Photon entanglement: why three angles?

[stevendaryl]

I need to confirm what exactly is meant by "angle", "dataset", and "partial dataset". Say Alice and Bob can turn their polarizers to 0, 20, and 30 degrees, and we are testing for these three combinations:

a= (0,20) = 20°
b= (30,0) = 30°
c= (30,20) = 10°

With relative angle a = 20° we get for example this dataset A = --, +-, ++, -+, ++
With relative angle b = 30° we get for example this dataset B = +-, ++, -+, -+, +-
With relative angle c = 10° we get for example this dataset C = ++, -+, +-, -+, --

Correct? What partial dataset are you talking about?

No, that's not what I mean. The assumption behind local hidden-variables theories is that each electron produced in EPR simultaneously has a spin component in EACH of the three directions a, b, and c. So associated with electron number $i$ is a triple of numbers $\langle R_{i,a}, R_{i,b}, R_{i,c} \rangle$, where $R_{i,a}$ is either +1 (to indicate spin-up in direction $a$) or -1 (to indicate spin-down). Analogously for $R_{i,b}$ and $R_{i,c}$.

So a complete dataset for the hidden variables $R_{i,j}$ would be a table consisting of one row for each electron produced, and each row would have three values, each of which is either $+1$ or $-1$.

Unfortunately, we can't measure the spin in more than one direction at a time. However, we can use the fact that in a twin-pair experiment, the spin of one particle in a particular direction is always the opposite of the spin of its twin in that direction. So that allows us to measure two of the three values for $R_{i,j}$. Alice can measure the spin in direction $a$ for one of the particles, and Bob can measure the spin in direction $b$ for the other particle. Since the two particles are anti-correlated, we just need to flip Bob's result to get the result that Alice would have measured if she had measured the spin in direction $b$. So we have two of the three angles covered. But we have no way to measure the spin in the third direction, $c$. So we leave that blank.

So suppose that in the first trial, Alice measures spin in the $a$ direction and gets spin-up. Bob measures spin in the $b$ direction and also gets spin-up, which means that Alice would[/itex] have gotten spin-down if she had measured in that direction. So the results of the first trial are written as the triple

$\langle +, -, ? \rangle$

In the second trial, Alice measures the spin in the $a$ direction again, and gets spin-down. Bob measures the spin in direction $c$ and gets spin-down, also, which means that Alice would have gotten spin-up. So the results of the second round are written as:

$\langle -, ?, + \rangle$

So the partial dataset might look like this:

$\left( \begin{array}\\ + & - & ? \\ - & ? & +\\ + & ? & - \\ ... \end{array} \right)$

 

[DrChinese]

Can you show an example QM dataset that can violate that inequality?

You have it backwards.

QM does not predict realism. The QM dataset consists of pairs in consonance with the predictions of QM, which violate the inequality BY DEFINITION. That is because the QM prediction is used to construct the inequality.

Please, stop and review the reference materials first. You are going around in circles. If nothing else, you are making me dizzy.

 

[billschnieder]

Apparently you do not understand a basic application of Malus, circa 1809.

A stream of Alice photons polarized at 0 degrees as + will have a 25% chance of being polarized + at 120 degrees. A stream of Alice photons polarized at 120 degrees as + will have a 25% chance of being polarized + at 240 degrees. A stream of Alice photons polarized at 0 degrees as + will have a 25% chance of being polarized + at 240 degrees. So if it passes the polarizer, it is matched.

It is matched with what? You need two things to do matching, don't you? You still haven't explained what you are matching the photon with, and what said matching has to do with malus at all.

 

[DrChinese]

It is matched with what? You need two things to do matching, don't you? You still haven't explained what you are matching the photon with, and what said matching has to do with malus at all.

This is off the subject of this thread, and I have already answered several times already.

