Non-locality

[Eise]

1 hour ago, Dalo said:

As a philosopher I can assure you that it is the most absurd affirmation you could ever express. Feel free to ask specialists in the philosophy of Physics.

Ah! On which university did you study philosophy?

[uncool]

There is one further main point, and that main point is the center of Bell's theorem. That difference is, approximately, that for a classical (non-quantum) theory, that certain probabilities must add in certain ways. 

[Dalo]

37 minutes ago, Eise said:

Ah! On which university did you study philosophy?

The Sorbonne, Paris, France. You?

[interested]

54 minutes ago, Dalo said:

  On 12/16/2017 at 2:25 PM, Dalo said:

1) both photons start with the same polarization,
2) both filters are identical.

In such a situation, we can say that the distance between both photons is totally irrelevant. What is important are the two identities I have just presented. Photon 2 and filter 2 can be considered as a local system in the sense that they could swap places with photon 1 and filter 1 without changing anything to the results of the experiment.

In other words, what happens with photon 2 and filter 2 is what would happen with photon 1 and filter 2 locally!

The following link is pretty simple, and explains your polarization problems, in none too technical terms. https://en.wikipedia.org/wiki/Polarization_(waves) 

Basically with no maths the first polarizing filter re-polarizes half of the non polarized light, via diffraction and delays in the propagation of the none polarized light passing through the filter.

[Dalo]

13 minutes ago, interested said:

The following link is pretty simple, and explains your polarization problems, in none too technical terms. https://en.wikipedia.org/wiki/Polarization_(waves) 

Basically with no maths the first polarizing filter re-polarizes half of the non polarized light, via diffraction and delays in the propagation of the none polarized light passing through the filter.

It is obvious you are not addressing the issue. The point is non-locality. Polarization is here an illustration of the problem. A mere example. It is often presented not with polarization but, for instance the spin of a particle. Limiting the discussion to polarization is meaningless.

https://en.wikipedia.org/wiki/Quantum_nonlocality

50 minutes ago, uncool said:

There is one further main point, and that main point is the center of Bell's theorem. That difference is, approximately, that for a classical (non-quantum) theory, that certain probabilities must add in certain ways. 

Yes, and that was also von Neumann's main argument. Only Bell limited the validity of this argument to local hidden variables. There was no way probabilities could  add up to the same statistical regularities everybody considers as valid (including me). The difference between von Neumann's and Bell's analysis is that Bell left the possibility open for non-local hidden variables.

https://en.wikipedia.org/wiki/Hidden_variable_theory

That does not change my claim which takes into account both these theories. See my first 5 or 6 posts.

Edited December 18, 2017 by Dalo

[interested]

12 minutes ago, Dalo said:

It is obvious you are not addressing the issue. The point is non-locality. Polarization is here an illustration of the problem. A mere example. It is often presented not with polarization but, for instance the spin of a particle. Limiting the discussion to polarization is meaningless.

https://en.wikipedia.org/wiki/Quantum_nonlocality

What level do you wish to pitch at, you expressed previously you wanted a none mathematical explanation of why 50% on none polarized light could pass the first filter. I gave you an explanation in as simple terms as is possible.

Here is a rather long drawn out link on non locality from utube the other links down the side are amusing and none mathematical also.

, the or

[uncool]

Dalo, I am curious. Have you heard of superdeterminism?

[Dalo]

14 minutes ago, interested said:

What level do you wish to pitch at, you expressed previously you wanted a none mathematical explanation of why 50% on none polarized light could pass the first filter. I gave you an explanation in as simple terms as is possible.

Here is a rather long drawn out link on non locality from utube the other links down the side are amusing and none mathematical also.

, the or

I do not need a lesson on non-locality. I had a very specific question, which was in fact off-topic, concerning polarization. Instead of trying to explain to me what non-locality is, maybe you could try and show me how my interpretation of it is faulty, and why therefore my claim is wrong. My whole claim says that the concept of non-locality is wrong and that entanglement is the result of a theoretical bias.

This is the point that you should tackle. I also said that I would not be surprised if somebody proved me wrong because I am aware of how pretentious my claim sound.

