Accelerator Facility
Mr. Robin Parsons,
In this particular thread I have not questioned the reality of the entanglement as a phenomenon, I am starting from the assumption that entanglement is the reality. To go from there to use an entanglement for faster than light communication. And faster than light communication is much more difficult goal to achieve than confirming the entanglement alike correlation between particles after results on both "ends" are compared.
As this excerpt from the article you posted is saying...
"Whatever the nature of the connection between entangled particles may be, nearly all physicists agree that it cannot be used to transmit messages faster than the speed of light. All it can do is assure that a random choice by one entangled particle is instantly echoed by its distant partner. This is not the same thing as transmitting information, the experts say, and therefore it does not violate relativity theory."
Anton
Mr. Robin Parsons,
In this particular thread I have not questioned the reality of the entanglement as a phenomenon, I am starting from the assumption that entanglement is the reality. To go from there to use an entanglement for faster than light communication. And faster than light communication is much more difficult goal to achieve than confirming the entanglement alike correlation between particles after results on both "ends" are compared.
As this excerpt from the article you posted is saying...
"Whatever the nature of the connection between entangled particles may be, nearly all physicists agree that it cannot be used to transmit messages faster than the speed of light. All it can do is assure that a random choice by one entangled particle is instantly echoed by its distant partner. This is not the same thing as transmitting information, the experts say, and therefore it does not violate relativity theory."
Anton
Hi Anton,
Your tests are very difficult to follow.
To simplify things.. maybe build up gradually ..
Can I suggest starting with a source of entangled photons (polarised at 90 degrees) .. then try to send pH to Alice and pV to Bob .. is this possible? Unless you detect a pH photon on the way to Alice you will need to block the pUnkown on the way to Bob .. which destroys the photon you were trying to send to Alice .. and so on.
Bestb wishes - C2.
Your tests are very difficult to follow.
To simplify things.. maybe build up gradually ..
Can I suggest starting with a source of entangled photons (polarised at 90 degrees) .. then try to send pH to Alice and pV to Bob .. is this possible? Unless you detect a pH photon on the way to Alice you will need to block the pUnkown on the way to Bob .. which destroys the photon you were trying to send to Alice .. and so on.
Bestb wishes - C2.
QUOTE (Confused2+Sep 27 2007, 01:21 PM)
Hi Anton,
Your tests are very difficult to follow.
To simplify things.. maybe build up gradually ..
Can I suggest starting with a source of entangled photons (polarised at 90 degrees) .. then try to send pH to Alice and pV to Bob .. is this possible? Unless you detect a pH photon on the way to Alice you will need to block the pUnkown on the way to Bob .. which destroys the photon you were trying to send to Alice .. and so on.
Bestb wishes - C2.
C2,
if it is difficult to follow then I will try to present the essence of proposal in the most simple way I can ( taking also into consideration that English is my second language ).
YES SIGNAL
The filter A is at 0/180 degrees position for "yes" signal. Therefore photons "b" arriving at filter B are either horizontally or vertically polarized for signal "yes".
The filter B is changing it's position for each particular arriving photon "b". So, if filter B was at 315/135 position for particular photon then for the next photon the filter is at 45/225 position, then for the next photon filter is at 315/135, and so on.
Considering that filter B is always either at 315/135 or 45/225 degrees position then horizontally or vertically polarized photons "b" of "yes" signal are always at 45 degrees in regard to the polarization of the filter B.
45 degrees= 50 % photons "b" passed the filter B.
50 % photons "b" passing the filter B= signal "yes"
NO SIGNAL
The filter A is at 315/45 degrees position for "no" signal. Therefore photons "b" arriving at filter B are either of 315/135 or 45/225 polarization for signal "no".
Now the essence of proposal. Photons "b" are hitting the filter B in random fashion in regard to their polarization ( as photons "b" of "yes" signal, of course ). If filter B would stay in the same position , for example in 315/135 degrees position, then at the end of the signal "no" the same number of 315/135 and 45/225 photons would hit the filter B ( as more photons hit the filter as more would total number be closer to the perfect 50 % for both polarizations). Which would result that half of those photons would pass the filter B, alike with "yes" signal.
However, if filter B is regularly changing it's position for each of arriving photons then
probability that unequal number of 315 or 45 photons would pass the filter B is increasing. If that discrepancy would be bigger than expected error/unbalance because some photons are lost etc, then with certainty ( which level of certainty/probability would have to be calculated ) we would assume/know that signal was "no".
