Encrypt me! A game-based approach to Bell inequalities and quantum cryptography
EEncrypt me! A game-based approach to Bell inequalities and quantum cryptography
Andrea López-Incera, Andreas Hartmann, and Wolfgang Dür Institute for Theoretical Physics, University of Innsbruck, A-6020 Innsbruck, Austria ∗ We present a game-based approach to teach Bell inequalities and quantum cryptography at highschool. The approach is based on kinesthetic activities and allows students to experience and discoverquantum features and their applications first-hand. We represent quantum states by the orientationof students, and mimic quantitative random behaviour and measurements using dice and apps.
I. INTRODUCTION
Quantum technologies are at the brink to be commer-cially utilized, and we are entering an era where quantummechanics is no longer just a topic that is relevant forscientists who study puzzling effects in their notebooksor labs, but also for engineers or even the general public.Quantum random number generators and quantum cryp-tography systems are already commercially available [1–3], and researchers and some companies have started tobuild prototypes of quantum computers or quantum sim-ulators [4–12], and to investigate their usage to solve rel-evant problems in science, economy and beyond. Quan-tum networks, or as some people call it a quantum in-ternet [13–15], might well be a reality within the nearor mid future, and may have a similar influence as thedevelopment of the classical internet which plays an im-portant role in society and everyone’s daily life. Moretechnical applications of quantum technologies, e.g. forhigh precision measurements (quantum metrology) arealso discussed, and there is a huge effort worldwide to fur-ther develop such quantum technologies as is expressedin flagship programs in various countries or areas.In order to further develop and use such technologies,the best minds are needed. This makes it an importantgoal to include such topics in the standard curriculumat high school, and to develop ways to teach them in amodern and interesting way. Raising the awareness of thepotential of such quantum technologies, together with abasic understanding of their functionality, can also beconsidered a goal for general education. This paper aimsat providing a treatment of the most advanced and imme-diate application of quantum technologies, namely quan-tum cryptography. At the same time, we treat a topicof fundamental interest that even goes beyond quantummechanics, namely Bell inequalities [16–19]. This is asubject that highlights the puzzling and counter-intuitivenature of our world, and contributes to the general themeof teaching on nature of science. While this is usuallynot part of a standard physics curriculum, we neverthe-less believe it is of importance and relevance, and showswhat natural sciences can tell us about the world we livein.In this work, we take a game based approach to thesetopics [20–23], where we follow Ref. [24]. There, a game ∗ [email protected] was introduced that allows students to take on the roleof quantum bits (qubits) and scientists, and discoverthe features and rules of the quantum world. The ba-sic principles of quantum mechanics including the super-position of states, the behavior under measurements aswell as entanglement can be treated together with ad-vanced topics such as decoherence. A key ingredient arekinesthetic activities [25–28] that allow students to di-rectly experience these features, which supports a betterunderstanding and helps assimilate concepts. One nicefeature is that students are supposed to discover under-lying rules from experimental data, thereby taking onthe role of true scientists. Here we adapt and extendthis setting to treat slightly more complex cases, includ-ing entanglement-based quantum cryptography [29–31]and Bell inequalities. While it suffices to consider onlytwo different bases ( z and x ) and hence four differentstates to understand basic features (and also quantumcryptography based on the BB84 protocol [32]), morestates and settings are required here. We hence slightlyadopt the representation of quantum states and measure-ments, identifying orientation in space with direction inthe Bloch circle (a reduced version of Bloch sphere wherecomplex coefficients are avoided). The behavior of thequantum qubits is probabilistic, and depends on the rel-ative orientation between the measurement axis (corre-sponding to the measurement basis) and the quantumstate, i.e., the orientation of the qubit (or student) inspace. In order to properly mimic the required proba-bility distributions we suggest the usage of dice (eitherstandard, tetrahedron or multiple ones). This can bedone using real dice or freely available apps. Together,this allows one not only to understand the underlyingprinciples of quantum cryptography, but also to expe-rience the puzzling features of quantum mechanics in aqualitative and quantitative way first-hand.The paper is organized as follows. In Section II wedescribe our main approach. We introduce the settingand representation we use for one qubit and entangledpairs of qubits. In Section III, we provide backgroundon Bell inequalities (CHSH [33] and a variant by Mer-min [34] that allows for a simplified treatment) and howwe can include Bell tests in the framework of our game.In Section IV we consider the application to quantumcryptography. Each of the mentioned sections is dividedinto the theoretical background section, which is neededto understand the physics behind our proposal, and the rules of the game section, where we specify the game and a r X i v : . [ phy s i c s . e d - ph ] O c t its rules, and how to implement it in class. We also pro-vide an estimate of the required statistics to observe thedesired effects (Appendix C). This is necessary as we areinterested in expectation values of random processes withintrinsic fluctuations. II. MAIN PROPOSAL
We propose a game to illustrate advanced concepts ofquantum mechanics such as entanglement, Bell tests orquantum cryptography. The students play the role ofboth qubits and scientists and the measurement resultsthat they obtain as scientists are identical to the onesthey would obtain in a real laboratory. Therefore, thestudents can experience first-hand the puzzling featuresof quantum mechanics and have to come up with conclu-sions and explanations for the results they obtain.In an initial work [24], a game is proposed in whichstudents play the role of both, scientists and qubits, insuch a way that quantum states are reproduced by dif-ferent positions of arms and legs and quantum measure-ments by body movements. The possible measurementprocesses that can be mimicked within this frameworkare measurements along the x and z directions.In this work, we present an alternative and comple-mentary approach that allows the scientists to measurethe qubits in more directions than just the orthogonal x and z . Thus, more advanced concepts and experimentssuch as the Bell test can be introduced without the needof a strong mathematical background.In this approach, the class is split into two groups: oneplays the role of qubits and the other one the role ofscientists. The goal of the scientist-students is to pre-pare qubit-students in certain states and measure themby hitting them with a ball. The qubit-students try toavoid the balls and they have to follow certain rules ofthe game (see Sections II A and II B for details) in orderto mimic the real qubits, including both the stochasticbehaviour and the state change after the measurement.Afterwards, the scientist-students analyze the measure-ment results, make conclusions and find theories that ex-plain what has been observed.The game is thus designed such that the students canlearn not only the theoretical features of quantum me-chanics, but also experience how to be a real scientistand what this implies. In particular, the analysis of theresults and the critical thinking that should be developedin order to come up with suitable explanations are of highimportance in our approach. Dealing with the puzzlingfeatures of quantum mechanics gives also an opportunityto enhance critical analysis since the experimental resultsdo not match the previous, classical intuition and knowl-edge that the students may have in advance.In the following sections, we explain the rules for thestudents in order to behave like actual qubits. The sci-entists can prepare either single-qubit states or a pair ofqubits in an entangled state. We thus describe each case separately. Depending on the activity, the teacher canchoose to work with only single-qubit states or with en-tangled states. For instance, only single-qubit states areneeded to play the BB84 cryptography protocol (Sec-tion IV A), but one needs entangled qubits to play theBell test (Section III) or the E91 protocol (Section IV B). A. Single qubit
1. Theoretical background
