Every of the 800 trials starts with the preparation of a pair. When the left going ball is about half way to Alice, Alice randomly rotates a setup, called her “crystal”, either so that it is at an “angle” *a*=0 or *a*=1. Similarly, before the ball going to the right is about to arrive at Bob’s crystal, Bob randomly puts his crystal either at angle *b*=1 or *b*=2; he never needs to pick *b*=0. (If it helps you to see this experiment simulated on a computer, here is a short computer program that does just that.)

After a ball enters a crystal, the ball exits through one of two exits out of that crystal: either exit “0” or exit “1”. This happens according to some set of rules considering the angle of the crystal and what is written on the ball.

Thus, for every pair, we obtain a result **RESULT = _{a},_{b}(A,B)**, where

*a*and

*b*tell us the angles that were chosen, namely

*a*= 0 or 1 and

*b*= 1 or 2, and (A,B) are the values for the exits that Alice’s and Bob’s crystals put their ball through: (A,B) = (0,0), or (0,1), or (1,0), or (1,1).

Example: We prepare a pair of balls, throw one ball to Alice and one to Bob, they pick the angles at random, say *a*=0 and *b*=2, then they look at how the balls exit, say out of exit “1” on both crystals, and so RESULT = _{0},_{2}(1,1) in this example.

*Penrose’s proof: There are more entertaining proofs, like the one by Penrose or the one presented very nicely also on Science2.0 recently. The one I present needs little that you need to take on mere trust. Much of the physics you can verify with polarized sunglasses, while experiments with electron spin are out of the reach of the layperson. We do not assume anything about the complexity of the supposed hidden reality nor just claim that certain probabilities are this or that. Following the proof here, you basically come up with all the important numbers yourself.*

The one important rule of the game is that if both Alice and Bob’s crystal are at the same angle, e.g. *a*=*b*=1, it *never* happens that both see their ball come out of the same exit, say that both emerge from exit 1. In other words, the number N that counts how often the result _{1},_{1}(1,1) occurs is zero:

**N _{1},_{1}(1,1)= 0**. Whenever the angles happen to be the same, the balls will not go both through the exits that have the same label, i.e. also

**N**.

_{1},_{1}(0,0) = 0This rule is to be ensured via the correct preparation of the pairs. If the left going ball for example has an instruction written on it that implies ‘*On encountering a crystal at angle 1, go trough exit 0*’, the right moving** **must** **have something complementary written on it, something that amounts to ‘*On encountering a crystal at angle 1, go through exit 1*.’

L*ocal* realism means that each ball is not only a real object, but it has all the necessary information locally with it. All information necessary to determine which exit to travel through is right here with the ball and the crystal and the decision does not have to consider which angle some other distant crystal is at. Remember, Alice and Bob both pick the angles at random just before the balls arrive, so there is no way in which a ball can know which angle the other ball of the pair has to deal with.

This models an important aspect of the real experiment with photons: Photons travel at the speed of light. Nothing travels faster than light, so the photons, if they were ‘real’ things, must know everything they need to know already and bring this information with them (locally) on their way. The local realist therefore thinks that photons carry* hidden information* that “tells” the photon what to do, somewhat like instructions written on the tennis balls.

Say the left going ball’s instructions prescribe “If a=1 then exit **0**”, meaning that if Alice picks angle 1 rather than 0, her ball goes through exit 0. Now the ball at Bob’s place does not know which angle Alice chooses! Even if it turns later out that she picked *a*=0, she *might* pick *a*=1, and if so, Bob’s ball cannot be also coming out of exit 0 if he happens to also pick *b*=1. Thus, the right ball *must* carry the complementary information “If *b*=1 then exit **1**”.

**Realism:** The physical ongoing depends on “really” existing, determined physical structures like a real hidden book written on a really existing ball ("counter factual definiteness" is a sophisticated term for this).

**Locality:** Nothing at Bob’s place can depend on what Alice did far away as long as the information about what she did cannot have possibly arrived at Bob’s place.

Local realism assumes that all behavior, also that of photons, is determined by locally present facts. Since the photons travel with the speed of light, that information must be predetermined when the photon pair is created, like as if you write something onto tennis balls.

