# Understanding Bell's inequality

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1. Feb 5, 2016

### entropy1

I am not sure if I am allowed to ask this, but here's my shot:

I find all the explanations of Bell's theorem summed up here, very different in interpretation and also (for me) incomprehensible. I have these simple questions:

How does the Bell inequality, stated as N(A, not B) + N(B, not C) ≥ N(A, not C), relate to the position of the detectors? Do A, B and C represent binary values or invertable continuous values? (What do they mean??) What is the most comprehensible source discussing Bell's inequality making use of this formula? (I fully understand the formula itself, but not its significance)

Put differently: in a Venn-diagram, what do A, B and C represent?

Thanks.

Last edited: Feb 5, 2016
2. Feb 5, 2016

### mathman

In a Venn diagram A, B, C are events. Fundamentally Bell's inequality is a mathematical theorem.

3. Feb 5, 2016

### entropy1

Such as measuring a value?

4. Feb 5, 2016

### Jilang

The areas in a Venn diagram made up of three areas:A, B and C which overlap to some degree.

5. Feb 5, 2016

### Mentz114

Here is yet another simple derivation of the correlation limit (attached). Using Venn diagram logic. I believe there is a simpler arithmetic version but you're probably fed up with these proofs !

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6. Feb 5, 2016

### Strilanc

A simple way to think of the Bell inequalities is as a game.

Two separated players will each flip a coin. Then, based on the coin flip, they each pick "yes" or "no". Their goal is for the choices to disagree when and only when the coins both land heads. They want to satisfy $x \land y = a \oplus b$, where $x$ and $y$ are the respective random coin flips while $a$ and $b$ are their respective choices.

Classically, the best strategies for this game win 75% of the time. Doing better requires cheating (e.g. communicating during the game, rigging the coin flips so they're not random, etc). This isn't too hard to intuit: $a$ can depend on $x$ but not $y$; similarly $b$ can depend on $y$ but not $x$. So really you're trying to pick $f$ and $g$ that maximize how often $x \land y = f(x) \oplus g(y)$ is satisfied. Probabilistic strategies are just weighted averages of the deterministic strategies, and the deterministic strategies all win 25% or 75% of the time, so classical strategies are stuck in the [0.25, 0.75] range.

But you can use quantum-entanglement-based strategies to push your win rate past 85%.

7. Feb 5, 2016

### DrChinese

"A" would be the + outcomes of a polarization (spin) measurement at some angle a. "Not A" would be the - outcomes (note that + and - are arbitrary labels, you could also label it 1 and 0, or H and V).

So "B" would be the + outcomes of a polarization (spin) measurement at some angle b. And so on.

With an entangled pair, you can only measure 2 of the 3 angles at a time. So you would need to run a suitable number of trials of the 3 pairs: N(A, not B), N(B, not C) and N(A, not C). That would give you averages that can be plugged in.

For a classical example: You have a room with 100 people in it. The A group are men, Not A are women. The B group are parents, Not B are people with no children. C are right handed, Not C are left handed. So we substitute into the formula N(A, not B) + N(B, not C) ≥ N(A, not C) and get:

[1] N(men, not parents) + N(parents, left handed) >= N(men, lefthanded)

The classical version will always be true, no matter how you select the people to go into the room. That is because the above inequality, with suitable rearrangement, becomes:

[2] N(men, not parents, right handed) + N(women, parents, left handed) >= 0

I can explain how I got from [1] to [2] if you like.

8. Feb 5, 2016

9. Feb 5, 2016

### entropy1

I have made a Venn-diagram for the equation:

N(A, not B) + N(B, not C) N(A, not C)

10. Feb 5, 2016

### DrChinese

For a classical example: You have a room with 100 people in it. The A group are men, Not A are women. The B group are parents, Not B are people with no children. C are right handed, Not C are left handed. So we substitute into the formula N(A, not B) + N(B, not C) ≥ N(A, not C) and get:

[1] N(men, not parents) + N(parents, left handed) >= N(men, lefthanded)

The classical version will always be true, no matter how you select the people to go into the room. That is because the above inequality, with suitable rearrangement, becomes:

[2] N(men, not parents, right handed) + N(women, parents, left handed) >= 0

How did I get from [1] to [2]?

