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  • April 2020

Perfect Discrimination in Approximate Quantum Theory of General Probabilistic Theories

Yuuya Yoshida1, Hayato Arai1, and Masahito Hayashi2,1,3,4 1 Graduate School of Mathematics, Nagoya University, Nagoya, Furo-cho, Chikusa-ku, 464-8602, Japan 2 Shenzhen Institute for Quantum Science and Engineering, Southern University of Science and Technology, Nanshan District, Shenzhen, 518055, China 3 Center for Quantum Computing, Peng Cheng Laboratory, Shenzhen, 518000, China 4 Centre for Quantum Technologies, National University of Singapore, 3 Science Drive 2, 117542, Singapore [email protected] [email protected] [email protected] [email protected]
Abstract

As a modern approach for the foundation of quantum theory, existing studies of General Probabilistic Theories gave various models of states and measurements that are quite different from quantum theory. In this paper, to seek a more realistic situation, we investigate models approximately close to quantum theory. We define larger measurement classes that are smoothly connected with the class of POVMs via a parameter, and investigate the performance of perfect discrimination. As a result, we give a sufficient condition of perfect discrimination, which shows a significant improvement beyond the class of POVMs.

Keywords: perfect discrimination, approximate quantum theory, negative eigenvalues, separable states, general probabilistic theories

1 Introduction

Quantum Theory (QT) is described by operators on Hilbert spaces, and the description is suitable to represent physical systems. Many researchers have tried to give a foundation of the mathematical description. A modern operational approach that starts with statistics of measurement outcomes is called General Probabilistic Theories (GPTs) [2, 13, 6, 7, 8, 3, 14, 4, 5, 1, 9, 10, 11, 12, 15, 16, 17, 18]. Simply speaking, a GPT is defined by state/measurement classes that satisfy the following postulate:

Non-negativity of probability: For each measurement and each state, the probability to obtain each measurement outcome is non-negative.

In QT, the state class and measurement class are given as density matrices and Positive-Operator Valued Measures (POVMs) respectively, which indeed satisfy non-negativity of probability. In this way, QT is a typical example of GPTs, and so is Classical Probability Theory (CPT). Unfortunately, there is no operational reason in the sense of GPTs why only QT and CPT describe physical systems. That is, no studies investigated how one denies an alternative realistic model of GPTs while it is known that there are superior models to QT/CPT with respect to information processing [13, 14, 15, 16, 17, 18].

Preceding studies of GPTs defined models by restricting a state class to a much smaller one than QT/CPT. Once restricting a state class, non-negativity of probability becomes a weaker condition, and the allowed measurement class becomes larger. Consequently, measurement classes of preceding studies are much larger than QT/CPT, and the classes sometimes show superiority of information processing. For example, the PR box, which is defined by restricting states to only convex combinations of four states, violates Bell’s inequality more strongly than QT, i.e., exceeds Tsirelson’s bound [16, 17]. Also, Ref. [18] focused on the case when available states are restricted to only separable states and all measurements with non-negativity of probability are allowed. The pair of these state/measurement classes is called SEP, and Ref. [18] showed that SEP has the superiority of perfect discrimination of bipartite separable pure states.

Refer to caption
Figure 1: The measurement classes s\mathcal{M}_{s} are smoothly connected with the class 0\mathcal{M}_{0} of POVMs via the parameter ss.

However, since the above models are too far from QT, the reality of these models is easily denied. Hence, we should consider measurement classes like s\mathcal{M}_{s} in Fig. 1 that are closer to the class of POVMs (0\mathcal{M}_{0} in Fig. 1) than the measurement class of SEP. If a measurement class is sufficiently close to the class of POVMs, it is hard for an experiment to deny the model because the difference between the experiment and model might be due to an experimental error. In this paper, in order to deny such an alternative measurement class theoretically, we investigate whether an extended measurement class drastically improves perfect discrimination of separable states even when it sufficiently approximates the class of POVMs. For this aim, we define slightly larger measurement classes than that of POVMs to satisfy the following three conditions, and investigate what happens in adopting them. (i) The measurement classes contain the class of POVMs. (ii) Non-negativity of probability holds for every separable state. (iii) The measurement classes are represented as a continuous one-parameter family with the parameter ss, and the case s=0s=0 is just the class of POVMs. For small s>0s>0, the measurement class can be regarded as an approximation of the class of POVMs.

We consider two types of measurement classes. The first one s\mathcal{M}_{s} is given by the restriction of negative eigenvalues of measurement elements while eigenvalues of POVM elements are restricted to be non-negative. The second one (𝒦s)\mathcal{M}(\mathcal{K}_{s}) is defined by the positive cone that is given as the sum of positive semi-definite matrices and positive partial transpose of restricted entangled vectors. As a result, we show that the performance of perfect discrimination is dramatically improved unless s=0s=0, which implies that the above approximate classes of POVMs are unlikely to exist.

The remainder of this paper is organized as follows. Section 2 describes QT in the framework of GPTs. Section 3 defines approximate QT on a bipartite system. Section 4 gives our main results and shows a drastic improvement of multiple-copy state discrimination. Section 5 is a conclusion.

2 Framework of GPTs

Throughout this paper, we only consider finite-dimensional systems. As already stated, there are state/measurement classes in a GPT. First, let us describe a state class. To handle both pure and mixed states, the set of all states must be a closed convex set. In GPTs, the set 𝒮(𝒦,u)\mathcal{S}(\mathcal{K},u) of all states is defined by the intersection of an affine hyperplane and a positive cone:

𝒮(𝒦,u)={x𝒦|x,u=1},\mathcal{S}(\mathcal{K},u)=\set{x\in\mathcal{K}}{\langle x,u\rangle=1},

where 𝒦\mathcal{K} is a positive cone of a finite-dimensional real Hilbert space 𝒱\mathcal{V} equipped with an inner product ,\langle\cdot,\cdot\rangle, and the unit effect uu is an interior point of the dual cone 𝒦\mathcal{K}^{\ast}. Here, 𝒦\mathcal{K} is called a positive cone if 𝒦\mathcal{K} is a closed convex set satisfying that

  • αx𝒦\alpha x\in\mathcal{K} for all α0\alpha\geq 0 and x𝒦x\in\mathcal{K};

  • 𝒦\mathcal{K} has non-empty interior;

  • 𝒦(𝒦)={0}\mathcal{K}\cap(-\mathcal{K})=\{0\}.

Also, for a positive cone 𝒦\mathcal{K}, the dual cone 𝒦\mathcal{K}^{\ast} is defined as

𝒦={y𝒱|x𝒦,x,y0},\mathcal{K}^{\ast}=\set{y\in\mathcal{V}}{\forall x\in\mathcal{K},\ \langle x,y\rangle\geq 0},

which is also a positive cone.

Next, let us describe a measurement class \mathcal{M} (of a GPT with a state class 𝒮(𝒦,u)\mathcal{S}(\mathcal{K},u)) by using non-negativity of probability. A measurement is given as a family {yi}i=1n\{y_{i}\}_{i=1}^{n}, where {1,2,,n}\{1,2,\ldots,n\} denotes the set of outcomes. If a state x𝒮(𝒦,u)x\in\mathcal{S}(\mathcal{K},u) is measured by a measurement {yi}i=1n\{y_{i}\}_{i=1}^{n}, then each outcome ii is obtained with probability x,yi\langle x,y_{i}\rangle. Therefore, we need the following postulate.

Postulate 1 (Non-negativity of probability).

