On the Possibility of Ontological Models of Quantum Mechanics

On the Possibility of Ontological Models of Quantum Mechanics D. J. Miller∗ and Matt Farr Centre for Time, University of Sydney NSW 2006, Australia (Dated: June 19, 2022) It is an unresolved question in quantum mechanics whether quantum states apply to individual quantum systems, or to ensembles of quantum systems. We show by way of a thought experiment that quantum states apply only to ensembles of quantum systems. A further unresolved question is whether quantum systems possess ontic states. If a quantum state is the state of an ensemble, as we claim, the answer to this question is that quantum states are not ontic. However, a notable recent result in quantum foundations shows that if there are any ontic states at all, then the quantum state must be ontic. Collectively, these two results imply that there are no ontic states. We examine the assumptions required for these results, and suggest that the retrospective effect on state preparations by entangling measurements provides good reason for relaxing the assumption of preparation independence at the ontic level. arXiv:1405.2757v1 [quant-ph] 12 May 2014 PACS numbers: 03.65.Ta,03.65.Ca,03.65.Ud I. INTRODUCTION There is a general category of interpretations of quantum mechanics, associated in one way another with Einstein [1, 2], Park [3], Ballentine [4] and others, in which a quantum state ψ is a property of an ensemble and individual systems are not assigned quantum states. This is often referred to as the statistical interpretation. However, this term is used generally in quantum mechanics, so we instead refer to the above category of interpretation as the ensemble-state interpretation (ESI). The more orthodox interpretation is that individual quantum systems may be assigned states; we refer to this as the quantumsystem interpretation (QSI). It follows that according to the ESI, there are no proper mixtures in quantum mechanics. In a proper mixture [6] one knows (perhaps from the way the ensemble was prepared) that a certain set of states ψi appears with probability pi among the constituents as a whole (although one may be ignorant of which constituent of the ensemble has which state). In an improper mixture, the constituents cannot be assigned their own states on any grounds. The most familiar example of an improper mixture is an ensemble whose state is the reduced density operator of an entangled state. Our aim is to show that the assumption that there are proper mixtures of quantum states leads to a contradiction and thereby prove (some version of) ESI is the only viable interpretation of quantum mechanics. We also consider this in relation to models that posit an ontic state λ that determines the outcomes of measurements. To demonstrate our result, we consider ensembles of qubits (the simplest quantum system is sufficient) and assume (it will turn out wrongly) a quantum state ψ can be identified with each qubit that is prepared. Then if two mixtures of qubits are paired off, each pair is in a product state and any ensemble (or sub-ensemble) formed from ∗ [email protected] them must be separable (a separable state is defined to be one which is not entangled). We aim to show that one can identify and measure a sub-ensemble in an entangled state within a separable ensemble formed by pairing-off two proper ensembles of qubits. There are two reasons that hint that this may be possible. Firstly, it is known from delayed-choice entanglement swapping experiments [7–9] that one can identify entangled sub-ensembles formed from pairing two improper ensembles of qubits (i.e. qubits whose state is the reduced density operator of an entangled state). Since this involves qubit ensembles that are not prepared in specific states, it does not decide anything directly about