Characterization of Entanglement, Nonlocality and the Role of Noncommutativity in Quantum Theory Debsuvra Mukhopadhyay Satyendra Nath Bose National Centre for Basic Sciences, Block-JD, Sector-III, Salt Lake, Kolkata-700098, India
∗
With the advent of quantum theory, we had been successful in expounding, or rather, providing a framework which is compatible with several features that have no place in classical physics. In this report, we have chosen to focus on three such aspects: entanglement, nonlocality and noncommutativity. We explore these concepts in some depth and at the same time, attempt to understand their relation. In relation to entanglement, we give a brief overview of the separability problem. Next, we move on to the aspect of nonlocality and try to characterize states which exhibit this feature. We also demonstrate the inequivalence between nonlocality and entanglement. Subsequently, we elucidate the implications of the noncommutativity of operators in quantum theory and justify how a violation of Bell’s inequality can be attributed to the existence of non-commuting observables. We eventually conclude by providing some motivations for further research on these intriguing features of quantum mechanics. The report assumes a working knowledge of basic quantum mechanics and the most commonly used notations and terminology of quantum information theory. Keeping in mind the aim of scope of the report, not every terminology used in the report will be defined before its usage. Furthermore, many of the theorems or corollaries, implicitly or explicitly made use of, and how they lead to the features stated will not be dealt with at any level of mathematical rigor, either.
Contents
Introduction
1
Classical Analogue of Entanglement? Classical Description of Composite Systems Quantum Theory and the Tensor Product Structure
1 1 2
The Separability Problem: How to know if a state is entangled? Pure states and the simplicity of entanglement characterization The Layman’s Technique Schmidt number as a signature of entanglement Von Neumann Entropy Mixed states: the problem gets tougher! Separability of Operators Exemplification: It isn’t all too intuitive Quantum Maps which are P but not CP The Reduction Criterion Positive Partial Transposition Criterion The method of Witnessing Entanglement: A Convex-Analytic Approach Characterizing Entanglement through Uncertainty Relations: A Non-Linear Technique Werner Class and Isotropic Class
3 3 3 3 4 4 5 5 6 7 7 8 9 11
Quantum Nonlocality Bell Locality and the Bell-CHSH Inequality Quantum Mechanics does not respect the CHSH Inequality! States maximally violating the CHSH Inequality: The “Eigenvalue” Approach “Nonlocality” of the Bell Operator Complementarity from Bell’s Inequality
11 12 14 15 16 18
1 Hardy’s Demonstration of Nonlocality
19
Types of Correlations and Nature’s “Intermediate” Choice No-Signalling Correlations Local Correlations Quantum Correlations: Nature Simply Loved This! The Interrelation: A Strict Inclusion Some properties of local, quantum and no-signalling correlations
19 20 20 20 21 22
Nonlocality and Entanglement: How are they Related? Gisin’s Theorem and its generalization Horodecki-criterion for violating Bell’s Inequality Comparing the Bell Nonlocality and Entanglement of the Werner Class: The Anomaly
22 23 23 24
Noncommutativity:Another Non-Classical Aspect of Quantum Theory Inapplicability of the Classical Measurement Rule: Von Neumann’s “Silly” Mistake Bell’s Inequality Violation as an indication of Noncommutativity The Tsirelson Operator and its “Quantumness” Derivation of Bell’s Inequality from the assumption of Commutativity A local realistic model
24 24 26 26 26 28
Epilogue Brief Summary Possible directions for further research from the foundational perspective
29 29 30
Acknowledgements
30
References
31
INTRODUCTION
Quantum Mechanics has been supremely successful as an operational theory of nature. No empirical evidence, so far, has been able to uproot its foundations. On the contrary, there are ever-increasing indications of the postulates of quantum mechanics as being the most precise description of nature. However, even though the fact that it works so well is definitely assuaging, its efficacy is emphatically surprising. Answering “why something works” has always been a nontrivial matter to address, but the same question when applied to quantum theory seems to be a bit too cryptic. When classical physics had held its sway, people usually took its foundations for granted. In fact, hardly anybody might have, given its simple postulates that possess a direct connection with the macro-world, really felt the need to investigate the bedrock of classical physics. Only as “non-classical” traits of nature came to be revealed in due course, physics needed a reformation at the fundamental level. With the genesis of quantum theory, all these “non-classical” features were somehow incorporated in its formalism.
CLASSICAL ANALOGUE OF ENTANGLEMENT?
This section basically provides a comparative picture between classical and quantum descriptions of composite systems. We see how the conjunction of the tensor product structure and the linear superposition principle gives rise to the unique feature of entanglement in quantum theory.
Classical Description of Composite Systems
In classical mechanics, the state of a particle is represented by a point in the phase space. The phase space has a dimension of d = 2n, where n is the total number of generalized coordinates qi ’s (i = 1, 2, ...n) required to specify the
2 physical location of the particle. Thus any point in the phase space is given by the 2n-tuple (q1 , q2 , ..., qn ; p1 , p2 , ..., pn ), where pi ’s (i = 1, 2, ..., n) are the corresponding conjugate momenta and this is what denotes the pure state of the classical particle. However, often there is an uncertainty over the exact state of the particle and an epistemic description is required. In this scenario, it might be possible to associate various points in phase space to the state of particle, compatible with the given physical situation [24] and therefore, we make use of a well-defined probability distribution ρ(~q, p~) representing the density of states in the phase space, which evolves in accordance with the Liouville0 s T heorem. So how do we describe the state of a system comprising of two (or more) particles? In classical physics, the phase space of the composite system is simply the cartesian product of the the phase spaces of the constituent particles, and the corresponding composite state is the cartesian product of the states of the latter. To make it clear, let us define cartesian product for the mathematically simplest case: Given two sets A and B, the cartesian product of the sets is expressed as A × B, which is the set of all ordered pairs (x, y) such that x ∈ A and y ∈ B.
FIG. 1: Cartesian Product of A and B
Obviously, the definition can be extended to the case of multipartite systems. From the definition, it is clear that the specification of the state of a composite mandates the specification of the states of its constituents. In the classical domain, any inadequacy of information about the state of any of its constituents gets manifested in our lack of knowledge about the state of the composite system. It is impossible to have exact knowledge of the composite system without having full knowledge of its constituents. Quantum Theory and the Tensor Product Structure
In quantum mechanics, pure states are represented by rays in a Hilbert space. Then, if HA and HB be the Hilbert spaces associated with two quantum mechanical systems A and B respectively, the appropriate Hilbert space associated with the composition of A and B is the tensor product of the respective spaces, i.e. HA ⊗ HB . This is clearly different from the classical composition rule. Now, for any ket |ψiA ∈ HA and |φiB ∈ HB representing the respective states of A and B, the corresponding tensor product |ψiA ⊗ |φiB (usually written simply as |ψiA |φiB ) is a ket belonging to the composite Hilbert space HA ⊗HB and represents the state of the composite system. However, there’s a caveat. Quantum Mechanics respects the superposition principle, i.e. if |ψi and |φi represents two possible state kets for a system, any normalized linear combination of them, α |ψi+β |φi represents an allowed 2 2 state of the system, where α, β ∈ C, satisfying |α| + |β| = 1 Now, if we assume that {|ei iA } and {|fj iB } are two orthonormal basis sets in HA and HA respectively with hei |ej iA = hfi |fj iB = δij , then {|ei iA |fj iB } is an orthonormal basis for HA ⊗ HB , where i = 1, 2, ..., dim(HA )) and j = 1, 2, ..., dim(HB )). Any generic state can, therefore, be expanded in this orthonormal basis in the following manner: X |ψiAB = cij |ei iA |fj iB (1) i,j
with the normalization condition assumed to be satisfied [25]. Now if |ψiAB can be expressed as a product |ψiA |ψiB (the tensor product, of course) for some kets |ψiA ∈ HA and |ψiB ∈ HB , then it is known as a product state. It
3 is evident that due to the superposition principle, not all bipartite states can be expressed as product states! For instance, the biqubit state a |01i + b |10i , a 6= 0, b 6= 0, where {|0i , |1i} represents an orthonormal basis for a single qubit, cannot be expressed in the product form. Any such state which does not admit a tensor product factorization is known as an entangled state. So if a bipartite system is in an entangled state, how do we assign states in the respective Hilbert spaces to the subsytems? Well, obviously, we can’t assign pure states(rays) to the subsystems in this case, but that makes it a really weird feature! Did we encounter such a possibility in classical physics? No! In classical physics, we did not have any superposition principle!. As a conclusion, we observe that quantum theory allows for pure bipartite states that do not necessarily correspond to pure states of their subsystems. In fact, whereas pure states corresponding to the subsytems ensure that the composite system is also in a pure state, the converse is barely guaranteed! And this is a manifestly non-classical feature observed in quantum theory. That entangled states indeed exist, have been experimentally verified as well. So we have to come to terms with it, no matter what! In the following section, we shall delineate some of the techniques which might be employed for the detection of entanglement.
THE SEPARABILITY PROBLEM: HOW TO KNOW IF A STATE IS ENTANGLED? Pure states and the simplicity of entanglement characterization
We have seen that whenever the state of a composite system cannot be written as a tensor product of the states of its constituents, it is entangled. Even though this is sort of a negative definition of entangled states, it seems like the problem of detecting it will be reasonably simple, since, by assuming a decomposable state for the composite system we must be able to arrive at a contradiction in case it is entangled. And yes, this brute-force method, though quite cumbersome for high-dimensional systems, is extremely simple for bi-qubit [26] systems. The Layman’s Technique
First we outline the brute-force method. Suppose we have a composite system of two qubits A and B associated with Hilbert spaces HA and HB respectively. Let us denote the Hilbert space of the biqubit system as HAB . Then we can write an arbitrary pure state in HAB as: X |ψi = aij |iji (2) i,j
where i, j ∈ {0, 1}. Assuming that it is a product state, we can write: |ψi = |φiA |χiB (3) P P for some |φiA ∈ HA and some |χiB ∈ HB . Let |φiA = i ri |ii and |χiB = i si |ii, where i ∈ {0, 1}. Since the product of these two is |ψi, we arrive at the following condition: aij = ri sj , ∀i, j ∈ {0, 1}
(4)
An immediate consequence of this condition is that the determinant of the coefficient matrix A = [aij ], i.e. (a00 a11 − a01 a10 ) vanishes. In fact, this is both a necessary and sufficient condition for |ψi to be a product state. Thus, a biqubit pure state is entangled iff the determinant of the coefficient matrix is non-zero. Pretty simple, right? Schmidt number as a signature of entanglement
For pure bipartite states, there exists a nice theorem attributed to the mathematician Erhard Schmidt. We state this theorem without proof [27] :
4
Theorem-1 (Schmidt Decomposition): Every pure bipartite state |ψiAB ∈ HAB can be expressed as a linear combination of some pertinent bi-orthonormal states in the following manner: |ψiAB =
R X
λi |ei iA |fi iB
(5)
i
where {|ei iA }/{|fi iB } designate an orthonormal system(not generally a basis!) of vectors in HA /HB , λi > 0 and PR satisfies i λi 2 = 1, and 1 ≤ R ≤ min(dA , dB ), dA and dB being the Hilbert space dimensions of the subsystems. R is called Schmidt Rank or Schmidt N umber, and λi ’s are the Schmidt coef f icients. This is a profound result because, while in general, we require “dA dB ” coefficients to write down the state, this theorem allows us to construct a bi-orthonormal system of R vectors in terms of which the state may be expanded, and R is bounded above by ”min(dA , dB )”! Furthermore, the Schmidt coefficients are all positive and therefore, real! It immediately follows from the above theorem that if |ψiAB is a product state, it has only one Schmidt coefficient and therefore, a Schmidt Rank of 1. And whenever the system is entangled, its Schmidt Rank must be at least 2. Thus, by virtue of the Schmidt rank one can reliably detect entanglement, in the pure bipartite case! We state some more important results in relation to this theorem: PR 1. Given the Schmidt decomposition, the reduced density operators are found to be: ρA = i λi 2 |ei iA hei | and PR 2 ρB = i λi |fi iB hfi |. Thus we find that the eigenvalues of ρA /ρB are squares of the Schmidt coefficients of |ψiAB and the corresponding eigenvectors constitute the orthogonal system of vectors {|ei iA }/{|fi iB }’s used in the original decomposition! If the eigenvalues are degenerate, the multiplicity gets reflected in the Schmidt coefficients as well. 2. The state is said to be maximally entangled if all the Schmidt coefficients are equal.
Von Neumann Entropy
We have seen that for a pure entangled state, the subsystems exist as mixtures. However, whenever the bipartite state is a pure product state, the subsytems are also in pure states. This difference can be exploited for the detection of entanglement by utilizing a mechanism by which the purity of any of the subsystems can be checked out. One such mechanism involves evaluating the V on N eumann Entropy of one of the subsystems. We give a formal definition of the V on N eumann Entropy below: Definition: The Von Neumann Entropy of a quantum state ρ is defined as S(ρ) = −T r(ρ ln ρ) . P In terms of the non-zero eigenvalues of ρ, we have: S(ρ) = − i λi ln λi . Clearly this is zero for a pure state and non-zero for a mixed one. Thus, by inspecting S(ρA ) (or S(ρB ) we can easily verify whether a pure bipartite state is entangled or not. The mixedness of A (and B) is both necessary and sufficient for the presence of entanglement in such a case. In the next section, we shall see how the problem of detecting entanglement gets significantly more onerous when we consider mixed bipartite states.
