# Question about Relativistic effects on Quantum entanglement [duplicate]

Let us consider the case where there are two entangled electrons, one with A and the other with B; whose spin Z can be measured. Let the state of the system be $$\frac{1}{\sqrt{2}}\left ( |00\rangle + |11\rangle \right)$$ Let X and Y correspond to the events in spacetime where A and B measure the spin of their own particles, respectively.

Consider a frame '1' in which X happens before Y and the wavefunction collapses to $$|00\rangle$$. Since simultaneity is not absolute, consider another frame '2' in which Y happens before X.

Is there a probability that the state now collapses to $$|11\rangle$$? Does that mean that the spins of particles are frame dependent?

PS: My thoughts: So far I think there are two possibilities (or neither :P)

1. The wavefunctions do differ from frame to frame. This is nothing new in Special Relativity as the observers will infact disagree on measurements. There won't be contradiction as the frames never decelerate.
2. The wavefunctions will collapse to the same state $$|00\rangle$$ or $$|11\rangle$$ in both frames, but with probability of half. So that half of the time both measure $$|00\rangle$$ and half the time, both measure $$|11\rangle$$.

The problem with the second explanation is that since the wavefunction seems to collapse to the same value irrespective of whether X happens first or not, there seems to be something else causing the collapse.

How to affirmatively resolve this issue?

• Entanglement here means there is zero probability of A and B measuring different spins. But viewing A's measurement of 0 as the "cause" of B measuring 0 or vice versa is an artifact of the formalism. Commented Mar 25, 2023 at 12:04
• ...and what @ConnorBehan said is what the linked answer explains... Commented Mar 25, 2023 at 13:02
• physics.stackexchange.com/a/756757/28512 Commented Mar 25, 2023 at 13:43