QQuantum PBR Theorem as a Monty Hall Game
Del Rajan ID and Matt Visser ID School of Mathematics and Statistics, Victoria University of Wellington, Wellington 6140, New Zealand. (Dated: L A TEX-ed September 20, 2019)The quantum Pusey–Barrett–Rudolph (PBR) theorem addresses the question of whether thequantum state corresponds to a ψ -ontic model (system’s physical state) or to a ψ -epistemic model(observer’s knowledge about the system). We reformulate the PBR theorem as a Monty Hall game,and show that winning probabilities, for switching doors in the game, depend whether it is a ψ -onticor ψ -epistemic game. For certain cases of the latter, switching doors provides no advantage. Wealso apply the concepts involved to quantum teleportation, in particular for improving reliability. Introduction:
No-go theorems in quantum foundationsare vitally important for our understanding of quantumphysics. Bell’s theorem [1] exemplifies this by showingthat locally realistic models must contradict the experi-mental predictions of quantum theory.There are various ways of viewing Bell’s theoremthrough the framework of game theory [2]. These arecommonly referred to as nonlocal games, and the bestknown example is the CHSH game; in this scenario theparticipants can win the game at a higher probabilitywith quantum resources, as opposed to having access toonly classical resources. There has also been work on therelationship between Bell’s theorem and Bayesian gametheory [3–5]; in a subset of cases it was shown that quan-tum resources provide an advantage, and lead to quan-tum Nash equilibria. In [6], it was shown that quantumnonlocality can outperform classical strategies in gameswhere participants have conflicting interests. In [7], anonlocal game was constructed where quantum resourcesdid not offer an advantage.Beyond Bell’s theorem, entropic uncertainty relationscan be viewed in the framework of a guessing game [8,9]; the uncertainty relation constraints the participant’sability to win the game. More broadly, the relationshipbetween quantum theory and game theory is investigatedin [10–12]. The Monty Hall game [13–16] has also beengeneralized into quantum versions [17–23].The Pusey–Barrett–Rudolph (PBR) theorem [24] is arelatively recent no-go theorem in quantum foundations.It addresses the question of whether the quantum statecorresponds to a ψ -ontic model (physical state of a sys-tem) or to a ψ -epistemic model (observer’s knowledgeabout the system) [25]. Notable developments on thePBR theorem and ψ -epistemic models have been carriedout in [26–34], including on the issue of quantum indis-tinguishability [35–37], as well being interpreted throughthe language of communication protocols [38, 39].Analogous to the game formulation of Bell’s theorem,a desirable construction is to view the PBR theoremthrough the lens of a game. One instantiation of this is inan exclusion game where the participant’s goal is to pro-duce a particular bit string [40, 41]; this has been shownto be related to the task of quantum bet hedging [42]. Furthermore, concepts involved in the PBR proof havebeen used for a particular guessing game [43].In this Letter, we reformulate the PBR theorem intoa Monty Hall game. This particular gamification of thetheorem highlights that winning probabilities, for switch-ing doors in the game, depend on whether it is a ψ -onticor ψ -epistemic game; we also show that in certain ψ -epistemic games switching doors provides no advantage.This may have consequences for an alternative experi-mental test of the PBR theorem. Furthermore, we shallalso use the concepts involved for modifying quantumteleportation [44, 45] to view it as a Monty Hall game.Using these notions, we develop an error-correcting strat-egy for unreliable teleportation which may be relevant forpractical quantum networks. PBR theorem:
