# Does the EPR paradox assume that entangled particles leave each other's light cones, or not?

So I was reading this post by Ethan Siegel, which introduces a recent talk by Lee Smolin. In the article, Siegel outlines the EPR paradox (or something like it) and mentions that it is based on the following suppositions:

1. You can create your pair of entangled particles at a particular location in space and time.
2. You can transport them an arbitrarily large distance apart from one another, all while maintaining that quantum entanglement.
3. Finally, you can make those measurements (or force those interactions) as close to simultaneously as possible.

Now, the "paradoxical" (or at least, "non-local") result of this that he mentions is that entanglement means you can know "information about a measurement... outside of your light cone".

But I was struck by the qualification in point 2 above. There, we're told that for the thought experiment to work, you have to be "maintaining" the entanglement, whilst simultaneously transporting them far apart. It seems to me, that for experimenter A with particle P to know that P is still entangled with particle Q throughout the journeys of both P and Q, then experimenter B (who has Q) would have to be continuously sending signals to A, letting them know that they haven't altered the entanglement of Q in some manner. If this is the case, wouldn't that mean Q would never actually leave P's light cone, thus contradicting the "paradoxical" non-locality result?

Based on the above, I have two questions:

1. Is the above a correct interpretation of the thought experiment, or have I misunderstood something somewhere?
2. If yes to the previous question, does this affect how the EPR paradox is usually explained/responded to by physicists?

The effect displayed in EPR-type experiments is most fully and convincingly displayed when the two entangled entities are spacelike separated. However many people would say that it is enough if the two entities are not interacting or signalling to one another, however that is brought about. My view is that it is a significant aspect of the effect that it applies to spacelike separated entities, and only in that scenario is the effect displayed in full. By "the effect" here, I mean that two parties establish a sequence of measurement outcomes which display correlations which break the Bell inequality.

Having said that, the two parties can never discover that they have such a sequence of outcomes until they either meet or signal to one another their results. Thus in order to show that a Bell inequality has been violated, there has to be this signalling after the measurements are done. It is relevent that the measurements themselves establish irreversible or 'classical' records, so that it does not make much sense to suppose that these records are somehow altered by the subsequent communication.

I don't find it helpful to describe such measurements using terminology in which A's measurement is said to influence B's results, or vice versa, (where A and B are the two spacelike-separated parties making measurements on entangled pairs sent to them). Rather, an entangled state is one in which a physical property such as spin direction is simply not appropriately ascribed to either particle on its own; rather what the particles share is a correlation, and the measurements allow this correlation to be observed.

Assuming no perturbations to the system, then there is no reason for P and Q to stop being entangled if no measurements are made. So experimenter A always knows that the particle is entangled throughout the journey.

But it does not actually matter. Entanglement simply implies that the measurements of P will be correlated to the measurements of Q. If P is not entangled with Q anymore, then this simply imply that Q was measured and so when A measures P, the results will be correlated with the results from Q.

The point is that the experimenters doesn't need to know if the particles are still entangled in that time to measure it. There is no need for continuous exchange of signal.

The only thing they need to do is to make sure that their clocks are synchronized; they agree that A is going to do a measurement in time $$t$$, B is going to do a measurement in time $$t+\delta t$$, then they go far from each other a distance greater than $$c\delta t$$, just to make sure that they are off of each other's light cone.

Being off of each other's light cone is exactly to make sure that information couldn't be exchanged between A's and B's measurements. So, B's physical state changes instantaneously after A's measurement (the wave function collapses in the whole space instantaneously, instead of "propagating with velocity lesser or equal to c"). That's the nonlocality shown by the experiment.

Just to be clear: they doesn't need to know about each other while measuring. They will just compare results later. There is no need for any exchange of information while or a little before measuring. The "maintaining that quantum entanglement" can be asserted later, when they got close again and discuss how each of them transported their particle and what happened in the measurement. If each of them took enough care to maintain the entanglement, it's okay, the result of the experiment can be trusted.

It's important to say that B cannot get information about A's measurement outside of the light cone; there is no information exchanged. B doesn't know (and in fact there is no way for him to know, if he is out of A's light cone) if A really took the measurement or not, until they come closer again and communicate.

As an example: suppose you have a system with 2 spins entangled in the following way:

$$\left | \psi \right > = \dfrac{\left | \uparrow \right >\left | \uparrow \right > + \left | \downarrow \right >\left | \downarrow \right >}{\sqrt2}$$

If A measures the first spin before B, A can get $$\left | \uparrow \right >$$ or $$\left | \downarrow \right >$$ with 50% chance each. If A got $$\left | \uparrow \right >$$, then B will necessarily get $$\left | \uparrow \right >$$. If A get $$\left | \downarrow \right >$$, B will necessarily get $$\left | \downarrow \right >$$. But notice that B have 50% chance of getting $$\left | \uparrow \right >$$ or $$\left | \downarrow \right >$$ anyway, so he does not know what happened with A's measurement. The wave function collapses instantly, but there is no information being propagated.