Mixed states: the problem gets tougher!
So far, we have seen that it required minimal effort to discern entanglement in pure bipartite states. For the simplest biqubit state, the problem becomes rather elementary. In fact, the simplicity of entanglement detection in these cases might tempt one into the presumption that even in mixed states, the problem should not be too tough either. So why
5 do we say it is a hard problem? Before answering this question, it is necessary to realize that we haven’t even defined entanglement for mixed states yet! And as a forethought, it is extremely important to note, at the very outset, that entanglement does not, in general, signify the inability to express ρAB as a (tensor) product of ρA and ρB ! Separability of Operators
Before venturing on to know about mixed state entanglement, one needs to get acquainted with the notion of separable operators. Definition: A positive semi-definite operator P acting on a Hilbert space HA ⊗ HB is said to be separable if and only if it admits the following decomposition: P =
n X
Ai ⊗ B i
(6)
i
where Ai ’s and Bi ’s are themselves positive semi-definite operators acting on the Hilbert spaces HA and HB respectively and n ∈ N. We denote the set of all such separable operators by S(HA , HB ). Any operator Q ∈ / S(HA , HB ) is said to be entangled. Now, we know that density operators are positive, semidefinite trace-class operators with trace equal to unity. Let us denote the set of all density operators on the composite Hilbert space by D(HA , HB ). Let us also represent the set of all “separable” density operators by DS (HA , HB ). Then one has the following relations: DS (HA , HB ) ⊂ D(HA , HB )
(7)
DS (HA , HB ) ⊂ S(HA , HB )
(8)
It then follows from the definition of separability, that a density operator is separable if f it can be written in the following form: X B ρAB = p i ρA (9) i ⊗ ρi i B where i pi = 1 (∵ T rρAB = 1) and ρA i ’s and ρi ’s are density operators corresponding to the subsystems A and B respectively. That is, the set DS (HA , HB ) is the convex hull of pure product density operators. How can we interprete this fact? Since any separable density operator can be written as a convex combination of pure product states, it means that one can look at it as being a mixture of pure product states and since any density operator that does not belong to DS (HA , HB ) cannot be written as a convex combination of pure product states, it cannot be looked upon as a mixture of pure product states - and that is all one needs to ensure entanglement in mixed bipartite states.
P
Having introduced the notion of separable and entangled density operators, we can now revert to the question of why detecting entanglement in the domain of mixed states is a tricky job. For one, the definition in this case is not as simple as in the case of pure states. In fact it is highly non-trivial. It is not easy to check if, given a mixed bipartite state, it is possible to resolve the state into a convex linear combination of pure product states! Another collateral reason is that even in those cases where things might look intuitively obvious, we cannot usually comment, straight-away, on the separability of states without a proper operational mechanism to validate it. This slippery aspect will be made clear through a few examples in the following section.
Exemplification: It isn’t all too intuitive
First we introduce the Bell states: 1 1 P 1. |φ± i = 2− 2 i (±1)i |i, ii) = 2− 2 (|00i ± |11i), i ∈ {0, 1} 1 P 1 2. |ψ ± i = 2− 2 i (±1)i |i, 1 − ii) = 2− 2 (|01i ± |10i), i ∈ {0, 1}
6 The Bell states are mutually orthogonal and form a basis, just like the computational basis states {|00i , |01i , |10i , |11i}. [28] Let us now look at the following examples of mixed biqubit states: Example.1 ρ = 2−1 (|φ+ i hφ+ | + |φ− i hφ− |). Clearly this is an equal mixture of |φ+ i and |φ− i, each of which is a maximally entangled state. Being a mixture of entangled states, it may physically appear to be entangled as well, except that it is not! But how do we draw such a drastic inference? Minor manipulation shows that the above state is exactly equal to 2−1 (|00i h00| + |11i h11|)! Then, according to the definition of separability, the state must be separable! So, even though we had a mixture of maximally entangled states to begin, we find that it can also be expressed as a mixture of the product states |00i and |11i! This is definitely astonishing! Example.2 ρ = 4−1 (|φ+ i hφ+ | + |φ− i hφ− | + |ψ + i hψ + | + |ψ − i hψ − |). This is an equal mixture of the four Bell states, all of which are maximally entangled. And here, again, the state is not entangled in reality. Since the Bell states form an orthonormal basis for the biqubit system, we conclude that 1AB (10) 4 represents the Identity Operator on the biqubit Hilbert space HAB . This allows us to recast the state as: ρ=
where 1AB
ρ = 4−1 (|00i h00| + |01i h01| + |10i h10| + |11i h11|)
(11)
Thus, once again, we have been able to express the state as a mixture of pure product states, and hence, the mixed state, in this case, too is separable! Example.3 ρ = p |φ+ i hφ+ | + (1 − p) |01i h01| , 1 > p > 0. This is a mixture of a Bell state, which is maximally entangled, and a computational basis state, which is a product state. It turns out that this is an entangled state, which we shall see later. In fact, a statistical mixture of an entangled and a product state is always entangled. Quantum Maps which are P but not CP
Valid quantum dynamical maps are linear, hermiticity-preserving, positivity-preserving, and also trace-preserving. Not only that, they have to be completely positive. Any map that preserves positivity is called a P-map and that which is completely positive is called a CP-map. Both these terminologies are defined below: 1. P-map: A map Λ is said to be a P ositive M ap iff it preserves positivity of an operator,i.e. if Q be a positive semi-definite operator, then Λ(Q) is also a positive semi-definite operator. 2. CP-map: Any map Λ is said to be a Completely P ositive M ap iff (1d ⊗ Λ) is a positive map, irrespective of the dimension d of the extension. From the above definitions it is clear that CP-maps are also P-maps, but the converse is obviously not true. The set of P-maps form a strict subset of CP-maps. Maps which are P but not CP are often called PnCP maps. These are not bonafide quantum operations, but can be extremely handy in the context of separability. It has been shown that a state ρ is separable iff for any PnCP map Λ (1 ⊗ Λ)ρ ≥ 0
(12)
where Λ acts on a subsystem. That separable states satisfy this criterion can be easily justified. A bipartite separable state ρ can be expressed as: X B ρ= pi ρA (13) i ⊗ ρi i
7 P B where i pi = 1 (∵ T rρ = 1) and ρA i ’s and ρi ’s are density operators corresponding to the subsystems A and B respectively. Applying the map Λ on Bob’s subsystem yields: X B (1 ⊗ Λ)ρ = p i ρA (14) i ⊗ Λ(ρi ) i
But Λ, being a positive map, preserves positivity, and from the definition of separability, we find that the resulting state represents a separable operator which is positive. Thus, separable states satisfy the above criterion. On the contrary, whenever we find that the condition is violated, we are free to conclude that the state is entangled!
The Reduction Criterion
This criterion is basically an application of PnCP maps. Consider the positive map Λ, which is not CP, and which acts on a generic positive operator Ω (defined on the relevant Hilbert space) in the following manner: Λ(Ω) = T rΩ1 − Ω
(15)
This is known as the Reduction Map. That this is indeed a positive map follows from the fact that none of the eigenvalues (which are all non-negative) of Ω can exceed Tr Ω. With the reduction map, the separability condition discussed in the previous section becomes: ρA ⊗ 1 − ρ ≥ 0
(16)
Equivalently, the dual criterion may be obtained by applying the composite map Λ ⊗ 1: 1 ⊗ ρB − ρ ≥ 0
(17)
We now revisit the example of section where we claimed that this state, i.e. ρ = p |φ+ i hφ+ | + (1 − p) |01i h01| is entangled for all values of p in the range 0 < p < 1. To acknowledge this, we apply the reduction criterion and are subsequently led to the following inequality: p 0 0 2 0 p 0 0 0 2 1−2p (18) 1−p ≥ 0 0 0 2 2 1 0 0 1−p 2 2 + where the inequality has been written in matrix form w.r.t the Bell i , |ψ − i , |φ+ i , basis:{|ψ 2 1 1 − 2p 1 − p |φ− i}. If we simply investigate the lower-right 2 × 2 block, i.e. we find that its determinant − p2 1 − p 1 2 is strictly negative unless p = 0. [29]
In fact, it should be added here, that for 2 × 2 and 2 × 3 systems, the reduction criterion is both necessary and sufficent for separability. For higher dimensions, this is just a necessary condition.
Positive Partial Transposition Criterion
Another example of a PnCP map is the T ranspose map. Suppose we have an orthonormal basis {|ii}, i ∈ {0, 1, 2, ..., d − 1} corresponding to some d-dimensional Hilbert space. The Transpose map acts in the following manner [30] : (|ii hj|)T = |ji hi| So, if we write down the state in the form ρ = ρT =
X i,j
P
i,j
(19)
ρij |ii hj|, applying transposition to this state results in:
ρij (|ii hj|)T =
X i,j
ρij |ji hi| =
X i,j
ρji |ii hj|
(20)
8 Thus, this necessarily amounts to transposing the density matrix! Since, the transpose of any matrix possesses the same spectrum, positivity is preserved. Now, we come to the notion of a partial transpose. Partial transposition of a bipartite state is obtained by tranposing just one of the subsystems. It is equivalent to applying the operation 1 ⊗ Tˆ where Tˆ signifies the T ranspose operator and nothing is done to the other subsystem. So, if we have a bipartite Hilbert space whose factor spaces corresponding to the subsystems have the orthonormal bases {|iiA }, i ∈ {0, 1, 2, ..., dA − 1} and {|jiB }, j ∈ {0, 1, 2, ..., dB − 1} respectively, and we write down the bipartite state in the following form: X ρAB = ρij,kl |iiA hk| ⊗ |jiB hl| (21) i,j,k,l
the partial transpositon with respect to the system A will be obtained as follows: X A ρTAB = ρkj,il |iiA hk| ⊗ |jiB hl|
(22)
i,j,k,l B Similarly we can obtain ρTAB . The most consequential aspect of transposition is that the partial transposition does not ensure positivity! Thus transposition is a PnCP map. States, which, under partial transposition, remain positive are known as PPT states. The rest are known as NPT states. An NPT state, under partial transposition, goes to an operator that has at least one negative eigenvalue and hence, does not remain positive anymore.
Once again let us apply the PPT criterion on the state ρ = p |φ+ i hφ+ | + (1 − p) |01i h01|. This leads to the following inequality: p 0 0 0 2 p 0 1 − p 2 0 ≥ 0 (23) p 0 0 0 2 p 0 0 0 2 where the matrix, in this case, has been evaluated in the computational basis. This condition, too, mandates that p = 0 and our claim is vindicated once again. Since transposition is a PnCP map, separable states are all PPT states! In this case too, for 2 × 2 and 2 × 3 systems, the PPT criterion is necessary as well as sufficient for separability. However, this is not the case for higher dimensions. In higher dimensions there exist PPT entangled states [31] too! The method of Witnessing Entanglement: A Convex-Analytic Approach
This is one method that does not involve coming up with some PnCP map. Instead the process of witnessing entanglement is an application of convex analysis. There is a theorem in functional analysis that goes by the name of Hahn-Banach T heorem or the Separating Hyperplane T heorem. The statement is as follows: Separating Hyperplane Theorem: Let S be a convex, compact set in a finite-dimensional Banach space [32] . Then if ρ ∈ / S, there exists a hyperplane separating ρ from S. In the same vein, since the set of separable density operators is a convex, compact set, we can construct a hyperplane that would separate a given entangled state from the set of separable density operators. Such a construction necessitates the detection of suitable hermitian operators. The criterion of entanglement is stated below: A state ρ is entangled iff there exists a hermitian operator W that satisfies T r(Wρ) < 0 and T r(Wσ > 0 for any σ belonging to set of separable states. The hyperplane is then identified by the equation T r(Wµ) = 0. The hermitian operator W is known as the witness operator that detects the entanglement. However, the detection of an appropriate witness operator may be a tricky job. Moreover, there exists no universal entanglement witness! If one witness operator can detect the entanglement of some state, there is no guarantee that it
9
FIG. 2: Visualization of Entanglement Witness
would be able to reliably detect entanglement in some other state. This is why we concern ourselves with the task of constructing an optimized witness operator with the ambition of identifying as many entangled states as practicable. Characterizing Entanglement through Uncertainty Relations: A Non-Linear Technique
So far, we have seen only linear methods of detecting entanglement. In this section, a non-linear mechanism will be introduced, which concerns a certain type of uncertainty relations. Whenever we think of uncertainty relations we tend to think of continuous variable systems and relations which are of the product form, as for example the well-known Heisenberg U ncertainty Relation (which is a weaker form of the Robertson-Schrodinger Uncertainty ~ Relation) which states that the product of uncertainties in position x ˆ and momentum pˆ is at least . However, 2 product-uncertainty relations are of no practical interest in finite-dimensional systems. Why so? Because, in these cases, all observable uncertainties are bounded and finite. So the minimum of the uncertainty product is trivially zero for any conjugate pair: whenever we have an eigenstate of one of the relevant observables, we end up having the product of uncertainties as zero. This is in contrast with continuous variable systems where zero-uncertainty of one of the variables necessarily implies an infinite uncertainty in its canonical conjugate. So in finite-dimensional systems, we resort to sum-type uncertainty relations. For a set of non-commuting observables {Ai }’s, we have a non-trivial [33] lower bound to the sum of their uncertainties, i.e. X ∆A2i > χ (24) i
where χ signifies a universal limitation to the measurement statistics pertaining to the above non-commuting set. In some cases, the value of χ might be fairly simple to calculate. Take for example the spin- 21 system and the non-commuting set of Pauli operators {σi }’s, i = {1, 2, 3}. Then it is not difficult to show that the following sumuncertainty relation holds: 3 X
∆σi2 > 2
(25)
i=1
P3 P3 2 Then, Λ2 = To show this, assume: Λ = i=1 ai σi , x = i=1 ai , where ai ∈ R, ∀i ∈ {1, 2, 3}.
