Noname manuscript No.(will be inserted by the editor)

A loophole of all ‘loophole-free’ Bell-type theorems

Marek Czachor

the date of receipt and acceptance should be inserted later

Abstract Bell’s theorem cannot be proved if complementary measurements haveto be represented by random variables which cannot be added or multiplied. Onesuch case occurs if their domains are not identical. The case more directly relatedto the Einstein-Rosen-Podolsky argument occurs if there exists an ‘element ofreality’ but nevertheless addition of complementary results is impossible becausethey are represented by elements from different arithmetics. A naive mixing ofarithmetics leads to contradictions at a much more elementary level than theClauser-Horne-Shimony-Holt inequality.

Keywords Bell inequality · logical loopholes · non-Kolmogorovian probability ·non-Diophantine arithmetics

1 Can we understand a theory that does not exist?

When one speaks of a physical system that violates the Bell inequality (Bell, 1964),what one really has in mind is a system that does not satisfy at least one assump-

tion needed for its proof. Some assumptions (locality of measurements, free willof observers, perfect detectors) are physically quite obvious. When it comes topostulating a joint probability measure for all random variables (Vorob’ev, 1962;Fine, 1982), the situation is much less clear. Is it just synonymous to realism,another explicit postulate of Bell? Does it mean that counterfactual probabilitiesare identical to the measurable ones, even in cases where the alternative measure-ments cannot be simultaneously performed in principle, because of purely classicallogical inconsistencies?

The latter is especially visible in the proof of the CHSH inequality (Clauser etal., 1969), which involves the following elementary step:

a0(x)b0(y) + a1(x)b0(y) =(a0(x) + a1(x)

)b0(y). (1)

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2 Marek Czachor

The value b0(y) of the random variable b0, measured by Bob, does not dependon the choice of a0 or a1, measured by Alice. This is precisely the assumption oflocality in the sense of Bell. In a nonlocal case we would have something like

a0(x, b0)b0(y, a0) + a1(x, b0)b0(y, a1)

so we could not further simplify the expression and proceed with the proof. Theproblem is easy to understand and is not a source of controversies.

However, the occurrence of a0(x) + a1(x) in (1) means that the sum of the twofunctions is well defined, which is not always the case in mathematics (think oflnx+ln(−x)). The subtlety is, of course, of a purely local nature. If a measurementof a0 can influence in some way the one of a1, the result of a0(x) + a1(x) might inprinciple depend on the order in which the two measurements are performed, butaddition is order independent. On the other hand, it can be shown on explicit ex-amples (see Czachor (1992) which extends the original argument of Aerts (1986))that in cases where the measurements of a0 and a1 influence each other, the actualprobabilities may differ from the counterfactual ones. One can prove a counterfac-tual inequality, but it may not apply to actually performed measurements. Theattempts of finding a local counterexample to Bell’s theorem (Kupczynski, 2017)are, in my opinion, based on this loophole.

The first version of the present paper was not meant for publication beyondarXiv.org. I just wanted to bring to a wider audience an old result of mine (Cza-chor, 1988), whose importance seemed to grow, but which was virtually unknown.Traces of my old idea could be found in several recent papers (Khrennikov, 2015;Christian, 2015; Masa, Ares & Luis, 2019), but their authors were clearly unawareof my 1988 little contribution. However, once I started to think again of my oldwork, the subject started to live its own life.

I believe the most important new element of the paper is a reformulation of theCHSH inequality in the context of non-Diophantine arithmetic. It naturally leadsto random variables whose addition or multiplication has to be done with greatcare. Since the goal of the Bell theorem is to eliminate all local hidden variabletheories, the ones based on non-standard arithmetics should not be excluded. Non-Diophantine arithmetics and non-Newtonian calculus seem counterintuitive onlyat a first encounter. The are as natural as non-Euclidean geometry, non-Booleanlogic, or non-Kolmogorovian probability.

