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Quantum entanglement estimation via symmetric measurement based positive maps

Jiaxin Li,1 Hongmei Yao,1,∗ Shao-Ming Fei,2,† Zhaobing Fan,1 and Haitao Ma1 1{1} Department of Mathematics, Harbin Engineering University, Harbin 150001, P.R. China
22 School of Mathematical Sciences, Capital Normal University, Beijing 100048, China
Correspondence to [email protected]
Correspondence to [email protected]
Abstract

We provide a class of positive and trace-preserving maps based on symmetric measurements. From these positive maps we present separability criteria, entanglement witnesses, as well as the lower bounds of concurrence. We show by detailed examples that our separability criteria, entanglement witnesses and lower bounds can detect and estimate the quantum entanglement better than the related existing results.

I I. Introduction

Quantum entanglement is the key resource in quantum information processing and plays an important role in quantum communication, quantum computing and other modern quantum technologies Hor2009 . Therefore, it is of significance to distinguish the entangled states from the separable ones and estimate degree of entanglement for the entangled states. However, generally the separability problem and the estimation of entanglement are very difficult and even NP-hard 2 . For low dimensional systems like 22\mathbb{C}^{2}\otimes\mathbb{C}^{2} (qubit-qubit) and 23\mathbb{C}^{2}\otimes\mathbb{C}^{3} (qubit-qutrit), the celebrated positive partial transposition (PPT) criterion is both necessary and sufficient for separability 3 ; 4 . For higher dimensional systems, more sophisticated methods are needed to detect the entanglement.

One separability criterion to detect entanglement is given by positive maps. A bipartite state ρ\rho is separable if and only if (𝕀Φ)(ρ)0(\mathbb{I}\otimes\Phi)(\rho)\geq 0 for any positive map Φ\Phi 5 , where 𝕀\mathbb{I} is the identity operator. Namely, ρ\rho is entangled if (𝕀Φ)(ρ)(\mathbb{I}\otimes\Phi)(\rho) has negative eigenvalues for some positive map Φ\Phi.

Entanglement can be also detected by entanglement witnesses. A Hermitian operator WW is called an entanglement witness if Tr(Wρsep)0\text{Tr}(W\rho_{sep})\geq 0 for all separable states ρsep\rho_{sep}, and Tr(Wρ)<0\text{Tr}(W\rho)<0 for some entangled sates ρ\rho 6 ; 7 . By Choi-Jamiołkowski isomorphism an entanglement witness is related to a positive but not completely positive map Φ\Phi. One kind of entanglement witnesses is the decomposable one, for which an entanglement witness can be written as W=A+BΓW=A+B^{\Gamma}, where A,B0A,B\geq 0 and BΓ=(𝕀T)BB^{\Gamma}=(\mathbb{I}\otimes T)B with TT denoting the transpose. However, the decomposable witness cannot detect the positive partial transpose (PPT) entangled states that are positive under partial transpose. The indecomposable witnesses can detect the PPT states 5 ; 9 ; 10 ; 11 ; 12 , which can be constructed by using realignment separability criterion 13 ; 14 ; 15 and covariance matrix criterion 16 ; 17 ; 18 . In 19 , the authors constructed a class of indecomposable witnesses by using MUBs (mutually unbiased bases). This method was extended to the one by using MUMs (mutually unbiased measurements) and SIC-POVMs (symmetric informationally complete measurements). New entanglement witnesses have been also obtained 20 ; 21 ; 22 . Recently, a new kind of measurements, called symmetric measurements, has been proposed in 23 . Based on the symmetric measurements a class of positive maps and entanglement witnesses are constructed in 24 .

To quantify the entanglement, many measures have been presented such as the entanglement of formation (EOF) 25 ; 26 and concurrence 27 ; 28 ; 29 . However, it is a challenge to evaluate the entanglement measures for general quantum states. Instead of analytic formulas, progress has been made toward the lower bounds of EOF and concurrence. Based on a positive map, a new lower bound of concurrence for arbitrary dimensional bipartite systems has been derived in 30 , which detects the entanglement that is not detected by the previous lower bounds 31 ; 32 .

In this paper, we first present a family of positive and trace-preserving maps based on symmetric measurements. Then we present new separability criteria and show that these separability criteria detect better entanglement of quantum states by a exact example. We then construct a series of entanglement witnesses which include some existing ones as special cases. These entanglement witnesses are shown to detect better entanglement including bound entanglements. At last we give a family of lower bounds of concurrence and demonstrate that the bounds estimate better the quantum entanglement than the existing ones.

II II. Positive maps and Separability criteria

A positive operator-valued measure (POVM) is given by a set of positive operators {EαEα0,αEα=𝕀}\{E_{\alpha}\mid E_{\alpha}\geq 0,~{}\sum_{\alpha}E_{\alpha}=\mathbb{I}\}. For a given state ρ\rho, the probability of the measurement outcome with respect to EαE_{\alpha} is pα=Tr(Eαρ)p_{\alpha}=\text{Tr}(E_{\alpha}\rho). Recently, a new POVM called symmetric measurement has been provided in 23 . A set of NN dd-dimensional POVMs {Eα,k|Eα,k0,k=1MEα,k=𝕀d}\{E_{\alpha,k}|\,E_{\alpha,k}\geq 0,\,\sum\limits_{k=1}^{M}E_{\alpha,k}=\mathbb{I}_{d}\}, α=1,,N\alpha=1,\ldots,N is called (N,M)(N,M)-POVM, which satisfies the following symmetry properties,

Tr(Eα,k)=dM,Tr(Eα,k2)=x,Tr(Eα,kEα,)=dMxM(M1),k,Tr(Eα,kEβ,)=dM2,βα,\displaystyle\begin{split}\text{Tr}(E_{\alpha,k})&=\frac{d}{M},\\ \text{Tr}(E_{\alpha,k}^{2})&=x,\\ \text{Tr}(E_{\alpha,k}E_{\alpha,\ell})&=\frac{d-Mx}{M(M-1)},\quad\ell\neq k,\\ \text{Tr}(E_{\alpha,k}E_{\beta,\ell})&=\frac{d}{M^{2}},\quad\beta\neq\alpha,\end{split} (1)

where

dM2<xmin{d2M2,dM}.\displaystyle\frac{d}{M^{2}}<x\leq\min\left\{\frac{d^{2}}{M^{2}},\frac{d}{M}\right\}. (2)

For any fixed dimension d<d<\infty, there are at least four different types of information complete (N,M)(N,M)-POVMs: i) M=d2M=d^{2} and N=1N=1 (general SIC POVM) 34 ; ii) M=dM=d and N=d+1N=d+1 (MUMs) 35 ; iii) M=2M=2 and N=d21N=d^{2}-1; iv) M=d+2M=d+2 and N=d1N=d-1. A general construction of informationally complete (N,M)(N,M)-POVMs has been presented by using orthonormal Hermitian operator bases {G0=𝕀d/d,Gα,k;α=1,,N,k=1,,M1}\{G_{0}=\mathbb{I}_{d}/\sqrt{d},\,G_{\alpha,k};\,\alpha=1,\ldots,N,\,k=1,\ldots,M-1\} with Tr(Gα,k)=0\text{Tr}(G_{\alpha,k})=0,

Eα,k=1M𝕀d+tHα,k,\displaystyle E_{\alpha,k}=\frac{1}{M}\mathbb{I}_{d}+tH_{\alpha,k}, (3)

where

Hα,k={GαM(M+1)Gα,k,k=1,,M1,(M+1)Gα,k=M,H_{\alpha,k}=\left\{\begin{aligned} &G_{\alpha}-\sqrt{M}(\sqrt{M}+1)G_{\alpha,k},\,k=1,\ldots,M-1,\\ &(\sqrt{M}+1)G_{\alpha},\,k=M,\end{aligned}\right. (4)

and Gα=k=1M1Gα,kG_{\alpha}=\sum\limits_{k=1}^{M-1}G_{\alpha,k}. The parameters tt and xx satisfy the following relation,

x=dM2+t2(M1)(M+1)2.\displaystyle x=\frac{d}{M^{2}}+t^{2}(M-1)(\sqrt{M}+1)^{2}. (5)

The optimal value xoptx_{\rm opt}, which is the greatest xx such that Eα,k0E_{\alpha,k}\geq 0, depends on the operator bases. There always exist informationally complete (N,M)(N,M)-POVMs for any integer dd.

In 24 the following class of positive maps have been presented,

Φ(X)=1b[aΦ0(X)+α=L+1NΦα(X)α=1LΦα(X)],\Phi(X)=\frac{1}{b}\Big{[}a\Phi_{0}(X)+\sum_{\alpha=L+1}^{N}\Phi_{\alpha}(X)-\sum_{\alpha=1}^{L}\Phi_{\alpha}(X)\Big{]}, (6)

where a=bN+2La=b-N+2L, b=(d1)M(xy)db=\frac{(d-1)M(x-y)}{d}, Φ0(X)=Tr(X)d𝕀d\Phi_{0}(X)=\frac{\text{Tr}(X)}{d}\mathbb{I}_{d} and

Φα(X)=Mdk,l=1M𝒪k(α)Eα,kTr(XEα,l)\Phi_{\alpha}(X)=\frac{M}{d}\sum_{k,l=1}^{M}\mathcal{O}^{(\alpha)}_{k\ell}E_{\alpha,k}\text{Tr}(XE_{\alpha,l}) (7)

are NN trace-preserving maps given by (N,M)(N,M)-POVMs {Eα,k}\{E_{\alpha,k}\} with {𝒪(α)|𝒪(α)=(𝒪k(α)),α=1,,N}\{\mathcal{O}^{(\alpha)}|\mathcal{O}^{(\alpha)}=(\mathcal{O}^{(\alpha)}_{k\ell}),\,\alpha=1,\cdots,N\} being a set of M×MM\times M orthogonal rotation operators that preserve the vector 𝐧=(1,,1)/d\mathbf{n}_{\ast}=(1,\ldots,1)/\sqrt{d}.

Using the maps (6), we have the following theorem.

Theorem 1. A bipartite state ρ\rho is entangled if (𝕀Φz)(ρ)0(\mathbb{I}\otimes\Phi_{z})(\rho)\ngeq 0, where

Φz(X)=(1z)Φ0(X)+zΦ(X),X𝕄d\Phi_{z}(X)=(1-z)\Phi_{0}(X)+z\Phi(X),~{}\forall X\in\mathbb{M}_{d} (8)

are positive and trace preserving linear maps with z[1,1]z\in[-1,1].

Proof.

