# FTL Quantum Entanglement

1. Feb 2, 2010

### Rafaelh

Hi everyone, first post here :)
Someone told me that information could be transmitted faster than light using quantum entanglement. I do not believe this, but I'm not really knowledgeable in this area so I started to investigate, and I haven't found an answer. I came here hoping you could help!

2. Feb 2, 2010

### nicksauce

Information **cannot** be transmitted FTL using entanglement. This is, as far as I know, an undisputed claim.

3. Feb 2, 2010

Well, you can . . . but at the same time you can never know if that is what in fact actually happened. Lets say you had two entangled spins and separated them by many light years. The spins are prepared in the superposition,
$$| \psi \rangle = \frac{1}{\sqrt{2}} ( | 1/2_A , -1/2_B \rangle + | -1/2_A , 1/2_B \rangle)$$

That is to say, there is a 50% chance for observer A to measure spin up and a 50% chance to measure spin down. However, as soon as observer A makes a measurement on one of the spins (resulting in either up or down), the superposition collapses and the spin of particle B is precisely anti-correlated.

$$\langle 1/2 |_A \otimes \hat{1}_B | \psi \rangle = | -1/2 \rangle_B$$
$$\langle -1/2 |_A \otimes \hat{1}_B | \psi \rangle = | 1/2 \rangle_B$$

Observer B has not made a measurement yet, but unbeknownst to him, the result has already been predetermined by Observer A's measurement.

So if we repeated this lets say, ten thousand times, both Observer A and Observer B will collect a bunch of data - half spin up , half spin down. To each observer the process looks random, but if they could somehow trade notes and compare, they would find that all of their measurements are perfectly anti-correlated.

Basically, to actually transmit data, you still need a classical information channel (slower than light) on which to share the results of the measurements.

Last edited: Feb 2, 2010
4. Feb 3, 2010

### edpell

But it is a great way to send an encryption key. If someone listens in they destroy the information. The receiver gets the wrong key and sends a message that no one can decrypt.

5. Feb 4, 2010

### Tao-Fu

This is the important bit. No information is actually transmitted until the two observers compare notes. Without information from the sender, I am simply making a random number of measurements in a random basis. There is no information content in that.

6. Feb 4, 2010

### Matterwave

No FTL transmission is further guaranteed by the No Clone theorem, which means you can't clone your particle a thousand times to see if your partner has observed his particle (in which case all 1000 particles would be either spin up or spin down), or not (in which case, on average 500 particles will be spin up and 500 spin down).

7. Feb 4, 2010

### SpectraCat

Ok, I have a (hopefully simple) question about the issue of using entangled pairs to transmit a signal. Say we have Alice and Bob, and are they measuring entangled photon pairs, with a normal communication channel to compare results. Say they conduct the following idealized experiment:

1) The source of entangled pairs is placed at some fixed location, with detection facilities for the two beams located one light minute in opposite directions along the beam path. There is also a normal communication line connecting the two facilities for comparison of results, although it will clearly take 2 minutes for signal transmission.

2) The source is configured such that it emits entangled pairs at a rate of 1 per second.

3) The detectors are "passive" .. by this I simply mean that they measure every particle when it arrives, so there is no question of delayed choice. I have in mind something like the (ideal) switched-polarization detectors described in the Aspect paper.

4) Alice and Bob synchronize their clocks while together at the source, leave at the same time, and travel at the same speed to their measurement locations.

5) At some agreed upon time measured relative to their synchronized starting time, they begin taking photon polarization measurements, and stop after precisely 1 minute, at which point they have 60 measurements.

Now, my question is, when they compare results over their normal communication line, what will they observe? On the one hand, the Aspect paper (and subsequent experiments) indicate that there should be a 100% correlation between their sets of measurements (assuming ideal equipment). On the other hand, doesn't relativity say that Alice and Bob cannot be sure that their sets of measurements are time-correlated? That is to say, Alice can't be sure that the measurements she took at t=1s and t=2s happened in the same relative order for Bob at his measurement site?