For that matter, this question of this thread has been answered multiple times already, and I will summarize it:

The reason for 3 angles & entanglement to demonstrate why local realism fails: it traces back to Bell's Theorem. Multiple variations on this have been presented, as well as Bell's original paper. Any subsequent answer will simply be yet another version of the same. If you haven't followed what has been presented so far, READ THE REFERENCES instead of asking the same question a different way.

 

[johana]

No, that's not what I mean. The assumption behind local hidden-variables theories is that each electron produced in EPR simultaneously has a spin component in EACH of the three directions a, b, and c. So associated with electron number $i$ is a triple of numbers $\langle R_{i,a}, R_{i,b}, R_{i,c} \rangle$, where $R_{i,a}$ is either +1 (to indicate spin-up in direction $a$) or -1 (to indicate spin-down). Analogously for $R_{i,b}$ and $R_{i,c}$.

So a complete dataset for the hidden variables $R_{i,j}$ would be a table consisting of one row for each electron produced, and each row would have three values, each of which is either $+1$ or $-1$.

I don't see how three orthogonal measurement axis in electron case compare with anything in entangled photons experiment. Billschnieder says A, B, C are outcomes, you describe them as potential outcomes. Normally one would think the outcome refers to both Alice and Bob data for a single entangled pair, but the outcome you are talking about seems to be taken from three entangled pairs and only on one side for either Alice or Bob.

Unfortunately, we can't measure the spin in more than one direction at a time. However, we can use the fact that in a twin-pair experiment, the spin of one particle in a particular direction is always the opposite of the spin of its twin in that direction.

This also doesn't seem to compare with entangled photons experiment. For photons 100% match/mismatch is reserved only for 0 and 90 degrees relative angles. Can we stick with photons since the whole thread was about photons so far?

 

[stevendaryl]

Note: I edited my answer to make it about photons, rather than electrons. It really doesn't make any difference to the argument.

I don't see how three orthogonal measurement axis in electron case compare with anything in entangled photons experiment.

It's almost exactly the same. Instead of measuring spin-up or spin-down relative to an axis, Alice and Bob either observe that the photon passed the filter, or the photon did not pass the filter relative to an axis. In both experiments, Alice and Bob pick an orientation, then they perform a measurement that has two possible values. The argument works exactly the same.

Alice has three possible axes to measure a photon's polarization: $a, b, c$. Similarly, Bob has three possible axes that he can measure: $a, b, c$. We convince ourselves through experiment, or by looking at the QM predictions, that for a pair of entangled photons, if Alice and Bob both measure the polarizations of entangled photons using the same axis, then they ALWAYS get the same results. (or they always get opposite results, depending on how the entangled photons are produced; let's assume that they always get the same results).

Since Alice and Bob ALWAYS get the same results for the same filter orientations, that means that Bob, by measuring his photon, can learn something about Alice's photon.

To Einstein (and whoever P and R were), that means that there must be a deterministic answer to the question: "What would the result be if Alice measured her photon's polarization relative to axis $a$?" It must be a deterministic answer, because Bob can predict it with 100% certainty by measuring his photon's polarization relative to axis $a$. So to E, P, and R, there must be, associated with each photon, a triple of numbers $\langle R_a, R_b, R_c\rangle$ telling whether Alice's photon will pass her filter or get blocked by her filter, should she set it at orientation $a$, $b$ or $c$.

She can only actually measure one of those three numbers, but the EPR reasoning implies that the three numbers exist, whether she can measure them or not. Putting Alice's measurement together with Bob's, it's possible to figure out what two of the three numbers are. To figure out $R_a$ and $R_b$, Alice measures polarization in direction $a$ and Bob measures polarization in direction $b$. Then they have to leave the answer for direction $c$ blank.

Billschnieder says A, B, C are outcomes, you describe them as potential outcomes.

Two of them are actual outcomes, and the third one is a "conterfactual": If Alice had oriented her filter at direction $c$, rather than $a$, her photon would have passed through (or would not have).

Normally one would think the outcome refers to both Alice and Bob data for a single entangled pair, but the outcome you are talking about seems to be taken from three entangled pairs and only on one side for either Alice or Bob.