In order to prove the positivity of Φz\Phi_{z}, we only need to prove Beng ,

Tr[(Φz(P))2]1d1\displaystyle\text{Tr}[(\Phi_{z}(P))^{2}]\leq\frac{1}{d-1} (9)

for every rank-1 projector P=|ψψ|P=|\psi\rangle\langle\psi|. By straightforward calculation we have

Tr[(Φz(P))2]=\displaystyle\text{Tr}[(\Phi_{z}(P))^{2}]= Tr{(1z)2Φ0(P)2+z2Φ(P)2\displaystyle\text{Tr}\Bigg{\{}(1-z)^{2}\Phi_{0}(P)^{2}+z^{2}\Phi(P)^{2}
+z(1z)(Φ0(P)Φ(P)+Φ(P)Φ0(P))}\displaystyle+z(1-z)(\Phi_{0}(P)\Phi(P)+\Phi(P)\Phi_{0}(P))\Bigg{\}}
=\displaystyle= (1z)2d(Tr(P))2+2zd(Tr(P))2(1z)\displaystyle\frac{(1-z)^{2}}{d}(\text{Tr}(P))^{2}+\frac{2z}{d}(\text{Tr}(P))^{2}(1-z)
+z2Tr(Φ(P)2)\displaystyle+z^{2}\text{Tr}(\Phi(P)^{2})
\displaystyle\leq (1z)2d+2z(1z)d+z2d1\displaystyle\frac{(1-z)^{2}}{d}+\frac{2z(1-z)}{d}+\frac{z^{2}}{d-1}
=\displaystyle= (z21)+dd(d1)1d1,\displaystyle\frac{(z^{2}-1)+d}{d(d-1)}\leq\frac{1}{d-1}, (10)

which completes the proof of positivity. The proof of trace preservation is obvious. The theorem follows from the separability criterion based on positive maps. \square

The map (8) is a linear but not convex combination of the map Φ\Phi and the completely depolarizing channel Φ0\Phi_{0} for z[1,0)z\in[-1,0). Namely, it is truly a new positive but not completely positive map. To illustrate the theorem let us consider the state 37 ,

ρ=\displaystyle\rho= 14diag(q1,q4,q3,q2,q2,q1,q4,q3,q3,q2,q1,q4,q4,\displaystyle\frac{1}{4}\text{diag}(q_{1},q_{4},q_{3},q_{2},q_{2},q_{1},q_{4},q_{3},q_{3},q_{2},q_{1},q_{4},q_{4}, (11)
q3,q2,q1)+q14i,j=1,6,11,16ijFi,j,\displaystyle q_{3},q_{2},q_{1})+\frac{q_{1}}{4}\sum_{i,j=1,6,11,16}^{i\neq j}F_{i,j},

where Fi,jF_{i,j} is a matrix with the (i,j)(i,j) entry 1 and rest entries 0, qm0q_{m}\geq 0, qm=1\sum q_{m}=1, m=1,2,3,4m=1,2,3,4. Set d=4d=4, N=L=5N=L=5, M=4M=4 and 𝒪(α)=I4\mathcal{O}^{(\alpha)}=I_{4} for any α[N]\alpha\in[N]. The (5,4)(5,4)-POVM are constructed from the Gell-Mann matrices (see Appendix A). From Theorem 1 we obtain that ρ\rho is entangled for 0.25<q1<10.25<q_{1}<1 by straightforward calculation, see Appendix C. The criterion given in 30 detects the entanglement when q1>q4q_{1}>q_{4}. Our criterion shows that when 0.25<q1<10.25<q_{1}<1, ρ\rho is entangled even if q1<q4q_{1}<q_{4}.

III III. Construction of entanglement witnesses from positive maps

By entanglement witnesses one can detect the entanglement of unknown quantum states experimentally. An entanglement witness WW can be obtained based on positive but not completely positive map Φ\Phi through Choi-Jamiołkowski isomorphism J ,

W=k,=1d|k|Φ[|k|],\displaystyle W=\sum_{k,\ell=1}^{d}|k\rangle\langle\ell|\otimes\Phi[|k\rangle\langle\ell|], (12)

where {|k}k=1d\{|k\rangle\}^{d}_{k=1} is an orthonormal basis in d\mathbb{C}^{d}. Therefore, by using the positive maps in Theorem 1, we get the following entanglement witnesses,

W=1b(awd𝕀d2+α=L+1NKαα=1LKα),\displaystyle W=\frac{1}{b}\left(\frac{aw}{d}\mathbb{I}_{d^{2}}+\sum_{\alpha=L+1}^{N}K_{\alpha}-\sum_{\alpha=1}^{L}K_{\alpha}\right), (13)

where

Kα=Mzdk,=1M𝒪k(α)E¯α,Eα,k\displaystyle K_{\alpha}=\frac{Mz}{d}\sum_{k,\ell=1}^{M}\mathcal{O}_{k\ell}^{(\alpha)}\overline{E}_{\alpha,\ell}\otimes E_{\alpha,k} (14)

with E¯α,\overline{E}_{\alpha,\ell} the conjugation of Eα,lE_{\alpha,l}. In particular, when z=1z=1 our witnesses includes the one given in 24 as a special case.

As the informationally complete (N,M)(N,M)-POVMs can be constructed by using an orthogonal basis {𝕀d/d,Gα,k}\{\mathbb{I}_{d}/\sqrt{d},G_{\alpha,k}\} of traceless Hermitian operators Gα,kG_{\alpha,k} for any dimension dd, we have the following entanglement witnesses,

W~=bt2W=d1d2M2(M+1)2𝕀d2+α=L+1NJαα=1LJα,\mathaccent 869{W}=\frac{b}{t^{2}}W=\frac{d-1}{d^{2}}M^{2}(\sqrt{M}+1)^{2}\mathbb{I}_{d^{2}}+\sum_{\alpha=L+1}^{N}J_{\alpha}-\sum_{\alpha=1}^{L}J_{\alpha}, (15)

where

Jα=Mzdk,=1M𝒪kl(α)H¯α,Hα,k\displaystyle J_{\alpha}=\frac{Mz}{d}\sum_{k,\ell=1}^{M}\mathcal{O}_{kl}^{(\alpha)}\overline{H}_{\alpha,\ell}\otimes H_{\alpha,k} (16)

with H¯α,\overline{H}_{\alpha,\ell} the conjugation of Hα,lH_{\alpha,l}. Note that these witnesses don’t depend on the parameter xx that characterizes the symmetric measurements. But W~\mathaccent 869{W} are relate to the number MM of operators in a single POVM. The larger the value of MM is, the larger the LL can be.

Notice that JαJ_{\alpha} in (16) can be directly represented by the operator basis {G0=𝕀d/d,Gα,k;α=1,,N,k=1,,M1}\{G_{0}=\mathbb{I}_{d}/\sqrt{d},G_{\alpha,k};\,\alpha=1,\ldots,N,\,k=1,\ldots,M-1\}, since Hα,kH_{\alpha,k} are directly given by Gα,kG_{\alpha,k}. By using (4), we further obtain

Jα=Mzdk,=1M1𝒬kl(α)G¯α,Gα,k,\displaystyle J_{\alpha}=\frac{Mz}{d}\sum_{k,\ell=1}^{M-1}\mathcal{Q}_{kl}^{(\alpha)}\overline{G}_{\alpha,\ell}\otimes G_{\alpha,k},

where

𝒬k(α)=\displaystyle\mathcal{Q}_{k\ell}^{(\alpha)}= M(𝒪MM(α)1)+M(M+1)2𝒪k(α)\displaystyle M(\mathcal{O}_{MM}^{(\alpha)}-1)+M(\sqrt{M}+1)^{2}\mathcal{O}_{k\ell}^{(\alpha)} (17)
M(M+1)(𝒪M(α)+𝒪kM(α)).\displaystyle-M(\sqrt{M}+1)(\mathcal{O}_{M\ell}^{(\alpha)}+\mathcal{O}_{kM}^{(\alpha)}).

Since

𝒬(α)T𝒬(α)=𝒬(α)𝒬(α)T=M2(M+1)4𝕀M1,\displaystyle\mathcal{Q}^{(\alpha)T}\mathcal{Q}^{(\alpha)}=\mathcal{Q}^{(\alpha)}\mathcal{Q}^{(\alpha)T}=M^{2}(\sqrt{M}+1)^{4}\mathbb{I}_{M-1},

𝒬(α)=(𝒬k(α))\mathcal{Q}^{(\alpha)}=(\mathcal{Q}_{k\ell}^{(\alpha)}) (α=1,,N)(\alpha=1,\cdots,N) are M×MM\times M rescaled orthogonal matrices. When 𝒪(α)=𝕀M\mathcal{O}^{(\alpha)}=\mathbb{I}_{M}, (17) can be rewritten as

𝒬k(α)=M(M+1)2δk.\displaystyle\mathcal{Q}_{k\ell}^{(\alpha)}=M(\sqrt{M}+1)^{2}\delta_{k\ell}. (18)

Next, we illustrate that W~\mathaccent 869{W} (15) are related to a well-known class of entanglement witnesses. Suppose the (N,M)(N,M)-POVM is informationally complete and L=NL=N. The corresponding witnesses is

W~=M2d(M+1)2[𝕀d2G0G0dM2(M+1)2α=1NJα],\displaystyle\mathaccent 869{W}=\frac{M^{2}}{d}(\sqrt{M}+1)^{2}\Bigg{[}\mathbb{I}_{d^{2}}-G_{0}\otimes G_{0}-\frac{d}{M^{2}(\sqrt{M}+1)^{2}}\sum_{\alpha=1}^{N}J_{\alpha}\Bigg{]}, (19)

where G0=𝕀d/dG_{0}=\mathbb{I}_{d}/\sqrt{d}. By a simple relabelling of the indices (α,k)μ(\alpha,k)\longmapsto\mu, we have

W~=dW~M2(M+1)2=𝕀d2μ,ν=0d21QμνGμTGν,\displaystyle\mathaccent 869{W}^{\prime}=\frac{d\mathaccent 869{W}}{M^{2}(\sqrt{M}+1)^{2}}=\mathbb{I}_{d^{2}}-\sum_{\mu,\nu=0}^{d^{2}-1}Q_{\mu\nu}G_{\mu}^{T}\otimes G_{\nu}, (20)

where QμνQ_{\mu\nu} are the entries of the following block-diagonal orthogonal matrix,

Q=1M(M+1)2[M(M+1)2z𝒬(1)Tz𝒬(2)Tz𝒬(N)T].\displaystyle Q=\frac{1}{M(\sqrt{M}+1)^{2}}\left[\begin{array}[]{c c c c c}M(\sqrt{M}+1)^{2}&&&&\\ &z\mathcal{Q}^{(1)T}&&&\\ &&z\mathcal{Q}^{(2)T}&&\\ &&&\ddots&\\ &&&&z\mathcal{Q}^{(N)T}\end{array}\right]. (26)

Therefore, the entanglement witnesses W~\mathaccent 869{W} constructed from symmetric measurements belong to a larger category of witnesses

W=𝕀d2μ,ν=0d21QμνGμTGν,\displaystyle W^{\prime}=\mathbb{I}_{d^{2}}-\sum_{\mu,\nu=0}^{d^{2}-1}Q_{\mu\nu}G_{\mu}^{T}\otimes G_{\nu}, (27)

which are related to the CCNR criterion Y . The GμG_{\mu} (27) are the elements of an arbitrary orthonormal Hermitian basis, and Q=(Qμν)Q=(Q_{\mu\nu}) is an arbitrary d2×d2d^{2}\times d^{2} orthogonal matrix with QTQ=𝕀d2Q^{T}Q=\mathbb{I}_{d^{2}} (in fact, QTQ𝕀d2Q^{T}Q\leq\mathbb{I}_{d^{2}} is sufficient).

