Derivation of the CHSH inequality

[Avodyne]

Why do you think you can multiply "photon A went left" with "photon B went right"?
... The integral doesn't imply multiplication of the two terms, but pairing, enumeration and counting.

I tried to explain, very slowly and carefully, exactly how this notation works, and why multiplying $A(a,\lambda)$ by $B(b,\lambda)$ is equivalent to counting the number of matches minus the number of mismatches. But you don't seem to be willing to make any effort to understand why this is true, and instead just keep repeating the same stuff over and over.

So I'm out.

 

[atyy]

Can we exclude step (5) and write the equation (6) without including +/- 1?Is $E(a,b) - E(a,b') = \int[D1 - D2]\rho(\lambda)d\lambda$ not the same thing as $E(a,b) - E(a,b') = E(a,b) - E(a,b')$?In terms of D variables:

A(a,λ)B(b,λ) = D1
A(a,λ)B(b′,λ) = D2
A(a′,λ)B(b′,λ) = D3
A(a′,λ)B(b,λ) = D4

this right hand side of the step (6):

goes like this:
$\int D1[1 \pm D3 ] - \int D2[1 \pm D4 ]$?

Can we not just pull that $1 \pm$ straight out of it? How did that +/- 1 even got there, what kind of mathematics is that?

The $1$s in $[1 \pm D3 ]$ and $[1 \pm D4 ]$ are what give you the $2$ in the final line of the CHSH derivation.

 

[morrobay]

Huh? You estimate $E_{xy}$ in an experiment based on the detection counts $N_{++}(x, y)$ and so on. It is a hypothesis, introduced by Bell and motivated by reasoning about relativistic causality, that $E_{xy}$ should have a theoretical value of the form $$E_{xy} = \int \mathrm{d}\lambda \, \rho(\lambda) \, A_{x}(\lambda) B_{y}(\lambda) \,. \qquad (1)$$ That is what is being tested in a Bell experiment. You don't seem to have understood this. The point of a Bell experiment is to try to show that the assumption (1) is wrong.

For clarification : wle showed in post #16 how E(a,b) = int A (a,lambda)B(b,lambda)p(lambda)dlambda .
And E_xy is the Null hypothesis with value < 2 that is to be dis-proven. Correct ?

 

[Alien8]

The $1$s in $[1 \pm D3 ]$ and $[1 \pm D4 ]$ are what give you the $2$ in the final line of the CHSH derivation.

Step (6) doesn't only introduce +/- 1, there is a whole new equation in step (6) into which +/- 1 is inserted. I would like to know what that equation is and understand according to what principle +/- 1 gets to be where it ended up to be.

D1(1 ± D3) − D2(1 ± D4)

What is the expression before insertion of +/- 1 and why was it not inserted into the left hand side of the equation or next to D1 and D2 like this: (1 ± D1)(1 ± D3) − (1 ± D2)(1 ± D4)? What does it even mean, did you ever seen "±" in the middle of any equation before?

where A and B are the average values of the outcomes. Since the possible values of A and B are −1, 0 and +1, it follows that:

At the very beginning it is already defined {-1, +1} is intrinsic limit to A and B. Expectation value E(a,b) already has its -1.0 to +1.0 range before any explicit insertion of heads and tails into the equation. It doesn't make any sense to take already existing intrinsic limits and then duplicate them explicitly in the same equation.

 

[atyy]

Step (6) doesn't only introduce +/- 1, there is a whole new equation in step (6) into which +/- 1 is inserted. I would like to know what that equation is and understand according to what principle +/- 1 gets to be where it ended up to be.

D1(1 ± D3) − D2(1 ± D4)

What is the expression before insertion of +/- 1 and why was it not inserted into the left hand side of the equation or next to D1 and D2 like this: (1 ± D1)(1 ± D3) − (1 ± D2)(1 ± D4)? What does it even mean, did you ever seen "±" in the middle of any equation before?

The +/- is just short hand for two sets of equations.