For each state x𝒮(𝒦,u)x\in\mathcal{S}(\mathcal{K},u) and each measurement {yi}i=1n\{y_{i}\}_{i=1}^{n}\in\mathcal{M}, the family {x,yi}i=1n\{\langle x,y_{i}\rangle\}_{i=1}^{n} is a probability distribution, i.e., x,yi0\langle x,y_{i}\rangle\geq 0 for all outcomes ii, and j=1nx,yj=1\sum_{j=1}^{n}\langle x,y_{j}\rangle=1.

Due to Postulate 1, each measurement element yiy_{i} must lie in 𝒦\mathcal{K}^{\ast}. The largest measurement class with Postulate 1 is given as

(𝒦,u)={{yi}i=1n|n,j=1nyj=u,yi𝒦(i)},\mathcal{M}(\mathcal{K}^{\ast},u)=\Set{\{y_{i}\}_{i=1}^{n}}{\begin{array}[]{c}n\in\mathbb{N},\ \sum_{j=1}^{n}y_{j}=u,\\ y_{i}\in\mathcal{K}^{\ast}\ (\forall i)\end{array}},

but we do not assume that \mathcal{M} is the largest one.

Now, let us describe the state/measurement classes of QT by using the above framework of GPTs. Assume that

  1. QT1.

    𝒱\mathcal{V} is the set 𝒯()\mathcal{T}(\mathcal{H}) of all Hermitian matrices on a finite-dimensional complex Hilbert space \mathcal{H};

  2. QT2.

    An inner product on 𝒯()\mathcal{T}(\mathcal{H}) is given by X,Y=TrXY\langle X,Y\rangle=\Tr XY;

  3. QT3.

    𝒦\mathcal{K} is the set 𝒯+()\mathcal{T}_{+}(\mathcal{H}) of all positive semi-definite matrices on \mathcal{H};

  4. QT4.

    uu is the identity matrix II on \mathcal{H};

  5. QT5.

    =(𝒯+(),I)\mathcal{M}=\mathcal{M}(\mathcal{T}_{+}(\mathcal{H}),I).

Then 𝒮(𝒯+(),I)\mathcal{S}(\mathcal{T}_{+}(\mathcal{H}),I) equals the set of all density matrices on \mathcal{H}, and \mathcal{M} equals the class of POVMs. Since these classes are usual ones in QT, it turns out that QT is a typical example of GPTs.

Perfect discrimination.—Let {xi}i=1n\{x_{i}\}_{i=1}^{n} be a family of nn states in 𝒮(𝒦,u)\mathcal{S}(\mathcal{K},u). We say that {xi}i=1n\{x_{i}\}_{i=1}^{n} is perfectly distinguishable if there exists a measurement {yj}j=1n\{y_{j}\}_{j=1}^{n}\in\mathcal{M} such that xi,yj=δij\braket{x_{i},y_{j}}=\delta_{ij}, where δij\delta_{ij} denotes the Kronecker delta. In this paper, we address the case n=2n=2 mainly.

3 Approximate QT

We consider a bipartite system of two finite-dimensional quantum systems A\mathcal{H}_{A} (Alice’s system) and B\mathcal{H}_{B} (Bob’s system), but the bipartite system is not necessarily QT. More precisely, we assume QT1, QT2, and QT4 for =AB\mathcal{H}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}, but do not necessarily assume QT3 or QT5. Let us consider such a bipartite system in the framework of GPTs. When Alice and Bob prepare quantum states ρA\rho^{A} and ρB\rho^{B} independently, the product states ρAρB\rho^{A}\otimes\rho^{B} is prepared on the bipartite system. Considering the convexity of a state class, we need the following postulate.

Postulate 2.

A state class of the bipartite system contains all separable states.

Hereinafter, 𝒯(A)\mathcal{T}(\mathcal{H}_{A}) is denoted by 𝒯(A)\mathcal{T}(A). The notations 𝒯+(A)\mathcal{T}_{+}(A), 𝒯(B)\mathcal{T}(B), 𝒯+(B)\mathcal{T}_{+}(B), 𝒯(AB)\mathcal{T}(AB), and 𝒯+(AB)\mathcal{T}_{+}(AB) are similarly defined. Also, since the unit effect of the bipartite system is always II, we denote (𝒦,I)\mathcal{M}(\mathcal{K},I) by (𝒦)\mathcal{M}(\mathcal{K}) simply. Let SEP(A;B)\mathrm{SEP}(A;B) be the set

{i=1nXiAXiB|n,XiA𝒯+(A),XiB𝒯+(B)(i)}.\Set{\sum_{i=1}^{n}X_{i}^{A}\otimes X_{i}^{B}}{\begin{array}[]{c}n\in\mathbb{N},\ X_{i}^{A}\in\mathcal{T}_{+}(A),\\ X_{i}^{B}\in\mathcal{T}_{+}(B)\ (\forall i)\end{array}}.

Ref. [18] used the largest measurement class (SEP(A;B))\mathcal{M}(\mathrm{SEP}(A;B)^{\ast}) to discriminate two separable pure states, but their measurement class is too far from the class of POVMs. Therefore, we need to define a measurement class that is sufficiently close to the class of POVMs. Moreover, for some state class with Postulate 2, the measurement class must satisfy Postulate 1. Let us define such measurement classes in two different ways here.

Definition 3 (Measurement class).

For s0s\geq 0, we define the measurement class s\mathcal{M}_{s} as

s={{Mi}i=1n|n,j=1nMj=I,MiSEP(A;B),neg(Mi)s(i)},\mathcal{M}_{s}=\Set{\{M_{i}\}_{i=1}^{n}}{\begin{array}[]{c}n\in\mathbb{N},\ \sum_{j=1}^{n}M_{j}=I,\\ M_{i}\in\mathrm{SEP}(A;B)^{\ast},\\ \operatorname{neg}(M_{i})\leq s\ (\forall i)\end{array}},

where for X𝒯(AB)X\in\mathcal{T}(AB) the value neg(X)\operatorname{neg}(X) is defined as

neg(X)={maxλ<0 eigenvalue of X|λ|if X has a negative eigenvalue,0otherwise.\operatorname{neg}(X)=\begin{dcases}\max_{\begin{subarray}{c}\lambda<0\\ \text{ eigenvalue of }X\end{subarray}}\left|\lambda\right|&\text{if $X$ has a negative eigenvalue},\\ 0&\text{otherwise}.\end{dcases}

To define another one-parameter family (𝒦s)\mathcal{M}(\mathcal{K}_{s}) of measurement classes, we define the following special positive cones.

Definition 4 (One-parameter family of positive cones).

For a vector vABv\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}, let sc(v)\operatorname{sc}(v) be the value

sc(v)={λ1λ2v0,0v=0,\operatorname{sc}(v)=\begin{cases}\lambda_{1}\lambda_{2}&v\not=0,\\ 0&v=0,\end{cases}

where λ1λ2λd\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{d}, d=min{dimA,dimB}d=\min\{\dim\mathcal{H}_{A},\dim\mathcal{H}_{B}\}, denote the Schmidt coefficients of v/vv/\|v\|. Then, for s[0,1/2]s\in[0,1/2], we define the positive cones 𝒦s(0)\mathcal{K}_{s}^{(0)} and 𝒦s\mathcal{K}_{s} as

𝒦s(0)\displaystyle\mathcal{K}_{s}^{(0)} =conv{|vv|vAB,sc(v)s},\displaystyle=\operatorname{conv}\{\ket{v}\!\bra{v}\mid v\in\mathcal{H}_{A}\otimes\mathcal{H}_{B},\ \operatorname{sc}(v)\leq s\},
𝒦s\displaystyle\mathcal{K}_{s} =𝒯+(AB)+Γ(𝒦s(0)),\displaystyle=\mathcal{T}_{+}(AB)+\Gamma(\mathcal{K}_{s}^{(0)}),

where conv(𝒳)\operatorname{conv}(\mathcal{X}) denotes the convex hull of a set 𝒳𝒯(AB)\mathcal{X}\subset\mathcal{T}(AB) and Γ\Gamma denotes the partial transpose on Bob’s system, i.e., Γ\Gamma is the linear map defined by the tensor product of identity map and transposition.