We provide a rough sketch of the PBRproof [24], and highlight crucial outcomes. Two quantumsystems are prepared independently, and each system isprepared in either state | (cid:105) or state | + (cid:105) = ( | (cid:105) + | (cid:105) ) / √ | Ψ (cid:105) = | (cid:105) ⊗ | (cid:105) , | Ψ (cid:105) = | (cid:105) ⊗ | + (cid:105) , | Ψ (cid:105) = | + (cid:105) ⊗ | (cid:105) , | Ψ (cid:105) = | + (cid:105) ⊗ | + (cid:105) . (1)The total system is brought together and measured inthe following entangled basis: | Φ (cid:105) = √ ( | (cid:105) ⊗ | (cid:105) + | (cid:105) ⊗ | (cid:105) ) , | Φ (cid:105) = √ ( | (cid:105) ⊗ |−(cid:105) + | (cid:105) ⊗ | + (cid:105) ) , | Φ (cid:105) = √ ( | + (cid:105) ⊗ | (cid:105) + |−(cid:105) ⊗ | (cid:105) ) , | Φ (cid:105) = √ ( | + (cid:105) ⊗ |−(cid:105) + |−(cid:105) ⊗ | + (cid:105) ) , (2)where |−(cid:105) = ( | (cid:105) − | (cid:105) ) / √ |(cid:104) Φ i | Ψ h (cid:105)| , where i, h = 1 , , ,
4, we have for i = h , |(cid:104) Φ i | Ψ i (cid:105)| = 0. Thismeans that for any value i , the outcome | Φ i (cid:105) never oc-curs when the system is prepared in quantum state | Ψ i (cid:105) .The PBR proof showed that in ψ -epistemic models thereis a non-zero probability q (whose value does not need tobe specified) that outcome | Φ i (cid:105) occurs when state | Ψ i (cid:105) isprepared, thereby contradicting the predictions of quan-tum theory; hence one can infer that the quantum statecorresponds to a ψ -ontic model. a r X i v : . [ qu a n t - ph ] S e p Classic Monty Hall:
A character named Monty hostsa game show. There are three closed doors respectivelylabelled { , , } . There is a prize behind one door, andgoats behind the remaining two. The prize door is de-noted A i where i takes one of the door labels, and thischoice of prize door is made by the producers of the show.We assume in the game that when a random choice needsto be made, all options are chosen with the same proba-bility. Hence, we have P ( A i ) = 1 / i . Thecontestant on the show, who doesn’t know which doorthe prize is behind, gets to pick a door; we label thisas B j where j takes door labels; given this is a randomchoice, we have P ( B j | A i ) = 1 /
3, for all values i, j . Next,Monty who knows where the prize is, has to open a goatdoor, C k where k takes one door labels. Monty’s decisionis constrained through the game rule that he can’t openthe door chosen by the contestant. Hence we have thefollowing conditional probabilities: P ( C k | B j ∩ A i ) = , if i = j (cid:54) = k, , if i (cid:54) = j (cid:54) = k, , otherwise . (3)Once a goat door is opened, Monty offers the contestantthe option to stick with the original choice or switch tothe other unopened door. By sticking, the contestant’sprobability of opening the prize door is 1 /
3. Counter-intuitively, by switching doors, the probability of winningincreases to 2 /
3. This can be seen by computing the non-zero joint probabilities for all events P ( A i ∩ B j ∩ C k ) = P ( C k | B j ∩ A i ) P ( B j | A i ) P ( A i ) , and then summing those values for the events where thecontestant would win by switching. This results in P (win if switch) = (cid:88) i (cid:54) = j (cid:54) = k P ( A i ∩ B j ∩ C k ) = 23 . (4) Ignorant Monty Hall:
Just as in the Classic case, wehave P ( A i ) = 1 / P ( B j | A i ) = 1 /
3, for all values i, j .But in this game, Monty doesn’t know what lies behindany of the doors. The only constraint is that Monty can’topen the door chosen by the contestant, hence we have P ( C k | B j ∩ A i ) = (cid:40) , if j = k, , otherwise , (5)There is now a probability that he will open up the prizedoor by accident, and thus ending the game: P (opens prize door) = (cid:88) i = k (cid:54) = j P ( A i ∩ B j ∩ C k ) = 13 . This implies that the probability that he opens a goatdoor is 2 /
3. The joint probability that Monty opens agoat door and the contestant wins by switching doors can be computed to be 1 /
