Photon entanglement: why three angles?

[Jabbu]

If the hidden variable $\lambda$ had value $\lambda_{+++}$, then the photon will pass through a filter oriented at $\theta = 0, \theta = 60, \theta =$. If it had value $\lambda_{++-}$, then the photon will pas through a filter oriented at $\theta = 0, \theta = 60$ but would be blocked by a filter at $\theta = 120$. Etc.

We can prove that there does not exist 8 probabilities
$P_{+++}, P_{++-}, ...$ (where $P_{+++}$ is the probability that $\lambda$ has value $\lambda_{+++}$, etc.) such that this sort of local realistic theory reproduces the predictions of QM.

Correlation is not calculated between readings for different angles. Experiments with three angles produce three separate data streams, one for each angle, and correlation is calculated independently for each data stream of AB pairs. For theta= 0, Alice and Bob readings should be 100% correlated just as if theta= 0 was tested independently by itself. Other angles have their completely separate data streams of AB pairs with their own probabilities and their own correlations according to QM's cos^2(theta).

Now, if we only have a single angle, $\theta=0$, then the predictions of QM are that
50% of the time, the photon passes, and 50% of the time, it is blocked. It's EASY to come up with a local realistic model for this case. In this case, there are two possible values for $\lambda$:

Proposed local classical equation must be confirmed by classical experiments, for any angle, it has to be able to actually work for photon-polarizer interaction in general, or it simply is not true. It has to give the same result as Malus law, it thus can not be anything else. Either Malus law works or the explanation is non-local, there is no other choice compatible with all the other classical experiments.

 

[RUTA]

Correlation is not calculated between readings for different angles. Experiments with three angles produce three separate data streams, one for each angle, and correlation is calculated independently for each data stream of AB pairs. For theta= 0, Alice and Bob readings should be 100% correlated just as if theta= 0 was tested independently by itself. Other angles have their completely separate data streams of AB pairs with their own probabilities and their own correlations according to QM's cos^2(theta).

"theta" is the difference between the two polarizer/SG magnet settings, so whatever the settings of the two detectors, if they are the same, theta = 0. Read the attached paper and go through all the calculations. There are some mistakes, but they're easy to find. After completing this homework assignment, you'll understand how an experiment showing the violation of Bell's inequality is actually done, to include data analysis. After you've finished, come back with your questions.

 

[stevendaryl]

Proposed local classical equation must be confirmed by classical experiments, for any angle, it has to be able to actually work for photon-polarizer interaction in general, or it simply is not true. It has to give the same result as Malus law, it thus can not be anything else. Either Malus law works or the explanation is non-local, there is no other choice compatible with all the other classical experiments.

I just showed you that a local hidden variables theory can reproduce exactly the predictions of QM for a single filter setting (both Alice and Bob have the filters at the same angle). So what you're saying is false: the predictions for a single angle does not demonstrate that anything nonlocal is going on. In contrast, when you have 3 filter settings to choose from, it is not possible to reproduce the predictions of QM by any local theory.

I don't know why you keep bringing up Malus' law. It is completely useless in predicting correlations for the case of entangled photons, because you can't apply it unless you know the photon polarizations.

 

[Jabbu]

"theta" is the difference between the two polarizer/SG magnet settings, so whatever the settings of the two detectors, if they are the same, theta = 0. Read the attached paper and go through all the calculations. There are some mistakes, but they're easy to find. After completing this homework assignment, you'll understand how an experiment showing the violation of Bell's inequality is actually done, to include data analysis. After you've finished, come back with your questions.

I see. You are not surprised correlation for theta= 0 is 100%, as if that is normal. Instead, what you find more convincing the interaction is non-local is that some combined thetas correlation function "S" is greater than 2. Is that it?

 

[RUTA]

I see. You are not surprised correlation for theta= 0 is 100%, as if that is normal. Instead, what you find more convincing the interaction is non-local is that some combined thetas correlation function "S" is greater than 2. Is that it?

Did you read the paper and reproduce all the calculations? I can help you understand the equations and how they are related to the experiment. What you choose to believe about those equations and the experiment is up to you.

 

[morrobay]

The first column of this example above should be read as "for the first pair if we measure the polarization on angle A the left-hand photon will pass and the right-hand one will not; if we measure on angles B or C the left-hand photon will not pass and the right-hand one will". The second column should be read as "for the second pair if we measure the polarization on angle B the left-hand photon will not pass and the right-hand one will; if we measure on angles A or C the left-hand one will pass and the right-hand one will not". This is exactly the local realistic theory I describe above - both photons are created with definite polarization values at all three angles.