For example, photons "b" of "no" signal, while 315/135 is "1" and 45/225 is "0".
Combination 1 and 0 or 0 and 1 means that filter B blocked photon with ortogonal polarization.
Combinatin of 1 and 1 or 0 and 0 means that filter B passed photon.
Polarizations of sequence of photons "b" of "no" signal arriving at filter B:
1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0
Polarization/position of the filter B:
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1
Therefore:
1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1
So, for example, 8 from 30 passing= "no" signal
Of course, this is simplification.
Am I clearer now?
Anton
Your tests are very difficult to follow.
To simplify things.. maybe build up gradually ..
Can I suggest starting with a source of entangled photons (polarised at 90 degrees) .. then try to send pH to Alice and pV to Bob .. is this possible? Unless you detect a pH photon on the way to Alice you will need to block the pUnkown on the way to Bob .. which destroys the photon you were trying to send to Alice .. and so on.
Bestb wishes - C2.
C2,
if it is difficult to follow then I will try to present the essence of proposal in the most simple way I can ( taking also into consideration that English is my second language ).
YES SIGNAL
The filter A is at 0/180 degrees position for "yes" signal. Therefore photons "b" arriving at filter B are either horizontally or vertically polarized for signal "yes".
The filter B is changing it's position for each particular arriving photon "b". So, if filter B was at 315/135 position for particular photon then for the next photon the filter is at 45/225 position, then for the next photon filter is at 315/135, and so on.
Considering that filter B is always either at 315/135 or 45/225 degrees position then horizontally or vertically polarized photons "b" of "yes" signal are always at 45 degrees in regard to the polarization of the filter B.
45 degrees= 50 % photons "b" passed the filter B.
50 % photons "b" passing the filter B= signal "yes"
NO SIGNAL
The filter A is at 315/45 degrees position for "no" signal. Therefore photons "b" arriving at filter B are either of 315/135 or 45/225 polarization for signal "no".
Now the essence of proposal. Photons "b" are hitting the filter B in random fashion in regard to their polarization ( as photons "b" of "yes" signal, of course ). If filter B would stay in the same position , for example in 315/135 degrees position, then at the end of the signal "no" the same number of 315/135 and 45/225 photons would hit the filter B ( as more photons hit the filter as more would total number be closer to the perfect 50 % for both polarizations). Which would result that half of those photons would pass the filter B, alike with "yes" signal.
However, if filter B is regularly changing it's position for each of arriving photons then
probability that unequal number of 315 or 45 photons would pass the filter B is increasing. If that discrepancy would be bigger than expected error/unbalance because some photons are lost etc, then with certainty ( which level of certainty/probability would have to be calculated ) we would assume/know that signal was "no".
For example, photons "b" of "no" signal, while 315/135 is "1" and 45/225 is "0".
Combination 1 and 0 or 0 and 1 means that filter B blocked photon with ortogonal polarization.
Combinatin of 1 and 1 or 0 and 0 means that filter B passed photon.
Polarizations of sequence of photons "b" of "no" signal arriving at filter B:
1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0
Polarization/position of the filter B:
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1
Therefore:
1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1
So, for example, 8 from 30 passing= "no" signal
Of course, this is simplification.
Am I clearer now?
Anton
QUOTE (Mate+Sep 27 2007, 09:04 AM)
(SNIP) As this excerpt from the article you posted is saying...
"Whatever the nature of the connection between entangled particles may be, nearly all physicists agree that it cannot be used to transmit messages faster than the speed of light. All it can do is assure that a random choice by one entangled particle is instantly echoed by its distant partner. This is not the same thing as transmitting information, the experts say, and therefore it does not violate relativity theory." Anton (SNoP)
"Whatever the nature of the connection between entangled particles may be, nearly all physicists agree that it cannot be used to transmit messages faster than the speed of light. All it can do is assure that a random choice by one entangled particle is instantly echoed by its distant partner. This is not the same thing as transmitting information, the experts say, and therefore it does not violate relativity theory." Anton (SNoP)
Yes in the circumstances that they are talking about, mine is simply facilitated by the use of a dual photon emitter (Not through a fiber cable) at the center, and the effecting operator is a crystal that causes simultaneous reactions- equal an opposite deviation - at either end.
If you can align it properly you can-could use the phenomenon to transmit information simply by waggling the crystal.