The instructions or ‘hidden variables’ maybe an infinite table or a formula with relatively few parameters. Whatever it is, it is equivalent to prescribing that if Alice for example gets exit 0 when measuring with angle 1, then Bob *must* get exit 1 if he also has his crystal at the angle 1. Thus, part of the information on Bob’s ball *can *be put into the form (B_{0}, B_{1}, B_{2}) which is listing the result for each angle b that Bob may choose. He never chooses 0, so B_{0} can be random as far as we care. For example, if Alice gets a ball (A_{0},A_{1}, A_{2}) with A_{1} = 1, Bob’s must be having hidden variable B_{1} = 0.

Those who argue desperately for local realism are quite religious people. They *believe* and will try anything to win the argument. Some claim that the hidden variables are spheres in four dimensional spaces. I do *not* assume that the hidden variables are ‘merely’ a table with zeros and ones. It could be any kind of complicated table or formula. The instructions may be as cunning as you can imagine and necessitating a super computer on every ball. For example, maybe there is the devious instruction: “*If you encounter Bob’s angle to be b=1, change B _{2} so that it becomes equal to Bob's shoe size*”. All this does not change the fact that the ball must locally have access to information that predetermines the actual outcome B

_{b}= 0 or 1 for all the angles

*b*= 1 or 2 that Bob

*may*choose – otherwise, forbidden coincidences with Alice’s measurements will occur. That is what we mean by that the hidden variables “

*can*be put into the form (B

_{0},B

_{1}, B

_{2})” for all we care.

So we have 800 pairs and each of them has hidden variables that can be very complicated but can also be summarized in the form (A_{0}, A_{1}, A_{2}) (B_{0}, B_{1},B_{2}) as far as we need to consider it. Thus, we can differentiate the total of 800 pairs into eight different cases. All balls may have little serial numbers and are thus otherwise distinguished. Nevertheless, as far as our game here is concerned, each single pair falls into one and only one of the following eight different cases:

N(0) occurrences of (1, 1, 1) (0, 0, 0)

N(1) occurrences of (1, 1, 0) (0, 0, 1)

N(2) occurrences of (1, 0, 1) (0, 1, 0)

N(3) occurrences of (1, 0, 0) (0, 1, 1)

N(4) occurrences of (0, 1, 1) (1, 0, 0)

N(5) occurrences of (0, 1, 0) (1, 0, 1)

N(6) occurrences of (0, 0, 1) (1, 1, 0)

N(7) occurrences of (0, 0, 0) (1, 1, 1)

The numbers add up to the total: N(0) + N(1)+ N(2) + N(3) + N(4) + N(5) + N(6) + N(7) = 800.

Actually, you do not need to write down A_{2} or B_{0}, since they are never measured. However, it is nicer to write numbers for them, because (4B_{0} + 2B_{1} + B_{2}) equals the new label:

For example N(5) counts the cases where (4B_{0}+ 2B_{1} + B_{2}) = (4 + 0 + 1) = 5.

If you even as much as suspect that this convenience may somehow bring in an unnecessary assumption through the back door, please feel free to write it all down instead with the A_{2} and B_{0} entries left empty: “N(0) occurrences of (1, 1, *) (*, 0, 0)” and so on.

Do not assume that all the N(0) and N(1) and so on must roughly equal 800/8 = 100. Realists are not embarrassed to claim that the N(i) may depend on the overall setup in super mysterious ways that deceive us about the *real* nature of quantum evilution, which is much like god who planted those fake fossils to test our faith. The variables are hidden and most of the entries will never be observed. For example, if Bob picks *b*=1, we will never be able to say what B_{2 }actually was. Imagine that balls are burned immediately after exiting a crystal. Even when Alice and Bob meet later and compare their notes, they cannot find out the actual numbers N(i). You personally may like to actually physically prepare 800 pairs of balls in order to prove me wrong. Feel free to write the information onto the balls so that N(0) = 760, N(1) = 40, and N(2) to N(7) are all zero, or whatever else you like!

