Maximally entangled state and Bell’s inequality in qubits

Maximally entangled state and Bell’s inequality in qubits

Annals of Physics 395 (2018) 183–195 Contents lists available at ScienceDirect Annals of Physics journal homepage: www.elsevier.com/locate/aop Maxi...

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Annals of Physics 395 (2018) 183–195

Contents lists available at ScienceDirect

Annals of Physics journal homepage: www.elsevier.com/locate/aop

Maximally entangled state and Bell’s inequality in qubits Su-Kuan Chu a , Chen-Te Ma b, *, Rong-Xin Miao c , Chih-Hung Wu b a

Joint Quantum Institute and Joint Center for Quantum Information and Computer Science, NIST/University of Maryland, College Park, MD 20742, USA b Department of Physics and Center for Theoretical Sciences, National Taiwan University, Taipei 10617, Taiwan, ROC c National Center for Theoretical Sciences, National Tsing-Hua University, Hsinchu 30013, Taiwan, ROC

article

info

Article history: Received 14 December 2017 Accepted 29 May 2018 Available online 4 June 2018 Keywords: Maximally entangled state Topological states of matter Topological field theories Bell’s inequality

a b s t r a c t A maximally entangled state is a quantum state which has maximum von Neumann entropy for each bipartition. Through proposing a new method to classify quantum states by using concurrences of pure states of a region, one can apply Bell’s inequality to study intensity of quantum entanglement of maximally entangled states. We use a class of seven-qubit quantum states to demonstrate the method, in which we express all coefficients of the quantum states in terms of concurrences of pure states of a region. When a critical point of an upper bound of Bell’s inequality occurs in our quantum states, one of the quantum state is a ground state of the toric code model on a disk manifold. Our result also implies that the maximally entangled states do not suggest local maximum quantum entanglement in our quantum states. © 2018 Elsevier Inc. All rights reserved.

1. Introduction Quantum entanglement is a physical phenomenon for that a quantum state of each particle cannot be described independently. When a quantum state in a quantum mechanical system has maximum von Neumann entropy for each bipartition, the entanglement entropy of a region A, SA ≡ −TrA ρA ln ρA ,

*

Corresponding author. E-mail addresses: [email protected] (S.-K. Chu), [email protected] (C.-T. Ma), [email protected] (R.-X. Miao), [email protected] (C.-H. Wu). https://doi.org/10.1016/j.aop.2018.05.016 0003-4916/© 2018 Elsevier Inc. All rights reserved.

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is equivalent to the Rényi entropy, Sα,A ≡ ln TrA ρAα /(1 −α ), of each order α , where the reduced density matrix of the region A satisfies ρA ≡ TrB ρ . (TrA is a partial trace operation on the region A and TrB is the same operation on the region B.) We call such state having only one effective entanglement quantity – entanglement entropy – a maximally entangled state. A maximally entangled state is useful for applications of quantum computing [1] and quantum algorithm [2] as well as topological entanglement entropy in a toric code model [3], which was already realized in [4] by using a geometric algebra procedure [5] or a combination of two-body interactions and radio-frequency pulses [5,6]. Topological entanglement entropy is a universal term of entanglement entropy in a toric code model [7]. The toric code model also has a suitable tensor product decomposition of a region to extract topological entanglement entropy [8]. When a region is not contractible, the topological entanglement entropy should be lowered, but the toric model in a disk manifold with holes does not [9]. When quantum entanglement occurs, it distinguishes a quantum system from a classical system. The quantum nature of entanglement can be manifested [10] through violation of Bell’s inequality [11], which states that correlations should satisfy the Bell’s inequality when local realism is realized. The local realism is that information transmission cannot be faster than the speed of light and that there should be a pre-existing value before any possible measurement. The violation of the Bell’s inequality is measured in a two-qubit system and is also theoretically studied in [12]. The maximum violation of Bell’s inequality in a two-qubit system is shown to be a √monotonic function with respect to the concurrence of a pure state of a region A [13], CA (ψ ) ≡ 2(1 − TrA ρA2 ), and the entanglement entropy of the region A, SA , through the two maximal eigenvalues of the 3 × 3 R-matrix Rij ≡ Tr(ρσi ⊗ σj ) [14,15], where i = 1, 2, 3 and j = 1, 2, 3, and σx , σy , σz are the Pauli matrices defined in [16]. For some n-qubit quantum states, an upper bound of Bell’s inequality is also a monotonic function with respect to the generalized concurrence of the pure state √ of a region A [9,17], CA (m, ψ ) ≡