You calculate the results by choosing any two of the three possible results because we only get to make two measurements, one on each photon. The challenge is to construct a data set that will lead to a violation of Bell's equality no matter which measurements we choose to make on each pair - and if you try it you'll find that it cannot be done. Therefore, no theory in which the results of measurements on all three angles are predetermined can match the experimental results.

The data set for above is for Alice:
a: ++-+--+
b: --+-++-
c: -+-++-+

I have found a violation from data set from above post. Converting above data to horizontal format including Bobs values:

..A....B
a b c...a b c
+ - -...- + +
+ -+...- + -
- + -...+ - +
+ - +...- + -
- + +...+ - -
- + -...+ - +
+ - +...- + -

n[a+c+] + n[b+c-] ≥ n[a+b+]

n=1 + n=1 ≥ n= 4 violation

This form of inequality is taken from http://math.ucr.edu/home/baez/physics/Quantum/bells_inequality.html

 

[Jabbu]

Did you read the paper and reproduce all the calculations? I can help you understand the equations and how they are related to the experiment. What you choose to believe about those equations and the experiment is up to you.

I did read it. I didn't reproduce calculations, but I promise I will. You neither confirmed nor denied what I said in my last two post. Please be more specific and at least tell me what are we talking about now.

2 < S = E(a,b) - E(a,b') + E(a',b) + E(a',b')

Is that it, the proof experiment data describe non-local interaction?

 

[RUTA]

I did read it. I didn't reproduce calculations, but I promise I will. You neither confirmed nor denied what I said in my last two post. Please be more specific and at least tell me what are we talking about now.

2 < S = E(a,b) - (a,b') + (a',b) + (a',b')

Is that it, the proof experiment data describe non-local interaction?

If all you're trying to understand are the "non-local" implications of entanglement, then any of Mermin's papers will do. Essentially, it comes down to "no instruction sets." The 1994 paper I posted earlier is the easiest to understand in my opinion because it doesn't involve statistics or inequalities.

If you're trying to understand how data is actually taken concerning "non-locality," then read and work through all the calculations in the last paper I posted. In that case, "non-locality" is represented by violating |S| <= 2. I posted that paper because you seemed confused as to what the cosine squared of theta actually means and it's explained nicely in that article.

In either case, all the material you need to understand what physicists mean by "quantum non-locality" has been posted and explained in this thread. If there is something about the equations you don't understand, people here will help you. If you want to argue about semantics or what physicists should or should not believe concerning ontology, the moderators will likely close the thread.

 

[DrChinese]

Is that it, the proof experiment data describe non-local interaction?

No. It is proof that it is non-local or non-realistic. Or both. No one knows for sure.

 

[RUTA]

No. It is proof that it is non-local or non-realistic. Or both. No one knows for sure.

Careful here, Jabbu. This is a more precise use of the term "non-local" as regards entanglement. The term "quantum non-locality" has come to mean "measurement outcomes with entangled particles that violate classical intuition." One way to characterize exactly what it is in classical intuition that is violated by these QM measurement outcomes is to say they are "non-local" and/or "non-realistic." In this sense, "non-local" means they involve "superluminal influences." In the first 1:30 of this talk by Weinstein , for example, he uses "outcome independence" and "parameter independence" instead of "non-local" and "non-real." He then says "the conjunction of these two is sometimes referred to as strong locality or just locality."

 

[Jabbu]

If you're trying to understand how data is actually taken concerning "non-locality," then read and work through all the calculations in the last paper I posted. In that case, "non-locality" is represented by violating |S| <= 2. I posted that paper because you seemed confused as to what the cosine squared of theta actually means and it's explained nicely in that article.

That. So I'm looking at page 907 and I'm trying to calculate their example:

(Ex) a= -45, a'= 0, b= -25.5, b'= 25.5

...they say result is obtained with equations 10, 20, and 21:

(10) Pvv(a,b) = cos^2(b-a) / 2

(20) E(a,b) = Pvv(a,b) + Phh(a,b) - Pvh(a,b) - Phv(a,b)

(21) S = E(a,b) - E(a,b') + E(a',b) + E(a',b')To evaluate equation 20 there is defined Pvv(a,b) in equation 10, but I'm still missing Phh(a,b), Pvh(a,b), and Phv(a,b). How do I get to these three? Is equation 20 supposed to equal cos^2(b-a)?

