WikiJournal of Science/A card game for Bell's theorem and its loopholes/Impossible correlations

Here we present a more conventional explanation of Bell's inequalites that begins with an inequality involving probabilities, and ends with correlation coefficients that seem "impossible".

Bell's inequality: Venn diagram
In The car and the goats it was argued that each player gains no advantage by changing strategies upon seeing their own question card. In other words, their best strategy appears to involve agreeing on all three answers before they part company. The fact that only two answers (even/odd) are available places a limit on how many answers are the same, and how many are different. For example, while it is possible for them to give the same answer (e.g., "even") to each question, at least two answers must be the same. They cannot give one answer to hearts, another answer to spades, and a third entirely different answer to diamonds. Hence there are only four possibilities: Either all three are the same, or one (and only one) answer is different from the other two.

Denote the three question (suits) by the letters $$\{a,b,c\}$$, and let $$P(i,j)$$ be the probability that the answer to $$i$$ is the same as the answer to $$j$$. Imagine that Bob and Alice keep track of their plans for each round, and place a small dot into the appropriate area $$(\alpha,\beta,\gamma,\delta )$$ of the Venn diagram (see quiz below). As you sum the probabilities, each dot in the center $$\Delta$$ will be counted three times. Therefore:


 * $$P(a,b)+P(b,c)+P(c,a)\ge 1$$

If it were possible to "count the dots" (i.e. know hypothetically answers to all possible questions), this inequality could be formally proven to hold for all experimentally obtained values of these probabilities. But this knowledge is not available to us due to the uncertainty principle. We shall later use this trick of "counting" actual events in a hypothetical experiment to derive the CHSH inequalities.

We shall use the dagger to denote cases where the answers are different: $$P(i,j^\dagger)$$ is the probability that the answers given by Alice and Bob are different. Obviously,
 * $$P(i,j^\dagger)+P(i,j)= 1$$, and $$P(i,j)=P(j,i).$$

The quantum correlation, which resembles but is distinctly different from the Pearson correlation coefficient is:
 * $$C(a,b)=P(a,b)-P(a,b^\dagger)=2P(a,b)-1$$

This sign reversal rule will be used later to generate other forms of Bell's inequality:
 * $$C(a,b)=C(a^\dagger,b^\dagger)=-C(a,b^\dagger)=-C(a^\dagger,b)$$

Substitution into the inequality for the three probabilities yields,


 * $$-C(a,b)-C(b,c)-C(c,a)\ge 1\,.$$

This version of Bell's inequality is appropriate for three symmetric polarization angles introduced in A card game for Bell's theorem and its loopholes, since all three correlations are negative. Typical experimental configurations do not orient the measurement angles in this fashion, and a different version of this inequality is more convenient. This will be addressed later in this essay.

Quiz

 * See QB/d Bell.Venn for a quiz that students and instructors can use and improve.

Quizbank is an effort to set education free by creating a large and open bank of exam questions. (Nearly) free education is achieved by writing software that allows schools and colleges to create exams from this bank. Students who are not good test takers (or have excess talent and/or enthusiasm) may may instead establish competence by developing quizzes or essays (see sample work by Annie Anonymous.)

One such quiz for this project is based on the assumption that Alice and Bob play five rounds, each time recording their intentions by placing a blue dot in one of the regions (&alpha;,&beta;,&gamma;,&Delta;) as shown in the figure. The measured probability that suits i and j receive the same answers is P(i,j), and C(i,j) is the quantum correlation between the answers. The following questions illustrate three questions that will be on the quiz.


 * 1) Calculate the measured value of P(&spades;,)
 * 2) Calculate the measured value of P(&spades;,) + P(&spades;,) + P
 * 3) Calculate the measured value of C(&spades;,) = &equiv; P(&spades;,) &minus; 2

If the first figure in the gallery shown above is selected, the answers to questions (1,2,3) are (3/5, 7/5, 1/5), respectively. The first 12 questions at QB/d Bell.Venn use only the two Venn diagrams shown above. But eight other images currently exist at this Venn diagram gallery. Students with easily learn to create other quizzes, including quizzes that introduce Bayes' theorem. A professor with access to the python code conceptual.py can convert ordinary text into a Quizbank quiz. The python code also allows instructors to create a "secret" cache of quiz questions that are not available to the public.