P3 P P 3 2 3 2 2 = x =⇒ hΛi 6 x =⇒ i ai hσi i 6 x. Choosing ai = hσi i, i=1 ai 1+ m,n=1 iam an mnr σr = x1. [34] Thus Λ
P3 P P P P3 √ 3 2 3 3 2 2 2 2 2 we have: x =⇒ i=1 ai 6 i=1 hσi i 6 1. Consequently, i=1 ∆σi = i=1 ( σi − hσi i ) = 3 − i=1 hσi i 6 2.
10 Equipped with this armoury, we would now like to implement this principle in the context of composite (in particular, bipartite) systems. Let us consider two pairs of non-commuting sets of local observables {Ai }’s and {Bi }’s, i = {1, 2, ...n} (where n ∈ N) defined for two individual systems A and B respectively. Then for each of this pair there exists a sum-uncertainty relation and a corresponding universal lower bound: n X
∆A2i > Ua
(26)
∆Bi2 > Ub
(27)
i=1 n X i=1
We suitably combine the above operators to form a new set of non-commuting joint-observables {Γi } = {Ai ⊗ 1b + 1a ⊗ Bi } for each i ∈ {1, 2, ..., n} [35] . Then, for any arbitrary state ρAB , we have: X X ∆Γ2i = [∆A2i + ∆Bi2 + 2(hAi Bi i − hAi i hBi i)] (28) i
i
For a pure product state (ρAB = ρA ⊗ ρB ),hAi Bi i = hAi i hBi i and thus, the following relation holds: X X ∆Γ2i = [∆A2i + ∆Bi2 ] > Ua + Ub i
(29)
i
P Further, for mixed separable states that have the form ρAB = k pk ρkA ⊗ ρkB the variance of any joint observable Γ satisfies the following inequality: X ∆Γ2 > pk ∆k Γ2 (30) k
where ∆k Γ signifies the variance of Γ in the k product state ρkA ⊗ ρkB appearing with probability pk in the mixture ρAB . Combining equations 29 and 30 , we infer that even for mixed separable states there exists the same lower bound, i.e. X ∆Γi 2 > Ua + Ub (31) 2
th
i
Such an uncertainty relation is known as a local uncertainty relation since it involves the sum of variances of local observables. Clearly any non-observance of this relation implies the presence of entanglement! 1
In order to exemplify the sufficiency of this condition let us consider the singlet state |ψ − i = 2− 2 (|01i − |10i). We have seen that the Pauli operators for a qubit satisfy a sum-uncertainty relation as depicted in equation 25 . So we consider the joint observable Γi = σA,i ⊗ 1B + 1A ⊗ σB,i . Then any biqubit separable state must satisfy the local uncertainty relation: X i
∆Γ2i =
3 X
2 2 (∆σA,i + ∆σB,i )>4
(32)
i=1
Now it turns out that the singlet state |ψ − i, maximally violates the above inequality, in the sense that the sum of the variances of Γi ’s in this state is zero, which is the trivial lower bound! This shows that the singlet state is indeed an entangled state [36] . There is, in fact, a straightforward way to construe the observation that the sum of variances is zero. The singlet state has an amazing property: it retains the same form regardless of the choice of an orthonormal product basis. That is, if we represent, by |0inˆ and |1inˆ , the eigenstates of ~σ · n ˆ with respective 1 eigenvalues +1 and −1, the singlet state has the form 2− 2 (|01inˆ − |10inˆ ) for any unit vector n ˆ . Simple observation tells us that the state is an eigenstate of (~σ · n ˆ ⊗ 1 + 1 ⊗ ~σ · n ˆ ) and this is, therefore, true for all unit vectors n ˆ! Thus the singlet state is an eigenstate of Γi (defined earlier) for each i = {1, 2, 3}, and no wonder the variance vanishes! In the following section, we shall state some applications of the separability techniques discussed earlier, in particular the reduction criterion and the PPT criterion, to the case of two important classes of mixed states known as the Werner class and the Isotropic class.
11 Werner Class and Isotropic Class
The Werner class of states [37] is characterized by the density operator given below: (1 − p) ρW = p ψ − ψ − + 12 ⊗ 12 4
(33)
where |ψ − i is the singlet state. This is an admixture [38] of the singlet state with white noise. As we have seen in the previous section, the singlet retains its form in any orthonormal product basis. Mathematically, this means that it remains invariant under any product unitary transf ormation of the form U ⊗ U : (U ⊗ U ) ψ − ψ − (U † ⊗ U † ) = ψ − ψ −
(34)
As a consequence, the Werner state also remains invariant under such a transformation: (U ⊗ U )ρW (U † ⊗ U † ) = ρW
(35)
Note that the Werner state can be written in the diagonal form as: ρW =
(1 + 3p) − − (1 − p) + + − − + + φ ) φ + φ ψ ψ + ψ + φ (ψ 4 4
(36)
From this form, it is clear that positivity requirement on the Werner class implies that − 31 6 p 6 1. It is a matter of simple exercise to show that both the reduction criterion and the PPT criterion lead, in this case, to the following relation: 1 (1 − 3p) >0⇒p6 4 3
(37)
Thus, the Werner states are separable as long as p 6 13 . As p goes up beyond this threshold value, ρW becomes entangled. Similarly, there are states which are invariant under product unitary transformations of the form U ⊗ U ∗ . These are the isotropic states. Biqubit isotropic states are characterized by the following density operator: (1 − α) 12 ⊗ 12 ρiso = α φ+ φ+ + 4
(38)
1
where |φ+ i is the Bell state 2− 2 (|00i + |11i). In this case, too, the positivity condition demands that − 13 6 α 6 1, and the state is separable if and only if α 6 13 .
QUANTUM NONLOCALITY
For a long time, the question whether quantum mechanics is local or non-local has piqued the physics community. And even today, there exists no unanimous answer. There are quite a few interpretations of quantum theory which view it as a non-local theory. However, it looks most likely (as claimed by many) that it is primarily a matter of interpretation if quantum mechanics should be looked upon as a local or non-local theory, and lot of it is dependent on the real semantic interpretation of the notion of “locality”. So what exactly are we going to understand by locality? One of the first attempts to qualify locality in the context of quantum theory was made by Einstein, Podolsky and Rosen. They, in their famous EPR paper, expressed dissent on the proclaimed completeness of quantum theory. In fact, they argued that unless we are ready to abnegate at least one of the seemingly natural notions of locality and realism, we must be willing to give up the idea that quantum theory provides a complete description of reality and look for means to supplement it with additional variables that serve to remove the indeterminacy of quantum theory. In the next section, we shall introduce the notion of Bell locality, which was the first serious attempt to understand what should be essentially expected of a local theory.
12
FIG. 3: Two spatially separate parties Alice and Bob performing local measurements on their individual qubits. Each of them has two choices of inputs: {x0 , x1 } for Alice and {y0 , y1 } for Bob. They can select any one of the inputs at a time. The respective outcomes are denoted by a and b, each of which can take a value either +1 or −1.
Bell Locality and the Bell-CHSH Inequality
It was John S. Bell who came up with a concrete proof that the additional “local” “realistic” hidden variables, as anticipated by Einstein didn’t exist! We shall provide a comprehensive introduction to the most commonly adduced version of Bell0 s T heorem - the CHSH Inequality. Before embarking on this journey, we provide the exact scenario in which the CHSH Inequality arises. [Figure 3 depicts the experimental set-up.] Once again, we need Alice and Bob’s cooperative assistance. There is a device that prepares, simultaneously, two qubits in some composite state. Following the production of any such pair, one of these particles goes to the possession of A (Alice) and the other goes to B (Bob) in such a way that both receive the particles at the same time. As soon as they receive the particle, each of them performs some local measurement on it. Let us denote the measurement settings of A and B by x and y and the outcomes by a and b respectively. Since we are dealing with qubits here, each of the observables measured is necessarily dichotomic (two-valued). Let the corresponding allowed outcomes be +1 and −1 respectively. So, a, b ∈ {+1, −1}. For the CHSH scenario, we also have: x, y ∈ {0, 1}[39]. Next, we define a joint probability distribution for the measurement outcomes of A and B: P (a, b|x, y) (or, for simplicity, P (ab|xy)) satisfying: X P (ab|xy) = P (a|xy) (39) b
X
P (ab|xy) = P (b|xy)
(40)
a
which are simply the marginals of the P (ab|xy). In quantum theory, one cannot factor P (ab|xy) as P (a|x)P (b|y), which illustrates that there exist statistical correlations between the outcomes of Bob and Alice! The existence of correlation between the outcomes of measurements carried out by the two parties on spatially separated objects was precisely the source of confusion that led to the EPR paradox and compelled Einstein to believe that the outcomes must be fundamentally pre-determined. Next we come to the pivotal assumption of local realism. We argue that just because the outcomes are correlated, one cannot conclude that the correlations are spooky (non-local). In principle, there must exist some hidden variable λ that is the “real” source of these correlations. In other words, λ is the “past factor” that exerts causal influences on both the observables. At no point does either of these observables causally affects the other! The quantum correlations are simply the manifestation of the fact that both these observables had been causally affected by the same hidden variable. So, in principle, if we are furnished with the knowledge of λ, we can factor out the joint probability: P (ab|xy, λ) = P (a|x, λ)P (b|y, λ)
(41)
13 The above equation simply reveals that no correlations exist between the measurement outcomes once λ is specified. Equation 41 is known as the locality assumption. [40] However, we seemed to have provided a description of local realism, but neither locality nor realism separately or independently. Therefore, it is necessary to explain what precisely is meant by realism in this context. That there exists a hidden variable λ which gives the most complete description of the reality of the composite state is equivalent to the assumption of realism. λ is called the ontic variable. But it is also essential to note that the existence of such a hidden state variable does not ensure locality by itself! There is no a priori reason to enforce that a realistic description of the state must satisfy equation 41! Only when such a hidden variable satisfies the equation 41 we are free to conclude that the variable is local realistic. Note, also, that the locality condition is defined particularly with respect to a given realist backdrop! This is, by no means, the most general definition of locality! In fact,going by this definition of locality, if we take it to be a general one, we should expect that the condition of realism must always precede the assumption of locality! That is why Bell locality is, strictly speaking, locality at the “realist” (ontological) level. Equation 41 can be equivalently expressed by the following two conditions: P (a|xy, λ, b) = P (a|x, λ)
(42)
P (b|xy, λ, a) = P (b|y, λ)
(43)
The basic meaning of the locality condition is that events at one region of space-time DO NOT influence those at other space-like separated regions (in compliance with special relativity). Note that not only are the outcomes statistically independent of one another, either of the outcomes are independent of the measurement settings chosen by the other party. If, for instance, Alice’s outcomes depended on the setting chosen by Bob, that would imply signalling. [41] Implication of the factorizability condition: The model we are dealing with has some intrinsic indeterminacy, nevertheless. We are not, for instance, assuming that specification of λ is enough to tell us about the outcomes completely (or deterministically). All we have assumed is that the measurement statistics obtained by A is independent of the statistics obtained by B. It is in this sense that there exist no nonlocal correlations between the measurement outcomes of Alice and Bob. Such a model is usually known as the stochastic, factorizable model. In addition, we make the assumption that the ontic space Λ that describes the space of hidden variables is a measureable space. Let us define the support Λρ corresponding to a given quantum state ρ as follows: Λρ = {λ ∈ Λ : ξ(λ|ρ) > 0}
(44)
where ξ(λ|ρ) represents the probability distribution of the variable λ corresponding to a fixed quantum state ρ. Clearly, then: Z dλ ξ(λ|ρ) = 1.[42] (45) Λρ
It should also be mentioned that P (a|x, λ) and P (b|y, λ) are known as the local response functions for Alice and Bob respectively, assumed to be properly normalized. [The stochastic model is called deterministic iff: p(a|x, λ), p(b|y, λ) ∈ {0, 1}, ∀a, b, x, y, λ. Here we shall assume a general stochastic model.] Then we can write, for a given state ρ: Z P (ab|xy) =
dλ P (λ|xy, ρ)P (ab|xy, λ)
(46)
dλ ξ(λ|ρ)P (a|x, λ)P (b|y, λ)
(47)
Λρ
Z = Λρ
where in the last equation we have assumed locality and also made use of the fact that the choice of measurement settings have no correlation, whatsoever, with the hidden variable λ (this is sometimes called the f ree will assumption).