We shall begin with a simple but rather formal example illustrating the mainidea. Observers have free will, measurements are local, there are no undetectedsignals, and yet a Bell-type inequality cannot be proved for the same reason itcannot be proved in the nonlocal case. I will then proceed with a more subtleexample based on non-Diophantine arithmetics. Here elements of reality in thesense of the EPR paradox are present, but the resulting complementary randomvariables cannot be added of multiplied, so that CHSH-type random variablescannot be automatically constructed.

2 A formal example

Alice and Bob are fans of two football teams, FC Aces and FC Bees, playing onSaturday nights in Acetown and Beetown. Whenever the Aces play in Acetown,Alice travels there and stays in a hotel. The hotels in Acetown are named Aα,

A loophole of all ‘loophole-free’ Bell-type theorems 3

Fig. 1 Random variables a0(x) = ±1 (full) and ln(16x(1−x)/3) (dashed). x is a room numberin hotels A0 or B0. x > 1 do not occur there and thus cannot be chosen by Alice or Bob.

where 0 ≤ α ≤ 1. Alice is free to choose any of them, but typically stays at A0 orA1. Rooms in Aα are numbered by x, α < x < α+ 1. An analogous system worksin Beetown. Unfortunately, the couple cannot travel together so if Alice supportsher Aces in Acetown, then Bob is with the Bees in Beetown. Bob typically stays inB0 or B1, but is also completely free to make his choice. A peculiarity of the hotelsystem is that the same room can belong to several hotels (in the same town), afact related to the structure of the local tax system, the antitrust law, and theunusual architecture typical of the region.

While leaving a hotel a visitor is asked to fill out a short questionnaire, reducingto a single question: Was it a nice visit? The answer is ‘+’ or ‘−’. Experience showsthat it is acceptable to stay in a room whose number is somewhere in the middle ofthe list: α+ 1/4 < x < α+ 3/4. Otherwise the noise made by fans who accompanythe guest team becomes unbearable. Since decibels correspond to a logarithmicscale, the statistics of positive and negative answers in Acetown hotels is welldescribed by the following random variable (Fig. 1 and Fig. 2):

a0(x) = sign(

ln(16x(1− x)/3)), (2)

aα(x) = a0(x− α). (3)

The same model works in Beetown.

Each time Alice and Bob leave their hotels they fill out the forms and producea pair of ‘results’ ± (each run of the experiment produces a pair of results, henceno ‘detection loophole’). During their travels Alice and Bob do not communicatewith each other (‘locality’). Once one knows the room number, the result ‘+’ or‘−’ is uniquely defined (‘deterministic hidden variables’). Alice and Bob are freeto choose the hotels (free will assumption). If Alice and Bob stayed in rooms withnumbers x in Aα, and y in Bβ , then random variables aα(x), bβ(y), and aα(x)bβ(y)are simultaneously defined.

So, can we prove Bell or CHSH inequalities? Let us try. Locality is satisfied,so apparently

a0(x)b0(y) + a1(x)b0(y) =(a0(x) + a1(x)

)b0(y). (4)

4 Marek Czachor

Fig. 2 Random variable aα(x) as a function of x and α. For a given α the domain of thefunction is (α, α+ 1).

Fig. 3 Random variable a0(x) + aα(x) as a function of x and α. Its domain shrinks withgrowing α and becomes empty for α = 1.

However, a look at Fig. 2 and Fig. 3 shows that a0(x) + a1(x) is undefined. Thedomain of the function is empty. It is not a zero function, but a function that doesnot exist at all. The putative proof got stuck already in its first line, in exactlythe same way it gets stuck in the nonlocal case. The same will happen if insteadof CHSH or Bell inequality one will try to prove the GHZ theorem (Greenberger,Horne & Zeilinger, 1989). Then, instead of the sum, one arrives at a product ofthe form a0(x)a1(x), which does not exist either.

