Unghostly entanglement

[uncool]

If you divide everything up into 1 degree increments there are 46,656,000 possibilities with the orientation of the 2 detectors and the pairs. If we hold one of the detectors still that cuts it down to a mere 129,600 possibilities. We are talking about 1 of them.

We're talking about 360 of them - with the orientation of the pairs.

In this case, when one detector accepts a wrong way particle the other will accept its partner and so no mismatch.

Why?

 

If the detectors are misaligned by 2 degrees, why should they accept the "wrong way particles" at the same time?

[Lazarus]

We're talking about 360 of them - with the orientation of the pairs.

Why?

 

If the detectors are misaligned by 2 degrees, why should they accept the "wrong way particles" at the same time?

 

They would get the same results under the conditions we are choosing. So would 3 degrees, 17 degrees and 359 degrees.

However that only accounts for 360 of the 129,600 possibilities. Other choices can get different results.

 

But the point still is that Bell's Theorem makes an umwarranted assumption.

[uncool]

They would get the same results under the conditions we are choosing.

What conditions, and why?

 

Why should misaligned detectors get the same results?

[Lazarus]

What conditions, and why?

 

Why should misaligned detectors get the same results?

 

 

The conditions we chose were that the detectors both start oriented vertically, the results would always be the same in that position, and we would look at only the vertical pairs of particles. That means when the detectors are misaligned to the same degree their results will agree.

 

Here is a different way at looking at this problem

Say with 4 pairs of particles, Bob’s detector detects 3 right way particles and 1 wrong way particles. Alice’s detector detects 2 right way particle and 2 wrong way particles.

 

Let's represent rights as 1's and wrongs as 0’s. Bob’s results are 0, 1, 1, 1. Alice’s results are 0, 0, 1.1. There are 16 different ways to match them up.

 

Bob’s/Alice’s

0/0 matched

0/0 matched

0/1 unmatched

0/1 unmatched

1/0 unmatched

1/0 unmatched

1/1 matched

1/1 matched

1/0 unmatched

1/0 unmatched

1/1 matched

1/1 matched

1/0 unmatched

1/0 unmatched

1/1 matched

1/1 matched

 

There is no way for Bell’s thing to predict which of the 16 possibilities is correct.

[uncool]

The conditions we chose were that the detectors both start oriented vertically, the results would always be the same in that position, and we would look at only the vertical pairs of particles. That means when the detectors are misaligned to the same degree their results will agree.

Again, why?

 

Here is a different way at looking at this problem

Say with 4 pairs of particles, Bob’s detector detects 3 right way particles and 1 wrong way particles. Alice’s detector detects 2 right way particle and 2 wrong way particles.

 

Let's represent rights as 1's and wrongs as 0’s. Bob’s results are 0, 1, 1, 1. Alice’s results are 0, 0, 1.1. There are 16 different ways to match them up.

 

Bob’s/Alice’s

0/0 matched

0/0 matched

0/1 unmatched

0/1 unmatched

1/0 unmatched

1/0 unmatched

1/1 matched

1/1 matched

1/0 unmatched

1/0 unmatched

1/1 matched

1/1 matched

1/0 unmatched

1/0 unmatched

1/1 matched

1/1 matched

 

There is no way for Bell’s thing to predict which of the 16 possibilities is correct.

Bell's theorem doesn't try to predict "which of the 16 possibilities is correct". It predicts the distribution - how often each possibility will happen.

[Lazarus]

Again, why?

 

 

Bell's theorem doesn't try to predict "which of the 16 possibilities is correct". It predicts the distribution - how often each possibility will happen.

 

Depending on which 4 of the 16 possibilties are chosen, the number of matches will vary from 1 to 3. Bell can't make the choices.

[uncool]

Flip two coins, one which lands heads 25% of the time, one which lands heads 50% of the time. Same objection applies.

 

Once again: Bell's theorem is a statistical theorem. It predicts the distribution.

[imatfaal]

 

Depending on which 4 of the 16 possibilties are chosen, the number of matches will vary from 1 to 3. Bell can't make the choices.

 

Your answers are showing that you clearly do not understand the Bell Inequality.

