Photon entanglement: why three angles?

[Nugatory]

T
I listened to your advice, but at the end I found what I was looking for in Wikipedia.

[The CHSH inequality]

This is it, no partial datasets or imaginary outcomes, and it actually applies to photon entanglement experiments we are talking about. With this beautiful definition my question becomes very simple and straight forward:

S = E(a,b)
S = 0

There it is equality QM violates all the way from -1 to 1, while according to standard local reality prediction S can not be different than zero. Only one relative arbitrary angle required, so what for do we need any more?

I'm sorry, but I do not understand what you're saying here.
First, CHSH is a four-angle inequality (a-b, a'-b, a-b',a'-b') in which as many as two of the possible results are "imaginary", so it doesn't do much to disprove the claim that a two-angle test is insufficient to falsify the local realist theories. I don't see how it applies to "one relative arbitrary angle".
Second, the local realist prediction is that the absolute value of S cannot cannot exceed 2, not that S is necessarily zero. The quantum mechanical prediction is that it can reach values as high as 2.82; the Weihs team measured values greater than 2.7 a few years ago.

 

[DrChinese]

Haha, I was just about to commend you for being so patient and answering the same questions over and over.

LOL, I blame myself...

You and stevendaryl and a few others have hung in there too. When these entanglement threads shoot past 100 replies, that's when you know circles* are being etched.

(*Or maybe one of your blockworld diagrams. )

 

[stevendaryl]

This is it, no partial datasets or imaginary outcomes, and it actually applies to photon entanglement experiments we are talking about.

But the question is: how do you DERIVE that inequality. It's derived by assuming the existence of a hidden-variables theory of the type I've been discussing, and showing that for every such theory, the inequality holds. So if the inequality is violated, then there can't be an explanation in terms of such a hidden-variables theory.

The assumption behind a hidden-variables theory is that

E(a,b) = \sum_\lambda P(\lambda) F_A(\lambda, a) F_B(\lambda, b)

where \lambda is the unknown hidden variable (in my post, it's the "type" of photon), and P(\lambda) is the probability of the hidden variable having value \lambda, and F_A and F_B are two unknown functions that always return +1 or -1.

Bell's inequality shows that there can't possibly be such functions P(\lambda), F_A, F_B.

Talking about data sets is a way of talking about the functions F_A and F_B. Each run i of the experiment gives us results A_i for Alice, at some filter angle \alpha_i, and a result B_i for Bob, at some filter angle \beta_i. The assumption of the hidden variables theory is that there is, associated with each run, is a hidden variable \lambda_i, and that

F_A(\lambda_i, \alpha_i) = A_i
F_B(\lambda_i, \beta_i) = B_i

We can convince ourselves that the functions F_A and F_B must be the same, since Alice and Bob always get the same answers on the same filter settings. So in each run, we know the numbers

F_A(\lambda_i, \alpha_i)
F_A(\lambda_i, \beta_i)

So we know F_A(\lambda_i, \alpha) for potentially two different values of \alpha. We can't know F_A(\lambda_i, \alpha) for three different values, because we can only test at most two different angles. So our information about F_A(\lambda_i, \alpha) is incomplete.

 

[stevendaryl]

LOL, I blame myself...

You and stevendaryl and a few others have hung in there too. When these entanglement threads shoot past 100 replies, that's when you know circles* are being etched.

(*Or maybe one of your blockworld diagrams. )

Well, it's hard to know when you've really run into a brick wall. If it becomes a disagreement about interpretation, or about what's plausible or implausible, or what's reasonable or wacky--that's a matter of opinion, and there's not much hope for progress. But if there is a misunderstanding about definitions, about what precisely is being claimed, about what precisely has been proved or empirically demonstrated, that's a disagreement that SHOULD be possible to resolve. johana and Jabbu before her seemed confused about points that ought to have definitive resolutions.

 

[stevendaryl]

With this beautiful definition my question becomes very simple and straight forward:

S = E(a,b)
S = 0

There it is equality QM violates all the way from -1 to 1, while according to standard local reality prediction S can not be different than zero. Only one relative arbitrary angle required, so what for do we need any more?

