Does violation of Bell's inequality imply faster-than-light communication?

[wywong]

In a traditional Bell-type experiment, although one set of measurements is correlated with the other set, no communication can be conveyed because each set in itself appears a random sequence. It requires both sets of measurements to reveal action at a distance.
What if entangled triplets are used? Is the correlation between the measurements of two beams of particles influenced by how the 3rd beam is measured? If so, the information (how the 3rd beam is measured) can be communicated from one place (where the 3rd beam is measured) to another place (where the other two beams are measured) faster than light! Although obtaining the correlation requires many measurements and considerable time, it can still in theory take less time than light can travel between the 2 places if they are distant enough.
Wai Wong (QM newbie)

 

[vanesch]

Is the correlation between the measurements of two beams of particles influenced by how the 3rd beam is measured?

No. There's a proof of that, using reduced density matrices. Call here (beam1 + beam2) = system1 and beam3 = system2, then you can show that the reduced density matrix of system1 (which determines ALL POSSIBLE expectation values of measurements on system1, so also those of correlations between beam1 and beam2) is invariant under a projection (in any basis of your choice) by a measurement local to system2 (beam3). I've typed the proof here already several times.

And the very explanation of this impossibility of having faster-than-light communication through entanglement in QM is made clear by the many worlds interpretation of QM. You can adhere to it or not, but the very fact that this interpretation exists, and where the results CAN be explained purely by local interaction, means that violation of Bell locality by itself is not a sufficient condition for faster-than-light interaction.

 

[wywong]

No. There's a proof of that, using reduced density matrices. Call here (beam1 + beam2) = system1 and beam3 = system2, then you can show that the reduced density matrix of system1 (which determines ALL POSSIBLE expectation values of measurements on system1, so also those of correlations between beam1 and beam2) is invariant under a projection (in any basis of your choice) by a measurement local to system2 (beam3).

Thank you very much. The mathematics is beyond me. Just one more question:
Let the error between the measurement of beam i and beam j be Eij. Using basic set theory, we have E12<=E13+E23 (the same logic as Bell's Theorem, but no HIDDEN variables here). Now suppose the angle between detectors 1 & 2 be 2x and the angle between detectors 3 and either of the other 2 detectors be x. The above inequality becomes
E12(2x)<=E13(x)+E23(x)
Assuming it possible to arrange the beams so that E12(x)=E13(x)=E23(x), the above inequality becomes
E(2x)<=2E(x)
Thus I arrive at Bell's Inequality WITHOUT the assumption of hidden variables, and violation of Bell's Inequality can only mean some of my steps are wrong. Could you point them out?
Wai Wong

 

[vanesch]

Thank you very much. The mathematics is beyond me. Just one more question:
Let the error between the measurement of beam i and beam j be Eij.

I don't understand what you mean here. Do you mean: the correlation between beam i and beam j ? (or better, the relative number of times when the clicks of beam i and beam j are NOT the same ?)
I'll start with that idea, correct me if I'm wrong:

E_ij = (number of "yes" in i and "no" in j + number of "no" in i and "yes" in j)/(number of (yes or no) in i and (yes or no) in j)

So if E_ij = 0, we have perfect correlation (each "yes" of i corresponds to a "yes" of j and same for the no's).
If E_ij = 1, then we have perfect anti-correlation.

Using basic set theory, we have E12<=E13+E23 (the same logic as Bell's Theorem, but no HIDDEN variables here).

Ok.

Now suppose the angle between detectors 1 & 2 be 2x and the angle between detectors 3 and either of the other 2 detectors be x. The above inequality becomes
E12(2x)<=E13(x)+E23(x)

Ok.

Assuming it possible to arrange the beams so that E12(x)=E13(x)=E23(x), the above inequality becomes
E(2x)<=2E(x)

This is what is not so simple. Well, it is possible, if you start from a totally symmetrical state, BUT BUT: E(x) is now NOT given anymore by the cos relationship. I explained this already somewhere else. Only in the case of TWO entangled particles, one can have the symmetrical, totally correlated state:

|psi12> = |z+>|z-> - |z->|z+>

This state has the mathematical property that it is 1) rotationally invariant (indeed, change z by x or n, and the expression remains the same) and 2) is totally anti-correlated: if you have |z+> for the first, you will have, with certainty, |z-> for the second.

