How can the quantum state of one particle depend on the other in entanglement without non-locality? [duplicate]

Question

There are many different interpretations of quantum mechanics. Some posit determinism and non-local influences between particles. Others do not, such as in the Copenhagen interpretation.

My question is in regards to the latter, especially since some of my searches on this site led me to seeing upvoted answers saying that QM does not technically violate locality.

So from what I know, Alice and Bob, when measuring the results of quantum entanglement, can either get a positive or a negative spin as a measurement outcome of an electron for example if they’re anti correlated.

The outcome of the system can only be (0,1) or (1,0). From Alice’s individual perspective, the outcome can be 0 or 1. So her outcome is effectively random.

Now what I don’t understand is how can this be possible without an interaction between the particles? I read that if you generate a pair of entangled particles with a total spin of zero, one of them will have a clockwise spin and another counterclockwise spin.

But each particle before measurement can be either clockwise or counterclockwise. They are not predefined as was thought by Einstein (according to the Copenhagen interpretation)

So if a) the spin is not predefined or predetermined and b) from each Alice and Bob’s perspective the spin is equivalent to flipping a coin but also c) the joint state (0,0) and (1,1) are impossible if they’re anti correlated, how can this be explained without an influence?

Wouldn’t one particle somehow have to “know” what the other particle will be measured as? How can it know this without somehow interacting with it?

Answer 1

Not to take anything away from Maple's correct answer, I would like to more closely address your following questions/statements:

A) ...especially since some of my searches on this site led me to seeing upvoted answers saying that QM does not technically violate locality.

B) ... So if a) the spin is not predefined or predetermined and b) from each Alice and Bob’s perspective the spin is equivalent to flipping a coin but also c) the joint state (0,0) and (1,1) are impossible if they’re anti correlated, how can this be explained without an influence?

C) Wouldn’t one particle somehow have to “know” what the other particle will be measured as? How can it know this without somehow interacting with it?

A) There is a great deal of controversy on this exact point. You are correct that there are many that say standard QM is local. This is often closely aligned with their preferred interpretation of QM. However, most physicists working with entangled systems would more likely say that QM is either clearly nonlocal or at the very least silent on nonlocality. If you look at papers in ARXIV, there are 5000+ with the words "nonlocal" or "nonlocality" in the title itself.

One key technical point: the quantum statistical prediction for standard Bell tests on polarization/spin entangled systems is dependent ONLY on the relative angles they are measured at. There are no other variables, and that is true regardless of distance and timing.

B) Your a) and b) points here are generally accepted as correct (but with some disagreement). c) is where interpretations come into play. The point you make is precisely the thing that Bell and others have been arguing: that there is "something" nonlocal occurring, even if the mechanism is currently unknown. Please be aware that an entangled system of 2 photons is sometimes referred to as a "biphoton" to highlight its nature as indivisible. It is not a system composed of 2 independent particles. A biphoton has spatiotemporal extent; meaning: it is inherently of a "size" that makes it quantum nonlocal.

C) Yes, it must "know" according to the QM prediction. This is called "contextuality". A future context - the final measurement setup/settings - is essential to the understanding of any experiment on entanglement.

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Now, just to add some experimental results to the equation that are almost universally ignored in discussions of quantum nonlocality:

1. You can entangle photons that have never existed within a common backward light cone - i.e. they have never even interacted. They are created separately without any initial correlation at all, and become entangled by a remote operation sufficiently distant that they never come close to each other. This sounds impossible, but check out:

High-fidelity entanglement swapping with fully independent sources "A successful entanglement swapping procedure will result in photons 1 and 4 being entangled, although they never interacted with each other." One of the authors of this paper shared the 2022 Nobel for this and other works.

2. And of course you can change measurement settings mid-flight so fast that there is no local signaling possible between the entangled particles to "know" how to act in advance:

Experimental loophole-free violation of a Bell inequality using entangled electron spins separated by 1.3 km "...the use of fast random basis selection and readout combined with a spatial separation of 1.3 km ensure the required locality condition..."

At the end of the day, you may form your own opinion based on arguments made in support of the different interpretations. But you should be aware of the many extremely sophisticated and varied experiments that have already been performed and confirmed. What I have presented is merely the tip of a very big iceberg around nonlocality in QM.

Answer 2

Most physicists understand this behavior as stemming from nonlocal effects in quantum physics. I will describe a Bell-type theorem that provides that main support for this view.

A Bell-type theorem involves two steps: (i) By assuming certain conditions of the universe, an inequality is derived that must be satisfied by the results of measurements. (ii) A quantum state preparation, transformation, and measurement scheme is identified that leads to a violation of the inequality. The take-home message of a Bell-type theorem is that the conditions you assumed about the universe are not compatible with quantum mechanics.

One of the conditions that goes into a Bell-type theorem is inspired by a notion of local causation. This condition is often called "factorizability," which I will now briefly describe (you can read more about this condition labeled (F) and found under Eq. (3) here). Consider two particles, denoted $A$ and $B$. Let $\lambda$ be some generic initial physical state of your system (quantum, classical, whatever). The measurement setting at one detector associated with particle $A$ is $a$ and the measurement setting for the other particle $B$ is $b$. The probability of getting outcome $s$ upon measuring particle $A$ with measurement setting $a$ while also getting outcome $t$ upon measuring particle $B$ with measurement setting $b$ is $$p_{a,b}(s,t|\lambda).$$ The factorizability condition is that this condition probability can be decomposed as $$p_{a,b}(s,t|\lambda) = p_a^A(s|\lambda) p_b^B(t|\lambda).$$ The intuition is that the measurement outcome of particle $A$ only depends on local physical objects which are manifested as the local measurement setting $a$ and the information about the initial state sent to the measurement apparatus and represented as $\lambda$. The same goes for the measurement outcome of particle $B$. Therefore, the joint probability should be factored in this way.

A Bell-type theorem shows that this assumption---stemming from local causality---is incompatible with quantum mechanics, so most physicists believe there are nonlocal interactions in quantum physics.

Now you ask "How does one particle 'know' the other particle's measurement outcome?" It "knows" this because of nonlocal causation due to quantum entanglement.