For any informationally complete (N,M)(N,M)-POVM, assume that 𝒪(α)=𝕀M\mathcal{O}^{(\alpha)}=\mathbb{I}_{M} and L=NL=N. According to (18)(\ref{Id}) we have Q=diag(1,z,z,,z)Q=\text{diag}(1,z,z,\cdots,z). The associated entanglement witnesses write

W~=𝕀d2G0G0zμ=1d21GμTGμ.\displaystyle\mathaccent 869{W}^{\prime}=\mathbb{I}_{d^{2}}-G_{0}\otimes G_{0}-z\sum_{\mu=1}^{d^{2}-1}G_{\mu}^{T}\otimes G_{\mu}. (28)

Therefore, it is possible to use different (N,M)(N,M)-POVMs to generate the same witnesses W~\mathaccent 869{W}^{\prime}, provided that the same Hermitian orthonormal basis is used.

Let M=2M=2. Then the rotation matrices can only be 𝒪(α)=𝕀2\mathcal{O}^{(\alpha)}=\mathbb{I}_{2} or 𝒪(α)=σ1=[0110]\mathcal{O}^{(\alpha)}=\sigma_{1}=\left[\begin{array}[]{c c}0&1\\ 1&0\end{array}\right]. In this case, all witnesses constructed from (N,2)(N,2)-POVMs have the following form,

W~=\displaystyle\mathaccent 869{W}^{\prime}= 𝕀d2G0G0+z(α=L+1NGαTGα\displaystyle\mathbb{I}_{d^{2}}-G_{0}\otimes G_{0}+z(\sum_{\alpha=L+1}^{N}G_{\alpha}^{T}\otimes G_{\alpha} (29)
α=1LGαTGα),\displaystyle-\sum_{\alpha=1}^{L}G_{\alpha}^{T}\otimes G_{\alpha}),

where Nd21N\leq d^{2}-1.

To show the advantages of our entanglement witnesses in detecting quantum entanglement, we compare our entanglement witnesses with the ones presented in 24 by three examples, which show that our entanglement witnesses can detect more entangled quantum states, see Appendix D.

IV IV. Lower bound of concurrence

Let H1H_{1} and H2H_{2} be dd-dimensional vector spaces. A bipartite quantum pure state |ψ|\psi\rangle in H1H2H_{1}\otimes H_{2} has a Schmidt form

|ψ=iαi|ei1|ei2,\displaystyle|\psi\rangle=\sum_{i}\alpha_{i}|e_{i}^{1}\rangle\otimes|e_{i}^{2}\rangle, (30)

where |ei1|e_{i}^{1}\rangle and |ei2|e_{i}^{2}\rangle are the orthonormal bases in H1H_{1} and H2H_{2}, respectively, and αi\alpha_{i} are the Schmidt coefficients satisfying iαi2=1\sum_{i}\alpha_{i}^{2}=1. The concurrence C(|ψ)C(|\psi\rangle) of the state |ψ|\psi\rangle is given by

C(|ψ)=2(1Trρ12)=2i<jαi2αj2,\displaystyle C(|\psi\rangle)=\sqrt{2(1-\text{Tr}\rho_{1}^{2})}=2\sqrt{\sum_{i<j}\alpha_{i}^{2}\alpha_{j}^{2}}, (31)

where ρ1=Tr2(|ψψ|)\rho_{1}=\text{Tr}_{2}(|\psi\rangle\langle\psi|) is the reduced state obtained by tracing over the second space C .

The concurrence is extended to mixed states ρ\rho by the convex roof,

C(ρ)=minipiC(|ψi),\displaystyle C(\rho)=\text{min}\sum_{i}p_{i}C(|\psi_{i}\rangle), (32)

where the minimum is taken over all possible pure state decompositions of ρ=ipi|ψiψi|\rho=\sum_{i}p_{i}|\psi_{i}\rangle\langle\psi_{i}|, where pi0p_{i}\geq 0 and ipi=1\sum_{i}p_{i}=1. Generally it is formidably difficult to calculate C(ρ)C(\rho). Instead, one considers the lower bound of C(ρ)C(\rho).

In 32 the authors presented a lower bound of C(ρ)C(\rho),

C(ρ)2d(d1)f(ρ),\displaystyle C(\rho)\geq\sqrt{\frac{2}{d(d-1)}}f(\rho), (33)

where f(ρ)f(\rho) is a real-valued and convex function satisfying

f(|ψψ|)2i<jαiαj\displaystyle f(|\psi\rangle\langle\psi|)\leq 2\sum_{i<j}\alpha_{i}\alpha_{j} (34)

for all pure states |ψ|\psi\rangle given by (30).

A lower bound (33) of concurrence can be obtained from a function ff satisfying (34) for arbitrary pure states. Nevertheless, it is still a problem to find such function ff. In fact, there are positive maps which can be used as separability criteria, but one has difficulties to use them to obtain lower bounds of concurrence by finding such functions ff. Based on the positive map defined in (8), we construct below new functions ff to obtain new lower bounds of concurrence C(ρ)C(\rho). Setting M=dM=d, L=N=d+1L=N=d+1 and using the (N,M)(N,M)-POVM constructed from the Gell-Mann matrices 35 in Φ\Phi, we have the following theorem, see the proof given in Appendix E.

Theorem 2. For any bipartite quantum state ρH1H2\rho\in H_{1}\otimes H_{2}, the concurrence C(ρ)C(\rho) satisfies

C(ρ)2d(d1)((𝕀dΦz)ρ1),\displaystyle C(\rho)\geq\sqrt{\frac{2}{d(d-1)}}(\|(\mathbb{I}_{d}\otimes\Phi_{z})\rho\|-1), (35)

where 𝕀d\mathbb{I}_{d} is identity operator, Φz\Phi_{z} is given in (8), \|\cdot\| stands for the trace norm.

It has been always a challenging problem to find new separability criteria which detect better entanglement, and new lower bounds of entanglement which are larger than the existing ones, at least for some quantum states. We have presented such separability criterion and lower bounds. We illustrate our results below by a detailed example.

Example 1. Let us consider the state (11). From (35) we have

C(ρ)16((𝕀4Φz)(ρ)1)=126(23q1z124z18+|23q1z124z18|).\displaystyle C(\rho)\geq\sqrt{\frac{1}{6}}(\|(\mathbb{I}_{4}\otimes\Phi_{z})(\rho)\|-1)=\frac{1}{2\sqrt{6}}(\frac{2}{3}q_{1}z-\frac{1}{24}z-\frac{1}{8}+|\frac{2}{3}q_{1}z-\frac{1}{24}z-\frac{1}{8}|). (36)

In 30 , a lower bound of the concurrence is given by

C(ρ)16((𝕀4Φ)(ρ)3)=146(q1q4+|q1q4|).\displaystyle C(\rho)\geq\sqrt{\frac{1}{6}}(\|(\mathbb{I}_{4}\otimes\Phi^{{}^{\prime}})(\rho)\|-3)=\frac{1}{4\sqrt{6}}(q_{1}-q_{4}+|q_{1}-q_{4}|). (37)

Fig. 1 shows the lower bounds of concurrence given in (36) for the state (11) versus parameters zz and q1q_{1}. We see that the lower bounds of concurrence are greater than 0 when 0.2<z10.2<z\leq 1, namely, the entanglement of states (11) are detected in this case. When z=1z=1 and 0.25<q1<10.25<q_{1}<1, the lower bound of concurrence is greater than 0. When z<1z<1, from Fig. 1 one sees the detected entanglement range of ρ\rho and the lower bound of concurrence decreases with zz. When z0.2z\leq 0.2, it can be seen from Fig. 1 that the lower bounds of concurrence becomes 0. The lower bound of (36) reaches the maximum at z=1z=1.

Our lower bounds of concurrence in (36) are better than the lower bounds of concurrence in (37) given in 30 at least for some states. Let us take z=1z=1 and q4=13q1+12q_{4}=-\frac{1}{3}q_{1}+\frac{1}{2}. Then (36) can be written as C(ρ)126(23q116+|23q116|)C(\rho)\geq\frac{1}{2\sqrt{6}}(\frac{2}{3}q_{1}-\frac{1}{6}+|\frac{2}{3}q_{1}-\frac{1}{6}|), while (37) can be written as C(ρ)146(43q112+|43q112|)C(\rho)\geq\frac{1}{4\sqrt{6}}(\frac{4}{3}q_{1}-\frac{1}{2}+|\frac{4}{3}q_{1}-\frac{1}{2}|). From Fig. 2 it is seen that our bound of concurrence (36) detects the entanglement for q1>0.25q_{1}>0.25, while the bound of concurrence in (37) detects entanglement for q1>0.375q_{1}>0.375.

Refer to caption
Figure 1: Lower bounds with respect to the parameters zz and q1q_{1}.
Refer to caption
Figure 2: Solid line: the lower bound given in (36). Dashed line: the lower bound given in (37).

V V. Conclusions and remarks

Based on symmetric measurements we have presented a family of positive and trace-preserving maps. From these maps we have obtained separability criteria which detect better the entanglement of quantum states. We have also constructed a series of entanglement witnesses which include some existing ones as special cases and detect even the entanglement of bound entangled states. We have derived a family of lower bounds of concurrence which are tighter than the related existing ones. Since our approach is based on the symmetric measurements, the entanglement of any known quantum states can be experimentally estimated. Moreover, our results may be also applied to the investigation on multipartite entanglement, and highlight on the detection of entanglement in optimal entanglement manipulations jpa .

VI acknowledgments

We thank referees for their valuable suggestions to improve the manuscript.This work is supported by JCKYS2021604SSJS002, JCKYS2023604SSJS017, G2022180019L, the National Natural Science Foundation of China (NSFC) under Grants 12075159 and 12171044, and the specific research fund of the Innovation Platform for Academicians of Hainan Province under Grant No. YSPTZX202215.