No, it's not three entangled pairs. For each entangled pair, Alice and Bob measure two of three possible angles. So for each entangled pair, they produce a triple of values: One value is computed by Bob's result. The other value is computed by Alice's result, and the third value is left "?", because nobody measures that one. So you end up with a list of triples, where each triple has two values that are $\pm 1$ and one value that is "?".

This also doesn't seem to compare with entangled photons experiment.

No, it's almost exactly the same. Instead of measuring "spin-up in direction a", they measure "passes the filter when the filter is oriented at direction a". We pick three axes: $a, b, c$. Alice measures photon polarization relative to axis $a$, and Bob measures photon polarization of the twin photon relative to axis $b$. Nobody measures polarization relative to axis $c$, so that one would be left "?".

For photons 100% match/mismatch is reserved only for 0 and 90 degrees relative angles. Can we stick with photons since the whole thread was about photons so far?

It doesn't make any difference. The argument is exactly the same.

 

[johana]

Alice has three possible axes to measure a photon's polarization: $a, b, c$. Similarly, Bob has three possible axes that he can measure: $a, b, c$. We convince ourselves through experiment, or by looking at the QM predictions, that for a pair of entangled photons, if Alice and Bob both measure the polarizations of entangled photons using the same axis, then they ALWAYS get the same results. (or they always get opposite results, depending on how the entangled photons are produced; let's assume that they always get the same results).

Ok.

Since Alice and Bob ALWAYS get the same results for the same filter orientations, that means that Bob, by measuring his photon, can learn something about Alice's photon.

Same filter orientation, ok.

To Einstein (and whoever P and R were), that means that there must be a deterministic answer to the question: "What would the result be if Alice measured her photon's polarization relative to axis $a$?" It must be a deterministic answer, because Bob can predict it with 100% certainty by measuring his photon's polarization relative to axis $a$.

100% match/mismatch certainty is reserved only for 0 and 90 degrees relative angles. Overall, the answer is rather probabilistic.

So to E, P, and R, there must be, associated with each photon, a triple of numbers $\langle R_a, R_b, R_c\rangle$ telling whether Alice's photon will pass her filter or get blocked by her filter, should she set it at orientation $a$, $b$ or $c$.

It doesn't work with 100% certainty for any arbitrary relative angle. Your premise started based on Alice and Bob having the same filter polarization.

 

[stevendaryl]

100% match/mismatch certainty is reserved only for 0 and 90 degrees relative angles. Overall, the answer is rather probabilistic.

Okay, you still don't quite get the local hidden variables assumption. $a$, $b$ and $c$ are NOT relative angles. They are three different directions in space. For example, $a$ might be the filter orientation in the x-y plane, with the filter slits running in the x-direction. $b$ might be again the x-y plane, with the filter slits running in the y-direction. $c$ might be again the x-y plane, with the filter slits running at a 45 degree angle relative to the x-direction. These are not relative angles.

The deterministic local hidden variables assumption is that there are 8 types of photons produced in the twin-pair experiment:

• Type 1: Passes through filters at orientations $a$, $b$ or $c$.
• Type 2: Passes $a$ and $b$, but blocked by $c$.
• Type 3: Passes $a$ and $c$, but blocked by $b$.
• Type 4: Passes $b$ and $c$, but blocked by $a$.
• Type 5: Blocked by $a$ and $b$, but passes $c$.
• Type 6: Blocked by $a$ and $c$, but passes $b$.
• Type 7: Blocked by $b$ and $c$, but passes $a$.
• Type 8: Blocked by $a$, $b$ or $c$

Since Alice and Bob always get the same answer to the same question, we assume that in every run of the experiment, Alice and Bob get photons of the same "type".