We consider qubits, i.e., two-level quantum mechan-ical systems with one degree of freedom that can havetwo possible values which we denote and . Physi-cally, this degree of freedom can be, for instance, thespin of a particle or the polarization of a photon. Oneof the main features that distinguishes quantum statesfrom classical ones is the superposition principle wherethe qubits can be in a superposition of the two states and , i.e. | ψ (cid:105) = α | (cid:105) + β | (cid:105) , where α and β are com-plex numbers with | α | + | β | = 1 . Thus, there existinfinitely many superposition states. All these superpo-sition states can be easily visualized as unit vectors in theBloch sphere [35–37]. In particular, the states | (cid:105) and | (cid:105) are the Bloch vectors pointing upwards and downwards,respectively, in the z direction. The superposition states | x (cid:105) = √ ( | (cid:105) + | (cid:105) ) , | x (cid:105) = √ ( | (cid:105) − | (cid:105) ) can be visu-alized as unit vectors pointing along the x direction. Ingeneral, any unit vector of the Bloch sphere correspondsto a quantum state that can be determined by the twoangles of the spherical coordinates (see Figure 1 (Left))as | ψ (cid:105) = cos (cid:18) θ (cid:19) | (cid:105) + e iϕ sin (cid:18) θ (cid:19) | (cid:105) (1)where θ ∈ [0 , π ] and ϕ ∈ [0 , π ) are the polar and az-imuthal angles, respectively. Figure 1. (Left) Bloch representation of quantum states interms of the polar θ and azimuthal ϕ angle. (Right) x − z plane of the Bloch sphere with ϕ = 0 and ϕ = π . States | (cid:105) , | (cid:105) , | x (cid:105) and | x (cid:105) are depicted whose spherical angles ( θ, ϕ ) are (0 , , ( π, , ( π/ , and ( π/ , π ) , respectively. For the purpose of this proposal it is enough to workonly with one Bloch plane. In particular, we considerthe x − z plane with azimuthal angles ϕ = 0 and ϕ = π (see Figure 1 (Right)). Given a generic direction a , sep-arated from the z -axis by an angle θ a , the state | a (cid:105) isdetermined by ϕ = 0 (right side of the plane) and θ a , i.e. | a (cid:105) = cos (cid:0) θ a (cid:1) | (cid:105) + sin (cid:0) θ a (cid:1) | (cid:105) . Analogously, the state | a (cid:105) is determined by ϕ = π (left side of the plane) and θ a , i.e. | a (cid:105) = cos (cid:0) θ a (cid:1) | (cid:105) − sin (cid:0) θ a (cid:1) | (cid:105) . If this math-ematical formalism is introduced at high school levels,one can alternatively consider only one real parameter θ ∈ [0 , π ) in order to avoid complex numbers.The next feature that characterizes qubits is the prob-abilistic nature of the measurement process. Accordingto Born’s rule, if one measures the state | ψ (cid:105) given by | ψ (cid:105) = α | (cid:105) + β | (cid:105) in the z direction, the probabilityof obtaining the state | (cid:105) ( | (cid:105) ) is p = | α | ( p = | β | ).Therefore, one obtains a deterministic outcome after ameasurement in z direction, i.e. α = 1 ( β = 1 ), only ifthe initial state is | ψ (cid:105) = | (cid:105) ( | ψ (cid:105) = | (cid:105) ). State | (cid:105) givesoutcome +1 , whereas state | (cid:105) gives outcome − . In al-gebraic terms, these two states are called eigenstates ofthe z basis with eigenvalues +1 and − . Considering therepresentation of | ψ (cid:105) in the Bloch sphere (Eq. (1)), theprobabilities p and p can be written in terms of thespherical angles θ and ϕ as p = cos θ , (2) p = sin θ . (3)The probability of getting either the outcome +1 or theoutcome − after a measurement in the z direction onlydepends on the angle θ , which is in this case the separa-tion between the measurement direction ( z -axis) and theBloch vector of the state | ψ (cid:105) . The smaller the angle θ ,the higher the probability p to obtain outcome +1 . Moregenerally, the measurement process can be performed inany direction a . Mathematically, the measurement is de-scribed by the observable O a , of the form O a = (+1) | a (cid:105) (cid:104) a | + ( − | a (cid:105) (cid:104) a | , (4)where | a (cid:105) and | a (cid:105) are the eigenstates of the a basis,with eigenvalues (outcomes) +1 and − , respectively,i.e. O a | a (cid:105) = (+1) | a (cid:105) and O a | a (cid:105) = ( − | a (cid:105) . Anal-ogously to Eq. (1), one can express an initial state | ψ (cid:105) in terms of the eigenstates {| a (cid:105) , | a (cid:105)} of any basis a .Thus, the probability p of getting the outcome +1 isgiven by the angle θ ψ,a between the Bloch vector of theinitial state | ψ (cid:105) and the state | a (cid:105) , that specifies the mea-surement direction a , i.e. p = cos θ ψ,a .Another crucial property of quantum mechanics isthat, once the measurement process is performed, theinitial state changes into one of the two eigenstates ofthe measurement direction a depending on the outcome.For instance, if the outcome was +1 , the final state is | a (cid:105) , and if it was − , the final state is | a (cid:105) . Therefore,no more information can be accessed from the originalstate since the state is no longer | ψ (cid:105) .
2. Rules of the game
As described in Section II A 1, the first feature ourgame should reproduce is the qubit states. We only con-sider the states that can be described within the x − z plane of the Bloch sphere and which allows us to repre-sent it by drawing a circle on the floor. Once the Blochplane is painted, the qubit-students mimic the Bloch vec-tor with their body orientation as shown in Figure 2. Asargued in Section II A 1, the angle θ is enough to describestates in this plane, where the angle ϕ = 0 , π is only re-sponsible for a phase. Therefore, given a generic direction a with angle θ a from the vertical axis, the qubit-studentjust needs to rotate their body by an angle θ a to representthe state | a (cid:105) (Figure 2 (Left)). Already in this position,if the student turns around and faces the other side ofthe a -axis, they now represent the state | a (cid:105) (Figure 2(Right)). Figure 2. The qubit-students themselves represent the Blochvector in the x − z Bloch plane painted on the floor. (Left) Astudent represents a qubit in the state | a (cid:105) when they rotateby an angle θ a from the vertical axis. (Right) The qubit-student faces towards the opposite side to represent the state | a (cid:105) along the a -axis. In addition to the states, we also need to reproducethe measurement process, which is stochastic due to theintrinsic nature of quantum mechanics. In order for thestudents to act stochastically the same way as qubits do,we propose that they use a biased die, which should beconstructed in such a way that the probabilities p and p (Equations (2) and (3)) can be obtained depending onthe measurement directions used in the activity. For ex-ample, if the scientists are only allowed to prepare statesand measure in the x and z directions, i.e. p = p = 1 / ( θ = π/ ), a standard die suffices: odd numbers mayindicate outcome +1 and even numbers outcome − . InSections III and IV B, several activities are proposed thatneed other measurement directions and thus more exam-ples of biased dice are given (see Appendix C 1).The game proceeds as follows: the scientist preparesthe qubits in an initial state. There are two possiblestates {| a (cid:105) , | a (cid:105)} per measurement direction a . Thequbit-students should orient their bodies with the anglethat corresponds to the state the scientists chose. Then,the scientist starts to measure the qubits in different di-rections. A qubit is measured when the scientist hitsthem with a ball, thus simulating the real measurementprocess of e.g. a photon "hitting" a particle. If the scien-tist succeeds, they choose a measurement direction andplace a compass stick on the floor to make the directionmore explicit and to mark which side of the axis is the"positive" side – since it corresponds to the outcome +1 – (see Figure 3). Once a qubit-student is hit, they havetwo possibilities:1. If the qubit-student is already aligned with themeasurement direction, they announce the outcomecorresponding to their state, i.e., either +1 (if theyare facing towards the positive sign of the compassstick) or − (if they are facing towards the negativesign of the compass sign).2. If the qubit-student is pointing towards a differentdirection, then they throw the corresponding biaseddie and, depending on the result, orient towards the | a (cid:105) (positive side of the a -axis) or | a (cid:105) (negativeside) state and announce the outcome (see Figure 3for an example). Figure 3. Measurement process. After the qubit-student ishit with a ball, the scientist announces a measurement di-rection, places the compass stick to establish the "positive"side of the axis more clearly (see text for details), and thequbit acts accordingly. Here, the scientist announces a direc-tion a that is different from the one that the qubit-student isaligned with ( z in this case). Thus, the qubit-student throwsa biased die, orients their body according to the die resultand announces the outcome corresponding to their final stateafter the measurement. In the example shown, the die resultis "Go to negative side" so the qubit-student aligns with thecompass stick direction and faces towards the negative signof the stick. Therefore, they announce the outcome − . B. Entangled pair of qubits
1. Theoretical background
We now consider two-qubit states, in particular, en-tangled states. The main feature of entangled states isthat the measurement outcomes of the two qubits are correlated in more than one measurement direction. Onerepresentative example of these entangled states is the socalled Bell state | ψ − (cid:105) that reads | ψ − (cid:105) = 1 √ | (cid:105) A ⊗ | (cid:105) B − | (cid:105) A ⊗ | (cid:105) B ) (5)where subindex A denotes the first qubit and subindex B the second one. In our game, we deal with this particularstate, where one scientist (let us call her Alice) measuresqubit A and another scientist (let us call him Bob) mea-sures qubit B . In actual experiments, these two qubitscan be arbitrarily separated in space and the measure-ment outcomes are still correlated, contrary to any (clas-sical) local theory. As in the single-qubit case, one canapply Born’s rule and notice that after Alice measuresher qubit in the z direction, she will get either the out-come +1 (state | (cid:105) A ⊗ | (cid:105) B after the measurement) withprobability / or the outcome − (state | (cid:105) A ⊗ | (cid:105) B )with probability / . Therefore, when Bob measures hisqubit in the same direction as Alice, the state that he getsis correlated to Alice’s, i.e., if Alice gets the outcome +1 ,Bob gets outcome − and if Alice gets the outcome − ,Bob gets +1 . This occurs not only in the z direction asexplained here, but also in any other direction , which isa genuinely quantum feature. Hence, measurement out-comes are always anti-correlated when both Alice andBob measure the state | ψ − (cid:105) in the same direction.
