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I was having a discussion about quantum physics with a friend, and we came to realise that we perceived the same situation in two very different ways. After searching online, we still weren't able to come to a conclusion on who is correct. Here is the situation:

Lets say we have two electrons that are quantum entangled, such that the spin of electron A is opposite to the spin of electron B.

You then measure the spin of electron A, and find it to be up spin.

My interpretation:

  • By observing electron A, you improve upon your knowledge about the second one. To someone who knows the spin of A, they can deduce the spin of B. To someone who doesn't know the spin of A, they have a 50-50 chance of guessing correctly. The entire time, ever since they were entangled, electron A was spin up and B was spin down. Simply no one knew it yet.

Their interpretation:

  • By observing electron A, you physically change the state of electron B. When you observed that A was spin up, electron B physically changed from being undetermined to become spin down, whereas before neither was individually a definite state, but in some combined state that exists in the tensor product space of A and B.

So my overall question: Whose interpretation is correct?

Are the electrons in an absolute state of having either up or down spin from the moment of being quantum entangled and we simply do not yet have enough information to know (until we observe A), or does observing electron A's spin physically change the spin of electron B, such that the results may have been different if electron B was measured first?

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  • $\begingroup$ 'Entangled' and 'absolute ' have opposite meaning . $\endgroup$
    – my2cts
    May 13, 2021 at 19:22

5 Answers 5

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In my view, both interpretations are incorrect (contradictory with the usual interpretations of QM). Before measurement the system of two electrons is in a superposition of $|\uparrow\downarrow\rangle$ and $|\downarrow\uparrow\rangle$. At that point, it doesn't really make sense to say that "electron $B$ is in a state", even undefinite. Measurement doesn't act on electron $B$, it acts on the entangled system of $2$ particles. After the measurement, the state of the $2$-particle system is un-entangled, so it makes sense to say "electron $A/B$ is in the state up/down".

  • The first interpretation is incorrect : the spin of $A$ is not defined before measurement. The act of measurement forces the system into a state in which the spin of $A$ has a definite value.

  • The second interpretation is incorrect as well : When you observe $A$ in the state "up", you acted physically on the whole system of $2$ particles. You didn't "change the state of $B$", since the state of $B$ was not defined before measurement

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There is no universally "correct" interpretation of what a measurement in quantum mechanics "really does" or what the quantum states "really mean". This is the measurement problem, and the various attempts to resolve it at are the different quantum interpretations. Crucially, these interpretations affect only how we think - or rather, tell stories - about quantum mechanics, they do not affect what we predict for the results of measurements. As such, they are metaphysical beliefs rather than integral parts of the physical formalism.

Your interpretation is that of a hidden variable theory - underlying the probabilistic notions of states of the ordinary formalism of quantum mechanics is a theory where all observables really do have well-defined quantities. Bell's theorem tells us that any such theory is necessarily non-local, i.e. must involve interactions that can affect the whole universe instantaneously. Note that this means that such interpretations, while perhaps at first glance somewhat easier to swallow, will likely involve other scenarios where they run counter to our usual conceptions of causality.

Your friend's interpretation is closer to how one would often phrase interpretations with instantaneous wave-function collapse. Measurement changes the state of the things being measured. Note that this in itself does not lead to issues with locality due to the no-communication theorem - no information can be transmitted only by measuring entangled states.

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    $\begingroup$ “Bell’s theorem tells us that any such theory is necessarily non local”...this is totally wrong. What it tells us is that any theory reproducing the results of Quantum Mechanics ( which itself )is non local, if you agree to the sensible notion locality being equivalent to Local Causality. $\endgroup$
    – Shashaank
    May 13, 2021 at 19:11
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    $\begingroup$ @Shashaank What do you think is wrong here, exactly? Bell's theorem says any interpretation of QM must be non-realist or non-local, and since hidden variable theories are realist, they must be non-local. $\endgroup$
    – ACuriousMind
    May 13, 2021 at 19:13
  • $\begingroup$ No Bell’s theorem doesn’t say that any interpretation of Quantum Mechanics must be either non realist or non local. It just says any interpretation (except Many Worlds where it doesn’t apply) of Quantum Mechanics is non local in the sense it violates Local Causality. So Bohemian Mechanics is realist and non local while Copenhagen is realist and still non local. Refer to CHSH inequalities or even the Einstein Boxes. Further the EPR argument is totally correct. Quantum Mechanics is non local even without hidden variables and you don’t need even an entangled pair to see that. $\endgroup$
    – Shashaank
    May 13, 2021 at 19:24
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    $\begingroup$ @Shashaank The conclusion of Bell's theorem is that at least one of the assumptions in the derivation of the inequality must be wrong because quantum mechanics (and experiments) violate it. ACuriousMind named the main assumptions. Oh and there is a difference between Bohmian and Bohemian ;-) $\endgroup$
    – M. Stern
    May 14, 2021 at 8:45
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    $\begingroup$ @Shashaank If you are saying that there is a derivation of the CHSH inequality that uses only a single assumption then please provide a link. If you give a name to a conjunction of assumptions then I don't find it helpful. $\endgroup$
    – M. Stern
    May 14, 2021 at 9:11
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Your discussion here closely mirrors early discussions by Einstein, Bohr, and others around the nature of quantum mechanics and how to interpret it. I hesitate to declare you or your friend more "correct" here, since there are elements of your interpretation that I think many would agree with.