The first set is:
$\begin{align} E(a,b) - E(a,b') &= \int[D1 - D2]\rho(\lambda)d\lambda \\ &= ∫[D1 - D2 + D1*D3 - D1*D3]\rho(\lambda)d\lambda \\ &= ∫[D1 - D2 + D1*D3 - D2*D4]\rho(\lambda)d\lambda \\ &= ∫[D1 + D1*D3 - D2 - D2*D4]\rho(\lambda)d\lambda \\ &= ∫[D1*(1 + D3) - D2*(1 + D4)]\rho(\lambda)d\lambda \\ \end{align}$

The second set is:
$\begin{align} E(a,b) - E(a,b') &= \int[D1 - D2]\rho(\lambda)d\lambda \\ &= ∫[D1 - D2 - D1*D3 + D1*D3]\rho(\lambda)d\lambda \\ &= ∫[D1 - D2 - D1*D3 + D2*D4]\rho(\lambda)d\lambda \\ &= ∫[D1 - D1*D3 - D2 + D2*D4]\rho(\lambda)d\lambda \\ &= ∫[D1*(1 - D3) - D2*(1 - D4)]\rho(\lambda)d\lambda \\ \end{align}$

 

[Alien8]

The +/- is just short hand for two sets of equations.

The first set is:
$\begin{align} E(a,b) - E(a,b') &= \int[D1 - D2]\rho(\lambda)d\lambda \\ &= ∫[D1 - D2 + D1*D3 - D1*D3]\rho(\lambda)d\lambda \\ &= ∫[D1 - D2 + D1*D3 - D2*D4]\rho(\lambda)d\lambda \\ &= ∫[D1 + D1*D3 - D2 - D2*D4]\rho(\lambda)d\lambda \\ &= ∫[D1*(1 + D3) - D2*(1 + D4)]\rho(\lambda)d\lambda \\ \end{align}$

Ok, I see now. Does the following hold true then as well:

$E(a,b) = \int D1 \rho(\lambda)d\lambda$
$E(a,b') = \int D2 \rho(\lambda)d\lambda$
$E(a',b') = \int D3 \rho(\lambda)d\lambda$
$E(a',b) = \int D4 \rho(\lambda)d\lambda$

$E(a,b) - E(a,b') = ∫[D1 + D1*D3 - D2 - D2*D4]\rho(\lambda)d\lambda$
$= ∫D1\rho(\lambda)d\lambda+ ∫D1\rho(\lambda)d\lambda*∫D3\rho(\lambda)d\lambda - ∫D2\rho(\lambda)d\lambda-∫D2\rho(\lambda)d\lambda*∫D4\rho(\lambda)d\lambda$
$= E(a,b) + E(a,b)*E(a',b') - E(a,b')-E(a,b')*E(a',b)$

$E(a,b) - E(a,b') = E(a,b) + E(a,b)*E(a',b') - E(a,b')-E(a,b')*E(a',b)$

 

[atyy]

Ok, I see now. Does the following hold true then as well:

$E(a,b) = \int D1 \rho(\lambda)d\lambda$
$E(a,b') = \int D2 \rho(\lambda)d\lambda$
$E(a',b') = \int D3 \rho(\lambda)d\lambda$
$E(a',b) = \int D4 \rho(\lambda)d\lambda$

$E(a,b) - E(a,b') = ∫[D1 + D1*D3 - D2 - D2*D4]\rho(\lambda)d\lambda$
$= ∫D1\rho(\lambda)d\lambda+ ∫D1\rho(\lambda)d\lambda*∫D3\rho(\lambda)d\lambda - ∫D2\rho(\lambda)d\lambda-∫D2\rho(\lambda)d\lambda*∫D4\rho(\lambda)d\lambda$
$= E(a,b) + E(a,b)*E(a',b') - E(a,b')-E(a,b')*E(a',b)$

$E(a,b) - E(a,b') = E(a,b) + E(a,b)*E(a',b') - E(a,b')-E(a,b')*E(a',b)$

No because $∫[D1*D3]\rho(\lambda)d\lambda \neq ∫D1\rho(\lambda)d\lambda*∫D3\rho(\lambda)d\lambda$.

 

[Alien8]

No because $∫[D1*D3]\rho(\lambda)d\lambda \neq ∫D1\rho(\lambda)d\lambda*∫D3\rho(\lambda)d\lambda$.

Aren't they both constant for each a,a',b,b' combination so that: $\int C * f(x)dx = C * \int f(x)dx$ ?