2. Rules of the game
In order to represent the entangled state | ψ − (cid:105) , the stu-dents playing the role of qubits hold hands facing eachother and start spinning at the center of the Bloch planeon the floor, which symbolizes the fact that they are en-tangled in all directions.The measurement process is illustrated in Figure 4.When the scientist that plays the role of Alice measuresone of the qubits, she announces the measurement di-rection, places the compass stick and tries to hit thequbit with the ball. Since the two qubits are holdinghands, once one qubit is hit, the whole pair keeps rotat-ing until they get to the direction marked by the compassstick. Alice’s qubit then announces out loud the outcome,which is +1 if she is facing towards the positive sign ofthe compass stick and − otherwise. Note that each ofthese outcomes has probability 1/2 to occur for Alice’s In an arbitrary direction a , this Bell state is written, up to aglobal phase, as | ψ − (cid:105) = √ ( | a (cid:105) A ⊗ | a (cid:105) B − | a (cid:105) A ⊗ | a (cid:105) B ) . Note that in this case, the scientist chooses a measurement di-rection before they actually managed to hit the qubit-student.Even if this is not completely realistic, we consider that this orderis clearer for the qubit-students, who have to rotate until theyget to the compass stick when measured. Placing the compassstick after the hitting-with-ball process would just increase thedifficulty for the qubit students to perform in the correct way.
Figure 4. (Up Left) Representation of an entangled state | ψ − (cid:105) by two students that hold hands and spin around the center ofthe Bloch plane. (Up right) Alice measures one of the qubitsof the entangled pair. She announces the measurement direc-tion and marks it with the compass stick on the floor, so thatthe positive side and the negative side of the axis are clear.Then, she tries to hit her qubit with the ball. (Down left)When Alice succeeds, both qubits keep rotating until theyreach the measurement direction marked with the compassstick (in this case, the z -axis), and the measured qubit an-nounces the corresponding result − (negative side). (DownRight) Bob measures the other qubit of the pair in the samedirection as Alice and obtains the anti-correlated result +1 ,since the qubit is facing opposite to Alice’s qubit. Note thatqubits are no longer entangled after Alice’s measurement. qubit, since the entangled pair is spinning when one ofthe qubit-students is hit, which can happen randomly atany point of the rotation, leading to an equal probabilityto get first to the positive or to the negative sign of thestick. If a second scientist (Bob) measures now the otherqubit of the entangled pair, the latter behaves as a nor-mal single qubit, i.e. as explained in Section II A 2, sincethe measurement of Alice’s qubit has resulted in bothqubits changing their initial state. In particular, if Bobmeasures in the same direction as Alice, it is clear thathe obtains the anti-correlated result since Bob’s qubitis already facing opposite to Alice’s (see Figure 4). Notethat after the measurement, the qubits are no longer en-tangled (they do not hold hands anymore), so any furthermeasurement does not lead to anti-correlated results, butto the qubits behaving independently of each other. If Alice gets +1 , Bob gets − and vice versa. These basic rules introduced in Sections II A and II Ballow for several games to be played. One possibility isto play the game as introduced in [24], where only thequbits know the rules they have to follow and the scien-tists should figure them out by making measurements andanalyzing the results. This game thus focuses on the fun-damental concepts of quantum mechanics (superpositionstates, measurement processes, Heisenberg’s uncertaintyrelation, entanglement, etc.).The other possibility is to play the cryptography games(see Sections IV A and IV B), which consist in perform-ing the BB84 and the E91 protocols for quantum keydistribution. In both cases, there are two scientists (orteams of scientists) whose goal is to share a secret mes-sage. They should use the qubits to create and share asecure key to encrypt and decrypt the message, so thatnobody else apart from them has access to it.
III. BELL TEST
The proposal presented in Section II B allows one tomeasure the qubits in any direction and to work withentangled qubits, in particular with those qubits in theBell state | ψ − (cid:105) . Thus, we already have all the necessarytools to mimic a Bell test.The Bell test is an experiment designed to test Bell’stheorem, introduced by John S. Bell [16, 38, 39]. Bell’stheorem was a response to the Einstein-Podolsky-Rosen(EPR) paradox [40] which was ought to show that themathematical description of quantum mechanics is in-complete. EPR argued that the probabilistic naturecomes from our ignorance of some degrees of freedom,called hidden variables, that we cannot access. The ar-gument in the EPR paradox is based on the assumptionof local realism, i.e., the state of one system cannot influ-ence the state of a separated system instantaneously (lo-cality) and the systems have fixed properties, no matterif the properties are measured or not (realism). On theother hand, Bell’s theorem states that no local, realistictheory can reproduce all the predictions of quantum me-chanics. More precisely, Bell’s theorem puts a bound onthe strength of correlations between the results of mea-surements performed on spatially separated systems, un-der the assumption that local realism holds. In turn,quantum mechanics predicts stronger correlations, whichhence implies that one (or both of the assumptions ofBell’s theorem - namely locality or realism - are wrong.Recent experiments have indeed shown that nature be-haves in the way quantum mechanics predicts, i.e., therehave been loophole free tests of Bell’s inequality [17–19]that show a clear violation. It should be stressed thoughthat Bell inequalities are not only about quantum me-chanics, but they are far more general. They apply to all local realist theories, and if we find in experiments thatNature indeed violates these inqualities, this shows onthe one hand that the predictions of quantum mechan-ics seem to be correct, and that quantum mechanics isindeed a theory where either reality or locality are not re-spected. What is more, it means that any future theorythat may eventually replace quantum mechanics cannotbe local realistic either - Nature simply does not behavethis way! This is what makes Bell’s theorem so inter-esting also from a fundamental point of view, as it tellsus something about the fundamental functionality of ourworld.In the following, we present two different approachesof testing Bell’s theorem in class; these two approachescan be performed within the framework of our proposal. A. CHSH Inequality
1. Theoretical background
One approach to experimentally test Bell’s theorem isto check the CHSH inequality [33] |(cid:104) ab (cid:105) − (cid:104) ab (cid:48) (cid:105) + (cid:104) a (cid:48) b (cid:105) + (cid:104) a (cid:48) b (cid:48) (cid:105)| ≤ , (6)where (cid:104)•(cid:105) denotes the expectation value operation and a, b, a (cid:48) , b (cid:48) are observables that have two possible measure-ment outcomes {± } (see Appendix A 1 for more details).Let us explain the details of this inequality in orderto understand the implications of its violation. Let usconsider two systems, one in Alice’s laboratory and theother one in Bob’s. In this situation, a and a (cid:48) denotetwo possible measurements that Alice can choose. Anal-ogously, Bob can measure either b or b (cid:48) . If one assumesthat the systems have fixed properties before any mea-surement (realism), there are two possible situations forAlice’s outcomes (considering that the two possible out-comes are ± ): a (cid:48) (cid:54) = a : a (cid:48) + a = 0 , a (cid:48) − a = ± , (7) a (cid:48) = a : a (cid:48) + a = ± , a (cid:48) − a = 0 . (8)If one also considers the results of Bob’s measurements,the quantity for S reads S = ( a (cid:48) + a ) b + ( a (cid:48) − a ) b (cid:48) = ± , (9)for both situations a = a (cid:48) and a (cid:54) = a (cid:48) with b, b (cid:48) ∈ {± } .Using the triangle inequality, it is easy to see that |(cid:104) S (cid:105)| ≤ (cid:104)| S |(cid:105) = 2 , (10)which is the CHSH inequality in Eq. (6). Thus, if a theoryviolates this inequality, the initial assumptions of eitherlocality or realism or both should be wrong. Therefore,only by making measurements in different directions andaveraging the results, one can experimentally test a the-oretical hypothesis such as local realism.It has been shown (see Appendix A 1 for details) thatquantum mechanics violates the CHSH inequality. Inparticular, the maximal violation ( |(cid:104) S QM (cid:105)| = 2 √ ) oc-curs in the case where Alice and Bob share a pair of max-imally entangled qubits (for example in the state | ψ − (cid:105) ) and the measurement directions a , b , a (cid:48) and b (cid:48) are in thesame plane and successively separated by an angle π/ (see figure 5(a)). This experiment is precisely what onecan perform in class with this proposal. Students canmeasure pairs of entangled qubits in any measurementdirection, as described in Section II B (see also Figure 4).Alice and Bob perform measurements in two different di-rections each of them can choose – denoted ( a, a (cid:48) ) forAlice and ( b, b (cid:48) ) for Bob – and write down the results. Inorder to check if the CHSH inequality is violated, theytake the outcomes of the cases where the two measure-ment directions do not coincide and compute the expec-tation values in Eq. (6) afterwards.
