Alice and Bob use Aravind's version of the Bell-Kochen-Specker theorem to fend off awkward questions about what exactly they were doing in Amsterdam last week

Anthony Sudbery

Kochen S.

English

arXiv

Sudbery, Anthony

White Rose Research Online

Alice and Bob get away with it: A playlet

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This is a repository copy of Alice and Bob get away with it: A playlet.White Rose Research Online URL for this paper:http://eprints.whiterose.ac.uk/3458/Article:Sudbery, Anthony (2007) Alice and Bob get away with it: A playlet. American Journal of Physics. pp. 720-723. ISSN 0002-9505 https://doi.org/10.1119/1.2750380eprints@whiterose.ac.ukhttps://eprints.whiterose.ac.uk/Reuse Items deposited in White Rose Research Online are protected by copyright, with all rights reserved unless indicated otherwise. They may be downloaded and/or printed for private study, or other acts as permitted by national copyright laws. The publisher or other rights holders may allow further reproduction and re-use of the full text version. This is indicated by the licence information on the White Rose Research Online record for the item. Takedown If you consider content in White Rose Research Online to be in breach of UK law, please notify us by emailing eprints@whiterose.ac.uk including the URL of the record and the reason for the withdrawal request. promoting access to White Rose research papers      Universities of Leeds, Sheffield and York http://eprints.whiterose.ac.uk/    This is an author produced version of a paper subsequently published in American Journal of Physics.   White Rose Research Online URL for this paper: http://eprints.whiterose.ac.uk/3458/    Published paper Sudbery, Anthony (2008) Alice and Bob get away with it: A playlet. American Journal of Physics, 75 (8) pp. 720-723.   White Rose Research Online eprints@whiterose.ac.uk  Alice and Bob get away with it: A playletAnthony SudberyDepartment of Mathematics, University of York,Heslington, York, England YO10 5DDEmail: as2@york.ac.ukAbstractAlice and Bob use Aravind’s version of the Bell-Kochen-Speckertheorem to fend off awkward questions about what exactly they weredoing in Amsterdam last week.1 IntroductionJacobs and Wiseman[1] use a story about a robbery to illustrate Bell’s the-orem, discussed in Mermin’s version[2] based on the Greenberger-Horne-Zeilinger (GHZ) state of three qubits [3]. They ask if a story can be con-structed around Aravind’s recent version [4] of the Bell-Kochen-Specker the-orem [5, 6]. Aravind’s proof is a particularly clear and simple demonstra-tion of the incompatibility between quantum mechanics and our intuitiveclassical ways of thinking about the physical world. He shows how an en-tangled state of four qubits (for example, the spin states of four electrons)can be used by two separated people, with no means of communication, toperform a joint task that would appear to be impossible without communi-cation. This and similar applications of quantum entanglement have becomeknown as “pseudo-telepathy.”[7] Although it cannot be used as an instanta-neous method of communication, pseudo-telepathy is possibly even closer tothe science-fiction idea of telepathy than quantum teleportation[8] is to itsscience-fiction namesake. It is therefore an attractive subject for describingthe peculiarities of quantum information to students and the general public,and one that is particularly well suited to illustration in story form.Such a story was told in a little play which was part of the MerchantAdventurers’ Science Discovery lecture, a public lecture given in York inMarch 2006. The script of this playlet is given in Sec. II. The theory of thedevice used by Alice and Bob is explained in Sec. III.12 The script: Alice and Bob Get Away WithItThe play was first performed in the Merchant Adventurers’ Hall, York, on 7March 2006. The part of Alice was played by Chris Higgins, that of Bob byGed Murray. The Lecturer was Tony Sudbery.The lecturer has just explained the EPR experiment and why it led Ein-stein, Podolsky and Rosen to conclude that each qubit has definite propertieswhich determine the results of all possible experiments on the qubit.Lecturer: But Einstein was wrong. John Bell showed that there are situ-ations where we can’t understand the effects of entanglement bysaying that each side has its own independent properties. Andthis is what gives entanglement its power: it gives rise to telepor-tation, [8] telepathy, [7] and —BOB (coming up from audience): Hey, wait a minute. You don’t expectus to believe this crap, do you? Parallel universes, telepathy,teleportation. . .We came here for a serious science lecture, andyou give us science fiction. This is the Merchant Adventurers’Hall, you know, not the City Screen.Lecturer: It’s Bob!Alice (coming up from audience): Just you be quiet, Bob Murray, and hearwhat he’s got to say. You might learn something if you give yourears a chance.Lecturer: Alice! It’s all coming true.Alice: Did you mention telepathy? That’s the only thing that would helpus in our current problem.Lecturer: What problem is that?Alice: We’ve just got to get our stories straight about last week.Bob: Won’t you ever give up? I’ve proved to you that we’re never going tobe able to do that the way you