 

[morrobay]

Can someone show the relationship and how E (a,b) and E_xy are equated in the derivation ?

 

[atyy]

Aren't they both constant for each a,a',b,b' combination so that: $\int C * f(x)dx = C * \int f(x)dx$ ?

No, because each $D$ term is not a constant, but does depend on $\lambda$ which is what you are integrating over. You can see this by looking at the definition of the $D$ terms in post #6.

 

[billschnieder]

I have no idea what you mean. A local hidden variable theory says that the result of a spin measurement on a particle carrying hidden variable $\lambda$ when the detector is set to $a$ is given by a function $A(a,\lambda)$ that takes on the values $+1$ and $-1$ only (which is what we choose to call the two possible results).

Yes, but the inequalities involve expectation values not $A(a, \lambda)$.

You said the counterfactual expectation values of the same set should be the same as the actual ones from a different set because hidden variable theories require that. And I'm saying it is wrong to think the combination of a mixture of actual and counterfactual expectation values from the same set is the same as the combination of actual expectation values from different sets, whether you have a hidden variable theory or not. The degree of freedom difference between two sets and 4 sets is not an issue that only applies to hidden variable theories. I gave an example of a local realistic coin for which the sum of actual and counterfactual expectation values was different from the sum of actual expectation values from two separate identical coins, and for each coin, the result I would have obtained by looking at the other side of the coin does exist.

 

[Nugatory]

I've removed a number of posts that were taking us away from the original topic (how the CHSH inequality is derived). Please try to keep posts on topic here, and start a new thread if you want to challenge the derivation instead of helping alien8 understand it.

 

[Alien8]

Post #41 dispersed the thickest fog around this derivation for me, thanks atyy. I'm unhappy about binary states like heads and tails being labeled as integers and subjected to ordinary arithmetic, and I don't see expressions like black - white = 2 and white - black = -2 are meaningful, but at least I know now what's going on and can investigate further on my own.

Questions I have left concern only a single expectation value and these two curves:

http://en.wikipedia.org/wiki/Local_hidden_variable_theory#Optical_Bell_tests

Q1.
I think the vertical axis is supposed to be marked "expectation value" instead of "correlation". Expectation value is: E = P(++) + P(--) - P(+-) - P(-+), that is ratio of matches - mismatches per total count, and therefore ranges from -1.0 to +1.0, for QM it's E = cos2(a-b). While correlation I think is only ratio of matches per total count or CORR = P(++) + P(--), therefore ranges from 0.0 to 1.0 and for QM it's CORR = cos^2(a-b). Right?

Q2.
http://en.wikipedia.org/wiki/CHSH_inequality#Derivation_of_the_CHSH_inequality : "We start with the standard assumption of independence of the two sides, enabling us to obtain the joint probabilities of pairs of outcomes by multiplying the separate probabilities..."

By "standard assumption of independence of the two sides" they mean "local theory" or "classical physics"? By "joint probabilities" they mean probabilities for each one of the four possible outcomes {++, --, +-, -+}, that is Pab(++), Pab(--), Pab(+-), Pab(-+)? And by "multiplying the separate probabilities" they mean P(A and B) = P(A)P(B), so that Pab(++) = Pa(+)Pb(+)?

Q3.
With common language established at Q1 and Q2, the actual question is how to obtain the integral or function which will plot each of those two curves. Let's start with the full line, a naive local prediction:

,
http://en.wikipedia.org/wiki/Local_hidden_variable_theory#Optical_Bell_tests

If dotted curve is cos2(a-b), shouldn't full line be 1/2 cos2(a-b), how did they get 1/8 + 1/4 cos^(a-b)?

Q4.
Isn't it strange the local theory predicts proportionally varying correlation relative to two supposedly independent measurements? It's not as much as is observed apparently, but how in the world can a local theory conclude there would be any correlation between independent events at all?

 

[stevendaryl]

I've removed a number of posts that were taking us away from the original topic (how the CHSH inequality is derived). Please try to keep posts on topic here, and start a new thread if you want to challenge the derivation instead of helping alien8 understand it.

I apologize for a complaint that was based on my confusion. I was getting this thread confused with another thread on roughly the same topic.

