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I am a little confused about the usual explanations about quantum entanglement.

In the Schrödinger picture, the state vector evolves in time.

In the Heisenberg picture, the state vector does not evolve in time.

In either case, there is no mention of the physical space, so the state vector is not dependent on space, clearly it belongs to an abstract Hilber space.

If it is not dependent on the physical space, it should be by definition a non-local object.

So why should we be surprised that something unrelated to space is connecting events that are spacelike separated?

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    $\begingroup$ not to mention $c$ does not appear in the Schrödinger equation. $\endgroup$
    – JEB
    Aug 28, 2023 at 16:15
  • $\begingroup$ Not that Schrödinger equation is any better for that. $\endgroup$
    – Mithoron
    Aug 30, 2023 at 13:57

5 Answers 5

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You make an excellent point: Since multi-particle wavefunctions aren't functions on space, there's no particular surprise when they exhibit spatially-nonlocal behavior. Indeed, this very non-spatial structure is what allows the non-locality in the mathematics.

But this still raises questions. I'll list a few:

  1. In general, non-local structures would be expected to permit non-local influences and signaling. If wavefunctions are generically non-local, then why can't we ever use them to signal faster than light? Preventing such signaling looks quite conspiratorial.

  2. Even if you forbid non-local signaling, in general non-local structures could in principle allow non-local correlations which are forbidden by quantum theory (as in the "PR-boxes"). Why would the particular non-local structures of quantum states not allow for these "supra-quantum" correlations?

  3. If Einstein was essentially correct, at least at some emergent level, about the unity of space+time as one four-dimensional manifold, then why would time survive while space disappeared at some fundamental level? (Looking curiously like the pre-relativistic viewpoint where time and space were assumed to be distinct.)

  4. If space is not fundamental, why do so many arguments and derivations in quantum theory rely on space? When building a Hamiltonian for a two-spin interaction, why would it be a spatial dot-product between those two spatially-oriented spins? Why would the logic for swapping the spatial location of identical particles work in the first place? Etc., etc.

  5. Our best description of a quantum state immediately after a localized measurement is a spatially-localized state, at least around the spatial region where the measurement occurred. (This is also where we have our most direct experimentally-verified knowledge about what is actually happening.). If you like time-symmetry, or don't like collapsing wavefunctions, there are lots of reasons to reasonably infer that just before the measurement, that part of the system must have also been spatially-localized, in contradiction to the usual viewpoint.

Each of the above concerns has various attempts to resolve them, but taken together I hope it's more evident why people are still grappling with entanglement and non-locality.

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    $\begingroup$ This is a very helpful overview, of the real problem with the non-local objects, and I'd say number 3 is the most puzzling one! $\endgroup$
    – VVM
    Aug 28, 2023 at 16:45
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    $\begingroup$ Multi-particle wave functions ARE functions of space! $\endgroup$ Aug 29, 2023 at 13:05
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    $\begingroup$ @flippiefanus: An N-particle wavefunction, as used in a multi-particle Schrodinger equation, looks like $\psi(\vec{x}_1,\vec{x}_2,...,\vec{x}_N,t)$. This is formally a function on 3N+1 dimensional configuration space, not 3+1D spacetime. True, the same sort of configuration space is used to describe states of knowledge in classical statistical mechanics, but the use of these functions in classical physics is to summarize a probability distribution over actual-space particle trajectories, $\vec{x}_1(t)$, $\vec{x}_2(t)$, etc., a move that's not possible in the usual quantum formalism. $\endgroup$ Aug 29, 2023 at 14:47
  • $\begingroup$ @KenWharton I appreciate your comment on this related question: physics.stackexchange.com/questions/778879/… $\endgroup$
    – VVM
    Sep 4, 2023 at 15:27
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    $\begingroup$ Many people think that one is looking for some sort of "principle" to explain the quantum limit. Five proposed principles that aim to explain "why" are listed on the relevant wikipedia page, but I'm sure I've seen even more than that in the literature. en.wikipedia.org/wiki/… $\endgroup$ Sep 15, 2023 at 3:40
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In addition to Ken Wharton's excellent answer I would add an extra point about what I see as the hierarchy of how these things fit together.