14 The stage is now set and we can finally derive the CHSH inequality. The inequality involves the following expression: X (−1)xy hax by i hBi = (48) x,y∈0,1
= ha0 b0 i + ha0 b1 i + ha1 b0 i − ha1 b1 i
(49)
where B is the Bell-CHSH operator, hBi represents the expectation value of B and hax by i stands for the expectation value of the product of Alice and Bob’s outcomes when their respective measurement settings are x and y. Now: X hax by i = abP (ab|xy) (50) a,b∈{+1,−1}
Z =
dλ ξ(λ|ρ) Λρ
abP (a|x, λ)P (b|y, λ)
(51)
a,b∈{+1,−1}
Z =
X
dλ ξ(λ|ρ) Λρ
X
aP (a|x, λ)
a∈{+1,−1}
X
bP (b|y, λ)
(52)
b∈{+1,−1}
Z = Λρ
dλ ξ(λ|ρ) hax iλ hby iλ
(53)
where equation 51 follows from 47 , and hax iλ and hby iλ denote the expectation values of a and b in the ontic state λ with the given measurement settings x and y respectively. Making use of equation 53 in 49, we obtain the following relation: Z hBi = dλ ξ(λ|ρ)[ha0 iλ (hb0 iλ + hb1 iλ ) + ha1 iλ (hb0 iλ − hb1 iλ )] (54) Λρ
Let us take χλ = ha0 iλ (hb0 iλ + hb1 iλ ) + ha1 iλ (hb0 iλ − hb1 iλ ) Then, since |hax iλ | and hby iλ are bounded above by 1, for all x,y and λ: |χλ | 6 |hb0 iλ + hb1 iλ | + |hb0 iλ − hb1 iλ |
(55)
6 2 · max(|hb0 iλ |, |hb1 iλ |)
(56)
=2
(57)
and this is true for all λ! Thus, what we have essentially found out is that |hBi| 6 2
(58)
for any local realistic model! Any theory which admits an underlying local realistic hidden variable model, must therefore respect the above inequality! Else, the theory cannot comply with the premise of local realism. The inequality 58 is the well-known CHSH Inequality. So, if we denote the premises of realism and locality by R and L respectively and the statement of CHSH inequality by BI, we have the following logical structure: R ∧ L =⇒ BI
(59)
∴ ¬BI =⇒ ¬(R ∧ L)
(60)
Quantum Mechanics does not respect the CHSH Inequality!
The main motivation behind the CHSH inequality was to verify if, in principle, quantum mechanics supports a local realistic hidden variable model as envisaged by Einstein. Bell took up the example of the EPR-Bohm state(singlet state) |ψ − i and came up with a scenario in which the CHSH inequality is violated. Let us denote the direction in which A measures the spin of her qubit by x ˆ and that in which Bob measures by yˆ. Then quantum mechanical calculations yield the following correlation function for the singlet state: hσxˆ ⊗ σyˆi = −ˆ x · yˆ
(61)
15 Then for this state, we have: |hBi| = |ˆ x0 · (ˆ y0 + yˆ1 ) + x ˆ1 · (ˆ y0 − yˆ1 )|
(62) √ Simple observation tells us that for x ˆ0 = ˆj, x ˆ1 = ˆi, yˆ0 = √12 (ˆi + ˆj) and yˆ1 = √12 (ˆi − ˆj), we have |hBi| = 2 2 a clear violation of the CHSH Inequality! Then, by 60 we conclude that no local realistic model can explain the observed correlations of the singlet state! This result was a huge jolt for those who, like Einstein, had reposed all their faith in an underlying classical or classical-like model! Note that this example makes use of an entangled state. It does not immediately say anything about a product state. For a product state which does not exhibit correlated outcomes, it is interesting to know if it would admit a LRHVM (Local Realistic Hidden variable model). Since product states ascribe independent states to each of the constituent particles, the joint probabilities are factorizable and hence, the locality condition is satisfied even at the quantum theoretic level. So, in this case, given the product state, we can immediately decompose the joint probabilities and the CHSH inequality must be satisfied! We need not even have to search for a hidden variable in the first place - the quantum state itself seems to “completely” describe the joint system! We can also convince ourselves via an explicit calculation, which yields: hBiψ⊗φ = ha0 iψ (hb0 iφ + hb1 iφ ) + ha1 iψ (hb0 iφ − hb1 iφ ) =⇒ hBiψ⊗φ 6 2!
(63) (64)
What about mixed separable states (or, separable states, in general)? Once again, since such states are mixtures of product states, it seems intuitive that they too must satisfy the CHSH inequality. And that is indeed true. Since any mixed separable state can be written as a convex combination of product states we have: X ρ= pλ ρλA ⊗ ρλB (65) λ
=⇒ hBiρ =
X
pλ hBiρA ⊗ρB
(66)
X X pλ hBiρA ⊗ρB 6 2 pλ = 2 =⇒ hBiρ 6
(67)
λ
λ
λ
Thus any separable state satisfies the CHSH inequality. Stated in another way, we have found out another suf f icient criterion for bi-qubit entanglement! But are there entangled states that satisfy the CHSH inequality? We would like to defer answering this question for the time being and address the aspect of necessity of this condition in a later section.
States maximally violating the CHSH Inequality: The “Eigenvalue” Approach
In quantum mechanics, measurement of any physical quantity is tantamount to a ”collapse” of the quantum state to an eigenstate of the associated observable (denoted by a self-adjoint operator) and simultaneous revelation of the corresponding eigenvalue as the measurement outcome. Thus, in the quantum mechanical formalism, the Bell operator we mentioned earlier is a self-adjoint operator having the following form: B = A0 ⊗ (B0 + B1 ) + A1 ⊗ (B0 − B1 )
(68)
Here Ai ’s and Bi ’s, for i ∈ {0, 1}, represent the local observables of Alice and Bob respectively and the index i has a one-to-one correspondence with the choice of measurement setting for every party. So, for instance, Ai (or Bi ) denotes an observable that is the quantum mechanical version of the physical quantity defined on the qubit in Alice’s (or Bob’s) possession corresponding to the ith measurement setting. Since quantum mechanical measurements are bound to reveal only the eigenvalues of the associated observable, the expectation value of any observable is an average of these eigenvalues weighted with their corresponding probabilities. Consequently, the maximum and minimum possible values of the expectation value of any observable are equal to the maximum and minimum eigenvalues of the same! So if the Bell operator B has a maximum eigenvalue of χmax , its maximum expectation
16 √ value must also be χmax ! Alternatively, χmax > 2 2! Our present task will be to justify this. To show the above condition, we construct the square of the Bell Operator: B 2 = 1 ⊗ (21 + {B0 , B1 }) + 1 ⊗ (21 − {B0 , B1 }) − [A0 , A1 ] ⊗ [B0 , B1 ] = 41 ⊗ 1 − [A0 , A1 ] ⊗ [B0 , B1 ]
(69) (70) (71)
The above derivation makes use of the fact that because the local observables are dichotomic and have eigenvalues +1 and −1 each, A2x = By2 = 1, ∀x, y ∈ {0, 1}.So if we let [A0 , A1 ] = 2iA2 and [B0 , B1 ] = 2iB2 [43] we have: B 2 = 41AB + 4A2 ⊗ B2
(72)
Writing Ax = ~σ · x ˆ and By = ~σ · yˆ, we find that A2 = ~σ · (xˆ0 × xˆ1 ) and B2 = ~σ · (yˆ0 × yˆ1 ). For simplicity we express (xˆ0 × xˆ1 ) as sin θxˆ2 where θ is the angle between xˆ0 and xˆ1 respectively, and xˆ2 is some unit vector. Similarly for the other cross product, we write yˆ0 × yˆ1 as sin φyˆ2 , where yˆ2 is another unit vector. Then B 2 becomes: B 2 = 41AB + 4 sin θ sin φ~σ · xˆ2 ⊗ ~σ · yˆ2
(73)
From this form, we can immediately conclude the eigenvalues and eigenvectors of B 2 : Eigenvalue 4(1 + sin θ sin φ) 4(1 − sin θ sin φ) Eigenvector(s) |00ixˆ2 ,yˆ2 , |11ixˆ2 ,yˆ2 |01ixˆ2 ,yˆ2 , |10ixˆ2 ,yˆ2 Here |ijixˆ2 ,yˆ2 represents an eigenstate of ~σ · xˆ2 ⊗ ~σ · yˆ2 . For instance, |01ixˆ2 ,yˆ2 is the simultaneous eigenstate of ~σ · xˆ2 ⊗ 1 and 1 ⊗ ~σ · yˆ2 with eigenvalues +1 and -1 respectively. 2 Since the √ eigenvalues of B are simply the square roots of the eigenvalues of B , the eigenvalues of B satisfy: |λ± | = 2 1 ± sin θ sin φ. From this expression, one thing is pretty clear: whatever be θ and φ, unless xˆ0 = ±xˆ1 or yˆ0 = ±yˆ1 or both, there exists two eigenvalues ±λ+ for which |hBi| exceeds 2! Thus, not considering the trivial case where at least one of the pairs of local measurement settings are parallel, we always have the possibility of violating CHSH inequality! Also it is clear that the maximal violation takes place √ when sin θ = sin φ = 1, i.e. when the local measurement settings are√mutually perpendicular, for which |λ | = 2 2 and therefore the CHSH expression + √ attains its maximum value of 2 2. Hence χmax = 2 2! In this case, it is easy to see that {A0 , A1 } = {B0 , B1 } = 0. Thus the violation noticed for the singlet state corresponding to the settings x ˆ0 = ˆj, x ˆ1 = ˆi, yˆ0 = √12 (ˆi + ˆj) and yˆ1 = √1 (ˆi − ˆj) was the maximal possible violation of the CHSH inequality! And we should also notice that the local 2
measurement settings in this case are mutually orthogonal! For any given pair of local settings that allow Bell’s inequality violation, the states that maximally violate the inequality are the eigenvectors of B with eigenvalues ±λ+ . These eigenvectors will be of the form |ψ0 i = c0 |00ixˆ2 ,yˆ2 + c1 |11ixˆ2 yˆ2 and |ψ1 i = c∗1 |00ixˆ2 ,yˆ2 − c∗0 |11ixˆ2 yˆ2 , since |00ixˆ2 ,yˆ2 and |11ixˆ2 ,yˆ2 are the eigenvectors of B 2 with the √ eigenvalue λ2± . [44] The maximal CHSH value of 2 2 is often called the T sirelson Bound, because it was Boris Tsirelson who first obtained this bound, though the approach we have adopted here is basically the one that was taken by Lawrence J. Landau in a later paper. “Nonlocality” of the Bell Operator
From the discussion in the last section, it is clear that the spectral form of the Bell operator that maximally violates the CHSH inequality is given by: √ B = 2 2(|ψ0 i hψ0 | − |ψ1 i hψ1 |) where |ψ0 i and |ψ1 i have been defined in the last section.