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APPENDIX

VI.1 A. Gell-Mann matrices

For d=4d=4, the Hermitian orthonormal basis is given by the following Gell-Mann matrices:

g01=12(0i00i00000000000),g10=12(0100100000000000),g20=12(0010000010000000),g30=12(0001000000001000),g11=12(1000010000000000),g02=12(00i00000i0000000),g12=12(000000i00i000000),g21=12(0000001001000000),g31=12(0000001001000000),g22=16(1000010000200000),g03=12(000i00000000i000),g13=12(0000000i00000i00),g23=12(00000000000i00i0),g32=12(0000000000010010),g33=123(1000010000100003),\displaystyle\begin{split}g_{01}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&-i&0&0\\ i&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix},\\ g_{10}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0&0\\ 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix},\\ g_{20}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&1&0\\ 0&0&0&0\\ 1&0&0&0\\ 0&0&0&0\end{pmatrix},\\ g_{30}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0&1\\ 0&0&0&0\\ 0&0&0&0\\ 1&0&0&0\end{pmatrix},\\ g_{11}=&\frac{1}{\sqrt{2}}\begin{pmatrix}1&0&0&0\\ 0&-1&0&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix},\end{split}\qquad\begin{split}g_{02}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&-i&0\\ 0&0&0&0\\ i&0&0&0\\ 0&0&0&0\end{pmatrix},\\ g_{12}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0&0\\ 0&0&-i&0\\ 0&i&0&0\\ 0&0&0&0\end{pmatrix},\\ g_{21}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&0\end{pmatrix},\\ g_{31}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&0\end{pmatrix},\\ g_{22}=&\frac{1}{\sqrt{6}}\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&-2&0\\ 0&0&0&0\end{pmatrix},\end{split}\qquad\begin{split}g_{03}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0&-i\\ 0&0&0&0\\ 0&0&0&0\\ i&0&0&0\end{pmatrix},\\ g_{13}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0&0\\ 0&0&0&-i\\ 0&0&0&0\\ 0&i&0&0\end{pmatrix},\\ g_{23}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0&0\\ 0&0&0&0\\ 0&0&0&-i\\ 0&0&i&0\end{pmatrix},\\ g_{32}=&\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&1&0\end{pmatrix},\\ g_{33}=&\frac{1}{2\sqrt{3}}\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&-3\end{pmatrix},\end{split}

and G0=𝕀4/4G_{0}=\mathbb{I}_{4}/\sqrt{4}. For the entanglement estimation with respect to the state (11), we fix the indices of Gα,kG_{\alpha,k} as follows,

G1,1=g01,G1,2=g02,G1,3=g03,G2,1=g10,G2,2=g12,G2,3=g13,G3,1=g20,G3,2=g21,G3,3=g23,G4,1=g30,G4,2=g31,G4,3=g32,G5,1=g11,G5,2=g22,G5,3=g33.\displaystyle\begin{split}&G_{1,1}=g_{01},\quad G_{1,2}=g_{02},\qquad G_{1,3}=g_{03},\\ &G_{2,1}=g_{10},\quad G_{2,2}=g_{12},\qquad G_{2,3}=g_{13},\\ &G_{3,1}=g_{20},\quad G_{3,2}=g_{21},\qquad G_{3,3}=g_{23},\\ &G_{4,1}=g_{30},\quad G_{4,2}=g_{31},\qquad G_{4,3}=g_{32},\\ &G_{5,1}=g_{11},\quad G_{5,2}=g_{22},\qquad G_{5,3}=g_{33}.\end{split} (A1)

For d=3d=3, the Hermitian orthonormal basis is given by the following Gell-Mann matrices:

g01=12(010100000),\displaystyle g_{01}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&0\end{pmatrix},\qquad g10=12(0i0i00000),\displaystyle g_{10}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&-i&0\\ i&0&0\\ 0&0&0\end{pmatrix},
g02=12(001000100),\displaystyle g_{02}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&1\\ 0&0&0\\ 1&0&0\end{pmatrix},\qquad g20=12(00i000i00),\displaystyle g_{20}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&-i\\ 0&0&0\\ i&0&0\end{pmatrix},
g12=12(000001010),\displaystyle g_{12}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0\\ 0&0&1\\ 0&1&0\end{pmatrix},\qquad g21=12(00000i0i0),\displaystyle g_{21}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0\\ 0&0&-i\\ 0&i&0\end{pmatrix},
g11=12(100010000),\displaystyle g_{11}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&0&0\\ 0&-1&0\\ 0&0&0\end{pmatrix},\qquad g22=16(100010002),\displaystyle g_{22}=\frac{1}{\sqrt{6}}\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&-2\end{pmatrix},

and G0=𝕀3/3G_{0}=\mathbb{I}_{3}/\sqrt{3}. For the entanglement witnesses in Example 2, we fix the indices of Gα,kG_{\alpha,k} as follows,

G1,1=g01,G1,2=g10,G2,1=g02,G2,2=g20,G3,1=g12,G3,2=g21,G4,1=g11,G4,2=g22.\begin{split}&G_{1,1}=g_{01},\quad G_{1,2}=g_{10},\qquad G_{2,1}=g_{02},\quad G_{2,2}=g_{20},\\ &G_{3,1}=g_{12},\quad G_{3,2}=g_{21},\qquad G_{4,1}=g_{11},\quad G_{4,2}=g_{22}.\end{split} (A2)

In the Example 4, we take

G1,1=g01,G1,2=g02,G1,3=g10,G1,4=g20,G2,1=g12,G2,2=g21,G2,3=g11,G2,4=g22.\displaystyle\begin{split}&G_{1,1}=g_{01},\quad G_{1,2}=g_{02},\quad G_{1,3}=g_{10},\quad G_{1,4}=g_{20},\\ &G_{2,1}=g_{12},\quad G_{2,2}=g_{21},\quad G_{2,3}=g_{11},\quad G_{2,4}=g_{22}.\end{split} (A3)

VI.2 B. Hermitian orthonormal basis from MUBs

Using the complete set of four mutually unbiased bases in d=3d=3 and the corresponding projectors

E1,1=(100000000),E1,2=(000010000),E1,3=(000000001),E2,1=13(111111111),E2,2=13(1ω2ωω1ω2ω2ω1),E2,3=13(1ωω2ω21ωωω21),E3,1=13(1ω2ω2ω11ω11),E3,2=13(1ω1ω21ω21ω1),E3,3=13(11ω11ωω2ω21),E4,1=13(1ωωω211ω211),E4,2=13(1ω21ω1ω1ω21),E4,3=13(11ω211ω2ωω1),\displaystyle\begin{split}E_{1,1}=&\begin{pmatrix}1&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix},\\ E_{1,2}=&\begin{pmatrix}0&0&0\\ 0&1&0\\ 0&0&0\end{pmatrix},\\ E_{1,3}=&\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&1\end{pmatrix},\end{split}\qquad\begin{split}E_{2,1}=&\frac{1}{3}\begin{pmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{pmatrix},\\ E_{2,2}=&\frac{1}{3}\begin{pmatrix}1&\omega^{2}&\omega\\ \omega&1&\omega^{2}\\ \omega^{2}&\omega&1\end{pmatrix},\\ E_{2,3}=&\frac{1}{3}\begin{pmatrix}1&\omega&\omega^{2}\\ \omega^{2}&1&\omega\\ \omega&\omega^{2}&1\end{pmatrix},\end{split}\qquad\begin{split}E_{3,1}=&\frac{1}{3}\begin{pmatrix}1&\omega^{2}&\omega^{2}\\ \omega&1&1\\ \omega&1&1\end{pmatrix},\\ E_{3,2}=&\frac{1}{3}\begin{pmatrix}1&\omega&1\\ \omega^{2}&1&\omega^{2}\\ 1&\omega&1\end{pmatrix},\\ E_{3,3}=&\frac{1}{3}\begin{pmatrix}1&1&\omega\\ 1&1&\omega\\ \omega^{2}&\omega^{2}&1\end{pmatrix},\end{split}\qquad\begin{split}E_{4,1}=&\frac{1}{3}\begin{pmatrix}1&\omega&\omega\\ \omega^{2}&1&1\\ \omega^{2}&1&1\end{pmatrix},\\ E_{4,2}=&\frac{1}{3}\begin{pmatrix}1&\omega^{2}&1\\ \omega&1&\omega\\ 1&\omega^{2}&1\end{pmatrix},\\ E_{4,3}=&\frac{1}{3}\begin{pmatrix}1&1&\omega^{2}\\ 1&1&\omega^{2}\\ \omega&\omega&1\end{pmatrix},\end{split} (B1)

where ω=exp(2πi/3)\omega=\exp(2\pi i/3), one finds the corresponding Hermitian orthonormal basis:

G1,1=13(1+3)(2300010001+3),\displaystyle G_{1,1}=\frac{1}{\sqrt{3}(1+\sqrt{3})}\begin{pmatrix}-2-\sqrt{3}&0&0\\ 0&1&0\\ 0&0&1+\sqrt{3}\end{pmatrix},\qquad G1,2=13(1+3)(1000230001+3),\displaystyle G_{1,2}=\frac{1}{\sqrt{3}(1+\sqrt{3})}\begin{pmatrix}1&0&0\\ 0&-2-\sqrt{3}&0\\ 0&0&1+\sqrt{3}\end{pmatrix},
G2,1=123(1+3)(0vvv0vvv0),\displaystyle G_{2,1}=\frac{1}{2\sqrt{3}(1+\sqrt{3})}\begin{pmatrix}0&-v^{\ast}&-v\\ -v&0&-v^{\ast}\\ -v^{\ast}&-v&0\end{pmatrix},\qquad G2,2=13(1+3)(0iviviv0iviviv0),\displaystyle G_{2,2}=\frac{1}{\sqrt{3}(1+\sqrt{3})}\begin{pmatrix}0&iv^{\ast}&-iv\\ -iv&0&iv^{\ast}\\ iv^{\ast}&-iv&0\end{pmatrix},
G3,1=123(1+3)(0uivu0vivv0),\displaystyle G_{3,1}=\frac{1}{2\sqrt{3}(1+\sqrt{3})}\begin{pmatrix}0&u^{\ast}&iv^{\ast}\\ u&0&-v^{\ast}\\ -iv&-v&0\end{pmatrix},\qquad G3,2=13(1+3)(0uvu0ivviv0),\displaystyle G_{3,2}=\frac{1}{\sqrt{3}(1+\sqrt{3})}\begin{pmatrix}0&u&-v^{\ast}\\ u^{\ast}&0&iv^{\ast}\\ -v&-iv&0\end{pmatrix},
G4,1=123(1+3)(0uivu0vivv0),\displaystyle G_{4,1}=\frac{1}{2\sqrt{3}(1+\sqrt{3})}\begin{pmatrix}0&u&-iv\\ u^{\ast}&0&-v\\ iv^{\ast}&-v^{\ast}&0\end{pmatrix},\qquad G4,2=13(1+3)(0uvu0ivviv0),\displaystyle G_{4,2}=\frac{1}{\sqrt{3}(1+\sqrt{3})}\begin{pmatrix}0&u^{\ast}&-v\\ u&0&-iv\\ -v^{\ast}&iv^{\ast}&0\end{pmatrix},

and G0=𝕀/3G_{0}=\mathbb{I}/\sqrt{3}, where u=(1i)(1+3)u=(1-i)(1+\sqrt{3}) and v=2+3+iv=2+\sqrt{3}+i. The entanglement witnesses in Example 3 is given by (29) with GμG_{\mu} grouped in the following way, {G1,G2,G3}={G1,2,G2,1,G2,2}\{G_{1},G_{2},G_{3}\}=\{G_{1,2},G_{2,1},G_{2,2}\} and {G4,G5,G6,G7,G8}={G1,1,G3,1,G3,2,G4,1,G4,2}\{G_{4},G_{5},G_{6},G_{7},G_{8}\}=\{G_{1,1},G_{3,1},G_{3,2},G_{4,1},G_{4,2}\}.