Bell's original inequality
To recover the customary version of Bell's inequality, we begin with this result from the Venn diagram:


 * $$-C(a,b)-C(b,c)-C(c,a)\ge 1$$

It is possible to replace one or more of the three variables $$(a,b,c)$$ by their compliments $$(a^\dagger,b^\dagger,c^\dagger)$$. This can be done by physically rotating the measurements polarization by 90&deg;, or more conveniently by redefining one of relations between "pass" and "even" in the original figure that defined the card game. It is sufficient to make only one such substitution, replacing $$b$$ by $$b^\dagger$$ and using the sign reversal discussed above:


 * $$C(a,b^\dagger)+C(b^\dagger,c)-C(c,a)\ge 1$$

Note that it is impossible to reverse the sign of all three correlations by any series of such substitutions.

It is worthwhile to ponder the "derivation" of an inequality that is routinely violated in realty. We have implicitly assumed Alice and Bob made a tentative plan while they were together, and that they did not change it subsequently because doing so would eventually cause them to give different answers to the same question. It would be considered cheating for Alice and Bob to collude with the referee in a classroom demonstration of the card game. In the next section we explore a simple case of quantum entanglement, which leads to the speculation that Alice, Bob and the referee might be able to "guess" each others actions with some degree of success. Is it possible that all the atoms in the room are somehow "entangled" so that the parent atom emits photons that match the measurement polarization angles? That mechanism is certainly consistent with how quantum systems must be modeled. It is also analogous to the partners in a classroom setting using their intuition to guess what choices the referee is likely to make, and use of such intuition about another person would certainly not be called "cheating". While causality prevents Alice and Bob from seeing each other's questions, or communicating during the experiment, there is nothing to prevent them from looking back to the Big Bang for clues as to how future events will evolve. But what makes this "spooky" is that photons from distant stars have been used to set the measurement polarization angles. Is it possible that we live in a universe in which certain outcomes are destined to occur?

Preliminary attempt at CHSH


In actual experiments, each electro-optic modulator switches between only two polarization angles. In the couples version of the card game, this is analogous to Alice and Bob each knowing that a certain question card will not be played, as illustrated in the figure to the left. In order to win, the partners must play different (even/odd) answer cards to the same suit, but play the same (even/odd) answer if they see the same suit. Here, they see the same suit only if both questions are spade &spades;.

This game represents an attempt to model a Bell's theorem experiment that ultimately fails because the partners are able achieve a average score of zero, matching the behavior of entangled photons when subject to three symmetric polarization measurements. Nevertheless, the game has the merit of giving instructors an excuse to teach students how to model all possible game outcomes on a spreadsheet using Boolean algebra. Also, Bojić did succeed in modifying this game so that it does achieve inequality that cannot be violated without [[w:Quantum computing|quantum computing].

Spreadsheet analysis
Microsoft Excel can accomplish analyze this version of the game using modulus-2 arithmetic and the function Mod(x,2)&equiv;MOD2(x). These operations should be familiar to students as the consequences of adding or multiplying even or odd integers. The results from using the spreadsheet in this manner are shown in the table to the right.

The nine cells in the upper-left corner associates (x,y) with the question posed to (Alice,Bob) as follows: For Alice, x=1 if the question is &spades;, while x=0 if the question is. For Bob y=1 also denotes the &spades;, but y=0 denotes the. It should be noted that the CHSH inequality does not require that Alice's &spades; and Bob's &spades; correspond to the same angle.

The last four columns to the right of Alice and Bob depict the possible combinations of questions available to the referee (, &spades;, &spades;, &spades;&spades;). In each case, the top row represents the modulus-base2 of the product, commonly written using the Boolean AND: xy &equiv; MOD2(xy) &equiv; $$x\wedge y$$.