Unluckily I have not read a single argument that took my claim and my arguments seriously. It was all about polarization, philosophy, ignorance of physics laws, or, last, in your reaction, a reference to what non-locality and entanglement mean. I know what they mean. I am trying to show that what they say cannot be true. Maybe I am wrong, and so, I am just waiting for somebody to show me why and how I am wrong. Can you do that?

I also said, it is not a matter of mathematics or physics, but a matter of interpretation. If it were a pure mathematical or physical problem I would have stayed away from it.

I understand this is a Physics forum and people are not used to approach some issues philosophically instead of mathematically or "physically", but I cannot help it that some problems do belong (also) to philosophy.

4 minutes ago, uncool said:

Dalo, I am curious. Have you heard of superdeterminism?

Yes, but then there are so many versions of it that you will have to be more specific.

Edited December 18, 2017 by Dalo

[uncool]

6 minutes ago, Dalo said:

Yes, but then there are so many versions of it that you will have to be more specific.

I do not believe that the version matters for my question. 

Does superdeterminism address your concerns?

Edited December 18, 2017 by uncool

[Dalo]

5 minutes ago, uncool said:

I do not believe that the version matters for my question. 

Does superdeterminism address your concerns?

in this case, no.

edit: what I am claiming is that both photons, in the polarization example can be considered as local. Each photon is reacting exactly as the other one would if it would have had to go successively though two polarizers having the same direction as the other two. 

In this sense, they are both local, and do not influence each other in any way. They react the same because they are identical and the filters they go through are also identical.

A very simple explanation. Maybe too simple. But nobody has proven it wrong yet.

Edited December 18, 2017 by Dalo

[uncool]

Just now, Dalo said:

in this case, no.

Why not?

[swansont]

On 12/16/2017 at 9:25 AM, Dalo said:

 Understood this way, the problem does not seem as hard if we assume that:
1) both photons start with the same polarization,
2) both filters are identical.

In such a situation, we can say that the distance between both photons is totally irrelevant. What is important are the two identities I have just presented. Photon 2 and filter 2 can be considered as a local system in the sense that they could swap places with photon 1 and filter 1 without changing anything to the results of the experiment.

In other words, what happens with photon 2 and filter 2 is what would happen with photon 1 and filter 2 locally!

The issue is that system 1 and system 2 are not local with respect to each other, and yet show correlations that should not be present.

[Dalo]

Just now, swansont said:

The issue is that system 1 and system 2 are not local with respect to each other, and yet show correlations that should not be present.

I have edited my previous post. I think that I answered your question there.

 

5 minutes ago, Dalo said:

in this case, no.

edit: what I am claiming is that both photons, in the polarization example can be considered as local. Each photon is reacting exactly as the other one would if it would have had to go successively though two polarizers having the same direction as the other two. 

In this sense, they are both local, and do not influence each other in any way. They react the same because they are identical and the filters they go through are also identical.

A very simple explanation. Maybe too simple. But nobody has proven it wrong yet.

 

[swansont]

12 minutes ago, Dalo said:

in this case, no.

edit: what I am claiming is that both photons, in the polarization example can be considered as local. Each photon is reacting exactly as the other one would if that would have had to go successively though two polarizers having the same direction as the other two. 

In this sense, they are both local, and do not influence each other in any way. They reach the same because they are identical and the filters they go through are also identical.

A very simple explanation. Maybe too simple. But nobody has proven it wrong yet.

If you place the polarizer at some angle and get a result, the probability that the other polarizer will detect a photon should have some value, depending on the angle. If the polarization is known (or there is a hidden variable which determines it), you can calculate the result.

e.g. the polarization is vertical, and your polarizer is at 45º. Your transmission probability is 50%. Only a quarter of the time will you expect to see both photons. Half the time you will see one in one detector only.

Experiments show, however, that the results are not consistent with this prediction. if you see a photon in one detector, you always see it in the other one. 

[uncool]

12 minutes ago, Dalo said:

in this case, no.

edit: what I am claiming is that both photons, in the polarization example can be considered as local. Each photon is reacting exactly as the other one would if it would have had to go successively though two polarizers having the same direction as the other two. 