(74)
17 In general, for arbitrary settings, the spectral decomposition has the form: p p B = 2 1 + sin θ sin φ(|ψ0 i hψ0 | − |ψ1 i hψ1 |) + 2 1 − sin θ sin φ(|ψ2 i hψ2 | − |ψ3 i hψ3 |)
(75)
where hψi |ψj i = δij ,∀i, j ∈ {0, 1, 2, 3}. Now, it is essential to note that the Bell operator is not, strictly speaking, a “local” operator, in the sense that it cannot be factorized into two local hermitian operators in its factor spaces HA and HB , i.e. it cannot be written in ˆ ⊗ Yˆ , for some X ˆ defined on HA and some Yˆ defined on HB . To show this, assume, on the contrary that the form X such a decomposition exists. So we can write B = (α1 + µ ~ · ~σ ) ⊗ (β1 + ~ν · ~σ )
(76)
where α, β ∈ R and µ ~ and ~ν are real vectors. Now, B is a traceless operator, which leads us to the conclusion that at least one of α and β must be zero! Suppose, β is zero [45]. Then B becomes: B = (α1 + µ ~ · ~σ ) ⊗ ~ν · σ 2
2
2
2
=⇒ B = ν [(α + µ )1 + 2α~ µ · ~σ ] ⊗ 1
(77) (78)
where µ = |~ µ|,ν = |~ν |. From the form of B 2 , it is apparent that any vector of the form |iiµˆ ⊗ |ψi , i ∈ {0, 1}, is an eigenvector of B 2 with eigenvalue ν 2 [α2 + β 2 ± 2αµ], [46] where |iiµˆ is an eigenstate of µ ~ · ~σ with the same eigenvalue, and this is true for all |ψi ∈ HB ! This has the essential implication that the eigenspace of B 2 corresponding to a given eigenvalue must necessarily comprise only product vectors, since any linear superposition of the eigenvectors, each having the form |iiµˆ ⊗ |ψi and the same eigenvalue, is an eigenvector of the form |iiµˆ ⊗ |φi with the same eigenvalue [47], for some |φi ∈ HB ! But in the previous section, we have seen that the eigenspace of B 2 corresponding to any given eigenvalue consists of only two orthogonal product vectors - the rest are all entangled states! For instance, the eigenspace corresponding to the eigenvalue 4(1 + sin θ sin φ) consists of linear combinations of |00ixˆ2 ,ˆy2 and |11ixˆ2 ,ˆy2 and all of them, except for the 2 product eigenstates, are entangled! This is in contradiction with our assumption that B can be written as a product of local hermitian operators defined on the factor spaces! ˆ ⊗ Yˆ is that How do we interprete the above observation? One decisive aspect of any observable of the form X we can always find at least one product eigenbasis, i.e. system of eigenvectors which are product states and form an orthonormal basis. Now, quantum mechanics treats product states on a similar footing as classical physics treats composite systems! Any description of the objective properties [48] of product states in quantum theory separates out the descriptions of the same corresponding to the individual subsystems, which essentially reflects the mutual statistical independence of the latter. And whenever such a decomposition exists, the locality assumption is automatically satisfied, even at the operational level [49] ! Thus, in such a scenario, since the expectation value of the product observable is bounded above by the maximum eigenvalue and there exists a product eigenstate (which can always be imparted an underlying local hidden variable structure) for this eigenvalue, we can, in principle, associate a local hidden variable model because the quantum mechanical bound does not exceed the bound predicted by a local √ model! For example, take the observable σx ⊗ (σx + σy ). Quantum mechanics predicts that |hσx ⊗ (σx + σy i)| 6 2. Any local hidden variable would predict an upper bound of 2, which is greater than the one predicted by quantum mechanics, and therefore, [50] quantum theory, in this case, would admit a local model. Thus, we can assert that the choice of the Bell operator is also instrumental in arriving at an inequality that actually depicts the incompatibility of quantum mechanics with local realism. If we had chosen an operator that could have been factored out we would not have been able to justify this incompatibility! It is in this sense that we have termed the Bell operator a suitably chosen “nonlocal” operator whose statistics can not be replicated by a local realistic model. Thus, at least two features are essential for unravelling a contradiction between quantum mechanics and the assumption of local realism: 1. A suitable [51] entangled state that exhibits correlations [52] which cannot be reproduced classically or by any local hidden variable theory. 2. A suitable hermitian operator which is non-factorizable and which does not have a product eigenbasis. [53]
18 But before jumping on to a conclusion that the above conditions should be enough, we ought to exercise some caution. We must, in parallel, ensure another extremely non-trivial criterion that the maximum eigenvalue of the chosen operator be greater than the bound predicted by a local model. For, if the maximum eigenvalue is below or equal to the latter, there is no scope of demonstrating the existence of quantum states which can violate the local bound! We must consider ourselves lucky that we know of the Bell operator that exhibits all these features.
Complementarity from Bell’s Inequality
Complementarity is the quantum mechanical principle enunciated by Niels Bohr, one of the founding fathers of quantum mechanics. The principle enumerates certain properties which cannot be gauged or measured simultaneously with arbitrary accuracy. Such properties are known as complementary properties. Examples of such properties are: wave and particle nature, position and momentum, spins along non-parallel directions, energy and time, etc. In this section, we show that the assumption of joint measureability of complementary properties (spin in different directions, in our case) implies Bell’s inequality! Recall that the construction of CHSH expression required 4 observables {A0 , A1 , B0 , B1 }, two of which (A0 , A1 ) are local to Alice and the rest (B0 , B1 ) [54] to Bob. Let us now assume complementarity, i.e. it is possible, in principle, to accurately measure non-commuting spins together. Consequently, there must exist a joint quadri-variate probability distribution for the outcomes of all the 4 observables - P (a0 , a1 , b0 , b1 |x0 , x1 , y0 , y1 ) which satisfies: X P (a0 , a1 , b0 , b1 |x0 , x1 , y0 , y1 ) = 1 (79) a0 ,a1 ,b0 ,b1
and returns the relevant joint probabilities of the experiment as its marginals: X P (a0 , a1 , b0 , b1 |x0 , x1 , y0 , y1 ) = P (a0 , b0 |x0 , y0 )
(80)
a1 ,b1
X
P (a0 , a1 , b0 , b1 |x0 , x1 , y0 , y1 ) = P (a0 , b1 |x0 , y0 )
(81)
P (a0 , a1 , b0 , b1 |x0 , x1 , y0 , y1 ) = P (a1 , b0 |x0 , y0 )
(82)
P (a0 , a1 , b0 , b1 |x0 , x1 , y0 , y1 ) = P (a1 , b1 |x0 , y0 )
(83)
a1 ,b0
X a0 ,b1
X a0 ,b0
Then the CHSH expression becomes: hBi = ha0 b0 i + ha0 b1 i + ha1 b0 i − ha1 b1 i X X X X = a0 b0 P (a0 , b0 ) + a0 b1 P (a0 , b1 ) + a1 b0 P (a1 , b0 ) − a1 b1 P (a1 , b1 ) a0 ,b0
=
X
a0 ,b1
a1 ,b0
[a0 (b0 + b1 ) + a1 (b0 − b1 )]P (a0 , a1 , b0 , b1 )
(84) (85)
a1 ,b1
(86)
a0 ,a1 ,b0 ,b1
where the last equality follows from the assumption of a joint probability distribution equipped with all the 4 relevant marginals. Now notice that since a0 , a1 , b0 , b1 ∈ {+1, −1}, one amongst (b0 + b1 ) and (b0 − b1 ) is zero while the other is either +2 or −2. Clearly then, the possible values of the expression in the parenthesis, i.e. a0 (b0 + b1 ) + a1 (b0 − b1 ) are ±2, and since the joint quadri-variate distribution is normalized, we must have |B| 6 2! Thus, any theory which ascribes joint measureability to spins of a particle in different (or more precisely, non-parallel) directions must obey the CHSH inequality! That quantum mechanics violates this inequality, thus, also goes to show that joint measureability of incompatible spin operators is unfeasible, and therefore, violation of Bell’s inequality can also be looked upon as a manifestation of the principle of complementarity in quantum theory. One might go further and ask: What is the exact relation between the premises of complementarity and local realism? Are they equivalent? Well, at least in this context, they turn out to be equivalent, as shown by Arthur Fine in one of his papers. In this paper, he argued that any theory that has an underlying local realistic structure admits
19 joint measureability [55] and vice-versa [56]. Thus the implications of Bell’s theorem are manifold and profound, and all of these seem to manifest the non-classicality [57] of nature. There are a few more premises from which Bell’s theorem can be derived and we shall discuss one of them later. Hardy’s Demonstration of Nonlocality
Lucien Hardy took an alternative approach and exposed the nonlocality of quantum theory in a unique manner. He too considered a situation where two parties Alice and Bob perform local measurements on dichotomic observables pertaining to their qubits. We label the choices of local measurement settings by {x0 , x1 } and {y0 , y1 } respectively, and the measurement outcomes by a and b, where a, b ∈ {+1, −1}. Now impose the following conditions: P (+1, +1|x0 , y0 ) = 0
(87)
P (+1, −1|x1 , y0 ) = 0
(88)
P (−1, +1|x0 , y1 ) = 0
(89)
Then if we assume that local realistic variables exist, we are led to the following equalities, for all values of λ: P (+1|x0 , λ)P (+1|y0 , λ) = 0
(90)
P (+1|x1 , λ)P (−1|y0 , λ) = 0
(91)
P (−1|x0 , λ)P (+1|y1 , λ) = 0
(92)
It is easy to show that the above conditions points to the following relation: P (+1|x1 , λ)P (+1|y1 , λ) = 0, ∀λ
(93)
=⇒ P (+1, +1|x1 , y1 ) = 0!
(94)
P (+1, +1|x1 , y1 ) is called the Hardy probability and is denoted by pHardy . It turns out that quantum mechanics violates condition 94. To realize this, consider an entangled state of the folowing form: |ψi = a(|01i + |10i) − b |11i
(95)
Both of them have the same measurement choices: 1. x0 , y0 : measurement in the computational basis. 2. x1 , y1 : measurement in the {cos θ |0i + sin θ |1i , sin θ |0i − cos θ |1i} basis, where the first state is associated with the outcome +1 and the next −1. Then it can be worked out that while conditions 87 , 88 and 89 are satisfied for a cos θ = b sin θ, pHardy equals b sin2 θ which is non-zero whenever b 6= 0(even though the state in that case is maximally entangled!) and sin θ 6= 0. It is a rather surprising observation that any entangled state except the maximally entangled one demonstrates this conflict with locality! This is, therefore, a good indication of the fact that nonlocality does not necessarily go up with √ entanglement! The maximal value of pHardy observed in quantum theory is 5 5−11 ≈ 0.0902. 2 TYPES OF CORRELATIONS AND NATURE’S “INTERMEDIATE” CHOICE
We shall now give a brief account of the principle types of correlations studied by physicists. Consider one again the familiar two-party scenario. Further, suppose that there are µ number of measurement settings per party and that there are δ number of possible outcomes for every local measurement. So x, y ∈ {1, 2, 3, ..., µ} and a, b ∈ {1, 2, 3, ..., δ}. Then P (ab|xy) characterizes the joint probability distribution for the outcomes. It also follows that there are a total of µ2 δ 2 joint probabilities. The set of all µ2 δ 2 joint probabilities is called a behavior. We call this set P. Then 2 2 P ∈ Rµ δ . Owing in a behavior satisfy: Pδto the positivity and normalization constraints, the joint probabilities 2 2 P (ab|xy) ≥ 0, and a,b=1 P (ab|xy) = 1. Then if P be the space of all behaviors, P ⊂ Rµ δ and dimP = (δ 2 −1)µ2 [58] . In the next few sections, we give a brief overview of the principle types of correlations that are studied by physicists.
20 No-Signalling Correlations
This is a special type of correlations that any legitimate physical theory is obligated to satisfy. Such correlations (or, behaviors) are characterized, in addition to the positivity and normalization requirement, by the following relations: δ X
P (ab|xy) =
b=1 δ X
δ X
P (ab|xy 0 ), ∀a, x, y, y 0
(96)
P (ab|x0 y), ∀b, x, x0 , y
(97)
b=1
P (ab|xy) =
a=1
δ X a=1
Means what? → A’s input setting does not exert any influence on the outcome obtained by B, and vice-versa. That is, the local marginals of A are independent of B’s choice of measurement setting, and vice-versa. Thus, no instantaneous signalling takes place and hence, the name. Had signalling existed, it would have implied the strongest possible violation of the principle of relativity which imposes that superluminal transfer of information is not allowed in nature. 2 2
Let us denote the set of all no-signalling behaviors by NS. Then NS is an affine subspace of Rµ δ and dimNS = dimP − 2µ(µ − 1)(δ − 1) = (δ − 1)2 µ2 + 2(δ − 1)µ [59] . For the binary outcome case, when a, b ∈ {+1, −1}, dim NS = µ2 + 2µ, and therefore, µ2 + 2µ independent quantities are required to specify a no-signalling behavior. These quantities are known as correlators. The most commonly used correlators are: hAx i,hBy i and hAx By i. Since x and y each take values from 1 to µ, it immediately follows that there are µ2 + 2µ of them in number. Any joint probability can be expressed in terms of these correlators as follows: P (ab|xy) =
1 (1 + a hAx i + b hBy i + ab hAx By i) 4
(98)
Due to positivity and normalizability of the joint probabilities we must have: − 1 6 a hAx i + hBy i + hAx By i 6 3
(99)
Local Correlations
The set of all local behaviors is denoted by L. Any behavior is called local if f the joint probabilities can be written in the following form: Z P (ab|xy) =
dλ ξ(λ)P (a|x, λ)P (b|y, λ)
(100)
Λ
R where Λ dλ ξ(λ) = 1 and P (a|x, λ) and P (b|y, λ) are the local response functions. What does this signify? This means that the outcome of any local measurement performed by A (or B) is independent of both the measurement setting chosen by B (or A) as well as the outcome obtained by B (or A), provided the value of the variable λ is known. λ is often called shared randomness because it is the common causal influence behind the measurement statistics at either stations. NOTE: For δ = 1, or µ = 1, any behavior is local.
Quantum Correlations: Nature Simply Loved This!
A behavior is said to be quantum iff it can be written as: P (ab|xy) = T r(ρAB χax ⊗ χby )
(101)
21 where χax and χby are POVM [60] elements on Hilbert spaces HA and HB respectively, satisfying χax ≥ 0, χby ≥ 0 Pδ Pδ and a=1 χax = 1A , b=1 χby = 1B . Without loss of generality, we can also assume the orthogonality of the POVM 0 0 elements, i.e. χax χax = δa,a0 χax and χby χbx = δb,b0 χby . The set of quantum behaviors is denoted by Q. Physicists are more or less convinced that natural correlations are all quantum in nature. In the following section, we shall discuss the relation between NS, L and Q. It turns out the Q is a set intermediate between the other two sets.