 

At no point are choices made nor are indivudual predictions required for Bell to be shown to be correct - the experiments work on the overal probibilities for multiple iterations of a test at different angles. Bell's breakthrough was to realise that the Classical/Hidden Variable vs Quantum Mechanical could be shown to produce different predictions for overall matches/mis-matches over multiple varying tests over numerous tests at different angles of receptors and that an already castiron mathematical/statistical inequality must apply. The experimenters then found ways to measure incidences of matches and mismatches - every time the quantum mechanical answers (of match/mismtch under certain circumstances) have been found AND by Bell's impeccable logic these answers CANNOT come from Hidden Variables

 

It would be better for the other members for you to read up on Bell's Inequality, then the Theorem, and the experiments which have come subsequently before claiming to challenge it

[Lazarus]

 

Your answers are showing that you clearly do not understand the Bell Inequality.

 

At no point are choices made nor are indivudual predictions required for Bell to be shown to be correct - the experiments work on the overal probibilities for multiple iterations of a test at different angles. Bell's breakthrough was to realise that the Classical/Hidden Variable vs Quantum Mechanical could be shown to produce different predictions for overall matches/mis-matches over multiple varying tests over numerous tests at different angles of receptors and that an already castiron mathematical/statistical inequality must apply. The experimenters then found ways to measure incidences of matches and mismatches - every time the quantum mechanical answers (of match/mismtch under certain circumstances) have been found AND by Bell's impeccable logic these answers CANNOT come from Hidden Variables

 

It would be better for the other members for you to read up on Bell's Inequality, then the Theorem, and the experiments which have come subsequently before claiming to challenge it

 

 

If someone would point to an experiment that actually does the two 1 degree detector rotation that gives the same results as one detector rotated 2% I will go away happy.

[Strange]

 

 

If someone would point to an experiment that actually does the two 1 degree detector rotation that gives the same results as one detector rotated 2% I will go away happy.

 

I haven't really understood what exactly you are objecting to, but there are several tests of Bell's Theorem described here:

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

https://scholar.google.co.uk/scholar?q=experimental+tests+bell's+theorem

[swansont]

If someone would point to an experiment that actually does the two 1 degree detector rotation that gives the same results as one detector rotated 2% I will go away happy.

 

That particular test may not have been done; it's a thought experiment. Thousands of experiments have been done that show the general relationships are correct. You can do some of them yourself if you have some basic equipment or you take the right physics class.

 

Personally I'm not a fan of the explanation you keep citing. The video in http://www.scienceforums.net/topic/87347-why-hidden-variables-dont-work/ does a much better job explaining Bell inequalities and entanglement.

[Lazarus]

 

Your answers are showing that you clearly do not understand the Bell Inequality.

 

At no point are choices made nor are indivudual predictions required for Bell to be shown to be correct - the experiments work on the overal probibilities for multiple iterations of a test at different angles. Bell's breakthrough was to realise that the Classical/Hidden Variable vs Quantum Mechanical could be shown to produce different predictions for overall matches/mis-matches over multiple varying tests over numerous tests at different angles of receptors and that an already castiron mathematical/statistical inequality must apply. The experimenters then found ways to measure incidences of matches and mismatches - every time the quantum mechanical answers (of match/mismtch under certain circumstances) have been found AND by Bell's impeccable logic these answers CANNOT come from Hidden Variables

 

It would be better for the other members for you to read up on Bell's Inequality, then the Theorem, and the experiments which have come subsequently before claiming to challenge it

 

 

Bell’s Theorem is said to be independent of the devices. I thank you for suggesting that I review the experiments because that made it clear that the concept of the CH74 experiments is quite similar to the concept of the rectangular hole bowling pin detector mentioned in posts 59 and 64.

 

That particular test may not have been done; it's a thought experiment. Thousands of experiments have been done that show the general relationships are correct. You can do some of them yourself if you have some basic equipment or you take the right physics class.

 

Personally I'm not a fan of the explanation you keep citing. The video in http://www.scienceforums.net/topic/87347-why-hidden-variables-dont-work/ does a much better job explaining Bell inequalities and entanglement.

 

The video you mentioned is very clear and makes the error in Bell’s Theorem easy to see. The correct number of combinations is 36 not 9 and the probability is 18/36 not 4/9. The video shows that we are only counting one of the two “spins”. None of the video choices counts the head down guy. For instance, a case we are not counting is the head down guy in position 1.

[swansont]

 

The video you mentioned is very clear and makes the error in Bell’s Theorem easy to see. The correct number of combinations is 36 not 9 and the probability is 18/36 not 4/9. The video shows that we are only counting one of the two “spins”. None of the video choices counts the head down guy. For instance, a case we are not counting is the head down guy in position 1.