In this sense, they are both local, and do not influence each other in any way. They react the same because they are identical and the filters they go through are also identical.

A very simple explanation. Maybe too simple. But nobody has proven it wrong yet.

How does this explain the correlations whatsoever?

[swansont]

We have a thread where entanglement, hidden variables, and these correlations are discussed. A nice video that explains it.

http://www.scienceforums.net/topic/87347-why-hidden-variables-dont-work/

 

[Dalo]

40 minutes ago, uncool said:

How does this explain the correlations whatsoever?

Finally! This The most important question of all that has been asked in this thread. 

Maybe explaining it as simply as I can will reveal any mistake in my reasoning and we can all go sleep tight and have nice dreams.

1) The photons, when they leave the atom and go each right or left, are already polarized!

2) In other words, when they go through the first filter, it is in fact equivalent to a non-polarized beam going through a second filter, after having been through the first.

3) If both the filters through which photon 1 and photon 2 have to go are oriented the same way, both photons will either pass or both will be absorbed.

4) let us now misalign both filters by an angle alpha. Photon 1 will be transmitted/absorbed by filter 1 in a certain ratio.

5) Photon 2 has now to go through filter 2 that is misaligned by an angle alpha relatively to fiilter 1.

6) But photon 2 has the same polarization as photon 1 before the latter went through filter 1.

7) That means that we could easily swap photon 1 and photon 2 before any passes their respective filter.

8) When photon 2 goes through filter 2, it reacts exactly as photon 1 would do if it also had to go through the same filter (2).

9) But since photon 1 has gone through filter 1, when you compare both you get the same correlations as if you had first put photon 1 through filter 1...

10) Then, magically, had undone all changes on photon 1, and passed him through filter 2.

11) comparing the result of photon 1 and filter 1 with the result of the same photon 1 with filter 2, and then the result with photon 2 through filter 2, you would find them identical.

 

I know, I feel a little bit dizzy too!

If you find my list too confusing remember the golden rule: both photons have the same polarization, and they will react therefore the same way to the same direction of a filter, or to a misalignment alpha.

Edited December 18, 2017 by Dalo

[Mordred]

Seriously Dalo The beginning of this thread I discussed correlation functions. Why dont you look back to the first page. Instead of claiming we haven't been attempting to answer your questions. 

Take the time to understand the answers provided in you numerous threads on the same topic. Albeit different routes.

Edited December 18, 2017 by Mordred

[Dalo]

Just now, Mordred said:

Seriously Dalo The beginning of this thread I discussed correlation functions. Why dont you look back to the first page.

Because it does not matter! It is a matter of logic, it has nothing to do with the actual correlations. The only important thing is that these correlations have to be the same as determined by quantic statistical regularities.

You insist on turning it into a technical matter that can be solved by technical means. 

I say it is a matter of logic. Once you have accepted that both situations are different, then you must accept entanglement and non-local variables as Bell did.

But my whole point is that both situations are in fact equivalent. And you can only prove that by logic and the empirical facts from which the whole situation starts:

1) both photons have the same polarization,

2) both filters are identical in the sense that they could be swapped without changing the experiment.

 

I am sorry, but I cannot make it clearer than this. And if you still think after my explanation that it still is a mathematical or a physics problem then all I can do is give up, because we will keep talking alongside each other, and that would be a waste of time for all of us.

[dimreepr]

20 minutes ago, Dalo said:

and that would be a waste of time for all of us.

 

finally, something we can all agree on...

[uncool]

Here is part of the problem. You are making statements that could have multiple meanings in the theory, some of which are correct, some are not.

52 minutes ago, Dalo said:

1) The photons, when they leave the atom and go each right or left, are already polarized!

Do you mean that both photons have a definite state (that is simply unknown), each time the experiment is run? Because if so, this is false, and in fact is exactly hat Bell's Theorem disproves. 

[swansont]

1 hour ago, Dalo said:

Finally! This The most important question of all that has been asked in this thread. 

