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I’m having a hard time wrapping my head around how the measurement of particle A does not affect the state of an entangled particle B even if no superluminal speeds exist.

Suppose Alice makes a measurement. Before she does, Bob’s particle is in an indefinite state. After she does, Bob’s particle is in a definite state. Let’s assume they’re anti correlated.

Regardless of whether or not Alice can communicate this information faster than the speed of light, a state change occurred, as long as we define an indefinite state to be a kind of state. If this state change occurred, how did Alice’s measurement not affect Bob’s particle’s state?

One cannot just say that this is merely an update in knowledge even if there is no realism. It is still going from an indefinite state to a definite state on the basis of that.

If we treat both particles as one indivisible but non local unit that has a combined indefinite state and then has a definite state when only either of them are measured, then and only then does it seem to imply no effects, correct? But isn’t this very treatment now assuming non locality? We are essentially considering two objects that could be light years apart as one.

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    You have to be more precise here. Yes, a measurement of Alice changes the overall state. But Bob's reduced state is unaltered (cf. no-communication theorem). As usual in QM, we only make statements and predictions in a statistical sense. It is also not clear what you mean with "indefinite state". – Tobias Fünke Nov 23 '23 at 18:15
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    Does this answer your question? Why is quantum entanglement considered to be an active link between particles? – Norbert Schuch Nov 23 '23 at 18:20
  • @TobiasFünke If the measurement of Alice changes the overall state, it is changing the state of both particles at once, even if the other hasn’t been measured yet, no? –  Nov 23 '23 at 18:21
  • Yes, but where is the problem? Note that if the events of measurements of both Alice and Bob are space-like separated, there is no objective notion of measurement order. QM predicts certain probabilities for the pairs of outcomes $(a,b)$; and these probabilities are invariant under change of measurement order, so everything is fine. Also, you cannot use entanglement to communicate information faster than light (again: no-communication theorem for a precise statement) – Tobias Fünke Nov 23 '23 at 18:28
  • in quantum mechanics, which is the basic theory where entanglement is defined and checked against measurements, every possible state is included in the wavefunction describing the system under study, but only the probability of the particular state is calculable. A measurement picks up one of those probable states, so for two particles when the quantum numbers for one of them is known by local measurement, the quantum numbers by conservation laws of the one away are known. – anna v Nov 23 '23 at 18:32
  • Sorry, I don't understand. And you just keep ignoring my comments. My last comment: In an entangled state,you cannot attribute a pure state to Alice (and neither to Bob). Rather, you might associate reduced density matrices to each of them. The no-communication theorem then shows that the reduced density matrix of Bob, conditioned on the fact that Alice performed a measurement, is the same as if Alice did nothing (so it equals Bob's reduced density matrix of the original state). And the RDM of Bob encapsulates all probability distributions of possible measurements he can do. – Tobias Fünke Nov 23 '23 at 18:34
  • @TobiasFünke All you are saying is that Bob’s measured state is not affected by whether Bob or Alice measures first statistically. It would be a 50% chance either way. But that wasn’t what was talked about in the post. I was referring to particle A’s measurement affecting the state of particle B (going from indefinite to definite) even before it is measured –  Nov 23 '23 at 18:37
  • The point of the entanglement issue is that it is very nuanced you have to be mathematically careful to discuss it without falling into interpretational issues. Sure the measurement "affects" Bob's particle but this is not an interaction in the usual sense, that's why some people prefer to call this weak locality or some other fancy name because it is not information conveying and yet it is not explainable using local hidden variables. – Mauricio Nov 23 '23 at 18:41
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    I can only and will just repeat myself (and ignoring your imprecise language): Where is the problem? – Tobias Fünke Nov 23 '23 at 18:42
  • your basic logical error is in thinking Alice and Bob are described by a simple wave function. Alice if it were possible to be modeled by a wavefunction has zillions of variables in it, and Bob too, so there is no way they can be described as a two particle quantum system and its quantum numbers that are subject to checking entanglement. – anna v Nov 23 '23 at 19:38
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    @annav if they’re entangled they have a joint non factorizable function Tobias Fünke the problem is that treating two particles divided by light years with no medium in between them as one unit with one joint wave function the collapse of which determines their states is non local, super luminal signalling or not –  Nov 23 '23 at 22:26
  • Reading over the conversation it seems you could center the discussion on "realism" or "action at a distance" is that is the case, please edit it to make it clear that it is about that and do not forget to define those terms – Mauricio Nov 23 '23 at 23:38
  • Addressing the remote state change of Bob's photon by Alice: Instead consider a 3 photon GHZ experiment. If locality holds, the specific result for any photon (Bob's) must not depend on which of 4 measurements is performed on the other two (by distant Alice), or on their outcome. Yet, an exact (not statistical) prediction can be made on what remote Bob will see which depends of Alice's decision. Further, the result is exactly the opposite of the local realistic prediction. So Alice steers Bob's photon remotely, and simultaneously rules out realism. drchinese.com/David/Bell-MultiPhotonGHZ.pdf – DrChinese Nov 24 '23 at 17:29

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