Using ordinary classical assumptions, I expand as follows:

[3]
N(men, not parents) = N(men, not parents, left handed) + N(men, not parents, right handed)
N(parents, left handed) = N(men, parents, left handed) + N(women, parents, left handed)
N(men, left handed) = N(men, parents, left handed) + N(men, not parents, left handed)

Make sense? Using those identities, I substitute into the original [1] and get:

[4]
N(men, not parents, left handed) + N(men, not parents, right handed) + N(men, parents, left handed) + N(women, parents, left handed)
>=
N(men, parents, left handed) + N(men, not parents, left handed)

I notice that the 2 right hand side terms both also appear on the left hand side as well. So I eliminate them from both sides. This gives me:

[5] N(men, not parents, right handed) + N(women, parents, left handed) >= 0

QED. Now, keep in mind that the classical assumption I made in [3] is subject to attack in the quantum world. Precisely because classical logic fails there, and the inequality is not valid in actual experiments.

11. Feb 5, 2016

### entropy1

Does the + vs the - mean pass vs not-pass the filter? (in case of polarisation)

And how does it apply to electron spin?

12. Feb 5, 2016

### DrChinese

The result formula below can be evaluated for angle settings instead of attributes of people.

[5] N(men, not parents, right handed) + N(women, parents, left handed) >= 0

If you pick almost any 3 angles randomly, there is an inequality similar to the above that would always be TRUE for a classical world; but is FALSE in an actual Bell test. Essentially:

Classical (the sum of expectation values of any 2 permutations must be greater than or equal to zero) :
[6] N(A, ~B, C) + N(~A, B, ~C) >=0

Quantum (yields a negative probability):
[7] N(A, ~B, C) + N(~A, B, ~C) < 0

Most people don't refer to Bell's Inequality in this fashion, but it is just as correct. The CHSH inequality is usually used for Bell tests instead.

13. Feb 5, 2016

### entropy1

How?

14. Feb 5, 2016

### DrChinese

Actual Bell tests usually use photons going to polarizing beam splitters, which have 2 outputs. They can be labeled any way you like. Not really sure how you would do it with electrons to be honest.

15. Feb 5, 2016

### entropy1

I don't know what 'A' and 'not A' means in terms of an experiment. Can anyone tell me?

For instance: is 'A' the set of photons passing the filter at setting A, and 'not A' the set of photons not passing the filter at setting A?

16. Feb 5, 2016

### DrChinese

I can give you one, it's a bit difficult to explain the mapping. Pick angles to be 0, 120 and 240 degrees for A, B and C. Label the outputs of the beam splitters as + and -. Then adapting to the original [1] we get:

[8] N(A+, B+) + N(B-, C-) >= N(A+, C-)

The quantum expectation value is for a match (++ or --) is cos^2(theta) where theta is the angle difference. Assuming the ++ case and the -- cases are equally likely, each would be half the total. For our 3 angles, the theta is 120 degrees for any pair. Cos^2(120 degress) is .25. The quantum expectation value is for a mismatch (-+ or +-) is sin^2(theta) which is .75. This evaluates to the quantum expectation values as follows :

[9] .125 + 125 >= .375

And the inequality is violated. I know this last step was a big jump, sorry bout that. Keep in mind that there is no quantum expectation value for something like:

[10] N(A+, B+, C+).

Work through the formulas a bit and hopefully you will get the idea.

17. Feb 5, 2016

### Strilanc

The inequality experiment that Dr. Chinese is explaining (which is different than the game I described, though they're examples of the same basic thing) is explained very clearly in this youtube video.

18. Feb 6, 2016

### entropy1

Last edited: Feb 6, 2016
19. Feb 6, 2016

### entropy1

Suppose hidden variables would exist. Would that mean that there would be a direct relation between the value of the variable and the value of the measurement?

It seems to me that the value of the measurement has to depend at least also on the factors that the beamsplitter or the polarizer represent...!

Last edited: Feb 6, 2016
20. Feb 6, 2016

### entropy1

I think I've got it: if the outcome of the measurement would not depend on the on the orientation of the filter, that would be just stupid. So it does. However, we can change the orientation of, say, filters FA and FB at the last moment, and still there is a correlation between the measurements. This would imply superluminal communication or something of the sort, since the measurements are now local.

So we could assume the outcomes of the measurements do not depend of the orientations of the filters, but rather on the hidden variables in a one-to-one fashion. And that is ruled out by the Bell inequality.

QED.

Last edited: Feb 6, 2016