 

[Nugatory]

The data set for above is for Alice:
a: ++-+--+
b: --+-++-
c: -+-++-+

I have found a violation from data set from above post. Converting above data to horizontal format including Bobs values:

..A....B
a b c...a b c
+ - -...- + +
+ -+...- + -
- + -...+ - +
+ - +...- + -
- + +...+ - -
- + -...+ - +
+ - +...- + -

n[a+c+] + n[b+c-] ≥ n[a+b+]
n=1 + n=1 ≥ n= 4 violation

Written in this form, the inequality applies to the results on one side (or the other), so we have for Alice $3+2\geq{0}$, for Bob $2+3\geq{0}$, and no violation.

That's just a trick of the notation: n[a+c+] means "Alice would measure + in the a and c directions" which means that either Alice measured a+ and Bob c-, or Alice measured c+ and Bob measured a-.

 

[RUTA]

That. So I'm looking at page 907 and I'm trying to calculate their example:

(Ex) a= -45, a'= 0, b= -25.5, b'= 25.5

...they say result is obtained with equations 10, 20, and 21:

(10) Pvv(a,b) = cos^2(b-a) / 2

(20) E(a,b) = Pvv(a,b) + Phh(a,b) - Pvh(a,b) - Phv(a,b)

(21) S = E(a,b) - E(a,b') + E(a',b) + E(a',b')


To evaluate equation 20 there is defined Pvv(a,b) in equation 10, but I'm still missing Phh(a,b), Pvh(a,b), and Phv(a,b). How do I get to these three? Is equation 20 supposed to equal cos^2(b-a)?

First, you caught the missing minus sign for b, good work.

You can get P_HH, P_VH, and P_VH from Eq (7) using <V| and <H| as needed:

P_HH = cos²(α-β)/2
P_HV = P_VH = sin²(α-β)/2.

Of course it must be the case that P_VV + P_HH + P_HV + P_VH = 1.

Now, from Eq (20) we have E(α-β) = cos²(α-β) - sin²(α-β) = cos2(α-β). This gives

E(a,b) = E(-45, -22.5) = cos(-45) = 1/√2
E(a,b') = E(-45, 22.5) = cos(-135) = -1/√2
E(a',b) = E(0, -22.5) = cos(45) = 1/√2
E(a',b') = E(0, 22.5) = cos(-45) = 1/√2.

Thus, Eq (21) gives S = E(a,b) - E(a,b') + E(a',b) + E(a',b') = 2√2 as shown in Eq (23)

 

[Jabbu]

P_HH = cos²(α-β)/2
P_HV = P_VH = sin²(α-β)/2.

Of course it must be the case that P_VV + P_HH + P_HV + P_VH = 1.

So those are counts of matching and mismatching pairs. Equation 20 is then the same what I called correlation formula: match - mismatch, and QM's cos^2(theta). It means two uniformly random binary sequences should be completely uncorrelated with equal number of matches and mismatches.

0.5:0.5 vs 0.5:0.5
chance for match: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
chance for mismatch: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
correlation: 0.5 - 0.5 = 0%

Now, from Eq (20) we have E(α-β) = cos²(α-β) - sin²(α-β) = cos2(α-β). This gives

Instead of theta = (a-b), shouldn't that be (b-a)? So E(a,b) = cos^2(b-a)?

E(a,b) = E(-45, -22.5) = cos(-45) = 1/√2
E(a,b') = E(-45, 22.5) = cos(-135) = -1/√2
E(a',b) = E(0, -22.5) = cos(45) = 1/√2
E(a',b') = E(0, 22.5) = cos(-45) = 1/√2.

How do you get theta = -45 from (-45, -25.5)? Shouldn't that be theta = (-25.5 - (-45)) = 25.5? Also theta = -135 from (-45, 22.5)?


This is what I get:

E(a,b) = E(-45, -25.5) = cos^2(19.5) = 0.89
E(a,b') = E(-45, 25.5) = cos^2(70.5) = 0.11
E(a',b) = E(0, -25.5) = cos^2(-25.5) = 0.82
E(a',b') = E(0, 25.5) = cos^2(25.5) = 0.82

S = 0.89 - 0.11 + 0.82 + 0.82 = 2.42

 

[Jabbu]

I just showed you that a local hidden variables theory can reproduce exactly the predictions of QM for a single filter setting (both Alice and Bob have the filters at the same angle). So what you're saying is false: the predictions for a single angle does not demonstrate that anything nonlocal is going on. In contrast, when you have 3 filter settings to choose from, it is not possible to reproduce the predictions of QM by any local theory.

