Issues on notation and concept of entanglement

[weirdoguy]

You assert that, you did not prove it.

Isn't that discussed in almost every textbook on QFT? Peskin & Shroeder for example, somewher in the first few chapter.

 

[AndreiB]

I still don't understand what you need to prove.

P1 speaks about commutativity, P2 about causation. In order to arrive at a contradiction you need to reformulate P1 so that it also speaks about causation, or reformulate P2 so that it speaks about commutativity. You did not do that.

As far as I understand commutativity means that the order of measurements do not matter (statistically, since the theory is not deterministic). Why is this incompatible with A causing B or B causing A?

 

[AndreiB]

Isn't that discussed in almost every textbook on QFT? Peskin & Shroeder for example, somewher in the first few chapter.

Can you provide a reference?

 

[vanhees71]

Just check a good textbook on relativistic QFT. The most lucid treatment of all these fundamental issues is in Weinberg, Quantum Theory of Fields, vol. 1.

 

[AndreiB]

Just check a good textbook on relativistic QFT. The most lucid treatment of all these fundamental issues is in Weinberg, Quantum Theory of Fields, vol. 1.

Can you provide a relevant quote? I do not have that book.

 

[Demystifier]

If you consider interaction without a mediator, which cannot be detected in principle, you are adopting a very strong form of antirealism.

No I'm not. If I consider something that cannot be detected, it means that I believe that things exist even when we don't measure them, which is the exact opposite of antirealism.

 

[martinbn]

No I'm not. If I consider something that cannot be detected, it means that I believe that things exist even when we don't measure them, which is the exact opposite of antirealism.

But you said that the mediator does not exist. As to the action it is not that you believe that it exist even when you cannot meausre it. It cannot be detected even if you do measurements of any sort.

 

[vanhees71]

Here's a very good online resource

https://www.damtp.cam.ac.uk/user/tong/qft.html

 

[Demystifier]

But you said that the mediator does not exist. As to the action it is not that you believe that it exist even when you cannot meausre it. It cannot be detected even if you do measurements of any sort.

So what? In Newtonian gravity planets exist, even though the mediator doesn't. In Bohmian mechanics particle positions exist, even though the mediator doesn't.

 

[Demystifier]

Can you provide a relevant quote? I do not have that book.

Here is the relevant part of Weinberg. Whether it confirms the claim of @vanhees71 or not, decide by yourself.

 

[martinbn]

So what? In Newtonian gravity planets exist, even though the mediator doesn't. In Bohmian mechanics particle positions exist, even though the mediator doesn't.

It is not the same. In the Newtonian gravity if Alice does something, Bob can detect the influence on his system. In QM you cannot. In your preferred interpretation you put realism in the particle trajectories and posit undetectable action with nonexistent mediator.

 

[AndreiB]

Here is the relevant part of Weinberg. Whether it confirms the claim of @vanhees71 or not, decide by yourself.

Thanks! Do you know what Weinberg means by "a measurement at point x should not be able to interfere with a measurement at point y"? What would an interference look like?

 

[Demystifier]

you put realism in the particle trajectories

Exactly, and that's why it's called realism, not antirealism.

 

[Demystifier]

Thanks! Do you know what Weinberg means by "a measurement at point x should not be able to interfere with a measurement at point y"? What would an interference look like?

He doesn't mean interference of probability amplitudes. He means interference in the sense of mutual influence.

It really means the following. Suppose that two quantum observables, A and B, commute. Furthermore, suppose that Alice measured A, that Bob knows that Alice measured it, but that he does not know the result of her measurement. Then, from this knowledge, Bob cannot conclude anything new about the probabilities of measurement outcomes of B.

 

[martinbn]

Exactly, and that's why it's called realism, not antirealism.

Of course, that is the standard terminology. But you also need nonexistant things. So you are just shifting where the antirealism will be. Some kind of Heisenberg cut for antirealism. To me your interpratation is equaliy unpalitable because it has antirealistic elements as well.

 

[AndreiB]

It really means the following. Suppose that two quantum observables, A and B, commute. Furthermore, suppose that Alice measured A, that Bob knows that Alice measured it, but that he does not know the result of her measurement. Then, from this knowledge, Bob cannot conclude anything new about the probabilities of measurement outcomes of B.

This seems compatible with A causing B, right?

 

[vanhees71]

No, it's not necessarily compatible with A causing B. An event at B can be causally influenced by A only if the event at B is in some future lightcome of an event at A. This holds for all special-relativistic dynamical models including standard local relativistic QFT.

 

[AndreiB]

An event at B can be causally influenced by A only if the event at B is in some future lightcome of an event at A.

How does this follow from what Weinberg says? There is no talk there about lightcones.

 

[Demystifier]

This seems compatible with A causing B, right?

It depends on what one means by "causing", but in the sense you mean it I would agree. Of course, adherents of orthodox QM by "causing" mean something else.

 

[Demystifier]

But you also need nonexistant things.

Which ones? I don't need mediator.

 

[AndreiB]

It depends on what one means by "causing", but in the sense you mean it I would agree. Of course, adherents of orthodox QM by "causing" mean something else.

By causing I mean that the spin at B is instantly forced to take the opposite value of A.

 

[Demystifier]

By causing I mean that the spin at B is instantly forced to take the opposite value of A.

Yes, in that sense I agree.