Why do you say by standard local reality, S can not be different than zero? That's simply not true.

 

[johana]

Why do you say by standard local reality, S can not be different than zero? That's simply not true.

I'm using this equation:

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


How do you get anything but zero?

 

[johana]

I'm sorry, but I do not understand what you're saying here.
First, CHSH is a four-angle inequality (a-b, a'-b, a-b',a'-b') in which as many as two of the possible results are "imaginary", so it doesn't do much to disprove the claim that a two-angle test is insufficient to falsify the local realist theories. I don't see how it applies to "one relative arbitrary angle".
Second, the local realist prediction is that the absolute value of S cannot cannot exceed 2, not that S is necessarily zero. The quantum mechanical prediction is that it can reach values as high as 2.82; the Weihs team measured values greater than 2.7 a few years ago.

I was referring to this miniature version:

S = E(a,b)
S = 0

...and asking why is that not sufficient. Do you know where did that number 2 you're talking about come from?

 

[stevendaryl]

I'm using this equation:

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


How do you get anything but zero?

Well, if N_{++} = 50, N_{--} = 50, N_{+-} = 0 N_{-+} = 0, then you get correlation 1, not zero. Why do you think it's going to be zero?

 

[johana]

Well, if N_{++} = 50, N_{--} = 50, N_{+-} = 0 N_{-+} = 0, then you get correlation 1, not zero. Why do you think it's going to be zero?

1 is QM prediction when a = b, it's total correlation. It think local reality predicts they will be equally random, converging to 25% each, because I read it somewhere. I'll look for it.

 

[DrChinese]

Well, it's hard to know when you've really run into a brick wall.

True. My view is that with such good advisors/moderators such as you, Nugatory, DaleSpam, and countless others, sometimes one person can get across the idea by saying it in a different manner.

On the other hand, most of these entanglement threads end up with thousands of views. This one is approaching 3,000 right now. So it is not just the person we are helping, we must consider also those who are learning by reading and are not yet ready to ask a question. So we must consider them, even though we don't know who they are.

 

[DrChinese]

It think local reality predicts they will be equally random...

Please keep in mind that there are several different local realistic models. The EPR local realistic model predicts perfect correlations as does QM. So assuming I understand the case you are modeling (questionable ) and it is the perfect correlation case, then stevendaryl's answer would be that case and would be correct for both QM and LR.

And also, as he pointed out: "Bell's inequality shows that there can't possibly be such functions P(λ),FA,FB." Of course, that's when other angles are considered. The 3 other angles.

 

[stevendaryl]

1 is QM prediction when a = b, it's total correlation. It think local reality predicts they will be equally random, converging to 25% each, because I read it somewhere. I'll look for it.

No, local reality does not predict zero correlation. That's the prediction for unentangled photons with random polarizations.

 

[stevendaryl]

So it is not just the person we are helping, we must consider also those who are learning by reading and are not yet ready to ask a question. So we must consider them, even though we don't know who they are.

That sounds like a hidden-variables theory to me.

 

[DrChinese]

That sounds like a hidden-variables theory to me.

Yes, or a conspiracy theory, not sure which...

 

[Nugatory]

Do you know where did that number 2 you're talking about come from?

There's a derivation of the CHSH inequality, including the number 2, in the wikipedia article that you yourself quoted from above.

 

[johana]

There's a derivation of the CHSH inequality, including the number 2, in the wikipedia article that you yourself quoted from above.

I see now, this:

That's absolute limit, it doesn't really say local reality prediction for E(x,y) is not zero. Kind of odd to compare a binary state in terms of numerical range, because binary state is exclusive of any other states, and +1/-1 are rather arbitrary. It could have been heads and tails, or ON and OFF, then stating |A| < ON is a bit undefined. And then they go on to apply the triangle inequality. Where is the triangle?

Never mind that, they end up with this statement:

(1) S = E(a, b) − E(a, b') + E(a', b) + E(a' b')
(2) − 2 ≤ S ≤ 2

...so given:

E(x,y) = cos^2(x-y) - sin^2(x-y)

...it doesn't actually hold true for these four:

a = 0; b= 22.5
a'= 45; b'= 67.5

E(a, b) − E(a, b') + E(a', b) + E(a' b')
= 0.71 + 0.71 + 0.71 + 0.71
= 2.84

The range of S goes all the way to 2.8, that's the actual theoretical limit. Different theories might suggest different functions for E(x,y), but is there anything in that inequality which says local hidden variable theory can not come up with E(x,y) that could just as well result in S going over that number 2?