What you showed are some statements that do not correspond to anything I've seen anywhere else. You used no any equations, you just asserted "we can prove" without showing any proof.

I don't know why you keep bringing up Malus' law. It is completely useless in predicting correlations for the case of entangled photons, because you can't apply it unless you know the photon polarizations.

Because that's how classical physics prediction is calculated. What do you think is classical physics prediction for theta = 0 degrees?

 

[RUTA]

So those are counts of matching and mismatching pairs. Equation 20 is then the same what I called correlation formula: match - mismatch, and QM's cos^2(theta). It means two uniformly random binary sequences should be completely uncorrelated with equal number of matches and mismatches.

0.5:0.5 vs 0.5:0.5
chance for match: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
chance for mismatch: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
correlation: 0.5 - 0.5 = 0%

I don't know what you're talking about, sorry. You have completely correlated outcomes for α-β = 0 (P_VV = P_HH = $\frac{1}{2}$cos²(0) = $\frac{1}{2}$ and P_VH = P_HV = $\frac{1}{2}$sin²(0) = 0) and completely anti-correlated outcomes for α-β=$\frac{∏}{2}$ (P_VV = P_HH = $\frac{1}{2}$cos²($\frac{∏}{2}$) = 0 and P_VH = P_HV = $\frac{1}{2}$sin²($\frac{∏}{2}$) = $\frac{1}{2}$).

Instead of theta = (a-b), shouldn't that be (b-a)? So E(a,b) = cos^2(b-a)?

It doesn't matter, the calculation is the same either way. Your equation for E(a,b) is wrong. E(a,b) = cos²(a-b) - sin²(a-b) = cos2(a-b). That's why your computations of E in the following are wrong.

How do you get theta = -45 from (-45, -25.5)? Shouldn't that be theta = (-25.5 - (-45)) = 25.5? Also theta = -135 from (-45, 22.5)?

This is what I get:

E(a,b) = E(-45, -25.5) = cos^2(19.5) = 0.89
E(a,b') = E(-45, 25.5) = cos^2(70.5) = 0.11
E(a',b) = E(0, -25.5) = cos^2(-25.5) = 0.82
E(a',b') = E(0, 25.5) = cos^2(25.5) = 0.82

S = 0.89 - 0.11 + 0.82 + 0.82 = 2.42

E(a,b) = cos2(a-b) = cos2(-45 + 22.5) = cos(-45) = 1/√2
Etc.

 

[DrChinese]

This is what I get:

E(a,b) = E(-45, -25.5) = cos^2(19.5) = 0.89
E(a,b') = E(-45, 25.5) = cos^2(70.5) = 0.11
E(a',b) = E(0, -25.5) = cos^2(-25.5) = 0.82
E(a',b') = E(0, 25.5) = cos^2(25.5) = 0.82

S = 0.89 - 0.11 + 0.82 + 0.82 = 2.42

You used the wrong angle settings. Instead of 25.5 it should be 22.5. Then the arithmetic works out.

Please stop referring to Malus and classical predictions for entanglement, a non-classical phenomena. This is your last warning on that.

At this point, it is completely unclear what question you are asking. You are switching from context to context so rapidly that none of the responders can assist. So ask a question clearly and immediately that can be addressed, or I suspect this thread will be reaching a conclusion. If you answer anything else instead, please expect the obvious to occur. This is not a forum for debate, or for you to put forth your (misinformed) opinions. We have been quite patient, Jabbu, but that patience is wearing thin. There are multiple professional physicists and science advisors spending time here, and you are going in circles.

 

[stevendaryl]

What you showed are some statements that do not correspond to anything I've seen anywhere else. You used no any equations, you just asserted "we can prove" without showing any proof.

The predictions of QM for a single angle $\theta=0$ using entangled photons is this:
(Both Alice and Bob keep their filters set at this angle, for each trial).

$50\%$ of the time, one photon passes through Alice's filter and the other passes through Bob's filter.[/itex]
$50\%$ of the time, one photon is blocked by Alice's filter and the other is blocked by Bob's filter.
It never happens that Alice and Bob get different results.

Do you understand that those are the predictions of QM for this case?

Do you really not understand how one could duplicate those predictions without nonlocal interactions?

Suppose that instead of photons, we have slips of paper that messages are written on. Instead of Alice's filter, we have a person who reads one of the slips of paper, and either says "Pass" or "Block". Do you really not see how you could generate slips of paper so that we could guarantee that Alice gets "Pass" 50% of the time, and Bob gets "Pass" 50% of the time, and they always get the same result?