VI.3 C. Calculation process of Section II

By direct computation,

(𝕀4Φz)(ρ)=12[A16q1z16q1z16q1zBCDD16q1zA16q1z16q1zBCCD16q1z16q1zA16q1zBBCD16q1z16q1z16q1zA],(\mathbb{I}_{4}\otimes\Phi_{z})(\rho)=\frac{1}{2}\left[\begin{array}[]{c c c c|c c c c|c c c c|c c c c}A&\cdot&\cdot&\cdot&\cdot&-\frac{1}{6}q_{1}z&\cdot&\cdot&\cdot&\cdot&-\frac{1}{6}q_{1}z&\cdot&\cdot&\cdot&\cdot&-\frac{1}{6}q_{1}z\\ \cdot&B&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&C&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot&D&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \hline\cr\cdot&\cdot&\cdot&\cdot&D&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ -\frac{1}{6}q_{1}z&\cdot&\cdot&\cdot&\cdot&A&\cdot&\cdot&\cdot&\cdot&-\frac{1}{6}q_{1}z&\cdot&\cdot&\cdot&\cdot&-\frac{1}{6}q_{1}z\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&B&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&C&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \hline\cr\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&C&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&D&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ -\frac{1}{6}q_{1}z&\cdot&\cdot&\cdot&\cdot&-\frac{1}{6}q_{1}z&\cdot&\cdot&\cdot&\cdot&A&\cdot&\cdot&\cdot&\cdot&-\frac{1}{6}q_{1}z\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&B&\cdot&\cdot&\cdot&\cdot\\ \hline\cr\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&B&\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&C&\cdot&\cdot\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&D&\cdot\\ -\frac{1}{6}q_{1}z&\cdot&\cdot&\cdot&\cdot&-\frac{1}{6}q_{1}z&\cdot&\cdot&\cdot&\cdot&-\frac{1}{6}q_{1}z&\cdot&\cdot&\cdot&\cdot&A\end{array}\right],

where

A\displaystyle A =z(38q1+524q4+524q3+524q2+1)+18+54z,\displaystyle=-z(\frac{3}{8}q_{1}+\frac{5}{24}q_{4}+\frac{5}{24}q_{3}+\frac{5}{24}q_{2}+1)+\frac{1}{8}+\frac{5}{4}z,
B\displaystyle B =z(524q1+38q4+524q3+524q2+1)+18+54z,\displaystyle=-z(\frac{5}{24}q_{1}+\frac{3}{8}q_{4}+\frac{5}{24}q_{3}+\frac{5}{24}q_{2}+1)+\frac{1}{8}+\frac{5}{4}z,
C\displaystyle C =z(524q1+524q4+38q3+524q2+1)+18+54z,\displaystyle=-z(\frac{5}{24}q_{1}+\frac{5}{24}q_{4}+\frac{3}{8}q_{3}+\frac{5}{24}q_{2}+1)+\frac{1}{8}+\frac{5}{4}z,
D\displaystyle D =z(524q1+524q4+524q3+38q2+1)+18+54z.\displaystyle=-z(\frac{5}{24}q_{1}+\frac{5}{24}q_{4}+\frac{5}{24}q_{3}+\frac{3}{8}q_{2}+1)+\frac{1}{8}+\frac{5}{4}z.

We have the following set of eigenvalues of (𝕀4Φz)(ρ)(\mathbb{I}_{4}\otimes\Phi_{z})(\rho): {12(A12q1z),12(A+16q1z),12(A+16q1z),12(A+16q1z),12B,12B,12B,12B,12C,12C,12C,12C,12D,12D,12D,12D}\{\frac{1}{2}(A-\frac{1}{2}q_{1}z),\frac{1}{2}(A+\frac{1}{6}q_{1}z),\frac{1}{2}(A+\frac{1}{6}q_{1}z),\frac{1}{2}(A+\frac{1}{6}q_{1}z),\frac{1}{2}B,\frac{1}{2}B,\frac{1}{2}B,\frac{1}{2}B,\frac{1}{2}C,\frac{1}{2}C,\frac{1}{2}C,\frac{1}{2}C,\frac{1}{2}D,\frac{1}{2}D,\frac{1}{2}D,\frac{1}{2}D\}. When 0<z10<z\leq 1, the negative minimum eigenvalue, 12(A12q1z)<0\frac{1}{2}(A-\frac{1}{2}q_{1}z)<0, implies that z16q1z+3<0z-16q_{1}z+3<0. We get q1>116+316zq_{1}>\frac{1}{16}+\frac{3}{16z}. From 0q110\leq q_{1}\leq 1 we get z[15,1]z\in[\frac{1}{5},1]. Therefore, our criterion detects the entanglement of ρ\rho for 0.25<q1<10.25<q_{1}<1.

VI.4 D. Examples about Entanglement witnesses

Example 2. Let us take N=4N=4 and M=3M=3, fix the operator basis Gα,kG_{\alpha,k} to be the Gell-Mann matrices (see Appendix A). For L=1L=1, take

𝒪(α)=(100010001)foranyα[N].\displaystyle\mathcal{O}^{(\alpha)}=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\ \text{for}\ \text{any}\ \alpha\in[N]. (D1)

The corresponding entanglement witnesses have the form,

W~1=(3+1)2[2z+23z3zz+2z+2z+23z2z+23zz+2z+2z+23z3z2z+2].\displaystyle\mathaccent 869{W}_{1}=(\sqrt{3}+1)^{2}\left[\begin{array}[]{c c c|c c c|c c c}2z+2&\cdot&\cdot&\cdot&-3z&\cdot&\cdot&\cdot&3z\\ \cdot&-z+2&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&-z+2&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \hline\cr\cdot&\cdot&\cdot&-z+2&\cdot&\cdot&\cdot&\cdot&\cdot\\ -3z&\cdot&\cdot&\cdot&2z+2&\cdot&\cdot&\cdot&3z\\ \cdot&\cdot&\cdot&\cdot&\cdot&-z+2&\cdot&\cdot&\cdot\\ \hline\cr\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&-z+2&\cdot&\cdot\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&-z+2&\cdot\\ 3z&\cdot&\cdot&\cdot&3z&\cdot&\cdot&\cdot&2z+2\end{array}\right]. (D11)

When z=1z=-1, it is verified that the entanglement of the following state can be detected,

ρ1=127[766111676111667].\displaystyle\rho_{1}=\frac{1}{27}\left[\begin{array}[]{c c c|c c c|c c c}7&\cdot&\cdot&\cdot&6&\cdot&\cdot&\cdot&6\\ \cdot&1&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&1&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ \hline\cr\cdot&\cdot&\cdot&1&\cdot&\cdot&\cdot&\cdot&\cdot\\ 6&\cdot&\cdot&\cdot&7&\cdot&\cdot&\cdot&6\\ \cdot&\cdot&\cdot&\cdot&\cdot&1&\cdot&\cdot&\cdot\\ \hline\cr\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&1&\cdot&\cdot\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&1&\cdot\\ 6&\cdot&\cdot&\cdot&6&\cdot&\cdot&\cdot&7\end{array}\right]. (D21)

When z=1z=1, it is the witness constructed in 24 which can not detect the entanglement of the state ρ1\rho_{1}.

Example 3. Let M=2M=2. Instead of the Gell-Mann matrices, we take the (N,2)(N,2)-POVM constructed from the orthonormal Hermitian basis presented in Appendix B. For N=7N=7 and L=4L=4, the corresponding witnesses W~2\mathaccent 869{W}_{2} are given by

W~2=16[4(1z)4zAz4Azz2(2+z)Bz4CzEz4Fz2(2+z)CzDz4Gz7z44BzCz2(2+z)4CzHzz4Dz4(1z)Dz4ziAzCz42(2+z)MzNz44EzGz4DzMz2(2+z)Az47zCz4Nz2(2+z)zFz4Hzzi44(1z)],\displaystyle\mathaccent 869{W}_{2}=\frac{1}{6}\left[\begin{array}[]{c c c|c c c|c c c}4(1-z)&\cdot&\cdot&4&z&Az&4&A^{*}z&z\\ \cdot&2(2+z)&\cdot&Bz&4&Cz&Ez&4&Fz\\ \cdot&\cdot&2(2+z)&C^{*}z&Dz&4&Gz&-7z&4\\ \hline\cr 4&B^{*}z&Cz&2(2+z)&\cdot&\cdot&4&C^{*}z&Hz\\ z&4&D^{*}z&\cdot&4(1-z)&\cdot&Dz&4&zi\\ A^{*}z&C^{*}z&4&\cdot&\cdot&2(2+z)&Mz&Nz&4\\ \hline\cr 4&E^{*}z&G^{*}z&4&D^{*}z&M^{*}z&2(2+z)&\cdot&\cdot\\ Az&4&-7z&Cz&4&N^{*}z&\cdot&2(2+z)&\cdot\\ z&F^{*}z&4&H^{*}z&-zi&4&\cdot&\cdot&4(1-z)\end{array}\right], (D31)

where

A\displaystyle A =12(3i),B=12(335i),C=(823i),\displaystyle=\frac{1}{2}(\sqrt{3}-i),\ B=\frac{1}{2}(3\sqrt{3}-5i),\ C=-(8-2\sqrt{3}i),
D\displaystyle D =12(53i),E=(8+23)i,F=12(53+3i),\displaystyle=\frac{1}{2}(5\sqrt{3}-i),\ E=-(8+2\sqrt{3})i,\ F=-\frac{1}{2}(5\sqrt{3}+3i),
G\displaystyle G =12(73+3i),H=12(33+11i),M=(73i)\displaystyle=\frac{1}{2}(7\sqrt{3}+3i),\ H=\frac{1}{2}(3\sqrt{3}+11i),\ M=-(7-\sqrt{3}i)
N\displaystyle N =12(3+3i).\displaystyle=-\frac{1}{2}(\sqrt{3}+3i).