2(1 − 2m−1 TrA ρA2 ), and the entanglement entropy of the region A

through the generalized R-matrix [9,17], Ri1 i2 ···in ≡ Tr(ρσi1 ⊗ σi2 ⊗ · · · ⊗ σin ) ≡ RIin , where iα = x, y, z and α = 1, 2, . . . , n are the site indices. The generalized R-matrix can be rewritten as a 3n−1 × 3 matrix RIin with the first index being a multi-index I = i1 i2 · · · in−1 and the second index being in . The n-qubit Bell’s operator Bn [18] is

( Bn ≡ A1 ⊗ A2 · · · ⊗ An−2 ⊗ An−1 ⊗



)

An + An

( ) + A′1 ⊗ A′2 · · · ⊗ A′n−2 ⊗ A′n−1 ⊗ An − A′n ,

(1)

where Ai ≡ ai · σ and A′i ≡ a′i · σ are the operators in the ith qubit with an and a′n being unit vectors and σ ≡ (σx , σy , σz ) being the vector of Pauli matrices. Therefore, the n-qubit Bell’s operator offers a way to qualitatively study entanglement entropy. In this letter, we discuss maximally entangled states in an n-qubit systems by using an upper bound of maximum violation of Bell’s inequality. Quantum entanglement indicates whether a quantum state of each particle can be factorized from the state of a total system. Consequently, quantum entanglement is a fact about a full density matrix rather than a reduced density matrix. One special property of quantum entanglement is that information of the whole system can be partially extracted by observing one single particle of the quantum state. The higher the intensity of quantum entanglement is, the stronger the correlation the system has. The upper bound of Bell’s inequality provides a way to quantify this concept. Thus, we use an upper bound of Bell’s inequality to discuss the intensity of quantum entanglement in an n-qubit maximally entangled state. Different maximally entangled states should have different intensity of quantum entanglement. One might also expect that in the Hilbert space the intensity of quantum entanglement of a maximally entangled state is at least a local maximum. We discover that this is a saddle point in our case. 2. Maximally entangled state and Bell’s inequality We use seven-qubit quantum states as an example to estimate intensity of quantum entanglement in the maximally entangled state through Bell’s inequality [11,18]. A ground state of a toric code

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model [3,4] on a disk manifold can be shown be occurred at a critical point of quantum entanglement in our quantum states through an upper bound of Bell’s inequality [9,14,17]. We also discuss relations between an upper bound of Bell’s inequality and concurrences of pure states of a region [13]. We choose the seven-qubit quantum states:

|ψ⟩7 ≡ α1 |000000⟩B |0⟩A + α2 |000111⟩B |1⟩A + α3 |111100⟩B |0⟩A + α4 |111011⟩B |1⟩A , 1 = α12 + α22 + α32 + α42 .

(2)

As indicated above, the system is separated into two regions A and B, where the region B contains six qubits and the region A contains one qubit. An upper bound of the Bell’s inequality can be obtained by expressing α12 , α22 , α32 and α42 in terms of concurrences of pure states of the region A [13] (see Appendix). Firstly, we can write α1 = sin(θ1 ), α2 = cos(θ1 ) sin(θ2 ), α3 = cos(θ1 ) cos(θ2 ) cos(θ3 ), α4 = cos(θ1 ) cos(θ2 ) sin(θ3 ), where 0 ≤ θ1 ≤ 2π , 0 ≤ θ2 , θ3 , θ4 ≤ π . When we crank up θ2 = π/2 and turn down θ3 = 0. Now we obtain new coefficients corresponding to the four product states β1 = sin(θ1 ), β2 = cos(θ1 ), β3 = β4 = 0. Consequently, we get TrA ρA2 ≡ TrA ρ12,A = β14 + β24 . We have

√ (



the freedom to pick β12 = 1/2 − 1 − C12,A (ψ )/2, where C1,A (ψ ) ≡ 2 1 − TrA ρ12,A . Thirdly, we still choose θ3 = 0 but restore the value of θ2 . We have another new coefficients 2 2 2 2 2 4 γ1 = sin(θ1 ), γ2 = cos(θ1 ) sin(θ2 ), γ( 3 = cos(θ1 ) cos(θ2 ), γ4 = 0 and Tr A ρA ≡ TrA ρ2,A = (γ1 +γ) 3 ) +γ2 . )/(