3. From the last two values, we cancalculate the conditional probability P (win if switch | opens goat door) = 1 / / . (6)This means if Monty opens a goat door, then the con-testant’s probability of winning is the same whether thecontestant chooses to switch the door or not. ψ -ontic Monty Hall game: Antidistinguishability [32,46, 47], where there is a measurement for which each out-come identifies that a specific member of a set of quantumstates was definitely not prepared, is highlighted in thePBR proof by |(cid:104) Φ i | Ψ i (cid:105)| = 0 for all i . We will exploitthis to construct our game, which can be thought of as aquantum Ignorant Monty Hall game.For state | Ψ (cid:105) in (1), we have |(cid:104) Φ | Ψ (cid:105)| = 0 , |(cid:104) Φ | Ψ (cid:105)| = 1 / , |(cid:104) Φ | Ψ (cid:105)| = 1 / , |(cid:104) Φ | Ψ (cid:105)| = 1 / . (7)For the other states in (1), the same probability distri-bution (0, 1 /
4, 1 /
4, 1 /
2) occur but across the differentoutcomes (2); hence we will focus our game on | Ψ (cid:105) , butsimilar constructions hold for the other states.The Monty Hall gamification is as follows: There arefour doors labelled { , , , } , and these correspond tothe different measurement outcomes listed in (2). Theprize door A i , where i takes one of the door labels, isthe outcome | Φ i (cid:105) that the state | Ψ (cid:105) collapses to uponmeasurement. For a ψ -ontic game, through the Bornprobabilities (7), we have P ( A i ) = |(cid:104) Φ i | Ψ (cid:105)| .The contestant on the show doesn’t know what statefrom (1) is used, and is only aware of the possible mea-surement outcomes (2). Based on this limited informa-tion, the contestant randomly picks one of the doorswhich we denote B j where j is the corresponding doorlabel; hence we have P ( B j | A i ) = 1 /
4, for all values i, j .Monty’s decision corresponds to the predictions ofquantum theory. He is aware that state | Ψ (cid:105) was used,and has access to the Born probabilities (7). The dooropened by Monty is denoted C k where k is one of the doorlabels. The main insight to construct this game is thatwhen Monty opens a goat door, he is opening a door thathas probability zero of having a prize in it. And for ourgame, a door that definitely does not have a prize in itcorresponds to outcome | Φ (cid:105) as P ( A ) = |(cid:104) Φ | Ψ (cid:105)| = 0.Hence in this game, Monty will open door C unless thecontestant has already chosen this door as their pick (asMonty can’t open the door chosen by the contestant);in that case Monty will open one of the other remainingdoors with equal probability, and there is a chance hemay open up the prize door as in the Ignorant MontyHall game. From these factors, one can compute, P ( C k | B j ∩ A i ) = , if j = 1 and k = 2 , , , , if j (cid:54) = 1 and k = 1 , , otherwise. (8)The probability that Monty opens the prize door is P (opens prize door) = (cid:88) i = k (cid:54) = j P ( A i ∩ B j ∩ C k ) = 112 . This implies that the probability that he opens a goatdoor is 11 /
12. Monty then offers the option to stick orswitch. Suppose the contestant always sticks with theinitial choice. Then the probability of winning if stickingand Monty opening a goat door is (cid:88) i = j (cid:54) = k P ( A i ∩ B j ∩ C k ) = 14 . With that, we can compute the conditional probability P (win if stick | opens goat door) = 1 / /
12 = 311 . (9)Suppose the contestant decides to always switch to oneof the other two unopened doors with equal probability1 /
2. Let | Φ l (cid:105) be the outcome switched to and let D l be the corresponding door. With that, we can compute P ( A i ∩ B j ∩ C k ∩ D j ) = P ( D l | C k ∩ B j ∩ A i ) P ( C k | B j ∩ A i ) P ( B j | A i ) P ( A i ). Hence, the probability of winning ifswitching and Monty opening a goat door is (cid:88) i = l (cid:54) = j (cid:54) = k P ( A i ∩ B j ∩ C k ∩ D j ) = 13 . (10)From that, one can calculate P (win if switch | opens goat door) = 1 / /
12 = 411 . (11)In a ψ -ontic game, switching provides an advantage. ψ -epistemic Monty Hall game: In the PBR proof, forthe ψ -epistemic model, there is a non-zero probability q that outcome | Φ (cid:105) occurs when state | Ψ (cid:105) is prepared.This implies that in a ψ -epistemic game, P ( A ) = q (cid:54) = 0.To allow for a comparison with the ψ -ontic game, let q = q + q + q , and with that let the other prize doorprobabilities take values P ( A ) = (1 / − q , P ( A ) =(1 / − q and P ( A ) = (1 / − q .As in the ψ -ontic game, P ( B j | A i ) = 1 /