If the latter statement is true, then it seems that the two measurement sets should be uncorrelated. However if they *are* correlated, then it seems like that might be developed into some sort of FTL communication device, using a some sort of clever protocol involving repeated measurements. (I have some ideas about how this might work, but I want to make sure my understanding of the experiment I described is correct.)

So, have I made a mistake in defining the parameters of the experiment? The thing I am most unsure about is the implied synchronization of the start times in step 5. Alice and Bob both use time measuring devices that travel with them at all times, so the synchronization seems to on the face of it to be possible under SR. However, I am not sure that SR allows Alice and Bob to be completely sure that they really started their measurements at the same time.

Or do I just have a fundamental misunderstanding(s) somewhere about what QM and relativity predict in these two cases.

8. Feb 4, 2010

### DrChinese

I think you will eventually determine that the experimental results do not change with the order of the measurements. The cos^2(theta) rule will apply. And that is fully independent of ordering.

9. Feb 4, 2010

### DrChinese

I think you will eventually determine that the experimental results do not change with the order of the measurements. The cos^2(theta) rule will apply. And that is fully independent of ordering.

10. Feb 5, 2010

### SpectraCat

Ok, at first I didn't understand what you were talking about, but I think I get it now. You are saying that the *pairs* of particles are always correlated, so in some absolute sense, it doesn't matter in which order you consider the *pairs* of measurements, right? But that is exactly the issue I was trying to get at with my question, so forgive me if this is naive, but please let me try to rephrase it a little more clearly.

At the end of step 5, Alice has measured the temporally ordered set of values, $$A=\left\{a_{1},a_{2},...,a_{60}\right\}$$, and Bob has measured the temporally ordered set of values, $$B=\left\{b_{1},b_{2},...,b_{60}\right\}$$. These sets of values are supposed to reflect the paired polarizations of the temporally ordered set of entangled pairs, $$S=\left\{\left(\alpha_{1};\beta_{1}\right),\left(\alpha_{2};\beta_{2}\right),...,\left(\alpha_{60};\beta_{60}\right)\right\}$$.

Q.M. says that the polarizations of the $$\alpha$$ and $$\beta$$ photons in each entangled pair in the set S should be 100% correlated, but it makes no predictions about the relative polarizations of photons from different entangled pairs. Therefore, when they compare results over the normal communication channel, in order to test the correlation of their measurements, Alice and Bob need to be certain that the orders in which they perceived the events in their respective measurement sets are the same. I am not sure that this last point is allowed by SR, since the sets of events A and B are outside the light-cones between Alice and Bob's measurement locations during the time-interval when the measurements were made.

That is the issue that I am trying to address with my question (including all of the other details from post #7. Does this make any sense at all, or have I just gotten something confused?

11. Feb 5, 2010

### DrChinese

Yes, I agree that members of different pairs are not correlated. But I believe that the order of the timings will be similar for A and B. In other words, both Alice and Bob see a member of pair 1 as being the first, a member of pair 2 being second, etc. They just line up their time tags until there is a (closely) matching set. The actual arrival times do vary and sometimes one of a pair arrives earlier or later. But that is within the coincidence time window.

12. Feb 5, 2010

### eaglelake

The difficulty here is that we are still trying to describe an EPR-like experiment in classical terms. This quantum experiment has certain characteristics, first discussed by Bohr, that make it radically different from a classical experiment.

First: Any quantum experiment requires a measurement result. This gives closure to the event. There is no experiment to discuss until the detector is triggered.

Second: If changes are made during the “run”, then the results, and the statistical distribution of those results, correspond to the experimental configuration in place at the instant the detector is triggered. It doesn’t matter when the changes are made. Quantum mechanics does not pretend to describe anything that might be happening before particle detection.

Third: The entire experimental apparatus, including the particle source, the preparation apparatus, the measuring device, and the measurement result is a single entity with no classical analog. The different parts of the experiment, including the two photons, do not act independently of each other. This property is known as the Non-Separability Principle of Quantum Mechanics.