the ESI versus QSI question. Secondly, it is possible that an ensemble composed of a proper mixture of entangled pairs is in a separable state overall. Timpson and Brown [10] have coined the term improper separability for those cases. Specifically, we show that one or more of the following statements is incorrect. The statements are sufficient for our purposes but some of them are not necessary and could be relaxed. [1.] Preparation of a pure state. A quantum system can be prepared in a pure state by measuring an observable with non-degenerate eigenvalues oi and corresponding eigenstates |ψi i. When oi is recorded by a classical system as being the result of the measurement, the qubit has been prepared in the state |ψi i. [2.] Persistence of a quantum state. A quantum system prepared in a known state |ψi i continues to be in that state no matter which ensemble the qubit is assigned to (if Û is the time-evolution operator over the relevant period we are, of course, assuming Û |ψi i = |ψi i for all i). [3.] Ensemble formation. Alice can prepare the states |ψm i, m = 1, M as in [1.] and keep a record of the states that are prepared by listing the run number i and the eigenvalue for each preparation. If Alice prepares a series of N qubits in states |ψm1 (1), i, |ψm2 (2), i, . . . , |ψmi (i)i, . . ., i = 1, N , the series forms an ensemble of pure states with the ensemble 2 state N 1 X |ψmi (i)ihψmi (i)|. N i=1 σ̂ = As an alternative to using experimental data, if one can establish by other means that (1) (2) Then one can form an ensemble of pairs of qubits with the state ρ̂ = N 1 X |ψmi (i)ihψmi (i)| ⊗ |φmi (i)ihφmi (i)|. (3) N 2 i=1 The ensemble is a proper mixture because its members are identifiable by the records of run numbers and eigenvalues. It is separable because the partial transpose of ρ̂ is the state itself and therefore has positive eigenvalues [11]. An ensemble formed in the above way will be called a proper, separable ensemble. [4.] Sub-ensemble formation. (a) By choosing a subset of runs of an experiment i ∈ {η} based on experimentally reproducible criteria, one can form a systematically selected sub-ensemble. (b) If this procedure is applied to the labelled qubits that Alice and Bob have prepared, the state of the subensemble is 1 X ρ̂{η} = |ψmi (i)ihψmi (i)| ⊗ |φmi (i)ihφmi (i)|(4) Nη2 i∈{η} where Nη is the number of members of the set {η}. All sub-ensembles formed in that way from Alice’s and Bob’s qubits are proper, separable ensembles. [5.] Quantum tomography. The state of a sub-ensemble of pairs of qubits with i ∈ {η} can be determined by quantum tomography (QT) [12] (provided the subensemble has a sufficiently large number of members). The state ρ̂{η} of a sub-ensemble of pairs of qubits can be expressed as [12] 1 XX Tr(ρ̂{η} σ̂j ⊗ σ̂k )σ̂j ⊗ σ̂k 4 j=0 3 3 ρ̂{η} = (5) k=0 ˆ the identity operator for a qubit, and σ̂j , where σ̂0 = I, j = 1, 2, 3 are the Pauli matrices σ̂x , σ̂y , σ̂z respectively. If ρ̂{η} is unknown, one can use the average values of measured QT data in the right-hand side of Eq. (5) 1 XX hσ̂j ⊗ σ̂k i{η} σ̂j ⊗ σ̂k . 4 j=0 3 ρ̂{η} = 3 k=0 (7) then from Eqs (5)–(7) one can conclude Bob can prepare the states |φm i, m = 1, M as in [1.] and keep a record similar to Alice’s. Let us assume Bob prepares the series |φm1 (1), i, |φm2 (2), i, . . . , |φmi (i)i, . . ., i = 1, N with the ensemble state N 1 X τ̂ = |φmi (i)ihφmi (i)|. N i=1 hσ̂j ⊗ σ̂k i{η} = Tr(τ̂ {η} σ̂j ⊗ σ̂k ) (6) ρ̂{η} = τ̂ {η} . (8) [6.] Standard quantum mechanics. The quantities needed for in Eq. (7) can be calculated from the procedures of standard quantum mechanics. In [4.] the requirement “experimentally reproducible criteria” for a “systematically selected sub-ensemble” means that if the whole experiment is repeated many times, one can identify a sub-ensemble in the same