The Interrelation: A Strict Inclusion
Lemma-1: Every local behavior satisfies no signalling condition. This follows from the definition of a local correlation: δ X
Z P (ab|xy) =
dλ ξ(λ)P (a|x, λ) Λ
b=1
δ X
P (b|y, λ)
b=1
Z =
dλ ξ(λ)P (a|x, λ) Λ
which is independent of y, and thus, consistent with the no-signalling condition. Similarly, the other condition can be justified. Lemma-2: Not all no-signalling behaviors are local. One simple way to show this to use the CHSH inequality and show that there exist no-signalling behaviors that violate the inequality. In fact, it so turns out the the no-signalling bound for this inequality is 4, which is the algebraic maximum! We can prove this by noting that the CHSH expression, under the assumption of no-signalling can be expressed in the following form: X hBi = 2 [P (a, a|x0 , y0 ) + P (a, −a|x0 , y1 ) + P (a, a|x1 , y0 ) + P (a, −a|x1 , y1 )] − 4 (102) a∈{+1,−1}
Now recall that since for the CHSH scenario, µ = δ = 2, there exist 4 independent joint probabilities in any no-signalling behavior. And in 102 we have exactly 4 joint probability functions. Consequently each of them can be independently regulated with the sole constraint that it takes values in the interval [0, 1]! Thus when all of them are 1, hBi=4. And it also follows from the same argument that any general N S behavior satisfies −4 6 hBi 6 4. Conclusion: L ⊂ NS. Lemma-3: Local behaviors admit quantum correlations. Given a local correlation, we can always mimic this using separable states. A local behavior characterP ized by P (ab|xy) = can be faithfully reproduced by a quantum behavior satisfying λ ξ(λ)P (a|x, λ)P (b|y, λ) P P (ab|xy) = T r(ρABP χax ⊗ χby ), if we choose ρAB = λ ξ(λ)ρλA ⊗ ρλB , T r(ρλA χax ) = P (a|x, λ) and T r(ρλB χby ) = P (b|y, λ). Here ξ(λ) > 0 and λ ξ(λ) = 1. Lemma-4: Not all quantum correlations are local. We have already seen this earlier. We know, for a fact, that quantum mechanics violates the CHSH inequality. Conclusion: L ⊂ Q.
22 Lemma-5: Quantum correlations respect the no-signalling constraints. δ X
P (ab|xy) = T r(ρAB χax ⊗
b=1
δ X
χby )
(103)
b=1
= T r(ρAB χax ⊗ 1B ) =
T r(ρA χax )
= P (a|x)
(104) (105) (106)
Lemma-6: There exist no-signalling behaviors which are not quantum. As already √ discussed earlier, no-signalling behavior can have a CHSH value of 4, whereas the quantum upper bound is 2 2. The Complete Relation: L ⊂ Q ⊂ NS. Thus, the local set is the most restrictive set and quantum behaviors are intermediate between the local and no-signalling sets.
Some properties of local, quantum and no-signalling correlations
Here we list some of the mathematical properties of L, Q and NS: 1. L, Q and NS are all closed, bounded and convex. Thus, if p~1 and p~2 both belong to any one of these sets, then µp~1 + (1 − µ)p~2 , with 0 ≤ µ ≤ 1 also belongs to the same set. 2. Since all of them are closed and convex, we can associate a hyperplane separation theorem with each of them. Every behavior ~p ∈ / L/Q/NS can be separated from the corresponding convex set L/Q/NS by a suitable hyperplane. P What’s more, if p~0 ∈ / K(L/Q/NS), there exists an inequality of the form ~s · ~p = a,b,x,y sa,b x,y p(ab|xy) ≤ SK that is satisfied by all ~p ∈ K but violated by p~0 . The CHSH inequality is one such inequality that distinguishes local behaviors from non-local ones. 3. L is the convex hull of a f inite number of points, and thus, it forms a polytope. The extremal points or vertices of the polytope are called deterministic behaviors since they satisfy: P (ab|xy, λ) ∈ {0, 1}∀a, b, x, y, λ. A behavior is local iff it can be written as a convex combination of deterministic behaviors. 4. For the local polytope, ~s · ~p ≤ SL . Then, F = {~p ∈ L : ~s · ~p = SL } is called a face of L. Faces of dimension dimF = dimL−1 are called facets of the polytope and the corresponding Bell inequalities as facet Bell inequalities or tight Bell inequalities. 5. The quantum set Q is not, in general, a polytope.
NONLOCALITY AND ENTANGLEMENT: HOW ARE THEY RELATED?
We have seen that entanglement is necessary to demonstrate non-local behavior. But even though one might expect a monotonic relation between nonlocality and entanglement, that is not the case. We have already encountered one such example while going through Hardy’s paradox. Further, we have seen that all separable states are local. However, we haven’t yet verified whether all entangled states are non-local.
23 Gisin’s Theorem and its generalization
Nicholas Gisin first came up with a proof that all pure entangled bi-qubit states violate Bell’s inequality. His proof is very simple. He considered a generic entangled state of the form |ψiAB = a |00i + b |11i,a, b 6= 0. It is a generic entangled state in the sense that it exhaustively represents all possible biqubit entangled states upto local unitary transformations. Local unitary transformations are transformations of the form UA ⊗ UB , where UA and UB are local unitaries defined on the subsystem Hilbert spaces. Local unitary transformations do not affect the amount of entanglement in a state. Moreover, under local unitary transformations, the maximum value of hBi remains the same, where the maximization is performed over all possible measurement settings. ˆ xˆ1 = sin γˆi + cosγ k, ˆ yˆ0 = sin βˆi + cosβ kˆ Choose the directions of local spin measurements as: xˆ0 = sin αˆi + cosαk, ˆ ˆ and yˆ1 = sin δ i + cosδ k. Then we have the result hxˆi · ~σ ⊗ yˆj · ~σ i = −2ab(sin α sin β + sin γ sin δ) for the state |ψiAB . We make it look even simpler by choosing α = 0, γ = −sgn(ab) π2 , where sgn(ab) is +1 if ab is positive and 1 −1 if ab is negative. Then it can be shown the maxβ,δ hBi = 2(1+4|ab|) 2 > 2, ∀a, b 6= 0, and Gisin’s theorem is proved. The above theorem has been generalized to higher dimensions as well, and even to the case of multipartite systems. Therefore, all pure entangled states are non-local, and in the territory of pure quantum states, there is an equivalence between Bell nonlocality and entanglement! In the following few sections, we shall examine if the equivalence continues to exist for mixed states as well. Horodecki-criterion for violating Bell’s Inequality
Note that any biqubit density operator can be written in the following form: ρAB =
3 X 1 tij σi ⊗ σj ] [1A ⊗ 1B + ~r · ~σ ⊗ 1B + 1A ⊗ ~s · ~σ + 4 i,j=1
(107)
Then hermiticity of ρAB entails that ~r,~s are real vectors and also that T = [tij ] is a real matrix. It is easy to see that tij = T r(ρAB σi ⊗ σj ). We now impose the condition that ρAB violates the CHSH inequality, i.e. |T r(ρAB B)| > 2. Taking B = xˆ0 · ~σ ⊗ (yˆ0 + yˆ1 ) · ~σ + xˆ1 · ~σ ⊗ (yˆ0 − yˆ1 ) · ~σ , we find: T r(ρAB B) = A˜0 T (B0 + B1 ) + A˜1 T (B0 − B1 ) xˆ01 xˆ11 yˆ0 1 yˆ1 1 where A0 = xˆ02 , A1 = xˆ12 , B0 = yˆ0 2 , B1 = yˆ1 2 are the matrices representing the unit vectors xˆ03 xˆ13 yˆ0 3 yˆ1 3 ˜ represents the transpose of X. Noting that (yˆ0 + yˆ1 ) ⊥ (yˆ0 − yˆ1 ) and that xˆ0 ,xˆ1 , yˆ0 and yˆ1 respectively, and X (yˆ0 + yˆ1 )2 + (yˆ0 + yˆ1 )2 = 4, we write (yˆ0 + yˆ1 ) = 2 cos θˆ z and (yˆ0 − yˆ1 ) = 2 sin θzˆ0 , where zˆ and zˆ0 are mutually orthogonal unit vectors. Then hBi becomes: hBi = 2 cos θA˜0 T Z + 2 sin θA˜1 T Z 0
(108)
where Z and Z 0 are the column matrices representing the vectors zˆ and zˆ0 respectively. Now, from Cauchy-Schwartz inequality, we have maxA0 ,A1 (A˜0 T Z) = kA0 kkT Zk = kT Zk, and maxA0 ,A1 (A˜1 T Z 0 ) = kA1 kkT Z 0 k = kT Z 0 k. Thus: maxA0 ,A1 hBi = 2 cos θkT Zk + 2 sin θkT Z 0 k q 2 2 =⇒ maxA0 ,A1 ,θ hBi = 2 kT Zk + kT Z 0 k
(109) (110) (111)
Next, we have to maximize over all possible mutually perpendicular unit vectors Z and Z 0 . By using the method 2 2 of Lagrange multipliers, one finds that maxZ,Z 0 (kT Zk + kT Z 0 k ) = λ1 + λ2 , where λ1 and λ2 are the two largest eigenvalues of T˜T . Consequently, we have: p (112) maxA0 ,A1 ,θ,Z,Z 0 hBi = 2 λ1 + λ2
24 Thus the condition that the state ρAB violates Bell inequality is: λ1 + λ2 > 1
(113)
If this condition is not fulfilled, Bell’s inequality is satisfied. Thus for a mixed biqubit state, we have derived the condition for which it is non-local. This criterion was first introduced by R. Horodecki et al in 1995. Comparing the Bell Nonlocality and Entanglement of the Werner Class: The Anomaly
Let us apply the Horodecki criterion on the Werner class of states, defined in . Then tij = T r(ρW σi ⊗ σj ) = pT r( ψ − ψ − σi ⊗ σj ) = −pδij
(114) (115) (116)
Thus T = −p13×3 =⇒ T˜T = p2 13×3 , which yields the following condition for violating Bell’s inequality: 2p2 > 1 1 =⇒ p > √ 2
(117) (118)
Thus, the Werner state violates Bell’s inequality if and only if p > √12 . But recall that while discussing about separability of the Werner state, we found out that it is entangled whenever p > 13 ! Thus, in the region 13 < p < √12 , the state is entangled but does not violate Bell’s inequality! In other words, we have finally found out that the relation between nonlocality and entanglement is not as simple for mixed states as it is for pure states! While all pure entangled states were found to be nonlocal, there exist mixed entangled states which are local(at least in the sense of Bell locality)! In fact, Werner himself, in 1989, showed that the Werner state admits a local model if p 6 21 , even though for p > 31 it is entangled. Entangled states supporting a local hidden variable model are called local states. Various other results have been obtained in relation to the locality of the Werner state in the region 31 < p < √12 . It was recently shown (in 2008) that the Werner state violates a particular type of Bell’s inequality whereby each party has 465 choices of measurement settings,if p ≥ 0.7056 < √12 . However, this ( 13 < p < √12 ) remains an extremely intriguing region and lots of problems are probed in relation to this region.