2. Rules of the game
In this section, we explain how Alice and Bob measurethe state | ψ − (cid:105) in directions separated by an angle π/ (see figure 5(a)) using our proposal of Section II B. Letus take the example of directions a (cid:48) and b whose measure-ment results are used to compute (cid:104) a (cid:48) b (cid:105) in Eq. (6). First,Alice chooses the measurement direction a (cid:48) , places thecompass stick in the orientation she wants, and hits herqubit-student with the ball (see Figure 5(b) (Left)). Theentangled pair which was spinning before being hit by theball, rotates until they reach the a (cid:48) -axis and stops there.Then, Alice’s qubit communicates the outcome out loud,i.e. +1 if he is facing towards the positive sign of the com-pass stick and − otherwise (see Figure 5(b) (Right)).The initial state | ψ − (cid:105) becomes either | a (cid:48) (cid:105) A ⊗| a (cid:48) (cid:105) B (out-come +1 ) or | a (cid:48) (cid:105) A ⊗ | a (cid:48) (cid:105) B (outcome − ) after the mea-surement. In the example of Figure 5(b), the final state is | a (cid:48) (cid:105) A ⊗| a (cid:48) (cid:105) B . Now, Bob measures his qubit in direction b , so he places his stick to mark it (see Figure 6 (Left)).Note that his qubit has changed its state due to Alice’smeasurement. In the example of Figure 6, the qubit is inthe state | a (cid:48) (cid:105) B . Since the measurement direction is ro-tated by π/ from the initial state direction | a (cid:48) (cid:105) B , Bob’squbit has to throw the biased die to decide if she rotatesto the positive or to the negative side of axis b (Figure 6).The biased die should be designed such that, with proba-bility p = cos (3 π/ ≈ . , the qubit-student rotatesto the positive side of the b -axis which is indicated bythe positive sign of the compass stick; and with proba-bility p = sin (3 π/ ≈ . , the qubit-student rotatesto the negative side of the b -axis (see Figure 6 (Left)).From this example, it is intuitively easy to see that thestudent has a higher probability of rotating to the sideof the measurement axis that is closer to them. Withthis intuitive idea, the die’s possible events can be "Goto the closer side" or "Go to the further side", which arecompletely general events that cover all possibilities. InAppendix C 1, several examples of biased dice are givenfor different separation angles between the measurementdirections a, b, a (cid:48) and b (cid:48) . To mimic the correspondingprobabilities in each case, we propose two alternativesto construct the dice, either to physically build k -sided (a) Measurement directions(b) Performance of the game Figure 5. [a] Measurement directions a , b , a (cid:48) and b (cid:48) are inthe same plane and successively separated by an angle π/ .[b] (Left) Alice measures her qubit in the a (cid:48) -direction. Thecompass stick on the floor indicates the orientation of a (cid:48) -axis.Once the qubit is hit with the ball, the pair rotates untilthey reach the a (cid:48) -axis. (Right) Alice’s qubit communicatesthe outcome out loud ( +1 in the picture –aligned with thecompass stick–, corresponding to the state | a (cid:48) (cid:105) in this case.Note that the initial state | ψ − (cid:105) becomes | a (cid:48) (cid:105) A ⊗ | a (cid:48) (cid:105) B afterthe measurement). dice or to make use of apps that allow one to choosethe number of sides. The creation of the dice can also beconsidered as an activity itself to work with probabilities.Once Bob’s qubit has rotated according to the die re-sult, the corresponding qubit-student communicates theoutcome out loud to Bob (Figure 6 (Right)). Alice andBob should repeat this measurement process with asmany entangled pairs as possible in order to get a sta-tistical reliable value of (cid:104) a (cid:48) b (cid:105) . Then, the same proce-dure should be done also for the combinations (cid:104) ab (cid:105) , (cid:104) ab (cid:48) (cid:105) and (cid:104) a (cid:48) b (cid:48) (cid:105) . In Appendix C 2, we present some estima-tions of the number of measurements that Alice and Bobshould do as well as estimations of the number of stu-dents needed to perform the whole process in order toget a statistical accurate value for the quantity |(cid:104) S (cid:105)| ofEq. (10).In order to make a clear description of the game, wehave explained the measurement process in such a waythat Alice measures first and Bob measures second, butthis order can be changed without loss of generality orchanging the results. We also remark that a crucial prop- Figure 6. (Left) Bob measures his qubit after Alice’s measure-ment. The state of Bob’s qubit after Alice’s measurement is | a (cid:48) (cid:105) . Since Bob measures in the b -direction, which is sepa-rated π/ from the a (cid:48) -axis, Bob’s qubit has to throw a biaseddie to determine how it rotates. In this case, the die is suchthat, with probability p ≈ . , the qubit rotates to the neg-ative side of the b -axis (indicated with the negative sign ofthe compass stick) and with probability p ≈ . , the qubitrotates to the positive side (see main text for details). (Right)After throwing the die, Bob’s qubit obtains "Go to the closerside", so it rotates to the negative side in this case and com-municates the outcome out loud to Bob. In this example, thefinal state for Bob’s qubit is | b (cid:105) . erty of real entangled particles is that they give corre-lated outcomes no matter how far away they are fromeach other, so Alice’s and Bob’s laboratories do not needto be in the same place. The game as we present it heredoes not mimic this property since the entangled studentshold hands, which forces them to be in the same place.A possible variant of the game could be introduced thatincludes such a property, namely the entangled studentscould imitate each other and perform as explained abovewhile they are placed in Bloch circles that are separatedfrom each other. In this case, if one of the qubit rotates,the other one also rotates and when one is measured,both keep rotating until they reach the measurement axismarked on the Bloch plane of the measured student .Although the activity can be presented to the studentsin numerous ways, we suggest to use it as described hereto enhance their critical thinking. Thus, we propose thefollowing unit (we assume that the initial concepts of su-perposition and entanglement have been already taughtbeforehand and we focus only on the Bell test):1. The ideas of locality and realism are presented. Thestudents should discuss their own initial concept ofreality based on the concepts of locality and real-ism. In this situation (entangled qubits are placed in two differentBloch circles), if both qubits are hit at the same time by theball, they stop spinning and rotate until both of them get tothe measurement direction chosen by the scientist who measuresfirst.
2. Considering the initial discussion, the dichotomousoutcomes are presented and the students shouldwork with the expression for the quantity S . Then,the CHSH inequality is presented.3. The students should perform all the necessary mea-surements on the entangled pairs in order to checkthe CHSH inequality (this part is performed withthe game introduced above).4. After gathering all the experimental data, studentsshould compare their initial intuition with the re-sulting data and analyze the implications of theviolation of the CHSH inequality.In addition to the CHSH approach, we also proposean alternative to test local realism. It is based on thework "Is the moon there when nobody looks? Realityand the Quantum theory", by N. David Mermin [34].This approach relies less than the CHSH inequality onthe mathematical background of the students, so thatone can directly apply it without any previous expla-nation. However, more measurements are needed to getgood statistics, which translates into larger groups of stu-dents needed (see Appendix C 2 for a detailed estimationon the number of students and measurements needed).The details of this alternative and a comparison to theCHSH approach are presented in Appendix B. IV. APPLICATION: QUANTUMCRYPTOGRAPHY
Considering our proposals from the previous sections,one may also use them to make a feasible approach toquantum cryptography, which is a direct application ofkey concepts in quantum mechanics such as superposi-tion states or entanglement. In this section, we showthat the main protocols in quantum cryptography canbe simulated in a realistic way with the tools presentedin Section II.
A. BB84
One of the most important cryptography protocols,which rely on the properties of quantum mechanics in-stead of the complexity of a mathematical problem, wasproposed originally by G. H. Bennett and G. Brassard in1984 and celebrated 30 years afterwards [32].
1. Theoretical background
In the BB84 protocol, a sender (Alice) wants to en-crypt a message and send it to a receiver (Bob) via asecure channel. This is done by generating a privatekey (password that consists of a sequence of bits to en-crypt the message) that only Alice and Bob know. The protocol is designed to generate this key, whose secu-rity relies on the properties of quantum mechanics. Itproceeds as follows: Alice and Bob can send and re-ceive their qubits in two bases, e.g. z and x , whichgives the set {| (cid:105) , | (cid:105) , | x (cid:105) , | x (cid:105)} of states in these bases.Here, | (cid:105) and | (cid:105) are the eigenstates in the z basis and | x (cid:105) = 12 ( | (cid:105) + | (cid:105) ) and | x (cid:105) = 12 ( | (cid:105) − | (cid:105) ) the eigenstatesin the x basis. The sender Alice sends the eigenstateof her qubit in a chosen basis to the receiver Bob, whochooses then in which basis he measures this receivedqubit. After this, Alice and Bob announce their cho-sen basis via a public channel. Alice and Bob repeatthis procedure multiple times and then collect the re-sults of the qubits for which the sent and measured basiscoincide (while the other results with different basis areneglected). This is the quantum key with which Aliceand Bob can encrypt messages. An intuitive exampleof applying the BB84 protocol and generating a securequantum key can be seen in Table I. A: 0 1 1 0 1 0 0 1A basis: z z x z x x x zB basis: z x x x z x z zB: 0 0 1 0 0 0 1 1Key: 0 1 0 1Table I. Example of generating a secure key based on quantumcryptography via the BB84 protocol.