In any case, why be so cautious and stop with the limiting range? If we want to express the inequality relative to local hidden variable prediction then we can be more exact and instead of some range we could narrow down S to a single number.

http://en.wikipedia.org/wiki/EPR_paradox#Locality_in_the_EPR_experiment

Locality in the EPR experiment
Whichever axis she uses, she has a 50% probability of obtaining "+" and 50% probability of obtaining "−", completely at random... Therefore, in the one measurement he is allowed to make, there is a 50% probability of getting "+" and 50% of getting "−", regardless of whether or not his axis is aligned with Alice's.

So if:

P_A(+) = P_A(-) = 0.5 and P_B(+) = P_B(-) = 0.5

...then:

P(++|−−) = (0.5 * 0.5) + (0.5 * 0.5) = 0.5
P(+−|−+) = (0.5 * 0.5) + (0.5 * 0.5) = 0.5

...thus:

E_{local}(x,y) = N_{++} + N_{−−} - N_{+−} - N_{−+} = 0.5 - 0.5 = 0.0

...and therefore:

S_{local} = E_{local}(x,y)
S_{local} = 0

 

[stevendaryl]

I see now, this:

That's absolute limit, it doesn't really say local reality prediction for E(x,y) is not zero.

Well, it's easy to come up with a local hidden-variables theory with E(x,y) nonzero.

(1) S = E(a, b) − E(a, b') + E(a', b) + E(a' b')
(2) − 2 ≤ S ≤ 2

...so given:

E(x,y) = cos^2(x-y) - sin^2(x-y)

...it doesn't actually hold true for these four:

a = 0; b= 22.5
a'= 45; b'= 67.5

E(a, b) − E(a, b') + E(a', b) + E(a' b')
= 0.71 + 0.71 + 0.71 + 0.71
= 2.84

The range of S goes all the way to 2.8, that's the actual theoretical limit. Different theories might suggest different functions for E(x,y), but is there anything in that inequality which says local hidden variable theory can not come up with E(x,y) that could just as well result in S going over that number 2?

Yes. That's what Bell's proof (or the similar CHSH proof) is all about. Quantum mechanics predictions a value for the expression E(a, b) − E(a, b&#039;) + E(a&#039;, b) + E(a&#039; b&#039;) that is larger than any local hidden variable theory predicts.

In any case, why be so cautious and stop with the limiting range? If we want to express the inequality relative to local hidden variable prediction then we can be more exact and instead of some range we could narrow down S to a single number.

No, different theories predict different values for S.

For example, we could try a semi-classical explanation for EPR: When a twin-photon is produced, the two photons are given the same randomly chosen polarization, \alpha. Then if a detector is oriented at angle \theta, then the photon passes through with probability cos^2(\alpha - \theta). This local theory predicts a correlation E(x,y) that (I think) is given by:

E(x,y) =\frac{1}{2} cos(2(x-y))

(which is exactly 1/2 of the quantum prediction of E(x,y) = cos(2(x-y)))

 

[johana]

No, different theories predict different values for S.

From the Wikipedia article I quoted it looks like locality in EPR experiments is defined exactly by the independence of the two data streams, which translates into prediction they should be completely random (50-50%) regardless of any absolute or relative polarizers rotation.

For example, we could try a semi-classical explanation for EPR: When a twin-photon is produced, the two photons are given the same randomly chosen polarization, \alpha. Then if a detector is oriented at angle \theta, then the photon passes through with probability cos^2(\alpha - \theta).

If random polarization is uniform then integrated average of cos^2 over 360° is 1/2, that's the same 50-50% Wikipedia is talking about. If the ratio is not 50-50% for any arbitrary (a-b) angle combination, then the theory is not local, or the experiment is not rotationally invariant.