want to.Alice (sarcastically): Mathematically, I suppose.Bob: Well, yes.Lecturer: Sounds interesting. Tell me more.Alice: We’re accountants, and last week we were in Amsterdam looking atthe accounts of three branches of our client’s company. Their head2office thinks there’s some funny business going on. Tomorrow we’llbe debriefed separately. A senior accountant will look at one ofthe branches with Bob, and our head of IT will look at the emailtraffic between the branches on one of the three days that we werethere. We’ve got all the data, but we’ll have to tell them exactlywhich branch we visited on which day.Lecturer: What’s the problem? Why don’t you just tell them what youdid?Bob and Alice exchange looks.Lecturer: I see. I didn’t know accountants had such interesting lives. Well,why can’t you get together beforehand and decide on your story?Alice: The thing is, it will make life much easier if we can say we did thingsin a particular way. On each day we should visit at least onebranch; maybe just one, but if we go on to a second one we mustalso visit the third on the same day, otherwise the first two willhave time to discuss things and warn the third one. We don’thave to have inspected all three branches, but if we do visit onewe must go twice – after the first visit we would look up ourrecords in the evening and then go back to check the next day.Bob: So when they ask Alice what we did on one of the three days, we wantto say we went to either one or three branches; and when theyask me what we did about one of the branches, I want to say thateither we didn’t visit it or we visited it on two days. But we can’tput together a schedule that does that! If we make either one orthree visits on each of three days, then altogether we would havemade an odd number of visits (three odd numbers add up to anodd number), but if we visit each branch either no times or twice,that makes an even number of visits in total.Lecturer: Surely your bosses can work that out just as well as you can.They can’t expect you to have done the impossible.Alice: No, but if we don’t do it this way they will want to probe a lotfurther. It would save a lot of awkward questions if we can besure that whatever they ask us, we have an answer that fits theserules. But we’ve got to be sure that when they compare notes ourstories are compatible — we must both say that we went to thesame branch on the same day.Lecturer: I’m not sure that I really understand why you want to do this —I don’t think I want to know. But yes, entanglement can do it for3you. As if you weren’t entangled enough already.Let me see if I’ve got it straight. You have to convince yourbosses that you undertook a schedule of visits to the three dif-ferent branches, A, B and C, on the three days that you were inAmsterdam. So you could fill in the grid in Table 1 by putting atick in the appropriate square if you visited a particular branchon a particular day. Alice will be asked to fill in one of the rows,and she wants to do so with an odd number of ticks; Bob will haveto fill in one of the columns, and he wants to put an even numberof ticks. But you want to be sure that you both put the samething in the square where Alice’s row and Bob’s column intersect.Right?Branch A Branch B Branch CTuesdayWednesdayThursdayTable 1: The grid to be filled in by Alice and BobAlice and Bob (together): Right.Lecturer: You’re right, you can’t do that with a complete schedule. What youneed is the Quantum Entangulator. Here you are – you each havea handset, cunningly disguised as a soft toy (a cat, as it happens).Each of these contains a pair of electrons, and a measuring devicewhich will do an experiment on the electrons and translate theresult into an answer to the question you have been asked, obeyingthe restrictions you’ve described. The electrons are all entangledtogether in such a way that the answers are guaranteed to fittogether.Let’s test it. Switch them on. Now, Alice, suppose you’reasked which branches you visited on Tuesday. Press button 1 tofind out what you should answer.Alice: It says B.Lecturer: And Bob, suppose you’re asked on which days you visited branchB. Press button B to find your answer.Bob: Tuesday and Wednesday.Lecturer: You see! They match up. And they always will. As Bob proved,the answers cannot be programmed in advance; there is no phys-ical connection between the two ends of the Entangulator — no4wires, no phone line, no radio; but they are guaranteed to givecompatible answers.Bob: Fantastic! The university should patent this!Lecturer: Not our idea, I’m afraid, Bob. [4]Alice: But have they realized what they’ve got here? When I got my diaryentry from my end of the Entangulator, Bob’s end immediatelygot the information about the branch they were asking about.No radio or anything — so it was going faster than light! You’vereally beaten Einstein in a big way here. It may not matter now,but think what it will mean in the future! Like, when spaceshipsare trying to send messages over huge distances — beating thespeed of light will be really important then.Lecturer: Calm down, Alice. I agree it looks good — but how are you goingto use it?Alice: Well, I put my message in the entangulator thingy, and it immediatelycomes out at Bob’s end.Lecturer: But Alice, when you use the entangulator tomorrow morning,you’ll never put anything into it. You only get something outof it — and you have no control over what that is. Bob gets acorresponding thing out, but there’s no way you can make thatbe the message you want to send.Other people have suggested that something was movingfaster than light in the entangulator, but they really ought toknow better.