 

[Alien8]

No, a local theory doesn't imply independence of the results, and it does not imply $P(A and B) = P(A)P(B)$. The reason why not is that even though $A$ can't influence $B$, and $B$ can't influence $A$, there might be a third cause that influences both. That's what the "local hidden variables" idea is all about: whether the correlations can be explained by assuming that there is a cause (the hidden variable) that influences both measurements.

A locally realistic model based on Malus' law is this: assume that in the twin-photon version of EPR, two photons are created with the same random polarization angle $\phi$. If Alice's filter is at angle $\alpha$ then she detects a photon with probability $cos^2(\alpha - \phi)$. Similarly, if Bob's filter is at angle $\beta$, then he detects a photon with probability $cos^2(\alpha - \phi)$. The correlation $E(\alpha, \beta)$ would then be:

$E(\alpha, \beta) = P_{++} + P_{--} - P_{+-} - P_{-+}$

where $P_{++}$ is the probability both Alice and Bob detect a photon, $P_{+-}$ is the probability Alice detects one and Bob doesn't, etc.

I read it like this: for the local theory based on Malus' law hidden variable $\lambda$ is just the common photon polarization $\phi$ shared by both photons in each entangled pair? Thus even though their interaction with separate analyzers are two independent events with separate independent probabilities P(A) and P(B), the ratio between their individual probabilities P(A) - P(B) would still be proportional to the ratio between their analyzer angle settings (a - b)?

For this model,
$P_{++} = \frac{1}{\pi} \int d\phi cos^2(\alpha - \phi) cos^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}cos^2(\alpha - \beta)$
$P_{+-} = \frac{1}{\pi} \int d\phi cos^2(\alpha - \phi) sin^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}sin^2(\alpha - \beta)$
$P_{-+} = \frac{1}{\pi} \int d\phi sin^2(\alpha - \phi) cos^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}sin^2(\alpha - \beta)$
$P_{--} = \frac{1}{\pi} \int d\phi sin^2(\alpha - \phi) sin^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}cos^2(\alpha - \beta)$

So
$E(\alpha, \beta) = \frac{1}{2}(cos^2(\alpha - \beta) - sin^2(\alpha - \beta)) = \frac{1}{2} cos(2 (\alpha - \beta))$

That's exactly 1/2 of the QM prediction.

So that equation I quoted from Wikipedia is just the probability for one of the four combinations, not really supposed to be marked as P(a,b), which I assumed is supposed to stand for E(a,b). That makes more sense.

I guess it doesn't make any difference, but why are you not integrating over 2pi or full 360 degrees? How do you convert that integral into the right hand side function for each of those probabilities, is there some simple principle behind it or you use some kind of integral calculator?

 

[Alien8]

For this model,
$P_{++} = \frac{1}{\pi} \int d\phi cos^2(\alpha - \phi) cos^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}cos^2(\alpha - \beta)$
$P_{+-} = \frac{1}{\pi} \int d\phi cos^2(\alpha - \phi) sin^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}sin^2(\alpha - \beta)$
$P_{-+} = \frac{1}{\pi} \int d\phi sin^2(\alpha - \phi) cos^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}sin^2(\alpha - \beta)$
$P_{--} = \frac{1}{\pi} \int d\phi sin^2(\alpha - \phi) sin^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}cos^2(\alpha - \beta)$

So
$E(\alpha, \beta) = \frac{1}{2}(cos^2(\alpha - \beta) - sin^2(\alpha - \beta)) = \frac{1}{2} cos(2 (\alpha - \beta))$

That's exactly 1/2 of the QM prediction.

Ok, I got that here: http://www.wolframalpha.com/input/?...n^2(a+−+phi)+*+cos^2(b+−+phi),++phi=0+to+2Pi+

http://www4a.wolframalpha.com/Calculate/MSP/MSP1161fh633fae46422g300001fc1f12g5d92ih4e?MSPStoreType=image/gif&s=58&w=532.&h=71. Only one question remains, the dotted curve, how does QM arrive to: cos2(a - b)?