We can try to construct a totally correlated state for THREE particles:

|psi123> = |z+>|z+>|z+> + |z->|z->|z->

or another variant on this theme. If we now have |z+> for the first, we know the result for the second and the third particle for sure. (it is z+).

But, but... this state is NOT rotationally invariant anymore! This state is not the same when you switch to, say y+ and y-.
So this means that the angular correlations are seen wrt Z, which is built into the state, and you cannot have E12(x) = E13(x) =... which requires rotational invariance.

On the other hand, you can construct a totally rotationally symmetric state with 3 particles, but THIS TIME YOU WILL NOT HAVE PERFECT CORRELATION OR ANTICORRELATION. So E12(x) will not go like cos^2(x).

cheers,
Patrick.

 

[wywong]

Vanesch,

Thank you very much for your enlightenment and patience!

Salute

Wai Wong

 

[CJames]

I wrote this recently in another thread:

I am thinking aloud right now, attempting to come up with a way of sending a signal via Bell non-locality.

So let's say we fire photon's through a beam-splitter and recombine them to generate an interference pattern. Now imagine we place a down-converter inside beams A and B, which generates signal beams A' and B' respectively. If A' or B' is detected, this tells us which beam photon P is a member of, thus destroying the interference pattern. Now imagine the detectors are one lightyear away. In attempt to return the interference pattern to the screen, somebody removes the detectors.

To me it appears that this act would not only send a signal faster than light, it would send a signal backward in time. The signal photons would be one year old, and so the interference pattern should appear a year before the detectors are removed.

My assumption here is that I have made a mistake. I have no knowledge of the math of quantum mechanics. I am merely relaying this because it is something that has puzzled me recently.

Vanesch, as of now my knowledge of math leaves your proof outside my grasp. I guess maybe I should just shut up, but this is really driving me nuts. I can't figure out what is wrong with this setup. It seems like either signal locality is violated or we measure the position of the photons and still get an interference pattern. Could you explain conceptually what would happen here?

 

[NateTG]

I wrote this recently in another thread:
Vanesch, as of now my knowledge of math leaves your proof outside my grasp. I guess maybe I should just shut up, but this is really driving me nuts. I can't figure out what is wrong with this setup. It seems like either signal locality is violated or we measure the position of the photons and still get an interference pattern. Could you explain conceptually what would happen here?

I'm not sure I understand how what you describe involves faster than light communication, and your description isn't entirely clear, so I'll try to go over the possibilties:

The effect of adding or removing the beam splitter 'propagates' to the detectors at the speed of light, so there isn't any faster than light communication possible from the splitter to the detectors. Since the detectors don't send out information, there isn't any communication between the detectors. Finally, communication from splitter a to splitter b using the light is only possible by having splitter b read the results from the detectors, which, is, at best, at light speed (if splitter a, a detector, and b are on a line).

 

[CJames]

I'm not sure I understand how what you describe involves faster than light communication, and your description isn't entirely clear, so I'll try to go over the possibilties:
The effect of adding or removing the beam splitter 'propagates' to the detectors at the speed of light, so there isn't any faster than light communication possible from the splitter to the detectors. Since the detectors don't send out information, there isn't any communication between the detectors. Finally, communication from splitter a to splitter b using the light is only possible by having splitter b read the results from the detectors, which, is, at best, at light speed (if splitter a, a detector, and b are on a line).

I am talking about exploiting wave/particle duality. In the "double-slit" experiment, a measurement of which slit the photon goes through forces the photon to behave like a particle and there is no interference pattern. So the simple act of measurement causes the interference pattern to disapear.

But what if the measurement is made a year later? Detecting the signal photon tells you which slit the photon traveled through, which should eliminate the interference pattern and force the photon to behave like a particle even before the measurement is made.

In this way it seems like you could send a binary message back in time. That is, if the signal photon never hits the detector, you will get an interference pattern, a one. If it does hit the detector, there will be no interference pattern, a zero.

I know that this couldn't work for a single photon, because one photon will not represent an interference pattern. But if several signal photons are detected and several are not, the ones which are detected will force the interference pattern to be destroyed a full year before the signal photons are detected.

You could manually remove the detector (not the beam splitter) to create an interference pattern one light year away and one year ago. This could be interpreted as a one. Then you could manually return the detector to destroy the interference pattern and send a zero.

Something isn't right.

 

[CJames]

Ok, I think I may have just finally solved this problem. Is it because, strictly speaking, the signal photons will always be "observed", even if it is not by the detectors? That is, no matter what, the experiment would yield no interference pattern, even if the detectors were removed, because the signal photon would at some point interact with an "observer", say an electron floating in deep space.