NONCOMMUTATIVITY:ANOTHER NON-CLASSICAL ASPECT OF QUANTUM THEORY
As we know, quantum mechanical measurements are associated with hermitian operators and the act of measurement reveals one of its eigenvalues as the outcome. The exact eigenvalue that turns up in a given measurement cannot be predicted with certainty before the measurement, unless the quantum state is an eigenstate of the observable being measured in which case the corresponding eigenvalue itself is revealed. This formalism allows for the existence of ˆ are said to be non-commuting iff [A, ˆ B] ˆ 6= 0. Even in a Hilbert space of operators that do not commute: Aˆ and B dimension as low as 2, almost all observables are non-commuting. The implications of noncommutativity are rather esoteric. Non-commuting observables do not have simultaneous eigenstates and there exists an uncertainty relation between them. This is another non-classical aspect of quantum theory since in classical physics, physical quantities have simultaneous existence and knowledge of one quantity does not preclude knowledge or predictability of another. Inapplicability of the Classical Measurement Rule: Von Neumann’s “Silly” Mistake
One important feature of classical measurements is that since all physical quantities have simultaneous existence, we can, in principle, measure them simultaneously with arbitrary accuracy. What’s more, if we wish to measure a physical quantity which is a function of two or more other physical quantities we can always do so by first measuring the latter and then evaluating the same function of these measured values. To elaborate it, let us assume that we have n [61] physical quantities A1 ,A2 , A3 ,...,An and we wish to measure the physical quantity f (A1 , A2 , A3 , ..., An ) where f denotes a
25 proper functional relation. Then, if measurements of A1 ,A2 ,A3 ,...,An yield the outcomes a1 ,a2 ,a3 ,...,an respectively, the classical measurement rule states that the value associated with the quantity f (A1 , A2 , A3 , ..., An ) is simply f (a1 , a2 , a3 , ..., an ). The rationale behind this rule is completely classical as it assigns well-defined realistic values to ˜ (Ai ), the measured value of Ai , the measured value of f (A1 , A2 , A3 , ..., An ) all physical quantities. If we denote, by M is given by: ˜ (f (A1 , A2 , A3 , ..., An )) = f (M ˜ (A1 ), M ˜ (A2 ), M ˜ (A3 ), ..., M ˜ (An )) M
(119)
However, there is no reason a priori to expect this rule to be applicable just because the theory is a realist one. So even though classical theory does not treat this as an independent physical postulate, one might as well elevate it to the status of a postulate. But what about quantum theory? Does this rule hold true in the quantum formalism as well? It turns out that in general, this is not applicable. Not only that, measuring a function of several observables has very little to do with the eigenvalues of all these observables; only the eigenvalues of the function of these observables are realizable and these do not have any connection [62] with the eigenvalues of the individual observables! But there’s a catch. In one particular scenario, the rule does go through - when all the involved observables are commuting! John Von Neumann, who was the first to come up with a confutation of the possibility of hidden variables in quantum mechanics, gave a very compelling and yet deceptive proof. His argument tacitly assumed the above measurement rule − something that is not faithfully obeyed in the quantum domain! Let us √ inspect the argument x + yˆ)], one that he presented: Consider the observable σx + σy . Since this is simply the operator 2[~σ · √12 (ˆ √ easily obtains the corresponding eigenvalues, which are ± 2.√ Now if we apply the above measurement rule, we expect that the sum of eigenvalues of σx and σy should be ± 2. Clearly that is not the case: the possible sum of their eigenvalues is either 0 or ±2, none of which coincides with any of the eigenvalues of σx + σy ! Von Neumann inferred from this observation that there cannot be any hidden variable in quantum mechanics, for, he believed that any hidden variable theory must satisfy the classical measurement rule. That is, given the hidden variable λ, ˜ (σx +σy ) = M ˜ (σx )+ M ˜ (σy ). there must be simultaneous, well-defined, realistic values of σx , σy and σx +σy satisfying M Clearly such a requirement of a hidden variable model is way too stringent and fundamentally erroneous, especially because it is well-acknowledged that quantum mechanics does not allow for the simultaneous existence of well-defined values corresponding to non-commuting observables! Thus, one should not even expect that a hidden variable model, if at all there exists one, should ascribe simultaneous realistic values to σx and σy . This is why Von Neumann’s assumption was dismissed by John Bell as a really silly one, when he finally came up with his own version of a no-go theorem. However, one could also argue that Bell’s own theorem essentially makes the same assumption (in addition to local realism) as invoked by Von Neumann in the above proof. One simple way to see this would require us to revisit the joint probability function used in the derivation of his inequality: Z P (ab|xy) = dλ ξ(λ)P (a|x, λ)P (b|y, λ), ∀a, b, x, y (120) Λρ
On close inspection, we can recognize that this is not strictly true. In general [Ax , Ax0 ] 6= 0, [By , By0 ] 6= 0 and thus, if we wish to respect the implications of noncommutativity we must choose disjoint ontic supports for noncommuting observables. What do we mean by ontic support here? If Λρ,xy be the ontic support for the composite observable Ax ⊗ By , then it basically means that the conditional joint probability of outcomes for a given ontic state λ ∈ Λρ , i.e. P (ab|xy, λ), is non-zero only for λ ∈ Λρ,xy and zero elsewhere (if defined!). Then, unless x = x0 and y = y 0 , we must have: Λρ,xy ∩ Λρ,x0 y0 = ∅ (121) R R Thus, taking Λρ,xy dλ ξ(λ)P (a|x, λ)P (b|y, λ) = Λρ dλ ξ(λ)P (a|x, λ)P (b|y, λ) alludes to the undeclared assumption that P (ab|xy, λ) is defined even outside Λρ,xy and therefore, zero in the region λcρ,xy , where λcρ,xy = Λρ − Λρ,xy . However, if for some λ0 ∈ Λcρ,xy , P (ab|xy, λ0 ) = 0, ∀a, b ∈ {+1, −1}, it means the for the pair of qubits in the ontic state λ0 , if Alice and Bob choose the settings x and y respectively, neither +1 or −1 outcome turns up for either of the local measurements! This is indubitably unfeasible - in any quantum state, whatever be the choice of input settings, the local outcomes will always assume values in the set {+1, −1}, at least forR an ideal set-up! Thus, we are forced into the conclusion that P (ab|xy, λ) is undefined for λ ∈ Λcρ,xy . The expression Λρ,xy dλ ξ(λ)P (a|x, λ)P (b|y, λ) R cannot be treated as being identical to Λρ dλ ξ(λ)P (a|x, λ)P (b|y, λ) (we cannot extend the integral beyond
26 the relevant ontic support), and unless we make this crucial assumption, we won’t arrive at the CHSH inequality! And since there exist so many noncommuting observables with disjoint supports, it is clear that Λρ,xy ⊂ Λρ , ∀x, y. Thus, even though we disregard the relevance of this assumption, we are fundamentally invoking it in addition to the usual more transparently expressed premises of realism and locality. And however “silly” might Von Neumann’s assumption have been, Bell’s derivation does not quite accomplish in getting around it either.
Bell’s Inequality Violation as an indication of Noncommutativity
As we have seen earlier, the derivation of Bell’s inequality does not require a particular premise only. It can be derived from various premises - local realism, existence of a quadrivariate joint probability distribution, the classical measurement rule discussed earlier, and many more. We shall see now that even the assumption of local commutativity, i.e. commutativity of Alice’s(or Bob’s) local observables, can lead to Bell’s inequality.
The Tsirelson Operator and its “Quantumness”
Recall the Tsirelson Operator defined in 70 : B 2 = 41A ⊗ 1B − [A0 , A1 ] ⊗ [B0 , B1 ]
(122)
We claim that the Tsirelson operator owes its “quantumness” to the presence of the two commutators [A0 , A1 ] and [B0 , B1 ]. What does this statement mean? Moreover, what do we mean by “quantumness” here? Note that, if the term involving the commutators was not present, we would have had: B 2 = 41AB
(123)
Any operator proportional to the Identity trivially has every state as its eigenstate. And even for an arbitrary mixed state, it returns, upon measurement, the same value equal to the proportionality constant, with certainty. A measurement of the Tsirelson operator, thus, would have produced an outcome of 4 for any state ρAB . Now consider the classical analogue of the Tsirelson operator: B 2 = [a0 (b0 + b1 ) + a1 (b0 − b1 )]2
(124)
= 2(1 + bo b1 ) + 2(1 − b0 b1 )
(125)
=4
(126)
Here we have used the fact that a2i = b2i = 1, ∀i ∈ {0, 1}. Thus, had there not been the term containing the commutators in the quantum Tsirelson operator, it would have exactly resembled its classical analogue. It is in this sense that we have made the claim that the Tsirelson operator derives its quantumness or non-classicality from the presence of the two commutators. This is not surprising because in classical physics, all physical quantities commute and therefore [a0 , a1 ] = [b0 , b1 ] = 0. Thus the generally non-zero contribution from the term [A0 , A1 ]⊗[B0 , B1 ] signifies a departure from classicality. However, it will also be interesting to know if it signifies a departure from local realism. If we look at it from the point of view of the CHSH inequality, it is indeed true that this additional√contribution is what precisely creates the difference between the local realistic bound (2) and the quantum bound (2 2). In absence of that term, we would have found that the eigenvalues of B are precisely ±2, as predicted by classical theory, and thus −2 6 hBi 6 2. So then, it provokes the question: is it possible to construct, theoretically, a local realistic model for the case when at least one of these commutators vanish?
Derivation of Bell’s Inequality from the assumption of Commutativity
The premise that Alice and Bob’s local observables commute seems, just like the classical measurement rule, questionable in the first place. Why should we fuss over a condition that is anyway not allowed in quantum theory? This is because we wish to have a clear picture of the role of noncommutativity in quantum mechanics and also, we
27 would like to have as complete an idea about the ramifications of Bell’s theorem in order to understand how we ought to modify our classical-like worldview. Another pertinent reason for pursuing this is the ambition to be in a better position to answer the question: Which quantum mechanical features are “responsible” for non-classicality and how? We shall consider different cases in this context for deriving Bell’s inequality: Case-1: Both [A0 , A1 ] and [B0 , B1 ] are zero. Since A0 commutes with A1 , the two operators are furnished with simultaneous eigenstates and therefore, simultaneous eigenvalues. Stated in another way, there exists an orthonormal basis in which both the observables are diagonal. Let us denote this basis as {|αii }, i ∈ {1, 2}. Similarly, since B0 commutes with B1 , there exists an orthonormal basis {|βii }, i ∈ {1, 2} in which both B0 and B1 are diagonal. Then it is easy to check that {|αii ⊗ |βij , i, j ∈ {1, 2}} forms a simultaneous eigenbasis of A0 ⊗ B0 , A0 ⊗ B1 , A1 ⊗ B0 and A1 ⊗ B1 . And it immediately follows that the former also forms an eigenbasis of B = A0 ⊗ (B0 − B1 ) + A1 ⊗ (B0 − B1 ). Now, from our assumption, A0 , A1 , B0 and B1 have simultaneous eigenstates, and therefore simultaneous eigenvalues associated with these eigenstates. So the eigenvalues of B are of the form a0 (b0 + b1 ) + a1 (b0 − b1 ), where ai , bj ∈ {+1, −1}, ∀i, j ∈ {0, 1}. And once again, it is simple enough to check that the eigenvalues, then, are ±2, which ensures |hBi| 6 2. Thus simultaneous commutativity of Alice and Bob’s local observables ensure that Bell inequalities hold. Case-2 Only one of [A0 , A1 ] and [B0 , B1 ] is zero. This is not as obvious a situation and there is no reason to expect that this should also satisfy Bell’s inequality. [63] Without loss of generality, we assume that [A0 , A1 ] 6= 0 and [B0 , B1 ] = 0. Thus there exists a basis |b0 , b1 i in which B0 and B1 are diagonal satisfying: B0 |b0 , b1 i = b0 |b0 , b1 i
(127)
B1 |b0 , b1 i = b1 |b0 , b1 i
(128)
where b0 , b1 ∈ {+1, −1}. So we can have two completely commuting sets of observables {A0 , B0 , B1 } and {A1 , B0 , B1 } with respective eigenbases {|ai0 |b0 , b1 i} and {|ai1 |b0 , b1 i} satisfying [64] : Ai |aii |b0 , b1 i = ai |aii |b0 , b1 i
(129)
Bj |aii |b0 , b1 i = bj |aii |b0 , b1 i
(130)
where i, j ∈ {0, 1}. Then we can expand an arbitrary pure state |ψi in either of the above bases: X X |ψi = µa,b0 ,b1 |ai0 |b0 , b1 i = νa,b0 ,b1 |ai1 |b0 , b1 i a,b0 ,b1
We now compute hBi. Note that, by choosing our basis appropriately, we have X X 2 2 |hA0 (B0 + B1 )i| = |µa,b0 ,b1 | a(b0 + b1 ) ≤ |µa,bo ,b1 | |b0 + b1 | a,b0 ,b1 a,b0 ,b1 X X 2 2 |hA1 (B0 − B1 )i| = |νa,b0 ,b1 | a(b0 − b1 ) ≤ |νa,bo ,b1 | |b0 − b1 |. a,b0 ,b1 a,b0 ,b1 Combining these two inequalities, we find: X X 2 2 |hBi| ≤ |µa,bo ,b1 | |b0 + b1 | + |νa,bo ,b1 | |b0 − b1 | a,b0 ,b1
(131)
a,b0 ,b1
(132)
(133)
(134)
a,b0 ,b1
(135)
28 Now, since one of |b0 + b1 | and |b0 − b1 | is zero and the other 2, we can rewrite the inequality as: X X 2 2 |νa,bo ,−b0 | |µa,bo ,b0 | + 2 |hBi| ≤ 2
(136)
a,b0
a,b0
= 2[p(b0 = b1 ) + p(b0 = −b1 )]
(137)
=2
(138)
Thus, even in this relaxed case which is, nevertheless, manifestly non-classical [65] , the Bell’s inequality holds. Thus, it is not enough to have one pair of non-commuting local observables in order to engender a violation of Bell’s inequality. It is absolutely necessary that both pairs of local observables are non-commuting!
A local realistic model
We shall follow the presentation of Robert Griffiths to show that the assumption of local commutativity ensures a local realistic model. We know, any quantum correlation can be expressed as: P (ab|xy) = T r(ρAB χax ⊗ χby )
(139)
where the symbols have meanings defined earlier. We now consider the familiar case of a biqubit system with Alice and Bob as the two participants in the experiment. Now assume that all local projectors of Alice commute [66] ,i.e. 0 [χax , χax0 ] = 0, ∀x, x0 , a, a0 . Then it is possible to construct a local model satisfying: X P (ab|xy) = ξ(λ)T rA (ρλA χax )T rB (ρλB χby ) (140) λ 0
a 0 0 a where λ ξ(λ) = 1. The question is: how? Well, since [χx , χx0 ] = 0, ∀x, x , a, a , there exists an orthonormal a basis |λi in which all projectors χx are diagonal. Then if we choose ξ(λ) = T r(ρAB |λi hλ| ⊗ 1B ), ρλA = |λi hλ|, T rA (ρAB |λi hλ| ⊗ 1B ) , we find that the locality decomposition rule is satisfied! To prove this, let us assume ρλB = ξ(λ) a that the projectors satisfy χx |λi = δaλ |λi , ∀x, a. Then equation 139 yields:
P
T r(ρAB χax ⊗ χby ) = T rB [T rA (ρAB χax ⊗ χby )] X = T rB [ hλ| ρAB χax ⊗ χby |λi]
(141) (142)
λ
= T rB [ha| ρAB |ai χby ]
(143)
whereas, with the choices that we have made, equation 140 becomes: X
ξ(λ)T rA (ρλA χax )T rB (ρλB χby ) =
λ
X
T rA (|λi hλ| χax )T rB [T rA (ρAB |λi hλ| ⊗ 1B )χby ]
(144)
λ
= T rB [ha| ⊗ 1B ρAB |ai ⊗ χby ] =
T rB [ha| ρAB |ai χby ]
(145) (146)
which is the same as 143 ! Thus, a local model underpinning the assumption of local commutativity of Alice’s operators has been constructed leading to the choices mentioned above. It is also easy to see that this model works even for the case where there is only one possible measurement setting for at least one of the two parties. In fact, the assumption of all local projectors of Alice commuting is equivalent to that of having effectively one measurement setting at her station.