Of course, you can do it. Half of the time, you write "Pass" on both pieces of paper. Half the time you write "Block" on both pieces of paper. That DOES it! It's trivial.

The weird fact is that if instead of one filter setting, Alice and Bob have a choice of three settings, then there is NO way to do it using messages written on pieces of paper.

 

[stevendaryl]

The predictions of QM for a single angle $\theta=0$ using entangled photons is this:
(Both Alice and Bob keep their filters set at this angle, for each trial).

$50\%$ of the time, one photon passes through Alice's filter and the other passes through Bob's filter.[/itex]
$50\%$ of the time, one photon is blocked by Alice's filter and the other is blocked by Bob's filter.
It never happens that Alice and Bob get different results.

Do you understand that those are the predictions of QM for this case?

Do you really not understand how one could duplicate those predictions without nonlocal interactions?

Suppose that instead of photons, we have slips of paper that messages are written on. Instead of Alice's filter, we have a person who reads one of the slips of paper, and either says "Pass" or "Block". Do you really not see how you could generate slips of paper so that we could guarantee that Alice gets "Pass" 50% of the time, and Bob gets "Pass" 50% of the time, and they always get the same result?

Of course, you can do it. Half of the time, you write "Pass" on both pieces of paper. Half the time you write "Block" on both pieces of paper. That DOES it! It's trivial.

The weird fact is that if instead of one filter setting, Alice and Bob have a choice of three settings, then there is NO way to do it using messages written on pieces of paper.

Jabbu, can you just say whether you agree that the thought experiment in terms of message passing on pieces of paper can be implemented to give the same statistical predictions as QM, in the case of a single filter setting? That conclusion is so trivial, that I really have no idea what you're asking for when you ask for a proof of it.

The harder proof is to show that it CAN'T be done with three filter settings.

 

[stevendaryl]

Because that's how classical physics prediction is calculated. What do you think is classical physics prediction for theta = 0 degrees?

This is not about classical physics. It's about quantum physics. Malus' law has nothing to do with the argument. Absolutely nothing. Your bringing up Malus' law over and over again means that you really, really don't understand what people are saying to you.

 

[Jabbu]

I don't know what you're talking about, sorry. You have completely correlated outcomes for α-β = 0 (P_VV = P_HH = $\frac{1}{2}$cos²(0) = $\frac{1}{2}$ and P_VH = P_HV = $\frac{1}{2}$sin²(0) = 0) and completely anti-correlated outcomes for α-β=$\frac{∏}{2}$ (P_VV = P_HH = $\frac{1}{2}$cos²($\frac{∏}{2}$) = 0 and P_VH = P_HV = $\frac{1}{2}$sin²($\frac{∏}{2}$) = $\frac{1}{2}$).

I'm talking about those terms in equation 20, in an experiment they are not calculate but counted.

It doesn't matter, the calculation is the same either way. Your equation for E(a,b) is wrong. E(a,b) = cos²(a-b) - sin²(a-b) = cos2(a-b). That's why your computations of E in the following are wrong.

If by "cos2(x)" you don't mean to say "cos^2(x)" I think 2 should go inside brackets like this "cos(2x)".

E(a,b) = cos2(a-b) = cos2(-45 + 22.5) = cos(-45) = 1/√2
Etc.

Ok. So how cos^2(theta) fits in? When a= -30 and b= +30 QM predicts correlation = cos^2(60) = 25%, right? So isn't E(-30,30) supposed to be that same "correlation" value?

 

[Jabbu]

You used the wrong angle settings. Instead of 25.5 it should be 22.5.

Thanks.

Please stop referring to Malus and classical predictions for entanglement, a non-classical phenomena.

How else do you compare experimental results with classical physics prediction?

At this point, it is completely unclear what question you are asking.

I'm talking to RUTA. See the paper he posted, or please ask specific questions about whatever it is unclear to you.

 

[Jabbu]

This is not about classical physics. It's about quantum physics. Malus' law has nothing to do with the argument. Absolutely nothing. Your bringing up Malus' law over and over again means that you really, really don't understand what people are saying to you.

To understand practical implications it is paramount to understand how experimental results differ from classical prediction. Please read the paper RUTA posted, or any other paper or article about Bell's theorem. Comparing experimental results with classical prediction is very important part of the analysis.

 

[Jabbu]

Jabbu, can you just say whether you agree that the thought experiment in terms of message passing on pieces of paper can be implemented to give the same statistical predictions as QM, in the case of a single filter setting?