state in the same way each time. A randomly chosen subensemble may appear to possess its own “state” but if that differs from the ensemble from which it is chosen, it is due merely to a statistical fluctuation entirely consistent with the sub-ensemble having the same state as the ensemble as a whole. A random selection from a repetition of the experiment would result in a different “state” for the randomly selected sub-ensemble in each repetition of the experiment and it is that which distinguishes a systematically selected sub-ensemble from a randomly selected sub-ensemble. II. THE EXPERIMENT We present the argument in terms of qubits to make things as simple as possible but the argument is readily generalised to higher dimensional Hilbert spaces. As shown in Fig. 1, Alice and Bob prepare ensembles of qubits as above (perhaps independently, e.g. at spacelike separation) in such a way that ˆ σ̂ = τ̂ = I/2 (9) where Iˆ is the identity operator in their two-dimensional Hilbert spaces. Alice at PA and Bob at PB subject each of their respective qubits i to one projective measurement suitable for QT chosen independently and randomly and record the run number i and the result and pass their qubit to the next stage irrespective of the outcome. Because the ensembles are maximally mixed, the state of each ensemble after these measurements continˆ ues to be σ̂ = τ̂ = I/2. Finally, an entangling measurement of the observable Q with the entangled eigenstates |Φn i , n = 1, 4 (e.g. the Bell states) is performed on each pair i. Let i ∈ {Φ1 } be the runs for which |Φ1 i is the result of the Q measurement. The runs with i ∈ {Φ1 } form a sub-ensemble of qubits prepared by Alice and Bob with the state 1 X ρ̂{Φ1 } = |ψmi (i)ihψmi (i)| NΦ2 1 i∈{η} ⊗|φmi (i)ihφmi (i)| (10) 3 is |Φ1 i = √12 (|0iA |0iB + |1iA |1iB ). It is shown in Appendix B, that for the sub-ensemble (i ∈ {1B , Φ1 }) for which Bob prepares |1iB and |Φ1 i is the result of the Q measurement, the QT data from PB give the pure state Q Φ! Φ! Φ! Φ! 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 FIG. 1: Runs i = 1 to i = 17 of a much longer series of runs of the experiment are shown. In each run, Alice chooses at random but with equal probability to prepare qubits in one of the orthogonal states |ψ1 i or |ψ2 i. Bob does the same for the orthogonal states |φ1 i or |φ2 i. Each qubit is subject to a quantum tomography measurement PA and PB respectively and then the qubits are entangled at Q. The runs i ∈ {Φ1 } for which |Φ1 i is obtained are noted and the PA and PB data for those runs are chosen to construct a state from the tomography data. which is a proper, separable ensemble (c.f. [4.]). We can also determine the state ρ̂{Φ1 } from the QT data. It is shown in Appendix A that the QT data can be readily calculated for this experiment by [6.] with the result that ρ̂Φ1 = P̂Φ1 = |Φ1 i hΦ1 | . (11) The reason for this simple result is that Alice’s and Bob’s qubits are in a maximally-mixed state. Consequently, the sub-ensemble made of the pairs i ∈ {Φ1 } is in the pure state |Φ1 i and therefore the sub-ensemble is in an entangled state. This contradicts the formation of the sub-ensemble given in Eq. (10) which, as noted above, is separable. This means one of the steps [1.] to [6.] is false. Steps [4.(a)], [5.] and [6.] are used elsewhere in quantum mechanics. For example, constructing the state of a subensemble from conditional probabilities on the whole ensemble via QT is standard procedure in delayed-choice entanglement swapping experiments [7–9]. Therefore it must be the steps [1.] and [2.] (and applying them in steps [3.] and [4.