Hence, one solution is cos2 (θ2 ) =

√ (

1 − C12,A (ψ ) +



1 − C22,A (ψ )

)

1+



1 − C12,A (ψ ) , where

C2,A (ψ ) ≡ 2 1 − TrA ρ22,A . Finally, we reinstate both θ2 and θ3 in an attempt to express θ3 in terms of concurrences of pure states of the region A. The purity of the region A, TrA ρA2 , is TrA ρA2 = (α12 + α32 )2 + (α22 + α42 )2 . Therefore, one solution is

)

cos2 (θ3 ) √



1−C12,A (ψ ) 1−C22,A (ψ )+1−C12,A (ψ )



=



1−C12,A (ψ )+ 1−C22,A (ψ )



1 − C12,A (ψ ) +



+



1 − CA2 (ψ )

1 − C22,A (ψ )

.

(3)

In so doing, we find that the quantum states can be classified by C12,A (ψ ), which is determined by β12 , C22,A (ψ ), which is determined by γ12 and γ22 , and CA2 (ψ ), which is determined by α12 , α22 and α32 . Here we do not give the most generic solution, but the most generic solution should be exactly solvable. We will demonstrate the upper bound of the Bell’s inequality for the seven-qubit quantum states and show that a critical point of the upper bound of the Bell’s inequality also corresponds to a ground state of the seven-qubit toric code model on a disk manifold since it is included in our solution. To compute the upper bound of the Bell’s inequality, which is defined by two largest eigenvalues of R† R, we introduce the lemma [9,15,17]:



Lemma 1. The maximum violation of the Bell’s inequalities γ ≡ maxBn Tr(ρ Bn ) ≤ 2 u21 + u22 , where u21



and u22 are the first two largest eigenvalues of R† R when n > 2 and γ = 2 u21 + u22 when n = 2. By invoking the states |ψ⟩7 , we can show that R† R only has diagonal( elements ( seven-qubit ) ( quantum ) ) (see Appendix): R† R xx = R† R yy = 16(α12 + α32 )(α22 + α42 ) + 48(α12 α42 + α22 α32 ) and R† R zz = 1 + 32α12 α32 + 32α22 α42 , and that a critical point of the upper bound of the Bell’s inequality occurs at

α12 = α22 = α32 = α42 = 1/4 (see Appendix). For our solution, the critical point of the Bell’s inequality corresponds to C12,A (ψ ) = C22,A (ψ ) = 3/4 and CA2 (ψ ) = 1, which also corresponds to maximally entangled states. The upper bound of the maximum violation of the Bell’s inequality turns out to √ be 4 5. The critical point also shows that the quantum states are not locally maximally entangled because the critical point of the upper bound of the Bell’s inequality is a saddle point.

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Fig. 1. We fix C12,A = 0.75 and choose C22,A = 0.75, 0.8, 0.85, 0.9, 0.95 to compute the upper bound of maximum violation of the Bell’s inequality. It is interesting to note that the upper bound of maximum violation of the Bell’s inequality is independent of C22,A when CA2 = 1.

If we consider C12,A (ψ ) = C22,A (ψ ) = CA2 (ψ ) = 1, the coefficients are α12 = α22 = 1/2 and

√ √ α = α42 = 0. The upper bound of the Bell’s inequality is 4 2 < 4 5, which can be shown from 2 3

the following theorem [9,14,17]:

Theorem 1. For an n-qubit quantum state |ψ⟩ = |u⟩B ⊗ λ+ |v⟩B ⊗ |1⟩A + λ− |˜v ⟩B ⊗ |0⟩A with λ+ |v⟩B ⊗|1⟩A +λ− |˜v ⟩B ⊗|0⟩A being a non-biseparable 2α -qubit state, where α is an integer, |u⟩B , |v⟩B , |˜v ⟩B being product states consisting of |0⟩’s and |1⟩’s, |v⟩B and |˜v ⟩B are orthogonal, and the maximum violation of the Bell’s inequality γ in an n-qubit system is γ ≡ maxBn Tr(ρ Bn ) = 2f2α (ψ ), in which the function √

(

f2α (ψ ) is defined as: f2α (ψ ) ≡

)

1 + 22α−2 CA2 (ψ ) when 22−2α ≥ CA2 (ψ ), f2α (ψ ) ≡ 2

22−2α ≤ CA2 (ψ ). The coefficients λ± satisfy λ2+ = 1 ±

(



1 − CA2 (ψ ) /2 and λ2− = 1 ∓

)

(

2α−1 2



CA (ψ ) when

1 − CA2 (ψ ) /2.