To be more specific instead of hypothetical piece of paper it's better to attribute hidden variables as properties of actual entities, in this case photons and polarizers. Then it's easier to realize real conditions these variables have to satisfy. So to answer your question, unless the paper says "cos^2(theta)" it would be refuted by every other experiment with known relative polarization different than 45 degrees.

But if you insist your hidden variable can fake experimental results for a single theta setting, then you need to realize S = E(a,b) - E(a,b') + E(a',b) + E(a',b') is defined by independent results of four separate theta settings from four separate experiments. Therefore, if your hidden variable can fake each of those experiments individually it will automatically fake the value of S.

http://en.wikipedia.org/wiki/CHSH_inequality

 

[atyy]

To be more specific instead of hypothetical piece of paper it's better to attribute hidden variables as properties of actual entities, in this case photons and polarizers. Then it's easier to realize real conditions these variables have to satisfy. So to answer your question, unless the paper says "cos^2(theta)" it would be refuted by every other experiment with known relative polarization different than 45 degrees.

Bell's theorem can also rule out any modification of the classical laws that are local, so it doesn't just rule out Malus's Law acting on classical photon pairs, but all local alternatives.

Nonetheless, if you would like to try to see if how close Malus's Law and classical photon pairs can come to mimicking the quantum entangled pairs, you can try this. First consider that each side receives 50% vertical and 50% horizontal photons, and that when their polarizers are both vertical, both sides always get the same result. In this case, you can imagine that this result is obtained because the classical source sends out 50% classical pairs with both photons vertically polarized, and 50% classical pairs with both photons horizontally polarized.

Then keeping the same assumption about the classical source, you can apply Malus's Law for other polarizer settings, and carry out the analysis for this classical case to compare with the quantum entangled case.

 

[stevendaryl]

To be more specific instead of hypothetical piece of paper it's better to attribute hidden variables as properties of actual entities, in this case photons and polarizers.

But that would be a silly thing to do. We know that the classical theory of polarization cannot explain the results of the EPR experiment for entangled photons. We already know that. Malus' law does not describe the situation. That's completely clear.

The issue is whether some other law describes the EPR experiment in terms of local interactions. That's the question that Bell was interested int.

I don't know why you keep bringing up Malus' law. We know that Malus' law doesn't work in the case of entangled photons.

Then it's easier to realize real conditions these variables have to satisfy. So to answer your question, unless the paper says "cos^2(theta)" it would be refuted by every other experiment with known relative polarization different than 45 degrees.

In the case of entangled photons, there is no polarization angle of the photons. It's completely unknown. QM actually says that the photons don't have a polarization.

As I said, you keep bringing up Malus' law when it has nothing to do with the argument that is being made. If you don't know the polarization angles, then you can't apply Malus' law.

What you can do is to assume that there is some variable associated with the pair of photons that determines whether it is absorbed or transmitted by a polarizing filter. In the case of a single polarization angle, that is possible. In the case of 3 different possible angles, it is not possible.

You ask a question: Why 3 angles. Then you don't pay any attention to the answer. It's very frustrating.

 

[DrChinese]

1. How else do you compare experimental results with classical physics prediction?

2. I'm talking to RUTA. See the paper he posted, or please ask specific questions about whatever it is unclear to you.

1. There are no classical predictions for entanglement. There is no such thing as entanglement in the classical world. EPR (1935). We have discussed this in the other threads that are now closed. In this thread, I showed you in great detail how any classical attempt to achieve the entangled state statistics would fail. Rather than stop there, you have kept on and wasted the time of others.

2. No, you are arguing with RUTA, me, atyy, Nugatory, bhobba, stevendaryl and probably a few others. I am sorry, but you are going around in circles again and clearly have no question out on the table to get answered.

 

[Jabbu]

Bell's theorem can also rule out any modification of the classical laws that are local, so it doesn't just rule out Malus's Law acting on classical photon pairs, but all local alternatives.

Yes, and look at it from the other side. As much as the inequality sets the boundaries from one side, so do classical physics laws set their own limits on the other side, as usual. The hidden variable is in a very tight spot, it has to sneak in between the both, and it doesn't really make sense to suggest local theories if they describe reality more bizarre than non-locality itself.

Nonetheless, if you would like to try to see if how close Malus's Law and classical photon pairs can come to mimicking the quantum entangled pairs, you can try this. First consider that each side receives 50% vertical and 50% horizontal photons, and that when their polarizers are both vertical, both sides always get the same result. In this case, you can imagine that this result is obtained because the classical source sends out 50% classical pairs with both photons are vertically polarized, and 50% classical pairs with both photons horizontally polarized.

