What I mean by this is, with wave function collapse,--and by extension, collapse between two entangled particles--being nonlocal (instantaneous across space), in what reference frame does the entanglement collapse nonlocally? Does it collapse instantaneously in the reference frame of the creation of the entanglement? Is it nonlocal in the frame of the initial measurement that collapses the state of the system? Is it nonlocal in some other reference frame? Essentially, as quantum entanglement means a correlation between two particle states, when will one state be correlated with the other? Is it when the proper time of each particle is the same? Or when one particle is measured in its reference frame? Or is it something else?
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1$\begingroup$ I am not very advanced when it comes to the mathematical formulation of quantum physics, but I am trying very hard to learn more and more. I want to br at an advanced level eventually, which is why I actively ask questions (i.e. why I came here) and consistently seek information. I do have a good grasp on the basics of both relativity and quantum physics, and have a good grasp of the math of special relativity. I understood the basics of the article. but did not quite grasp the math yet. However, please be patient and help me learn so that I may improve myself and fully grasp this information. $\endgroup$– SciencemasterCommented Dec 23, 2019 at 16:50
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$\begingroup$ You're asking a good question. One little thing that might help is thinking in terms of the Heisenberg picture, where the wavefunction is time-independent. Testable predictions are the same whether derived using the Schrödinger picture or the Heisenberg picture, so switching pictures (back and forth, interchangeably) can help distill what it is about the wavefunction that really matters physically. (Non-relativistic QM is traditionally presented in the Schrödinger picture, and relativistic QFT is usually presented in the Heisenberg picture, but the pictures are interchangeable.) $\endgroup$– Chiral AnomalyCommented Dec 26, 2019 at 1:00
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2 Answers
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Sciencemaster,
Entanglement implies a certain correlation between some measurements you may perform on the subsystems. The way this correlation is "enforced" is not known but there are two main possibilities:
The wavefunction is a real, physical entity. In this case the collapse is a non-local process. The only way to avoid paradoxes is to reject Einstein's view of relativity and go back to a Newtonian absolute frame of reference. I cannot tell you how to find such a frame, but you may explore the de Broglie-Bohm interpretation where this kind of question is being studied.
The wavefunction as an incomplete, statistical description of the system. It's collapse can be understood as a change of available information/knowledge. In this case nothing non-local needs to happen. If you have two distant boxes, A and B, and you know that there is a coin in one of them but you don't know in each one, the probability of finding the coin in A is 50%. If you look in A and find the coin there, the probability "collapses" to 100% in A and 0% in B.
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$\begingroup$ Andrei, First off, hasn't the wave function and superposition been shown repeatedly in experiments (such as interference in particles with mass)? This being the case, scenario 1 would have to be the case, as uncertainty is a physical superposition of states rather than a lack of knowledge of a system. Is this correct? Second, what specifically contradicts wave funtion collapse in relativity? The particle isn't technically accelerating beyond C, as all that is happening is that some states are being instantaneously "destroyed" as the result of a quantum interation. Am I making a mistake here? $\endgroup$ Commented Dec 9, 2019 at 14:05
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1$\begingroup$ The wavefunction+Born's rule correctly predicts all experiments, including interference, no doubt about it. However, there is no logical argument (or at least I don't know any) that can lead you to conclude that the wavefunction must be a real physical entity. Think about the center of mass. It's a useful concept, gives correct prediction for mechanical problems, yet if you were to look at that location you will find nothing. For a sphere, the center of mass is just vacuum. The wavefunction might be similar in this respect. $\endgroup$– AndreiCommented Dec 10, 2019 at 5:09
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$\begingroup$ You may also consider properties like pressure or temperature. They describe well the behavior of a gas, yet they are not fundamental entities, just mathematical abstractions. So, it is possible that the wavefunction is an abstract concept that for some yet to be discovered reason, gives a correct statistical description of the behavior of quantum particles. Given the strong evidence we have for locality, I think (2) is the most reasonable position $\endgroup$– AndreiCommented Dec 10, 2019 at 5:15
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$\begingroup$ . "what specifically contradicts wave function collapse in relativity?" The collapse is instantaneous, so a measurement at A, under the assumption that the wavefunction is real, will have instantaneous consequences at B no matter how far B is from A. That should not be possible in relativity, where all physical effects, not only mass transfer, should be limited at C. $\endgroup$– AndreiCommented Dec 10, 2019 at 5:20
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$\begingroup$ I see your first point now about the wave function being a mathematical description that just predicts outcomes directly. However, my issue with this is that the wave function has direct effects that are present in classical wave mechanics that don't seem to have any other classical analogue outside of wave mechanics, such as interference of waves/wavefunctions, and electron orbitals being created from the only wavelengths around an atom that reinforce rather than destructively interfere with each other. How would these occur if the wave fucntion is just a representation of our knowledge of a $\endgroup$ Commented Dec 23, 2019 at 16:34
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The answer to your question is that QM is a theory that has been proven experimentally, but the way we interpret it can be different. I am writing about one of these interpretations.