29 EPILOGUE
In this report, we have identified three strongly non-classical features of quantum mechanics. These are interesting aspects in themselves and extensive research is carried out on each of these topics, both from a foundational perspective as well as an information-theoretic perspective. These topics are central to the foundations of quantum theory and even after quite a few decades of research, they continue to be conceptually obscure notions. Brief Summary
A gist of all that have been described in this report is listed below: 1. Entanglement is a highly intriguing feature of quantum theory and no analogue exists in the classical realm. While in classical physics, complete specification of the state of a composite system necessitates the specification of the states of the constituent subsystems, such a feature seems to be missing in quantum theory! Quantum mechanics allows for pure composite states for which the subsystems are not necessarily in pure states. Entanglement arises naturally in quantum theory as a consequence of the tensor product structure and the linear superposition principle. 2. Pure bipartite entangled states can be easily detected. A Schmidt number of at least 2, corresponding to a pure state, testifies to the presence of entanglement. Alternatively, the purity of any of the subsystems can also be checked and connected to the question of separability of the pure bipartite state. 3. Mixed entangled states cannot be so easily detected. However, in low dimensional systems, there exist some simple criteria, such as the Reduction criterion, PPT criterion, etc. which are both necessary and sufficient for separability. The method of entanglement witness also works extremely well, provided we can come up with an optimized witness operator. This method works faithfully on Hilbert spaces of all finite dimensions. 4. The imposition of realism coupled with locality leads to the Bell’s (CHSH) inequality. Quantum mechanics does not respect this inequality, demonstrating its incompatibility with any local realistic hidden variables, unlike classical statistical mechanics. However, all separable states satisfy the inequality. Entanglement turns out to be a necessary condition for the violation of the inequality. 5. The maximum violation of the inequality can be obtained by evaluating the maximum eigenvalue of the Bell √ operator. While the local realistic bound is 2, the maximum eigenvalue of the Bell operator turns out to be 2 2. Also, the maximum violation takes place when both the local observables of Alice and Bob are anti-commuting. 6. The Bell operator is a non-factorizable operator that cannot be expressed as a product of two operators in the subsystem Hilbert spaces. Furthermore, it does not have a product eigenbasis. 7. Hardy’s conspicuous illustration of nonlocality does not involve any inequality and provides an example of a more direct conflict of quantum theory with local realism. 8. Physicists usually identify three main types of correlations - No-Signalling, Local and Quantum. Out of these, Local correlations are the most restrictive and No-signalling the most relaxed. Correlations in quantum theory are intermediate between the other two. There is ample proof that correlations in nature are decidely quantum. 9. The relation between Bell nonlocality and entanglement is not a monotonic one. In fact, there exist entangled states which abide by the CHSH inequality. For example, the Werner state, for a particular range of the parameter p, is entangled but does not violate the CHSH inequality. 10. In the quantum mechanical formalism, most observables pertaining to a single system are non-commuting. Noncommutativity makes the classical measurement rule generally inapplicable. 11. If at least one pair of local observables in the CHSH experiment happens to commute, there is no possibility of a violation of Bell’s inequality. In fact, a local realistic model can always be constructed in such a case. To have the possibility of violation of Bell’s inequality, both Alice and Bob’s local observables must be non-commuting.
30 Possible directions for further research from the foundational perspective
Finally, we attempt to motivate some problems which could be meticulously investigated, with the anticipation that probing these might lead to novel insights in quantum theory. 1. In our report, whenever we have talked of locality we implied locality at the realist level. In fact, when we say quantum mechanics is non-local we mean that it that does not comply with “local realism”. But whether quantum mechanics violates locality or realism, or both, still remains a matter of debate. Many inequalities have been derived in the spirit of Bell inequalities but none of them have been able to single out which one we ought to relinquish. One major complication associated with such no-go theorems is that all of them appear to be constrained to make multiple assumptions - a single assumption usually gets us nowhere. For example, the Leggett-Garg inequality assumes, in addition to realism, non-invasive measurability - something which is strictly enjoined by quantum mechanics. Thus the problem of coming up with an inequality or any mathematical condition violated by quantum mechanics but based on a single premise (either locality or realism) [67] still remains open. [68] 2. We have seen that both locality [69] and commutativity leads to Bell’s inequality. Thus any violation of Bell’s inequality implies both local realism as well as commutativity. But are these premises equivalent? As in, does local realism and commutativity go hand in hand? And can the absence of local realism in quantum mechanics be attributed to the noncommutativity in its formalism? Moreover, can we relate, quantitatively, locality and the “degree” of commutativity? 3. The exact relation between entanglement and nonlocality is still not known. While separability is a sufficient condition for locality, it is not necessary. For, we can have entangled states which admit local models. This is a surprising feature observed for some mixed entangled states. And yet we know that all entangled states exhibit correlations which cannot be simulated classically. One indication that this provides is that classical correlations form a strict subset of local correlations, and thus even more restrictive. That is, while classicality implies locality, the converse is not true. Local realism is not a “singular” attribute of classical correlations. This provokes the question: what are the classical features not implied by local realism? For example, just like certain mixed entangled states admit LHVM, which non-classical features are compatible with LHVM? [Recall that we encountered another such example in the penultimate section where we showed that the assumption of commuting local projectors for just one of the parties can give rise to a local model.] 4. The definition of separable states is a tricky one. Even if we have a statistical mixture of entangled states we cannot guarantee entanglement. For instance, we found out that an equal mixture of the Bell states is a separable state. What light does this throw on the density operator and its interpretation? What do mixtures really signify in the quantum domain as opposed to classical statistical mixtures? 5. We have seen that even in the derivation Bell’s theorem, we do require the existence of overlapping ontic supports for all noncommuting observables. Somehow the assumption of commutativity seems to be a necessary ingredient for conjuring a no-go theorem. So, if we might dare to be ambitious, is noncommutativity the strongest nonclassical feature observed in quantum mechanics, which possibly acts as a cover for everything non-classical about quantum theory? The list of open problems related to entanglement, nonlocality and noncommutativity is pretty much myriad. We can only hope that in the times to come, we will successfully untangle more and more of these puzzles and will be truly able to “make peace” with the quantum world.
Acknowledgements
I am extremely grateful to my thesis supervisor Prof. Archan S. Majumdar for his constructive tutelage and for having provided me complete flexibility in exploring my interests. This had rendered me the license for learning and cultivating the topic of my choice, to the best of my convenience. Besides, I am also indebted to all the professors, including Prof. Majumdar, for the educational lectures that they had delivered on “Quantum Foundations; Correlations, Information Processing and Applications” at the summer school organized at Indian Statistical Institute,
31 Kolkata, which I was privileged to have attended, from May-22 to July-3. Much of what I have gathered in the past two months, in relation to these topics, is due to my participation in this school.
∗
Electronic address:
[email protected] [1] Michael A. Nielsen, Isaac L. Chuang, Quantum Computation and Quantum Information. Cambridge University Press, 2000. [2] John S. Bell, Physics 1 195, 1964. [3] J.F. Clauser, M.A. Horne, A. Shimony, R.A. Holt, Phys. Rev. Lett. 23 (15): 8804, 1969. [4] B.S Tsirelson, Lett. Math. Phys. 4 93,1980. [5] Lawrence J. Landau, doi:10.1016/0375-9601(87)90075-2,1987. [6] A.Fine, Phys. Rev. Lett. 48, 291.,1982. [7] M Revzen, M Lokajcek and A Mannz, Quantum Semiclass. Opt. 9, 1996. [8] Robert Griffiths, arXiv:1304.4425, 2013. [9] Tim Maudlin, Quantum Non-Locality and Relativity, Volume 3d ed. Wiley-Blackwell, New York,2011. [10] S. L. Braunstein, A. Mann, and M. Revzen, Phys. Rev. Lett. 68, 3259, 1992. [11] James D. Malley, Phys. Rev. A 69, 022118, 2004. [12] Holger F. Hofmann and Shigeki Takeuchi, Phys. Rev. A 68, 032103, 2003. [13] James D. Malley and A. Fine, doi:10.1016/j.physleta.2005.06.032, 2005. [14] Nicolas Brunner, Daniel Cavalcanti, Stefano Pironio, Valerio Scarani and Stephanie Wehner, Rev. Mod. Phys. 86, 419, 2014. [15] M. Horodecki, P. Horodecki, R. Horodecki, doi:10.1016/S0375-9601(96)00706-2,1996. [16] Asher Peres, Phys. Rev. Lett. 77, 1413, 1996. [17] Dagmur Bruss, arXiv:quant-ph/0110078, 2002. [18] M. Horodecki, P.Horodecki, arXiv:quant-ph/9708015, 1998. [19] R. Horodecki, P. Horodecki, M. Horodecki, K. Horodecki, Rev. Mod. Phys. 81, 865, 2009. [20] Reinhard. F. Werner, Phys. Rev. A 40, 4277, 1989. [21] A. Acin, R. Gill, N. Gisin, arXiv:quant-ph/0506225, 2005. [22] N. Gisin, doi:10.1016/0375-9601(91)90805-I, 1991. [23] R. Horodecki, P. Horodecki, M. Horodecki, doi:10.1016/0375-9601(95)00214-N , 1995. [24] such as, a given value of energy [25] Following this, whenever we express a state, generic or specific, as a superposition of some basis states, we shall implicitly assume the normalization constraint. [26] Henceforth, I shall work with bi-qubit systems only, unless otherwise explicitly mentioned. [27] This can be derived from the Singular V alue Decomposition of linear algebra. [28] Note that the computational basis states are product states whereas the Bell states are entangled states - in fact they are maximally entangled. [29] and we know that the state is a product state when p=0. [30] However, it should be noted that for two arbitrary states |ψi and |φi, (|ψi hφ|)T = |φ∗ i hψ ∗ | . [31] such states are also called Bound Entangled States (though the definition of bound entanglement is more general and these are not the only bound entangled states). [32] A Hilbert space is necessarily also a Banach space. [33] non-zero [34] am an mnr σr is antisymmetric in m,n. [35] that these form a non-commuting set can be easily checked and follows from the fact that {Ai }’s and {Bi }’s are both non-commuting sets. [36] and, in some sense, that it is maximally entangled. [37] We shall restrict ourselves with the case of bi-qubit systems only and not talk about the general d × d dimensional Werner states. [38] or a pseudo-mixture if p ≤ 0. [39] This does not mean that there can, in principle, be only two possible measurement settings. It is only in the context of the CHSH inequality that we need two settings per party. [40] Other terms like local causality and Bell locality are also used. [41] More about signalling will be discussed later. [42] Here we are taking λ to be a continuous variable. One might as well work with a discrete set. [43] the commutator of two hermitian operators is an anti-hermitian operator. [44] The exact values of the coefficients can be determined, in principle, but since that would be cumbersome, we shall not take the plunge. [45] One might also assume that α is zero - a similar kind of contradiction would appear in that case. [46] + when i = 0 and − when i = 1. [47] same eigenvalue implies that the i-index is unaltered.
32 [48] [49] [50] [51] [52] [53]
[54] [55] [56] [57] [58] [59] [60] [61] [62] [63] [64] [65] [66] [67] [68] [69]
the measurement statistics at the level of quantum mechanics even though it leaves open the possibility of further probing the discrepancy in the two bounds, we haven’t yet answered the question whether any entangled state would suffice or not. Such correlations are often called nonlocal correlations. We shall see that later. Clearly the constraint on factorizability is not enough by itself, for we can always add a scalar times the Identity operator to any factorizable operator and we can still have a product eigenbasis. The Tsirelson operator B2 is an example of such an operator which is non-factorizable but has a product eigenbasis. we have used first bracket so as to ward off any possibleconfusion with the anti-commutator. in the sense that there exists a joint quadri-variate probability distribution returning the experimental probabilities as its marginals. His paper actually justifies the mutual equivalence between a whole set of premises in the context of Bell’s theorem. since these have no existence in the classical domain. there are µ2 constraints due to normalization. each of the sets of equations defining the no-signalling condition implies µ(µ − 1)(δ − 1) independent constraints. Positive-Operator Valued Measure where n could be an arbitrarily large natural number. at least no trivial connection. One could, of course, argue by considering the form of the Tsirelson operator. Here we have dropped the tensor product notation for convenience. because one pair of local observables are non-commuting. no doubt it is an absurd assumption. if at all that is possible It may be stated in this context that a section of physicists have argued in favor of locality and ruled out realism. local realism