2. Generating a key
As we have shown in the previous Section IV A 1, onlythe four states {| (cid:105) , | (cid:105) , | x (cid:105) , | x (cid:105)} and the two measure-ment directions x and z are needed to reproduce theBB84 protocol.In this section, we propose a specific activity that re-lies on the rules of the game explained in Section II A 2.The goal of the activity is to communicate a secret mes-sage successfully. To do so, the students form groups,each of which consists of one student playing the roleof Alice, another one playing Bob and the rest will bequbits. First, Alice and Bob generate the key for the en-cryption of the message using the BB84 protocol. Then,Alice chooses the message she wants to send, encryptsit with that key and sends it to Bob. Finally, Bob usesthe key to decrypt the message. They win if the messageis transmitted successfully. Let us illustrate this activitywith an example.In order to make it easier, let us assign a number toeach letter of the alphabet, i.e. A-1, B-2, ..., Z-26. Ifthese numbers are written in binary, each letter will cor-respond to a binary code of 5 digits, e.g. E = (00101) .In our example, Alice wants to send the message "EY"( E = 00101 , Y = 11001 in binary) to Bob. First, sheneeds to encrypt it with a password or key , so that onlya person that knows the key will be able to decrypt itand get the message. One simple way to encrypt it is bysumming the message and the key in binary (mod 2) asshown for example in Table II. This method of encryptionis called one-time pad and it is designed in such a waythat the encrypted message cannot be directly (withoutknowing the key) deciphered by taking advantage of e.g.language patterns such as the high frequency of certainletters. In particular, if the key is only used once, it is atleast as long as the message and it is truly random, theresulting encryption after the sum will also be random(thus no patterns can be extracted). Therefore, Aliceand Bob just need to ensure that only the two of themknow the key in order to be sure that the message re-mains secret. Once both have the key, Alice can publiclysend the encrypted message and only Bob will be able todecrypt it (he just needs to sum [mod 2] the key and theencrypted message to get the real message). Message: 0 0 1 0 1 1 1 0 0 1Key: 1 0 0 1 1 0 0 0 0 1Encryption: 1 0 1 1 0 1 1 0 0 0Table II. Example of message encryption.
Let us illustrate how Alice and Bob generate the keywith the BB84 protocol according to the rules of ourgame.As explained in Section IV A 1, Alice has to prepare thequbits in certain states that Bob will afterwards measure.Since only two measurement directions are needed, theBloch plane painted on the floor only needs to have twoperpendicular axes (vertical for the z -axis and horizontalfor the x -axis). Before the protocol starts, Alice and Bobshould agree on which is the positive side of each axisand keep the same criterion throughout the whole pro-tocol. In order for the qubits to know this criterion andfor them to remember it, they can place compass sticksbesides the axis on the floor (see Figure 3). Then, theprotocol begins. Alice can prepare the qubits in the states {| (cid:105) , | (cid:105) , | x (cid:105) , | x (cid:105)} , whose corresponding body orienta-tions are depicted in Figure 7.Bob measures Alice’s qubits by following the rules ex-plained in Section II A 2. In this case, a normal die issufficient (odd numbers may indicate outcome +1 andeven numbers outcome -1) to determine the qubit’s ori-entation in the measurement process (see Table C.1 inAppendix C 1). Bob writes down both the measurementdirection he chose and the outcome he obtained, in or-der to create a table analogous to Table I. Note thatAlice should also write down the prepared state (both ⊕ , ⊕ , ⊕ , ⊕ Figure 7. Students’ positions corresponding to the states {| (cid:105) , | (cid:105) , | x (cid:105) , | x (cid:105)} respectively (from left to right). the axis and the side of the axis towards which studentis oriented). However, in order to create a key which isa string of binary digits (0 or 1), one cannot use directlythe outcomes +1 and − because they are not binary. Asimple transformation is enough to solve this: wheneverthe outcome is +1 , the bit is (in analogy to | (cid:105) and | x (cid:105) , which are the states corresponding to the outcome +1 ); whereas the outcome − becomes the bit .Once Bob finishes measuring the qubits Alice prepares,both say the list of preparation and measurement axes(but not the bits!) out loud –i.e. they share it through apublic channel–. Then, Bob checks which of those coin-cide with his measurement directions and take the corre-sponding outcome bits as the key for decryption. Oncethe key is ready, Alice encrypts her message with it andsends it to Bob publicly (she says the encrypted digitstring out loud). Bob decrypts the message with the keyand they win the game if the message has been commu-nicated successfully.
3. Eavesdropping
So far we have presented an illustrative example ofjust the key generation and the process of encryption anddecryption. However, this is not a realistic scenario sincethe main purpose of a secure cryptography protocol is tobe able to detect and prevent other parties from knowingthe message. Thus, in this section we assume that there is an Eavesdropper that tries to intercept the messageAlice sends to Bob and we propose some strategies todetect it.One of the main features of the BB84 protocol is thatit provides an easy way, based on the quantum propertiesof qubits, to detect interventions in the key generation.As we have seen in Section II A, the initial state of quan-tum systems may change after the measurement process, Note that only half of the directions will coincide, so Alice shouldprepare at least twice as many qubits as the number of digits ofthe message. If this is so, the key is secure. Ifnot, they abort the protocol and start the process again.However, since the measurement process is stochastic,the Eavesdropper has some probability to go unnoticed.For the sake of argument, let us assume for now that theEavesdropper gets access to Alice’s qubit. If, by chance,the Eavesdropper measures in the same direction as Al-ice’s preparation axis, the state does not change at all, sothere is no way to detect the external intervention. Onthe other hand, if the Eavesdropper’s measurement di-rection does not coincide with Alice’s, the prepared stateis modified and Bob obtains a different outcome with 1/2probability. In total, the Eavesdropper is detected withprobability 1/4 at each measurement of one qubit (onebit of the key). Therefore, the more bits are used tocheck the security of the key, the less probable it is thatthe Eavesdropper goes unnoticed. The phase of checkingdescribed above will be referred to as security test in thefollowing.In order to play the full version of the BB84 proto-col, we propose that some of the students in each groupplay the role of Eavesdroppers and they have to comeup with strategies to intercept the message without be-ing discovered. Then, the goal of the game is for themto discover the key; and for Alice and Bob to send themessage successfully and secretly.We now propose a few possible eavesdropping strate-gies in case the teacher wants to suggest them to thestudents in advance. The preparation of states and the In order to compare the subset of bits, they can use a publicchannel (say it out loud in our game). Note that this subset canno longer be used to encrypt the message. An example of this situation is the following: Alice prepares thestate | x (cid:105) in the x direction. Then, the Eavesdropper choosesto measure in the z direction, which transforms the qubit stateinto either | (cid:105) or | (cid:105) . In both cases, if Bob then measures thequbit in Alice’s direction ( x ), he does not obtain the state | x (cid:105) deterministically –as it would be the case without the externalintervention–, but with 1/2 probability. This is so because Bob’smeasurement direction ( x ) and the direction of the state he gets( z ) are separated θ = π/ , so the probability that he obtains theoutcome +1 is p = 1 / , given by Eq. (2). measurement processes are always performed followingthe rules of the game detailed in Section II A 2.