)

Because the first three sites of the quantum states are not entangled and other sites of the quantum states are maximally entangled, the upper bound of the Bell’s inequality is equivalent to the maximum upper bound of the maximum violation of the Bell’s inequality of a four-qubit system. It is worthy to note that even though the concurrences of the pure states C12,A (ψ ), C22,A (ψ ), and CA2 (ψ ) are maximal, the



upper bound of the Bell’s√inequality 4 2 is smaller than that in the previous case of the upper bound of the Bell’s inequality 4 5, in which all coefficients of the quantum state have equal amplitude. Note that the upper bound of the Bell’s inequality is independent of C22,A when the seven-qubit



quantum states satisfy 1 − 2 1 − C12,A (ψ ) −



1 − CA2 (ψ ) = 0 and Rxx ≥ Rzz . The property is demon-



strated in Fig. 1. When the seven-qubit quantum states satisfy 1 − 2 1 − C22,A (ψ ) +



1 − CA2 (ψ ) = 0

and Rxx ≥ Rzz , the upper bound of the Bell’s inequality is also independent of C12,A . The property is also demonstrated in Fig. 2. When we prepare the maximally entangled state α12 = α22 = α32 = α42 = 1/4 in an experiment [4], this property can give an additional consistency check. Although entanglement entropy of the region A is also a monotonic function with respect to CA2 (ψ ) as in the case of two qubits [14], the upper bound of the Bell’s inequality in general is not. This is shown in Figs. 1 and 2. The non-monotonic property reflects that a system with a higher number of qubits can have a more interesting entanglement structure than a two-qubit quantum system. When the seven-qubit quantum state |ψ⟩7 demonstrates a critical point of the upper bound of the Bell’s inequality, one quantum state is a ground state of a seven-qubit toric code model on a disk manifold. Hamiltonian of the seven-qubit toric code model on a disk manifold is defined as ∑6 The ∑ 2 H7td ≡ − i=1 Ai − j=1 Bj , where A1 = σz ⊗ σz ⊗ 1 ⊗ 1 ⊗ 1 ⊗ 1 ⊗ 1, A2 = σz ⊗ 1 ⊗ σz ⊗ 1 ⊗ 1 ⊗ 1 ⊗ 1, A3 = 1 ⊗ σz ⊗ 1 ⊗ σz ⊗ σz ⊗ 1 ⊗ 1, A4 = 1 ⊗ 1 ⊗ σz ⊗ σz ⊗ 1 ⊗ σz ⊗ 1, A5 = 1 ⊗ 1 ⊗ 1 ⊗ 1 ⊗ σz ⊗ 1 ⊗ σz , A6 = 1 ⊗ 1 ⊗ 1 ⊗ 1 ⊗ 1 ⊗ σz ⊗ σz , B1 = σx ⊗ σx ⊗ σx ⊗ σx ⊗ 1 ⊗ 1 ⊗ 1, B2 = 1 ⊗ 1 ⊗ 1 ⊗ σx ⊗ σx ⊗ σx ⊗ σx . The sites of the lattice are labeled as in Fig. 3.

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Fig. 2. We fix C22,A = 0.75 and choose C12,A = 0.75, 0.8, 0.85, 0.9, 0.95, 1 to compute the upper bound of maximum violation of the Bell’s inequality. It is interesting to note that the upper bound of maximum violation of the Bell’s inequality is independent of C12,A when CA2 = 1.

Fig. 3. The sites of the qubits of the seven-qubit toric code model on a disk manifold are labeled by the integers. The boundaries are not identified on the disk manifold.

3. Outlook Our goal is to show the intensity of quantum entanglement in a maximally entangled state by using violation of Bell’s inequality [9,17,18]. This should help us to know the physical meaning for the definition of the maximum entangled state. Our theoretical result can also be realized and examined in experiments [4]. The ground states of a topologically ordered system can be prepared in the Nuclear Magnetic Resonance (NMR) quantum simulator [4]. The preparation procedure is rather indirect. This begins from an initial pseudo pure state and uses the NMR Hamiltonian to adiabatically generate the ground states [4]. By exploring this method, the ground states of the double Fibonacci, double semion and toric code models had been successfully prepared [4]. Acknowledgments Chen-Te Ma would like to thank Nan-Peng Ma for his encouragement. Su-Kuan Chu was supported by AFOSR, NSF QIS, ARL CDQI, ARO MURI, ARO and NSF PFC at JQI and Rong-Xin Miao was supported in part by NCTS and the grant MOST 105-2811-M-007-021 of the Ministry of Science and Technology of Taiwan. Appendix. The coefficients of the seven-qubit quantum states We choose a class of seven-qubit quantum states which can be parametrized as

|ψ⟩7 ≡ α1 |000000⟩B |0⟩A + α2 |000111⟩B |1⟩A + α3 |111100⟩B |0⟩A + α4 |111011⟩B |1⟩A .