As discussed in "A card game for Bell's theorem and its loopholes", absent any "psychic" abilities, humans gain no advantage by changing their strategy after one of them has seen a question card. The choices that the team can make are listed by the base-2 integers from 0-15 (0000-1111) that occupy the bottom 16 rows of the table. As shown in Fig. 1, "even" or "odd" and odd card numbers are associated with 0 and 1, respectively. Whether the team "wins" or "loses" is ascertained by first performing the Boolean XOR operation on the answers a and b: a+b &equiv; MOD2(a+b) &equiv; $$a\oplus b$$

This table shows the best 8 out of the 16 stragegies, each with a 3/4 probability of winning. All other strategies are inferior to those shown here because each of the other eight contains less than a 75% chance of winning, and none of them offer a chance to win that is missing from the list above. The asterisks in the center show player strategies in which they never incur the penalty for showing different answers to the same suit. We conclude that humans can play this game as well as can entangled particles. Nothing spooky here!

Proving the CHSH inequality
The table to the right was copied from an Excel spreadsheet. The first four columns show all possible combinations of answers that Alice and Bob might plan, given that each faces one of two possible polarization orientations. After a space, the cross-correlations for each pair of possible questions is calculated. The column labeled &Sigma;+ sums all four correlations, while the &Sigma;&minus;column sums the first three but subtracts the fourth (a'b') correlation. By labeling the answers as &minus;1 and +1 Instead of "even/odd", students can calculate all 64 possible correlations in the four columns automatically by using the spreadsheet to multiplying the corresponding pairs of answers (a,a',b,b'). While the previous version of the CHSH card game made for a useful student activity with Boolean algebra and a spreadsheet, it failed to recover the "impossible" correlations that entangled particles manage to achieve. This derivation of the CHSH inequality achieves that aim, without resorting to calculus

The Venn diagram proof of the original Bell's inequality described above labeled the three orientations as $$(a,b,c)$$ with the understanding Alice and Bob had pre-arranged to give the same (or opposite) answers to each of the three questions. This time each partner faces only one of two possible questions, which we shall label as primed and unprimed, reserving the letter to denote which partner receives which answer: Alice sees a question selected from $$\{a,a'\}$$ while Bob sees one selected from $$\{b,b'\}$$. In contrast to the original original Bell's inequality, the responses to the same orientations are neither required to be fully correlated nor anti-correlated. .

As with the Venn diagram proof shown above, we need only keep track of whether the answers are the same or different. And, we also make the "hidden variable" assumption that Alice and Bob are prepared to give specific answers to both question cards (even though they are required to submit an answer to only one). The quiz mentioned in a previous section used dots and a Venn diagram to keep track of all the answers the partners were prepared to give. For the CHSH inequality, we shall use a spreadsheet to investigate all possible combinations of answers. From the &Sigma;&minus; column, we observe that,

$$C(ab)+C(a'b)+C(ab')-C(a'b')=\pm2\,.$$

This does not imply that the experimentalist will always obtain &plusmn;2 when the observed quantum correlations are measured and summed to create an average. We may make no assumptions regarding how often Alice and Bob select from each of the 16 strategies available to them. But, if the referee randomly selects the questions, then the results will be a random selection among the four columns (ab, a'b, ab', a'b') for whichever row that Alice and Bob select.

Suppose for example, that Alice and Bob select a strategy for which the linear combination of correlations, &Sigma;&minus; = +2 with a probability of $$P$$, and therefore the probability of selecting a strategy that yields a &minus;2 correlation is $$(1-P)$$. Then if the selection of two questions by the referee is random, the average linear combination of correlations is:


 * $$\overline{C_{ab}+C_{a'b}+C_{ab'}-C_{a'b'}} = 2P -2(1-P) = 4P-1$$

Since $$0\le P\le 1$$, we have,


 * $$\left|\overline{C_{ab}+C_{a'b}+C_{ab'}-C_{a'b'}}\right| =

\left| \overline{C_{ab}}+\overline{C_{a'b}}+\overline{C_{ab'}}-\overline{C_{a'b'}} \right| \le 2\,,$$

and the CHSH inequality is recovered. The linearity of measured quantum correlations is essential to this proof. The correlation for N measurements is obtained by averaging $$C_{ij}$$ for all occasions in which any given pair of orientations are selected for measurement.

Of course, if the referee does not select equally among the four columns (ab,a'b,ab,ab'), or if the partners could otherwise anticipate which combination of questions will occur, this inequality could be violated by a proper selection among the 16 rows available to the partners.