- #1
DevilsAvocado
Gold Member
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I truly believe Einstein was right; Faster-than-light (FTL) communication is not possible.
And this far, nobody has proven Einstein wrong, not even “Italian Massive Mountains” could do it.
There are hypothetical particles like the Tachyon that (in theory) always moves faster than light, but I don’t like it, and the obvious skeptic reason is unsolvable and chaotic Causality paradoxes that comes by default with FTL. Furthermore we have the No-communication theorem.
In Bell test experiments no real usable information is propagated, only random “quantum information” and later (< c) established correlations between Alice & Bob, as a consequence of the shared wavefunction between the two entangled photons A & B.
Therefore I know something must be wrong with my thought experiment.
Could someone please tell me what it is?
* EPR-Bell FTL Gedankenexperiment *
Typical Bell Test Setup
This is the typical setup for Bell test experiments, which can be performed in the undergraduate lab:
The yellow symbols are photon detectors
For additional info, watch this short movie on the basic setup for testing Bell inequalities:
https://www.youtube.com/watch?v=c8J0SNAOXBg
http://www.youtube.com/watch?v=c8J0SNAOXBg&hd=1
A complete description can be found in the paper Entangled photons, nonlocality and Bell inequalities in the undergraduate laboratory by D. Dehlinger and M. W. Mitchell.
The Measurement Problem
This Gedankenexperiment has the potential to goof right here, so watch out for errors. My assumption is that the shared wavefunction between the two entangled photons A & B does not decohere/collapse (or split the MWI-universe) when passing the Polarizing Beam Splitters (PBS).
AFAIK most will agree. As an example; one prominent SA in this forum once quoted Professor Joseph H. Eberly, pointing out:
The global quantum object characterized by one state vector (i.e. entanglement), is finally and irreversibly affected once the photon enters the detector and is actually measured. The PBS is not acting as a choice-device for “this or that way” – the single photon (wavefunction) takes both paths with its [itex]\left|H\right\rangle[/itex] and [itex]\left|V\right\rangle[/itex] superposition components.
An argument for this line of reasoning is the good old Double-slit experiment. We know that the two slits do not destroy the (single) wavefunction, hence interfering with itself behind the slits:
Other beamsplitter examples are the Mach–Zehnder interferometer and the Delayed choice quantum eraser.
The EPR-Bell FTL Setup
Please don’t laugh. This will not work, but I’m just too ignorant to find out why.
There’s actually nothing strange about this setup. In fact it’s almost identical to the undergraduate lab setup, except for the extended (wind up & asymmetrical) optical fibers (red).
Bob is an old fashion guy, he always calls the cards first (10 km to detectors D3 & D4 vs. 15 km to detectors D1 & D2) and thereby imposing Alice a restricted “quantum freedom”, and in some cases like perfect correlation – she has no choice at all – if Bob get 1V Alice will get 1V as well.
Alice is deeply dissatisfied with current chauvinistic situation and has therefore forced Bob out of the lab – 10 km out in the wilderness – with only one detector D3 (i.e. a highly realistic quantum drama ).
Despite these “quantum-drama-modifications”, the setup ought to generate exactly the same result as a standard Bell test in the undergraduate lab.
But the excitement doesn’t end here – Bob feels lonely & grumpy – and starts disconnecting detector D3 every now and then, to let photon B pass without any detection. Payback time!
The Big Question
What will Alice see in the lab? Will she notice anything? My claim is that she will definitely get the message – there’s a “Bob Revolution” going on out there in the wilderness.
Why??
Well, let’s do the math. In the setup we use a SPDC Type I down converter that along orthogonal axes will always result in the same outcome (i.e. perfect correlation).
Let’s assume we measure 100 entangled photons with a 90° relative angle between Alice & Bob:
sin^2(90°) = 100% = 100 correlated matches, i.e. [1V, 1V] or [1H, 1H]
We derive the result from:
sin^2(90°) = 100% x 50 = 50 correlated matches
+ 50% x 50 = 25 = A total of 75 matches
This is clearly enough to confirm there’s a “Bob Revolution” going on 10 km out in the wilderness.
(In fact, as soon as Alice detects A at D2 and B is not detected at D4 we know something is going on.)
Where’s FTL!?
Well, once we established that Bob could use his detector D3 to signal a change to the lab 10 km away, it’s easy to build a system for Morse code, or better, a binary transmitter. In the movie above on the undergraduate lab we get 500 entangled measurements per second. If 100 photons is enough to confirm that Bob did a change, that mean we could send 5 bit/s.