(A.1)

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We have two constraints for the coefficients of the seven-qubit quantum states. The first constraint of the seven-qubit quantum states is

α12 + α22 + α32 + α42 = 1

(A.2)

TrA ρA = 1.

(A.3)

from

The second constraint of the seven-qubit quantum states is TrA ρA2 = (α12 + α32 )2 + (α22 + α42 )2 .

(A.4)

Solving the first constraint of the seven-qubit quantum states (A.2), we obtain:

α1 = sin(θ1 ), α2 = cos(θ1 ) sin(θ2 ), α3 = cos(θ1 ) cos(θ2 ) cos(θ3 ), α4 = cos(θ1 ) cos(θ2 ) sin(θ3 ),

(A.5)

where 0 ≤ θ1 ≤ 2π ,

0 ≤ θ2 , θ3 , θ4 ≤ π.

(A.6)

We are going to express the coefficients α1 , α2 , α3 , and α4 in terms of concurrences. To solve the algebra, we first choose:

θ2 =

π

, θ3 = 0. 2 but leave the value of θ1 unchanged. Then we can get the new coefficients of the states: β1 = sin(θ1 ),

β2 = cos(θ1 ),

β3 = β4 = 0.

(A.7)

(A.8)

Thus, we obtain: TrA ρA2 = TrA ρ12,A = β14 + β24 = 2β14 − 2β12 + 1.

(A.9)

We can choose:

β = 2 1

β22 =

1−



1 − C12,A (ψ ) 2

1+

,



1 − C12,A (ψ ) 2

,

(A.10)

where C1,A (ψ ) ≡

√ (

2 1 − TrA ρ12,A .

)

(A.11)

Now we choose θ3 = 0, but restore the original value of θ2 . Thus, we have another new coefficients:

γ1 = sin(θ1 ),

γ2 = cos(θ1 ) sin(θ2 ),

γ3 = cos(θ1 ) cos(θ2 ), and TrA ρ22,A

= (γ12 + γ32 )2 + γ24

γ4 = 0,

(A.12)

S.-K. Chu et al. / Annals of Physics 395 (2018) 183–195

= sin2 (θ1 ) + cos2 (θ1 )cos2 (θ2 )

(

)2

189

+ cos4 (θ1 )sin4 (θ2 )

= sin4 (θ1 ) + cos4 (θ1 )cos4 (θ2 ) + 2cos2 (θ1 )sin2 (θ1 )cos2 (θ1 ) + cos4 (θ1 )sin4 (θ2 ) = cos4 (θ1 )cos4 (θ2 ) + 2cos2 (θ1 )sin2 (θ1 )cos2 (θ2 ) ( )2 + cos4 (θ1 ) 1 − cos2 (θ2 ) + sin4 (θ1 ) ( ) = cos4 (θ1 )cos4 (θ2 ) + 2cos2 (θ1 ) 1 − cos2 (θ1 ) cos2 (θ2 ) ( ) + cos4 (θ1 ) cos4 (θ2 ) − 2cos2 (θ2 ) + 1 + sin4 (θ1 ) = 2cos4 (θ1 )cos4 (θ2 ) ( ) + −4cos4 (θ1 ) + 2cos2 (θ1 ) cos2 (θ2 ) ( ) + 2cos4 (θ1 ) − 2cos2 (θ1 ) + 1 .

(A.13)

Hence, one solution is:

( cos (θ2 ) = 2

2cos2 (θ1 ) − 1 +

)

2cos2 (θ1 ) − 1 +

)

=



1 − C22,A (ψ )

2cos2 (θ1 )



1 − C12,A (ψ ) + 1+

sin2 (θ2 ) =

−1 + 2TrA ρ22,A

2cos2 (θ1 )

(

=



1−





1 − C22,A (ψ )



1 − C12,A (ψ )

−1 + 2TrA ρ22,A

2cos2 (θ1 ) 1−

= 1+



1 − C22,A (ψ )



1 − C12,A (ψ )

1−

=

,



1 − C22,A (ψ )

2cos2 (θ1 )

,

(A.14)

where C2,A (ψ ) ≡

√ (

2 1 − TrA ρ22,A .

