There is no apparent faster-than-light communication involved in quantum entanglement.Sayestu said:
Not more than any other interpretation.Sayestu said:
That's a broad question!Sayestu said:
Well, yes we do understand it perfectly well - assuming QM is correct. Bell's Theorem places constraints on any other proposed alternative, or extended, theory to replace QM. Loosely speaking you need instant communication or measurements which do not provide a definite outcome. "Every possibility happening at once" for instance. So you could argue that the apparent FTL communication is an artefact of rejecting indefiniteness. Not that the required indefiniteness is any easier to swallow than FTL.Sayestu said:
Sure. By postulating that the wave function doesn't collapse, MWI leaves multiple possible outcomes superposed. These develop into the separate worlds. This co-existence of outcomes satisfies one of the constraints of Bell's Theorem, indefiniteness. So there is no need to suppose there's any FTL communication. You don't need both.Sayestu said:
Not specifically. MWI was developed to get rid of the collapse postulate. As such it affects all of QM.Sayestu said:
Sort of. According to MWI, spooky action at a distance is just an illusion. Fundamentally, there is no action at a distance because there is no distance. According to MWI, we don't really live in a 3-dimensional world (that's just an illusion), so there is no distance in a 3-dimensional sense.Sayestu said:
This is valid up to a certain extent. It does not explain why we we do not have faster than light communication.Demystifier said:
It's up to you to explain why you think we might have FTL communication involved in quantum entanglement.pines-demon said:
We do not have FTL communication. Call it whatever you want weak nonlocality, influence, or correlations, but it is not FTL signalling or FTL communication of energy/information.PeroK said:
Do you mean the multi-dimensional world of phase space or something arcane, esoteric and ineffable?Demystifier said:
or just Hilbert space.kered rettop said:Do you mean the multi-dimensional world of phase space or something arcane, esoteric and ineffable?
Not exactly, but you are close. It can be the multi-dimensional world of configuration space (that's where the wave function in the position representation lives), or it can be the quantum Hilbert space.kered rettop said:Do you mean the multi-dimensional world of phase space or something arcane, esoteric and ineffable?
MWI explains quite easily why entanglement does not allow faster than light communication. It is because the agent who is supposed to communicate cannot choose at will which of the many branches of the wave function will be his branch.pines-demon said:
I would just like to reassure pines-demon that, if he feels that Demystifier is missing out a few vital steps in the argument, he is not alone.Demystifier said:
Those are left as an exercise for the reader.kered rettop said:I would just like to reassure pines-demon that, if he feels that Demystifier is missing out a few vital steps in the argument, he is not alone.
Sorry, I can't read.Demystifier said:Those are left as an exercise for the reader.
Do you know why entanglement cannot be used for FTL signaling in standard QM?kered rettop said:
kered rettop said:I would just like to reassure pines-demon that, if he feels that Demystifier is missing out a few vital steps in the argument, he is not alone.
I was just going to go deep into the semantics of MWI but I can live with this answer.Demystifier said:Those are left as an exercise for the reader.
I sense a trap! I am tempted to say, The No-Communication Theorem, but as your question presumably harks back to post #12, I'll just say no.Demystifier said:
This is one of those cases where you should yield to temptation.kered rettop said:
The question specifically asked about "standard QM", and the no communication theorem is the answer for that. Different interpretations will tell different stories about why the no communication theorem is true (and post #12 gives a brief version of the story that the MWI might tell), but that's not "standard QM", that's interpretations.Where have I heard that before?PeterDonis said:
Of course. Everybody knows that...PeterDonis said:
... or maybe not everybody.pines-demon said:
Does any interpretation single out entanglements as being "broken" by a measurement? I would have thought that they must assert that either any superposition is or that none is. Serious question.PeterDonis said:
Any interpretation in which measurements have single outcomes will, since the resulting state will be a product state.kered rettop said:
That's why I said "single out entanglements"! To my simple mind, "breaking" an entanglement is exactly the same as "collapse" of the wave function. Why have two words for the same thing, unless they are interpreted differently?PeterDonis said:
</>Drat the system timeout! I really wanted to have another go at clarifying what I meant. Well, never mind. I'm clear on the physics and I don't want to go hair-splitting or getting bogged down in semantics. But I do feel that talking about "breaking the entanglement" suggests that something weird and wonderful is going on unrelated to "collapse of the wavefunction" - which is merely weird and wearisome. So I was wondering whether there are any interpretations, preferably within the canon of acceptable sources on PF, which tell a different story about state reduction in different scenarios? Probably not!kered rettop said:
You can model the breaking of entanglement with unitary evolution. You just need to include the pointer degrees of freedom.kered rettop said:
Of course. Same as collapse of the wavefunction. Make the pointer big enough and you've got MWI. So was that a yes or a no?Morbert said:
Still looking for straight answers in the interpretations subforum, I see? Sorry, they're not on offer there. :)kered rettop said:
No, it isn't, because you can have wave function collapse even in cases where you are not measuring an entangled system. For example, if I measure the spin of a single free electron that isn't entangled with anything, I still have wave function collapse (in an interpretation where there is such a thing) when I detect the electron in one or the other output arm of the Stern-Gerlach device.kered rettop said:
That doesn't break entanglement, it just spreads it out to include the pointer degrees of freedom. (And then to include the environment degrees of freedom, once you take decoherence into account.) I already posted about this.Morbert said:
I wouldn't say they're exactly "on offer" anywhere else on PFNugatory said:
Still, point taken.Thanks. It's nice to be told you've got something right, once in a while.Nugatory said:
Ah, just when I thought I was following you! What's wrong with "we have two particles"? It may be that in QFT, the number of particles can be uncertain but surely not in good old vanilla QM?Nugatory said:
Didn't I mention this? Must have edited it out thinking it was too obvious to be worth saying. Life being short and all that. Yes, obviously, the words do mean different things and you can't break the entanglement of a system that is not entangled because that would be silly. But they still refer to the same putative physical process.PeterDonis said:
In the math we don’t have two particles. We have one quantum system described by one wave function, it just so happens that the two possible measurements (spin/polarization at one detector, spin/polarization at the other) we might perform on this system happen at different places. Because of the spatial separation our classical intuition demands that we think in terms of two distinguishable particles in two different places… and it’s a short step from there to spooky action at a distance and all the other entanglement misunderstandings that show up in our B-level threads.kered rettop said:
I am referring to the entanglement in the microscopic system. E.g. Consider a two-particle system in the Bell state ##|\Phi^+_{12}\rangle\langle\Phi^+_{12}|## and a joint (and ideal and nondestructive) spin measurement represented by ##\frac{\hbar^2}{4}\sigma^1_x\sigma^2_z## (I.e. a measure of spin-x on particle 1 and spin-z on particle 2). If we include the pointer degrees of freedom, a unitary evolution through the measurement process and a trace over the pointer degrees of freedom will yield the two-particle system in the unentangled state ##\rho_1\otimes\rho_2##.PeterDonis said:
Unitary evolution doesn't break that either. The microscopic entanglement will no longer be maximal, because other degrees of freedom are involved, but it doesn't go away either.Morbert said:
Throws away precisely the information you need to see that the entanglement between the microscopic degrees of freedom is still there. To properly evaluate this question you need to look at the actual pure state of the entire system; you can't trace out anything.Morbert said: