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We have 2 EPR experiments running in parallel, with Alice having one leg of each (a1,a2) and Bob the other leg of each (b1,b2). Thus (a1,b1) are anticorrelated, as are (a2,b2). Thus also (a1,a2) are uncorrelated as are (b1,b2). Now Alice locally entangles (a1,a2), and Bob measures b1 and b2. After repeating the entire experiment (including setting up the initial entanglements) many times, does Bob see consistent correlation or anticorrelation between his measured b1 and b2?

How Alice accomplishes this final entanglement (a1,a2) is either via entanglement swapping, or via the method described in Yurke and Stoler, Phys. Rev. A46, 2229 (1992): "Bell’s-inequality experiments using independent-particle sources".

Put another way, it's clear that b1 and b2 will show no correlation if Alice does not entangle (a1,a2), since the two EPR experiments are independent. Will this situation change for Bob as a result of Alice's entanglement of (a1,a2)?

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It's clear from no signalling--- by entangling $a_1$ and $a_2$, Alice uses local operators which necessarily commute with the spin operators on $b_1$ and $b_2$, so the reduced density matrix for $b_1$ and $b_2$ stays completely random. It makes no difference what method Alice uses, unless it involves mucking around with Bob's electrons.

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Well spotted; I was trying to sneak by no-signalling. The only loose end for me is to see the maths that demonstrates your assertion. A link that makes this clear mathematically would be great. – Andrew Palfreyman Aug 27 '12 at 13:07
I tried this because, since (a1,a2) are now correlated - i.e. are distinguishable from random - I inferred that (b1,b2) would behave similarly, and thus would be distinguishable from random. – Andrew Palfreyman Aug 27 '12 at 13:15

This is a subtle sort of thing that often leads to papers that make correct statements in the body of the paper but have titles and abstracts alluding to sending information superluminally or back in time or future events influencing the past. That's all nonsense and there is always a "trick" that makes it not what you were hoping for. Every time it makes me curious as to whether the authors realize they are pulling one over on you when they are writing the title and abstract or whether they are pulling one over on themselves without realizing it.

The question you ask is a very good example of such a thing. Yes, Alice can do a joint measurement on (a1,a2) that will cause a correlation or anti-correlation of Bob's particles (b1,b2). However, Bob won't know whether (b1,b2) are correlated or whether they are anti-correlated until Alice tells him the outcome of her measurement.

Now here is something that you may find shocking: none of this requires quantum mechanics at all! When you think you may have spotted some quantum magic, the first thing to do is substitute the words "probability distribution" wherever you used the words "quantum state", and "corellation" where you used "entanglement". Here is how it goes in the present case. Let a1 and a2 be random and independent. Let b1=a1 and b2=a2. So initially, b1 and b2 are uncorrelated. Now Alice measures whether a1 and a2 are the same. If they are, then instantaneously (faster than the speed of light!) Bob's values (b1,b2) go from being uncorrelated to being correlated! And if Alice sees that a1 and a2 are not the same, then Bob's particles instantaneously become anti-correlated.

Does this sound like magic? Probably not. Does information travel faster than the speed of light here? It depends on how you define things. But in my opinion, no. There are various tricks that sometimes get thrown into the mix. One is post-selection. The way that post-selection works is that Alice checks whether a1=a2 and you only consider the cases where that check turns up positive. This way, after Alice's measurement, Bob's particles are guaranteed to always be correlated. Post-selection has important theoretical uses in some cases. In other cases, such as in problems resembling the one here, it seems to mainly be used for confusing the audience (or even the author) in my somewhat cynical opinion.

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If it's the case that Alice's entanglement (a1,a2) causes (b1,b2) to become correlated (and never anti-correlated) then Bob can distinguish this from randomness and receive one binary bit superluminally. But Ron is saying that Alice cannot guarantee the sign of the correlation between successive trials, and so Bob cannot build up any (1 - 2^-N) fidelity of measurement. At least, I believe that's Ron's message. – Andrew Palfreyman Aug 28 '12 at 8:14
Yes, Ron is right. After Alice's measurement, Bob's particles will be correlated or anticorrelated but he doesn't know which (although Alice knows). Since Alice cannot choose the outcome of her measurement, she can't guarantee that it will be correlated and not anticorrelated. Unless we allow post-selection, which is cheating. In the quantum case there is also a phase thrown in, but the basic principles are the same. – Dan Stahlke Aug 28 '12 at 12:10
But that's signalling, isn't it? Define binary '0' as uncorrelated, and binary '1' as either correlated or anticorrelated. Unless the fact that we're dealing with an ensemble makes all three cases indistinguishable? – Andrew Palfreyman Jul 5 at 2:47
For any given run of the experiment, "either correlated or anticorrelated" is the same thing as "uncorrelated". Think about it: if Bob has two binary numbers, they will always be either correlated or anticorrelated (i.e. they are either the same or different). Alice doesn't get to choose which of those two cases happens. So there is no way Alice can use this to send Bob a message. Therefore it is non-signalling. – Dan Stahlke Jul 5 at 23:29

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