)

(A.15)

Finally, we restore θ2 and θ3 in an effort to solve the constraints of the seven-qubit quantum states (A.2) and (A.4) to obtain θ3 through the concurrences of the pure states of the region A, C1,A , C2,A and CA , [13]: TrA ρA2

= (α12 + α32 )2 + (α22 + α42 )2 ( )2 = sin2 (θ1 ) + cos2 (θ1 )cos2 (θ2 )cos2 (θ3 ) ( )2 + cos2 (θ1 )sin2 (θ2 ) + cos2 (θ1 )cos2 (θ2 )sin2 (θ3 ) √ ( 1 − 1 − C 2 (ψ ) 1,A = 2 √ √ ) 1 − C12,A (ψ ) + 1 − C22,A (ψ ) + cos2 (θ3 ) 2 2 √ ( 1 − 1 − C 2 (ψ ) 2,A + 2

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S.-K. Chu et al. / Annals of Physics 395 (2018) 183–195

√ +

1 − C12,A (ψ ) +

1−

+

1 − C22,A (ψ )

2

( =



sin2 (θ3 )

)

2

C12,A (ψ ) + C22,A (ψ ) 2

)





1 − C12,A (ψ ) 1 − C22,A (ψ ) cos4 (θ3 )

( + −1 +

C12,A (

ψ) −



1−

C12,A (



ψ) 1 −

ψ)

C22,A (

× cos2 (θ3 ) 2 − C12,A (ψ ) + .

)

(A.16)

2

Therefore, one particular solution is: √



1−C12,A (ψ ) 1−C22,A (ψ )+1−C12,A (ψ )



cos (θ3 ) = √ 2



1−C12,A (ψ )+ 1−C22,A (ψ )

1 − C12,A (ψ ) +



1 − C22,A (ψ )

√ −1 + 2TrA ρA2 √ +√ 1 − C12,A (ψ ) + 1 − C22,A (ψ ) √



1−C12,A (ψ ) 1−C22,A (ψ )+1−C12,A (ψ )



= √



1−C12,A (ψ )+ 1−C22,A (ψ )

1 − C12,A (ψ ) +



1 − C22,A (ψ )

√ +√ √

1 − CA2 (ψ ) 1 − C22,A (ψ )

,



1−C12,A (ψ ) 1−C22,A (ψ )+1−C22,A (ψ )



sin (θ3 ) = √



1 − C12,A (ψ ) +

2



1−C12,A (ψ )+ 1−C22,A (ψ )

1 − C12,A (ψ ) +



1 − C22,A (ψ )

√ −1 + 2TrA ρA2 √ −√ 1 − C12,A (ψ ) + 1 − C22,A (ψ ) √



1−C12,A (ψ ) 1−C22,A (ψ )+1−C22,A (ψ )



= √



1−C12,A (ψ )+ 1−C22,A (ψ )

1 − C12,A (ψ ) +



1 − C22,A (ψ )

√ −√

1 − CA2 (ψ )

1 − C12,A (ψ ) +



1 − C22,A (ψ )

.

(A.17)

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191

For convenience, we list all coefficients of the quantum state from the particular solution:

α12 = α22 =

1−



1 − C12,A (ψ )

√ 2 1 − 1 − C22,A (ψ ) 2

√ α32 =

1 − C12,A (ψ ) 1 − C22,A (ψ ) + 1 − C12,A (ψ )

(√



α42 =

1 − C12,A (ψ ) +

)



1 − C22,A (ψ )

1 − CA2 (ψ ) 2

, √

1 − C12,A (ψ ) 1 − C22,A (ψ ) + 1 − C22,A (ψ )

(√

2

√ −

,



2

+ √

,

1−

C12,A (

1 − CA2 (ψ ) 2

ψ) +



1−

C22,A (

)

ψ)

,

(A.18)

where 0 ≤ α12 , α22 , α32 , α42 ≤ 1.