A loophole of all ‘loophole-free’ Bell-type theorems 5

There is completely no problem with measuring experimentally an expectationvalue of a0(x)b0(y), so there exists a probability distribution ρ00 such that

〈a0b0〉 =

∫ 1

0

dx

∫ 1

0

dy a0(x)b0(y)ρ00(x, y). (5)

The two functions, a0(x)b0(y) and ρ00(x, y), are defined on the same domain (0, 1)×(0, 1). Similarly,

〈a1b0〉 =

∫ 2

1

dx

∫ 1

0

dy a1(x)b0(y)ρ10(x, y), (6)

with a1(x)b0(y) and ρ10(x, y) defined on (1, 2)× (0, 1).Note that the domains are disjoint, but there are no undetected signals. In

1988, pressed by a referee while resubmitting (Czachor, 1988), I wrote that disjointdomains imply undetected signals (if x 6∈ (α, α+ 1) then aα ‘produces no result’),although already at that time I was not quite convinced that the referee was right.Of course, one way of modeling non-ideal detectors is to use the trick with non-identical domains (this is what Pearle (1970) does) but the inverse implicationis not true. Another argument of the referee was that my random variables werethree-valued: ±1 and 0, where zero occured when there ‘was no result’ (again,this is what Pearle (1970) assumes; the latter interpretation has been recentlychallenged by Christian (2019)). This type of argumenation does not apply here:Logarithm is not ‘zero’ if its argument is negative. The same applies to aα(x) forx 6∈ (α, α+ 1).

The only CHSH-type inequality one can find in our example is the trivial one,

|〈a0b0〉+ 〈a1b0〉+ 〈a0b1〉 − 〈a1b1〉| ≤ 4. (7)

Formally, having four different ραβ , it is not difficult to give examples that saturatethe right-hand side of (7), while maintaining

〈a0〉 = 〈a1〉 = 〈b0〉 = 〈b1〉 = 0. (8)

An example of a theory that reproduces quantum probabilities, and which is im-plicitly based on the trick with disjoint domains of complementary observablescan be found in Pitowsky’s monograph (Pitowsky, 1989). When stripped of ab-stract details (decomposition of a unit interval into infinitely many disjoint non-measurable sets), it turns out that what is essential for the Pitowsky constructionis the disjointness and not the non-measurability. So, instead of decomposing [0, 1]into non-measurable sets, one can replace [0, 1] by [0, 1] × [0, 1] and the effect isthe same.

In the next section I will give further examples of random variables that cannotbe carelessly added or multiplied.

3 A subtler example: Non-Diophantine random variables

One sometimes encounters physical quantities whose arithmetic is not the standardone. Velocities in special relativity are added by means of

β1 ⊕ β2 = tanh(

tanh−1(β1) + tanh−1(β2))

= f−1(f(β1) + f(β2)), (9)

6 Marek Czachor

with f(x) = tanh−1(x). Note that an analogous multiplication,

β1 � β2 = f−1(f(β1) · f(β2))

= tanh(

tanh−1(β1) tanh−1(β2)), (10)

does not seem to occur in the literature. A parallel configuration of resistors is aresistor whose resistance is computed by means of the harmonic addition

R1 ⊕R2 = 1/(1/R1 + 1/R2) = f−1(f(R1) + f(R2)), (11)

with f(x) = 1/x. Here multiplication is unchanged,

R1 �R2 = f−1(f(R1) · f(R2))

= 1/(1/R1 · 1/R2) = R1R2. (12)

The rules (9)–(12) can be regarded as particular examples of the so-called pro-jective non-Diophantine arithmetic, with ‘projection’ f and ‘coprojection’ f−1

(Burgin, 1977, 1997, 2010; Burgin & Meissner, 2017), constructed as follows.Let x, x′ ∈ X and assume there exists a bijection fX : X → R which defines all

the four arithmetic operations in X,

x⊕X x′ = f−1

X(fX(x) + fX(x′)

), (13)

xX x′ = f−1

X(fX(x)− fX(x′)

), (14)

x�X x′ = f−1

X(fX(x) · fX(x′)

), (15)

x�X x′ = f−1

X(fX(x)/fX(x′)

). (16)

Elements of X are ordered: x ≤X x′ if and only if fX(x) ≤ fX(x′). The operationsare commutative and associative, and �X is distributive with respect to ⊕X. Thisis so because the one-to-one map fX makes the arithmetic of X isomorphic to thestandard Diophantine arithmetic of R.