The assumption is that some unknown fraction of the time, call it $P_1$, type 1 photons are produced. Some other fraction of the time, $P_2$ type 2 photons are produced. Etc. So the probabilities, according to the hidden variables theory, don't come in the probability that a SPECIFIC photon will pass through a filter at a specific angle. The probabilities are assumed to be due to the fact that the type of photon, Type 1 through Type 8, is chosen randomly, according to a certain probability distribution.

So that's the hidden-variables theory: EACH photon has an associated "type". The type answers the question "Will this photon pass through a filter oriented at angle $\alpha$?" for each possible value for $\alpha$. It's assumed that in a twin-pair experiment, both Alice and Bob get the same type photon. If Alice's photon passes at angle $a$, and Bob's photon is blocked at angle $b$, then that means that their photons must have been Type 3 or Type 7 (according to the numbering above). If both photons pass, that means their photons must have been Type 1 or Type 2.

So we can reason as follows:

• Since 50% of the time when the filter is at setting $a$, the photon passes, we conclude that $P_1 + P_2 + P_3 + P_7 = \frac{1}{2}$. That's because if it passes through at angle $a$, then it must be a photon of type 1, 2, 3 or 7, according to the list above.
• Since the probability of passing $a$ and also passing $b$ is $\frac{1}{2} cos^2(\theta_{a, b})$, we conclude that $P_1 + P_2 = \frac{1}{2} cos^2(\theta_{a, b})$, where $\theta_{a,b}$ is the angle between $a$ and $b$
• etc.

That's the hidden-variables theory for twin-pair photons. The only problem with it is that the numbers don't work out. There are no solutions to the probabilities $P_1$ through $P_8$ that satisfy all the statistical predictions of quantum mechanics.

 

[RUTA]

This question of this thread has been answered multiple times already, and I will summarize it:

The reason for 3 angles & entanglement to demonstrate why local realism fails: it traces back to Bell's Theorem. Multiple variations on this have been presented, as well as Bell's original paper. Any subsequent answer will simply be yet another version of the same. If you haven't followed what has been presented so far, READ THE REFERENCES instead of asking the same question a different way.

Haha, I was just about to commend you for being so patient and answering the same questions over and over. Someone posted a link to Mermin's 1985 paper "Is the moon there when nobody looks?" https://cp3.irmp.ucl.ac.be/~maltoni/PHY1222/mermin_moon.pdf that answers the original question directly, yet I've seen that same question asked afterwards. Here is an excerpt from p 9

"Alas, this explanation –the only one, I maintain, that someone not steeped in quantum mechanics will ever be able to come up with (though it is an entertaining game to challenge people to try)- is untenable. It is inconsistent with the second feature of the data: There is no conceivable way to assign such instruction sets to the particles from one run to the next that can account for the fact that in all runs taken together, without regard to how the switches are set, the same colors flash half the time. Pause to note that we are about to show that “something one cannot know anything about” –the third entry in an instruction set- cannot exist. For even if instruction sets did exist, one could never learn more than two of the three entries (revealed in those runs where the switches ended up with two different settings)."

He then goes on to give the argument. Note that the title of the paper is making exactly this point, i.e., the third entry -- the one that doesn't get measured (not looked at) -- "cannot exist." So the answer to the title question is "The moon is not there when nobody looks," where "when nobody looks" means "not interacting with anything else in the universe." If someone reads that paper and still doesn't see the answer to the OP, I'm not sure you can help them here, despite your heroic efforts

 

[johana]

Please, stop and review the reference materials first. You are going around in circles. If nothing else, you are making me dizzy.

The origin of the three angles within inequality derivation seems to be a different question than the original question which was about experiments, but I did think they are the same question. Maybe I should open a new thread about the derivation?

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

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

The usual form of the CHSH inequality is:

(1) − 2 ≤ S ≤ 2,

where
(2) S = E(a, b) − E(a, b′) + E(a′, b) + E(a′ b′).

a and a′ are detector settings on side A, b and b′ on side B, the four combinations being tested in separate subexperiments.

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?

 

[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 $E(a, b) − E(a, b') + E(a', b) + E(a' b')$ 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.