Every of these eight different classes of pairs is further divided into sub-categories according to the angles that Alice and Bob randomly pick. For example, the N(5) occurrences of (0, 1, 0) (1, 0, 1) are further divided according to the four configurations of possible angles:

N(5) = N_{0},_{1}(5) + N_{0},_{2}(5)+ N_{1},_{1}(5) + N_{1},_{2}(5)

Starting to lose track? OK, slowly: N_{1},_{2}(5) counts all the cases when a pair (0, 1, 0) (1, 0, 1) encountered with its left part [that is (0, 1, 0)] the angle *a*=1 at Alice’s place, leading to the measurement A = A_{1} = 1, and with its right part the angle *b*=2 at Bob’s end, leading to the measurement B = B_{2} = 1. The combined measurement is (A,B) = (1,1). In other words:

**The N _{1},_{2}(5) cases all result in (A,B) = (1,1)!**

Now you should verify for yourself at least a few of the following; just read off the entries as listed; we will later only need the ones in bold fond:

The N_{0},_{1}(0) cases led to (A,B) = (1,0).

The N_{0},_{2}(0) cases led to (1,0).

The N_{1},_{1}(0) cases led to (1,0).

The N_{1},_{2}(0) cases led to (1,0).

The N_{0},_{1}(1) cases led to (1,0).

The N_{0},_{2}(1) cases led to (1,1).

The N_{1},_{1}(1) cases led to (1,0).

**The N _{1},_{2}(1) cases led to (1,1).**

The N_{0},_{1}(2) cases led to (1,1).

The N_{0},_{2}(2) cases led to (1,0).

The N_{1},_{1}(2) cases led to (0,1).

The N_{1},_{2}(2) cases led to (0,0).

The N_{0},_{1}(3) cases led to (1,1).

The N_{0},_{2}(3) cases led to (1,1).

The N_{1},_{1}(3) cases led to (0,1).

The N_{1},_{2}(3) cases led to (0,1).

The N_{0},_{1}(4) cases led to (0,0).

The N_{0},_{2}(4) cases led to (0,0).

The N_{1},_{1}(4) cases led to (1,0).

The N_{1},_{2}(4) cases led to (1,0).

The N_{0},_{1}(5) cases led to (0,0).

**The N _{0},_{2}(5) cases led to (0,1).**

The N_{1},_{1}(5) cases led to (1,0).

**The N _{1},_{2}(5) cases led to (1,1).**

**The N _{0},_{1}(6) cases led to (0,1).**

The N_{0},_{2}(6) cases led to (0,0).

The N_{1},_{1}(6) cases led to (0,1).

The N_{1},_{2}(6) cases led to (0,0).

**The N _{0},_{1}(7) cases led to (0,1).**

**The N _{0},_{2}(7) cases led to (0,1).**

The N_{1},_{1}(7) cases led to (0,1).

The N_{1},_{2}(7) cases led to (0,1).

This enumerates all possible cases exhaustively! The enumeration is according to angles and *hidden variables*. These are **32 **numbers N_{a},_{b}(i) where Alice picked angle a, Bob picked b, and the i is the for us important part of the hidden variables written on the balls.

**N _{0},_{2}(0,1)** is the number of all cases where Alice picked angle

*a*=0 and Bob picked

*b*=2 and the measurement was (A,B) = (0,1). Try to find these in all the cases above. You will

*only*find

**N**and

_{0},_{2}(5)**N**. All others have either different angles or different outcomes (A,B) instead of (0,1). The

_{0},_{2}(7)**16**numbers N

_{a},

_{b}(A,B) count how often out of 800 trials the 16 different results

_{a},

_{b}(A,B) each occurred. Obviously, these 16 numbers and the 32 just listed are related as follows:

N_{0},_{1}(0,0) = N_{0},_{1}(4) + N_{0},_{1}(5)

**N _{0},_{1}(0,1)** =

**N**

_{0},_{1}(6) + N_{0},_{1}(7)N_{0},_{1}(1,0) = N_{0},_{1}(0) + N_{0},_{1}(1)

N_{0},_{1}(1,1) = N_{0},_{1}(2) + N_{0},_{1}(3)

N_{0},_{2}(0,0) = N_{0},_{2}(4) + N_{0},_{2}(6)

**N _{0},_{2}(0,1) = N_{0},_{2}(5) + N_{0},_{2}(7)**

N_{0},_{2}(1,0) = N_{0},_{2}(0) + N_{0},_{2}(2)

N_{0},_{2}(1,1) = N_{0},_{2}(1) + N_{0},_{2}(3)