The paper posted by RUTA says this:

- "In each pair, the signal and idler photon have the same polarization Ls = Li = L. As successive pairs are produced, L changes in an unpredictable manner that uniformly covers the whole range of possible polarizations. The quantity L is the ‘‘hidden variable,’’ a piece of information that is absent from quantum mechanics."

It seems L (lambda) directly corresponds only to unknown photon polarization, and since they say it uniformly covers the whole range, that should be the same as unpolarized light. Malus can make predictions only if photon polarization relative to the polarizer is known, or known to be uniformly random, as the paper suggest, in which case the probability is 50-50% as you said. That's 0% correlation for any Alice and Bob polarizer settings. Classically then, there is simply no reason for two separate uniformly random events to be correlated, at all.

 

[stevendaryl]

But if you insist your hidden variable can fake experimental results for a single theta setting, then you need to realize S = E(a,b) - E(a,b') + E(a',b) + E(a',b') is defined by independent results of four separate theta settings from four separate experiments. Therefore, if your hidden variable can fake each of those experiments individually it will automatically fake the value of S.

No, that's not true. I'm getting very frustrated with you, because you keep asking questions, and don't seem interested in the answers.

Let me try one more time to see if I can explain to you why 4 angles (or 3) makes a difference.

We can come up with a hidden-variables model that reproduces the QM predictions for angles $a$ and $b$. We can come up with a DIFFERENT model that reproduces the predictions for angles $a$ and $b'$. We can come up with a third model that reproduces the predictions for angles $a'$ and $b$. We can come up with a fourth model that reproduces the predictions for $a'$ and $b'$.

What we can't do is to come up with a SINGLE[/itex] model that works in all 4 cases. That's the point! There is no way to combine the 4 models into one.

The way that a hidden-variables model of EPR would work is the following:

1. When the pair of photons is created, there is some hidden information $\lambda$ associated with each photon. This extra information is NOT polarization angle, because it's easy to see that Malus' law is not sufficient to explain EPR.
2. When Alice's photon reaches her filter, her filter is at some angle $\theta_A$.
3. We assume that whether it passes or not is a function: $f_A(\lambda, \theta_A)$ that depends on both the extra information in the photon, $\lambda$, and on the angle of Alice's filter, $\theta_A$. The function $f_A$ returns either $+1$ (meaning the photon passes) or $-1$ (meaning the photon is blocked).
4. Similarly, when Bob's photon reaches his filter, whether it passes or not is dependent on a different function $f_B(\lambda, \theta_B)$

The point is that the hidden variable $\lambda$ has to be chosen BEFORE[/itex] it is known what angles Alice and Bob will choose. That makes it a lot harder. If Alice and Bob's angles are known ahead of time, then it's easy to come up with a $\lambda$ that works.

That's why multiple angles is harder: You have to choose $\lambda$ that works with any possible choice made by Alice and Bob.

 

[Jabbu]

...and clearly have no question out on the table to get answered.

We are not going in circles, we came all the way to talk about these two equations:

E(a,b) = Pvv(a,b) + Phh(a,b) - Pvh(a,b) - Phv(a,b)

S = E(a,b) - E(a,b') + E(a',b) + E(a',b')


My questions are clearly marked with question marks: - So how cos^2(theta) fits in? When a= -30 and b= +30 QM predicts correlation = cos^2(60) = 25%, right? So isn't E(-30,30) supposed to be that same "correlation" value?

 

[stevendaryl]

We are not going in circles, we came all the way to talk about these two equations:

E(a,b) = Pvv(a,b) + Phh(a,b) - Pvh(a,b) - Phv(a,b)

S = E(a,b) - E(a,b') + E(a',b) + E(a',b')


My questions are clearly marked with question marks: - So how cos^2(theta) fits in? When a= -30 and b= +30 QM predicts correlation = cos^2(60) = 25%, right? So isn't E(-30,30) supposed to be that same "correlation" value?

No, $E(a,b)$ is defined to be (you've got the definition right there):

$E(a,b) = P_{vv}(a,b) +P_{hh}(a,b) - P_{hv}(a,b) - P_{hv}(a,b)$

The relationship with $cos^2(\theta)$ is this:
$P_{vv}(a,b) +P_{hh}(a,b) = cos^2(a - b)$
$P_{vh}(a,b) +P_{hv}(a,b) = 1 - cos^2(a - b) = sin^2(a-b)$

So $E(a,b) = cos^2(a-b) - sin^2(a-b)$

 

[atyy]

Yes, and look at it from the other side. As much as the inequality sets the boundaries from one side, so do classical physics laws set their own limits on the other side, as usual. The hidden variable is in a very tight spot, it has to sneak in between the both, and it doesn't really make sense to suggest local theories if they describe reality more bizarre than non-locality itself.