(A.19)

A.1. The upper bound of the Bell’s inequality of the seven-qubit quantum states We start from the seven-qubit quantum states

|ψ⟩7 = α1 |0000000⟩ + α2 |0001111⟩ + α3 |1111000⟩ + α4 |1110111⟩, where α1 , α2 , α3 and α4 are arbitrary real constants. Thus, the density matrix of the seven-qubit quantum states is given by

ρ7 = α12 |0000000⟩⟨0000000| + α22 |0001111⟩⟨0001111| + α32 |1111000⟩⟨1111000| + α42 |1110111⟩⟨1110111| + α1 α2 (|0000000⟩⟨0001111| + |0001111⟩⟨0000000|) + α1 α3 (|0000000⟩⟨1111000| + |1111000⟩⟨0000000|) + α1 α4 (|0000000⟩⟨1110111⟩ + |1110111⟩⟨0000000|) + α2 α3 (|0001111⟩⟨1111000| + |1111000⟩⟨0001111|) + α2 α4 (|0001111⟩⟨1110111| + |1110111⟩⟨000111|)

(A.20)

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+ α3 α4 (|1111000⟩⟨1110111| + |1110111⟩⟨1111000|).

(A.21)

The upper bound of the Bell’s inequality [9,14,17], which is defined by two largest eigenvalues of the matrix R† R, can be computed from the generalized R-matrix RIin [9,15,17] through the lemma [9,15,17]: Now we compute the generalized R-matrix of the seven-qubit quantum states: Rxxxzxxx = Rxyyzyyx = Ryyxzyyx Rxyyzxxx Rxyxzyxx

Ryyyzyxx Rzzzxxxx Rzzzyyxx Rzzzxyyx Ryxxzxxy Ryyyzxxy Ryyxzyxy Rxxxzyxy Rzzzyxxy Rzzzxyxy Rzzzyyyy Rzzzzzzz Rxxxxzzz Ryyxxzzz Ryxxyzzz

= = = = = = = = = = = = = = = = = = = = = = = =

Ryxyzyyx = 2(α1 α4 − α2 α3 ), Rxxxzyyx = Ryyxzxxx Ryxyzxxx = −2(α1 α4 − α2 α3 ), Ryxxzyxx = Rxyxzxyx Rxxyzyxx = Rxxyzxyx = Ryxxzxyx −2(α1 α4 + α2 α3 ), Ryyyzxyx = 2(α1 α4 + α2 α3 ), 2(α1 α2 − α3 α4 ), Rzzzyxyx = −2(α1 α2 + α3 α4 ), −2(α1 α2 − α3 α4 ), Ryyyzyyy = Rxyxzxxy Rxxyzxxy = −2(α1 α4 + α2 α3 ), Rxxyzyyy = Ryxxzyyy Rxyxzyyy = 2(α1 α4 + α2 α3 ), Ryyxzxyy = Ryxyzyxy = Ryxyzxyy Rxyyzyxy = Rxyyzxyy = 2(α1 α4 − α2 α3 ), Rxxxzxyy = −2(α1 α4 − α2 α3 ), −2(α1 α2 + α3 α4 ), Rzzzxxyy = −2(α1 α2 − α3 α4 ), 2(α1 α2 + α3 α4 ), α12 + α22 + α32 + α42 , 2(α1 α3 − α2 α4 ), Ryxyxzzz = Rxyyxzzz = −2(α1 α3 − α2 α4 ), Rxyxyzzz = Rxxyyzzz = −2(α1 α3 + α2 α4 ),

Ryyyyzzz = 2(α1 α3 + α2 α4 ), and other components of the generalized R-matrix vanish. We could find that R† R only has diagonal elements:

(

R† R

) xx

( ) = R† R yy = 32(α12 α42 + α22 α32 − 2α1 α2 α3 α4 ) + 32(α12 α42 + α22 α32 + 2α1 α2 α3 α4 ) + 8(α12 α22 + α32 α42 + 2α1 α2 α3 α4 ) + 8(α12 α22 + α32 α42 − 2α1 α2 α3 α4 ) = 64α12 α42 + 64α22 α32 + 16α12 α22 + 16α32 α42 = 16(α12 + α32 )(α22 + α42 ) + 48(α12 α42 + α22 α32 ) = 16(α12 + α32 )(1 − α12 − α32 ) ( ) + 48 α12 (1 − α12 − α22 − α32 ) + α22 α32

(A.22)

S.-K. Chu et al. / Annals of Physics 395 (2018) 183–195

193

= 16(α12 − α14 − 2α12 α32 + α32 − α34 ) + 48(α12 − α14 − α12 α22 − α12 α32 + α22 α32 ) = 16(4α12 − 4α14 − 5α12 α32 + α32 − α34 − 3α12 α22 + 3α22 α32 ), R† R

(

) zz

= α14 + α24 + α34 + α44 + 2α12 α22 + 34α12 α32 + 2α12 α42 + 2α22 α32 + 34α22 α42 + 2α32 α42 = 1 + 32α12 α32 + 32α22 α42 = 1 + 32α12 α32 + 32α22 (1 − α12 − α22 − α32 ) = 1 + 32α22 + 32α12 α32 − 32α12 α22 − 32α24 − 32α22 α32 ,

(A.23)

in which we used

α12 + α22 + α32 + α42 = 1.