If you can align it properly you can-could use the phenomenon to transmit information simply by waggling the crystal.
C2,
yesterday I made a mistake in attempt to clarify the proposal.
Yesterday I said...
Combination 1 and 0 or 0 and 1 means that filter B blocked a photon with orthogonal polarization.
Combinations of 1 and 1 or 0 and 0 means that filter B passed photon.
Polarizations of sequence of photons "b" of "no" signal arriving at filter B:
1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0
Polarization/position of the filter B:
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
Therefore:
1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
So, it is not that 8 photons "b" from 30 are passing, but 22 photons "b" from 30 are passing.
Sorry about that mistake in typing. However regardless of that mistake the result is the same.
NO signal.
Anton
yesterday I made a mistake in attempt to clarify the proposal.
Yesterday I said...
Combination 1 and 0 or 0 and 1 means that filter B blocked a photon with orthogonal polarization.
Combinations of 1 and 1 or 0 and 0 means that filter B passed photon.
Polarizations of sequence of photons "b" of "no" signal arriving at filter B:
1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0
Polarization/position of the filter B:
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
Therefore:
1 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
So, it is not that 8 photons "b" from 30 are passing, but 22 photons "b" from 30 are passing.
Sorry about that mistake in typing. However regardless of that mistake the result is the same.
NO signal.
Anton
BTW you might wish to note that the idea of instantaneous (Across what we see as time) is indicative of a Background-realm-space of 'no time' - a background of 'no time' is what would be expected to be seen if the background was-is the infinite, as the Infinite has no time.
Neat
Neat
QUOTE (Mr. Robin Parsons+Sep 30 2007, 01:54 AM)
BTW you might wish to note that the idea of instantaneous (Across what we see as time) is indicative of a Background-realm-space of 'no time' - a background of 'no time' is what would be expected to be seen if the background was-is the infinite, as the Infinite has no time.
Neat
I do not think that human mind can really understand the term infinite.
Of course, we are all using the term, it can be applied in limited meaning as a description of something, but none really understand infinity.
Anton
Neat
I do not think that human mind can really understand the term infinite.
Of course, we are all using the term, it can be applied in limited meaning as a description of something, but none really understand infinity.
Anton
QUOTE (Mate+Sep 30 2007, 06:04 AM)
I do not think that human mind can really understand the term infinite. Of course, we are all using the term, it can be applied in limited meaning as a description of something, but none really understand infinity. Anton
Yes (I) would agree, as (I) have, on other occasions, noted that it is in-conceivable, un-thinkable.
Then again, as we would exist within it, (for lack of a better way to put that) we would be sensate of it/to it.......
Yes (I) would agree, as (I) have, on other occasions, noted that it is in-conceivable, un-thinkable.
Then again, as we would exist within it, (for lack of a better way to put that) we would be sensate of it/to it.......
Now that is an intelligent statement. I really wish people could understand that one little word as it would make everything in their lives so much easier. especially for mathematicians. I think there are those that do understand it the problem is is that it is such a small group of people that truely understand the meaning it has certainly been lost on the masses.
In the meantime I had a conversation with one person which comments convinced me that FTL communication could not be achieved in the manner as I have proposed it.
Then today I am not so sure about that.
Anyway, I would be grateful if someone here could say why exactly, preferably in a detailed argumentation/explanation, why following proposal for FTL communication would not work.
Here is the last version...
If signal is "no" then filter A is at 315/135 degrees position and therefore photons "b" arriving at filter B are either of 315/135 or 45/225 degrees of polarization, and if signal is "yes" then filter A is at 0/180 degrees position and therefore photons "b" arriving at filter B are either of 0/180 or 270/90 degrees of polarization.
The initial position of the filter B is at 315/135 degrees at the beginning of the each signal.
We know the time interval between two photons "b" arriving at the filter B.
When first photon "b" at the beginning of the signal has passed the filter B at 315/135 degrees, to hit the detector behind the filter, we take that as the beginning of that particular signal.
Filter B stays at 315/135 position as long as photons are passing in the expected intervals through the filter.
As soon as one photon have not passed the filter, while according to the time passed since last photon passed the filter we know that that photon has been blocked by the filter , then software gives command to the filter to switch on the 45/225 degrees position.
If next photon would pass the filter B set up at 45/225 degrees then filter stays at that position as long as photons are passing through the filter in expected time intervals. As it is described above.