Let us stress that the only assumption we make about X is that its cardinalityequals that of the continuum R (otherwise the bijection would not exist). X can bea highly nontrivial set, for example a fractal (Czachor, 2016; Aerts et al., 2016a,b,2018; Czachor, 2019), or a subset of Rn.

The first two natural numbers, ‘zero’ and ‘one’ are the neutral elements ofaddition and multiplication. Denoting them by 0X and 1X we find

0X = f−1X (0), 1X = f−1

X (1). (17)

Indeed, x ⊕X 0X = x, x �X 1X = x, x X x = 0X, and x �X x = 1X (for x 6= 0X). Anegative of x is Xx = 0X X x = f−1

X(− fX(x)

). An arbitrary natural number nX

is obtained by adding 1X an appropriate number of times,

nX = 1X ⊕X · · · ⊕X 1X︸ ︷︷ ︸n times

= f−1X (n). (18)

Then, in particular, nX ⊕mX = (n+m)X. An n-th power of x,

xnX = x�X · · · �X x︸ ︷︷ ︸n times

= f−1X(fX(x)n

), (19)

satisfies xnX �X xmX = xnX⊕XmX . Finally, fractions are defined by nX �X mX. For

example

(1X �X 2X)⊕X (1X �X 2X) = 1X = (1/2)X ⊕X (1/2)X, (20)

A loophole of all ‘loophole-free’ Bell-type theorems 7

where rX = f−1X (r), now for any r ∈ R.

Probabilities are any numbers pj satisfying 0X ≤X pj ≤X 1X and

⊕Xnj=1pj = p1 ⊕X · · · ⊕X pn = 1X, (21)

which is equivalent to

n∑j=1

fX(pj) = 1. (22)

For example, p1 = 1X �X 7X, p2 = 2X �X 7X, p3 = 4X �X 7X are probabilities.Averages are defined analogously,

〈a〉 = ⊕Xnj=1aj �X pj (23)

= f−1X

(∑j

fX(aj)fX(pj)). (24)

The form (24) is just the well known Kolmogorov-Nagumo average with respect toprobabilities Pj = fX(pj) (Kolmogorov, 1930; Nagumo, 1930; Czachor & Naudts,2002; Naudts, 2011)

As a particular example of fX consider X = R, fX(x) = x2 sgn(x), f−1X (x) =√

|x| sgn(x), 0X = 0, (±1)X = ±1. Multiplication and division are unchanged. Forx ≥ x′ ≥ 0 we get

x⊕X x′ =

√x2 + x′2, (25)

xX x′ =

√x2 − x′2, (26)

x�X x′ = x · x′, (27)

x�X x′ = x/x′. (28)

The average reads

〈a〉 = ⊕Xnj=1aj �X pj = ±

√∣∣∣∑j

f(aj)p2j

∣∣∣, (29)

where ± is the sign of∑j fX(aj)p

2j . It is clear that for this concrete form of arith-

metic the non-Diophantine probabilities play essentially the role of real probabilityamplitudes.

Once one knows how to add, subtract, multiply, and divide, one can constructan entire calculus. Historically, the first example of such a ‘non-Newtonian calculus’was given by M. Grossman and R. Katz (Grossman & Katz, 1972; Grossman, 1979,1983). The idea was rediscovered by E. Pap in his g-calculus (Pap, 1993, 2008)and still later, but in its currently most general form, by myself (Czachor, 2016;Aerts et al., 2018; Czachor, 2019). Certain old problems of fractal analysis (aFourier transform on an arbitrary Cantor set, wave propagation along a Kochcurve) have found simple solutions in the non-Newtonian framework (Aerts et al.,2016b; Czachor, 2019).

