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Please consider the following thought experiment. The conclusion surprises me, I don't know if it is correct.

A and B are particles with entangled spin: Measure the up-down spin of A and B then the results are the same. Measure the up-down spin of A and the left-right spin of B then the results are uncorrelated. Measure A twice, without measuring B, then the results are the same.

Now consider we measure u-d of A at spacetime event A1 and A2, and measure l-r of B at event B1. First attempt the events are at the same position in our reference frame and succeeding times A1, B1 then A2. It is uncontroversial to conclude measurements at A1 and A2 are uncorrelated due to B1.

Now move B1 outside the light cones of both A1 and A2:It is no longer possible for action at B1 to affect A2, so A1 and A2 must be the same. (Otherwise we could use this as a means of sending a decision to measure at B1 faster than light to A2)

This is weird in several ways, for example a small displacement of B could disrupt the correlation in A potentially from a very large distance, without any signalling. Is it really like this?

spacetime diagram

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  • $\begingroup$ The first of your several errors is here: "It is uncontroversial to conclude measurements at A1 and A2 are uncorrelated". In the absence of forces, the measurements at A1 and A2 are identical; in the presence of forces the first measurement determines the statistics of the second in a way that you can calculate, and in general yields a nonzero correlation. $\endgroup$ Commented Nov 12 at 12:35
  • $\begingroup$ No, can't be correlated - measuring B1 collapses the entangled state either left or right, then measuring A2 is either up or down with 50% probability independent of the B1 measurement. $\endgroup$ Commented Nov 12 at 16:23
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    $\begingroup$ Scratch the last comment, I see my error - simply stated A1 reduces the entangled state to a product state, and B1 then has no effect on A2. $\endgroup$ Commented Nov 12 at 16:48

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No. You have made an assumption that is unjustified.

  1. An up down measurement on A is uncorrelated to a left right measurement on the entangled B. That’s correct. But a subsequent up down measurement on B will not be correlated to A in any way. And yes, a second up down measurement of A will produce the same results as previously for A. Note: a left right measurement on A after a previous up down measurement on A will be completely uncorrelated to B.

  2. All of the above remains true, regardless of where B is when B is measured initially. Regardless of measurement times or distances or sequence, A and B will be correlated only if they are measured on the same basis while entangled.

In spin tests on entangled particle pairs generally: A and B are separated and outside each other‘s light cones. Whether you do or do not believe there’s any action at a distance, distance plays no observable role in outcomes whatsoever.

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  • $\begingroup$ You don't seem to understand the experiment as stated - B is measured just once at B1. A1 and B1 will be uncorrelated no matter the space time connection, that is stated in the first part. The question is the effect of B1 on the A1-A2 relation depending on whether B1 is outside the light cones of both A1 and A2 or not $\endgroup$ Commented Nov 12 at 10:39
  • $\begingroup$ @D.Menzies There is no known or observable effect on the A1-A2 relationship regardless of when or where or how the B1 measurement occurs. If A2 is performed on a different (non-commuting) basis than A1, there will be no correlation with either A1 or B1. If A1, B1 and A2 are all done on the same basis, all will be correlated. Again: normal entanglement tests are performed with the A and B measurements outside of each other’s light cones. It is certainly “possible” that a measurement of one of a pair affects the other. But it’s not an absolute requirement, and direction of action is ambiguous. $\endgroup$ Commented Nov 12 at 20:20
  • $\begingroup$ @D.Menzies I just noticed the hand drawn diagram. The selection to perform B1 or B1’ has no observable effect on A1 or A2 or their relative relationship under any choice of measurement bases. $\endgroup$ Commented Nov 12 at 20:29
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    $\begingroup$ Yes I realised my basic mistake in a comment further up - after A1 the state is no longer entangled $\endgroup$ Commented Nov 12 at 21:44
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In reality there is no way to use quantum entanglement to communicate non-locally and there is a non communication theorem to that effect, which is entirely uncontroversial, see Section II E of this review:

https://arxiv.org/abs/quant-ph/0212023

I could leave it at that but I will explain a bit more assuming that the equations of motion of quantum theory are used to describe what's going on including measurements.

A measurement is an interaction that produces a record of some property of a system. In your proposal there are two systems $A$ and $B$ in the state $$|\phi\rangle_{AB}=\tfrac{1}{\sqrt{2}}(|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B)$$ where $|0\rangle,|1\rangle$ are the $+1,-1$ eigenstates of $\sigma_z$. The corresponding eigenstates for $\sigma_x$ are $|0\rangle_x=\tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle),|1\rangle_x=\tfrac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$.

A perfect measurement of an observable $\hat{A}$ with eigenvectors $|j\rangle_S$ of a system $S$ onto a measurement device $M$ in a ready state $|0\rangle_M$ has the effect: $$|j\rangle|0\rangle_M\to|j\rangle|j\rangle_M$$ and doesn't change any system other than the measuring device and the measured system. The measurement operator would be $$U_{SM}=\sum_{j=1}^d|j\rangle|j+k\rangle_M\langle j|\langle k|_M$$ where $+$ stands for addition modulo $d$, which is the dimension of the Hilbert spaces of $S$ and $M$.

So if the initial state is $$|\Psi(0)\rangle=|\phi\rangle_{AB}|0\rangle_{MA_1}|0\rangle_{MA_2}|0\rangle_{MB}$$ where $MA_1,MA_2,MB$ denote the measurement devices used to do the measurements on $A_1,A_2,B$ respectively.

The evolution operators for the $A_1$ measurement is then $$U_{AMA_{1}}\otimes I_{MA_2}\otimes I_{B}\otimes I_{MB}$$ where $I_{whatever}$ is the identity operator on that system. So the $A_1$ measurement changes $A$ and $MA_{1}$ and nothing else. And you can write down the measurement operators for the other systems and reach similar conclusions for the other measurements. Doing the maths might help you understand the situation better.

It is commonly claimed that measurements have non-local effects even if they can't be used for communication but this isn't an implication of the equations of motion of quantum theory. Using those equations of motion the local flow of information in the Heisenberg picture, see

https://arxiv.org/abs/quant-ph/9906007

https://arxiv.org/abs/1109.6223

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  • $\begingroup$ Thanks for the info $\endgroup$ Commented Nov 12 at 21:46

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