(b)]) which are in error. In other words the error was due to assigning a state to a qubit in the first place, i. e. adopting the QSI is not viable. Before discussing this conclusion, we mention a variation of the experiment in Fig. 1. We take a particular example for simplicity but many other combinations of states could be used instead. Let Bob prepare the states |0iB or |1iB with equal probability, so ˆ that his ensemble state is I/2, and measure PB as before. Let Alice do the same but omit any PA measurement. The entangling Q measurement is performed on Alice and Bob’s qubits such that one of the outcomes {1 ,Φ1 } ρ̂B B = P̂|1iB = |1i B B h1| . (12) That is, as expected, the state of the sub-ensemble i ∈ {1B , Φ1 } of Bob’s qubits is the pure state |1iB . However, for the sub-ensemble of those runs i ∈ {0A , Φ1 } for which Alice prepares |0iA and |Φ1 i is the result of the Q measurement, the QT data from PB give the pure state (see Appendix B) {0 ,Φ1 } ρ̂B A = P̂|0iB = |0i B B h0| . (13) That is, state of the sub-ensemble i ∈ {0A , Φ1 } of Bob’s qubits is the pure state |0iB . As shown in Appendix B, some runs belong to both sub-ensembles (i ∈ {0A &Φ1 } & i ∈ {1B &Φ1 }). If Bob’s qubits possessed states that would mean the qubits were in the pure, orthogonal states |0iB and |1iB at the same time. That is impossible because an ensemble in a pure state cannot be decomposed into other states [13]. Once again the conclusion is that qubits cannot be assigned states according to [1.]Ȯur argument can easily be generalised to higher dimensional Hilbert spaces and so we can say quantum systems cannot be assigned states according to [1.]. Once again, the QSI is ruled out. To relate this conclusion to other work we must examine another way of classifying quantum models that has been explained by Harrigan and Spekkens [5]. In an ontological model, the primitives of the theory are the properties of microscopic systems specified by ontic states λ. When a system is prepared in the pure state ψ, this may determine only a distribution µψ (λ) over the ontic states λ. In some sense ψ represents our imperfect knowledge of the λ so if the µψ (λ) for different ψ overlap, the model is a ψ-epistemic model. If the µψ (λ) for different ψ do not overlap, we can deduce ψ from λ and hence ψ can be regarded as part of the λ; this is called a ψ-ontic (ontological) model. A ψ-ontic model clearly fits a QSI and the interest usually is in the relationship between ψ-ontic and ψ-epistemic models, so we will reserve the term ψ-epistemic also to apply to an individual system so that, in our usage at least, a ψ-epistemic model fits a QSI. On the ESI, the quantum state applies to an ensemble of systems, not the systems themselves. There is no reason why the individual members of the ensemble could not have ontic states λ given by the distribution µψ (λ) where ψ is the state of the ensemble, not of each member of the ensemble. When there are λ’s and µψ (λ) involves an ensemble state ψ, we will call the model a ψ-ensemble model to distinguish it from a ψ-epistemic model we have defined above. Clearly a ψ-ensemble model is an example of an ESI while a ψ-epistemic model is an example of a QSI. The difference between them is important for our purposes. In a ψ-ensemble model, the ensemble is a 4 proper mixture at the ontological λ level but an improper mixture at the quantum ψ level, while in a ψ-epistemic model an ensemble is a proper mixture at both levels. Since our results rule out the QSI, our result eliminates ψ-ontic and ψ-epistemic models. What we have called ψensemble models are still possible. PBR [18] have eliminated ψ-epistemic models and, according to our classification and understanding, also ψ-ensemble models. Therefore the two results together (PBR and ours) eliminate any sort of model. The only way of avoiding that conclusion is to negate one of the assumptions required for the present and/or PBR results. The first assumption to question is preparation independence. In the PBR case, that is the