Ha, 5 bit/s! Grandpa’s bicycle is faster!
True, but the communication is instantaneous. Change km to ly and it will definitely be FTL.
Wait! Something Is Missing!
I know some of you are ROFL by now. There’s one “little” thing missing... do you know what?
We need a tool to separate the rare entangled photons (1 pair in every 10^12) from the noise, and the only way to do this is timing and coincidence counting, meaning Bob has to transmit this data thru classical channels (< c) back to the lab. There goes FTL down the drain. Period. Case closed. End of story.
But still, I can’t believe that our current inability to generate entanglement in a deterministic process is the only thing between us and FTL communication?? This can’t be true?? (How are we going to build quantum computers without 'controlled' entanglement?)
AFAICT we now have the following “non FTL” options:
Regards
DA
UPDATE:
If I’ve done the thinking right, we don’t need perfectly controlled/deterministic entanglement. All that’s needed is improved efficiency. With an efficiency of 1 entangled pair in every 10 it would be possible to see a statistical difference when “FLT” Bob is on/off.
Is there anything in QM preventing us from higher efficiency?
UPDATE2:
I know some of you will dismiss the whole “FLT” setup by relativity, and argue that it’s impossible to say that Bob always does his measurement first because; to a moving observer Alice will be the one ‘breaking’ the shared wavefunction/entanglement. This is of course correct; however remember detectors D1, D2 and D4 are all at the same location in the “lab frame”, where D4 will always “flash” before D2, no matter how fast or in which direction an observer travels (or?). Since the wavefunction always takes both paths to D3 & D4 with its (decomposed) superposition components, it will probably be a quite advanced relativity assignment to explain how the same QM superposition could dissolve/exist in different frame of reference. Maybe it’s possible... I have absolutely no idea...
And this far, nobody has proven Einstein wrong, not even “Italian Massive Mountains” could do it.
There are hypothetical particles like the Tachyon that (in theory) always moves faster than light, but I don’t like it, and the obvious skeptic reason is unsolvable and chaotic Causality paradoxes that comes by default with FTL. Furthermore we have the No-communication theorem.
In Bell test experiments no real usable information is propagated, only random “quantum information” and later (< c) established correlations between Alice & Bob, as a consequence of the shared wavefunction between the two entangled photons A & B.
Therefore I know something must be wrong with my thought experiment.
Could someone please tell me what it is?
* EPR-Bell FTL Gedankenexperiment *
Typical Bell Test Setup
This is the typical setup for Bell test experiments, which can be performed in the undergraduate lab:
The yellow symbols are photon detectors
For additional info, watch this short movie on the basic setup for testing Bell inequalities:
https://www.youtube.com/watch?v=c8J0SNAOXBg
http://www.youtube.com/watch?v=c8J0SNAOXBg&hd=1
A complete description can be found in the paper Entangled photons, nonlocality and Bell inequalities in the undergraduate laboratory by D. Dehlinger and M. W. Mitchell.
The Measurement Problem
This Gedankenexperiment has the potential to goof right here, so watch out for errors. My assumption is that the shared wavefunction between the two entangled photons A & B does not decohere/collapse (or split the MWI-universe) when passing the Polarizing Beam Splitters (PBS).
AFAIK most will agree. As an example; one prominent SA in this forum once quoted Professor Joseph H. Eberly, pointing out:
“Quantum theory says, therefore, there is no physical sense to an intermediate polarization”
The global quantum object characterized by one state vector (i.e. entanglement), is finally and irreversibly affected once the photon enters the detector and is actually measured. The PBS is not acting as a choice-device for “this or that way” – the single photon (wavefunction) takes both paths with its [itex]\left|H\right\rangle[/itex] and [itex]\left|V\right\rangle[/itex] superposition components.
An argument for this line of reasoning is the good old Double-slit experiment. We know that the two slits do not destroy the (single) wavefunction, hence interfering with itself behind the slits:
Other beamsplitter examples are the Mach–Zehnder interferometer and the Delayed choice quantum eraser.
The EPR-Bell FTL Setup
Please don’t laugh. This will not work, but I’m just too ignorant to find out why.
There’s actually nothing strange about this setup. In fact it’s almost identical to the undergraduate lab setup, except for the extended (wind up & asymmetrical) optical fibers (red).
Bob is an old fashion guy, he always calls the cards first (10 km to detectors D3 & D4 vs. 15 km to detectors D1 & D2) and thereby imposing Alice a restricted “quantum freedom”, and in some cases like perfect correlation – she has no choice at all – if Bob get 1V Alice will get 1V as well.