 

[vanhees71]

Even if you agree with that, the cause is not a faster-than-light interaction, at least not within a local relativistic QFT, but it's due to the correlation described by entanglement.

 

[martinbn]

Which ones? I don't need mediator.

And CI doesn't need values for observables that have not been measured. It is exactly the same. You also need something (the action) that is there, but no way you can tell even in principle, it is like it isn't there.

 

[Demystifier]

You also need something (the action) that is there

Define "action"!

 

[Demystifier]

Even if you agree with that, the cause is not a faster-than-light interaction, at least not within a local relativistic QFT, but it's due to the correlation described by entanglement.

The cause I have in mind is of course a faster-than-light interaction, which of course is incompatible with orthodox relativistic QFT, which of course is why I have in mind an unorthodox relativistic QFT. The unorthodox relativistic QFT uses all the equations of orthodox QFT and makes the same measurable predictions, but the narrative is slightly different.

 

[vanhees71]

Well, but then you contradict the mathematical formulations. How can an interpretation make sense if it contradicts the mathematical foundations, and if all measurable predictions are the same what sense does it make to invent inconsistencies between interpretation on the one hand as well as the mathematical description and empirical facts?

 

[Demystifier]

Well, but then you contradict the mathematical formulations. How can an interpretation make sense if it contradicts the mathematical foundations, and if all measurable predictions are the same what sense does it make to invent inconsistencies between interpretation on the one hand as well as the mathematical description and empirical facts?

There are no inconsistencies. You assume that only one interpretation (the orthodox one) is consistent with the undisputed equations, but that's wrong. There are many interpretations of quantum theory consistent with the equations of quantum theory.

 

[martinbn]

Define "action"!

https://www.physicsforums.com/threa...-concept-of-entanglement.1007463/post-6545856

 

[vanhees71]

There are no inconsistencies. You assume that only one interpretation (the orthodox one) is consistent with the undisputed equations, but that's wrong. There are many interpretations of quantum theory consistent with the equations of quantum theory.

How can an interpretation claiming the existence of ftl interactions be consistent for a theory which excludes them in its mathematical formulation?

 

[martinbn]

How can an interpretation claiming the existence of ftl interactions be consistent for a theory which excludes them in its mathematical formulation?

By using different terminology. Interaction means something else, not what you might have guessed. Just like non-local means something else, not the usual.

 

[Demystifier]

How can an interpretation claiming the existence of ftl interactions be consistent for a theory which excludes them in its mathematical formulation?

By extending the set of interacting objects. Relativistic QFT excludes ftl interactions in a law for evolution of the state in the Hilbert space $\psi$. But minimal relativistic QFT says nothing about other possible interacting objects $\lambda$ that are not given by $\psi$. Bohmian interpretation is an extension of minimal relativistic QFT. It does not change the evolution of $\psi$, but it makes a concrete proposal for $\lambda$ and postulates a nonlocal law for evolution of $\lambda$. Since minimal and Bohmian QFT agree on equations for $\psi$, and since minimal QFT says nothing mathematical about the evolution of $\lambda$, there is no any mathematical contradiction between minimal and Bohmian QFT. The contradiction is only philosophical, because minimal QFT uses some philosophical arguments to argue that there is no $\lambda$ to begin with.

Or schematically:
Minimal QFT: $\psi$ local, period.
Bohmian QFT: $\psi$ local, $\lambda$ nonlocal.

 

[vanhees71]

By using different terminology. Interaction means something else, not what you might have guessed. Just like non-local means something else, not the usual.

Yes, and this is what makes this philosophy inclined topic so useless for science. You use the words not in their clear scientific meaning but distort them as long as it has no meaning anymore at all. Within the exact sciences one should say interaction when one means an interaction and correlation when one means correlation and should keep this strictly separated with a well-defined meaning. You save a lot of time for real discussions rather than defining again and again the same words.

 

[vanhees71]

By extending the set of interacting objects. Relativistic QFT excludes ftl interactions in a law for evolution of the state in the Hilbert space $\psi$. But minimal relativistic QFT says nothing about other possible interacting objects $\lambda$ that are not given by $\psi$. Bohmian interpretation is an extension of minimal relativistic QFT. It does not change the evolution of $\psi$, but it makes a concrete proposal for $\lambda$ and postulates a nonlocal law for evolution of $\lambda$. Since minimal and Bohmian QFT agree on equations for $\psi$, and since minimal QFT says nothing mathematical about the evolution of $\lambda$, there is no any mathematical contradiction between minimal and Bohmian QFT. The contradiction is only philosophical, because minimal QFT uses some philosophical arguments to argue that there is no $\lambda$ to begin with.

Or schematically:
Minimal QFT: $\psi$ local, period.
Bohmian QFT: $\psi$ local, $\lambda$ nonlocal.

Standard QFT uses physical arguments and doesn't interoduce fictitious enigmatic entities called $\lambda$ in the first place. I guess you mean $\lambda$ as in Bell's original paper on his local deterministic HV models, and this is disproven by experiment. We can move on with new physics for at least 10-20 year now!

 

[Demystifier]

I guess you mean $\lambda$ as in Bell's original paper on his local deterministic HV models, and this is disproven by experiment.

I mean Bell's $\lambda$ which can be either local or nonlocal. The nonlocal Bell's $\lambda$ is of course not disproven.