 

Please explain how you calculate your probability. You had an error earlier which was pointed out to you, and you never acknowledged it. Not this time.

[Lazarus]

 

Please explain how you calculate your probability. You had an error earlier which was pointed out to you, and you never acknowledged it. Not this time.

 

 

 

Yes, I have made lots of errors. Most of them have been corrected, usually thanks to your pointing them out.

 

By only counting the head up guy’s positions you get 3 positions for the first detector. Combined with the 2nd detector that makes 9 combinations. That does not cover all possibilities. In each position there are 2 possibilities because if you rotate the guys 180 degrees that is a separate possibility. With 6 choices the combined choices of the 2 detectors is 36 not 9. That makes the split 18 out of 36.

[swansont]

 

Yes, I have made lots of errors. Most of them have been corrected, usually thanks to your pointing them out.

 

By only counting the head up guy’s positions you get 3 positions for the first detector. Combined with the 2nd detector that makes 9 combinations. That does not cover all possibilities. In each position there are 2 possibilities because if you rotate the guys 180 degrees that is a separate possibility. With 6 choices the combined choices of the 2 detectors is 36 not 9. That makes the split 18 out of 36.

 

 

That's not the calculation he's doing, but I'm not sure what the objection is. 9 combinations is what he looks at. There is no rotation of the whole system 180 degrees — that's not the problem he's solving.

 

I don't see where you get 18/36 instead of 4/9. You still haven't explained that.

[Lazarus]

 

 

That's not the calculation he's doing, but I'm not sure what the objection is. 9 combinations is what he looks at. There is no rotation of the whole system 180 degrees — that's not the problem he's solving.

 

I don't see where you get 18/36 instead of 4/9. You still haven't explained that.

 

The video said that it considered all the possibilities of alignment of the 2 guys. That is not true. They only considered the 3 possibilities by counting the head up guy with his head pointing out.. There are 3 unconsidered possibilities. They should also count the possibilities with his head pointing in. Rather than 3 possibilities there are actually 6. Multiply the possibilities of the 2 detectors as they did when they claimed only 3 possibilities. Six times six is 36. That makes the final result 18/36 rather that 5/9. The 18/36 result agrees with experiment and their result does not.

[swansont]

The video said that it considered all the possibilities of alignment of the 2 guys. That is not true. They only considered the 3 possibilities by counting the head up guy with his head pointing out..

Because that's how you do a conditional probability calculation. The question being asked is what is the probability of a result in detector B given a particular result in detector A. The probability of what happens at detector A is 1. So, given that detector A gives the result at that location, the calculation is what happens at detector B.

 

There are 3 unconsidered possibilities. They should also count the possibilities with his head pointing in. Rather than 3 possibilities there are actually 6. Multiply the possibilities of the 2 detectors as they did when they claimed only 3 possibilities. Six times six is 36. That makes the final result 18/36 rather that 5/9. The 18/36 result agrees with experiment and their result does not.

 

WHERE DO YOU GET 18? You have shown no calculation, or given any explanation, at all.

 

There is no possible result with the head pointing in. Your only choices are up and down.

 

The video goes through two iterations of the hidden variable case, with defined spins for the 3 orientations in detector A. For each one, there are 3 cases to check in detector B. That's nine. Any other iteration for detector A is going to give the same result: 5/9, or higher. He does not claim there are only 3 possibilities. Apparently you missed the part where he says that all other combinations are mathematically equivalent. That's why he only investigates the two cases. The rest will give the same result.

[Lazarus]

Because that's how you do a conditional probability calculation. The question being asked is what is the probability of a result in detector B given a particular result in detector A. The probability of what happens at detector A is 1. So, given that detector A gives the result at that location, the calculation is what happens at detector B.

 

 

WHERE DO YOU GET 18? You have shown no calculation, or given any explanation, at all.

 

There is no possible result with the head pointing in. Your only choices are up and down.

 

The video goes through two iterations of the hidden variable case, with defined spins for the 3 orientations in detector A. For each one, there are 3 cases to check in detector B. That's nine. Any other iteration for detector A is going to give the same result: 5/9, or higher. He does not claim there are only 3 possibilities. Apparently you missed the part where he says that all other combinations are mathematically equivalent. That's why he only investigates the two cases. The rest will give the same result.

 

6 - Count em.

[swansont]

 

6 - Count em.6pair.JPG

 

And if the particle at detector A is down, you get the same set of correlation results as if it was up, so the math is the same. 10/18 is the same as 5/9.