Alice is deeply dissatisfied with current chauvinistic situation and has therefore forced Bob out of the lab – 10 km out in the wilderness – with only one detector D3 (i.e. a highly realistic quantum drama ).
Despite these “quantum-drama-modifications”, the setup ought to generate exactly the same result as a standard Bell test in the undergraduate lab.
But the excitement doesn’t end here – Bob feels lonely & grumpy – and starts disconnecting detector D3 every now and then, to let photon B pass without any detection. Payback time!
The Big Question
What will Alice see in the lab? Will she notice anything? My claim is that she will definitely get the message – there’s a “Bob Revolution” going on out there in the wilderness.
Why??
Well, let’s do the math. In the setup we use a SPDC Type I down converter that along orthogonal axes will always result in the same outcome (i.e. perfect correlation).
Let’s assume we measure 100 entangled photons with a 90° relative angle between Alice & Bob:
sin^2(90°) = 100% = 100 correlated matches, i.e. [1V, 1V] or [1H, 1H]
We derive the result from:
- 50% of the time photon B is detected at D4 and Alice will then detect photon A at D2.
- 50% of the time photon B is not detected at D4 and Alice will then detect photon A at D1 (and we assume Bob detected photon B at D3).
- 50% of the time photon B is detected at D4 and Alice will then detect photon A at D2.
- 50% of the time photon B is not detected at all.
- When photon B is not detected, Alice measurement will be 100% random, i.e. 50/50 chance for photon A to be measured as [itex]\left|H\right\rangle[/itex] or [itex]\left|V\right\rangle[/itex].
sin^2(90°) = 100% x 50 = 50 correlated matches
+ 50% x 50 = 25 = A total of 75 matches
This is clearly enough to confirm there’s a “Bob Revolution” going on 10 km out in the wilderness.
(In fact, as soon as Alice detects A at D2 and B is not detected at D4 we know something is going on.)
Where’s FTL!?
Well, once we established that Bob could use his detector D3 to signal a change to the lab 10 km away, it’s easy to build a system for Morse code, or better, a binary transmitter. In the movie above on the undergraduate lab we get 500 entangled measurements per second. If 100 photons is enough to confirm that Bob did a change, that mean we could send 5 bit/s.
Ha, 5 bit/s! Grandpa’s bicycle is faster!
True, but the communication is instantaneous. Change km to ly and it will definitely be FTL.
Wait! Something Is Missing!
I know some of you are ROFL by now. There’s one “little” thing missing... do you know what?
We need a tool to separate the rare entangled photons (1 pair in every 10^12) from the noise, and the only way to do this is timing and coincidence counting, meaning Bob has to transmit this data thru classical channels (< c) back to the lab. There goes FTL down the drain. Period. Case closed. End of story.
But still, I can’t believe that our current inability to generate entanglement in a deterministic process is the only thing between us and FTL communication?? This can’t be true?? (How are we going to build quantum computers without 'controlled' entanglement?)
AFAICT we now have the following “non FTL” options:
- Entanglement is (by theory?) forever ‘doomed’ to be generated only from completely random sources like quantum fluctuations, and this will “save us” from FTL.
- The Polarizing Beam Splitters are not that innocent, they do act as a “this or that way” measurement after all.
- The Polarizer acts as a measurement.
- The combination Polarizer + PBS act as a measurement.
- I missed something else in the Gedankenexperiment ...
Regards
DA
UPDATE:
If I’ve done the thinking right, we don’t need perfectly controlled/deterministic entanglement. All that’s needed is improved efficiency. With an efficiency of 1 entangled pair in every 10 it would be possible to see a statistical difference when “FLT” Bob is on/off.
Is there anything in QM preventing us from higher efficiency?
UPDATE2:
I know some of you will dismiss the whole “FLT” setup by relativity, and argue that it’s impossible to say that Bob always does his measurement first because; to a moving observer Alice will be the one ‘breaking’ the shared wavefunction/entanglement. This is of course correct; however remember detectors D1, D2 and D4 are all at the same location in the “lab frame”, where D4 will always “flash” before D2, no matter how fast or in which direction an observer travels (or?). Since the wavefunction always takes both paths to D3 & D4 with its (decomposed) superposition components, it will probably be a quite advanced relativity assignment to explain how the same QM superposition could dissolve/exist in different frame of reference. Maybe it’s possible... I have absolutely no idea...