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Imagine the usual twin paradox, a twin stays on Earth, the other twin leaves in a rocket reaches $c(1-\epsilon)$ speed for small $\epsilon$ and the twin turns around and goes back to Earth at the same speed. The twin that left Earth and came back is now younger than the twin that stayed on Earth.

Now suppose before leaving, each of them has a pair of spin entangled particles and can keep the quantum state long enough. Would the two particle be entangled when measured after the rocket round trip?

Suppose that we repeat this experiment many times (let's say the twin does not travel long). Would the two particles violate Bell inequalities? Or does the state gain some weird phase due to time mismatch?

I know that twin paradox can be solved with special relativity, so I was wondering if relativistic quantum theory could solve this too without needing quantum gravity (GR).

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  • $\begingroup$ The gist of the EPR argument is already that spacelike measurements are causally disconnected, so that there can be no effective notion of time ordering at play when making sense of the correlations. What difference does your setup make? $\endgroup$
    – Stéphane Rollandin
    Commented Feb 8 at 11:02
  • $\begingroup$ @StéphaneRollandin I agree. I was just trying to add "some GR" to entanglement to see if something different could be said about quantum mechanics in curved spacetime $\endgroup$
    – Mauricio
    Commented Feb 8 at 11:05

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Each particle evolves unitarily. If they are acted on by unitary operators $U_1$ and $U_2$, then the pair is acted on by $U_1\otimes U_2$.

Easy exercise: Show that an operator of this form cannot transform an entangled system to an unentangled system, or vice versa.

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  • $\begingroup$ Well, to be fair, it is also an (more or less implicit) assumption here on the Hamiltonian. In other words, you already assume that the form of the Hamiltonian is $H=H_1+H_2$. From the axioms of QM you cannot prove or disprove something related to SR and so on...This point is also often ignored in the discussion of the no-communication theorem in NRQM... $\endgroup$
    – Tobias Fünke
    Commented Feb 8 at 9:49
  • $\begingroup$ Don't this need Schrödinger's picture which is at the same time? $\endgroup$
    – Mauricio
    Commented Feb 8 at 10:49

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