 

But you somehow got 18/36, which is what I asked about (and I think I was quite clear on this) and you still haven't addressed that calculation. If it was a calculation.

[Lazarus]

 

And if the particle at detector A is down, you get the same set of correlation results as if it was up, so the math is the same. 10/18 is the same as 5/9.

 

But you somehow got 18/36, which is what I asked about (and I think I was quite clear on this) and you still haven't addressed that calculation. If it was a calculation.

 

The 18 was not calculated but the additional possibilities should be the reverse of the discussed ones. It is all moot if there are not 6 possibilities rather than 3.

 

These are the 6 possible positions of the guy that started heads up.

 

 

If both detectors have number 1, that is a match. if one detector has a 1 and the other has a 4, that is a mismatch.

 

There have to be 2 possibilities in that case and 6 possibilites total.

Also, here is why the result is 18 unmatched of of 36 possibilities.

 

In the 6 possibilities numbered in post number 120, numbers 1, 2 and 3 are always detected but 4, 5 and 6 are never detected. There are 36 combinations of the 6 different choices for each detector. The nine combinations of 1, 2, and 3 are all matched. Similarly, the nine combinations of 4, 5,and 6 are all matched. The nine combinations of 1, 2, 3 vs 4, 5, 6 are all unmatched. The nine combinations of 4, 5, 6 vs 1, 2, 3 are all matched. The net result is 18 unmatched out of 36 possibilities.

Edited December 16, 2015 by Lazarus

[swansont]

These are the 6 possible positions of the guy that started heads up.

 

6ways.JPG

 

To say he "started heads" up means you are discussing the hidden variable approach (or this points to your misunderstanding of the situation). In standard QM, his state is undetermined.

 

But if he is heads up, then he cannot be detected heads down. The probability of state 4 is zero.

 

In the case being analyzed, we are looking at the correlation of detector B. So if we look at the 0º orientation of detector A, we get heads up. We want to know the three possibilities of the three orientations of detector B.

 

When we look at the 120º case, there is again only one possible answer — that's the whole meaning of having a hidden variable. If the hidden variable says he is heads up, then he will never be detected heads down, and vice-versa. So there is only one possible answer for detector A at 120º, and three answers for detector B. Same for 240º. That's where the nine results come from.

 

If you do the analysis for getting spin down at detector A, you will get nine more results, but the answer will be the same (5/9, or 1, depending on the hidden variable correlation)

 

If both detectors have number 1, that is a match. if one detector has a 1 and the other has a 4, that is a mismatch.

 

There have to be 2 possibilities in that case and 6 possibilites total.

 

Also, here is why the result is 18 unmatched of of 36 possibilities.

 

In the 6 possibilities numbered in post number 120, numbers 1, 2 and 3 are always detected but 4, 5 and 6 are never detected. There are 36 combinations of the 6 different choices for each detector. The nine combinations of 1, 2, and 3 are all matched. Similarly, the nine combinations of 4, 5,and 6 are all matched. The nine combinations of 1, 2, 3 vs 4, 5, 6 are all unmatched. The nine combinations of 4, 5, 6 vs 1, 2, 3 are all matched. The net result is 18 unmatched out of 36 possibilities.

You aren't analyzing the problem that was given to you. You are analyzing something else. (I'm not sure what, but it's not the Bell test)

 

The particle at A and particle at B must always be anti-correlated if detected in the same orientation, in the example in the video. The angular momentum (spin) is conserved. So some of your 36 possibilities don't actually exist. It is not possible to detect the same orientation of the particle for detectors with the same orientation. You will never see both detectors giving you state 1. It's physically impossible, so you can't call it a possible result.

[Lazarus]

 

To say he "started heads" up means you are discussing the hidden variable approach (or this points to your misunderstanding of the situation). In standard QM, his state is undetermined.

But if he is heads up, then he cannot be detected heads down. The probability of state 4 is zero.

In the case being analyzed, we are looking at the correlation of detector B. So if we look at the 0º orientation of detector A, we get heads up. We want to know the three possibilities of the three orientations of detector B.

When we look at the 120º case, there is again only one possible answer — that's the whole meaning of having a hidden variable. If the hidden variable says he is heads up, then he will never be detected heads down, and vice-versa. So there is only one possible answer for detector A at 120º, and three answers for detector B. Same for 240º. That's where the nine results come from.