Let us now consider a collection of different sets Xk equipped with differentarithmetics, defined by different bijections fXk

: Xk → R. To be concrete, letX1 = R+ (positive reals), X2 = −R+ (negative reals), fX1

(x) = lnx, f−1X1

(r) = er,

8 Marek Czachor

fX2(x) = ln(−x), f−1

X2(r) = −er. There are reasons to believe that arithmetics of

this type are employed by human and animal nervous systems (Czachor, 2020).This is why decibels and star magnitudes correspond to logarithmic scales.

Assume that we have randomly selected a number r ∈ R, say r = 1. This isequivalent to randomly selecting f−1

X1(1) = 1X1

= e ∈ X1 and f−1X2

(1) = 1X2= −e ∈

X2. Following the logic of the Bell theorem we could consider the sum (anyway,1X1

and 1X2are just real numbers),

1X1+ 1X2

= e+ (−e) = 0 = f−1X1

(−∞) = X1∞X1

, (30)

which, if correct, would imply the inconsistent result

e = −1X2= 1X1

⊕X1∞X1

=∞X1= f−1

X1(∞) = e∞ =∞. (31)

The inconsistency occurs because I have recklessly mingled three different arith-metics. The problem is that one is not allowed to naively add elements from X1 tothose from X2 (see the next Section), so the whole calculation is meaningless. Inspite of the fact that 1X1

+1X2is zero, from the perspective of X1 or X2 the expres-

sion is ambiguous. Different arithmetics behave analogously to different maximalsets of jointly measurable quantities in quantum mechanics, an analogy worthy ofa separate study. There are arguments that problems with dark energy and darkmatter may result from an analogous mismatch of arithmetics (Czachor, 2017,2020).

If the readers are not yet convinced, consider the case where X1 is the doublecover of the Sierpinski set discussed in (Aerts et al., 2018), while X2 is the Cantorline from (Czachor, 2016). In the first case, 1X1

= (1, 0)+ belongs to the positiveside of an oriented R2. In the second case, 1X2

= 1 ∈ R. What should be meant bytheir sum?

Very little is known about tensor products of arithmetics, an issue related tothe difficult problem of tensor products of fields (in the algebraic sense of thisword). Even less is known about probabilistic correlation experiments involvingmeasurements mathematically described by different arithmetics. Yet less is knownabout hidden-variable theories involving different arithmetics. The standard Belltheorem tells us virtually nothing about the subject.

So, let us try. But first a digression.

4 Back to quantum mechanics (a digression)

In quantum mechanics random variables are represented by self-adjoint operators.Values of the random variables are represented by eigenvalues of the operators.Commuting operators represent random variables that can be measured simul-taneously. Tensor products of operators represent products of random variablesassociated with independent measurements. In the context of the CHSH inequal-ity one deals with

AB = a · σ ⊗ b · σ, (32)

where a, b are unit vectors in R3. In general, the operator a · σ has eigenvalues±|a|. This is why it is justified to treat AB as a random variable whose values ±1are products of the eigenvalues ±1 of a · σ and b · σ.

A loophole of all ‘loophole-free’ Bell-type theorems 9

The problem with the Bell theorem is that in the course of its proof one assumesthat

A+A′ = (a · σ + a′ · σ)⊗ I, (33)

A−A′ = (a · σ − a′ · σ)⊗ I, (34)

represent random variables whose values are 0, ±2. However, since the eigenvaluesof (33) and (34) are ±|a + a′| and ±|a − a′| the observables are two-valued andnot three-valued, so 0 is not an eigenvalue unless a′ = ±a. A contradiction withquantum mechanics is obtained already at this stage. Nonlocality is not needed.One can also show that a quantum CHSH inequality holds if and only if both[A,A′] and [B,B′] are nonzero. This follows from the fact (Wolf, Perez-Garcia &Fernandez, 2009; Khrennikov, 2019) that the Bell operator

C = A⊗B +A⊗B′ +A′ ⊗B −A′ ⊗B′ (35)

violates the inequality if and only if

C2 = 4 + [A,A′]⊗ [B,B′] > 4 (36)

and this is possible if both commutators are nonzero (effectively, if operators B,B′ with eigenvalues ±1 commute, then B′ = ±B, and two of the four terms in C

cancel out; then the argument on eigenvalues of A±A′ does not work since eitherA or A′ is missing). The conditions [A,A′] 6= 0 and [B,B′] 6= 0 are, of course,of a local nature. The argument based on noncommutativity does not have anexact analogue in the arithmetic framework and is a peculiarity of the quantumformalism.