The value sc(v)\operatorname{sc}(v) is closely related to negative eigenvalues of Γ(|vv|)\Gamma(\ket{v}\!\bra{v}): for every vABv\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}

v2sc(v)=neg(Γ(|vv|)).\|v\|^{2}\operatorname{sc}(v)=\operatorname{neg}(\Gamma(\ket{v}\!\bra{v})). (1)

Eq. (1) follows from the fact that, if v0v\not=0, the set of all eigenvalues of Γ(|vv|)/v2\Gamma(\ket{v}\!\bra{v})/\|v\|^{2} is {±λiλj,λk2|1i<jd, 1kd}\set{\pm\lambda_{i}\lambda_{j},\,\lambda_{k}^{2}}{1\leq i<j\leq d,\ 1\leq k\leq d}, where λ1λd\lambda_{1}\geq\cdots\geq\lambda_{d} denote the Schmidt coefficients of v/vv/\|v\|. Hence, the inequality sc(v)s\operatorname{sc}(v)\leq s is a restriction of negative eigenvalues of elements of 𝒦s\mathcal{K}_{s}. Since the Schmidt coefficients of a unit vector vABv\in\mathcal{H}_{A}\otimes\mathcal{H}_{B} represent the amount of entanglement about the pure state |vv|\ket{v}\!\bra{v}, one can also regard the inequality sc(v)s\operatorname{sc}(v)\leq s as a restriction of entanglement on the inside of the partial transpose Γ\Gamma. Also, once the parameter ss increases, the positive cones 𝒦s(0)\mathcal{K}_{s}^{(0)} and 𝒦s\mathcal{K}_{s} become larger. Thus, the following inclusion relations hold:

SEP(A;B)=𝒦0(0)𝒦s(0)𝒦1/2(0)=𝒯+(AB),\displaystyle\mathrm{SEP}(A;B)=\mathcal{K}_{0}^{(0)}\subset\mathcal{K}_{s}^{(0)}\subset\mathcal{K}_{1/2}^{(0)}=\mathcal{T}_{+}(AB),
𝒯+(AB)=𝒦0𝒦s𝒦1/2SEP(A;B).\displaystyle\mathcal{T}_{+}(AB)=\mathcal{K}_{0}\subset\mathcal{K}_{s}\subset\mathcal{K}_{1/2}\subset\mathrm{SEP}(A;B)^{\ast}.

Note that the classes 0\mathcal{M}_{0} and (𝒦0)\mathcal{M}(\mathcal{K}_{0}) are the class of POVMs. Also, since 𝒦s(0)\mathcal{K}_{s}^{(0)} satisfies local unitary invariance, i.e., (UAUB)𝒦s(0)(UAUB)=𝒦s(0)(U_{A}\otimes U_{B})\mathcal{K}_{s}^{(0)}(U_{A}\otimes U_{B})^{\dagger}=\mathcal{K}_{s}^{(0)} for all unitary matrices UAU_{A} and UBU_{B}, no positive cones 𝒦s\mathcal{K}_{s} depend on an orthonormal basis of B\mathcal{H}_{B} that defines the partial transpose Γ\Gamma.

4 Perfect discrimination in approximate QT

Let us consider perfect discrimination of separable pure states by measurements of s\mathcal{M}_{s} and (𝒦s)\mathcal{M}(\mathcal{K}_{s}). First, for separable pure states that are parameterized, we give concrete measurements in the case dimA=dimB=2\dim\mathcal{H}_{A}=\dim\mathcal{H}_{B}=2. Let ρ1\rho_{1} and ρ2\rho_{2} be the separable pure states given as

ρ1=[1000][1000],ρ2=[1α1β1β1α1][1α2β2β2α2],\rho_{1}=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}\otimes\begin{bmatrix}1&0\\ 0&0\end{bmatrix},\quad\rho_{2}=\begin{bmatrix}1-\alpha_{1}&\beta_{1}\\ \beta_{1}&\alpha_{1}\end{bmatrix}\otimes\begin{bmatrix}1-\alpha_{2}&\beta_{2}\\ \beta_{2}&\alpha_{2}\end{bmatrix}, (2)

where α1,α2[0,1]\alpha_{1},\alpha_{2}\in[0,1] and βi=αi(1αi)\beta_{i}=\sqrt{\alpha_{i}(1-\alpha_{i})}. If the relations s[0,1/2]s\in[0,1/2] and

(1α1)(1α2)4s2α1α2(1-\alpha_{1})(1-\alpha_{2})\leq 4s^{2}\alpha_{1}\alpha_{2} (3)

hold, then ρ1\rho_{1} and ρ2\rho_{2} are perfectly distinguishable by some measurement {Ti+Γ(Ti)}i=1,2s\{T_{i}+\Gamma(T_{i})\}_{i=1,2}\in\mathcal{M}_{s}. The measurement {Ti+Γ(Ti)}i=1,2s\{T_{i}+\Gamma(T_{i})\}_{i=1,2}\in\mathcal{M}_{s} is given below except for the trivial cases α1=1\alpha_{1}=1 and α2=1\alpha_{2}=1: If γα1+α2>1\gamma\coloneqq\alpha_{1}+\alpha_{2}>1, then

2γT1=γ|v1v1|+(γ1)|v2v2|+(γ1)|v3v3|,\displaystyle 2\gamma T_{1}=\gamma\ket{v_{1}}\!\bra{v_{1}}+(\gamma-1)\ket{v_{2}}\!\bra{v_{2}}+(\gamma-1)\ket{v_{3}}\!\bra{v_{3}},
v1=[10][10]β1β2α1α2[01][01],\displaystyle v_{1}=\begin{bmatrix}1\\ 0\end{bmatrix}\otimes\begin{bmatrix}1\\ 0\end{bmatrix}-\frac{\beta_{1}\beta_{2}}{\alpha_{1}\alpha_{2}}\begin{bmatrix}0\\ 1\end{bmatrix}\otimes\begin{bmatrix}0\\ 1\end{bmatrix},
v2=[1β1/α1][01],v3=[01][1β2/α2],\displaystyle v_{2}=\begin{bmatrix}1\\ -\beta_{1}/\alpha_{1}\end{bmatrix}\otimes\begin{bmatrix}0\\ 1\\ \end{bmatrix},\quad v_{3}=\begin{bmatrix}0\\ 1\\ \end{bmatrix}\otimes\begin{bmatrix}1\\ -\beta_{2}/\alpha_{2}\end{bmatrix},
T2=(UAUB)T1(UAUB),\displaystyle T_{2}=(U_{A}\otimes U_{B})T_{1}(U_{A}\otimes U_{B})^{\dagger},
UA=1α1[β1α1α1β1],UB=1α2[β2α2α2β2];\displaystyle U_{A}=\frac{1}{\sqrt{\alpha_{1}}}\begin{bmatrix}\beta_{1}&\alpha_{1}\\ \alpha_{1}&-\beta_{1}\end{bmatrix},\quad U_{B}=\frac{1}{\sqrt{\alpha_{2}}}\begin{bmatrix}\beta_{2}&\alpha_{2}\\ \alpha_{2}&-\beta_{2}\end{bmatrix};

if γ=1\gamma=1, then

T1=12[1001000000001001],T2=12[0000011001100000].T_{1}=\frac{1}{2}\begin{bmatrix}1&0&0&-1\\ 0&0&0&0\\ 0&0&0&0\\ -1&0&0&1\end{bmatrix},\quad T_{2}=\frac{1}{2}\begin{bmatrix}0&0&0&0\\ 0&1&1&0\\ 0&1&1&0\\ 0&0&0&0\end{bmatrix}.

When a measurement class is (𝒦s)\mathcal{M}(\mathcal{K}_{s}), Eq. (3) turns to

(1α1)(1α2)tα1α2,(1-\alpha_{1})(1-\alpha_{2})\leq t\alpha_{1}\alpha_{2}, (4)

where s=t/(1+t)s=\sqrt{t}/(1+t) and t[0,1]t\in[0,1].

A simple calculation ensures that the above measurements indeed discriminate the states (2) perfectly. Thus, we only have to examine whether the above measurements are contained in s\mathcal{M}_{s} and (𝒦s)\mathcal{M}(\mathcal{K}_{s}). For details, see supplemental material.

Next, let us consider the general case dimA,dimB2\dim\mathcal{H}_{A},\dim\mathcal{H}_{B}\geq 2. Let ρ1=ρ1Aρ1B\rho_{1}=\rho_{1}^{A}\otimes\rho_{1}^{B} and ρ2=ρ2Aρ2B\rho_{2}=\rho_{2}^{A}\otimes\rho_{2}^{B} be separable pure states. We can take orthonormal bases of A\mathcal{H}_{A} and B\mathcal{H}_{B} such that ρ1\rho_{1} and ρ2\rho_{2} are expressed as (2), i.e., their representation matrices are given by the direct sums of the matrices (2) and the zero matrix. Therefore, the general case is reduced to the case dimA=dimB=2\dim\mathcal{H}_{A}=\dim\mathcal{H}_{B}=2, and we obtain the following theorems.

Theorem 5 (Perfect discrimination with s\mathcal{M}_{s}).

If xTrρ1Aρ2Ax\coloneqq\Tr\rho_{1}^{A}\rho_{2}^{A} and yTrρ1Bρ2By\coloneqq\Tr\rho_{1}^{B}\rho_{2}^{B} satisfy the relations s[0,1/2]s\in[0,1/2] and

xy4s2(1x)(1y),xy\leq 4s^{2}(1-x)(1-y), (5)

then ρ1\rho_{1} and ρ2\rho_{2} are perfectly distinguishable by some measurement of s\mathcal{M}_{s}.

Theorem 6 (Perfect discrimination with (𝒦s)\mathcal{M}(\mathcal{K}_{s})).

If xTrρ1Aρ2Ax\coloneqq\Tr\rho_{1}^{A}\rho_{2}^{A} and yTrρ1Bρ2By\coloneqq\Tr\rho_{1}^{B}\rho_{2}^{B} satisfy the relations s=t/(1+t)s=\sqrt{t}/(1+t), t[0,1]t\in[0,1], and

xyt(1x)(1y),xy\leq t(1-x)(1-y), (6)

then ρ1\rho_{1} and ρ2\rho_{2} are perfectly distinguishable by some measurement of (𝒦s)\mathcal{M}(\mathcal{K}_{s}).

Using Theorems 5 and 6, we find the following drastic improvement of multiple-copy state discrimination. Let σ1\sigma_{1} and σ2\sigma_{2} be distinct pure states on a single quantum system. In QT, the non-trivial nn-copies σ1n\sigma_{1}^{\otimes n} and σ2n\sigma_{2}^{\otimes n} never be perfectly distinguishable, where we say that the nn-copies σ1n\sigma_{1}^{\otimes n} and σ2n\sigma_{2}^{\otimes n} are non-trivial if σ1\sigma_{1} and σ2\sigma_{2} are distinct and non-orthogonal. However, it is known [18] that, for some finite nn, the non-trivial nn-copies σ1n\sigma_{1}^{\otimes n} and σ2n\sigma_{2}^{\otimes n} are perfectly distinguishable by some measurement of (SEP(A;B))\mathcal{M}(\mathrm{SEP}^{\ast}(A;B)). Surprisingly, the same statement is true for the measurement classes s\mathcal{M}_{s} and (𝒦s)\mathcal{M}(\mathcal{K}_{s}) that are slightly larger than the class of POVMs. To see this fact, regarding the 2n2n-copies σ12n=σ1nσ1n\sigma_{1}^{\otimes 2n}=\sigma_{1}^{\otimes n}\otimes\sigma_{1}^{\otimes n} and σ22n=σ2nσ2n\sigma_{2}^{\otimes 2n}=\sigma_{2}^{\otimes n}\otimes\sigma_{2}^{\otimes n} as bipartite separable pure states, we apply Theorem 5 to them. Assume s(0,1/2]s\in(0,1/2]. Since

x=y=Trσ1nσ2n=(Trσ1σ2)nn0,x=y=\Tr\sigma_{1}^{\otimes n}\sigma_{2}^{\otimes n}=(\Tr\sigma_{1}\sigma_{2})^{n}\xrightarrow{n\to\infty}0,

a sufficiently large nn satisfies the inequality (5). Thus, for some finite nn, the 2n2n-copies σ12n\sigma_{1}^{\otimes 2n} and σ22n\sigma_{2}^{\otimes 2n} are perfectly distinguishable by some measurement of s\mathcal{M}_{s}. Also, since a sufficiently large nn satisfies the inequality (6), the same statement is true for (𝒦s)\mathcal{M}(\mathcal{K}_{s}). We summarize these facts as the following corollary and Table 1.

Corollary 7 (Multiple-copy state discrimination).

Assume s(0,1/2]s\in(0,1/2]. Then, for some finite nn\in\mathbb{N}, the 2n2n-copies σ12n=σ1nσ1n\sigma_{1}^{\otimes 2n}=\sigma_{1}^{\otimes n}\otimes\sigma_{1}^{\otimes n} and σ22n=σ2nσ2n\sigma_{2}^{\otimes 2n}=\sigma_{2}^{\otimes n}\otimes\sigma_{2}^{\otimes n} are perfectly distinguishable by some measurement of s\mathcal{M}_{s}. The same statement is true for (𝒦s)\mathcal{M}(\mathcal{K}_{s}).

Table 1: Finite-copy perfect discrimination for each measurement class. Assume s(0,1/2]s\in(0,1/2] here.
Measurement class (𝒯+(AB))\mathcal{M}(\mathcal{T}_{+}(AB)) s\mathcal{M}_{s} (𝒦s)\mathcal{M}(\mathcal{K}_{s}) (SEP(A;B))\mathcal{M}(\mathrm{SEP}(A;B)^{\ast})
Perfect discrimination Impossible Possible Possible Possible
of non-trivial nn-copies for finite nn for finite nn for finite nn

The domains of (x,y)(x,y) in Theorems 5 and 6 are illustrated as Fig. 2. Once the parameter s[0,1/2]s\in[0,1/2] decreases, the domains of (x,y)(x,y) in Theorems 5 and 6 becomes smaller. However, the origin is an interior point of the domain of (x,y)(x,y) (as a subspace of the square [0,1]2[0,1]^{2}) unless ss is zero. This fact is important to understand Corollary 7. To see this importance, recall that the value xn=yn=(Trσ1σ2)nx_{n}=y_{n}=(\Tr\sigma_{1}\sigma_{2})^{n} converges to zero as nn\to\infty. As long as the origin is an interior point of the domain of (x,y)(x,y), for some finite nn the point (xn,yn)(x_{n},y_{n}) lies in the domain of (x,y)(x,y). Since the origin is an interior point of the domains of (x,y)(x,y) in Theorems 5 and 6, we can check Corollary 7 again.

Refer to caption
Figure 2: The domains of (x,y)(x,y) in Theorem 5 (resp. Theorem 6) for s=0,1/4,1/2s=0,1/4,1/2 (resp. (s,t)=(0,0),(2/5,1/4),(1/2,1)(s,t)=(0,0),(2/5,1/4),(1/2,1)).

5 Conclusion

To investigate the performance of perfect discrimination in models approximately close to QT, we have defined the two measurement classes s\mathcal{M}_{s} and (𝒦s)\mathcal{M}(\mathcal{K}_{s}) that are smoothly connected with the class of POVMs (s=0s=0). As a result, unless s=0s=0, the performance of perfect discrimination is drastically improved for both the measurement classes s\mathcal{M}_{s} and (𝒦s)\mathcal{M}(\mathcal{K}_{s}). More precisely, their measurements enable us to discriminate non-trivial 2n2n-copies perfectly for some finite nn. This result suggests that the approximate measurement classes s\mathcal{M}_{s} and (𝒦s)\mathcal{M}(\mathcal{K}_{s}) are unlikely to exist.

Although we have shown perfect discrimination of non-trivial 2n2n-copies for some finite nn, it is interesting to investigate the converse: Let \mathcal{M} be a measurement class. If there never exist nn\in\mathbb{N} and non-trivial nn-copies such that the nn-copies are perfectly distinguishable by some measurement of \mathcal{M}, then is \mathcal{M} contained in POVMs? It is a future work. Also, to consider another problem, assume that there exist pure states σ1\sigma_{1} and σ2\sigma_{2} such that for every nn\in\mathbb{N} the nn-copies σ1n\sigma_{1}^{\otimes n} and σ2n\sigma_{2}^{\otimes n} are not perfectly distinguishable by any measurements of \mathcal{M}. Then it is also interesting to examine the error probability in discriminating σ1n\sigma_{1}^{\otimes n} and σ2n\sigma_{2}^{\otimes n}. If \mathcal{M} is the class of POVMs, the error probability is exponentially decreasing and the exponential decreasing rate is known [19, Section 3]. However, we are interested in the case where \mathcal{M} is general. It is another future work.

YY was supported by Japan Society for the Promotion of Science (JSPS) Grant-in-Aid for JSPS Fellows No. 19J20161. MH was supported in part by JSPS Grant-in-Aids for Scientific Research (A) No. 17H01280 and for Scientific Research (B) No. 16KT0017, and Kayamori Foundation of Information Science Advancement.

Appendix: Proofs of technical lemmas

In this appendix, we prove technical lemmas, which yield Theorems 5 and 6.

Lemma 8 (Perfect discrimination with s\mathcal{M}_{s}).

Let ρ1\rho_{1} and ρ2\rho_{2} be the separable pure states given as

ρ1=[1000][1000],ρ2=[1α1β1β1α1][1α2β2β2α2],\rho_{1}=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}\otimes\begin{bmatrix}1&0\\ 0&0\end{bmatrix},\quad\rho_{2}=\begin{bmatrix}1-\alpha_{1}&\beta_{1}\\ \beta_{1}&\alpha_{1}\end{bmatrix}\otimes\begin{bmatrix}1-\alpha_{2}&\beta_{2}\\ \beta_{2}&\alpha_{2}\end{bmatrix}, (7)

where α1,α2[0,1]\alpha_{1},\alpha_{2}\in[0,1] and βi=αi(1αi)\beta_{i}=\sqrt{\alpha_{i}(1-\alpha_{i})}. If the relations s[0,1/2]s\in[0,1/2] and

(1α1)(1α2)4s2α1α2(1-\alpha_{1})(1-\alpha_{2})\leq 4s^{2}\alpha_{1}\alpha_{2} (8)

hold, then ρ1\rho_{1} and ρ2\rho_{2} are perfectly distinguishable by some measurement {Ti+Γ(Ti)}i=1,2s\{T_{i}+\Gamma(T_{i})\}_{i=1,2}\in\mathcal{M}_{s}. The measurement {Ti+Γ(Ti)}i=1,2s\{T_{i}+\Gamma(T_{i})\}_{i=1,2}\in\mathcal{M}_{s} is given below except for the trivial cases α1=1\alpha_{1}=1 and α2=1\alpha_{2}=1: If γα1+α2>1\gamma\coloneqq\alpha_{1}+\alpha_{2}>1, then

2γT1=γ|v1v1|+(γ1)|v2v2|+(γ1)|v3v3|,\displaystyle 2\gamma T_{1}=\gamma\ket{v_{1}}\!\bra{v_{1}}+(\gamma-1)\ket{v_{2}}\!\bra{v_{2}}+(\gamma-1)\ket{v_{3}}\!\bra{v_{3}}, (9)
v1=[10][10]β1β2α1α2[01][01],\displaystyle v_{1}=\begin{bmatrix}1\\ 0\end{bmatrix}\otimes\begin{bmatrix}1\\ 0\end{bmatrix}-\frac{\beta_{1}\beta_{2}}{\alpha_{1}\alpha_{2}}\begin{bmatrix}0\\ 1\end{bmatrix}\otimes\begin{bmatrix}0\\ 1\end{bmatrix}, (10)
v2=[1β1/α1][01],v3=[01][1β2/α2],\displaystyle v_{2}=\begin{bmatrix}1\\ -\beta_{1}/\alpha_{1}\end{bmatrix}\otimes\begin{bmatrix}0\\ 1\\ \end{bmatrix},\quad v_{3}=\begin{bmatrix}0\\ 1\\ \end{bmatrix}\otimes\begin{bmatrix}1\\ -\beta_{2}/\alpha_{2}\end{bmatrix}, (11)
T2=(UAUB)T1(UAUB),\displaystyle T_{2}=(U_{A}\otimes U_{B})T_{1}(U_{A}\otimes U_{B})^{\dagger}, (12)
UA=1α1[β1α1α1β1],UB=1α2[β2α2α2β2];\displaystyle U_{A}=\frac{1}{\sqrt{\alpha_{1}}}\begin{bmatrix}\beta_{1}&\alpha_{1}\\ \alpha_{1}&-\beta_{1}\end{bmatrix},\quad U_{B}=\frac{1}{\sqrt{\alpha_{2}}}\begin{bmatrix}\beta_{2}&\alpha_{2}\\ \alpha_{2}&-\beta_{2}\end{bmatrix}; (13)

if γ=1\gamma=1, then

T1=12[1001000000001001],T2=12[0000011001100000].T_{1}=\frac{1}{2}\begin{bmatrix}1&0&0&-1\\ 0&0&0&0\\ 0&0&0&0\\ -1&0&0&1\end{bmatrix},\quad T_{2}=\frac{1}{2}\begin{bmatrix}0&0&0&0\\ 0&1&1&0\\ 0&1&1&0\\ 0&0&0&0\end{bmatrix}.
Lemma 9 (Perfect discrimination with (𝒦s)\mathcal{M}(\mathcal{K}_{s})).

Let ρ1\rho_{1} and ρ2\rho_{2} be the separable pure states given as (7). If the relations s=t/(1+t)s=\sqrt{t}/(1+t), t[0,1]t\in[0,1], and

(1α1)(1α2)tα1α2,(1-\alpha_{1})(1-\alpha_{2})\leq t\alpha_{1}\alpha_{2}, (14)

hold, then ρ1\rho_{1} and ρ2\rho_{2} are perfectly distinguishable by the measurement {Ti+Γ(Ti)}i=1,2s\{T_{i}+\Gamma(T_{i})\}_{i=1,2}\in\mathcal{M}_{s} given in Lemma 8.

We prove Lemmas 9 and 8 in this order.

Proof of Lemma 9.

Assume that s=t/(1+t)s=\sqrt{t}/(1+t), t[0,1]t\in[0,1], and (14). All we need is to show that

  1. 1.

    T1+T2+Γ(T1+T2)=IT_{1}+T_{2}+\Gamma(T_{1}+T_{2})=I,

  2. 2.

    Ti𝒦s(0)T_{i}\in\mathcal{K}_{s}^{(0)} for all i=1,2i=1,2,

  3. 3.

    Trρ1T2=Trρ2T1=0\Tr\rho_{1}T_{2}=\Tr\rho_{2}T_{1}=0.

Indeed, if (i) and (ii) hold, then {Ti+Γ(Ti)}i=1,2(𝒦s)\{T_{i}+\Gamma(T_{i})\}_{i=1,2}\in\mathcal{M}(\mathcal{K}_{s}). Also, if (i) and (iii) hold, then the equations Γ(ρi)=ρi\Gamma(\rho_{i})=\rho_{i}, i=1,2i=1,2, imply that Trρi(Tj+Γ(Tj))=2TrρiTj=δij\Tr\rho_{i}(T_{j}+\Gamma(T_{j}))=2\Tr\rho_{i}T_{j}=\delta_{ij} for all i,j{1,2}i,j\in\{1,2\}. Therefore, if (i)–(iii) hold, then Lemma 8 follows. Also, note that (1α2)(1α1)tα1α2α1α2(1-\alpha_{2})(1-\alpha_{1})\leq t\alpha_{1}\alpha_{2}\leq\alpha_{1}\alpha_{2} thanks to t[0,1]t\in[0,1] and (14). Thus γ=α1+α21\gamma=\alpha_{1}+\alpha_{2}\geq 1. If α1α2=0\alpha_{1}\alpha_{2}=0, then α1=1\alpha_{1}=1 or α2=1\alpha_{2}=1, which is a trivial case. Therefore, without loss of generality, we may assume α1α2>0\alpha_{1}\alpha_{2}>0.

Proof of (i). First, assume γ=1\gamma=1. Then

T1+T2+Γ(T1+T2)=12[1001011001101001]+12[1001011001101001]=I.T_{1}+T_{2}+\Gamma(T_{1}+T_{2})=\frac{1}{2}\begin{bmatrix}1&0&0&-1\\ 0&1&1&0\\ 0&1&1&0\\ -1&0&0&1\\ \end{bmatrix}+\frac{1}{2}\begin{bmatrix}1&0&0&1\\ 0&1&-1&0\\ 0&-1&1&0\\ 1&0&0&1\\ \end{bmatrix}=I.

Next, assume γ>1\gamma>1. Put wi=(UAUB)viw_{i}=(U_{A}\otimes U_{B})v_{i} for i=1,2,3i=1,2,3. Then wiw_{i}, i=1,2,3i=1,2,3, can be calculated as follows:

w1\displaystyle w_{1} =1α1α2([β1α1][β2α2]β1β2α1α2[α1β1][α2β2])\displaystyle=\frac{1}{\sqrt{\alpha_{1}\alpha_{2}}}\biggl{(}\begin{bmatrix}\beta_{1}\\ \alpha_{1}\end{bmatrix}\otimes\begin{bmatrix}\beta_{2}\\ \alpha_{2}\end{bmatrix}-\frac{\beta_{1}\beta_{2}}{\alpha_{1}\alpha_{2}}\begin{bmatrix}\alpha_{1}\\ -\beta_{1}\end{bmatrix}\otimes\begin{bmatrix}\alpha_{2}\\ -\beta_{2}\end{bmatrix}\biggr{)}
=1α1α2([β1β2β1α2α1β2α1α2]β1β2α1α2[α1α2α1β2β1α2β1β2])=1α1α2[0β1β2γ1],\displaystyle=\frac{1}{\sqrt{\alpha_{1}\alpha_{2}}}\Biggl{(}\begin{bmatrix}\beta_{1}\beta_{2}\\ \beta_{1}\alpha_{2}\\ \alpha_{1}\beta_{2}\\ \alpha_{1}\alpha_{2}\end{bmatrix}-\frac{\beta_{1}\beta_{2}}{\alpha_{1}\alpha_{2}}\begin{bmatrix}\alpha_{1}\alpha_{2}\\ -\alpha_{1}\beta_{2}\\ -\beta_{1}\alpha_{2}\\ \beta_{1}\beta_{2}\end{bmatrix}\Biggr{)}=\frac{1}{\sqrt{\alpha_{1}\alpha_{2}}}\begin{bmatrix}0\\ \beta_{1}\\ \beta_{2}\\ \gamma-1\end{bmatrix},
w2=1α1α2[01][α2β2]=α2α1v3,w3=1α1α2[α1β1][01]=α1α2v2.w_{2}=\frac{1}{\sqrt{\alpha_{1}\alpha_{2}}}\begin{bmatrix}0\\ 1\end{bmatrix}\otimes\begin{bmatrix}\alpha_{2}\\ -\beta_{2}\end{bmatrix}=\sqrt{\frac{\alpha_{2}}{\alpha_{1}}}v_{3},\quad w_{3}=\frac{1}{\sqrt{\alpha_{1}\alpha_{2}}}\begin{bmatrix}\alpha_{1}\\ -\beta_{1}\end{bmatrix}\otimes\begin{bmatrix}0\\ 1\end{bmatrix}=\sqrt{\frac{\alpha_{1}}{\alpha_{2}}}v_{2}.

Thus, putting ξ=β1β2/α1α2\xi=\beta_{1}\beta_{2}/\alpha_{1}\alpha_{2}, we have

T1+T2=12(|v1v1|+|w1w1|)+γ12γ(|v2v2|+|v3v3|+|w2w2|+|w3w3|)\displaystyle\quad T_{1}+T_{2}=\frac{1}{2}(\ket{v_{1}}\!\bra{v_{1}}+\ket{w_{1}}\!\bra{w_{1}})+\frac{\gamma-1}{2\gamma}(\ket{v_{2}}\!\bra{v_{2}}+\ket{v_{3}}\!\bra{v_{3}}+\ket{w_{2}}\!\bra{w_{2}}+\ket{w_{3}}\!\bra{w_{3}})
=12(|v1v1|+|w1w1|)+γ12γ(|v2v2|+|v3v3|+α2α1|v3v3|+α1α2|v2v2|)\displaystyle=\frac{1}{2}(\ket{v_{1}}\!\bra{v_{1}}+\ket{w_{1}}\!\bra{w_{1}})+\frac{\gamma-1}{2\gamma}\Bigl{(}\ket{v_{2}}\!\bra{v_{2}}+\ket{v_{3}}\!\bra{v_{3}}+\frac{\alpha_{2}}{\alpha_{1}}\ket{v_{3}}\!\bra{v_{3}}+\frac{\alpha_{1}}{\alpha_{2}}\ket{v_{2}}\!\bra{v_{2}}\Bigr{)}
=12(|v1v1|+|w1w1|)+γ12(1α2|v2v2|+1α1|v3v3|)\displaystyle=\frac{1}{2}(\ket{v_{1}}\!\bra{v_{1}}+\ket{w_{1}}\!\bra{w_{1}})+\frac{\gamma-1}{2}\Bigl{(}\frac{1}{\alpha_{2}}\ket{v_{2}}\!\bra{v_{2}}+\frac{1}{\alpha_{1}}\ket{v_{3}}\!\bra{v_{3}}\Bigr{)}
=12[100ξ00000000ξ00ξ2]+12α1α2[00000β12β1β2(γ1)β10β1β2β22(γ1)β20(γ1)β1(γ1)β2(γ1)2]\displaystyle=\frac{1}{2}\begin{bmatrix}1&0&0&-\xi\\ 0&0&0&0\\ 0&0&0&0\\ -\xi&0&0&\xi^{2}\end{bmatrix}+\frac{1}{2\alpha_{1}\alpha_{2}}\begin{bmatrix}0&0&0&0\\ 0&\beta_{1}^{2}&\beta_{1}\beta_{2}&(\gamma-1)\beta_{1}\\ 0&\beta_{1}\beta_{2}&\beta_{2}^{2}&(\gamma-1)\beta_{2}\\ 0&(\gamma-1)\beta_{1}&(\gamma-1)\beta_{2}&(\gamma-1)^{2}\end{bmatrix}
+γ12α1α2([α1β1β11α1][0001]+[0001][α2β2β21α2])\displaystyle\quad+\frac{\gamma-1}{2\alpha_{1}\alpha_{2}}\biggl{(}\begin{bmatrix}\alpha_{1}&-\beta_{1}\\ -\beta_{1}&1-\alpha_{1}\end{bmatrix}\otimes\begin{bmatrix}0&0\\ 0&1\end{bmatrix}+\begin{bmatrix}0&0\\ 0&1\end{bmatrix}\otimes\begin{bmatrix}\alpha_{2}&-\beta_{2}\\ -\beta_{2}&1-\alpha_{2}\end{bmatrix}\biggr{)}
=12[100ξ00000000ξ00ξ2]+12α1α2[00000β12β1β2(γ1)β10β1β2β22(γ1)β20(γ1)β1(γ1)β2(γ1)2]\displaystyle=\frac{1}{2}\begin{bmatrix}1&0&0&-\xi\\ 0&0&0&0\\ 0&0&0&0\\ -\xi&0&0&\xi^{2}\end{bmatrix}+\frac{1}{2\alpha_{1}\alpha_{2}}\begin{bmatrix}0&0&0&0\\ 0&\beta_{1}^{2}&\beta_{1}\beta_{2}&(\gamma-1)\beta_{1}\\ 0&\beta_{1}\beta_{2}&\beta_{2}^{2}&(\gamma-1)\beta_{2}\\ 0&(\gamma-1)\beta_{1}&(\gamma-1)\beta_{2}&(\gamma-1)^{2}\end{bmatrix}
+γ12α1α2[00000α10β100000β101α1]+γ12α1α2[0000000000α2β200β21α2].\displaystyle\quad+\frac{\gamma-1}{2\alpha_{1}\alpha_{2}}\begin{bmatrix}0&0&0&0\\ 0&\alpha_{1}&0&-\beta_{1}\\ 0&0&0&0\\ 0&-\beta_{1}&0&1-\alpha_{1}\end{bmatrix}+\frac{\gamma-1}{2\alpha_{1}\alpha_{2}}\begin{bmatrix}0&0&0&0\\ 0&0&0&0\\ 0&0&\alpha_{2}&-\beta_{2}\\ 0&0&-\beta_{2}&1-\alpha_{2}\end{bmatrix}.

When tijt_{ij} denotes the (i,j)(i,j)-th entry of T1+T2T_{1}+T_{2}, it follows that t11=1/2t_{11}=1/2, t12=t21=t13=t31=0t_{12}=t_{21}=t_{13}=t_{31}=0, t14=t41=ξ/2t_{14}=t_{41}=-\xi/2, t23=t32=ξ/2t_{23}=t_{32}=\xi/2,

t24=t42=(γ1)β12α1α2(γ1)β12α1α2=0,t34=t43=(γ1)β22α1α2(γ1)β22α1α2=0,t_{24}=t_{42}=\frac{(\gamma-1)\beta_{1}}{2\alpha_{1}\alpha_{2}}-\frac{(\gamma-1)\beta_{1}}{2\alpha_{1}\alpha_{2}}=0,\quad t_{34}=t_{43}=\frac{(\gamma-1)\beta_{2}}{2\alpha_{1}\alpha_{2}}-\frac{(\gamma-1)\beta_{2}}{2\alpha_{1}\alpha_{2}}=0,
t22\displaystyle t_{22} =β122α1α2+(γ1)α12α1α2=1α12α2+γ12α2=1/2,\displaystyle=\frac{\beta_{1}^{2}}{2\alpha_{1}\alpha_{2}}+\frac{(\gamma-1)\alpha_{1}}{2\alpha_{1}\alpha_{2}}=\frac{1-\alpha_{1}}{2\alpha_{2}}+\frac{\gamma-1}{2\alpha_{2}}=1/2,
t33\displaystyle t_{33} =β222α1α2+(γ1)α22α1α2=1α22α1+γ12α1=1/2,\displaystyle=\frac{\beta_{2}^{2}}{2\alpha_{1}\alpha_{2}}+\frac{(\gamma-1)\alpha_{2}}{2\alpha_{1}\alpha_{2}}=\frac{1-\alpha_{2}}{2\alpha_{1}}+\frac{\gamma-1}{2\alpha_{1}}=1/2,
t44\displaystyle t_{44} =ξ22+(γ1)22α1α2+(γ1)(1α1)2α1α2+(γ1)(1α2)2α1α2\displaystyle=\frac{\xi^{2}}{2}+\frac{(\gamma-1)^{2}}{2\alpha_{1}\alpha_{2}}+\frac{(\gamma-1)(1-\alpha_{1})}{2\alpha_{1}\alpha_{2}}+\frac{(\gamma-1)(1-\alpha_{2})}{2\alpha_{1}\alpha_{2}}
=(1α1)(1α2)2α1α2+(γ1)22α1α2+(γ1)(2γ)2α1α2\displaystyle=\frac{(1-\alpha_{1})(1-\alpha_{2})}{2\alpha_{1}\alpha_{2}}+\frac{(\gamma-1)^{2}}{2\alpha_{1}\alpha_{2}}+\frac{(\gamma-1)(2-\gamma)}{2\alpha_{1}\alpha_{2}}
=1γ+α1α22α1α2+γ12α1α2=1/2.\displaystyle=\frac{1-\gamma+\alpha_{1}\alpha_{2}}{2\alpha_{1}\alpha_{2}}+\frac{\gamma-1}{2\alpha_{1}\alpha_{2}}=1/2.

Therefore,

T1+T2+Γ(T1+T2)=12[100ξ01ξ00ξ10ξ001]+12[100ξ01ξ00ξ10ξ001]=I.T_{1}+T_{2}+\Gamma(T_{1}+T_{2})=\frac{1}{2}\begin{bmatrix}1&0&0&-\xi\\ 0&1&\xi&0\\ 0&\xi&1&0\\ -\xi&0&0&1\end{bmatrix}+\frac{1}{2}\begin{bmatrix}1&0&0&\xi\\ 0&1&-\xi&0\\ 0&-\xi&1&0\\ \xi&0&0&1\end{bmatrix}=I.