assumption that if two pure states ψi and ψj are prepared independently (e.g. at space-like separation), the probability distributions for the corresponding ontic states satisfy µψi ,ψj (λi , λj ) = µψi (λi )µψj (λj ). (14) In our case, the independence assumption is implicit in [1.], namely that a qubit can be prepared in the state ψi by measuring an observable on the qubit (no other qubit being involved). It is significant that the independence assumption is at the λ level in the case of PBR but at the ψ level in our case. Abandoning preparation independence for ψ would mean a change in the formalism of standard quantum mechanics which assigns quantum states based on local laboratory preparations. We will not explore this option further. The possibility of abandoning preparation independence at the λ level (required for PBR) has been considered [18–20]. Alternative assumptions leading to constraints on ψ-epistemic models are being investigated actively (see [19, 21–24] and references therein) but it seems from that work that non-trivial assumptions [19] will always be required to rule out ψ-epistemic models. Here we further investigate only the preparation independence assumption. If one is to relax preparation independence at the λ level, it should be to the minimum extent required in the first instance. Measurements of quantum systems that have been entangled by a prior measurement are not independent. One option is to assume that preparation independence does not hold for quantum systems that are entangled by a later measurement. The idea that measurements can have retrospective (backwards-in-time) influences has been suggested in other contexts within quantum mechanics. These include nonlocal correlations for pairs in entangled states (for a discussion, see [14]) and as a quantum information link [15–17] in quantum teleportation and other experiments. Further support for the idea that entanglement can have a retrospective, as well as the standard, prospective effect is shown by an extension of our previous thought experiment. Fig. 2 shows the preparation of an ensemble of two independently produced qubits in the maximally-mixed ˆ independent QT measurements PA and PB on state I/2, each, an entangling measurement Q, as in Fig. 1 but now followed by independent QT measurements RA and RB on each qubit emerging from the Q measurement. 14 13 12 11 10 9 8 7 6 5 4 3 2 1 FIG. 2: Alice’s and Bob’s qubits are as in Fig. 1. After the entangling measurement Q, each emerging qubit is subject to another quantum tomography measurement R1 or R2 respectively. The runs i ∈ {Φ1 } for which |Φ1 i is obtained are noted and a state is constructed from the PA and PB tomography data, and also from the R1 or R2 tomography data, for those runs. If one selects a sub-ensemble for which the Q outcome is the entangled state |Φ1 i and considers the subensemble of runs i ∈ Φ1 , we know from Eq. (11) that the QT at PA and PB reveal the state |Φ1 i and obviously the QT at RA and RB reveal the state |Φ1 i also. This is to be expected from a ψ-ensemble model since there is no difference between choosing an ensemble from future criteria and choosing an ensemble from past criteria [7]. The only way to accommodate this result within a ψ-ontic model would be to assign the states of the qubits at the time of their preparation to be the entangled state that they are going to be found in at the Q measurement. ˆ Note that if the initial states of the qubits are not I/2, the symmetry of Fig. 2 is destroyed because the qubits ˆ state because of the entanwill still end up in the I/2 gling operation at Q (even when selected only on the outcome |Φ1 i). In that case (initial states of the qubits ˆ not I/2), the PA and PB correlations are changed (perhaps no longer showing non-classical correlations) but the RA and RB