Maybe explaining it as simply as I can will reveal any mistake in my reasoning and we can all go sleep tight and have nice dreams.

1) The photons, when they leave the atom and go each right or left, are already polarized!

2) In other words, when they go through the first filter, it is in fact equivalent to a non-polarized beam going through a second filter, after having been through the first.

3) If both the filters through which photon 1 and photon 2 have to go are oriented the same way, both photons will either pass or both will be absorbed.

No, they will not, as I have previously explained.

"e.g. the polarization is vertical, and your polarizer is at 45º. Your transmission probability is 50%. Only a quarter of the time will you expect to see both photons. Half the time you will see one in one detector only."

 

Quote

4) let us now misalign both filters by an angle alpha. Photon 1 will be transmitted/absorbed by filter 1 in a certain ratio.

5) Photon 2 has now to go through filter 2 that is misaligned by an angle alpha relatively to fiilter 1.

6) But photon 2 has the same polarization as photon 1 before the latter went through filter 1.

7) That means that we could easily swap photon 1 and photon 2 before any passes their respective filter.

8) When photon 2 goes through filter 2, it reacts exactly as photon 1 would do if it also had to go through the same filter (2).

9) But since photon 1 has gone through filter 1, when you compare both you get the same correlations as if you had first put photon 1 through filter 1...

You don't do the experiment in a way where the results are ambiguous; that would be pointless. 

You do it in a way where one model predicts one result, and a different model predicts a different result. 

1 hour ago, Dalo said:

Because it does not matter! It is a matter of logic, it has nothing to do with the actual correlations. 

It is ludicrous to conclude that the experimental results have no bearing on the correctness of the physics. 

Maybe that's how you do it in philosophy, but in science experiment is the decider. If an idea disagrees with experiment, then the idea is wrong. Logic can go take a hike, because the logic or premise is in some way flawed.

Your flaw here is the implicit assumption that the polarizers and photon polarization are aligned. What if they aren't? The only thing you know is that the two photons have the same polarization. But nowhere is it stated that you know what this polarization is. So tell me, how do you guarantee that when a photon passes through one filter, its twin passes through the other, when the orientations are not known?

[Mordred]

2 hours ago, Dalo said:

 

1) both photons have the same polarization,

 

Wrong again so much for your vaulted logic. 

2 hours ago, Dalo said:

Because it does not matter! It is a matter of logic, it has nothing to do with the actual correlations. The only important thing is that these correlations have to be the same as determined by quantic statistical regularities.

You insist on turning it into a technical matter that can be solved by technical means. 

 

Of course I do how can you accurately apply logic without understanding the technical details. Wild blooming guess work?

Thank you for wasting my time

Edited December 18, 2017 by Mordred

[Dalo]

10 hours ago, Mordred said:

Wrong again so much for your vaulted logic. 

Congratulations. You have put the finger on the weak spot of my argumentation.

Those were my two fundamental assumptions:

1) When both photons leave the atom, they have the same polarization;

2) both filters can be considered as identical.

There are some problems though with this objection. It would be a hidden variable theory and therefore would negate the possibility of entanglement. The entanglement conclusion only works with both assumptions.

That is also the opinion of Maudlin. I have already posted one or more quotes where he emphatically presents the photons as having the same polarization at the outstart.

My assumptions therefore are not something I have dreamed up, they are an essential part of any entanglement experiment.

Once you accept that, the problem becomes very simple:

Both photons and both filters are interchangeable. It would be like if we had one single photon and two filters. It would therefore not be surprising that the same correlations would manifest themselves in both cases.

10 hours ago, Mordred said:

Of course I do how can you accurately apply logic without understanding the technical details. Wild blooming guess work?

I can do that because they are irrelevant to my case. Polarization is the example Maudlin have taken, and I followed him in that. It could have been the spin of an electron, the logic would not have changed.

 

It comes down to this

Edited December 19, 2017 by Dalo

[Mordred]

Well try this for a non technical image. The two polarizations involved in the EPR experiment will be specifically the Linear vertical and horizontal.

The superposition state will involve a combination of the two.

If you had just one polarization state there would be no need for Superposition of the entangled particles.