As soon as some photon would not passed the filter B in expected time interval then software again gives a command to the filter B to switch on the 315/135 degrees. And so on.
So, if we have, for example, a sequence of 10 photons of "yes" signal with the same polarization and sequence of 10 photons of "no" signal with the same polarization, that sequence of photons "no" has 100% probability that it will pass the filter as a whole, while probability that that sequence of photons "yes" would pass the filter as a whole is smaller. Parts of the sequence which are made of photons of "yes" signal, and which are of the same polarization, would pass/have-greater-probability of passing the filter in more uniformed distribution then parts of the sequence which are made of photons of "no" signal, and which are also of the same polarization. That because , in the context of the proposed movement of the filter B , every part of the sequence of the signal "no" which is made of two or more photons of the same polarization would pass the filter B as a streak of consecutive photons.
In short, with parts of the sequence of photons of the same polarization of "yes" signal we have "just" a probability that those photons would pass the filter B in a streaks, while with parts of the sequence of photons of the same polarization of "no" signal we have a certainty that those photons would pass the filter B in streaks.
Of course, it may happen that some parts of the sequence of photons of "yes" signal with same polarization would also pass in streaks , and it may also happen that some parts of the sequence of photons of "yes" signal , which are of alternating polarization, would also pass in streaks. So the question is are all of these possibilities "canceling" each other so we cannot distinguish "yes" from "no" signals on the basis of characteristic pattern of distributions of photons passing the filter B, in the context of proposed movement of the filter B.
In this proposal described expected characteristic pattern of distribution of the signal "yes" passing the filter B is more unformed one with less consecutive streaks of photons passing the filter B , while expected characteristic pattern of distribution of the signal "no" passing the filter B is less uniformed one with more consecutive streaks. While total percentage of photons passing the filter B is the same/very similar for both signals, close to 50 % for both signals.
So, why I am wrong?
Anton
Then today I am not so sure about that.
Anyway, I would be grateful if someone here could say why exactly, preferably in a detailed argumentation/explanation, why following proposal for FTL communication would not work.
Here is the last version...
If signal is "no" then filter A is at 315/135 degrees position and therefore photons "b" arriving at filter B are either of 315/135 or 45/225 degrees of polarization, and if signal is "yes" then filter A is at 0/180 degrees position and therefore photons "b" arriving at filter B are either of 0/180 or 270/90 degrees of polarization.
The initial position of the filter B is at 315/135 degrees at the beginning of the each signal.
We know the time interval between two photons "b" arriving at the filter B.
When first photon "b" at the beginning of the signal has passed the filter B at 315/135 degrees, to hit the detector behind the filter, we take that as the beginning of that particular signal.
Filter B stays at 315/135 position as long as photons are passing in the expected intervals through the filter.
As soon as one photon have not passed the filter, while according to the time passed since last photon passed the filter we know that that photon has been blocked by the filter , then software gives command to the filter to switch on the 45/225 degrees position.
If next photon would pass the filter B set up at 45/225 degrees then filter stays at that position as long as photons are passing through the filter in expected time intervals. As it is described above.
As soon as some photon would not passed the filter B in expected time interval then software again gives a command to the filter B to switch on the 315/135 degrees. And so on.
So, if we have, for example, a sequence of 10 photons of "yes" signal with the same polarization and sequence of 10 photons of "no" signal with the same polarization, that sequence of photons "no" has 100% probability that it will pass the filter as a whole, while probability that that sequence of photons "yes" would pass the filter as a whole is smaller. Parts of the sequence which are made of photons of "yes" signal, and which are of the same polarization, would pass/have-greater-probability of passing the filter in more uniformed distribution then parts of the sequence which are made of photons of "no" signal, and which are also of the same polarization. That because , in the context of the proposed movement of the filter B , every part of the sequence of the signal "no" which is made of two or more photons of the same polarization would pass the filter B as a streak of consecutive photons.
In short, with parts of the sequence of photons of the same polarization of "yes" signal we have "just" a probability that those photons would pass the filter B in a streaks, while with parts of the sequence of photons of the same polarization of "no" signal we have a certainty that those photons would pass the filter B in streaks.
Of course, it may happen that some parts of the sequence of photons of "yes" signal with same polarization would also pass in streaks , and it may also happen that some parts of the sequence of photons of "yes" signal , which are of alternating polarization, would also pass in streaks. So the question is are all of these possibilities "canceling" each other so we cannot distinguish "yes" from "no" signals on the basis of characteristic pattern of distributions of photons passing the filter B, in the context of proposed movement of the filter B.