[9] When you activate your side of it, it seems tochange Bob’s side, it’s true, because it changes the possible an-swers that can be flashed up on his screen; some answers be-come impossible. But Bob doesn’t know that. Just because hedoesn’t get a particular answer on a single occasion, he can’t de-duce that it’s become impossible. He doesn’t get any informationuntil you’ve told him independently what answer you got.In general, there’s a theorem that entanglement can never beused by itself to send signals. [10] In particular, it can’t commu-nicate faster than light. So if you want to send Bob a kiss in thequantum wonderland, you’ll just have to do it the old-fashionedway.53 The ExplanationThe entangulator contains four qubits, two in Alice’s cuddly toy and two inBob’s, in the state|Ψ〉 = |00〉A|00〉B + |01〉A|01〉B + |10〉A|10〉B + |11〉A|11〉B, (1)which can be written as|Ψ〉 = |E1〉A|E1〉B + |E2〉A|E2〉B + |E3〉A|E3〉B + |E4〉A|E4〉B (2)where |E1〉, . . . |E4〉 is any orthonormal basis of two-qubit states that havereal coefficients when expanded in the basis |00〉, |01〉, |10〉, |11〉. Hence ifAlice measures her two qubits in any such basis and obtains the result |Ei〉,Bob’s qubit will be projected into the same state |Ei〉. The measurementsmade by Alice and Bob are shown in Table 2:Branch A Branch B Branch CTuesday I ⊗ σz σz ⊗ I −σz ⊗ σzWednesday σx ⊗ I I ⊗ σx −σx ⊗ σxThursday σx ⊗ σz σz ⊗ σx −σy ⊗ σyTable 2: The measurements made by Alice and BobEach row and each column of Table 2 contains a set of three commutingtwo-qubit observables. Each row therefore determines a basis of two-qubitstates, namely the simultaneous eigenvectors of the observables in that row.The transformation matrices between these bases are purely real, so any ofthe bases could serve as |E1〉, . . . , |E4〉 in Eq. (2).When she is asked about a particular day, Alice measures the commutingobservables in the corresponding row of the table, and converts the result intoan answer to the question about which branches they visited that day: if theobservable under branch X is measured to have the value −1, she says thatthey visited branch X that day; if the value is +1 she says they didn’t visit it.This measurement projects Bob’s pair of qubits onto the same simultaneouseigenstate. When he is asked when they visited branch X, he measures thecommuting observables in the corresponding column, and converts the resultinto an answer to the question of when they visited this branch, in a similarway to Alice. If the observable in the Tuesday row has the value −1, he saysthey visited branch X on Tuesday; otherwise they didn’t. Because his stateis now an eigenstate of the observable that is common to his set and Alice’s,he is bound to obtain the same value for this observable. He therefore givesthe same answer as Alice to the question in the square common to his column6and her row. Thus, they both agree as to whether they visited his branch onher day.The product of the observables in each row is −I ⊗ I, while the productof the observables in each column is I⊗I. The result of Alice’s measurementtherefore includes −1 an odd number of times — she says that they visited anodd number of branches on each day — while the result of Bob’s measurementincludes −1 an even number of times – he says that they visited each branchan even number of times. Whatever they were doing in Amsterdam, theyget away with it.References[1] K. Jacobs and H. M. Wiseman, “An entangled web of crime: Bell’s the-orem as a short story,” Am. J. Phys. 73, 932–937 (2005).[2] N. D. Mermin, “Quantum mysteries revisited,” Am. J. Phys. 58, 731–734(1990).[3] D. M. Greenberger, M. A. Horne, and A. Zeilinger. Bell’s theorem withoutinequalities. Am. J. Phys., 58, 1131–1143, 1990.[4] P. K. Aravind, “Bell’s theorem without inequalities and only two distantobservers,” Found. Phys. Lett. 15, 397–405 (2002); quant-ph/0206070.[5] J. S. Bell. On the problem of hidden variables in quantum mechanics.Rev. Mod. Phys., 38, 447–452, 1966.[6] S. Kochen and E. Specker. The problem of hidden variables in quantummechanics. J. Math. Mech., 17, 59–87, 1967.[7] G. Brassard, A. Broadbent, and A. Tapp, “Quantum pseudo-telepathy,”Found. Phys. 35, 1877–1907 (2005); quant-ph/0407221.[8] C. H. Bennett, G. Brassard, C. Crépeau, R. Jozsa, A. Peres, and W.Wootters, “Teleporting an unknown state via dual classical and EPRchannels,” Phys. Rev. Lett. 70, 1895–1899 (1993).[9] V. Scarani, W. Tittel, H. Zbinden, and N. Gisin, “The speed of quantuminformation and the preferred frame: Analysis of experimental data,”Phys. Lett. A 276, 1–7 (2000); quant-ph/0007008.[10] G. C. Ghirardi, A. Rimini, and T. Weber, “A general argument againstsuperluminal transmission through the quantum-mechanical measure-ment process,” Lett. Nuovo Cimento 2, 293–298 (1980).7

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Alice and Bob get away with it: A playlet

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