 

[Nugatory]

From the Wikipedia article I quoted it looks like locality in EPR experiments is defined exactly by the independence of the two data streams...

Yes, if by "independence" you mean that the results at polarizer A can be described by a probability distribution that is independent of the setting of polarizer B and vice versa. However...

which translates into prediction they should be completely random (50-50%) regardless of any absolute or relative polarizers rotation.

No, this does not follow. If the results at polarizer A depend on the setting of polarizer A and also some property of the photon at polarizer A, and the results at polarizer B depend on the setting of polarizer B and some property of the photon at polarizer B, then the two streams are independent in the sense that I described above; there's no a in the B function and no b in the A function, and the individual streams will be completely random. Nonetheless, there can be a strong correlation between the streams if the two photons both acquired some properties when the pair was created. With entangled pairs created by PDLC or atomic cascade processes, conservation of angular momentum requires that if one photon is horizontally polarized the other will be vertically polarized, and this allows a correlation between the two streams even though each one is isolation is completely random.

 

[johana]

No, this does not follow.

It's given here as explicit description (requirement) for EPR locality:
http://en.wikipedia.org/wiki/EPR_paradox#Locality_in_the_EPR_experiment

If the results at polarizer A depend on the setting of polarizer A and also some property of the photon at polarizer A, and the results at polarizer B depend on the setting of polarizer B and some property of the photon at polarizer B, then the two streams are independent in the sense that I described above; there's no a in the B function and no b in the A function, and the individual streams will be completely random. Nonetheless, there can be a strong correlation between the streams if the two photons both acquired some properties when the pair was created.

Yes, but after they are created they have yet to negotiate with their polarizers about the actual outcome. Interaction of photon A with polarizer A and interaction of photon B with polarizer B in classical physics are supposed to be two independent probabilistic events, even with exactly the same odds the two outcomes should still be no more correlated than two separate coin toss outcomes, due to microscopic differences between the polarizers and the point of impact relative to the polarizer molecular structure.

 

[Avodyne]

Yes, but after they are created they have yet to negotiate with their polarizers about the actual outcome. Interaction of photon A with polarizer A and interaction of photon B with polarizer B in classical physics are supposed to be two independent probabilistic events, even with exactly the same odds the two outcomes should still be no more correlated than two separate coin toss outcomes, due to microscopic differences between the polarizers and the point of impact relative to the polarizer molecular structure.

This is wrong. In classical physics, the interaction of EM radiation with a polarizer is completely deterministic; "microscopic differences between the polarizers and the point of impact relative to the polarizer molecular structure" plays no role at all: http://en.wikipedia.org/wiki/Polarizer

 

[johana]

This is wrong. In classical physics, the interaction of EM radiation with a polarizer is completely deterministic; "microscopic differences between the polarizers and the point of impact relative to the polarizer molecular structure" plays no role at all: http://en.wikipedia.org/wiki/Polarizer

A function which outputs an average, such as Malus' law, is not deterministic but probabilistic.

...where I0 is the initial intensity, and θi is the angle between the light's initial polarization and the axis of the polarizer.

It means if you send a beam of light with intensity of 100,000 photons polarized at 30° towards a polarizer with polarization axis at 60°, you can expect cos^2(30) = 75% of them will pass through. Malus' law is a probability function with the domain {0°, 90°} and sample space {heads, tails}, or {+1,-1} if you prefer.

 

[atyy]

A function which outputs an average, such as Malus' law, is not deterministic but probabilistic.

...where I0 is the initial intensity, and θi is the angle between the light's initial polarization and the axis of the polarizer.

It means if you send a beam of light with intensity of 100,000 photons polarized at 30° towards a polarizer with polarization axis at 60°, you can expect cos^2(30) = 75% of them will pass through. Malus' law is a probability function with the domain {0°, 90°} and sample space {heads, tails}, or {+1,-1} if you prefer.

If the probability to pass through on each side is 0.75, what is the probability that if both photons are polarized at 30°, they will both pass through?

 

[johana]

If the probability to pass through on each side is 0.75, what is the probability that if both photons are polarized at 30°, they will both pass through?