When z=1z=-1, it can detect the entanglement of the following state,

ρ2=175[722944944494272494449449227].\rho_{2}=\frac{1}{75}\left[\begin{array}[]{c c c|c c c|c c c}7&\cdot&\cdot&\cdot&2&\cdot&\cdot&\cdot&2\\ \cdot&9&\cdot&\cdot&\cdot&-4&-4&\cdot&\cdot\\ \cdot&\cdot&9&-4&\cdot&\cdot&\cdot&-4&\cdot\\ \hline\cr\cdot&\cdot&-4&9&\cdot&\cdot&\cdot&-4&\cdot\\ 2&\cdot&\cdot&\cdot&7&\cdot&\cdot&\cdot&2\\ \cdot&-4&\cdot&\cdot&\cdot&9&-4&\cdot&\cdot\\ \hline\cr\cdot&-4&\cdot&\cdot&\cdot&-4&9&\cdot&\cdot\\ \cdot&\cdot&-4&-4&\cdot&\cdot&\cdot&9&\cdot\\ 2&\cdot&\cdot&\cdot&2&\cdot&\cdot&\cdot&7\end{array}\right]. (D32)

For z=1z=1, these witnesses reduce to the one given in 24 , which can not detect the entanglement of the state ρ2\rho_{2}.

It is well-known that indecomposable witness is a very important kind of entanglement witnesses, but difficult to be constructed. A witness WW is decomposable if it can be written as W=A+BΓW=A+B^{\Gamma} with AA and BB being positive operators and Γ=𝕀T\Gamma=\mathbb{I}\otimes T denoting a partial transpose. Otherwise the WW is indecomposable. Next we give an example of indecomposable witnesses obtained from symmetric measurements.

Example 4. Consider the (1,5)(1,5)-POVM constructed from the orthonormal Hermitian operator basis of the Gell-Mann matrices. Let L=1L=1 and

𝒪(1)=(0000110000010000010000010).\displaystyle\mathcal{O}^{(1)}=\begin{pmatrix}0&0&0&0&1\\ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\end{pmatrix}. (D33)

From (20) we get the following entanglement witnesses,

W~3=16[4BzCzDzBz4Az30zi430ziAzAz30zi4Bz4Cz430ziAz4Dz4Bz4],\displaystyle\mathaccent 869{W}^{\prime}_{3}=\frac{1}{6}\left[\begin{array}[]{c c c|c c c|c c c}4&\cdot&\cdot&\cdot&B^{\ast}z&C^{\ast}z&\cdot&D^{\ast}z&B^{\ast}z\\ \cdot&4&\cdot&A^{\ast}z&\cdot&\cdot&-30zi&\cdot&\cdot\\ \cdot&\cdot&4&30zi&\cdot&\cdot&-A^{\ast}z&\cdot&\cdot\\ \hline\cr\cdot&Az&-30zi&4&\cdot&\cdot&\cdot&\cdot&\cdot\\ Bz&\cdot&\cdot&\cdot&4&\cdot&\cdot&\cdot&\cdot\\ Cz&\cdot&\cdot&\cdot&\cdot&4&\cdot&\cdot&\cdot\\ \hline\cr\cdot&30zi&-A^{\ast}z&\cdot&\cdot&\cdot&4&\cdot&\cdot\\ Dz&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&4&\cdot\\ Bz&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&4\end{array}\right], (D43)

where

A\displaystyle A =15(1i)(2i+5),\displaystyle=15(1-i)(2-i+\sqrt{5}),
B\displaystyle B =15(1i)(2+i+5),\displaystyle=15(1-i)(2+i+\sqrt{5}),
C\displaystyle C =305(2+5),\displaystyle=-30\sqrt{5}(2+\sqrt{5}),
D\displaystyle D =30(12i)(2+5).\displaystyle=30(1-2i)(2+\sqrt{5}).

Consider the following state,

ρ3=181[9794i94i4+i99794+i999].\displaystyle\rho_{3}=\frac{1}{81}\left[\begin{array}[]{c c c|c c c|c c c}9&\cdot&\cdot&\cdot&\cdot&-7&\cdot&\cdot&\cdot\\ \cdot&9&\cdot&4-i&\cdot&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&9&\cdot&\cdot&\cdot&-4-i&\cdot&\cdot\\ \hline\cr\cdot&4+i&\cdot&9&\cdot&\cdot&\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot&\cdot&9&\cdot&\cdot&\cdot&\cdot\\ -7&\cdot&\cdot&\cdot&\cdot&9&\cdot&\cdot&\cdot\\ \hline\cr\cdot&\cdot&-4+i&\cdot&\cdot&\cdot&9&\cdot&\cdot\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&9&\cdot\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&\cdot&9\end{array}\right]. (D53)

It is directly verified that ρ3\rho_{3} is PPT state. Take z=1z=-1. From (D43) we have that the state ρ3\rho_{3} is entanglement. Hence the entanglement witness (D43) is an indecomposable witness when z=1z=-1. For z=1z=1, these witnesses reduce to the one given in 24 and one has Tr(W~3ρ2)0\text{Tr}(\mathaccent 869{W}^{\prime}_{3}\rho_{2})\geq 0, i.e., it cannot detect the entanglement of ρ3\rho_{3}.

From the above examples, we see that the entanglement witnesses we presented cover the ones in 24 , and can detect more entangled states including bound entangled ones.

VI.5 E. Proof of Theorem 2

Let f(|ψψ|)=(𝕀dΦz)|ψψ|1f(|\psi\rangle\langle\psi|)=\|(\mathbb{I}_{d}\otimes\Phi_{z})|\psi\rangle\langle\psi|\|-1. Obviously f(|ψψ|)f(|\psi\rangle\langle\psi|) is convex as the trace norm is convex. What we need to prove is that for any pure state in Schmidt form (30), the inequality (34) holds.

Since the trace norm does change under local coordinate transformation, we take |ψ=(α1,0,,0,0,α2,,0,0,0,α3,,0,,0,,0,αd)T|\psi\rangle=(\alpha_{1},0,\ldots,0,0,\alpha_{2},\ldots,0,0,0,\alpha_{3},\ldots,0,\ldots,0,\ldots,0,\alpha_{d})^{T}, where T denotes transpose and the Schmidt coefficients satisfy 0α1,α2,α3,,αd1,i=1dαi2=1.0\leq\alpha_{1},\alpha_{2},\alpha_{3},\ldots,\alpha_{d}\leq 1,\ \sum\limits_{i=1}^{d}\alpha_{i}^{2}=1.

By direct computation, we have

(𝕀dΦz)(|ψψ|)=\displaystyle(\mathbb{I}_{d}\otimes\Phi_{z})(|\psi\rangle\langle\psi|)=
1d(d1)[(d1+z)α12dzα1α2dzα1αddzα1α2(d1+z)α22dzα2αddzα1αddzα2αd(d1+z)αd2](d1+z)α12Id1(d1+z)αd2Id1,\displaystyle\frac{1}{d(d-1)}\left[\begin{array}[]{c c c c}(d-1+z)\alpha_{1}^{2}&-dz\alpha_{1}\alpha_{2}&\cdots&-dz\alpha_{1}\alpha_{d}\\ -dz\alpha_{1}\alpha_{2}&(d-1+z)\alpha_{2}^{2}&\cdots&-dz\alpha_{2}\alpha_{d}\\ \vdots&\vdots&\ddots&\vdots\\ -dz\alpha_{1}\alpha_{d}&-dz\alpha_{2}\alpha_{d}&\cdots&(d-1+z)\alpha_{d}^{2}\end{array}\right]\oplus(d-1+z)\alpha_{1}^{2}I_{d-1}\oplus\cdots\oplus(d-1+z)\alpha_{d}^{2}I_{d-1},

The matrix (𝕀dΦz)(|ψψ|)(\mathbb{I}_{d}\otimes\Phi_{z})(|\psi\rangle\langle\psi|) has dd singular values with the multiplicity d1d-1: 1d(d1)(d1+z)α12,1d(d1)(d1+z)α22,,1d(d1)(d1+z)αd2\frac{1}{d(d-1)}(d-1+z)\alpha_{1}^{2},\,\frac{1}{d(d-1)}(d-1+z)\alpha_{2}^{2},\ldots,\frac{1}{d(d-1)}(d-1+z)\alpha_{d}^{2}, and the remaining dd values are the singular values of the following matrix PP:

P\displaystyle P =1d(d1)[(d1)(1z)α12dzα1α2dzα1αddzα1α2(d1)(1z)α22dzα2αddzα1αddzα2αd(d1)(1z)αd2]\displaystyle=\frac{1}{d(d-1)}\left[\begin{array}[]{c c c c}(d-1)(1-z)\alpha_{1}^{2}&-dz\alpha_{1}\alpha_{2}&\cdots&-dz\alpha_{1}\alpha_{d}\\ -dz\alpha_{1}\alpha_{2}&(d-1)(1-z)\alpha_{2}^{2}&\cdots&-dz\alpha_{2}\alpha_{d}\\ \vdots&\vdots&\ddots&\vdots\\ -dz\alpha_{1}\alpha_{d}&-dz\alpha_{2}\alpha_{d}&\cdots&(d-1)(1-z)\alpha_{d}^{2}\end{array}\right]
=dzd(d1)[tα12α1α2α1αdα1α2tα22α2αdα1αdα2αdtαd2]dzd(d1)H,\displaystyle=\frac{dz}{d(d-1)}\left[\begin{array}[]{c c c c}t\alpha_{1}^{2}&-\alpha_{1}\alpha_{2}&\cdots&-\alpha_{1}\alpha_{d}\\ -\alpha_{1}\alpha_{2}&t\alpha_{2}^{2}&\cdots&-\alpha_{2}\alpha_{d}\\ \vdots&\vdots&\ddots&\vdots\\ -\alpha_{1}\alpha_{d}&-\alpha_{2}\alpha_{d}&\cdots&t\alpha_{d}^{2}\end{array}\right]\triangleq\frac{dz}{d(d-1)}H,