A similar difficulty will nevertheless occur.

5 Arithmetics of Alice, Bob, and Eve

Let us begin with the simplest problem of adding two averages. Assume Alicemeasures two random variables, a and a′, with values in arithmetics A and A′,respectively. After n measurements of a and n′ measurements of a′ the results are(a1, . . . , an) ⊂ A× · · · × A, (a′1, . . . , a

′n′) ⊂ A′ × · · · × A′. She computes averages,

〈a〉 = ⊕Anj=1aj �A nA, (37)

〈a′〉 = ⊕A′n′

j=1a′j �A′ n′A′ . (38)

What should be meant by their sum, even if the experimental samples have equallength n = n′? Can we write 〈a〉+ 〈a′〉 = 〈a+ a′〉? In general, no.

The problem remains even if A ⊂ R and A′ ⊂ R. The random variables arereal, but rules of their addition and division are different. More strikingly, even ifaj = rA = f−1

A (r), a′j = r′A′ = f−1A′ (r), so that r ∈ R behaves analogously to an

‘element of reality’ common to aj and a′j , the sum of aj and a′j is as ambiguous as(31). This seems to be the first example (outside of quantum mechanics) where onecannot add two random variables even though they may be regarded as possessinga common element of reality in the sense of the EPR argument.

Let us go further. Let Alice work with arithmetics A or A′, and Bob with Bor B′. They perform binary measurements with the results 1X or X1X, where

10 Marek Czachor

X is the corresponding arithmetic. Although numbers can be consistently addedor multiplied only within a single arithmetic, results from one arithmetic can beuniquely translated into another arithmetic by means of the isomorphism with R.For example, if

x⊕BA x′ = f−1

A(fA(x) + fB(x′)

), (39)

xBA x′ = f−1

A(fA(x)− fB(x′)

), (40)

x�BA x′ = f−1

A(fA(x) · fB(x′)

), (41)

x�BA x′ = f−1

A(fA(x)/fB(x′)

). (42)

then

2A ⊕BA 2B = f−1

A(2 + 2

)= 4A, (43)

2A ⊕AB 2B = f−1

B(2 + 2

)= 4B. (44)

In the logarithmic example, A = R+, B = −R+, fA(x) = lnx, f−1A (r) = er,

fB(x) = ln(−x), f−1B (r) = −er, we will of course conclude that ‘two plus two

equals four’, but the result looks as follows:

2A ⊕BA 2B = e4 = 4A = 2A �B

A 2B, (45)

2A ⊕AB 2B = −e4 = 4B = 2A �A

B 2B. (46)

Alice and Bob would agree that they have obtained identical results, namely ‘four’,but Eve might disagree. In her opinion their results were exactly opposite: e4 and−e4.

The other two arithmetics needed for a CHSH random variable are, in principle,completely independent. For example, B′ = R+, A′ = −R+, fB′(x) = (lnx)3,

f−1B′ (r) = er

1/3

, fA′(x) =(

ln(−x))3

, f−1A′ (r) = −er

1/3

.Now, consider the following arithmetic operations

a⊕ABX b = fABX

−1(fA(a) + fB(b)), (47)

aABX b = fABX

−1(fA(a)− fB(b)), (48)

a�ABX b = fABX

−1(fA(a) · fB(b)), (49)

a�ABX b = fABX

−1(fA(a)/fB(b)), (50)

where a ∈ A, b ∈ B, but X and the bijection fABX : X → R are yet unspecified.Recall that for r ∈ R we denote rA = f−1