For θ = 30°, cos^2(30) = 0.75:

P_A(+)= 0.75, P_A(-)= 0.25
P_B(+)= 0.75, P_B(-)= 0.25

P_{AB}(+ and +) = 0.75 * 0.75 = 56.3% <- both go through
P_{AB}(- and -) = 0.25 * 0.25 = 6.3%
P_{AB}(+ and -) = 0.75 * 0.25 = 18.8%
P_{AB}(- and +) = 0.25 * 0.75 = 18.8%

 

[atyy]

For θ = 30°, cos^2(30) = 0.75:

P_A(+)= 0.75, P_A(-)= 0.25
P_B(+)= 0.75, P_B(-)= 0.25

P_{AB}(+ and +) = 0.75 * 0.75 = 56.3% <- both go through
P_{AB}(- and -) = 0.25 * 0.25 = 6.3%
P_{AB}(+ and -) = 0.75 * 0.25 = 18.8%
P_{AB}(- and +) = 0.25 * 0.75 = 18.8%

So in the classical case, one doesn't have to get that P_{AB}(+ and +) = P_{AB}(- and -) = P_{AB}(+ and -) = P_{AB}(- and +) = 0.25. In the case you calculated the results are correlated, because both photons were prepared in the same state.

 

[stevendaryl]

From the Wikipedia article I quoted it looks like locality in EPR experiments is defined exactly by the independence of the two data streams, which translates into prediction they should be completely random (50-50%) regardless of any absolute or relative polarizers rotation.

If that's the impression you got from the Wikipedia article, then it needs to be rewritten, because that's absolutely not true.

If random polarization is uniform then integrated average of cos^2 over 360° is 1/2, that's the same 50-50% Wikipedia is talking about. If the ratio is not 50-50% for any arbitrary (a-b) angle combination, then the theory is not local, or the experiment is not rotationally invariant.

That's not correct. Here's a local realistic model: You generate a pair of photons that are polarized at angle \alpha, where \alpha is chosen randomly. Then, the probability of passing through a filter is cos^2(\alpha - \theta) where \theta is the orientation of the filter. Then the correlation E(a,b) will be given by:

E(a,b) = \frac{1}{2\pi}\int d\alpha (cos^2(\alpha - a) cos^2(\alpha - b) + sin^2(\alpha - a) sin^2(\alpha - b) - cos^2(\alpha - a) sin^2(\alpha - b) - sin^2(\alpha - a) cos^2(\alpha - b))

The positive terms, cos^2(\alpha - a) cos^2(\alpha - b) + sin^2(\alpha - a) sin^2(\alpha - b), give the probability of both filters having the same result--either they both pass, or they both are blocked. The negative terms, cos^2(\alpha - a) sin^2(\alpha - b) - sin^2(\alpha - a) cos^2(\alpha - b)) give the probability that the two filters get different results--one passes and the other is blocked.

You can go through it yourself, if you know trigonometry. The answer is:

E(a,b) = \frac{1}{2} cos(2(a-b))

which is definitely not zero, except in the case where a-b = \frac{\pi}{4}

 

[johana]

So in the classical case, one doesn't have to get that P_{AB}(+ and +) = P_{AB}(- and -) = P_{AB}(+ and -) = P_{AB}(- and +) = 0.25. In the case you calculated the results are correlated, because both photons were prepared in the same state.

Yes, but it depends on rotational invariance which is prerequisite for local theory prediction of 50-50%. So if rotational invariance is assumed (theory), or guaranteed (experiment), then cos^2(θ) is uniformly integrated over 360° which averages out to 50-50%, regardless of any absolute or relative polarizers rotation, which then combined with the other side splits up into 25% for each of 4 possible combinations.

 

[atyy]

Yes, but it depends on rotational invariance which is prerequisite for local theory prediction of 50-50%. So if rotational invariance is assumed (theory), or guaranteed (experiment), then cos^2(θ) is uniformly integrated over 360° which averages out to 50-50%, regardless of any absolute or relative polarizers rotation, which then combined with the other side splits up into 25% for each of 4 possible combinations.

The integration needs some work. So let's first try a simpler scenario to get the 50-50. Let's say both polarizers are oriented at 0°. For half the pairs, both photons are polarized at 0°; the other half of the pairs have both photons polarized at 90°. What are the probabilities P(++), P(--), P(+-), P(-+)?