where t=(d1)(1z)dzt=\frac{(d-1)(1-z)}{dz}. As PP is Hermitian and real, its singular values are simply given by the square roots of the eigenvalues of P2P^{2}. In fact we need only to consider the absolute values of the eigenvalues of PP. The eigenpolynomial equation of HH is

h(x)=|xIdH|=xdtxd1+(t1)(t+1)(i<jαi2αj2)xd2(t2)(t+1)2(i<j<kαi2αj2αk2)xd3+(t3)(t+1)3(i1<i2<i3<i4αi12αi22αi32αi42)xd4++(1)d2(td+3)(t+1)d3(i1<i2<<id2αi12αi22αid22)x2+(1)d1(td+2)(t+1)d2(i1<i2<<id1αi12αi22αid12)x+(1)d(td+1)(t+1)d1(α12α22αd2)=0.\displaystyle\begin{split}h(x)=|xI_{d}-H|&=x^{d}-tx^{d-1}+(t-1)(t+1)(\sum_{i<j}\alpha_{i}^{2}\alpha_{j}^{2})x^{d-2}-(t-2)(t+1)^{2}(\sum_{i<j<k}\alpha_{i}^{2}\alpha_{j}^{2}\alpha_{k}^{2})x^{d-3}\\ &+(t-3)(t+1)^{3}(\sum_{i_{1}<i_{2}<i_{3}<i_{4}}\alpha_{i_{1}}^{2}\alpha_{i_{2}}^{2}\alpha_{i_{3}}^{2}\alpha_{i_{4}}^{2})x^{d-4}+\cdots\\ &+(-1)^{d-2}(t-d+3)(t+1)^{d-3}(\sum_{i_{1}<i_{2}<\cdots<i_{d-2}}\alpha_{i_{1}}^{2}\alpha_{i_{2}}^{2}\cdots\alpha_{i_{d-2}}^{2})x^{2}\\ &+(-1)^{d-1}(t-d+2)(t+1)^{d-2}(\sum_{i_{1}<i_{2}<\cdots<i_{d-1}}\alpha_{i_{1}}^{2}\alpha_{i_{2}}^{2}\cdots\alpha_{i_{d-1}}^{2})x\\ &+(-1)^{d}(t-d+1)(t+1)^{d-1}(\alpha_{1}^{2}\alpha_{2}^{2}\cdots\alpha_{d}^{2})\\ &=0.\end{split} (E2)

Let x1,x2,x3,,xdx_{1},x_{2},x_{3},\ldots,x_{d} denote the dd roots of (E2). By using the relations between the roots and the coefficients of the polynomial equation, one has

i=1dxi=t,i=1dxi=(td+1)(t+1)d1(α12α22αd2).\sum_{i=1}^{d}x_{i}=t,\ \prod_{i=1}^{d}x_{i}=(t-d+1)(t+1)^{d-1}(\alpha_{1}^{2}\alpha_{2}^{2}\cdots\alpha_{d}^{2}). (E3)

From (VI.5) and that i=1dαi2=1\sum\limits_{i=1}^{d}\alpha_{i}^{2}=1, the inequality (34) that needs to be proved now has the form,

f(|ψψ|)=(IdΦz)|ψψ|1=dzd(d1)i=1d|xi|+d1d(d1)(d1+z)12(i<jαiαj).\begin{split}f(|\psi\rangle\langle\psi|)&=\|(I_{d}\otimes\Phi_{z})|\psi\rangle\langle\psi|\|-1\\ &=\frac{dz}{d(d-1)}\sum_{i=1}^{d}|x_{i}|+\frac{d-1}{d(d-1)}(d-1+z)-1\\ &\leq 2(\sum_{i<j}\alpha_{i}\alpha_{j}).\end{split} (E4)

Next, consider the eigenpolynomial equation (E2). Set β=i=1dαi2\beta=\prod_{i=1}^{d}\alpha_{i}^{2}. Since t=(d1)(1z)dzt=\frac{(d-1)(1-z)}{dz}, when z(0,1]z\in(0,1], we get t[0,+)t\in[0,+\infty), and when z(1,0]z\in(-1,0], we have t(,(22d))t\in(-\infty,-(2-\frac{2}{d})).

(II) When td2t\geq d-2:

(a) If β=0\beta=0, then h(0)=0h(0)=0, where 0 is an eigenvalue of HH. From the derivative of h(x)h(x) with respect to xx,

h(x)=dxd1t(d1)xd2+(d2)(t1)(t+1)(i<jαi2αj2)xd3(d3)(t2)(t+1)2(i<j<kαi2αj2αk2)xd4++2(1)d2(td+3)(t+1)d3×(i1<i2<<id2αi12αi22αid22)x+(1)d1(td+2)(t+1)d2×(i1<i2<<id1αi12αi22αid12),\displaystyle\begin{split}h^{{}^{\prime}}(x)&=dx^{d-1}-t(d-1)x^{d-2}\\ &+(d-2)(t-1)(t+1)(\sum_{i<j}\alpha_{i}^{2}\alpha_{j}^{2})x^{d-3}\\ &-(d-3)(t-2)(t+1)^{2}(\sum_{i<j<k}\alpha_{i}^{2}\alpha_{j}^{2}\alpha_{k}^{2})x^{d-4}\\ &+\cdots+2(-1)^{d-2}(t-d+3)(t+1)^{d-3}\\ &\times(\sum_{i_{1}<i_{2}<\cdots<i_{d-2}}\alpha_{i_{1}}^{2}\alpha_{i_{2}}^{2}\cdots\alpha_{i_{d-2}}^{2})x\\ &+(-1)^{d-1}(t-d+2)(t+1)^{d-2}\\ &\times(\sum_{i_{1}<i_{2}<\cdots<i_{d-1}}\alpha_{i_{1}}^{2}\alpha_{i_{2}}^{2}\cdots\alpha_{i_{d-1}}^{2}),\end{split} (E5)

we have that if dd is even, h(x)<0h^{{}^{\prime}}(x)<0 when x<0x<0. Therefore, h(x)h(x) is a monotonically decreasing function for x<0x<0. Taking into account that h(0)=0h(0)=0, we see that there exist no negative roots of (E2) in this case. When dd is odd, h(x)h(x) is a monotonically increasing function for x<0x<0. There are also no negative roots of (E2).

The inequality (E4) that needs to be proved now has the form

dzd(d1)i=1dxi+d1d(d1)(d1+z)12(i<jαiαj).\frac{dz}{d(d-1)}\sum_{i=1}^{d}x_{i}+\frac{d-1}{d(d-1)}(d-1+z)-1\leq 2(\sum_{i<j}\alpha_{i}\alpha_{j}). (E6)

According to the relations in (E3) and t=(d1)(1z)dzt=\frac{(d-1)(1-z)}{dz}, the left hand of the inequality (E6) is zero. Hence the inequality (E4) holds.

(b) If β0\beta\neq 0, we have h(0)=(1)d(td+1)(t+1)d1(α12α22αd2)h(0)=(-1)^{d}(t-d+1)(t+1)^{d-1}(\alpha_{1}^{2}\alpha_{2}^{2}\cdots\alpha_{d}^{2}).

When t(d1,+)t\in(d-1,+\infty), we have h(0)>0h(0)>0.

If dd is even, since h(x)h(x) is a monotonically decreasing function for x<0x<0, there exist no negative roots of (E2) in this case.

If dd is odd, i=1dxi=(td+1)(t+1)d1(α12α22αd2)>0\prod_{i=1}^{d}x_{i}=(t-d+1)(t+1)^{d-1}(\alpha_{1}^{2}\alpha_{2}^{2}\cdots\alpha_{d}^{2})>0. then (E2) has no negative roots or even number negative roots. Since h(x)h(x) is monotonically increasing when x<0x<0, it has at most one negative root. Therefore, the eigenpolynomial equation (E2) has no negative roots.

This case is similar to (a) and can be shown to satisfy (E4).

When t[d2,d1)t\in[d-2,d-1), we have i=1dxi=(td+1)(t+1)d1(α12α22αd2)<0\prod_{i=1}^{d}x_{i}=(t-d+1)(t+1)^{d-1}(\alpha_{1}^{2}\alpha_{2}^{2}\cdots\alpha_{d}^{2})<0. Therefore, there exists at least one negative root, say x1<0x_{1}<0, such that h(x1)=0h(x_{1})=0.

If dd is even, then h(0)<0h(0)<0 and h(x)h(x) is a monotonically decreasing function when x<0x<0. Thus, x1<0x_{1}<0 is the only negative root. Hence the inequality (E4) needing to be proved becomes

dzd(d1)(i=2dxix1)+d1d(d1)(d1+z)12(i<jαiαj).\frac{dz}{d(d-1)}(\sum_{i=2}^{d}x_{i}-x_{1})+\frac{d-1}{d(d-1)}(d-1+z)-1\leq 2(\sum_{i<j}\alpha_{i}\alpha_{j}). (E7)

From (E3) we only need to prove that x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}). From the definition of h(x)h(x), we have h(d1z(i<jαiαj))=|d1z(i<jαiαj)IdH|=|d1z(i<jαiαj)Id+H|0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=|-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}-H|=|\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}+H|\geq 0, where in the last step the property of the diagonally dominant matrix d1z(i<jαiαj)Id+H\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}+H has been used. Since h(x1)=0h(d1z(i<jαiαj)h(x_{1})=0\leq h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}) and h(x)h(x) is a monotonically decreasing function when x<0x<0, we have that x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}).

If dd is odd, then h(0)>0h(0)>0 and h(x)h(x) is a monotonically increasing function when x<0x<0. Similarly, h(x)h(x) only has one negative root. Hence, we still only need to prove the inequality (E7). From (E3) we need to prove that x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}). From the definition of h(x)h(x), we have h(d1z(i<jαiαj))=|d1z(i<jαiαj)IdH|=|d1z(i<jαiαj)Id+H|0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=|-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}-H|=-|\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}+H|\leq 0, where in the last step the property of the diagonally dominant matrix d1z(i<jαiαj)Id+H\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}+H has been used. Since h(x1)=0h(d1z(i<jαiαj)h(x_{1})=0\geq h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}) and h(x)h(x) is a monotonically increasing function when x<0x<0, we have that x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}).