1. The Eavesdroppers measure the qubit be-fore Bob.
One way of getting the key is to interceptAlice’s qubit, measure it and send it on to Bob so thathe does not notice any difference. Equivalently in ourgame, the Eavesdroppers try to hit Alice’s qubit withthe ball before Bob but, since the qubit student doesnot move away from the Bloch circle, Bob can measureit after the Eavesdroppers. Note that the state of thequbit may have changed after the Eavesdroppers’ mea-surement, depending on which basis they choose.The Eavesdroppers proceed as Bob, i.e. they measurethe qubit in one direction ( x or z ) and then, when the listof preparation axes is made public by Alice and Bob, theyonly take the measurement outcomes of the coincidentaxes. In this way, the Eavesdroppers have access to thekey and can decrypt the secret message. However, theyget the correct bits with probability
2. The Eavesdroppers keep Alice’s qubit.
Thisstrategy is slightly different from the previous one. In-stead of measuring Alice’s qubit first, the Eavesdropperstake the qubit and store it without measuring it. Theyreplace Alice’s qubit with another one, prepared in a ran-dom state, so that Bob can still measure one qubit. Inour game, the Eavesdroppers take the qubit student outof the Bloch circle and replace them with another qubitstudent prepared in one of the states {| (cid:105) , | (cid:105) , | x (cid:105) , | x (cid:105)} at random (see Figure 7). The original qubit preparedby Alice is guarded by one of the Eavesdroppers. OnceAlice announces the list of preparation axes, the Eaves-droppers measure the guarded qubit (see Section II A 2)in the same direction as Alice’s preparation axis to alwaysobtain exactly the same bits of the key that Alice has.However, since Bob gets a random state, he obtains thesame outcome as Alice only with probability 1/2, evenwhen he measures in Alice’s directions. Therefore, theEavesdroppers are detected with probability 1/2. As forthe previous strategy, Alice and Bob can do the securitytest to detect Eavesdroppers’ action.
3. The Eavesdroppers try to decrypt the mes-sage by guessing the key.
The Eavesdroppers haveaccess to the encrypted message since it is sent via apublic channel (said out loud in our game), so they cantry and guess the key to decrypt it. With this simplestrategy, students can work with probability theory (e.g. Sum of the probabilities that Eavesdroppers choose the same di-rection as Alice and Bob and that they choose the other directionbut still they get the same outcome, i.e. + · = . n ).If Alice and Bob detect the action of the Eavesdrop-pers, they abort the protocol and do not send any mes-sage. The Eavesdroppers win if they are able to decipherthe secret message without being noticed. On the otherhand, Alice and Bob win if they are able to communicatethe message secretly and successfully. B. E91
1. Theoretical background
The goal in this case is the same as for the BB84 pro-tocol, i.e., to generate a secure encryption key. However,the E91 protocol developed by Artur K. Ekert in 1991[41] proposes a different procedure to generate a securekey. In this case, an entangled pair of qubits is sent froma source S to two receivers, Alice and Bob. Alice and Bobchoose their measurement basis randomly and indepen-dently among the ones shown in Figure 8 and announcetheir chosen orientation via a public channel. The mea-surements are divided into two groups and there are twopossibilities of outcomes: (i) Alice and Bob have chosenthe same measurement basis, i.e., due to the propertiesof entangled qubits, every time Alice gets an outcome 0(or 1), Bob gets the complementary outcome 1 (or 0).Alice and Bob use these bits for the key. (ii) Alice andBob have chosen different measurement basis and theyneed to compute the quantity S of the CHSH inequal-ity with their data to check whether the correlation ofthe outcomes of the entangled pair of qubits has beenclassical ( S ≤ and thus CHSH inequality fulfilled) orquantum mechanical ( S > and thus violates the CHSHinequality). If the CHSH inequality is fulfilled, the com-munication has been eavesdropped by a third party andthus Alice and Bob have to create a new secure key. In case letters are encrypted as binary numbers as presented here,this probability is / n . Figure 8. Measurement directions for the E91 protocol. Thepossible directions that Alice and Bob can choose are denotedas { a , a , a } and { b , b , b } respectively. Note that Aliceand Bob both measure in the same direction if they choose,respectively, ( a , b ) or ( a , b ) . In order to perform a Belltest, they measure in the directions depicted in bold.
2. Rules of the game
In this case, we also need students playing the roles ofAlice, Bob, Eavesdroppers and qubits. The qubits startas entangled pairs (see Figure 4 (Up left)).The goal of this activity is exactly the same as in Sec-tion IV A, i.e., sending a secret message successfully. Theonly difference is how Alice and Bob generate a securekey. Therefore, we refer the reader to Section IV A 2 formore information on how to encode and decode the mes-sage with a key.In this case, Alice and Bob share a pair of entangledqubits in the state | ψ − (cid:105) and they are going to mea-sure them in different directions. In particular, theyperform a Bell test to check if the transmission of thekey is really secure, so among their measurement direc-tions, some of them should be separated by π/ (see Sec-tion III A 2). All possible measurement directions (Fig-ure 8) are painted on the Bloch plane on the floor andthe positive side of each axis is marked with the com-pass stick by Alice and Bob at the beginning of the game(they have to agree on this so that the positive side ofeach axis is the same for both).As explained in Section IV B 1, in order to create asecure key, Alice and Bob measure several pairs of en-tangled qubits in the state | ψ − (cid:105) . For each pair, Alicechooses to measure her qubit in one out of the threedirections { a , a , a } (see Figure 8). She writes downboth the measurement direction she chose and the out-come she obtained. We refer the reader to Section III A 2–Figure 5(b) in particular– for details on the rules ofAlice’s measurement process. Then Bob does the same,i.e. he randomly chooses a measurement direction among { b , b , b } and writes down his axis choice and the out-come. If the chosen direction was the same as Alice’s,the detailed procedure for our game can be seen in Fig-ure 4. Otherwise, the measurement axes are separatedby π/ , i.e. the situation is the same as for a Bell test2(see Section III A 2 and Figure 6).After both of them have performed all the measure-ments, they publicly compare (say out loud) the list ofmeasurement directions. In order to create the key bitstring, Alice and Bob take the outcomes of coincidentmeasurement axis. They have to agree on how to trans-form the outcomes +1 , − into , . Since they alwaysget anti-correlated results when both measure the state | ψ − (cid:105) in the same direction (see Section II B), the easiestway is: when Alice gets outcome +1 ( − ), she writes thebit ( ). Bob does the opposite, i.e., when he gets +1 ( − ), he writes the bit ( ). Hence, both get the samebit of the key when they measure in the same direction.But, how can the Eavesdroppers act in this case? Likein the BB84 protocol, the Eavesdroppers can measure(hit with the ball) Bob’s qubit before him and thensend the qubit to him, hoping that he would not real-ize that his qubit has already been measured. However,the Eavesdroppers’ measurement broke the initial entan-glement, so the qubit Bob receives is no longer entangled.Thus, it is easy for Alice and Bob to detect eavesdroppingby just performing a Bell test and checking the CHSH in-equality (see Eq. (6)). After the measurement process,they just need to take the outcomes of the non-coincidentmeasurement directions, compute |(cid:104) S (cid:105)| and check if itis larger than 2, in which case they know that their qubitswere perfectly entangled and there was no Eavesdropper.Otherwise, they discard the key. Once they have the se-cure key, Alice sends (say out loud) an encrypted messageand Bob can decrypt it. Alice and Bob win if they areable to send the secret message successfully without theEavesdroppers getting access to it, so they have to makesure that their key is secure. The goal of the Eavesdrop-pers is to intercept the message.This whole protocol is based on Alice and Bob sharingpairs of entangled particles. Real entangled particles canbe separated by an arbitrarily large distance and theyare still entangled, which makes the protocol even moreuseful, since the key can be generated among parties thatare far apart from each other. If the teacher wants toemphasize this property, the variant of the game whereentangled qubits are placed in two different Bloch circlescan be played (see Section III A 2). V. SUMMARY