The entangled pair, the entire apparatus, and the detection facilities are non-separable. Time and distance are irrelevant parameters! We should not be discussing the “two (detection) facilities” as if they were independent of each other. In fact, all the information obtained from this “entanglement” experiment requires only one detector.

Assume that we are measuring linear polarization. There are two possible results:
Result 1: particle A is x polarized and particle B is y polarized.
Result 2: particle A is y polarized and particle B is x polarized.
Result 1, and Result 2 also, is a single result. If particle A is x polarized, then we know automatically that particle B is y polarized, even without measuring it! Particle A - x polarized and particle B - y polarized always go together as a single result. Bob could be at lunch with his detector turned off. It doesn’t matter. If Alice had an ideal measuring device that yielded Result 1, then it would display “x” for particle A and “y” for particle B simultaneously at the instant her detector is triggered.

To do the coincidence measurement required to test relativity, you must somehow do a time of flight measurement on each photon in the pair. This means that each photon must be treated separately and have its own trajectory. But, it is now well established that quantum particles do not have trajectories like classical particles do. Wheeler has insisted that a quantum experiment does not describe the behavior of any photons traveling from the source to the detector. But, if we still demand that each photon does travel in such a classical way, as is done in the original EPR paper, then our calculations yield erroneous results in disagreement with real experiments.

The detectors need not trigger simultaneously. Nor do we need two detectors to obtain all the information yielded by this experiment.
The only time of importance to us is the instant that the first detector is triggered. That first detection event fixes everything. The experiment is over and done. If you then record a later event in the second detector, then this will only confirm the first result, but in no way does it change the result already obtained.

We cannot test the special theory of relativity with EPR-like experiments and it is now generally accepted that quantum mechanics is incompatible with such experiments. The fallacy lies in our attempt to apply classical principles to a non-classical experiment where they do not apply.

Best wishes.

13. Feb 5, 2010

### DrChinese

I agree with most of what you are saying. But the above statements involve complicating factors that I thought I might mention.

1. You are correct that the context of the entire experiment is to be considered. But there is certainly "evidence" (depending on interpretation) that the entire result is NOT fixed by the first measurement... because that would then conflict with the idea that the entire setup must be considered, which we just agreed was necessary.

So the "evidence" is delayed choice quantum erasers (DCQE). In these experiments, the statistical results appear to depend on events that occur AFTER the first measurement. Again, this is interpretation dependent.

2. There have been entire threads about this subject, so I won't go into the pros and cons here. Suffice it to say that there are a variety of opinions about what ways QM and SR are in conflict, if any. Some of the discussion is rather philosophical since the two can coexist in almost all cases.

14. Feb 5, 2010

### SpectraCat

Thank you for the detailed reply. I appreciate you taking the time to put your remarks in context.

However, I fail to see how my example described anything in classical terms ... I tried to be quite careful to do the opposite. I agree that the Q.M. description says that one of the detectors I describe is redundant, since measuring either member of a pair will tell us the result for the other member. However, that doesn't change the fact that the detection events will have an apparent order to both Alice and Bob, and that a comparison between the paired events can only be accurate if they can agree on the order in which the events occurred.

Anyway, after some more thinking about this, I think the whole light cone thing I was worried about is a non-issue. SR allows for synchronization of clocks with space-like separation, which is all that is required to nail down the correlation I was worried about. Furthermore, nothing else in my example would constitute a test of SR in any way that I can see.

I do see that I need to be careful to re-consider my understanding of these experiments in light of your comments however, particularly the ones that deal with the propagation of the photons between the source and detector.

15. Feb 5, 2010

### DrChinese

You may find the following useful to consider. Route Alice and Bob along equal fiber path lengths (or unequal as desired) and return them to the same place. Then measure and record time tags. Although they are not spacelike separated in strict terms, who cares? You really don't have to "prove" there are no sub-c effects at play in a case like this. And now you don't have to worry about synchronization. Instead, you can focus on how changing the measurement order (which you can now adjust precisely) might affect the outcome. (Of course, I don't think it changes anything which can be observed as such. But who knows?)