In this proposal described expected characteristic pattern of distribution of the signal "yes" passing the filter B is more unformed one with less consecutive streaks of photons passing the filter B , while expected characteristic pattern of distribution of the signal "no" passing the filter B is less uniformed one with more consecutive streaks. While total percentage of photons passing the filter B is the same/very similar for both signals, close to 50 % for both signals.
So, why I am wrong?
Anton
QUOTE (Mate+Sep 14 2007, 12:20 PM)
We have two points in space, point A and point B, which are distant from each other, for example, 180 000 000 km.
At point A is just an ordinary detector which just detects and absorbs photons.
At point B is a barrier with two slits, like a barrier with two slits in double slits experiment. Behind a barrier with two slits is a detector, like a detector in double slit experiment.
Between point A and point B is a source which is releasing pairs of entangled photons at very very high frequency, as highest frequency technologically possible in order that we can use as many entangled pairs of photons as possible for each of the "yes" or "no" signals, then source makes small break between signals in order to discern between signals, than again the same for the next signal. So series of entangled photons goes into direction of the point A, and series of photons from theirs pair goes into direction of the point B for one signal, then small break, than again for the next signal.
So...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector
Let us assume that intensity of gravity at the point A, B and the source is the same or almost the same , and let us assume that positions of the points A, B and the source are stationary in regard to each other as it is practically possible.
Now, in the, so called, Illustration above the source is at the equal distance in regard to the points A and B.
But we would place the barrier with slits and detector at point B closer to the source, alike this...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbb>>slits/B>>detector
The distance for which the barrier and detector at point B would be moved closer to the source is the one , "minimum one", which would guarantee with certainty that entangled photons "b" are hitting the slits before their entangled "a" photons from their pairs are hitting the detector at point A. That "minimum distance" must be big enough to compensate eventual movement of the points A, B and the source in regard to each other because the system cannot practically be at the absolute rest and the same distance from each other.
Now, instead having the classical DS barrier with slits and the detector at fixed place at point B, we instead that have the barrier with slits and the detector which are moving for each of the signal ( with certain speed, I do not know what would that speed be considering the whole context of the proposal ) in direction away from the source for each of the signal. For each signal for 2 "minimum distances". Then barrier with detector goes back at their starting position, and the detector resets itself for next signal.
So, for each signal this is starting position, and following movements of the barrier with slits and the detector...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (starting position )
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (1 "minimum distance" )
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (2 "minimum distances")
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector ( reseting, the starting position for next signal ).
Now, we want to send signal "yes" from point A to point B. We already agreed that pure interference pattern of hits on the detector at point B would be "yes" signal. For signal "yes" the detector at point A is removed so photons "a" stay unobserved by passing by point A undetected/unobserved for that "yes" signal. At point B photons "b" are hitting the slits in the mixed state, and consequently the detector behind would register a pure interference pattern of hits no matter that barrier is moving away from photons "b" because photons "b" would hit the silts with the same speed of light regardless are splits with detector moving away or not.
Now, we want to send signal "no" from point A to point B. For signal "no" the detector at point A stays at it's place. At the beginning of the signal "no" because the slits at point B are closer to the source than detector at point A, a photons "b" are hitting the slits in mixed state, which is making an interference pattern of hits on the detector behind the splits.
However, as the barrier and the slits and detector are moving away from the source at one moment we would have the situation when entangled photons "a" would hitting the detector at point A just before or in the exact moment when entangled photons "b" from their pairs would be at slits.
Then the barrier would move a bit more from the source and photons "a" would hitting the detector at point A before photons "b" are near the slits.
Then the systems resets and goes back on the starting position for the next signal.
Here is the crucial assumption. In the described situation when entangled photons "a" are hitting the detector at point A just before or in the exact moment when entangled photons "b" would be at slits, detector/absorption of the photons "a" would influence their photons "b" from their pairs to switch from mixed state onto after an observation state. If photons "b" are not at the slits ( or very close to the slits ) in the mixed state then those photons "b" cannot be at both slits at the same time in the sense of Copenhagen interpretation of quantum mechanics. And the consequence is that those "b" photons cannot pass both slits simultaneously , to interfere with itself, to hit the detector at point B in an interference pattern of hits. Those photons "b" would hit the detector at point B in a particle pattern of hits.