 

[atyy]

Post #41 dispersed the thickest fog around this derivation for me, thanks atyy. I'm unhappy about binary states like heads and tails being labeled as integers and subjected to ordinary arithmetic, and I don't see expressions like black - white = 2 and white - black = -2 are meaningful, but at least I know now what's going on and can investigate further on my own.

The expression is not "black - white = 2". Rather, we assign a value to black and a value to white. For example, "value of black = 1" and "value of white = -1". Then "value of black - value of white = 2." We could choose other values, but this is the choice that is made when people refer to CHSH.

The derivation given by Aspect http://arxiv.org/abs/quant-ph/0402001 is different from the one in Wikipedia, but both are correct. If we use Aspect's approach, he explains the reason for assigning these values to particular outcomes in section 3.1. In particular, in Eq 11 of section 3 he relates it to his definition of E(a,b) = P₊₊(a,b) + P_--(a,b) - P_+-(a,b) - P_-+(a,b).

 

[stevendaryl]

I guess it doesn't make any difference, but why are you not integrating over 2pi or full 360 degrees? How do you convert that integral into the right hand side function for each of those probabilities, is there some simple principle behind it or you use some kind of integral calculator?

Well, for the particular integral I was talking about, the integrand is symmetric between $0 < \phi < 180$ and $180 < \phi < 360$. So you get the same results if you integrate over 360 and divide by $2 \pi$, or just integrate over 180 and divide by $\pi$

 

[stevendaryl]

Ok, I got that here: http://www.wolframalpha.com/input/?i=1/(2Pi) * integral cos^2(a − phi) * cos^2(b − phi) + sin^2(a − phi) * sin^2(b − phi) - cos^2(a − phi) * sin^2(b − phi) - sin^2(a − phi) * cos^2(b − phi), phi=0 to 2Pi

http://www4a.wolframalpha.com/Calculate/MSP/MSP1161fh633fae46422g300001fc1f12g5d92ih4e?MSPStoreType=image/gif&s=58&w=532.&h=71. Only one question remains, the dotted curve, how does QM arrive to: cos2(a - b)?

In QM, for EPR with correlated photons, the probability that Alice detects a photon is 1/2. The probability that Bob detects a photon given that Alice detects a photon is $cos^2(a - b)$. So for QM:

$E(a,b) = P_{++} + P_{--} - P_{+-} - P_{-+} = 1/2 cos^2(a-b) + 1/2 cos^2(a-b) - 1/2 sin^2(a-b) -1/2 sin^2(a-b)$
$= cos^2(a-b) - sin^(a-b) = cos(2(a-b))$

It's not an integral in the QM case.

 

[Alien8]

In QM, for EPR with correlated photons, the probability that Alice detects a photon is 1/2. The probability that Bob detects a photon given that Alice detects a photon is $cos^2(a - b)$. So for QM:

$E(a,b) = P_{++} + P_{--} - P_{+-} - P_{-+} = 1/2 cos^2(a-b) + 1/2 cos^2(a-b) - 1/2 sin^2(a-b) -1/2 sin^2(a-b)$
$= cos^2(a-b) - sin^(a-b) = cos(2(a-b))$

It's not an integral in the QM case.

Ok, but that only moves the question to the probability that Bob detects a photon given that Alice detects a photon: what is "$cos^2(a - b)$" given by?

 

[stevendaryl]

Ok, but that only moves the question to the probability that Bob detects a photon given that Alice detects a photon: what is "$cos^2(a - b)$" given by?

What do you mean "what is it given by"? Do you mean, how is it derived?

 

[Alien8]

What do you mean "what is it given by"? Do you mean, how is it derived?

Yes, how, where from, or based on what it is derived.

 

[atyy]

Yes, how, where from, or based on what it is derived.

You'll have to learn quantum mechanics to understand the quantum prediction. The quantum prediction is derived in chapter 4 of John Preskill's lecture notes. http://www.theory.caltech.edu/people/preskill/ph229/#lecture

To learn some quantum mechanics you can start with Braam Gaasbeek's http://arxiv.org/abs/1007.4184 or a standard text like Zettili https://www.amazon.com/dp/0470026790/?tag=pfamazon01-20 or Griffiths https://www.amazon.com/dp/0131118927/?tag=pfamazon01-20 or Rae https://www.amazon.com/dp/1584889705/?tag=pfamazon01-20.