Your idea sounds like a hidden-variable theory, where a particle's physically observable properties exist prior to our measurement and the act of measurement merely reveals them to us. This is I think an extremely natural and intuitive reaction to a description of quantum entanglement, so much so that it was explored quite thoroughly (most famously the EPR paper). John Bell later showed that many such hidden-variable theories are not merely alternate interpretations but are incompatible with quantum theory, and subsequent experimental confirmations of Bell's theorem seemed to finish brushing hidden-variable theories into the dustbin of physics history, although there are definitely versions of it still around to some degree.

Your friend's interpretation I think has the correct attitude on the indeterminacy prior to measurement being part of the physical nature of the system, but there is some issue with the characterization of the collapse of the wavefunction of one electron actively causing the other to collapse in a complementary way; this violates the principle of locality and is the very "spooky action at a distance" that was the thrust of the EPR experiment. This is where I think elements of your interpretation have value, in that the measurement of the one particle doesn't physically act on the other and merely changes our knowledge about its state. See this excellent answer for a better explanation than I could give.

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    $\begingroup$ This is a totally wrong answer. Refer to my comment on A curious mind’s answer. You have understood Bell’s theorem totally wrong $\endgroup$
    – Shashaank
    May 13, 2021 at 19:26
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    $\begingroup$ "In a theory in which parameters are added to quantum mechanics to determine the results of individual measurements, without changing the statistical predictions, there must be a mechanism whereby the setting of one measuring device can influence the reading of another instrument, however remote" (Bell 1964). I maybe was being reductive and should have specified that by "many such HVTs" I meant ones that obey locality, but I don't understand your particularly forceful protest. $\endgroup$
    – Sam Tempel
    May 14, 2021 at 4:24
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    $\begingroup$ I agree what you quote and indeed it is diffuse to understand what Bell is saying,but I would like to refer you Bell’s later writings where he develops a Local Causality principle. What you quote is just Parameter independence. But local causality = parameter independence + outcome independence. Bohmian mechanics violates parameter independence and preserves outcome independence thus violates Local Causlaity whereas Copenhagen violates Outcome independence while preserving parameter independence thus violates local causality. Bell’s theorem says that any theory.. $\endgroup$
    – Shashaank
    May 14, 2021 at 7:20
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    $\begingroup$ whose predictions of measurement outcomes are the same as Quantum Mechanics violates Local Causality.So whether it be Bohmian Mechanics or standard Quantum Mechanics interpretation both are non local in the sense that both violate Local causality. You are misinterpreting Bell(1964) by understanding that he says that measurement settings affecting each other at spacelike separation is the only form of non locality. Of course this is non locality and violating Parameter independence is non locality but even violating outcome independence (like Copenhagen) is non locality and you don’t.... $\endgroup$
    – Shashaank
    May 14, 2021 at 7:21
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    $\begingroup$ even need an entangled pair to see Copenhagen Interpretation is non local. Einstein boxes argument is sufficient to show it is non local. $\endgroup$
    – Shashaank
    May 14, 2021 at 8:15
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When we say that a pair of particles are entangled, we mean that their joint state cannot be expressed as a product of single particle states; i.e. there is not a separate state of $A$ and state of $B$. For the case of a pair of spin-$\frac12$ particles with opposite spins, we can say that they are in the spin state $\frac{1}{\sqrt{2}}(\lvert\uparrow\downarrow\rangle-\lvert\downarrow\uparrow\rangle)$. Note that this cannot be expressed as a product of single particle states. $$\frac{1}{\sqrt{2}}(\lvert\uparrow\downarrow\rangle-\lvert\downarrow\uparrow\rangle) \ne (\alpha\lvert\uparrow\rangle+\beta\lvert\downarrow\rangle)_A(\gamma\lvert\uparrow\rangle+\delta\lvert\downarrow\rangle)_B\quad\forall\alpha,\beta,\gamma,\delta$$

In this state, the particles do not have definite spins, so your interpretation is incorrect. If they did have definite spins, such as in the state $\lvert\uparrow\downarrow\rangle$, they would not actually be entangled, since $\lvert\uparrow\downarrow\rangle=\lvert\uparrow\rangle_A\lvert\downarrow\rangle_B$. Your friend is correct that measuring the spin of $A$ does change the state, but it cannot change the "state of $B$," since there is no separate state for $B$. Instead, the measurement changes the state of the system consisting of $A$ and $B$. If you measure the spin of $A$ to be up, then the system transitions to the state $\lvert\uparrow\downarrow\rangle$, meaning that you would measure the spin of $B$ to be down. While this may seem like you changed the spin of $B$, it doesn't really make sense to say that, since that implies that $B$ was in some spin state $\gamma\lvert\uparrow\rangle+\delta\lvert\downarrow\rangle$ prior to the measurement, which we know to not be the case.

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    $\begingroup$ This seems to miss the interpretational issue here - we can very well talk about the state of $A$ or $B$ in the entangled state: They would just be mixed states, not pure states (see e.g. this answer of mine). The measurement of $A\otimes B$ changes these mixed states to pure states - it breaks the entanglement (at least in $\psi$-ontic interpretations; $\psi$-epistemic interpretations might phrase this differently). $\endgroup$
    – ACuriousMind
    May 13, 2021 at 19:08
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RulerOfTheWorld,

Both interpretations are possible, yet the second violates the principle of locality , since the effect of A on B is instantaneous. Therefore, your first interpretation is most likely the correct one since it is in agreement with the rest of physics, like relativity.

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