 

[atyy]

Yes, but it depends on rotational invariance which is prerequisite for local theory prediction of 50-50%. So if rotational invariance is assumed (theory), or guaranteed (experiment), then cos^2(θ) is uniformly integrated over 360° which averages out to 50-50%, regardless of any absolute or relative polarizers rotation, which then combined with the other side splits up into 25% for each of 4 possible combinations.

If you want to try the integration, under the assumption that all polarizations are equally likely, but that both photons always have the same polarization, and that both polarizers are oriented at the same constant angle, then to get P(++) you should integrate cos²(θ)cos²(θ) = cos⁴(θ) over 360°, where θ is the angle between a photon and a polarizer. Similarly to get P(--) you should integrate sin⁴(θ) over 360°. I'm not sure I did this right, but I get the following.

P(+) = 1/2
http://www.wolframalpha.com/input/?i=integrate+(1/(2*pi))+*+(cos(x))^2+dx+from+x+=+0+to+2*pi

P(-) = 1/2
http://www.wolframalpha.com/input/?i=integrate+(1/(2*pi))+*+(sin(x))^2+dx+from+x+=+0+to+2*pi

P(++) = 3/8
http://www.wolframalpha.com/input/?i=integrate+(1/(2*pi))+*+(sin(x))^4+dx+from+x+=+0+to+2*pi

P(--) = 3/8
http://www.wolframalpha.com/input/?i=integrate+(1/(2*pi))+*+(sin(x))^4+dx+from+x+=+0+to+2*pi

P(+-) = P(-+) = 1/8
http://www.wolframalpha.com/input/?...*+(cos(x))^2+(sin(x))^2+dx+from+x+=+0+to+2*pi

So P(++) + P(--) - P(+-) - P(-+) = 6/8-2/8 = 1/2, which is another example of classical correlations built in at the source producing correlated outcomes.

 

[Avodyne]

A function which outputs an average, such as Malus' law, is not deterministic but probabilistic.

Étienne-Louis Malus died in 1812. He never heard of quantum mechanics, and certainly thought of his law as a deterministic formula for the intensity of EM radiation passing through a polarizer.

It is true that, in quantum electrodynamics, Malus' Law becomes probabilistic. According to quantum electrodynamics, this probability has a purely quantum mechanical origin, and does not arise "due to microscopic differences between the polarizers and the point of impact relative to the polarizer molecular structure".

That's the conventional QM story. In hidden-variable theory, some underlying hidden variables are supposed to determine what happens to each and every photon, in a fully deterministic way. But no such theory can reproduce the predictions of QM.

 

[johana]

That's not correct. Here's a local realistic model: You generate a pair of photons that are polarized at angle \alpha, where \alpha is chosen randomly. Then, the probability of passing through a filter is...

http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties

A beam of unpolarized light can be thought of as containing a uniform mixture of linear polarizations at all possible angles. Since the average value of cos^2 \theta is 1/2, the transmission coefficient becomes \frac {I}{I_0} = \frac {1}{2}.

\int_0^{2\pi} \frac{cos^2(x)}{2\pi} dx = 1/2

Wolfram: integrate 1/(2pi) * cos^2(x) dx, x = 0 to 2pi

 

[atyy]

http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties\int_0^{2\pi} \frac{cos^2(x)}{2\pi} dx = 1/2

Wolfram: integrate 1/(2pi) * cos^2(x) dx, x = 0 to 2pi

Sure, that's just P(+) and P(-). To get P(++), you have to multiply the probability the probability to get + on each side for each angle and integrate over all angles, ie. integrate cos⁴(x) in the case that both polarizers are set to the same angle. In fact stevendaryl gave the general answer in the same post you quoted:

That's not correct. Here's a local realistic model: You generate a pair of photons that are polarized at angle \alpha, where \alpha is chosen randomly. Then, the probability of passing through a filter is cos^2(\alpha - \theta) where \theta is the orientation of the filter. Then the correlation E(a,b) will be given by:

E(a,b) = \frac{1}{2\pi}\int d\alpha (cos^2(\alpha - a) cos^2(\alpha - b) + sin^2(\alpha - a) sin^2(\alpha - b) - cos^2(\alpha - a) sin^2(\alpha - b) - sin^2(\alpha - a) cos^2(\alpha - b))

The positive terms, cos^2(\alpha - a) cos^2(\alpha - b) + sin^2(\alpha - a) sin^2(\alpha - b), give the probability of both filters having the same result--either they both pass, or they both are blocked. The negative terms, cos^2(\alpha - a) sin^2(\alpha - b) - sin^2(\alpha - a) cos^2(\alpha - b)) give the probability that the two filters get different results--one passes and the other is blocked.

You can go through it yourself, if you know trigonometry. The answer is:

E(a,b) = \frac{1}{2} cos(2(a-b))

which is definitely not zero, except in the case where a-b = \frac{\pi}{4}

.

 

[stevendaryl]

http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties


\int_0^{2\pi} \frac{cos^2(x)}{2\pi} dx = 1/2

Wolfram: integrate 1/(2pi) * cos^2(x) dx, x = 0 to 2pi

And what do you think that number is showing? That's the probability of any single filter passing a photon. It doesn't tell you anything about the correlation between two different filters.

To compute the correlation of two filters, one oriented at angle a, and one oriented at angle b, you have to consider the following four numbers:

1. P(a|\alpha) = cos^2(a-\alpha) the probability that a photon with polarization \alpha passes through a filter at angle a.
2. P(b|\alpha) = cos^2(b-\alpha) the probability that a photon with polarization \alpha passes through a filter at angle b.
3. \bar{P}(a|\alpha) = sin^2(a-\alpha) the probability that a photon with polarization \alpha does not pass through a filter at angle a.
4. \bar{P}(b|\alpha) = sin^2(b-\alpha) the probability that a photon with polarization \alpha does not pass through a filter at angle b.

Then the correlation E(a,b) is given by:
\frac{1}{2\pi} \int d\alpha (P(a|\alpha) P(b|\alpha) +\bar{P}(a|\alpha) \bar{P}(b|\alpha)<br /> -P(a|\alpha) \bar{P}(b|\alpha) - \bar{P}(a|\alpha) P(b|\alpha))

That number is E(a,b) = \frac{1}{2}cos(2(a-b))

 

[Nugatory]

Johana, I said this in you other thread, but repeating it here:

Here is a question for you (and it is not a rhetorical question):

Have you read and understood the EPR paper and Bell's paper? If you haven't read them, you're wasting your time and ours. If you have read them, and there are parts of the arguments that you don't follow, ask and we can have a more focused and productive discussion.

 

[DrChinese]

\int_0^{2\pi} \frac{cos^2(x)}{2\pi} dx = 1/2

Wolfram: integrate 1/(2pi) * cos^2(x) dx, x = 0 to 2pi

As stevendaryl and Nugatory and billschnieder have been saying: the reason things are going in circles is because the compass has been lost. What relevance is the above?

We all are familiar with the math of Bell, entangled photons, etc. There are a lot of very similar looking formulae, and the key is to keep things labeled and moving in a direction.

The issue in this thread is that it takes 3 angles, a/b/c, to get the Bell outcome. There are a variety of different candidate local realistic theories that can be tested against this backdrop, and then shown not to match the predictions of QM. As we have said repeatedly, the approach you are taking gives a prediction that is substantially at odds with QM (and experiment). No surprise there, that's Bell. The part none of us follow is: do you see why? Because it doesn't matter if you present a formula and integrate it if you don't know where you are going.

Fact 1: all entangled photon pairs will yield 100% correlated (or anti-correlated depending on type) results when measured at the same angle.

Fact 2: entangled photon pairs act and are best described as single systems of 2 particles, not 2 systems of 1 particle. QM and experiment match.

Fact 3: all local realistic theories are predicated on the idea that entangled photons are fully independent and separable entities, and there is no ongoing physical connection. Bell says no such local realistic theory can yield predictions consistent with QM.


Do you understand these 3 things? If you do not, please let us know which you don't.

 

[Nugatory]

Closed - this discussion is no longer adding any value.