Proof of (ii). First, assume γ=1\gamma=1. Then t[0,1]t\in[0,1] and (14) implies that

α1α2=γ=1(1α2)(1α1)(14)tα1α2t[0,1]α1α2,\alpha_{1}\alpha_{2}\overset{\gamma=1}{=}(1-\alpha_{2})(1-\alpha_{1})\overset{\text{\eqref{supp-mat-S2}}}{\leq}t\alpha_{1}\alpha_{2}\overset{t\in[0,1]}{\leq}\alpha_{1}\alpha_{2},

whence t=1t=1 and s=1/2s=1/2. Since it is easily checked that Ti𝒦1/2(0)T_{i}\in\mathcal{K}_{1/2}^{(0)} for all i=1,2i=1,2, we obtain (ii). Next, assume γ>1\gamma>1. Since the function t/(1+t)\sqrt{t^{\prime}}/(1+t^{\prime}), t[0,1]t^{\prime}\in[0,1], is increasing, from (10) and (11), it follows that sc(v2)=sc(v3)=0\operatorname{sc}(v_{2})=\operatorname{sc}(v_{3})=0 and

sc(v1)=β1β2/α1α21+(β1β2/α1α2)2(14)t1+t=s.\operatorname{sc}(v_{1})=\frac{\beta_{1}\beta_{2}/\alpha_{1}\alpha_{2}}{1+(\beta_{1}\beta_{2}/\alpha_{1}\alpha_{2})^{2}}\overset{\text{\eqref{supp-mat-S2}}}{\leq}\frac{\sqrt{t}}{1+t}=s.

Thus T1𝒦s(0)T_{1}\in\mathcal{K}_{s}^{(0)}. Thanks to (12), we also have T2𝒦s(0)T_{2}\in\mathcal{K}_{s}^{(0)}. Therefore, (ii) holds.

Proof of (iii). First, assume γ=1\gamma=1. Then it is easily checked that Trρ1T2=0\Tr\rho_{1}T_{2}=0 and

Trρ2T1=(1α1)(1α2)+α1α22β1β2=γ=1α1α2+α1α22α1α2=0,\Tr\rho_{2}T_{1}=(1-\alpha_{1})(1-\alpha_{2})+\alpha_{1}\alpha_{2}-2\beta_{1}\beta_{2}\overset{\gamma=1}{=}\alpha_{1}\alpha_{2}+\alpha_{1}\alpha_{2}-2\alpha_{1}\alpha_{2}=0,

which are just (iii). Next, assume γ>1\gamma>1. Since the equations (9), ρ2|v2=ρ2|v3=0\rho_{2}\ket{v_{2}}=\rho_{2}\ket{v_{3}}=0, and

v1|ρ2|v1\displaystyle\braket{v_{1}}{\rho_{2}}{v_{1}} =(1α1)(1α2)+(β1β2α1α2)2α1α2β1β2α1α22β1β2\displaystyle=(1-\alpha_{1})(1-\alpha_{2})+\Bigl{(}\frac{\beta_{1}\beta_{2}}{\alpha_{1}\alpha_{2}}\Bigr{)}^{2}\alpha_{1}\alpha_{2}-\frac{\beta_{1}\beta_{2}}{\alpha_{1}\alpha_{2}}\cdot 2\beta_{1}\beta_{2}
=(1α1)(1α2)(β1β2)2α1α2=0\displaystyle=(1-\alpha_{1})(1-\alpha_{2})-\frac{(\beta_{1}\beta_{2})^{2}}{\alpha_{1}\alpha_{2}}=0

hold, we have

Trρ2T1=12v1|ρ2|v1+γ12γv2|ρ2|v2+γ12γv3|ρ2|v3=0.\Tr\rho_{2}T_{1}=\frac{1}{2}\braket{v_{1}}{\rho_{2}}{v_{1}}+\frac{\gamma-1}{2\gamma}\braket{v_{2}}{\rho_{2}}{v_{2}}+\frac{\gamma-1}{2\gamma}\braket{v_{3}}{\rho_{2}}{v_{3}}=0.

Moreover, since ρ2=(UAUB)ρ1(UAUB)\rho_{2}=(U_{A}\otimes U_{B})^{\dagger}\rho_{1}(U_{A}\otimes U_{B}) holds, we obtain

Trρ1T2\displaystyle\Tr\rho_{1}T_{2} =(12)Trρ1(UAUB)T1(UAUB)\displaystyle\overset{\text{\eqref{supp-mat-local-uni}}}{=}\Tr\rho_{1}(U_{A}\otimes U_{B})T_{1}(U_{A}\otimes U_{B})^{\dagger}
=Tr(UAUB)ρ1(UAUB)T1=Trρ2T1=0.\displaystyle=\Tr(U_{A}\otimes U_{B})^{\dagger}\rho_{1}(U_{A}\otimes U_{B})T_{1}=\Tr\rho_{2}T_{1}=0.

Therefore, (iii) holds. ∎

Proof of Lemma 8.

Assume that s[0,1/2]s\in[0,1/2] and (8). All we need is to show that

  1. 1.

    T1+T2+Γ(T1+T2)=IT_{1}+T_{2}+\Gamma(T_{1}+T_{2})=I,

  2. 2.

    neg(Ti+Γ(Ti))s\operatorname{neg}(T_{i}+\Gamma(T_{i}))\leq s for all i=1,2i=1,2,

  3. 3.

    Trρ1T2=Trρ2T1=0\Tr\rho_{1}T_{2}=\Tr\rho_{2}T_{1}=0,

by the same reason as the proof of Lemma 8. Since (i) and (iii) have been already proved, we show only (ii). Also, the inequality γ=α1+α21\gamma=\alpha_{1}+\alpha_{2}\geq 1 holds, and we may assume α1α2>0\alpha_{1}\alpha_{2}>0, by the same reason as the proof of Lemma 8.

Proof of (ii). First, assume γ=1\gamma=1. Then s[0,1/2]s\in[0,1/2] and (8) implies that

α1α2=γ=1(1α2)(1α1)(8)4s2α1α2s[0,1/2]α1α2,\alpha_{1}\alpha_{2}\overset{\gamma=1}{=}(1-\alpha_{2})(1-\alpha_{1})\overset{\text{\eqref{supp-mat-S1}}}{\leq}4s^{2}\alpha_{1}\alpha_{2}\overset{s\in[0,1/2]}{\leq}\alpha_{1}\alpha_{2},

whence s=1/2s=1/2. Since it is easily checked that neg(Ti+Γ(Ti))=1/2\operatorname{neg}(T_{i}+\Gamma(T_{i}))=1/2 for all i=1,2i=1,2, we obtain (ii). Next, assume γ>1\gamma>1. Then

neg(T1+Γ(T1))\displaystyle\operatorname{neg}(T_{1}+\Gamma(T_{1})) neg(Γ(T1))(11)neg(Γ(12|v1v1|))\displaystyle\leq\operatorname{neg}(\Gamma(T_{1}))\overset{\text{\eqref{supp-mat-eq01''}}}{\leq}\operatorname{neg}\Bigl{(}\Gamma\Bigl{(}\frac{1}{2}\ket{v_{1}}\!\bra{v_{1}}\Bigr{)}\Bigr{)}
=(1)12v12sc(v1)=(10)β1β22α1α2(8)s.\displaystyle\overset{\text{\eqref{sc-neg}}}{=}\frac{1}{2}\|v_{1}\|^{2}\operatorname{sc}(v_{1})\overset{\text{\eqref{supp-mat-eq01'}}}{=}\frac{\beta_{1}\beta_{2}}{2\alpha_{1}\alpha_{2}}\overset{\text{\eqref{supp-mat-S1}}}{\leq}s.

Since (12) holds, we have neg(T2+Γ(T2))neg(Γ(T2))=neg(Γ(T1))s\operatorname{neg}(T_{2}+\Gamma(T_{2}))\leq\operatorname{neg}(\Gamma(T_{2}))=\operatorname{neg}(\Gamma(T_{1}))\leq s. Therefore, (ii) holds. ∎

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