How does QFT handle non-locality?

[ShayanJ]

This assumption is of course unnecessary, because the outcome of B's measurement is not affected by the collapse. The probabilities for finding a certain polarization state at Bob's place are given as well by the initial entangled state, in which the biphoton has been prepared, including the non-classical correlations violating Bell's inequality. This is the minimal interpretation, and the collapse even unobservable. So why should I assume it to happen?

Correct me if I'm wrong, but what I understand from this is that when we send two spins to Alice and Bob, Alice is left with a spin in a state described by the density matrix $\rho_A=\frac{1}{2}(|\downarrow\rangle\langle \downarrow |+|\uparrow\rangle\langle \uparrow |)$ regardless of the fact that Bob has made any measurement or not. When a system is in such a state, we know that there is no axis that when Alice measures her spin along that axis, she gets +1 with certainty. So if we do this experiment over and over again, she'll get 50-50 distribution of ups and downs for any axis she chooses. But if collapse is correct, after Bob has measured his spin, Alice's spin will end up in one of the states $|\uparrow \rangle$ or $|\downarrow \rangle$, which means if we do this experiment over and over again, Alice is able to find an axis that continues to give her the same result +1 every time she measures her spin. This seems to me an experimental way to settle the issue whether collapse is really there or not, or maybe I'm just misunderstanding something!(Or maybe its not that much easy to say whether there exists such an axis as described above or not!)

 

[mfb]

But are there? As far as I know, Copenhagen (in one flavour or another) is the only consensus interpretation.

There is certainly no consensus interpretation involving collapses.
All interpretations have some problems.

I think the minimal interpretation (which I consider to be a flavor of the Copenhagen interpretation) is the only one which is really consensus among physicists.

Okay, depends on the definition of "Copenhagen interpretation". The description of collapses and Copenhagen are often combined.

 

[Demystifier]

No, I didn't forget that posting, but there you didn't mention the collapse

You obviously did forget a lot about that posting, since I did mention the collapse, several times, in items 4. and 6.

Explaining quantum non-locality takes a several steps. To understand it, one has to be able to have all the steps in one's mind at once.

 

[atyy]

There is certainly no consensus interpretation involving collapses.
All interpretations have some problems.

There is a consensus interpretation, and it involves collapse. This is why the textbooks have collapse. Can you give a consensus source for any interpretation without a collapse?

 

[mfb]

Can you give a consensus source for any interpretation without a collapse?

Of course not, because there is no consensus.
If there would be, this discussion and hundreds of papers discussing different interpretations would not exist.

 

[atyy]

Of course not, because there is no consensus.
If there would be, this discussion and hundreds of papers discussing different interpretations would not exist.

But if there isn't, then unless one believes the textbooks are wrong, the textbook version of Copenhagen is the only consensus interpretation (it's not even an interpretation, it's simply QM).

 

[mfb]

But if there isn't, then unless one believes the textbooks are wrong, the textbook version of Copenhagen is the only consensus interpretation

Where do the textbooks claim that Copenhagen with collapses is a consensus interpretation?
If they would claim that, they would be wrong, but they don't. They just do not cover all interpretations, and they do not have to.

 

[atyy]

Where do the textbooks claim that Copenhagen with collapses is a consensus interpretation?
If they would claim that, they would be wrong, but they don't. They just do not cover all interpretations, and they do not have to.

But all other interpretations have problems, to the point where it is unclear if they even work as scientific theories. So Copenhagen simply has no viable competitors. Can you name any viable interpretations except Copenhagen?

 

[mfb]

But all other interpretations have problems, to the point where it is unclear if they even work as scientific theories.

Collapse has collapse as problem. "We let fields evolve with a unitary, local, deterministic evolution. Then (at some arbitrary, unmeasurable point in time with unclear definition) a magical fairy comes and changes the wavefunction in some ill-defined, nonlocal, nondeterministic way, to the point that we suddenly have elements that are not described with a wave function any more but have to be treated in a macroscopic way."

Can you name any viable interpretations except Copenhagen?

All major interpretations are viable, and Wikipedia has a list.

Take any survey about favorite interpretations of scientists: collapses find a sizeable number of votes, but not the absolute majority. And consensus would be far more than an absolute majority. Claiming consensus where there is none is just wrong.