I hope that's right. Anybody know?

 

[wywong]

CJames,

Could you tell me, in your setup, how long are the paths for the two beams from the splitter to the interference pattern? If both are one light-year long, and there are no detectors along the way, you can expect to get an interference pattern one year after the beam passes through the splitter. In a thought experiment, you can ignore the unwanted molecules in space. You can destroy the interference pattern by detecting one beam along its way within that one year span, but the effect will only be observed at the end of that one year, and there is nothing going backwards in time. You don't need quantum non-locality to work that out; you just need classical wave theory.

Regards

Wai Wong

 

[CJames]

Alright, I can see that people are having trouble visualizing the setup. Let me see if I can explain it better. First, a laser fires through a beam splitter, creating beams A and B. Placed inside each beam is a down converter.

Down converters turn each photon into two photons of half the frequency. So every photon C that passes through the down converter is split into two photons, so that every photon which hits the screen has a corresponding signal photon. I called these secondary beams, carrying the signal photons, A' and B'. If a photon is detected in A', it means that the corresponding photon C is a member of beam A. In this way, the measurement of photon A' dictates where C was located. But if photon A' and B' are not detected, that means photon C was in a superposition.

If C was in a superposition, an interference pattern forms. If C was in A or B, then there is no interference pattern.

My proposition was that beams A and B are only meters long, but signal beams A' and B' are a lightyear long.

Can you visualize where I am going with this?

 

[vanesch]

small note on EPR and density matrices.

Ok, I wrote up (hopefully once and for all) in a small LaTeX document, the proof that no matter how you try, there's no way of using entanglement to signal anything between the different parts of the system in quantum theory.

You can find it here as an attachment...

cheers,
Patrick.

 

[NateTG]

Alright, I can see that people are having trouble visualizing the setup. Let me see if I can explain it better. First, a laser fires through a beam splitter, creating beams A and B. Placed inside each beam is a down converter.
Down converters turn each photon into two photons of half the frequency. So every photon C that passes through the down converter is split into two photons, so that every photon which hits the screen has a corresponding signal photon. I called these secondary beams, carrying the signal photons, A' and B'. If a photon is detected in A', it means that the corresponding photon C is a member of beam A. In this way, the measurement of photon A' dictates where C was located. But if photon A' and B' are not detected, that means photon C was in a superposition.
If C was in a superposition, an interference pattern forms. If C was in A or B, then there is no interference pattern.
My proposition was that beams A and B are only meters long, but signal beams A' and B' are a lightyear long.
Can you visualize where I am going with this?

So, you're talking about something where:
Beam A goes into a down converter and produces beams A1 and A2.
Beam B goes into a down converter and produces beams B1 and B2
Where A1 and B1 go to forming the interference pattern, and A2 and B2 might be observed?

 

[CJames]

So, you're talking about something where:
Beam A goes into a down converter and produces beams A1 and A2.
Beam B goes into a down converter and produces beams B1 and B2
Where A1 and B1 go to forming the interference pattern, and A2 and B2 might be observed?

Yes, that is the setup. And if A2 and B2 are observed, it should cause A1 and B1 to not form an interference pattern. But if A2 and B2 are not observed, then there should be an interference pattern.

So to put it in chronological order, if we are looking at the screen where A1 and B1 meet, and we see that there is an interference pattern, then we know that a year from now the corresponding signal photons in A2 and B2 will not be detected. If they are, it means that each photon in A1 and B1 was EITHER in A1 OR B1, meaning there couldn't have been an interference pattern. If, on the other hand, we see no interference pattern, then we know that a year from now the corresponding signal photons will be detected. So I'm having trouble reconciling this.

If the experiment is set up so that there is a high probability that the signal photons will be detected, and yet we look at the interference pattern and see that they will not be detected, and there is explicit instructions to whoever is at the detectors a year from now that an interference pattern is a 1 and no interference pattern is a 0, then it seems to me that a message could be sent simply by removing and replacing the detectors, or even turning them off and on.

Vanesch, I'm sure you're right. In fact, I'm almost POSITIVE you are right. I will look at your proof but odds are I will not understand it. The reason I posted this is not to claim I have found some FTL communication that will revolutionize physics and the world. I'm sure there is something wrong with my setup. I just don't know what it is, and it's driving me nuts. So I want to know what will actually happen in this setup and the basic reasoning behind it.