correlations remain the same. Of course the same applies if the sub-ensemble is changed by eliminating from consideration some of the final states so that ˆ the final state of each qubit is no longer I/2. Then the RA and RB correlations are changed but the PA and PB correlations still show the entanglement as before. III. CONCLUSION The main result is that the thought experiment shown in Fig. 1 proves that quantum states apply at the ensemble level and cannot apply to individual systems. As mentioned in the Introduction, this conclusion supports a long-held interpretation of quantum mechanics. 5 We have related our result to the options for models of quantum mechanics that have been discussed recently. The first alternative is between an ontological model, in which the primitives of the theory are the properties λ at the microscopic level, and a non-ontological (or instrumentalist) model, in which the primitives of the theory are preparations and measurements and one is only concerned with correlations between the two. We can say nothing directly about non-ontological models. Recent results have important consequences for ontological models. On the basis of the PBR and present results taken together, if both the quantum states ψ and ontic states λ of systems can be prepared independently, then no ontological model is possible. The PBR [18] result rules out ψ-epistemic and ψ-ensemble models and the present result rules out ψ-ontic and ψ-epistemic models. Putting it another way, PBR requires ψ to be ontic and the present results forbid ψ being ontic because ψ can only apply to ensembles, not individual systems. Stopping at this point would mean that one is forced to adopt a non-ontological (instrumentalist) stance. If one relaxes the preparation independence assumption by allowing the ontic states λ to be correlated even when the preparations are independent, then PBR imposes no restrictions, i.e. PBR then allows ψ-epistemic and ψ-ensemble models but the present results are unaffected: ψ-ontic and ψ-epistemic models are still ruled out. The option for ontological models at this point is that ψ is the state of an ensemble of systems each of which may possess an ontic state λ. If one relaxes the preparation independence assumption by allowing the quantum states ψ to be correlated even when the preparations are independent, then the present results impose no restrictions. We have suggested that the most plausible reason why preparation independence at the ontic level might not apply is that future entangling measurements have retrospective effects on the preparation of states. Finally, ψ-epistemic (and ψ-ensemble, in our terms) models have been ruled out or brought into question on the basis of different assumptions [19, 21–25] than preparation independence. Some of the assumptions involve only a single quantum system and entanglement is not involved. It will be interesting to investigate those assumptions, especially to see if the retrospective effects of non-entangling measurements could avoid their consequences for ψ-epistemic and ψ-ensemble models. Appendix A: Quantum tomography calculations 1. Pair of qubits In relation to Eq. (5), we can define [12] parameters {η} Sjk {η} Sjk = Tr(ρ̂{η} σ̂j ⊗ σ̂k ) = hσ̂j ⊗ σ̂k i{η} (A1) and if ρ̂{η} is unknown, the point of QT is that an estimate of them can be obtained from the average values of suitable measurements performed on the sub-ensemble with i ∈ {η} hSjk i{η} = hσ̂j ⊗ σ̂k i{η} + (A2a) − = (P{η} [|j i] − αj P{η} [|j i]) × (P{η} [|k + i] − αk P{η} [|k − i]) (A2b) where α0 = −1, αj = 1, j = 1, 3 and P{η} [|j ± i] is the probability of obtaining the state |j ± i from measurements of σ̂j . The states are |0+ i = |3+ i = |0i , |0− i = |3− i = |1i (A3a) 1 1 |1± i = √ (|0i ± |1i , |2± i = √ (|0i ± i |1i(A3b) . 