N_{1},_{1}(0,0) = 0

N_{1},_{1}(0,1) = N_{1},_{1}(2) + N_{1},_{1}(3) + N_{1},_{1}(6) + N_{1},_{1}(7)

N_{1},_{1}(1,0) = N_{1},_{1}(0) + N_{1},_{1}(1) + N_{1},_{1}(4) + N_{1},_{1}(5)

N_{1},_{1}(1,1) = 0

N_{1},_{2}(0,0) = N_{1},_{2}(2) + N_{1},_{2}(6)

N_{1},_{2}(0,1) = N_{1},_{2}(3) + N_{1},_{2}(7)

N_{1},_{2}(1,0) = N_{1},_{2}(0) + N_{1},_{2}(4)

**N _{1},_{2}(1,1) = N_{1},_{2}(1) + N_{1},_{2}(5)**

Try to verify at least two of them. We only need the three in bold fond again:

**I) N _{0},_{1}(0,1)** =

**N**

_{0},_{1}(6) + N_{0},_{1}(7)**II) N _{0},_{2}(0,1) = N_{0},_{2}(5) + N_{0},_{2}(7)**

**III)N _{1},_{2}(1,1) = N_{1},_{2}(1) + N_{1},_{2}(5)**

Alice and Bob compare their laboratory logbooks and confirm that all choices of angles occurred about equally often. Moreover, the balls have not yet arrived when Alice and Bob pick the angles, so the balls cannot bias the random choice of angles. Hence, for example the N(5) occurrences of pairs are further subdivided into four subdivisions of *roughly equal size* according to the four configurations _{a},_{b} of possible angles. In other words, the N_{1},_{2}(5) of equation III) will be roughly equal to N_{0},_{2}(5) in equation II) and they will be both about N(5) divided by four. Whatever you may like to write on the balls, however large or small you may cunningly choose N(5) to be, you cannot influence Alice and Bob choosing angles at random, so N_{1},_{2}(5) ~ N(5)/4 is assured.

The same for the other numbers, i.e.

N_{0},_{2}(7) ~ N_{0},_{1}(7) ~ N(7)/4

N_{1},_{2}(1) ~ N(1)/4

N_{0},_{1}(6) ~ N(6)/4

Hence we know that:

**N _{0},_{1}(0,1)** ~

**[N(6) + N(7)]/4**

**N _{0},_{2}(0,1) ~ [N(5) + N(7)]/4**

**N _{1},_{2}(1,1) ~ [N(1) + N(5)]/4**

Bell’s famous inequality [1] states something totally obvious, namely that N(5) + N(7) alone is **smaller** or at most equal to N(6) + N(7) and N(1) + N(5) all combined. In other words, we expect:

**N _{0},_{2}(0,1) < N_{0},_{1}(0,1) + N_{1},_{2}(1,1)**

In fact, with random numbers you would expect N(6) and N(7) and so on all to be about 800 divided by 8, that is 100, which would bring the inequality to 50 < 50 + 50. (My computer simulation resulted in 53 < 49 + 50.)

Again: **N _{0},_{2}(0,1) being smaller or equal than N_{0},_{1}(0,1) plus N_{1},_{2}(1,1) **is expected; this is pure mathematics and there is no other way in a locally realistic world – period! However, this is not what is observed! If you observe the numbers with actual experiments as explained in “Disproving Local Realism”, then “The important end result is that

**N**alone is by more than 20 occurrences

_{0},_{2}(0,1) ~ 85**larger**than

**N**and

_{0},_{1}(0,1)**N**combined, which only sum to 15 +50 = 65.”

_{1},_{2}(1,1)

Let me discuss the conclusions that can be drawn in another post.

-----------------------------

[1] J.S. Bell, "On the Einstein Podolsky Rosen paradox," *Physics,* **1**(3) 1964 pp. 195-200.Reprinted in J. S. Bell, *Speakable and Unspeakable in Quantum Mechanics*,2nd ed., Cambridge: Cambridge University Press, 2004; S. M. Blinder, *Introduction to Quantum Mechanics*,Amsterdam: Elsevier, 2004 pp. 272-277.

More appetite for reading about that reality does in a sense not exist? Here you go:

Quantum Perspective of the Nonexistence of Light

The World is not woven from Real Stuff

Why There is Something Instead of Nothing

If Schrödinger's Cats All Die, Do the Alive ones go to Hell?

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