 

[dr_syed_ameen]

dec.08.2005

There is a problem with optical experiments which in case of Bell's creates questions,whether such an experiment could be accurate or precise enough in prctical measurements?

Dr.Syed Ameen (Ph.D.)

 

[vanesch]

Vanesch, I'm sure you're right. In fact, I'm almost POSITIVE you are right. I will look at your proof but odds are I will not understand it. The reason I posted this is not to claim I have found some FTL communication that will revolutionize physics and the world. I'm sure there is something wrong with my setup. I just don't know what it is, and it's driving me nuts. So I want to know what will actually happen in this setup and the basic reasoning behind it.

Ok, I've looked a bit at what your setup looks like, and the problem is the following. A photon can only give an interference pattern with itself!
The incoming photon |in> goes into a beamsplitter, and this photon gets into a superposed state:
|in-left> + |in-right>
If you make it interfere again, a bit later, then there's no problem because this is one and the same photon that is recombined, and you'll get your interference pattern.
But now, you put a downconverter in each path. This means that the single-photon state |in-left> will be converted into a 2-photon state |idle-left>|signal-left> ; while the |in-right> state will be converted into the two-photon state |idle-right>|signal-right>.
As such, our quantum state is now:
|idle-left>|signal-left> +|idle-right>|signal-right>
and there is no way to make idle-left interfere with idle-right. The only way to do so, would be by COMBINING the signal-left and the signal-right and factor them out. This can be done:
Imagine you turn |signal-left> into |s-up> + |s-down> ; and that you turn |s-right> into |s-up> - |s-down>. This can be done by having the signal-left and the signal-right beam interfere (with a beamsplitter for instance).
The two terms have to remain orthogonal because we take it that signal-left and signal-right are orthogonal and a unitary transformation preserves orthogonality.
So after this mixing of signal-left and signal-right, we obtain:
|idle-left>(|s-up> + |s-down>)+|idle-right>(|s-up> - |s-down>)
If we now measure s-up/s-down, then we will have, in the case we measure |s-up>:
(|idle-left> + |idle_right>) |s-up>
And NOW idle-left and idle-right can interfere.
But we will also have |s-down> as result, and then we will have:
(|idle-left> - |idle_right>) |s-down>
Here, again, idle-left and idle-right will interfere, BUT IT WILL BE THE COMPLEMENTARY INTERFERENCE PATTERN.
So what to conclude from this ?
If we mix idle-left and idle-right in a beamsplitter, overall, we will NOT get an interference pattern.
If we mix signal-left and signal-right in a beamsplitter, and we look at the outcome, the we can SUBSELECT a subsample of the idle-left and idle-right interactions, by requiring coincidence with a s-up event, and we will notice an interference pattern in the subsample.
We could also require a coincidence with an s-down event, and we will ALSO notice an interference pattern in this other subsample. However, this interference pattern will be COMPLEMENTARY to the previous one. So we understand that if we do NOT subselect anything, we add the two interference patterns together, and find no interference.
Note also that in order for us to find the right trigger to find a subselection of samples that give us an interference pattern, it was necessary to mix signal-left and signal-right, and to detect s-up. As such, WE HAVE LOST ALL POSSIBILITY TO SAY WHETHER WE HAD signal-left or signal-right.
If we decide to re-demix s-up and s-down we can, with yet another beamsplitter, to recover s-left, and s-right, then we are NOT able anymore to find the right trigger to subselect the interference pattern. However, this time, we can say whether we had signal-left, or signal-right.
With the lightyear of difference, there is no problem. The data collected now, from idle-left and idle-right, do not show any interference. We NEED to use the s-up or s-down trigger to subselect it. As long as we didn't do so, there's no way for us to subselect a subsample with interference.
If we decide to detect signal-left or right, we destroyed for ever the possibility of finding the trigger that allows us to subselect the interference pattern (but we can say now what way the photon went). If we combine signal-left with signal-right, to detect s-up, we DO have now our trigger (so we can retro-actively look into our old data to see with what clicks that corresponded, and subselect an interference pattern), but we have now destroyed all means to say whether the photon went left or right.
BTW, this kind of experiment has been performed.

 

[ukmicky]

Faster than light communication can never happen because information can't travel faster than c. quantum entanglement/effects moving at superluminal speeds run into the uncertainty principle making it impossible to control any signals.