First, you are completely correct that from either Schrodinger or Heisenberg pictures a quantum state is nonlocal, and this can be seen as the source of the nonlocaltiy in experimental tests like Bell tests. While this is the right way of thinking within the theory, we need to approach things in the opposite direction when we are justifying the theory.

Imagine someone comes over to you and says "Your theory with state vectors in Hilbert space has this crazy feature where things are nonlocal. Why did you put that in your theory? Why don't we use a different theory which does not have that nonlocality?"

In my view the best answer to this is "Because we measured the nonlocality in experiments."

So you are right: Given that we have decided to include nonlocal entities in our description of reality it is indeed unsurprising that our description of reality is nonlocal.

But: In an important sense the observations come first and the theory (state vectors in Hilbert space) builds up on top of them. The fact that the theory then makes those observations seem unsurprising is more of an insight into how well the theory has been tailored to fit around those observations than it is into how surprising the observations might be in themselves.

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  • $\begingroup$ What about cosmological constant problem, where the theory and observation diverge enormously? $\endgroup$
    – VVM
    Aug 29, 2023 at 17:44
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    $\begingroup$ @VVM I am not knowledgeable about the cosmological constant problem. However, if theory and observation diverge enormously then I think it is safe to say that (1) the observations are surprising and (2) the theory is bad, and in need of replacement. This is how quantum theory came into existence: the pre-quantum theories disagreed with observation for things like the photoelectric effect (so these observations were surprising), so those theories were replaced. $\endgroup$
    – Dast
    Aug 30, 2023 at 10:20
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Just because the notion of different pictures don't always "mention" the spatial degrees of freedom does not mean that the state is independent of the spatial degrees of freedom. In fact, the emphasis on time in these pictures is because, in the context of non-relativistic QM, the theories are based on the Hamiltonian, which is the time-time component of the energy-momentum tensor.

In different scenarios where one is more interested in the spatial evolution (say along $z$) of the state (like when a quantum state is propagating through a random medium) then the system is modeled in terms of the propagation operator, instead of the Hamiltonian. In such a scenario, the different "pictures" would specify whether the state depends on $z$ or not.

All these pictures are formal devices to aid our calculations. The physical system that is being modeled by these theories doesn't care in which picture we model it.

So if we think of the physical system for multiple quanta of some field, and we want to model it with a wave function, then that wave function will depend on all the degrees of freedom, both spatial and temporal degrees of freedom. As @KenWharton correctly mentions in his comment, we can write such a wave function as $\psi(\mathbf{x}_1,\mathbf{x}_2,...,\mathbf{x}_N,t)$ for $N$ particles. What such a wave function represents is the probability amplitude to find those $N$ quanta at the all those positions at time $t$. The modulus square of the wave function gives probability density.

The term "non-locality" is ambiguous, because it can mean different things. Sometimes it just means that we can measure correlations at different points in space. Such a situation can simply be explained as the consequence of some event that happened in the past and then propagated to those to points. However, it can also mean "spooky action at a distance" (an action at one point instantaneously causes an effect at another point) for which there is no scientific evidence. Entanglement does not imply "spooky action at a distance."

The basic concept of entanglement is well understood (excluding all the different interpretations, which is not really science). However, there are numerous consequences that is still being researched actively.