16. Feb 5, 2010

### JesseM

In relativity all observers agree on the order of timelike-separated events, which would include any two events on Bob's worldline, so there will be no disagreement about which of two measurements made by Bob happened first. The only disagreement will be about stuff like whether Alice's third measurement happened before or after Bob's second measurement and so forth, but when doing correlations I think they only need to worry about the sequence for each observer without worrying about simultaneity--i.e. Alice's first measurement should be correlated with Bob's first measurement, Alice's second measurement should be correlated with Bob's second measurement, etc. In case they don't actually detect every particle that gets sent, they could also use the speed of the particle (easiest if they're photons) combined with the position of the emitter to backtrack and figure out which pair of measurements correspond to particles that were emitted at the same point on the emitter's worldline.

By the way, one general proof that QM can't be used for FTL communication is known as Eberhard's theorem. I posted some other papers discussing Eberhard's theorem in post #22 of this thread.

Last edited: Feb 5, 2010
17. Feb 6, 2010

### Frame Dragger

This is fascinating, so essentially FTL communication in any scheme requires a double-device to confirm results with Bob (or Alice), until which point all that is being done is a steady measurement of random quantities.

As each FTL 'device' would suffer an identical problem, you'd never be able to confirm that the behaviour you observed was random, or the 'signal'. AND YET, as edpell said there are properties of this sub-c that have an impact on future communications and cryptology. Fascinating.

18. Feb 6, 2010

### SpectraCat

Thank very much for the link. I knew this proof and paper existed, but I hadn't been able to find it yet, because I didn't recall the name of the theorem. Just to be clear, I want to make sure I understand the implications of the proof. Suppose Alice and Bob have a space-like separation along the beam path, and are making measurements in some similar setup to the one I have described. What Eberhard's theorem effectively says is that, although the state of one entangled particle is known immediately once the state of the other particle is measured, the information is confined to the world line where the measurement occured, and thus by definition cannot propagate to another observer at super-luminal speeds. Is this correct?

Stated another way, Alice can't be sure that Bob made a measurement until he tells her so over another channel, right?

So, all that is needed for a counterexample (and I am not claiming that I can provide one), if for Bob to somehow make a change to his end of the line in such a way that Alice could be sure that he had done something within a time-like interval shorter than the time-like interval separating them along the beam path (not sure about the phrasing of that, but hopefully you get the gist). Right?

19. Feb 6, 2010

### JesseM

What do you mean by "state"? Are you talking about the particle's quantum state vector immediately prior to measurement (which is altered by the measurement, so after the first measurement the results of subsequent measurements will no longer be correlated with the other particle), or just the state of some particular variable like momentum or the spin on a particular axis? If you're talking about the latter, note that Bob makes a measurement and gets a certain result, he doesn't what result was found by Alice unless she measured the same property of her particle. If Bob measures spin on the x-axis and gets spin-up while Alice measures spin on the y-axis, then Bob doesn't know what result Alice got until he gets an ordinary message from her. He also can't say that Alice's particle must have been spin-up on the x-axis if Alice measures some other axis, because that would constitute a hidden-variables theory.
Yes, FTL communication would involve Bob making a measurement in such a way that Alice could infer something (at least in a probabilistic way) about what type of measurement Bob had made, based on her own measurement which has a spacelike separation from Bob's measurement. But I believe Eberhard's theorem shows this is impossible as long as orthodox quantum physics is correct (though one of the articles I linked to in post #22 mentioned that if you modify QM with a nonlinear term, in that case FTL communication using entanglement would be possible).

20. Feb 7, 2010

### Frame Dragger

Hell, didn't I just say you'd need TWO FTL communicators to make this scheme work? Logically you'd then need an arbitrarily large progression of this method to confirm the initial state change.

Oh, and for the change to be meaningful, like a one-time cypher pad, you'd have to use non-FTL means to communicate the means by which changes in a given observed property in a given time should be interpretated.

In other words... causality still lives.