Now, if I am right so far the only thing to discern "yes" from "no" signal is to calibrate experimentally the detector at point B so that detector register and analyze ( by software for the sake of speeding the system ) and immediately recognizes hits on the detector where photons never hit while hitting in a pure interference pattern. Which means signal "no".
What do you think? Does this concept allows faster than light communication? Of course in improved version, if there is the room for that.
Anton
It is exceptionally hard to get some information about physics. Finally I got the answer about this proposal. I thought, wrongly, that if two particles are in the entangled state that absorption of one entangled particles is forcing other entangled particles from the pair to switch from mixed unobserved state into after an observation state.
Let me ask this. If one entangled particle is observed passing one of the slits in double slit apparatus, what would happen with other entangled particle from the pair which is in that very moment at slits on other double split apparatus?
Does entangled particle ever lose an "ability" to pass through both slits ( in the sense of Copenhagen interpretation of quantum mechanics ) if other entangled particle from the pair is been observed and localized while passing by ( not being absorbed ) ?
Anton
At point A is just an ordinary detector which just detects and absorbs photons.
At point B is a barrier with two slits, like a barrier with two slits in double slits experiment. Behind a barrier with two slits is a detector, like a detector in double slit experiment.
Between point A and point B is a source which is releasing pairs of entangled photons at very very high frequency, as highest frequency technologically possible in order that we can use as many entangled pairs of photons as possible for each of the "yes" or "no" signals, then source makes small break between signals in order to discern between signals, than again the same for the next signal. So series of entangled photons goes into direction of the point A, and series of photons from theirs pair goes into direction of the point B for one signal, then small break, than again for the next signal.
So...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector
Let us assume that intensity of gravity at the point A, B and the source is the same or almost the same , and let us assume that positions of the points A, B and the source are stationary in regard to each other as it is practically possible.
Now, in the, so called, Illustration above the source is at the equal distance in regard to the points A and B.
But we would place the barrier with slits and detector at point B closer to the source, alike this...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbb>>slits/B>>detector
The distance for which the barrier and detector at point B would be moved closer to the source is the one , "minimum one", which would guarantee with certainty that entangled photons "b" are hitting the slits before their entangled "a" photons from their pairs are hitting the detector at point A. That "minimum distance" must be big enough to compensate eventual movement of the points A, B and the source in regard to each other because the system cannot practically be at the absolute rest and the same distance from each other.
Now, instead having the classical DS barrier with slits and the detector at fixed place at point B, we instead that have the barrier with slits and the detector which are moving for each of the signal ( with certain speed, I do not know what would that speed be considering the whole context of the proposal ) in direction away from the source for each of the signal. For each signal for 2 "minimum distances". Then barrier with detector goes back at their starting position, and the detector resets itself for next signal.
So, for each signal this is starting position, and following movements of the barrier with slits and the detector...
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (starting position )
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (1 "minimum distance" )
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector (2 "minimum distances")
A/detector<<aaaaaaaaaaaaaaaaaaaaaaaaaaa<<source>>bbbbbbbbbbbbbbbbbbbb>>slits/B>>>>detector ( reseting, the starting position for next signal ).
Now, we want to send signal "yes" from point A to point B. We already agreed that pure interference pattern of hits on the detector at point B would be "yes" signal. For signal "yes" the detector at point A is removed so photons "a" stay unobserved by passing by point A undetected/unobserved for that "yes" signal. At point B photons "b" are hitting the slits in the mixed state, and consequently the detector behind would register a pure interference pattern of hits no matter that barrier is moving away from photons "b" because photons "b" would hit the silts with the same speed of light regardless are splits with detector moving away or not.
Now, we want to send signal "no" from point A to point B. For signal "no" the detector at point A stays at it's place. At the beginning of the signal "no" because the slits at point B are closer to the source than detector at point A, a photons "b" are hitting the slits in mixed state, which is making an interference pattern of hits on the detector behind the splits.
However, as the barrier and the slits and detector are moving away from the source at one moment we would have the situation when entangled photons "a" would hitting the detector at point A just before or in the exact moment when entangled photons "b" from their pairs would be at slits.
Then the barrier would move a bit more from the source and photons "a" would hitting the detector at point A before photons "b" are near the slits.
Then the systems resets and goes back on the starting position for the next signal.