4, for all values i, j . Monty as a character corresponds to the predictionsof quantum theory (7); he will assume C is definitely agoat door since |(cid:104) Φ | Ψ (cid:105)| = 0. This means the prob-abilities in (8) apply in this game as well. Hence, theprobability that Monty opens the prize door P (opens prize door) = (cid:88) i = k (cid:54) = j P ( A i ∩ B j ∩ C k ) = 112 + 2 q . This implies that the probability that Monty opens a goatdoor is (11 / − (2 q/ (cid:88) i = j (cid:54) = k P ( A i ∩ B j ∩ C k ) = 14 . From this we compute P (win if stick | opens goat door) = 311 − q . (12)If a switching strategy is adopted then: (cid:88) i = l (cid:54) = j (cid:54) = k P ( A i ∩ B j ∩ C k ∩ D j ) = 13 − q , (13) P (win if switch | opens goat door) = 4 − q − q . (14)Thus the probabilities depend on whether the game isa ψ -ontic or ψ -epistemic game. For value q = 1 /
4, wecan calculate that P (win if switch | opens goat door) = P (win if stick | opens goat door); hence for certain ψ -epistemic games, switching offers no advantage. Experimental implications:
Comparing a ψ -ontic gameto a ψ -epistemic game, Monty opens the prize door lessoften. This corresponds to certain probabilities in thePBR proof being zero; some work on the experimentaltests [24, 48–51] of PBR discuss this exact zero probabil-ity as an experimental difficulty. Through our game, weprovide another viewpoint; the difference in the proba-bilities of winning conditioned that a goat door is openedare simply different for the two physical scenarios. Thismay provide insights to alternative experimental designsto test PBR. Quantum teleportation:
Consider the standard pro-tocol [45]. Alice wants to send some unknown state | ψ (cid:105) = α | (cid:105) + β | (cid:105) to Bob. They each possess a mem-ber of the Bell state | β (cid:105) = √ ( | (cid:105) + | (cid:105) ). The initialstate is | ψ (cid:105) ⊗ | β (cid:105) . Alice applies a CNOT gate to herqubits, followed by a Hadamard gate to her first qubit.The resulting state can be written as12 (cid:16) | (cid:105) ( α | (cid:105) + β | (cid:105) ) + | (cid:105) ( α | (cid:105) + β | (cid:105) )+ | (cid:105) ( α | (cid:105) − β | (cid:105) ) + | (cid:105) ( α | (cid:105) − β | (cid:105) ) (cid:17) . (15)When Alices measures her qubits she gets one of the re-sults on the left in (16). Bob would then apply the cor-responding Pauli operator on his qubit to obtain | ψ (cid:105) :00 → Does nothing , → Applies σ x = | (cid:105) (cid:104) | + | (cid:105) (cid:104) | , → Applies σ z = | (cid:105) (cid:104) | − | (cid:105) (cid:104) | , → Applies σ z σ x . (16)Bob receives the two bits from Alice in (16) through aclassical channel. This protocol has been extended toprobabilistic cases [52–54] and noisy cases [55–59]. Monty Hall teleportation:
For our first application, wewant to modify the standard teleportation protocol intoa Monty Hall game. Alice can be viewed as Monty, andBob as the contestant. The four doors are respectivelylabelled (00 , , , ab , and whatBob would need get the desired state | ψ (cid:105) . The contes-tant’s initial choice of door would be equivalent to whatBell state was used at the start of the protocol. In thismodification, the contestant is allowed to choose any ofthe four doors (00 , , , xy . Thisevent coincides with using Bell state | β xy (cid:105) = 1 √ | (cid:105) | y (cid:105) + ( − x | (cid:105) | ¯ y (cid:105) ) , (17)where ¯ y is the negation of y . As an example, if the con-testant chooses door 01, then a way to implement thisis that Bob applies the operator ( σ ⊗ σ z ) | β (cid:105) = | β (cid:105) ,and communicates that to Alice; the last step would beanalogous to Monty being aware of what door the contes-tant chooses. In this modified protocol, the initial state is | ψ (cid:105) | β xy (cid:105) . After Alice applies a CNOT gate to her qubitsfollowed by a Hadamard gate the resulting state is12 (cid:16) | (cid:105) ( α | y (cid:105) + β ( − x | ¯ y (cid:105) ) + | (cid:105) ( α ( − x | ¯ y (cid:105) + β | y (cid:105) )+ | (cid:105) ( α | y (cid:105) − β ( − x | ¯ y (cid:105) ) + | (cid:105) ( α ( − x | ¯ y (cid:105) − β | y (cid:105) ) (cid:17) . At this step, Alice measures her qubits to get her result.If Alice’s result is ab = xy , meaning it coincides with theBell state used | β xy (cid:105) , then Bob has to do nothing and hehas the desired state | ψ (cid:105) (the exception is if the initialBell state used was | β (cid:105) in which case Bob has to applyoperator ( − σ ) to get | ψ (cid:105) if result is 11). This is why thecontestant’s initial choice relates to the Bell state used.In this Monty Hall protocol, Alice sends Bob two bitsas in (16) with the following modification: she choosestwo bits denoted cd (ie goat door) that are not xy (iecontestant’s initial choice) and are not ab (ie prize door).Should Bob do nothing, or apply one of the possible op-erators (which depend on what Bell state was used) toget | ψ (cid:105) ie should the contestant stick or switch?To answer this, let B xy be the door chosen by con-testant. For this example, assume we use | β (cid:105) , hence P ( B ) = 1. Let A ab be the prize door and due to Bornprobabilities we have P ( A ab ) = 1 /
4. Let C cd be the goatdoor opened by Monty whose probabilities, from the pro-tocol description, work out as: P ( C cd | B ∩ A ab ) = , if 00 = ab (cid:54) = cd, , if 00 (cid:54) = ab (cid:54) = cd, , otherwise . (18)If Bob always does nothing (ie, stick strategy), then P (win if stick) = (cid:88) ab =00 (cid:54) = cd P ( A ab ∩ B ∩ C cd ) = 28 . (19)Suppose Bob decides to always apply one of the two op-erators (ie, switch strategy). Then there are one of two possibilities which we denote ef and given its a randomchoice, each occur with probability 1 /
2. Let D ef repre-sent that door, and P (win if switch) is (cid:88) ab = ef (cid:54) = cd (cid:54) =00 P ( A ab ∩ B ∩ C cd ∩ D ef ) = 38 . (20)This means Bob should apply one of the two operators(switch) rather than do nothing (stick) to get state | ψ (cid:105) . Unreliable teleportation:
For our second application,consider the standard teleportation protocol with the fol-lowing unreliability: one of the two bits (either the firstor second) Alice sends to Bob in (16) is received but theother is lost; each event occurs with probability 1 /
2. Ifthe initial Bell state is | β (cid:105) and Alice’s result is 00, thenBob can do nothing. But in this scenario, if Bob re-ceives the single bit as 1, then the possible options are01 , , or 11; in this case he should apply one of the op-erators (switch). If Bob receives bit 0, then his optionsare 00 , ,
10. Should he stick (to 00) or switch (to 01 or10)? To answer this, let us use the notation developed.We have P ( B ) = 1 and P ( A ab ) = 1 / d in C d be the single bit received by Bob; based on the sce-nario described above, we have P ( C | B ∩ A ) = 1 ,P ( C | B ∩ A ) = 1 /
2, and P ( C | B ∩ A ) = 1 / P (received bit 0) = (cid:88) ab (cid:54) =11 P ( C ∩ B ∩ A ab ) = 12 . If Bob decides to always do nothing then this would belike a sticking strategy. The probability that bit 0 isreceived and Bob wins by sticking is P ( A ∩ B ∩ C ) =1 /
4. Hence we can compute the conditional probability: P (win if stick | received bit 0) = 1 / / . (21)If an always switching strategy is adopted, then there aretwo possibilities (01 or 10) each occuring with probability1 /
2. In this case probability of winning if switched andbit 0 is received is P ( A ∩ B ∩ C ∩ D ) + P ( A ∩ B ∩ C ∩ D ) = 1 /
8. With that we compute, P (win if switch | received bit 0) = 1 / / . (22)It is an advantage to stick ie Bob should do nothing.This strategy may used as an error-correcting design forreliability issues in practical quantum networks [60, 61] Conclusions:
We have reformulated PBR theorem intoa Monty Hall game. We argue that future investigationof Monty Hall concepts applied to antidistinguishabilityscenarios will lead to novel quantum protocols.
Acknowledgments:
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