 

If you do the analysis for getting spin down at detector A, you will get nine more results, but the answer will be the same (5/9, or 1, depending on the hidden variable correlation)

You aren't analyzing the problem that was given to you. You are analyzing something else. (I'm not sure what, but it's not the Bell test)

The particle at A and particle at B must always be anti-correlated if detected in the same orientation, in the example in the video. The angular momentum (spin) is conserved. So some of your 36 possibilities don't actually exist. It is not possible to detect the same orientation of the particle for detectors with the same orientation. You will never see both detectors giving you state 1. It's physically impossible, so you can't call it a possible result.

 

 

Here are all 6 possibilities. Which ones are impossible?

 

A1 Up B1 Down

A1 Down B1 Up

A2 Up B2 Down

A2 Down B2 Up

A3 Up B3 Down

A3 Down B3 Up

[swansont]

 

 

 

Here are all 6 possibilities. Which ones are impossible?

 

A1 Up B1 Down

A1 Down B1 Up

A2 Up B2 Down

A2 Down B2 Up

A3 Up B3 Down

A3 Down B3 Up

 

6 possibilities? What happened to 36? (with 18 matches)

 

You said "If both detectors have number 1, that is a match. if one detector has a 1 and the other has a 4, that is a mismatch." But here, you have omitted the possibility of both having the same spin, i.e. what you called a match.

 

You don't seem to be making a consistent argument.

 

You also haven't addressed any of the other possibilities discussed in the video, such as measuring the A1 and B2 correlation, or any of the others where the detector orientations differ. You don't seem to be analyzing the same experiment as is described in the video.

 

[Lazarus]

 

6 possibilities? What happened to 36? (with 18 matches)

 

You said "If both detectors have number 1, that is a match. if one detector has a 1 and the other has a 4, that is a mismatch." But here, you have omitted the possibility of both having the same spin, i.e. what you called a match.

 

You don't seem to be making a consistent argument.

 

You also haven't addressed any of the other possibilities discussed in the video, such as measuring the A1 and B2 correlation, or any of the others where the detector orientations differ. You don't seem to be analyzing the same experiment as is described in the video.

 

 

Yes, I made an obvious error in reversing the matched / unmatched label.

 

Here are all 36 possibilities with the 2 detectors. Which ones are impossible?

 

The first 18 are matched.

 

A1 up B1 down

A1 down B1 up

A1 up B2 down

A1 down B2 up

A1 up B3 down

A1 down B3 up

 

A2 up B1 down

A2 down B1 up

A2 up B2 down

A2 down B2 up

A2 up B3 down

A2 down B3 up

 

A3 up B1 down

A3 down B1 up

A3 up B2 down

A3 down B2 up

A3 up B3 down

A3 down B3 up

 

The other 18 are unmatched.

 

A1 up B1 up

A1 down B1 down

A1 up B2 up

A1 down B2 down

A1 up B3 up

A1 down B3 down

 

A2 up B1 up

A2 down B1 down

A2 up B2 up

A2 down B2 down

A2 up B3 up

A2 down B3 down

 

A3 up B1 up

A3 down B1 down

A3 up B2 up

A3 down B2 down

A3 up B3 up

A3 down B3 down

[swansont]

 

Yes, I made an obvious error in reversing the matched / unmatched label.

 

Here are all 36 possibilities with the 2 detectors. Which ones are impossible?

 

The first 18 are matched.

 

A1 up B1 down

A1 down B1 up

A1 up B2 down

A1 down B2 up

A1 up B3 down

A1 down B3 up

 

A2 up B1 down

A2 down B1 up

A2 up B2 down

A2 down B2 up

A2 up B3 down

A2 down B3 up

 

A3 up B1 down

A3 down B1 up

A3 up B2 down

A3 down B2 up

A3 up B3 down

A3 down B3 up

 

The other 18 are unmatched.

 

A1 up B1 up

A1 down B1 down

A1 up B2 up

A1 down B2 down

A1 up B3 up

A1 down B3 down

 

A2 up B1 up

A2 down B1 down

A2 up B2 up

A2 down B2 down

A2 up B3 up

A2 down B3 down

 

A3 up B1 up

A3 down B1 down

A3 up B2 up

A3 down B2 down

A3 up B3 up

A3 down B3 down

 

If you have a hidden variable, i.e. the spin is already determined before it hits the detector, then half of your choices are not possible for detector A. You cannot detect a particle as spin down if it really is spin up.

 

That makes any further critique moot. Did you even watch the whole video?