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Roger Penrose introduced the Andromeda Paradox as a thought experiment that delves into the implications of relativity and quantum mechanics on our understanding of simultaneity and reality. The scenario involves two observers walking past each other, potentially experiencing different "present moments" due to relativistic effects.

https://en.wikipedia.org/wiki/Rietdijk%E2%80%93Putnam_argument

Let's now take this a step further by introducing a "quantum version" of the paradox.

In this quantum scenario, an observer walking toward Andromeda might measure a particle (which is on Andromeda) in a definite state (e.g., spin up), while an observer walking away might observe the same particle in a superposition of states until he makes his measurement.

Considering the time elapsed between the first measurement by the first observer and the second observer's measurement, many years have passed on Andromeda. Does this imply that the measurement otucome of the second observer was already predetermined during this time interval? How do we reconcile the determinism implied by the following paradox with quantum mechanics?

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  • $\begingroup$ I guess that, like in the normal Andromeda Paradox, both observers are on earth. Is the measured particle on Andromeda? $\endgroup$
    – A. P.
    Commented Dec 21, 2023 at 10:22
  • $\begingroup$ Yes, i forgot to mention it. Thank you for pointing it out. $\endgroup$ Commented Dec 21, 2023 at 11:14
  • $\begingroup$ oh boy. now some holliwood director reads this and go on to make "The Andromeda Paradox" movie... $\endgroup$
    – drgetwrekt
    Commented Dec 21, 2023 at 11:50
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    $\begingroup$ I dont see the quantum version adding anything to the paradox. I mean how do you reconcile determinism in the original one. $\endgroup$
    – lalala
    Commented Dec 21, 2023 at 12:43

2 Answers 2

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On the one hand this whole area is quite subtle so to understand how modern physics deals with it you need quite a lot of learning (university level and beyond). Having said that, modern physics does offer a perfectly coherent account of all such phenomena so they do not present paradoxes or puzzles in that sense.

The chief thing I would offer to someone asking who has not learned quantum measurement theory or quantum field theory is that you have to beware of the loose usage associated with terms such as "measurement" and "observe" in popular accounts of science. Rather than those terms it is better, in the first instance, to speak about interactions and entanglement. For an example, one physical system (called by us a measuring device) interacts with another physical system (called by us a particle) and the two become entangled. What happens next depends on the individual case. It might be that the two subsequently become disentangled, or it might be that instead the larger one then interacts with further systems, spreading the entanglement out into many degrees of freedom, such that the disentanglement never occurs. In the latter case one finds that all the predictions of quantum physics are just as if a collapse of the wavefunction had occurred, but it is not necessary to pick any particular moment when the collapse takes place.

In hopes of clarifying a little, consider a sum such as $$ f = a + e^{i \theta} b $$ where $i^2 = -1$ and to follow this you will need either to know about complex numbers or just take it on trust. Suppose we wish to know the value of $|f|^2$. That's easy, it is $$ |f|^2 = |a|^2 + |b|^2 + \left(a b^* e^{-i\theta} + a^* b e^{i\theta}\right) $$ Now consider what happens when this $|f|^2$ is in fact a probability, as opposed to something like a length or a mass or a time. How do you measure a probability? You can't! Not in one go, at least. Rather you have to try some method such as run an experiment many times and take the average. But now the answers will depend on what happens with $\theta$. If $\theta$ always has the same value (and so do $a$ and $b$) then the average of $|f|^2$ is $$ \langle|f|^2\rangle = |a|^2 + |b|^2 + \left(a b^* e^{-i\theta} + a^* b e^{i\theta}\right) \tag{1} $$ but if $\theta$ varies randomly then the average of $|f|^2$ is $$ \langle|f|^2\rangle = |a|^2 + |b|^2. \tag{2} $$ The difference between (1) and (2) is what people are talking about when they discuss things like 'collapse of the wavefunction'. The reason is that any physical description leading to case (2) can here be replaced by a different physical description making the same prediction. The different description is that $|a|^2$ is the probability for one case and $|b|^2$ is the probability for another, and all we are doing is adding those probabilities.

When two different physical descriptions lead to the same physical predictions then you get debates about which physical description is the more elegant, and the debate cannot be resolved by experimental test. This does not mean the debate is without value, because it concerns things like elegance and coherence of ideas and these are important in science.

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Marco Fabbri,

The only way to explain the EPR experiment in a local, relativistic way is to introduce deterministic hidden variables. This means that both measurements need to have predetermined results.

If you are not familiar with this argument I'll describe it below.

You have a pair of spin-entangled particles which are sent to two distant locations, A and B. At both locations you are detectors fixed on the same axis, say Z.

When A measures its particle (say it's spin-UP on Z) the spin of particle B has to be spin-Down, this is what QM says. You have now two options:

  1. You assume locality.

This implies that the measurement at A cannot change the B particle in any way. Since after the A measurement b is in a spin-Down on Z state, and that state did not change, it logically follows that, even before the A measurement, B particle was in a spin-Down on Z state.

Once we established that the Z-spin of B particle was predetermined it also follows that the Z spin of A particle was predetermined so that the perfect anticorrelation is preserved.

  1. You reject locality.

In this case you might say that the result of the "first" measurement is random, and then the spin of B is instantly collapsed to the opposite value. Such a scenario is not compatible with relativity and you need to introduce an absolute frame of reference to decide which measurement is the first one.

To answer your question, there is no incompatibility/paradox between QM and determinism, on the contrary, determinism is the only way to make sense of QM in a relativistic setup.

It is also the case that observers always agree. The observer walking away in your example does not actually observe anything. He cannot observe the particle in a superposition, he may just assume that because he has no information about the particle. So, one observer has better information than the other, it's not really a paradox.

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  • $\begingroup$ Both forks you presented (1 and 2) are incorrect. The 1 commentary ignores Bell. Bell shows clearly that locally predetermined outcomes are incompatible with the correct predictions of QM. The 2 commentary is wrong because there is no relevant order in measuring entanglement. The predictions of QM do not specify a causal direction. You are free to assume one, but relativity has nothing to do with the quantum version of this. $\endgroup$
    – DrChinese
    Commented Dec 21, 2023 at 20:19
  • $\begingroup$ @DrChinese, Those forks are the only logically consistent positions you can have in the light of the argument presented here. The argument was not about Bell, but even in that case (1) survives in the form of superdeterminism. The logical principle of the excluded middle implies that the (1) and (2) options are the only possible ones (either locality is true or it is not). But if you go with 2 you need to introduce an order so that you can distinguish between cause and effect, otherwise the theory is incomplete. $\endgroup$
    – Andrei
    Commented Dec 22, 2023 at 6:15
  • $\begingroup$ Whether you call QM incomplete or not is, today, is more a definitional or philosophical question. You are living in the EPR of 1935. I would not agree with your characterization of there being only 2 forks, as the experimental proofs of quantum nonlocality demonstrate something that doesn’t quite fit into either of your buckets. At any rate, dismissal of Bell undermines almost everything you wrote as an answer. Mentioning superdeterminism is flawed because there is no such theory or interpretation at this time. And discussion of it belongs in a separate question, not in an answer to the OP. $\endgroup$
    – DrChinese
    Commented Dec 22, 2023 at 17:03

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