2 2 The state of the ensemble constructed from the tomographic data is then 1 XX = hSjk i{η} σ̂j ⊗ σ̂k 4 j=0 3 ρ̂ {η} 3 (A4) k=0 For the case we are interested in, all the runs of pairs i ∈ {Φ1 } and for only those pairs the result |Φ1 i of the Q measurement was obtained. Therefore the probabilities like the following which are required for QT (see Eq. (A2b)) can be calculated from P{Φ1 } [|j + i]P{Φ1 } [|k + i] = Prob[|j + i & |k + i | |Φ1 i] (A5) where the probability on the right is for the whole ensemble conditional on Φ1 being the result of the Q measurement. From [6.], i.e. applying the usual formalism of quantum mechanics, Prob[|j + i & |k + i & |Φ1 i] = Tr((P̂|j + i ⊗ P̂|k+ i )ρ̂ AB (A6a) (P̂|j + i ⊗ P̂|k+ i )P̂Φ1 )(A6b) where the projection operator P̂|j + i = |j + i hj + |, etc. ˆ ⊗ I/2, ˆ Since we constructed ρ̂AB = I/2 Prob[Φ1 ] = 1/4 and so Acknowledgments Prob[|j + i & |k + i | |Φ1 i] = Tr(P̂Φ1 (P̂|j + i ⊗ P̂|k+ i )). (A7) This type of expression applies to all of the terms in Eq. (A2b). Therefore The authors thank Eric Cavalcanti for a very illuminating discussion and helpful comments. hSjk i{Φ1 } = Tr(P̂Φ1 σ̂j ⊗ σ̂k ) (A8) 6 From Eqs (A12) and (A15b), and from Eq. (A4) 1 XX Tr(P̂Φ1 σ̂j ⊗ σ̂k )σ̂j ⊗ σ̂k 4 j=0 3 ρ̂Φ1 = 3 ρ̂{0A ,Φ1 } = k=0 ρ̂Φ1 = P̂Φ1 = |Φ1 i hΦ1 | Single qubit {η} ρ̂B 1 = 2 {η} Tr(ρ̂B σ̂jB )σ̂jB j=0 1 = 2 3 X ρ̂{0A ,Φ1 } = P̂|0iB = |0i B B h0| (A10) We wish to construct the state of Bob’s qubits from the PB measurements when Alice’s qubits are prepared (we are assuming the QSI) in the state |0iA and the entangled state is |Φ1 i = √12 (|0iA |0iB +|1iA |1iB ). The state of the Bob’s qubits in any sub-ensemble of runs with i ∈ {η} can be expressed in the QT form for single qubits [12] as 3 X SjB σ̂jB {1 ,Φ1 } B = Tr(P̂|Φ1 i (IˆB ⊗ P̂|1iB )P̂|Φ1 i P̂|j + iB )(A18a) (A11) = Tr(P̂|1iB P̂|j + iB ) j=0 1X B hS i{η} σ̂jB 2 j=0 j (A18b) and hence hSjB i{1A ,Φ1 } = Tr(P̂|1iB (P̂|j + iB − αj P̂{Φ1 } ))(A19a) 3 {η} (A17) That is the sub-ensemble ρ̂{0A ,Φ1 } is in the pure state |0iB . On the other hand, if we select a sub-ensemble i ∈ {1B } for which Bob had prepared (assuming the QSI) his qubits in the state |1iB and the entangled state is |Φ1 i, we can calculate the probabilities required find hSj i{1B ,Φ1 } from the conditional probability P[|j ± iB | |1iB & |Φ1 i] for the whole ensemble P|j +Bi which defines the parameters Sj . Bob can therefore esti{η} mate ρ̂B from his QT data by ρ̂B = (A16) which means which means, from Eqs (A9) and (5) 2. 1X Tr(P̂|0iB σj )σ̂j 2 j=0 3 (A9) = Tr(P̂|1iB σ̂j ). (A12) (A19b) From Eqs (A12) and (A19b), where 1X Tr(P̂|1iB σj )σ̂j 2 j=0 3 hSjB i{η} = P{η} [|j + iB ] − αj P{η} [|j − iB ] and, as before, P{η} [|j ± iB ] is the probability of obtaining the state |j ± iB from the measurements of σ̂j performed on the sub-ensemble formed from i ∈ {η}. As mentioned, we require the probabilities for the runs in which Alice’s qubits are prepared in the state |0iA and the measured entangled state is |Φ1 i, i.e. for the sub-ensemble of runs with i ∈ {|0iA & |Φ1 i}. We can calculate the probabilities required for Eq. (A13) from the corresponding conditional probabilities like P[|j + iB | |0iA &Φ1 ] for the whole ensemble P{0A ,Φ1 } [|j ± iB ] = Tr(P̂|j ± iB (P̂|0iA ⊗ IˆB )P̂|j ± iB P̂|Φ1 i ) (A14a) = Tr(P̂|0iB P̂|j ± iB ) ρ̂{1B ,Φ1 } = (A13) which means ρ̂{1B ,Φ1 } = P̂|1iB = |1i B B h1| (A21) That is the sub-ensemble ρ̂{1B ,Φ1 } is in the pure state |1iB . Because of the intervening measurement PB the probability P[0A &1B ] is not necessarily zero even when the result |Φ1 i is obtained + P[0A &1B |jB &Φ1 ] = Tr(P̂|j + iB (P̂|0iA ⊗ P̂|1iB ) ×P̂|j + iB P̂|Φ1 i ) (A14b) + 2 + (A22a) 2 = |B h1|j iB | |B hj |0iB | (A22b) and hence hSjB i{0A ,Φ1 } = Tr(P̂|0iB (P̂|j + iB − αj P̂{j + }B )) (A15a) = Tr(P̂|0iB σ̂j ). 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