We have introduced a game to teach advanced conceptsof quantum mechanics such as Bell tests and quantumcryptography. Students play the role of both scientistsand qubits so they can make experiments and analyze theoutcomes in the same way as if they were in a real lab-oratory, getting even the same results! This kinesthetic The CHSH inequality reads |(cid:104) a b (cid:105)−(cid:104) a b (cid:105) + (cid:104) a b (cid:105) + (cid:104) a b (cid:105)| ≤ in this particular case. approach allows the students not only to experience first-hand the properties of quantum mechanics, but also tolearn how it is to be a scientist and to develop skills suchas critical thinking and creativity.The Bloch sphere is represented as a circle on thefloor (Bloch plane) and the students themselves repre-sent Bloch vectors by orienting their bodies in the corre-sponding directions. Pairs of entangled qubits are playedby two students holding hands and spinning around thecenter of the circle. 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The CHSH inequality was first mentioned by Clauseret al. in 1969 [33] and gives constraints on theories whichare based on so called local hidden variables [39] and thusare subject to local realism. These local hidden variablesarised through the assumption of Einstein, Podolsky andRosen that quantum mechanics may not be a completetheory and thus become complete by the introduction ofthese local hidden variables. The CHSH inequality gener-alizes Bell’s theorem [38, 39] to experimental realizations[42, 43] and reads in its general form S = | E ( a, b ) − E ( a, b (cid:48) ) + E ( a (cid:48) , b ) + E ( a (cid:48) , b (cid:48) ) | , S ≤ (A.1)where the correlation E ( a, b ) is defined to be the expec-tation value of the product of the outcomes A ( a ) and B ( b ) at measurement settings a and b , i.e., the statisti-cal average of A ( a ) · B ( b ) . The mathematical formulationof the theory of quantum mechanics predicts a maximumvalue for the quantity S of S QM = 2 √ ≈ . which isgreater than 2 that one would maximally obtain for the-ories based on locality and realism. A violation of (A.1)thus shows that this tested theory, in particular quantummechanics, does not follow the rules of local realism.This maximal violation of the CHSH inequality dueto the unique features of quantum mechanics can beachieved by considering one of the maximally entan-gled Bell states, e.g. | ψ − (cid:105) , as used for our approachin the main text. For this calculation, we first definean operator ˆ A ( a ) = (cid:126)r ( a ) (cid:126) ˆ σ = sin( a )ˆ σ x + cos( a )ˆ σ z on theBlochsphere at angle φ = 0 , i.e., our vector (cid:126)r ( a ) rotatesin the x − z plane. This quantum mechanical expecta-tion value E ( a, b ) for two configurations a and b in thestate | ψ − (cid:105) reads E ( a, b ) = (cid:104) ψ − | ˆ A ( a ) ⊗ ˆ B ( b ) | ψ − (cid:105) = − cos ( a − b ) (A.2)where ⊗ denotes the tensor product.For the specific set a = π/ , a (cid:48) = 0 , b = π/ and b (cid:48) = − π/ of measurement angles, these quantum mechanicalexpectation values read E ( a, b ) = E ( a (cid:48) , b ) = E ( a (cid:48) , b (cid:48) ) = − E ( a, b (cid:48) ) = − cos( π/
4) = − / √ . Plugging these valuesinto Eq. (A.1), the quantum mechanical quantity S QM reads S QM = |− · / √ | = 2 √ ≈ . which gives themaximal violation of the CHSH inequality by consideringthe unique features of quantum mechanics. Bell’s derivation
Although Clauser et al. first proposed this inequality,Bell [44] gave a general derivation of this formula in 1971 Here, the variable a denotes the angle θ a in the x − z plane. which will be described here.We first start by defining the measurement outcomes ofan operator in certain directions (cid:126)a and (cid:126)b , where (cid:126)a = a · (cid:126)σ and (cid:126)b = b · (cid:126)σ with (cid:126)σ and (cid:126)σ being normalized vectors,can take the values A ( a ) = ± and B ( b ) = ± . Nowwe introduce a single continuous local hidden variable,called λ . In general (cid:126)λ could be a multi-component vec-tor; yet for the simplicity of the derivation, we will omitthe vector properties. Bell thus – in the spirit of theEPR paradox – introduced this originally proposed lo-cal hidden variable that would make the theory of quan-tum mechanics complete. The measurement outcomes oftwo quantities A and B that now depend on both, themeasurement direction and the additional local hiddenvariable, read A ( a, λ ) = ± , B ( b, λ ) = ± . (A.3)Furthermore, we assume locality, i.e., measurement of A is independent on the setting b and measurement of B isindependent on a , respectively.Let p ( λ ) be the normalized probability distribution of λ , i.e. (cid:82) R p ( λ ) dλ = 1 . Since a source emits entangledparticles in two distant directions in a physical mannerindependent on the parameters a and b , p ( λ ) is not de-pendent on a and b either.According to Ref. [39], we define the expectation valueof the inner product of A and B as the correlation func-tion E ( a, b ) = (cid:90) Γ A ( a, λ ) B ( b, λ ) p ( λ ) dλ (A.4)where Γ is the set of all possible values of λ .With a (cid:48) and b (cid:48) being alternative measurement settings,we obtain E ( a, b ) − E ( a, b (cid:48) )= (cid:90) Γ ( A ( a, λ ) B ( b, λ ) − A ( a, λ ) B ( b (cid:48) , λ )) p ( λ ) dλ = (cid:90) Γ A ( a, λ ) B ( b, λ )[1 ± A ( a (cid:48) , λ ) B ( b (cid:48) , λ )] p ( λ ) dλ − (cid:90) Γ A ( a, λ ) B ( b (cid:48) , λ )[1 ± A ( a (cid:48) , λ ) B ( b, λ )] p ( λ ) dλ. (A.5)Now, we take the absolute value of (A.5) and by usingthe fact that | A | , | B | ≤ , we further obtain | E ( a, b ) − E ( a, b (cid:48) ) | ≤ = (cid:90) Γ | ± A ( a (cid:48) , λ ) B ( b (cid:48) , λ ) | p ( λ ) dλ + (cid:90) Γ | ± A ( a (cid:48) , λ ) B ( b, λ ) | p ( λ ) dλ = 2 ± (cid:90) Γ | A ( a (cid:48) , λ ) B ( b (cid:48) , λ ) + A ( a (cid:48) , λ ) B ( b, λ ) | p ( λ ) dλ = 2 ± | E ( a (cid:48) , b (cid:48) ) + E ( a (cid:48) , b ) | . (A.6)5We can rewrite (A.6), by using symmetry properties ofthe absolute value and the triangle inequality, as | E ( a, b ) − E ( a, b (cid:48) ) + E ( a (cid:48) , b ) + E ( a (cid:48) , b (cid:48) ) | < | E ( a, b ) − E ( a, b (cid:48) ) | + | E ( a (cid:48) , b ) + E ( a (cid:48) , b (cid:48) ) | ≤ (A.7)which describes the CHSH inequality of Eq. (6) in themain text. Appendix B: Mermin’s approach
In this section, we present an alternative approach totest local realism based on the work "Is the moon therewhen nobody looks? Reality and the Quantum theory",by N. David Mermin [34]. The main idea is the same,i.e., the quantum properties of the entangled state | ψ − (cid:105) contradict the classical concepts of locality and realism.However, in this case there is not a Bell inequality as theCHSH inequality to check. Instead, a different experi-mental setup is proposed.There are three different measurement directions sepa-rated by an angle of π/ instead of π/ as in the CHSHapproach highlighted in Section III A. As before, if Aliceand Bob both measure the state | ψ − (cid:105) in the same di-rection, they always obtain anti-correlated results. How-ever, if they measure in directions separated by π/ , theprobability of getting anti-correlated results is . If one considers a data set containing the measure-ment results of Alice and Bob measuring the state | ψ − (cid:105) in all possible combinations of measurement directions,the probability to find two outcomes that are different is p diff = p same dir · p diff + p diff dir · p diff (B.1) = 13 · ·
14 = 12 , (B.2)i.e., the probability that both measure in the same di-rection and obtain different results or they measure indifferent directions and obtain different results. These probabilities are obtained by using the projection operator P ( n, ± ) = ( ± (cid:126)n · (cid:126)σ ) onto the states | n (cid:105) ( + ) and | n (cid:105) ( − ) ofa generic direction (cid:126)n . (cid:126)σ denotes the vector ( σ x , σ y , σ z ) , where σ x,y,z are the Pauli matrices. If one denotes Alice’s measurementdirection as a and Bob’s as b , the resulting probabilities are: p (+ A , + B ) = (cid:104) ψ − | P A ( a, +) P B ( b, +) | ψ − (cid:105) = 14 (1 − cos α ) ,p ( − A , − B ) = (cid:104) ψ − | P A ( a, − ) P B ( b, − ) | ψ − (cid:105) = 14 (1 − cos α ) ,p (+ A , − B ) = (cid:104) ψ − | P A ( a, +) P B ( b, − ) | ψ − (cid:105) = 14 (1 + cos α ) ,p ( − A , + B ) = (cid:104) ψ − | P A ( a, − ) P B ( b, +) | ψ − (cid:105) = 14 (1 + cos α ) , where p (+ A , + B ) is the probability that Alice gets the out-come +1 and Bob gets − , etc. and α is the angle betweenthe measurement directions a and b . Thus, the probabilityof getting anti-correlated results when α = 2 π/ is p diff = p (+ A , − B ) + p ( − A , + B ) = 1 / . Note that for the same mea-surement direction ( α = 0 ): p diff = 1 . Let us now consider the case in which we assume re-alism in our theory, i.e., the properties of the state arefixed before the measurement. Since the results from Al-ice and Bob by measurement in the same direction areanti-correlated, one can conceive two possible situationsin which the qubits have fixed properties and give suchresults:1. Alice’s qubit has fixed properties such that it al-ways gives +1 when measured in direction a , +1 when measured in a and +1 when measured in a . Thus, Bob’s qubit should have fixed propertiessuch that it gives ( − , − , − when measured indirections ( b , b , b ) , respectively, in order for theoutcomes to be anti-correlated. In this case, Aliceand Bob always get different outcomes no mattertheir choice of measurement direction.2. Alice’s qubit has fixed properties such that it al-ways gives +1 when measured in direction a , − when measured in a and +1 when measured in a ; so Bob’s qubit should give ( − , +1 , − aftermeasurement in ( b , b , b ) , respectively. Thus, theoutcomes are different in the cases where Alice andBob measure in the following directions: a − b , a − b , a − b , a − b and a − b , i.e. in / ofthe times. All other possible sets of fixed outcomesare analogous to this case