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  • $\begingroup$ "...it can also mean "spooky action at a distance" (an action at one point instantaneously causes an effect at another point) for which there is no scientific evidence." : Perhaps you missed the thousands* of papers demonstrating quantum nonlocality, such as this from a team by a 2022 Nobel winner arxiv.org/abs/quant-ph/0201134 You are welcome to dismiss evidence proving things you don't like with a hand wave, but saying there is none is far-fetched. *You can check this here: arxiv.org/search/advanced with "nonlocal" in the title. I got over 5,000 hits. $\endgroup$
    – DrChinese
    Aug 30, 2023 at 22:15
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    $\begingroup$ @DrChinese, as I explained, the term nonlocal can mean different things. The nonlocality found in literature is not in general the one referring to spooky action at a distance. $\endgroup$ Aug 31, 2023 at 3:48
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The most fundamental variants of quantum theory are quantum field theories that have equations of motion that are local in space and time such as the Dirac equation. One answer above sez that the Schrodinger equation doesn't mention the speed of light. This is true, but entirely irrelevant since the Schrodinger equation is the non-relativistic limit of relativistic quantum theories so it doesn't apply when relativistic effects are relevant.

Bell's theorem doesn't imply non-locality. It sez that if we have a theory where observable quantities are represented by stochastic variables, then that theory must be non-local unless the measurement devices are somehow correlated with the measured systems before the experiment is performed. But in quantum theory measurable quantities are described by Hermitian operators not by stochastic variables so it doesn't satisfy the conditions required for non-locality.

There is an entirely local explanation of the EPR correlations, teleportation and so on in terms of locally inaccessible quantum information being carried in decoherent channels:

https://arxiv.org/abs/quant-ph/9906007v2

https://arxiv.org/abs/1109.6223

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  • $\begingroup$ "Bell's theorem doesn't imply non-locality. It sez that if we have a theory where observable quantities are represented by stochastic variables, then that theory must be non-local unless the measurement devices are somehow correlated with the measured systems before the experiment is performed. " This is not the other fork of Bell's Theorem. The measurement device settings can be changed mid flight and outside of the light cones of the systems being measured, so there cannot be such correlations before the experiment is performed. Time ordering is not relevant to the predictions of QM. $\endgroup$
    – DrChinese
    Aug 30, 2023 at 22:23
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    $\begingroup$ @DrChinese According to QM the measurement devices aren't correlated before the measurement results are compared and their settings can be changed in mid flight. Bell's theorem is about what has to be true for a theory to match the predictions of QM with stochastic variables. Superdeterminism claims the measurement devices are somehow correlated with the measured systems in advance. BT doesn't strictly rule this out, but there are other criticisms of superdeterminism physics.stackexchange.com/questions/106725/… $\endgroup$
    – alanf
    Sep 1, 2023 at 6:33
  • $\begingroup$ QFT is only local when it comes to constructing the states. To make real physical predictions from those states, it must necessarily use the Born rule on them. Then, it stops being local. QM is fundamentally non local and so is QFT. Only a fraction of it, when looked in isolation, is local. $\endgroup$
    – Juan Perez
    Sep 24, 2023 at 23:01
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As I cannot find this point mentioned in the above answers, it gives me an opportunity to contribute.

It is not totally true that the state of a system is not dependent on physical space. The state $$ |\psi \rangle \in \mathcal{H} \,,$$ where $\mathcal{H}$ is a Hilbert space - a (complete complex) vector space with an inner product.

Mathematically, $\mathcal{H}$ is defined as $\mathcal{H} = L^2(\mathbb{R}^3)$, i.e., it is a space of all square-integrable functions defined on the physical space $\mathbb{R}^3$.

In ordinary words, this means that $| \psi \rangle$ can be seen as a function $\psi(x,y,z)$ where $(x,y,z) \in \mathbb{R}^3$ which satisfies certain properties we would like to assign to a quantum state. As such, the state depends on $\mathbb{R}^3$.

However, if the particle defined by state $|\psi \rangle$ is a free particle, then the dependence of $\mathbb{R}^3$ gives us no new information. But for a bound system, say an electron in an infinite potential well $|\psi \rangle$ is really dependent on $\mathbb{R}^3$. By that, I mean that $|\psi \rangle$ is non-zero only at certain values of $\mathbb{R}^3$.

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