(A.24)

Now we show an extreme value of the upper bound of the Bell’s inequality. If we have the inequality

(

R† R

) xx

( ) ≥ R† R zz ,

(A.25)

we can find that an extreme value of the upper bound of the Bell’s inequality satisfies: 8α12 + 3α22 + 5α32 = 4,

α12 = α32 ,

5α12 − 3α22 + 2α32 = 1.

(A.26)

Hence, the extreme value occurs at

α12 = α22 = α32 = α42 =

1 4

,

(A.27)

which also satisfies the inequality

(

R† R

) xx

( ) ≥ R† R zz .

(A.28)

From a second derivative test, the critical point of the upper bound of the Bell’s inequality is a saddle point of the Bell’s inequality. If we consider the inequality

(

R† R

) xx

( ) ≤ R† R zz ,

(A.29)

we can find that an extreme value of the upper bound of the Bell’s inequality satisfies: 8α12 + 5α22 + 3α32 = 4, 5α12 + 4α22 − α32 = 2, 3α12 − α22 + 2α32 = 1.

(A.30)

The solution is given by

α12 = α22 = α32 = α42 =

1 4

,

(A.31)

but it does not satisfy the inequality

(

R† R

) xx

( ) ≤ R† R zz .

(A.32)

194

S.-K. Chu et al. / Annals of Physics 395 (2018) 183–195

Thus, we conclude that the upper √bound of the Bell’s inequality at the critical points of the upper bound of the Bell’s inequality is 4 5 and it occurs at

α12 = α22 = α32 = α42 =

1 4

.

(A.33)

The above analysis is valid for all solutions. Now we substitute (A.18) into (A.23): R† R

(

) xx

( ) = R† R yy = 16(α12 + α32 )(α22 + α42 ) + 48(α12 α42 + α22 α32 ) = 4CA2 (ψ ) + 48(α12 α42 + α22 α32 ) = 4CA2 (ψ ) √ [ ( 1 − 1 − C 2 (ψ ) ) A + 48 α12 − α22 2 √ )] ( 1 + 1 − C 2 (ψ ) A 2 2 − α1 + α2 2

=

4CA2 (

ψ) [√

√ + 12 1 − C12,A (ψ ) + 1 − C22,A (ψ ) √ √ − 2 1 − C12,A (ψ ) 1 − C22,A (ψ ) (√ ) √ + 1 − C12,A (ψ ) − 1 − C22,A (ψ ) ] √ 2 × 1 − C A (ψ ) , R† R

(

) zz

= 1 + 32α12 α32 + 32α22 α42 √ [ ( 1 + 1 − C 2 (ψ ) ) A 2 2 = 1 + 32 α1 − α1 2 √ ( 1 − 1 − C 2 (ψ ) )] A + α22 − α22 2 ( ) = −15 + 8 C12,A (ψ ) + C22,A (ψ ) ) [(√ √ 1 − C12,A (ψ ) + 1 − C22,A (ψ ) +8 (√ ) √ 2 2 + 1 − C2,A (ψ ) − 1 − C1,A (ψ ) ] √ × 1 − CA2 (ψ ) ,

(A.34)

in which we used:

α +α = 2 1

2 3

α22 + α42 =

1+



1 − CA2 (ψ )

√ 2 1 − 1 − CA2 (ψ ) 2

, .

(A.35)

S.-K. Chu et al. / Annals of Physics 395 (2018) 183–195

195

We note that the upper bound of the Bell’s inequality does not depend on C22,A when



1 − 2 1 − C12,A (ψ ) −



1 − CA2 (ψ ) = 0,

Rxx ≥ Rzz

(A.36)

and the upper bound of the Bell’s inequality does not depend on C12,A when



1 − 2 1 − C22,A (ψ ) + Rxx ≥ Rzz .



1 − CA2 (ψ ) = 0, (A.37)

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