and also give differentoutcomes / of the time.Thus, in order to check their theory for local real-ism, Alice and Bob just need to make measurements ofthe state | ψ − (cid:105) in all possible combinations of measure-ment directions and then count the number of times theyobtain different outcomes. If they obtain different out-comes / of the time or more local realism is still valid,whereas if they obtain different outcomes / of the time,then local realism cannot be assumed. In order to dis-tinguish between the probabilities p cl = 1 / . and p qm = 5 / ≈ . , there should be enough measure-ments to get good statistics and a reliable result. Anestimation of the necessary number of students and mea-surements per student is given in Appendix C.Within the framework of our proposal, the measure-ment process of the entangled pair of qubits is exactlythe same as the one detailed in Section III A in the maintext. The only difference in this second approach is thatthe measurement directions are separated by an angleof π/ instead of π/ . Alice measures first in the ex-act same way as described in Figure 5(b). Then, if Bobmeasures in directions that are separated by π/ fromAlice’s direction, the corresponding qubit should throwthe biased die. In this case, the die should be designedsuch that with a probability of the qubit rotates tothe closer side of the axis and with it rotates to thefurther side (see Figure 6).The advantage of this approach is that there is no needto introduce any mathematical concept such as the CHSHinequality in advance. Therefore, we suggest that the ac-tivity is conducted as follows: first, the concept of local6realism is presented and the students should discuss theirown intuitions and ideas with respect to it. Then, theycompute the probabilities for the different cases basedon local realism hypothesis (see points 1 and 2 in theenumeration above). Finally, the students perform themeasurements with the rules of the game we propose andanalyze the probabilities they obtain to compare and con-trast their initial intuitions with the "real" experiment. Comparison of approaches
We have proposed two different ways of testing localrealism to give the teachers the opportunity to choosehow the content is presented. On the one hand, theCHSH approach has the advantage that fewer measure-ments are needed to get good statistics and compute thequantity S (see Appendix C). However, the CHSH in-equality is harder to explain to students without goinginto much mathematical depth. On the other hand, Mer-min’s approach provides the possibility to directly com-pare the initial intuition of the students, presumably infavor of local realism, to the quantum non-intuitive re-sults. Thus, they can experience the whole process oftesting an hypothesis (local realism) with experimentalresults and elaborating explanations after the contradic-tory results. In this case, the activity should be presentedwithout previous lectures on local realism and the differ-ence with the quantum case, so that the students canexperience it by themselves. The downside of this lat-ter approach is that the number of measurements neededto get good statistics is much larger than for the CHSHapproach (see Appendix C). Appendix C: Statistics1. Examples of biased dice
Here, we depict for some specific sets of measurementsettings a , b , a (cid:48) and b (cid:48) examples of the biased die thatthe students should throw when perfoming the game forthe different approaches.For the Mermin’s approach of Appendix B, an angle of π/ between each measurement setting of a and b giveswith Eqs. (2) and (3) from the main text the probabilities p = cos ( π/
3) = 1 / and p = sin ( π/
3) = 3 / . Thus,the students can use a tetrahedron to perform the gameas proposed. A tetrahedron can also be easily built bythe students beforehand as an external activity.For the CHSH inequality of Section III A in the maintext, an angle of π/ between each measurement set-ting is used. Here, the corresponding probabilities read p = cos ( π/ ≈ . and p = sin ( π/ ≈ . . Inthis case, a possible biased die could be an icosahedron,i.e. 20-sided die. Here, 17 of 20 sides correspond to prob-ability p , 3 of 20 sides correspond to p . The studentsshould build the dice first by themselves as an external Game angle p p dieMermin π/ π/ ≈ ≈ x − z basis π/ a, b, a (cid:48) and b (cid:48) . activity. Since the probabilities are not precisely 0.85 and0.15, respectively, there will be a statistical relative errorof (0 . − . / . ≈ . that cummulates withthe number of dice thrown.To circumvent (or at least shrink) this statistical error,freely available apps for smartphones or computers maybe used as an alternative multi-faced die (for example,RNG Plus for Android or Roll Dice Online) where onecan define the number of sides of the die. For the casedescribed above, for example a 1000-sided die may bethrown. If the number of the simulated thrown die isbetween 1 and 854, the event with probability p takesplace; for a number between 855 and 1000, the event withprobability p takes place.For the last case of Section II A 2 in the main textwhere the x and z directions are the measuring basis,i.e. an angle of π/ between each measurement setting,the probabilities are p = p = 0 . which can be realisedby throwing a standard die (where e.g. odd numberscorrespond to p and even numbers to p ) or a standardcoin.An overview over the proposed dice in the differentapproaches can be seen in Table C.1.
2. Statistical error
The two approaches introduced in the main text re-quire multiple measurements of each quantity whosestatistics we describe here. We present some estimationsof the number of measurements of the quantity S in theCHSH approach and probability p in the Mermin’s ap-proach, respectively, such that the students obtain a sta-tistical reliable answer to the question whether the rulesof the game were based on the theory of quantum me-chanics or local realism.As a measure of distinguishability of the students’ mea-surement outcome between quantum mechanics and lo-cal realism, we consider the statistical standard deviationof each quantity. The standard deviation of a quantitywhich is measured multiple times is defined as σ = 1 √ N (C.1)7where N is the total number of measurements.For the CHSH approach presented in Section III A inthe main text, the goal of the proposed game is that thestudents shall violate the CHSH inequality, i.e., obtaininga value for the quantity S that is larger than 2. This thustells that the game the students played – and in particularthe theory of quantum mechanics – does not follow localrealism.We say that we accept the outcome (either CHSH in-equality is violated or is not violated) when we can statis-tically claim the correctness and reliability of the value ofour quantity S , i.e., we want to statistically distinguishthe two outcomes with a high certainty. Here, two errorscan occur: Either our null hypothesis H (here, quantummechanics does not follow the theory of local realism) iscorrect but we erroneously refuse it (type one error) dueto our statistics, or our null hypothesis is wrong but weerroneously accept it (type two error). For our signifi-cance level α as the probability of error given for the nullhypothesis H , we allow a value of , i.e., our confi-dence of the correct result is . In general, the typeone error is the graver error which we thus want to avoid.In the game based on the Mermin’s approach of Ap-pendix B, the students have to distinguish between theprobabilities p qm = 1 / . for a game that is based onthe features of quantum mechanics and p cl = 5 / ≈ . for a theory based on local realism. Our motivation is todistinguish between these two probabilities p qm and p cl .We assume a Gaussian distribution for both probabili-ties for which we need to calculate the standard devia-tion σ such that they are perfectly distinguishable (notoverlapping). For a one-tailed test with Gaussian distri-bution, of the obtained results lay in the interval I qm = [ p qm ; p qm + 1 . σ ] . Therefore, we set our upperborder I r = 5 / such that the obtained result was dueto a game based on the unique features of quantum me-chanics with success probability and consequently with significance level α = 0 . , i.e., | . − / | = 1 . σ and thus the standard deviation reads σ ≈ . . Sub-sequently and according to Eq. (C.1), the total num-ber of measurements to be taken by the students is N ≈ . Alternatively, one can calculate this num-ber for N by making the ansatz that the right border ofthe interval I qm just equals the left border of the interval I qm = [ p cl − . σ ; p cl ] , i.e. p qm + 1 . σ = p cl − . σ and solve for N and which leads to the same result.For the CHSH approach of Section III A in the maintext, the calculation of the statistics is similar. In theCHSH game [45], the classical win probability p cl for Al-ice and Bob to answer at least one of two questions askedby a neutral referee correctly is p cl = 0 . . On the otherhand, when Alice and Bob share an entangled pair ofqubits, the winning percentage rises up to a maximumvalue of p qm = 0 . . / √ ≈ . . Again we assumea Gaussian distribution of the measured values for thewin probability p around its mean value p qm ≈ . asthe maximal winning percentage (thus we have a one-tailed hypothesis test where of the obtained re-sults lay in the interval I = [ p qm − . σ ; p qm ] . There-fore, | . − . | = 1 . σ gives the standard deviation σ ≈ . and thus N ≈250