In the present paper, we shall treat quantum superpositions exactly as in the Copenhagen interpretation; not the wave functions but only the probabilities are physical realities.
https://arxiv.org/abs/1112.1811
Contrary to popular belief, the wavefunction is not a physical wave, and there is no physical collapse. The wave function is rather a collection of numbers, our knowledge about a physical system.
Dear Jack, there is no physical phenomenon that could be called the collapse. The collapse of the wave function, as first emphasized by Werner Heisenberg and then many others, is just the event when we learn something about a physical property of a physical system. It is a collection of numbers that summarizes our knowledge about the physical system and that can be used to make predictions.
On the nature of the collapse of the wave function
Now in your case, two entangled particles create a QM system, and these two particles are in some ways indistinguishable, the wavefunction describes them both, they have a common wavefunction.
Thus, there is nothing instantaneous between the two particles, there is no information, no particles traveling between them instantaneously. The wavefuntion describes both particles, their properties, and the probabilities of measuring both particles' certain properties. This is the only way to understand and avoid nonlocality.
When things are measured, one of the outcomes is just realized. To simplify our reasoning, we may forget about the possibilities that will no longer happen because we already know what happened with the first particle. But this step, in which the original overall probabilities for the second particle were replaced by the conditional probabilities that take the known outcome involving the first particle into account, is just a change of our knowledge - not a remote influence of one particle on the other. No information may ever be answered faster than light using entangled particles. Quantum field theory makes it easy to prove that the information cannot spread over spacelike separations - faster than light. An important fact in this reasoning is that the results of the correlated measurements are still random - we can't force the other particle to be measured "up" or "down" (and transmit information in this way) because we don't have this control even over our own particle (not even in principle: there are no hidden variables, the outcome is genuinely random according to the QM-predicted probabilities).
Why is quantum entanglement considered to be an active link between particles?
answered Dec 8, 2019 at 22:38
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2$\begingroup$ "The wave function is rather a collection of numbers, our knowledge about a physical system" $-$ so you're saying that $\psi$-epistemic interpretations are the only viable ones, despite the well-known no-go theorems that show that the only viable such theories are superdeterminism and similar veins? Precisely what interpretation are you advocating for here? $\endgroup$ Commented Dec 8, 2019 at 22:49
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2$\begingroup$ ... which is to say, it is indeed possible to present a well-sustained argument, well-grounded in the literature, for at least some of what you're claiming here $-$ and I can't wait for you to fill it out in detail. It is also possible to come up with this type of shallow argument by stringing together badly-digested bits of pop-sci, though, which is why it's crucial that you present a solid base of argumentation (and withdraw the claims that you cannot defend solidly). $\endgroup$ Commented Dec 8, 2019 at 22:52
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$\begingroup$ @EmilioPisanty are you saying that Luboš Motl is wrong? physics.stackexchange.com/a/10070/132371 $\endgroup$ Commented Dec 9, 2019 at 0:37
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3$\begingroup$ Luboš has a very idiosyncratic view of quantum foundations that most people working on the field strongly disagree with. That answer is a very simplified version of that view, and Luboš knows precisely what is being left out and what type of criticism the answer is liable to get. Do you? If you don't, but you still want to take snippets from that text and parrot then mindlessly in other contexts where they're no longer appropriate, then that's wrong and an abuse of this site. $\endgroup$ Commented Dec 9, 2019 at 7:37
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2$\begingroup$ So, again: your answer is interpretation-dependent. Precisely what quantum interpretation are you using as a basis here? If you cannot provide that basis in full then this answer is somewhere between noise and misinformation and needs to be withdrawn. $\endgroup$ Commented Dec 9, 2019 at 7:39