 

[Alien8]

You'll have to learn quantum mechanics to understand the quantum prediction. The quantum prediction is derived in chapter 4 of John Preskill's lecture notes. http://www.theory.caltech.edu/people/preskill/ph229/#lecture

To learn some quantum mechanics you can start with Braam Gaasbeek's http://arxiv.org/abs/1007.4184 or a standard text like Zettili https://www.amazon.com/dp/0470026790/?tag=pfamazon01-20 or Griffiths https://www.amazon.com/dp/0131118927/?tag=pfamazon01-20 or Rae https://www.amazon.com/dp/1584889705/?tag=pfamazon01-20.

I'm not ready for all that, too many unknowns. Let's get back to the integral for the classical prediction:

Shouldn't we be puzzled by this similarity between 1/2 cos2(a-b) and cos2(a-b)? Local theory ended up with (a-b) term even though a and b should be oblivious to one another. It's interesting, the same function only squashed in half. I mean, whatever non-local magic QM describes the mechanics of it is somehow captured by this Malus' law integral. Are we sure we integrated that right? If it's integrated from 0 to 2Pi but only over 1/Pi then the curve stretches out to cos2(a-b):

http://www5a.wolframalpha.com/Calculate/MSP/MSP140920b44a6h65g2f07h00004d93g0a279789h43?MSPStoreType=image/gif&s=40&w=516.&h=68.

I'm not sure what would that practically mean, but isn't it possible that might actually be the proper way to integrate it?

 

[atyy]

I'm not ready for all that, too many unknowns. Let's get back to the integral for the classical prediction:

I'll let stevendaryl handle that, he's the expert. I'll help out if he doesn't reply, but I'm not so familiar with this particular "classical prediction". I think it is important to keep in mind that this "classical prediction" is just one example of a local variable theory, and there could be many others. A violation of CHSH shows that no local variable theory - even those we haven't explicitly constructed - will not work (except for some bizarre exceptions which we can worry about after you understand the main idea of the inequality).

 

[atyy]

Shouldn't we be puzzled by this similarity between 1/2 cos2(a-b) and cos2(a-b)? Local theory ended up with (a-b) term even though a and b should be oblivious to one another. It's interesting, the same function only squashed in half. I mean, whatever non-local magic QM describes the mechanics of it is somehow captured by this Malus' law integral.

If I understand stevendaryl's model correctly, this is in fact the quantum mechanical prediction for unentangled pairs, ie. without entanglement quantum mechanics does not violate the Bell inequality, and its predictions can be reproduced by a local theory. A local theory, and quantum mechanics without entanglement, is able to produce correlations between distant locations, because of correlations in the source. The famous example is that if I prepare a pair of socks, each pair having a different random colour, but both socks in a pair having the same colour, and send one to Alice and the other to Bob, they will receive socks with random colours, but their colours will always be correlated. In stevendaryl's example, the orientation of each pair of unentangled photons from the source is random from trial to trial, but within one trial both photons always have the same polarization. So the presence of correlation alone is not enough to rule out a local model. It must be correlation that violates a Bell inequality.

Are we sure we integrated that right? If it's integrated from 0 to 2Pi but only over 1/Pi then the curve stretches out to cos2(a-b):

http://www5a.wolframalpha.com/Calculate/MSP/MSP140920b44a6h65g2f07h00004d93g0a279789h43?MSPStoreType=image/gif&s=40&w=516.&h=68.

I'm not sure what would that practically mean, but isn't it possible that might actually be the proper way to integrate it?

Yes, stevendaryl integrated it right. You can change the nomalization to $\frac{1}{\pi}$ but you must also change the upper limit of the integral to $\pi$. In fact, since photon polarizations only vary from $0$ to $\frac{\pi}{2}$, his integral will work if you normalize with $\frac{1}{\pi/2}$ and change the upper limit to $\frac{\pi}{2}$.

 

[Alien8]

If I understand stevendaryl's model correctly, this is in fact the quantum mechanical prediction for unentangled pairs, ie. without entanglement quantum mechanics does not violate the Bell inequality, and its predictions can be reproduced by a local theory. A local theory, and quantum mechanics without entanglement, is able to produce correlations between distant locations, because of correlations in the source.