 

[atyy]

Collapse has collapse as problem. "We let fields evolve with a unitary, local, deterministic evolution. Then (at some arbitrary, unmeasurable point in time with unclear definition) a magical fairy comes and changes the wavefunction in some ill-defined, nonlocal, nondeterministic way, to the point that we suddenly have elements that are not described with a wave function any more but have to be treated in a macroscopic way."

Yes, but that is not a problem since Copenhagen acknowledges that it needs magical fairies.

All major interpretations are viable, and Wikipedia has a list.

Take any survey about favorite interpretations of scientists: collapses find a sizeable number of votes, but not the absolute majority. And consensus would be far more than an absolute majority. Claiming consensus where there is none is just wrong.

That is not true. Whether Bohmian Mechanics, for example, can work for all relativistic quantum theories is still a matter of research. Similarly, major proponents of MWI acknowledge that it has problems. Copenhagen is consensus in the sense that if these other interpretations work, then they must derive Copenhagen.

 

[mfb]

Yes, but that is not a problem since Copenhagen acknowledges that it needs magical fairies.

Okay, if you acknowledge that you have a problem it is not a problem any more?.

Similarly, major proponents of MWI acknowledge that it has problems.

Wait, the acknowledgment trick is fine for collapses, but not for MWI?

Copenhagen is consensus in the sense that if these other interpretations work, then they must derive Copenhagen.

That is not what "consensus" means at all.

Sorry, this discussion is getting too ridiculous, I'm out.

 

[atyy]

Okay, if you acknowledge that you have a problem it is not a problem any more?.

Yes, because the problem is not a technical problem, ie. the theory makes sense if there are magical fairies. Also, the magical fairies have been observed.

Wait, the acknowledgment trick is fine for collapses, but not for MWI?

In MWI the problems are technical, it is not clear whether any magical fairies can save MWI.

That is not what "consensus" means at all.

Sorry, this discussion is getting too ridiculous, I'm out.

One again, I am only defending textbook QM. If you are right, then there is no QM at all.

 

[Demystifier]

There is a consensus interpretation, and it involves collapse. This is why the textbooks have collapse. Can you give a consensus source for any interpretation without a collapse?

There is, indeed, a consensus that collapse is a useful bookkeeping tool. However, there is no consensus whether the collapse is anything more than that.

 

[vanhees71]

Correct me if I'm wrong, but what I understand from this is that when we send two spins to Alice and Bob, Alice is left with a spin in a state described by the density matrix $\rho_A=\frac{1}{2}(|\downarrow\rangle\langle \downarrow |+|\uparrow\rangle\langle \uparrow |)$ regardless of the fact that Bob has made any measurement or not. When a system is in such a state, we know that there is no axis that when Alice measures her spin along that axis, she gets +1 with certainty. So if we do this experiment over and over again, she'll get 50-50 distribution of ups and downs for any axis she chooses. But if collapse is correct, after Bob has measured his spin, Alice's spin will end up in one of the states $|\uparrow \rangle$ or $|\downarrow \rangle$, which means if we do this experiment over and over again, Alice is able to find an axis that continues to give her the same result +1 every time she measures her spin. This seems to me an experimental way to settle the issue whether collapse is really there or not, or maybe I'm just misunderstanding something!(Or maybe its not that much easy to say whether there exists such an axis as described above or not!)

I'd formulate the first sentence slightly differently: We prepare a spin-entangled two-particle state, and A and B at (perhaps far) distant locations measure a spin component of the particles.

No matter what Alice measures, without communicating with Bob there is no way for her predicting which outcome her spin-component measurement will have, regardless of the orientation of her Stern-Gerlach apparatus. If both experimenters take an accurate record of the time of their spin measurements and if they orient their SG apparati in the same direction they will find a 100% correlation when comparing their measurement protocols. It doesn't matter in which temporal order they do their measurements (they could be even spacelike separated, i.e., not having any time order at all). For me that's a clear indication that the local measurements of the spins at A's and B's places do not affect each other but that the correlation of the outcome of spin-component measurements is inherent in the preparation of the two-particle state in the given entangled way, and it's not possible to empirically justify or disprove the claim that a state collapse has occured.

Of course, you have to perform the experiment very often, because the predictions of QT are probabilistic, and this you can test empirically only by preparing a lot of such particle pairs stochastically independently and perform the measurement on a sufficiently large ensemble to get the statistical significance you want (for discovery in the HEP community you must aim for at least $5 \sigma$ significance before you can cry "heureka").