A (r), etc. The product has the followingproperties:

xA �ABX yB = fABX

−1(xy) = (xy)A �ABX 1B = 1A �AB

X (xy)B

= (xA �A yA)�ABX 1B = 1A �AB

X (xB �B yB), (51)

xA �ABX (yB ⊕B zB) = fABX

−1(x(y + z))

= fABX−1(fA(xA)fB(yB) + fA(xA)fB(zB)

)= fABX

−1(fA ◦ f−1

A(fA(xA)fB(yB)

)+ fB ◦ f−1

B(fA(xA)fB(zB)

))= fABX

−1(fA(xA �B

A yB)

+ fB(xA �A

B zB))

= (xA �BA yB)⊕AB

X (xA �AB zB) (52)

A loophole of all ‘loophole-free’ Bell-type theorems 11

Since

xA �ABX yB = yA �AB

X xB, (53)

we immediately also get

(xA ⊕A yA)�ABX zB = (xA �B

A zB)⊕ABX (yA �A

B zB) (54)

These are all the properties needed for a tensor product. Still, how should we addaA �AB

X bB ∈ X and aA �AB′

X b′B′ ∈ X? The most natural addition is the one intrinsicto X,

(aA �ABX bB)⊕X (aA �AB′

X b′B′) = f−1X

(fX(aA �AB

X bB) + fX(aA �AB′

X b′B′))

(55)

= f−1X

(fX ◦ fABX

−1(ab) + fX ◦ fAB′

X−1(ab′)

). (56)

It cannot be further simplified, since fX ◦ fABX−1 and fX ◦ fAB

′

X−1 are in general

different functions, so we are back to our original argument. Bell’s theorem cannotbe proved if inconsistent sets of data have to be combined in a nontrivial way.

Let us now consider two extreme cases: (i) X = R, fX(x) = x, and (ii) a

nontrivial X but fAB′

X = fABX = fA′B

X = fA′B′

X = fX independent of the arithmeticsof Alice and Bob. In the first case the CHSH random variable

fABR−1(ab) + fAB

′

R−1(ab′) + fA

′BR−1(a′b)− fA

′B′

R−1(a′b′) (57)

cannot be further simplified in general. In the second case the random variable isf−1X (±2), but in general differs from ±2. The available structures are here much

richer than in the standard approach. The slogan ‘Bell’s theorem is a mathematicaltheorem’ appears here in all its weaknesses and limitations.

Finally, thinking in quantum cryptographic terms, how to formulate a test foreavesdropping? The intermediate arithmetics that define products of the resultsobtained by Alice and Bob involve additional arithmetics associated with the pro-cess of combining the data. This additional arithmetic may, in principle, involvean eavesdropper (or hacker) Eve.

6 Remarks on non-classical correlations beyond quantum mechanics

Examples where Bell-type inequalities are violated beyond quantum mechanicscan be found in the literature. Which loopholes of the proof are there employed?

In the model invented by Diederik Aerts (Aerts, 1986), after the first mea-surement of a random variable A a hidden variable λ, an argument of B(λ), isfound with nonzero probability in a state whose probability would be zero if Ahad not been measured. The measurement of A nontrivially and ‘actively’ changesthe state of B. This is not the standard conditioning by ‘getting informed’. Oneshould not confuse this type of conditioning with the one that leads to the Borelparadox (Czachor, 1992). The model avoids the Borel paradox for the price ofnonlocality in the sense of Bell. The example explains why nonclassical probabil-ities can occur in classical systems, but it does not contradict the conclusion ofBell. The other examples studied by the Brussels group, such as the two connectedvessels of water (Aerts, 1982), ‘Bertlmann wearing no socks’ (Aerts & Sassoli de

12 Marek Czachor

Bianchi, 2018), or Sven Aerts’ mechanical model that simulates a non-local PRbox (Aerts, 2005), are also nonlocal in this sense. In all these cases a state λ inB(λ) is actively influenced by the choice of A or A′ in the correlated system, andthe mechanisms responsible for these influences are explicitly described. Informa-tion about the choice made by Alice has enough time to reach Bob, even thoughone cannot use this information to communicate. Action at-a-distance is present,but there is nothing ‘spooky’ about it.