"Bizarre" is a matter of taste. Also, maybe one day quantum mechanics will be falsified, and we will have to look for a new theory. Bell's theorem says that even if one finds quantum mechanics too bizarre, or if quantum mechanics is experimentally falsified, the theory that replaces quantum mechanics must also be nonlocal. (Caveat: there remain a couple of loopholes in the experimental implementations of the Bell tests.)

The paper posted by RUTA says this:

- "In each pair, the signal and idler photon have the same polarization Ls = Li = L. As successive pairs are produced, L changes in an unpredictable manner that uniformly covers the whole range of possible polarizations. The quantity L is the ‘‘hidden variable,’’ a piece of information that is absent from quantum mechanics."

It seems L (lambda) directly corresponds only to unknown photon polarization, and since they say it uniformly covers the whole range, that should be the same as unpolarized light. Malus can make predictions only if photon polarization relative to the polarizer is known, or known to be uniformly random, as the paper suggest, in which case the probability is 50-50% as you said. That's 0% correlation for any Alice and Bob polarizer settings. Classically then, there is simply no reason for two separate uniformly random events to be correlated, at all.

In that paper http://arxiv.org/abs/quant-ph/0205171, the local hidden variable theory they propose does not assume Malus's Law, instead they replace it with their own law (Eq 18), because they want to propose a local hidden variable theory that achieves the limit allowed by one of the Bell inequalities (Eq 22). If we used Malus's Law (instead of Eq 18), then we would have P_V(γ,λ) = cos²(γ-λ). When I put Malus's Law into their Eq 19 for both polarizers vertically oriented, I get P_VV(0,0) = 3/8. If there were no correlation between Alice and Bob, then I would expect P_VV(0,0) = 1/4. I'm not sure I did that right, but it seems that with their local hidden variable theory does produce more than 0% correlation between Alice and Bob. This correlation is due to a local hidden variable, which is built in at the source - although the polarization of the photons is random, both photons in each pair have the same correlation.

 

[RUTA]

I'm talking about those terms in equation 20, in an experiment they are not calculate but counted.

Yes, there are coincidence counts for data in the next part of the paper, but I didn't see that what you wrote had anything to do with experimental data of that type.

If by "cos2(x)" you don't mean to say "cos^2(x)" I think 2 should go inside brackets like this "cos(2x)".

I meant cos(2x) per the trig identity I showed you.

Ok. So how cos^2(theta) fits in? When a= -30 and b= +30 QM predicts correlation = cos^2(60) = 25%, right? So isn't E(-30,30) supposed to be that same "correlation" value?

No, you're confusing P_VV with E.

 

[johana]

What we can't do is to come up with a SINGLE[/itex] model that works in all 4 cases. That's the point! There is no way to combine the 4 models into one.

That doesn't answer why would they test 4 cases in the same experiment, instead of individually in four separate experiments.

The point is that the hidden variable $\lambda$ has to be chosen BEFORE[/itex] it is known what angles Alice and Bob will choose. That makes it a lot harder. If Alice and Bob's angles are known ahead of time, then it's easy to come up with a $\lambda$ that works.

Let so called "Bell test angles" be known ahead of time: 0°, 45°, 22.5° and 67.5°. How that makes it any easier to come up with a single $\lambda$ function that works for each combination: E(a,b), E(a,b'), E(a',b), and E(a',b')? What time has to do with any of it?

 

[Nugatory]

That doesn't answer why would they test 4 cases in the same experiment, instead of individually in four separate experiments.

We're comparing coincidence rates at various angles. Thus we need a setup that gives us a series of measurements in which everything is the same except the angles. The most practical way of doing that is to run one experiment in which only the angle varies.

Let so called "Bell test angles" be known ahead of time: 0°, 45°, 22.5° and 67.5°. How that makes it any easier to come up with a single $\lambda$ function that works for each combination: E(a,b), E(a,b'), E(a',b), and E(a',b')?

it doesn't, and that's why we use multiple angles. It's easy to come up with a $\lambda$ function that matches the quantum mechanical prediction if we know the two angles that any given photon pair will encounter, but impossible if we only know up front that the pair will encounter some combination of two of those four angles.