Yes, plain old unpolarized and nonentangled photons, just every two in each pair having the same polarization. Not sure why you call it QM prediction when it's standard Malus' law integration for unpolarized light vs one polarizer:

http://www5a.wolframalpha.com/Calculate/MSP/MSP40581d9f66hfh79ab24500005e9bi8g5bd9fdgi4?MSPStoreType=image/gif&s=53&w=203.&h=40.

Therefore average photon probability for unpolarized light to mark "+" event is P(+) = 0.5 or 50%.

Individually then per each pair we have:

$P_a(+) = cos^2(a - \phi)$
$P_b(+) = cos^2(b - \phi)$

Two independent events, so their joint probability is: $P_{ab}(++) = P_a(+)P_b(+)$, thus for each combination:

$P_{ab}(++) = cos^2(a - \phi) * cos^2(b - \theta)$
$P_{ab}(--) = sin^2(a - \phi) * sin^2(b - \theta)$
$P_{ab}(+-) = cos^2(a - \phi) * sin^2(b - \theta)$
$P_{ab}(-+) = sin^2(a - \phi) * cos^2(b - \theta)$

...which plugged into that same integral for ordinary unpolarized light yields expectation value based on average probability:

$E_{Malus}(a,b) = P_{ab}(++) + P_{ab}(--) - P_{ab}(+-) - P_{ab}(-+) =$

Yes, stevendaryl integrated it right. You can change the nomalization to $\frac{1}{\pi}$ but you must also change the upper limit of the integral to $\pi$. In fact, since photon polarizations only vary from $0$ to $\frac{\pi}{2}$, his integral will work if you normalize with $\frac{1}{\pi/2}$ and change the upper limit to $\frac{\pi}{2}$.

Yes, we should really normalize over the whole limit, but still, it's the same freaking function... only half of it, but exactly half of it! Originally the impression I had is that local and non-local interpretations would be inherently and utterly incompatible, so I find their direct proportionality quite surprising.

 

[Nugatory]

Originally the impression I had is that local and non-local interpretations would be inherently and utterly incompatible, so I find their direct proportionality quite surprising.

On the contrary, they make the same predictions in most cases. This is why it took decades after the EPR paper before someone (John Bell) discovered the conditions under which they make different predictions.

 

[atyy]

Yes, plain old unpolarized and nonentangled photons, just every two in each pair having the same polarization. Not sure why you call it QM prediction when it's standard Malus' law integration for unpolarized light vs one polarizer:

http://www5a.wolframalpha.com/Calculate/MSP/MSP40581d9f66hfh79ab24500005e9bi8g5bd9fdgi4?MSPStoreType=image/gif&s=53&w=203.&h=40.

Well, QM can also deal with plain old unpolarized and nonentangled photons, for which we do get Malus's law. In fact, we can even say that Malus's law applies in some sense to entangled photons if we use a technical tool in QM called the "reduced density matrix".

Yes, we should really normalize over the whole limit, but still, it's the same freaking function... only half of it, but exactly half of it! Originally the impression I had is that local and non-local interpretations would be inherently and utterly incompatible, so I find their direct proportionality quite surprising.

Me too! But in hindsight, the situation with unpolarized and unentangled photons is such that probabilities for detecting a photon are the same for all polarizer angles, and the probability of detecting a photon on both sides depends only on the difference between the polarizer angles. In other words, the source has rotational symmetry. The entangled state that is used in the quantum calculation is the Bell state, which also has rotational symmetry in the sense that it can be thought of as a pair of photons having the same polarization at some angle $\theta$ and simultaneously a pair of photons having the same polarization at the orthogonal angle $\theta + \frac{\pi}{2}$. This particular quantum state has rotational symmetry because one can use any angle for $\theta$ without changing the quantum state. Perhaps the rotational symmetry of this particular classical example and this particular entangled quantum example makes it plausible that the unentangled and entangled curves the same shape, but with larger correlations for the entangled case.

For reference, the rotational symmetry of the quantum state is seen in Eq 1 and 3 of http://arxiv.org/abs/quant-ph/0205171. I should stress that there are entangled states without this high degree of symmetry.