(IIII) When t[d3,d2)t\in[d-3,d-2):

Set

p0=1,p1=t,p2=(t1)(t+1)(i<jαi2αj2),p3=(t2)(t+1)2(i<j<kαi2αj2αk2),p4=(t3)(t+1)3(i1<i2<i3<i4αi12αi22αi32αi42),pd2=(1)d2(td+3)(t+1)d3(i1<i2<<id2αi12αi22αid22),pd1=(1)d1(td+2)(t+1)d2(i1<i2<<id1αi12αi22αid12),pd=(1)d(td+1)(t+1)d1(α12α22αd2).\begin{split}p_{0}&=1,\\ p_{1}&=-t,\\ p_{2}&=(t-1)(t+1)(\sum_{i<j}\alpha_{i}^{2}\alpha_{j}^{2}),\\ p_{3}&=-(t-2)(t+1)^{2}(\sum_{i<j<k}\alpha_{i}^{2}\alpha_{j}^{2}\alpha_{k}^{2}),\\ p_{4}&=(t-3)(t+1)^{3}(\sum_{i_{1}<i_{2}<i_{3}<i_{4}}\alpha_{i_{1}}^{2}\alpha_{i_{2}}^{2}\alpha_{i_{3}}^{2}\alpha_{i_{4}}^{2}),\\ \vdots\\ p_{d-2}&=(-1)^{d-2}(t-d+3)(t+1)^{d-3}(\sum_{i_{1}<i_{2}<\cdots<i_{d-2}}\alpha_{i_{1}}^{2}\alpha_{i_{2}}^{2}\cdots\alpha_{i_{d-2}}^{2}),\\ p_{d-1}&=(-1)^{d-1}(t-d+2)(t+1)^{d-2}(\sum_{i_{1}<i_{2}<\cdots<i_{d-1}}\alpha_{i_{1}}^{2}\alpha_{i_{2}}^{2}\cdots\alpha_{i_{d-1}}^{2}),\\ p_{d}&=(-1)^{d}(t-d+1)(t+1)^{d-1}(\alpha_{1}^{2}\alpha_{2}^{2}\cdots\alpha_{d}^{2}).\end{split} (E8)

If ρ=|ψψ|\rho=|\psi\rangle\langle\psi| is an entangled pure state, there are at most d2d-2 Schmidt coefficients that are zero. We can assume that:

(a) If β0\beta\neq 0, except that pd2p_{d-2} has the same sign as pd1p_{d-1}, one has p0>0,p1<0,p2>0,p3<0,p_{0}>0,\,p_{1}<0,\,p_{2}>0,\,p_{3}<0,\cdots. The sign of the polynomial coefficients {pi}i=0d\{p_{i}\}_{i=0}^{d} changes V({pi}i=0d)=d1V(\{p_{i}\}_{i=0}^{d})=d-1 times. By the Descartes rule of signs for the polynomial which only has real roots 38 , there are V({pi}i=0d)=d1V(\{p_{i}\}_{i=0}^{d})=d-1 positive roots of h(x)h(x). Since there is no zero root of h(x)h(x), we have that there is only one negative root of h(x)h(x), say x1<0x_{1}<0, such that h(x1)=0h(x_{1})=0. Therefore, we still only need to prove the inequality x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}).

When dd is even, h(d1z(i<jαiαj))=|d1z(i<jαiαj)IdH|=|d1z(i<jαiαj)Id+H|0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=|-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}-H|=|\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}+H|\geq 0. If h(d1z(i<jαiαj))=0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=0, x1=d1z(i<jαiαj)x_{1}=-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}) since h(x)h(x) only has one negative root. If h(d1z(i<jαiαj))>0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))>0, let us suppose x1<d1z(i<jαiαj)<0x_{1}<-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})<0. Because h(0)<0h(0)<0 and h(x)h(x) is continuous, by zero point theorem, there exists another root between d1z(i<jαiαj)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}) and 0, which is contradict with the fact that h(x)h(x) has only one negative root. Hence, x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}).

When dd is odd, h(d1z(i<jαiαj))=|d1z(i<jαiαj)IdH|=|d1z(i<jαiαj)Id+H|0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=|-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}-H|=-|\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}+H|\leq 0. If h(d1z(i<jαiαj))=0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=0, x1=d1z(i<jαiαj)x_{1}=-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}) since h(x)h(x) only has one negative root. If h(d1z(i<jαiαj))<0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))<0, suppose x1<d1z(i<jαiαj)<0x_{1}<-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})<0. Because h(0)>0h(0)>0 and h(x)h(x) is continuous, by zero point theorem there exists another root between d1z(i<jαiαj)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}) and 0, which is contradict with the fact that h(x)h(x) only has one negative root. Hence, x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}).

(b) If β=0\beta=0, we set α1==αK=0\alpha_{1}=\ldots=\alpha_{K}=0 and αK+1,,αd0\alpha_{K+1},\ldots,\alpha_{d}\neq 0, where 1Kd21\leq K\leq d-2. Then pdK+1==pd=0p_{d-K+1}=\ldots=p_{d}=0, and there exist KK zero roots of h(x)h(x). The sign of the polynomial coefficients V({pi}i=0d)V(\{p_{i}\}_{i=0}^{d}) changes V({pi}i=0d)=dKordK1V(\{p_{i}\}_{i=0}^{d})=d-K\,or\,d-K-1 times. Then there is either no negative roots or only one negative root of h(x)h(x). The case that h(x)h(x) has no negative roots can be proved as the case (II) (a). When h(x)h(x) has only one negative root, say x1<0x_{1}<0, such that h(x1)=0h(x_{1})=0, we still only need to prove x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}).

When dd is even, h(d1z(i<jαiαj))=|d1z(i<jαiαj)IdH|=|d1z(i<jαiαj)Id+H|0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=|-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}-H|=|\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}+H|\geq 0. If h(d1z(i<jαiαj))=0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=0, x1=d1z(i<jαiαj)x_{1}=-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}) since h(x)h(x) only has one negative root. If h(d1z(i<jαiαj))>0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))>0, from the derivative of h(x)h(x) with respect to xx,

h(x)=\displaystyle h^{{}^{\prime}}(x)= dp0xd1+(d1)p1xd2+(d2)p2xd3\displaystyle dp_{0}x^{d-1}+(d-1)p_{1}x^{d-2}+(d-2)p_{2}x^{d-3} (E9)
++kpdkxk1,\displaystyle+\cdots+kp_{d-k}x^{k-1},

the sign of the polynomial coefficients of h(x)h^{{}^{\prime}}(x) changes dKd-K or dK1d-K-1 times and there are KK zero roots of h(x)h^{{}^{\prime}}(x). Hence, h(x)h^{{}^{\prime}}(x) has no negative roots or only has one negative root. Since h(x1)=h(0)=0h(x_{1})=h(0)=0 and h(x)h(x) is continuous, according to Rolle’s Mean value theorem, there exists a ξ(x1,0)\xi\in(x_{1},0) such that h(ξ)=0h^{{}^{\prime}}(\xi)=0. Thus, h(x)h^{{}^{\prime}}(x) must have only one negative root. Since h(x)h^{{}^{\prime}}(x)\rightarrow-\infty when xx\rightarrow-\infty, h(x)<0h^{{}^{\prime}}(x)<0 when x<ξx<\xi. According to h(d1z(i<jαiαj))>0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))>0, d1z(i<jαiαj)(,x1)(ξ,0)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})\in(-\infty,x_{1})\cup(\xi,0). Suppose d1z(i<jαiαj)(ξ,0)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})\in(\xi,0), according to that h(ξ)<0h(\xi)<0 and h(x)h(x) is continuous, by zero point theorem we have that there exists another negative root between ξ\xi and d1z(i<jαiαj)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}), which is contradict with the fact that h(x)h(x) only has one negative root. Therefore, d1z(i<jαiαj)(,x1)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})\in(-\infty,x_{1}), i.e., x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}).

When dd is odd, h(d1z(i<jαiαj))=|d1z(i<jαiαj)IdH|=|d1z(i<jαiαj)Id+H|0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=|-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}-H|=-|\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})I_{d}+H|\leq 0. If h(d1z(i<jαiαj))=0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))=0, x1=d1z(i<jαiαj)x_{1}=-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}) since h(x)h(x) only has one negative root. If h(d1z(i<jαiαj))<0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))<0, from (E9) the sign of the polynomial coefficients of h(x)h^{{}^{\prime}}(x) changes dKd-K or dK1d-K-1 times and there are KK zero roots of h(x)h^{{}^{\prime}}(x). Hence, h(x)h^{{}^{\prime}}(x) has no negative roots or only has one negative root. Since h(x1)=h(0)=0h(x_{1})=h(0)=0 and h(x)h(x) is continuous, according to the Rolle’s Mean value theorem, we get that there exists a ξ(x1,0)\xi\in(x_{1},0) such that h(ξ)=0h^{{}^{\prime}}(\xi)=0. Thus, h(x)h^{{}^{\prime}}(x) must have only one negative root. Since h(x)+h^{{}^{\prime}}(x)\rightarrow+\infty when xx\rightarrow-\infty, h(x)>0h^{{}^{\prime}}(x)>0 when x<ξx<\xi. According to that h(d1z(i<jαiαj))<0h(-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}))<0, one has d1z(i<jαiαj)(,x1)(ξ,0)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})\in(-\infty,x_{1})\cup(\xi,0). Suppose d1z(i<jαiαj)(ξ,0)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})\in(\xi,0). Accounting to that h(ξ)>0h(\xi)>0 and h(x)h(x) is continuous, by the zero point theorem we get that there exists another negative root between ξ\xi and d1z(i<jαiαj)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}), which is contradict with the fact that h(x)h(x) only has one negative root. Hence, d1z(i<jαiαj)(,x1)-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j})\in(-\infty,x_{1}), i.e., x1d1z(i<jαiαj)x_{1}\geq-\frac{d-1}{z}(\sum_{i<j}\alpha_{i}\alpha_{j}).

Similarly, we can prove that Theorem 2 holds when t[d4,d3),[d5,d4),,[0,1)t\in[d-4,d-3),[d-5,d-4),\ldots,[0,1).

(IIIIII) When t(,(22d))t\in(-\infty,-(2-\frac{2}{d})):

We have h(0)=(1)d(td+1)(t+1)d1(α12α22αd2)0h(0)=(-1)^{d}(t-d+1)(t+1)^{d-1}(\alpha_{1}^{2}\alpha_{2}^{2}\cdots\alpha_{d}^{2})\geq 0. From (E5) we have h(x)>0h^{{}^{\prime}}(x)>0 when x>0x>0. Taking into account that h(0)0h(0)\geq 0, we see that there exist no positive roots of (E2) in this case. The inequality (34) that we need to prove also has the same form as (E6) and holds as well.

(IVIV) When z=0z=0:

f(|ψψ|)=(𝕀dΦz)|ψψ|1=0f(|\psi\rangle\langle\psi|)=\|(\mathbb{I}_{d}\otimes\Phi_{z})|\psi\rangle\langle\psi|\|-1=0. The inequality (34) also holds. \square