Here is the crucial assumption. In the described situation when entangled photons "a" are hitting the detector at point A just before or in the exact moment when entangled photons "b" would be at slits, detector/absorption of the photons "a" would influence their photons "b" from their pairs to switch from mixed state onto after an observation state. If photons "b" are not at the slits ( or very close to the slits ) in the mixed state then those photons "b" cannot be at both slits at the same time in the sense of Copenhagen interpretation of quantum mechanics. And the consequence is that those "b" photons cannot pass both slits simultaneously , to interfere with itself, to hit the detector at point B in an interference pattern of hits. Those photons "b" would hit the detector at point B in a particle pattern of hits.
Now, if I am right so far the only thing to discern "yes" from "no" signal is to calibrate experimentally the detector at point B so that detector register and analyze ( by software for the sake of speeding the system ) and immediately recognizes hits on the detector where photons never hit while hitting in a pure interference pattern. Which means signal "no".
What do you think? Does this concept allows faster than light communication? Of course in improved version, if there is the room for that.
Anton
It is exceptionally hard to get some information about physics. Finally I got the answer about this proposal. I thought, wrongly, that if two particles are in the entangled state that absorption of one entangled particles is forcing other entangled particles from the pair to switch from mixed unobserved state into after an observation state.
Let me ask this. If one entangled particle is observed passing one of the slits in double slit apparatus, what would happen with other entangled particle from the pair which is in that very moment at slits on other double split apparatus?
Does entangled particle ever lose an "ability" to pass through both slits ( in the sense of Copenhagen interpretation of quantum mechanics ) if other entangled particle from the pair is been observed and localized while passing by ( not being absorbed ) ?
Anton
Space, EPR experiment and quantum potential
The idea of a physical space where time exists only as a duration of events throws new light in the explanation and interpretation of Einstein-Podolski-Rosen (EPR) experiment and thus of quantum non-locality.
space itself "informs" both of particles about their spin.
space is a direct information medium
The idea of a physical space where time exists only as a duration of events throws new light in the explanation and interpretation of Einstein-Podolski-Rosen (EPR) experiment and thus of quantum non-locality.
space itself "informs" both of particles about their spin.
space is a direct information medium
QUOTE (fivedoughnut+Aug 12 2007, 04:12 AM)
Fivedoughnuts simple guide to hyperspacial communication ..... a sketch of 4 easy to follow steps.
Righty ho folks, listen good & proper....
Step 1: Collect positrons and electrons from gamma photon decays (lots are recommended)
Step 2: Entrap them cryogenically in pre-purged magnetic containment devices (ultra vacuum/particle density near zero) to minimize possibility of annihilation.
Step 3: Separate the electron and positron containment devices (at any distance you choose fit).
Step 4: finally mobilise (by affecting containment field) either the electrons or positrons ..... then by the magic of what used to be called quantum interconnectivity, the corresponding anti-particles will anti-wobble, thus producing an inductive effect on their containment coils which are simply amplified as a signal.
I'd like the 1st hyperspacial transmission to be the song "I've got a lovely bunch of coconuts".
Damn fine technology, same concept allows for hyperspacial processing too!
you've tried to teach us that theory at least a hundred times....we get it....you like to showoff!!!!!
end of story.....turn off your computer......stand up,step back.and clean up your room!!!
Righty ho folks, listen good & proper....
Step 1: Collect positrons and electrons from gamma photon decays (lots are recommended)
Step 2: Entrap them cryogenically in pre-purged magnetic containment devices (ultra vacuum/particle density near zero) to minimize possibility of annihilation.
Step 3: Separate the electron and positron containment devices (at any distance you choose fit).
Step 4: finally mobilise (by affecting containment field) either the electrons or positrons ..... then by the magic of what used to be called quantum interconnectivity, the corresponding anti-particles will anti-wobble, thus producing an inductive effect on their containment coils which are simply amplified as a signal.
I'd like the 1st hyperspacial transmission to be the song "I've got a lovely bunch of coconuts".
Damn fine technology, same concept allows for hyperspacial processing too!
you've tried to teach us that theory at least a hundred times....we get it....you like to showoff!!!!!
end of story.....turn off your computer......stand up,step back.and clean up your room!!!
I think the error is in your assumption regarding the alignment of the two sets of slits. Would they not need to be 'perfectly aligned'? In this regard you could not get the required signal because you could not get them lined up with each other, which would destroy the entanglement.
If you used a cloud of entangled particles that each traveler took with them then wouldn't you have some difficulty in choosing the requisite 'pair' for measurement?
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