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Quantifying the entanglement of quantum states under the geometric method

Xian Shi [email protected] College of Information Science and Technology, Beijing University of Chemical Technology, Beijing 100029, China LMIB(Beihang University), Ministry of Education, and School of Mathematical Sciences, Beihang University, Beijing 100191, China    Lin Chen [email protected] (corresponding author) LMIB(Beihang University), Ministry of Education, and School of Mathematical Sciences, Beihang University, Beijing 100191, China International Research Institute for Multidisciplinary Science, Beihang University, Beijing 100191, China    Yixuan Liang [email protected] (corresponding author) LMIB(Beihang University), Ministry of Education, and School of Mathematical Sciences, Beihang University, Beijing 100191, China
Abstract

Quantifying entanglement is an important issue in quantum information theory. Here we consider the entanglement measures through the trace norm in terms of two methods, the modified measure and the extended measure for bipartite states. We present the analytical formula for the pure states in terms of the modified measure and the mixed states of two-qubit systems for the extended measure. We also generalize the modified measure from bipartite states to tripartite states.

pacs:
03.65.Ud, 03.67.Mn

I Introduction

Quantum entanglement is an essential feature of quantum mechanics. It plays an important role in quantum information and quantum computation theory horodecki2009quantum , such as superdense coding bennett1992communication , teleportation bennett1993teleporting and the speedup of quantum algorithms shimoni2005entangled .
How to quantify the amount of entanglement for a multipartite system is important in quantum information theory. As it is linked with many areas of the entanglement theory, such as entanglement distillability bennett1996mixed , the transformation of quantum states vedral1997quantifying ; jonathan1999minimal ; rains1999bound , monogamy of entanglement coffman2000distributed ; bai2014general ; zhu2014entanglement ; de2014monogamy ; shi2021multilinear , quantum speed limit rudnicki2021quantum and so on. Due to the importance of this issue, it has been investigated almost since the end of the last century shimony1995degree ; bennett1996mixed ; vedral1997quantifying . In 1996, Bennett etet al.al. proposed the distillable entanglement and entanglement cost, as well as their operational interpretations bennett1996mixed . In 1997, Vedral presented three necessary conditions that an entanglement measure should satisfy vedral1997quantifying . And Vidal presented a general method to quantify the entanglement for the bipartite entangled states vidal2000entanglement . This method is built on the function of pure states in the bipartite system and generalized to the mixed states under the convex roof extended methods on the bipartite system. Other than the generic conditions and the building methods of entanglement measures, some meaningful thoughts to bulid the entanglement measures are presented. The robustness of an entangled state tells us how many degrees of the separable states needed to make the state separable vidal1999robustness . The squashed entanglement was motived by the intrinsic information tucci2002entanglement ; christandl2004squashed , it is monogamous for arbitrary dimensional systems christandl2004squashed and faithful brandao2011faithful . Moreover, the way to measure the entanglement can also be defined based on the geometrical ways, such as, the geomeric measure of entanglement barnum2001monotones ; wei2003geometric , the quantum relative entropy vedral1997quantifying and the fidelity streltsov2010linking . However, there are few results of the measure based on the distance between an entangled state and the set of separable states under the trace norm.

Trace norm not only can quantify the entanglement for a multipartitie mixed state, but also it provides a way to quantify the discord paula2013geometric ; montealegre2013one ; ciccarello2014toward ; roga2016geometric , the measurement-induced nonlocality hu2015measurement , asymmetry marvian2014extending , steering sainz2018formalism and coherence baumgratz2014quantifying ; rana2016trace ; chen2018notes . It is also helpful in quantum communication pirandola2017fundamental ; pirandola2020advances and quantum algorithms gebhart2021quantifying ; bai2022quantum .

This article is organized as follows. In section II, first we present the preliminary knowledge needed. We also present some properties of the modified entanglement measure. In section \@slowromancapiii@, we consider the properties of the entanglement mesurement in terms of the trace norm under two methods. Specifically, in section III.1, we present the solutions of the modified measure for pure states, and then we generalize it to a measure for tripartite mixed states Dfsep()D^{{}^{\prime}}_{fsep}(\cdot), which tells the distance between a tripartite state and the cone of full separable states. In section III.2, we consider the measure based on the trace norm under the method proposed in gour2020optimal ; shi2021extension . In section IV, we end with a conclusion.

II Preliminary Knowledge

In the following, we denote 𝒟A\mathcal{D}_{A} as the set of states on A\mathcal{H}_{A} with finite dimensions. Next for a state ρAB𝒟AB\rho_{AB}\in\mathcal{D}_{AB}, if it can be written as ρAB=ipi|ψiAψi||ϕiBϕi|\rho_{AB}=\sum_{i}p_{i}|\psi_{i}\rangle_{A}\langle\psi_{i}|\otimes|\phi_{i}\rangle_{B}\langle\phi_{i}|, then it is separable, otherwise, it is an entangled state. We denote SEPAB\mathop{\rm SEP}_{AB} as the set of separable states on AB\mathcal{H}_{AB} with finite dimensions. And we may leave out the systems in the absence of ambiguity in the following.

Assume |ψAB|\psi\rangle_{AB} is a bipartite pure state in AB\mathcal{H}_{AB}, |ψAB=iλi|ii|\psi\rangle_{AB}=\sum_{i}\sqrt{\lambda_{i}}|ii\rangle, then its concurrence is defined as

C(|ψAB)=2(1TrρA2),\displaystyle C(|\psi\rangle_{AB})=\sqrt{2(1-\mathop{\rm Tr}\rho_{A}^{2})}, (1)

here ρA\rho_{A} is the reduced density matrix of ρAB.\rho_{AB}. When ρAB\rho_{AB} is a mixed state, then

C(ρAB)=min{pi,|ψAB}ipiC(|ψiAB),\displaystyle C(\rho_{AB})=\mathop{\rm min}_{\{p_{i},|\psi\rangle_{AB}\}}\sum_{i}p_{i}C(|\psi_{i}\rangle_{AB}), (2)

where the minimum takes over all the decompositions {pi,|ψi}\{p_{i},|\psi_{i}\rangle\} of ρAB\rho_{AB}.

Next we recall the definition of a distance measure in terms of trace norm for a bipartite mixed state. Assume ρ𝒟AB\rho\in\mathcal{D}_{AB}, its distance measure in terms of trace norm is defined as,

Dsep(ρ)=minβSEPρβ1,\displaystyle D_{sep}(\rho)=\mathop{\rm min}_{\beta\in\mathop{\rm SEP}}\lVert\rho-\beta\rVert_{1}, (3)

here β\beta takes over all the separable states in the system AB.\mathcal{H}_{A}\otimes\mathcal{H}_{B}.

Then we recall E:𝒟(AB)+E:\mathcal{D}(\mathcal{H}_{AB})\rightarrow\mathbb{R}^{+} is an entanglement measure vedral1997quantifying if it satifies:

  • (I)

    E(ρAB)=0E(\rho_{AB})=0 if ρAB𝒮(AB)\rho_{AB}\in\mathcal{S}(\mathcal{H}_{AB}).

  • (II)

    EE does not increase under the local operation and classical communication (LOCC),

    E(Ψ(ρAB))E(ρAB),\displaystyle E(\Psi(\rho_{AB}))\leq E(\rho_{AB}),

    here Ψ\Psi is LOCC.

In vidal2000entanglement , the author presented that when EE satisfies the following two conditions, EE is an entanglement monotone,

  • (III)

    E(ρ)kpkE(σk),E(\rho)\geq\sum_{k}p_{k}E(\sigma_{k}), here σk=i,k(ρAB)pk,\sigma_{k}=\frac{\mathcal{E}_{i,k}(\rho_{AB})}{p_{k}}, pk=Tri,k(ρAB)p_{k}=\mathop{\rm Tr}\mathcal{E}_{i,k}(\rho_{AB}), i,k\mathcal{E}_{i,k} is any unilocal quantum operation performed by any party AA or B,B, and ii are on behave of the party AA or B.B.

  • (IV)

    For any decomposition {pk,ρk}\{p_{k},\rho_{k}\} of ρAB\rho_{AB}

    E(ρ)kpkE(ρk).\displaystyle E(\rho)\leq\sum_{k}p_{k}E(\rho_{k}).

The condition (IV) can also be regarded as the convexity of an entanglement measure. Next we recall the simplified conditions when the function satisifies the LOCC monotone horodecki2005simplifying : For a convex function ff does not increase under LOCC if and only if

  • 1

    . ff satisfies local unitary invariance (LUI)

    f(UAUBρABUAUB)=f(ρAB)\displaystyle f(U_{A}\otimes U_{B}\rho_{AB}U_{A}^{\dagger}\otimes U_{B}^{\dagger})=f(\rho_{AB}) (4)
  • 2

    . ff satisfies

    f(ipiρABi|iXi|)=ipif(ρABi),\displaystyle f(\sum_{i}p_{i}\rho_{AB}^{i}\otimes|i\rangle_{X}\langle i|)=\sum_{i}p_{i}f(\rho_{AB}^{i}), (5)

    for X=A,BX=A^{{}^{\prime}},B^{{}^{\prime}}, where |i|i\rangle are local orthogonal flags.

The properties (I), (II) and (IV) are satisfied by Dsep()D_{sep}(\cdot) vedral1997quantifying ; chen2014comparison , however, the property (III) for the entanglement measure Dsep()D_{sep}(\cdot) is not valid qiao2017activation . Then we recall the modified version Dsep()D_{sep}^{{}^{\prime}}(\cdot) of the entanglement measure Dsep()D_{sep}(\cdot),

Dsep(ρAB)=minσSEP,λ>0ρλσ1.\displaystyle D^{{}^{\prime}}_{sep}(\rho_{AB})=\mathop{\rm min}_{\sigma\in SEP,\lambda>0}\lVert\rho-\lambda\sigma\rVert_{1}. (6)

Due to the properties of 1\lVert\cdot\rVert_{1} and the definition of Dsep(),D_{sep}^{{}^{\prime}}(\cdot), if we could prove

Dsep(ρ)kpkDsep(ρk),\displaystyle D_{sep}^{{}^{\prime}}(\rho)\geq\sum_{k}p_{k}D^{{}^{\prime}}_{sep}(\rho_{k}), (7)
pk=Trk(ρk),ρk=k(ρk)/pk,\displaystyle p_{k}=\mathop{\rm Tr}\mathcal{E}_{k}(\rho_{k}),\hskip 8.53581pt\rho_{k}=\mathcal{E}_{k}(\rho_{k})/p_{k},

here {k}\{\mathcal{E}_{k}\} is a quantum operation on the party AA at some stage (kk labels different outcomes if at some stage A performs a measurement), Dsep()D^{{}^{\prime}}_{sep}(\cdot) satisfies the property (III). Next we recall the Naimark theorem paulsen2002completely :

Assume {Fi}i=1n\{F_{i}\}_{i=1}^{n} is a POVM acting on a Hilbert space A\mathcal{H}_{A} with dimension dAd_{A}, then there exists a projection-valued measures (PVM) {Πi}i=1n\{\Pi_{i}\}_{i=1}^{n} acting on a Hilbert space A\mathcal{H}_{A^{{}^{\prime}}} with dimension dAd_{A^{{}^{\prime}}}, such that

Fi=VΠiV,\displaystyle F_{i}=V^{\dagger}\Pi_{i}V, (8)

A method is via the direct sum, requiring

Tr[Fiρ]=Tr[Pi(ρ0)],\displaystyle\mathop{\rm Tr}[F_{i}\rho]=\mathop{\rm Tr}[P_{i}(\rho\oplus 0)], (9)

here ρ\rho is an arbitrary operator in A\mathcal{H}_{A}, 0 is the zero matrix with dimension dAdAd_{A}^{{}^{\prime}}-d_{A}. As the local operations \mathcal{E}, addition of an uncorrelated ancilla system and the dismissal of a local part of the whole system satisfy

Dsep(ρ)Dsep((ρ)),\displaystyle D^{{}^{\prime}}_{sep}(\rho)\geq D^{{}^{\prime}}_{sep}(\mathcal{E}(\rho)),

the above inequality is due to the definition of Dsep()D_{sep}^{{}^{\prime}}(\cdot) and the monotonicity of the trace norm under .\mathcal{E}. Combing the Naimark theorem, we only need to prove (7) is valid for the PVM, that is, we need to prove that

Dsep(ρ)kpkDsep(ρk),\displaystyle D_{sep}^{{}^{\prime}}(\rho)\geq\sum_{k}p_{k}D^{{}^{\prime}}_{sep}(\rho_{k}), (10)

here pk=TrΠkρ,p_{k}=\mathop{\rm Tr}\Pi_{k}\rho, ρk=ΠkρΠkpk\rho_{k}=\frac{\Pi_{k}\rho\Pi_{k}}{p_{k}}, {Πk}\{\Pi_{k}\} is a PVM on the AA-party.

Dsep(ρ)=\displaystyle D_{sep}^{{}^{\prime}}(\rho)= ρσ1\displaystyle\lVert\rho-\sigma\rVert_{1}
\displaystyle\geq k(ΠkρΠkΠkσΠk)1\displaystyle\lVert\sum_{k}(\Pi_{k}\rho\Pi_{k}-\Pi_{k}\sigma\Pi_{k})\rVert_{1}
=\displaystyle= kΠkρΠkΠkσΠk1\displaystyle\sum_{k}\lVert\Pi_{k}\rho\Pi_{k}-\Pi_{k}\sigma\Pi_{k}\rVert_{1}
=\displaystyle= kpkΠkρΠkpkσk1\displaystyle\sum_{k}p_{k}\lVert\frac{\Pi_{k}\rho\Pi_{k}}{p_{k}}-\sigma_{k}\rVert_{1}
\displaystyle\geq kpkDsep(ρk).\displaystyle\sum_{k}p_{k}D^{{}^{\prime}}_{sep}(\rho_{k}). (11)

Here in the first equality, σ\sigma is the optimal in terms of Dsep()D^{{}^{\prime}}_{sep}(\cdot) for ρ\rho. In the first inequality, {Πk}\{\Pi_{k}\} is a PVM on the AA-party of ρAB\rho_{AB}, kΠk=I\sum_{k}\Pi_{k}=I, then due to the contractive under the trace-preserving quantum operations, the first inequality is valid. In the third equality, pk=TrΠkρΠk,p_{k}=\mathop{\rm Tr}\Pi_{k}\rho\Pi_{k}, σk=ΠkσΠkpk,\sigma_{k}=\frac{\Pi_{k}\sigma\Pi_{k}}{p_{k}}, here the trace of σk\sigma_{k} may be not 1, that is, σk\sigma_{k} may be not a state. As σk\sigma_{k} a separable state, and due to the definition of Dsep(),D_{sep}(\cdot), the last inequality is valid. We finish the proof of the inequality (11)(\ref{m2}).

At last, we prove the convexity of Dsep()D_{sep}^{{}^{\prime}}(\cdot). Assume {pk,ρk}\{p_{k},\rho_{k}\} is an arbitrary decompostion of ρAB\rho_{AB}, σk\sigma_{k} is the optimal for the state ρk\rho_{k} in terms of Dsep()D^{{}^{\prime}}_{sep}(\cdot), then

kpkDsep(ρk)=\displaystyle\sum_{k}p_{k}D^{{}^{\prime}}_{sep}(\rho_{k})= kpkρkσk1\displaystyle\sum_{k}p_{k}||\rho_{k}-\sigma_{k}||_{1}
\displaystyle\geq k(pkρkpkσk)1\displaystyle\lVert\sum_{k}(p_{k}\rho_{k}-p_{k}\sigma_{k})\rVert_{1}
\displaystyle\geq ρσ1=Dsep(ρ),\displaystyle\lVert\rho-\sigma\rVert_{1}=D^{{}^{\prime}}_{sep}(\rho), (12)

the first equality is due to the definition of (6),(\ref{mem}), the first inequality is due to the convexity of the 1\lVert\cdot\rVert_{1}. In the second inequality, we denote σ\sigma as the optimal in terms of Dsep()D_{sep}^{{}^{\prime}}(\cdot) for ρ\rho, and the second inequality is due to the definition of Dsep().D_{sep}^{{}^{\prime}}(\cdot). Then we prove the property (IV) is satisfied by the entanglement measure Dsep()D_{sep}(\cdot).

III Main Results

In this section, we will consider the entanglement measures based on trace norm in terms of two methods, the modified distance measure and the extended distance measure. First we present the analytical formula of a class of entangled states in terms of the modified entanglement measure, we also present some properties of the modified measure. At last, we consider the extended measure generated under the method presented in gour2020optimal ; shi2021extension and show this measure is monogamous for nn-qubit systems .

III.1 The modified distance measure

In this section, we first rewrite the definition of modified distance measure (6)(\ref{mem}) as follows,

Dsep(ρ)=minσρσ1,\displaystyle D_{sep}^{{}^{\prime}}(\rho)=\mathop{\rm min}_{\sigma}\lVert\rho-\sigma\rVert_{1}, (13)

here σ\sigma takes over all the elements in the set {σ|σ=iqiωiAζiB,qi0,ωiAD(A),ζiBD(B)}\{\sigma|\sigma=\sum_{i}q_{i}\omega_{i}^{A}\otimes\zeta_{i}^{B},q_{i}\geq 0,\omega_{i}^{A}\in D(\mathcal{H}_{A}),\zeta_{i}^{B}\in D(\mathcal{H}_{B})\}. In yu2016alternative , the authors presented the method to quantify the coherence in this way. In johnston2018modified , they showed that the modified distance of almost all pure states is a constant, and they numerically showed that a similar result occurs for states with a fixed rank. Here we showed that the same result occurs for the entangled pure states.

Theorem 1

Assume |ψ=iλi|ii|\psi\rangle=\sum_{i}\lambda_{i}|ii\rangle is an arbitrary pure state, here λiλi+1,\lambda_{i}\geq\lambda_{i+1},then the modified distance measure of |ψ|\psi\rangle is

Dsep(|ψ)={1,λ112,2λ11λ12,λ1>12.\displaystyle D^{{}^{\prime}}_{sep}(|\psi\rangle)=\left\{\begin{aligned} 1\hskip 19.91692pt&,&\lambda_{1}\leq\frac{1}{\sqrt{2}},\\ 2\lambda_{1}\sqrt{1-\lambda_{1}^{2}}&,&\lambda_{1}>\frac{1}{\sqrt{2}}.\end{aligned}\right. (14)

We present the proof of this theorem in Sec. VI.

Remark 1

In regula2019one , the authors considered the same problem for the pure states in terms of Dsep()D_{sep}^{{}^{\prime}}(\cdot). The method there is due to the convex analysis, ours is different from there. Moreover, comparing with the result in regula2019one , ours is more apparent. And we can also generalize our methods to a class of mixed states.

Corollary 1

Assume ρ=a0|0000|+a1|0011|+a1¯|1100|+a2|1111|\rho=a_{0}|00\rangle\langle 00|+a_{1}|00\rangle\langle 11|+\overline{a_{1}}|11\rangle\langle 00|+a_{2}|11\rangle\langle 11| is a two-qubit mixed state, then Dsep(ρ)=2|a1|D_{sep}^{{}^{\prime}}(\rho)=2|a_{1}|.

The main proof of the above corollary is the same as the proof for the pure states above, and by combing the theorem 1 in johnston2018modified , we can get the corollary. Next we present some properties of the modified distance measure.

Theorem 2

Assume ρAB\rho_{AB} and σAB\sigma_{AB} are two mixed states with ρABσABϵ\lVert\rho_{AB}-\sigma_{AB}\rVert\leq\epsilon, then we have

|Dsep(ρAB)Dsep(σAB)|ϵ\displaystyle|D^{{}^{\prime}}_{sep}(\rho_{AB})-D^{{}^{\prime}}_{sep}(\sigma_{AB})|\leq\epsilon (15)
Proof.

Here we can always assume Dsep(ρAB)Dsep(σAB)0D^{{}^{\prime}}_{sep}(\rho_{AB})-D^{{}^{\prime}}_{sep}(\sigma_{AB})\geq 0, then we have

Dsep(ρAB)Dsep(σAB)\displaystyle D^{{}^{\prime}}_{sep}(\rho_{AB})-D^{{}^{\prime}}_{sep}(\sigma_{AB})
\displaystyle\leq ρABαAB1σABαAB1\displaystyle\lVert\rho_{AB}-\alpha_{AB}\rVert_{1}-\lVert\sigma_{AB}-\alpha_{AB}\rVert_{1}
\displaystyle\leq ρABσAB1ϵ.\displaystyle\lVert\rho_{AB}-\sigma_{AB}\rVert_{1}\leq\epsilon. (16)

Here αAB\alpha_{AB} is the optimal for σAB\sigma_{AB}. The first inequality is due to the definition of Dsep()D^{{}^{\prime}}_{sep}(\cdot), as αAB\alpha_{AB} may not be the optimal for ρAB\rho_{AB}, the second inequality is due to the triangle inequality

ρABαAB1σABαAB1\displaystyle\lVert\rho_{AB}-\alpha_{AB}\rVert_{1}-\lVert\sigma_{AB}-\alpha_{AB}\rVert_{1}
\displaystyle\leq ρABαABσAB+αAB1\displaystyle\lVert\rho_{AB}-\alpha_{AB}-\sigma_{AB}+\alpha_{AB}\rVert_{1}
=\displaystyle= ρABσAB1\displaystyle\lVert\rho_{AB}-\sigma_{AB}\rVert_{1}

\sqcap\sqcup

At the end of this subsection, we generalize the measure Dsep(ρ)D^{{}^{\prime}}_{sep}(\rho) to a quantity Dfsep()D^{{}^{\prime}}_{fsep}(\cdot) for a tripartite system which tells the distance from a tripartite state to the set of fully separable states. Next we recall the fully separable states. A pure state |ψABC|\psi\rangle_{ABC} is fully separable if it can be written as |ψABC=|ϕ1A|ϕ2B|ϕ3C|\psi\rangle_{ABC}=|\phi_{1}\rangle_{A}|\phi_{2}\rangle_{B}|\phi_{3}\rangle_{C}. If a mixed state ρABC\rho_{ABC} can be written as

ρABC=i=1kpiρAiρBiρCi,\displaystyle\rho_{ABC}=\sum_{i=1}^{k}p_{i}\rho_{A}^{i}\otimes\rho_{B}^{i}\otimes\rho_{C}^{i},

then ρABC\rho_{ABC} is a fully separable state. The definition of Dfsep(ρABC)D^{{}^{\prime}}_{fsep}(\rho_{ABC}) is defined as follows,

Dfsep(ρABC)=infγFSEPργ1,\displaystyle D^{{}^{\prime}}_{fsep}(\rho_{ABC})=\inf_{\gamma\in FSEP^{{}^{\prime}}}\lVert\rho-\gamma\rVert_{1}, (17)

here the infimum takes over all the elements in {γ|γ=i=1kpiρAiρBiρCi,pi0,ρAi𝒟(A),ρBi𝒟(B),ρCi𝒟(C),i=1,2,,k.}\{\gamma|\gamma=\sum_{i=1}^{k}p_{i}\rho_{A}^{i}\otimes\rho_{B}^{i}\otimes\rho_{C}^{i},p_{i}\geq 0,\rho_{A}^{i}\in\mathcal{D}(\mathcal{H}_{A}),\rho_{B}^{i}\in\mathcal{D}(\mathcal{H}_{B}),\rho_{C}^{i}\in\mathcal{D}(\mathcal{H}_{C}),i=1,2,\cdots,k.\}.

In the following, we will compute the values Dfsep(|W)D^{{}^{\prime}}_{fsep}(|W\rangle) and Dfsep(|GHZ).D^{{}^{\prime}}_{fsep}(|GHZ\rangle).

Example 1

Assume

|W=\displaystyle|W\rangle= 13(|001+|010+|100),\displaystyle\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle),
|GHZ=\displaystyle|GHZ\rangle= 12(|000+|111).\displaystyle\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle).

Then Dfsep(W)=1,Dfsep(GHZ)=1.D^{{}^{\prime}}_{fsep}(W)=1,D^{{}^{\prime}}_{fsep}(GHZ)=1.

We place the proof of Example 1 in Sec. VI.

III.2 The extended distance measure

Here we consider another entanglement measure based on the trace norm under the method presented in gour2020optimal ; shi2021extension . There the authors presented a way to generate the entanglement measure E¯\overline{E} for the mixed states from the entanglement measure EE on pure states. Assume ρABB(𝒜)\rho_{AB}\in B(\mathcal{H_{A}\otimes H_{B}}),

E¯(ρAB)=inf|ψAB(ρAB)E(|ψAB),\displaystyle\overline{E}(\rho_{AB})=\inf_{|\psi\rangle_{AB}\in\mathcal{R}(\rho_{AB})}E(|\psi\rangle_{AB}), (18)

here the infimum takes over all the pure states in the set

(ρAB)={ψAB\displaystyle\mathcal{R}(\rho_{AB})=\{\psi_{AB} AB|ρAB=Λ(ψAB),Λ𝒯,\displaystyle\in\mathcal{H}_{AB}|\rho_{AB}=\Lambda(\psi_{AB}),\Lambda\in\mathcal{T},
𝒯 is a class of quantum channels. },\displaystyle\textit{$\mathcal{T}$ is a class of quantum channels. }\},

Here we specifically consider the properties of the entanglement measures based on the trace norm under the method above,

D¯sep(ρAB)=\displaystyle\overline{D}_{sep}(\rho_{AB})= inf|ψAB(ρAB)Dsep(|ψAB),\displaystyle\inf_{|\psi\rangle_{AB}\in\mathcal{R}(\rho_{AB})}D_{sep}(|\psi\rangle_{AB}),
(ρAB)=\displaystyle\mathcal{R}(\rho_{AB})= {ψABAB|ρAB=Λ(ψAB),ΛLOCC.}.\displaystyle\{\psi_{AB}\in\mathcal{H}_{AB}|\rho_{AB}=\Lambda(\psi_{AB}),\Lambda\in LOCC.\}. (19)

Then we show the above quantity is an entanglement measure. When ρAB\rho_{AB} is a mixed state, |ψAB|\psi\rangle_{AB} is the optimal pure state in (19), assume Λ\Lambda is LOCC, then

D¯sep(ρAB)=\displaystyle\overline{D}_{sep}(\rho_{AB})= inf|ψAB{Dsep(|ψAB)|ρAB=Λ(ψAB),ΛLOCC.}\displaystyle\inf_{|\psi\rangle_{AB}}\{D_{sep}(|\psi\rangle_{AB})|\rho_{AB}=\Lambda(\psi_{AB}),\Lambda\in LOCC.\}
\displaystyle\geq inf|ψAB{Dsep(|ψAB)|(ρAB)=Λ(ψAB),ΛLOCC.}\displaystyle\inf_{|\psi\rangle_{AB}}\{D_{sep}(|\psi\rangle_{AB})|\mathcal{M}(\rho_{AB})=\mathcal{M}\circ\Lambda(\psi_{AB}),\Lambda\in LOCC.\}
\displaystyle\geq inf|ψAB{Dsep(|ψAB)|(ρAB)=Θ(ψAB),ΘLOCC.}=D¯sep((ρAB)),\displaystyle\inf_{|\psi\rangle_{AB}}\{D_{sep}(|\psi\rangle_{AB})|\mathcal{M}(\rho_{AB})=\Theta(\psi_{AB}),\Theta\in LOCC.\}=\overline{D}_{sep}(\mathcal{M}(\rho_{AB})),

here ()\mathcal{M}(\cdot) is LOCC. Hence, the property (II) is valid. As a separable pure state can be transformed into a separable mixed state through LOCC, then the quantity D¯sep()\overline{D}_{sep}(\cdot) of a separable state is 0. Hence, it is an entanglement measure.

Next we consider the mixed states on two-qubit systems in terms of D¯sep().\overline{D}_{sep}(\cdot).

Theorem 3

When ρAB\rho_{AB} is a two-qubit mixed state, D¯sep(ρAB)\overline{D}_{sep}(\rho_{AB}) is an entanglement monotone. Furthermore,

D¯sep(ρAB)=C(ρAB)=min{pi,|ψiAB}ipiD¯sep(|ψiAB),\displaystyle\overline{D}_{sep}(\rho_{AB})=C(\rho_{AB})=\mathop{\rm min}_{\{p_{i},|\psi_{i}\rangle_{AB}\}}\sum_{i}p_{i}\overline{D}_{sep}(|\psi_{i}\rangle_{AB}),

where the minimum takes over all the decompositions {pi,|ψiAB}\{p_{i},|\psi_{i}\rangle_{AB}\} of ρAB.\rho_{AB}. Furthermore, assume ρA|B1B2Bn1\rho_{A|B_{1}B_{2}\cdots B_{n-1}} is an nn-qubit mixed state, then

D¯sep2(ρA|B1B2Bn1)i=1n1D¯sep2(ρABi).\displaystyle\overline{D}_{sep}^{2}(\rho_{A|B_{1}B_{2}\cdots B_{n-1}})\geq\sum_{i=1}^{n-1}\overline{D}_{sep}^{2}(\rho_{AB_{i}}). (20)
Proof.

Assume ρAB\rho_{AB} is a two-qubit mixed state, then

D¯sep(ρAB)=\displaystyle\overline{D}_{sep}(\rho_{AB})= Dsep(|ψAB)\displaystyle D_{sep}(|\psi\rangle_{AB})
=\displaystyle= C(|ψAB)\displaystyle C(|\psi\rangle_{AB})
=\displaystyle= C¯(ρAB)\displaystyle\overline{C}(\rho_{AB})
=\displaystyle= min{pi,|ψiAB}ipiD¯sep(|ψiAB),\displaystyle\mathop{\rm min}_{\{p_{i},|\psi_{i}\rangle_{AB}\}}\sum_{i}p_{i}\overline{D}_{sep}(|\psi_{i}\rangle_{AB}),

where the minimum in the last equality takes over all the decompositions {pi,|ψiAB}\{p_{i},|\psi_{i}\rangle_{AB}\} of ρAB.\rho_{AB}. In the first equality, |ψAB|\psi\rangle_{AB} is the optimal for ρAB\rho_{AB} in terms of D¯sep()\overline{D}_{sep}(\cdot), and combing Dsep(|ψAB)=C(|ψAB)D_{sep}(|\psi\rangle_{AB})=C(|\psi\rangle_{AB}) chen2016quantifying , we have the second equality is valid. As for any two-qubit pure state |ϕAB,|\phi\rangle_{AB}, C(|ϕAB)=Dsep(|ϕAB)C(|\phi\rangle_{AB})=D_{sep}(|\phi\rangle_{AB}) chen2016quantifying , and due to the definition of D¯sep()\overline{D}_{sep}(\cdot) and C¯()\overline{C}(\cdot), the third equality is valid. Next we show the last inequality,

D¯sep2(ρA|B1B2Bn1)=\displaystyle\overline{D}^{2}_{sep}(\rho_{A|B_{1}B_{2}\cdots B_{n-1}})= Dsep2(|ψA|B1B2Bn1)\displaystyle D^{2}_{sep}(|\psi\rangle_{A|B_{1}B_{2}\cdots B_{n-1}})
=\displaystyle= C2(|ψA|B1B2Bn1)\displaystyle C^{2}(|\psi\rangle_{A|B_{1}B_{2}\cdots B_{n-1}})
\displaystyle\geq C2(ρA|B1B2Bn1)\displaystyle C^{2}(\rho_{A|B_{1}B_{2}\cdots B_{n-1}})
\displaystyle\geq i=1n1C2(ρABi)\displaystyle\sum_{i=1}^{n-1}C^{2}(\rho_{AB_{i}})
=\displaystyle= i=1n1D¯sep2(ρABi).\displaystyle\sum_{i=1}^{n-1}\overline{D}_{sep}^{2}(\rho_{AB_{i}}).

In the first equality, we denote |ψ|\psi\rangle as the optimal in terms of D¯sep().\overline{D}_{sep}(\cdot). As |ψA|B1B2Bn1|\psi\rangle_{A|B_{1}B_{2}\cdots B_{n-1}} is a 2n2\otimes n pure state, then we have Dsep2(ψA|B1B2Bn1)=C2(ψA|B1B2Bn1)D^{2}_{sep}(\psi_{A|B_{1}B_{2}\cdots B_{n-1}})=C^{2}(\psi_{A|B_{1}B_{2}\cdots B_{n-1}}) chen2016quantifying . The first inequality is due to that ρA|B1B2Bn1\rho_{A|B_{1}B_{2}\cdots B_{n-1}} can be transformed by ψA|B1B2Bn1\psi_{A|B_{1}B_{2}\cdots B_{n-1}} under LOCC. The last inequality is proved in osborne2006general . As when ρAB\rho_{AB} is a two-qubit mixed state, C(ρAB)=D¯sep(ρAB)C(\rho_{AB})=\overline{D}_{sep}(\rho_{AB}), we have the last equality is valid.     \sqcap\sqcup

IV Conclusion

In this paper, we have considered the entanglement measures in terms of the trace norm under two methods. We have presented the modified distance measure satisfies convexity for arbitrary dimensional systems. And we also have presented the analytical expressions for some classes of states. Then we have shown the extended measure is an entanglement monotone for the 2-qubit systems and monogamous for nn-qubit systems. We think the above results may be not valid for larger systems, although we havenot found a counterexample. At last, we believe that our results are helpful in the study of monogamy relations for multipartite entanglement systems. And we hope our work could shed some light on related studies.

V Acknowledgement

This work was supported by the NNSF of China (Grant No. 11871089), and the Fundamental Research Funds for the Central Universities (Grant Nos. KG12080401 and ZG216S1902).

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VI Appendix

Here we present the proof of Example 1 and Theorem 1.

VI.1 The proof of Example 1

ExampleExample 1: Assume

|W=13(|001+|010+|100),\displaystyle|W\rangle=\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle),
|GHZ=12(|000+|111).\displaystyle|GHZ\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle).

Then Dfsep(W)=1,Dfsep(GHZ)=1.D_{fsep}(W)=1,D_{fsep}(GHZ)=1.

Proof.

As |W|W\rangle satisfies the following property:

Uθ3|WW|Uθ3=|WW|,\displaystyle U^{\otimes 3}_{\theta}|W\rangle\langle W|{U_{\theta}^{\dagger}}^{{\otimes 3}}=|W\rangle\langle W|, (21)

here Uθ=diag(1,eiθ)U_{\theta}=\mathop{\rm diag}(1,e^{i\theta}), then

|WW|σ1\displaystyle\lVert|W\rangle\langle W|-\sigma\rVert_{1}
=\displaystyle= Uθ3(|WW|σ)Uθ31\displaystyle\lVert U^{\otimes 3}_{\theta}(|W\rangle\langle W|-\sigma){U_{\theta}^{\dagger}}^{\otimes 3}\rVert_{1}
=\displaystyle= 12π02πUθ3(|WW|σ)Uθ31𝑑θ\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}\lVert U^{\otimes 3}_{\theta}(|W\rangle\langle W|-\sigma){U_{\theta}^{\dagger}}^{\otimes 3}\rVert_{1}d\theta
\displaystyle\geq 12π02π[|WW|Uθ3σUθ3]𝑑θ1\displaystyle\lVert\frac{1}{2\pi}\int_{0}^{2\pi}[|W\rangle\langle W|-U^{\otimes 3}_{\theta}\sigma{U^{\dagger}_{\theta}}^{\otimes 3}]d\theta\rVert_{1}
=\displaystyle= |WW|σd1,\displaystyle\lVert|W\rangle\langle W|-\sigma_{d}\rVert_{1}, (22)

here σd\sigma_{d} can be written as

σd\displaystyle\sigma_{d}
=\displaystyle= m0|000000|+m1|WW|+m2|W¯W¯|+m3|111111|,\displaystyle m_{0}|000\rangle\langle 000|+m_{1}|W\rangle\langle W|+m_{2}|\overline{W}\rangle\langle\overline{W}|+m_{3}|111\rangle\langle 111|,
|W¯=\displaystyle|\overline{W}\rangle= 13(|011+|101+|110),\displaystyle\frac{1}{\sqrt{3}}(|011\rangle+|101\rangle+|110\rangle),

From (22), we have

Dfsep(|W)=minσDDFSEP|WW|σ1,\displaystyle D_{fsep}(|W\rangle)=\mathop{\rm min}_{\sigma\in DD\cap FSEP}\lVert|W\rangle\langle W|-\sigma\rVert_{1},

here the set DD={δ=m0|000000|+m1|WW|+m2|W¯W¯|+m3|111111||m0,m1,m2,m30.}.DD=\{\delta=m_{0}|000\rangle\langle 000|+m_{1}|W\rangle\langle W|+m_{2}|\overline{W}\rangle\langle\overline{W}|+m_{3}|111\rangle\langle 111||m_{0},m_{1},m_{2},m_{3}\geq 0.\}. Combing the Theorem 1 in yu2016separability , the above can be written as the following,

Dfsep(|W)=infmM|m0|+|m2|+|m3|+|1m1|,\displaystyle D_{fsep}(|W\rangle)=\inf_{\vec{m}\in M}|m_{0}|+|m_{2}|+|m_{3}|+|1-m_{1}|,
M={(m0,m1,m2,m3)|m0m2m12/3,m1m3m22/3.},\displaystyle M=\{(m_{0},m_{1},m_{2},m_{3})|m_{0}m_{2}\geq m_{1}^{2}/3,m_{1}m_{3}\geq m_{2}^{2}/3.\}, (23)

from computation, we have Dfsep(|W)=1.D_{fsep}(|W\rangle)=1.

Next we present the result on |GHZ|GHZ\rangle. In eltschka2012entanglement , the authors presented that |GHZ|GHZ\rangle satisfies the following properties:

  • i.

    it is a symmetric state,

  • ii.

    σx3|GHZ=|GHZ,\sigma_{x}^{\otimes 3}|GHZ\rangle=|GHZ\rangle,

  • iii.

    U(ϕ1,ϕ2)|GHZ=|GHZ,U(\phi_{1},\phi_{2})|GHZ\rangle=|GHZ\rangle,
    here U(ϕ1,ϕ2)=eiϕ1σzeiϕ2σzei(ϕ1+ϕ2)σz,U(\phi_{1},\phi_{2})=e^{i\phi_{1}\sigma_{z}}\otimes e^{i\phi_{2}\sigma_{z}}\otimes e^{-i(\phi_{1}+\phi_{2})\sigma_{z}}, ϕ1,ϕ2\phi_{1},\phi_{2} are arbitrary values.

As |GHZ|GHZ\rangle is a symmetric state, that is,

Vσ|GHZGHZ|=|GHZGHZ|Vσ=|GHZGHZ|,\displaystyle V_{\sigma}|GHZ\rangle\langle GHZ|=|GHZ\rangle\langle GHZ|V_{\sigma^{{}^{\prime}}}=|GHZ\rangle\langle GHZ|,

here σ,σ\sigma,\sigma^{{}^{\prime}} are arbitrary permutations σ,σΩN,\sigma,\sigma^{{}^{\prime}}\in\Omega_{N}, ΩN\Omega_{N} isthe group of all permutations of an NN-element set, and Vσ|ψ1|ψ2|ψN=|ψσ(1)|ψσ(2)|ψσ(N).V_{\sigma}|\psi_{1}\rangle|\psi_{2}\rangle\cdots|\psi_{N}\rangle=|\psi_{\sigma(1)}\rangle|\psi_{\sigma(2)}\rangle\cdots|\psi_{\sigma(N)}\rangle. Assume θ\theta is the state attaining the minimum, then

|GHZGHZ|θ+VσθVσ21\displaystyle\lVert|GHZ\rangle\langle GHZ|-\frac{\theta+V_{\sigma}\theta V_{\sigma^{{}^{\prime}}}}{2}\rVert_{1}
=\displaystyle= 12(|GHZGHZ|θ)+12(Vσ(|GHZGHZ|θ)Vσ)1\displaystyle\lVert\frac{1}{2}(|GHZ\rangle\langle GHZ|-\theta)+\frac{1}{2}(V_{\sigma}(|GHZ\rangle\langle GHZ|-\theta)V_{\sigma^{{}^{\prime}}})\rVert_{1}
\displaystyle\leq 12(|GHZGHZ|θ1+Vσ(|GHZGHZ|θ)Vσ1\displaystyle\frac{1}{2}(\lVert|GHZ\rangle\langle GHZ|-\theta\rVert_{1}+\lVert V_{\sigma}(|GHZ\rangle\langle GHZ|-\theta)V_{\sigma^{{}^{\prime}}}\rVert_{1}
=\displaystyle= |GHZGHZ|θ1,\displaystyle\lVert|GHZ\rangle\langle GHZ|-\theta\rVert_{1},

here σ,σ\sigma,\sigma^{{}^{\prime}} are arbitrary permutations. As ss and ss^{{}^{\prime}} are arbitrary, θ\theta is a symmetric state. By the similar method, θ\theta should satisfy the same properties (i),(ii) and (iii). That is, the matrix of θ\theta should be the following: (m0000000n0m100000000m100000000m100000000m100000000m100000000m10n000000m0)\begin{pmatrix}m_{0}&0&0&0&0&0&0&n\\ 0&m_{1}&0&0&0&0&0&0\\ 0&0&m_{1}&0&0&0&0&0\\ 0&0&0&m_{1}&0&0&0&0\\ 0&0&0&0&m_{1}&0&0&0\\ 0&0&0&0&0&m_{1}&0&0\\ 0&0&0&0&0&0&m_{1}&0\\ n&0&0&0&0&0&0&m_{0}\end{pmatrix}

Then by simple computation, we have Dfsep(GHZ)=1.D_{fsep}(GHZ)=1.     \sqcap\sqcup

VI.2 The proof of Theorem 1

TheoremTheorem 1: Assume |ψ=iλi|ii|\psi\rangle=\sum_{i}\lambda_{i}|ii\rangle is an arbitrary pure state, here λiλi+1,\lambda_{i}\geq\lambda_{i+1},then the modifies distance measure of |ψ|\psi\rangle is

Dsep(|ψ)={1,λ112,2λ11λ12,λ1>12.\displaystyle D^{{}^{\prime}}_{sep}(|\psi\rangle)=\left\{\begin{aligned} 1\hskip 19.91692pt&,&\lambda_{1}\leq\frac{1}{\sqrt{2}},\\ 2\lambda_{1}\sqrt{1-\lambda_{1}^{2}}&,&\lambda_{1}>\frac{1}{\sqrt{2}}.\end{aligned}\right. (24)
Proof.

Assume |ψ|\psi\rangle is a pure state, |ψ=iλi|ii|\psi\rangle=\sum_{i}\sqrt{\lambda_{i}}|ii\rangle, then we have (UU¯)|ψψ|(UU¯)=|ψψ|(U\otimes\overline{U})|\psi\rangle\langle\psi|(U\otimes\overline{U})^{\dagger}=|\psi\rangle\langle\psi|, here UU is a diagonal unitary matrix. Next assume σ\sigma is an arbitrary positive operator which can be written as ρ=iωiAθiB\rho=\sum_{i}\omega_{i}^{A}\otimes\theta_{i}^{B}, here ωiA\omega_{i}^{A} and θiB\theta_{i}^{B} are positive operators on the systems AA and BB, respectively.

|ψψ|ρ1=\displaystyle\lVert|\psi\rangle\langle\psi|-\rho\rVert_{1}= U𝔻(U)(UU¯)(|ψψ|ρ)(UU¯)1𝑑U\displaystyle\int_{U\in\mathbb{D}(U)}\lVert(U\otimes\overline{U})(|\psi\rangle\langle\psi|-\rho)(U\otimes\overline{U})^{\dagger}\rVert_{1}dU
\displaystyle\geq U𝔻(U)(UU¯)(|ψψ|ρ)(UU¯)𝑑U1\displaystyle\lVert\int_{U\in\mathbb{D}(U)}(U\otimes\overline{U})(|\psi\rangle\langle\psi|-\rho)(U\otimes\overline{U})^{\dagger}dU\rVert_{1}
=\displaystyle= |ψψ|U𝔻(U)(UU¯)ρ(UU)𝑑U1\displaystyle\lVert|\psi\rangle\langle\psi|-\int_{U\in\mathbb{D}(U)}(U\otimes\overline{U})\rho(U\otimes U)^{\dagger}dU\rVert_{1} (25)

Then let σ\sigma^{{}^{\prime}} be the optimal in terms of the modified distance measure Dsep()D_{sep}^{{}^{\prime}}(\cdot) for |ψ|\psi\rangle, we have

Dsep(|ψAB)=\displaystyle D_{sep}^{{}^{\prime}}(|\psi\rangle_{AB})= |ψψ|σ1\displaystyle\lVert|\psi\rangle\langle\psi|-\sigma^{{}^{\prime}}\rVert_{1}
\displaystyle\geq |ψψ|U𝔻(U)(UU¯)σ(UU¯)𝑑U1,\displaystyle\lVert|\psi\rangle\langle\psi|-\int_{U\in\mathbb{D}(U)}(U\otimes\overline{U})\sigma^{{}^{\prime}}(U\otimes\overline{U})^{\dagger}dU\rVert_{1}, (26)

as σ\sigma^{{}^{\prime}} is the optimal, and U𝔻(U)(UU¯)σ(UU¯)\int_{U\in\mathbb{D}(U)}(U\otimes\overline{U})\sigma^{{}^{\prime}}(U\otimes\overline{U})^{\dagger} is an element in the set of λSEP\lambda SEP, then we have the inequality in (26)(\ref{s2}) is an equality. Let

ϑ=\displaystyle\vartheta= U𝔻(U)(UU¯)ϱ(UU¯)𝑑U\displaystyle\int_{U\in\mathbb{D}(U)}(U\otimes\overline{U})\varrho(U\otimes\overline{U})^{\dagger}dU
(ϑ)ijkl=\displaystyle(\vartheta)_{ijkl}= zizj¯zlzk¯ϱijkl𝑑zi𝑑zj𝑑zk𝑑zl\displaystyle\int\int\int\int z_{i}\overline{z_{j}}z_{l}\overline{z_{k}}\varrho_{ijkl}dz_{i}dz_{j}dz_{k}dz_{l}
=\displaystyle= {ϱijkl,if (i,l)=(j,k),ϱijkl,if (i,l)=(k,j),0,otherwise.\displaystyle\left\{\begin{aligned} \varrho_{ijkl},\hskip 8.53581pt\textit{if $(i,l)=(j,k)$,}\\ \varrho_{ijkl},\hskip 8.53581pt\textit{if $(i,l)=(k,j)$,}\\ 0,\hskip 31.29802ptotherwise.\end{aligned}\right.

here 𝔻(U)\mathbb{D}(U) is the set consisting of all diagnoal unitary matrices, ϱijkl\varrho_{ijkl} denotes the coefficient of the state ϱ\varrho in the basis |ijkl||ij\rangle\langle kl|. Hence we only need to compute the minimum of the following equality, here the minimum takes over all the positive operators ϱ\varrho that can be written as ϱ=iωiAθiB,\varrho=\sum_{i}\omega_{i}^{A}\otimes\theta_{i}^{B}, ωiA\omega_{i}^{A} and θiB\theta_{i}^{B} are positive operators on the systems AA and BB, respectively.

|ψψ|ijϱiijj|iijj|i,jϱijij|ijij|1\displaystyle\lVert|\psi\rangle\langle\psi|-\sum_{i\neq j}\varrho_{iijj}|ii\rangle\langle jj|-\sum_{i,j}\varrho_{ijij}|ij\rangle\langle ij|\rVert_{1}
=\displaystyle= i,jλiλj|iijj|ijϱiijj|iijj|iϱiiii|iiii|1+ijϱijij|ijij|1\displaystyle\lVert\sum_{i,j}\sqrt{\lambda_{i}\lambda_{j}}|ii\rangle\langle jj|-\sum_{i\neq j}\varrho_{iijj}|ii\rangle\langle jj|-\sum_{i}\varrho_{iiii}|ii\rangle\langle ii|\rVert_{1}+\lVert\sum_{i\neq j}\varrho_{ijij}|ij\rangle\langle ij|\rVert_{1}
=\displaystyle= i(λiϱiiii)|iiii|+ij[(λiλjϱiijj)|iijj|]1+ijϱijij|ijij|1\displaystyle\lVert\sum_{i}(\lambda_{i}-\varrho_{iiii})|ii\rangle\langle ii|+\sum_{i\neq j}[(\sqrt{\lambda_{i}\lambda_{j}}-\varrho_{iijj})|ii\rangle\langle jj|]\rVert_{1}+\lVert\sum_{i\neq j}\varrho_{ijij}|ij\rangle\langle ij|\rVert_{1}
=\displaystyle= i(λiϱiiii)|iiii|+ij[(λiλjϱiijj)|iijj|]1+ij|ϱijij|\displaystyle\lVert\sum_{i}(\lambda_{i}-\varrho_{iiii})|ii\rangle\langle ii|+\sum_{i\neq j}[(\sqrt{\lambda_{i}\lambda_{j}}-\varrho_{iijj})|ii\rangle\langle jj|]\rVert_{1}+\sum_{i\neq j}|\varrho_{ijij}|
\displaystyle\geq i(λiϱiiii)|iiii|+ij[(λiλjϱiijj)|iijj|]1+i<j(ϱiijj|iijj|+ϱiijj¯|jjii|)1\displaystyle\lVert\sum_{i}(\lambda_{i}-\varrho_{iiii})|ii\rangle\langle ii|+\sum_{i\neq j}[(\sqrt{\lambda_{i}\lambda_{j}}-\varrho_{iijj})|ii\rangle\langle jj|]\rVert_{1}+\lVert\sum_{i<j}(\varrho_{iijj}|ii\rangle\langle jj|+\overline{\varrho_{iijj}}|jj\rangle\langle ii|)\rVert_{1}
\displaystyle\geq i(λiϱiiii)|iiii|+ijλiλj|iijj|1\displaystyle\lVert\sum_{i}(\lambda_{i}-\varrho_{iiii})|ii\rangle\langle ii|+\sum_{i\neq j}\sqrt{\lambda_{i}\lambda_{j}}|ii\rangle\langle jj|\rVert_{1}
=\displaystyle= |ψψ|iϱiiii|iiii|1\displaystyle\lVert|\psi\rangle\langle\psi|-\sum_{i}\varrho_{iiii}|ii\rangle\langle ii|\rVert_{1}
=\displaystyle= |ϕϕ|iϱiiii|ii|1.\displaystyle\lVert|\phi\rangle\langle\phi|-\sum_{i}\varrho_{iiii}|i\rangle\langle i|\rVert_{1}. (27)

In the first equality, as |ψ=iλi|ii|\psi\rangle=\sum_{i}\sqrt{\lambda_{i}}|ii\rangle, the coefficients of |ψψ||\psi\rangle\langle\psi| in the basis |ijij||ij\rangle\langle ij| are 0. The first inequality is due to that the positive operator ϱ\varrho satisfies the PPT condition, then |ϱiijj||ϱijijϱjiji||ϱijij|+|ϱjiji|2.|\varrho_{iijj}|\leq\sqrt{|\varrho_{ijij}\varrho_{jiji}|}\leq\frac{|\varrho_{ijij}|+|\varrho_{jiji}|}{2}. The last inequality is due to the triangle inequality of the trace norm. In the last equality, |ϕ=iλi|i.|\phi\rangle=\sum_{i}\sqrt{\lambda_{i}}|i\rangle. Next in johnston2018modified , the authors showed that the optimal coherence states for a pure state in terms of the modified trace norm of coherence is a diagonal matrix, and in the last equality of (27)(\ref{s3}), this problem is turned into the modified trace norm of coherence for a pure state, then we finish the theorem.     \sqcap\sqcup