In another class of models the results A and B are created at the very momentthe product AB is computed. This is what happens in various cognitive-scienceexamples where B becomes a context for A (Aerts & Gabora, 2005; Aerts, Gabora& Sozzo, 2013); see also the polemic (Dzhafarov & Kujala, 2013; Hampton, 2013;Aerts, 2014). Contextuality of cognitive models is extensively studied, both theo-retically and experimentally, also in the works of Khrennikov (2010). Exactly thesame effect occurs in the rock-paper-scissors game, which is ubiquitously presentin many natural systems (Friedman & Sinervo, 2016). ‘Rock’ in itself does nothave a value +1 (win) or −1 (lose). However, the pairs (rock,paper) = (−1,+1)and (rock,scissors) = (+1,−1) are uniquely defined. All these models are nonlocalin the sense of Bell, but there is nothing unphysical about their nonlocality.

The controversial model by Christian (Christian, 2014) is also explicitly basedon the fact that there is a nontrivial relation between A, B, and AB. Tensorproduct is there replaced by geometric product, a possibility considered in similarcontexts also by other authors (Doran et al., 1996; Doran & Lasenby, 2013; Aerts& Czachor, 2007; Czachor, 2007; Aerts & Czachor, 2008)). Operators such asPauli matrices are interpreted in the standard geometric-algebra way, namely as amatrix representation of an orthonormal basis in R3 (Hestenes, 1966), so the wholecalculation can be performed in purely geometric terms. On a philosophical side,there is some similarity between the way Christian treats binary variables with theone I have described above in the context of 1X and X1X. For Christian, ‘plus’and ‘minus’ denote some kind of abstract handedness, defined at a quaternioniclevel. However, in spite of all his efforts, I believe the probability interpretation ofmultivector random variables is not yet sufficiently understood, a fact obscuringthe physical meaning of the result.

What I find most intriguing and worthy of further studies is the link with theoriginal model of Pearle (1970). Pearle’s hidden variable is essentially an element ofthe rotation group SO(3), whereas Christian works with SU(2). In both cases thehidden variables are classical but noncommutative (rotations in three dimensionsdo not commute). The random variables of Pearle are three-valued (0 and ±1),where 0 is interpreted as no detection. So, Pearle’s model formally looks like a spin-1 system. The model of Christian is essentially a spin-1/2, where the 0 eigenvalueis missing. Does it mean, as suggested by Christian, that undetected signals canbe in principle eliminated if one considers hidden-variables that belong to thecovering space of the Pearle model? In principle, it cannot be excluded becausemany structures occurring in quantum information theory have their space-timeanalogues (e.g. the two-qubit Bell basis is formally identical to the Minkowskitetrad (Czachor, 2008)). This whole new research area is basically unexplored.

A loophole of all ‘loophole-free’ Bell-type theorems 13

7 Summary

Bell’s theorem cannot be proved if complementary measurements have to be repre-sented by random variables that cannot be added or multiplied. Complementaritymeans here that there exists a fundamental logical reason why one cannot performthe two measurements simultaneously. In quantum mechanics one represents com-plementarity by noncommutativity. Alternatively, two quantum random variablesare complementary if eigenvalues of their sum are not the sums of their eigenvalues.There is no proof that analogous properties cannot be found in non-quantum sys-tems, a fact of fundamental importance for proofs of security in quantum cryptog-raphy. For example, random variables with values in non-Diophantine arithmeticsare analogous to quantum observables belonging to different maximal sets of si-multaneously measurable quantities. Similarly to eigenvalues of non-commutingoperators, it is not allowed to naively add or multiply elements from differentnon-Diophantine arithmetics, even if the arithmetics are defined in subsets of R.Non-Diophantine arithmetics provide the first examples of non-quantum modelswhere elements of reality in the sense of the EPR paradox are present, but theresulting complementary random variables cannot be added or multiplied, so thatCHSH-type random variables cannot be automatically constructed. What I haveshown in the paper is just a very preliminary study that should evolve into a largerresearch project.

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