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Acta Mathematica Vietnamica

, Volume 43, Issue 4, pp 661–674 | Cite as

Generalizing Choi-Like Maps

  • Dariusz Chruściński
  • Marcin Marciniak
  • Adam Rutkowski
Open Access
Article
  • 265 Downloads

Abstract

A problem of further generalization of generalized Choi maps Φ[a,b,c] acting on \(\mathbb M_{3}\) introduced by Cho, Kye, and Lee is discussed. Some necessary conditions for positivity of the generalized maps are provided as well as some sufficient conditions. Also, some sufficient conditions for decomposability of these maps are shown.

Keywords

Positive map Indecomposable Choi matrix PPT Separable state 

Mathematics Subject Classification (2010)

Primary 46L60 15B48 81P40 Secondary 81Q10 46L05 

1 Introduction

The paper concerns linear maps acting between matrix algebras. For \(n\in \mathbb N\), by \(\mathbb M_{n},\) we denote the algebra of square n × n-matrices X = (xij) with complex coefficients. Let \({\Phi }:\mathbb M_{m}\to \mathbb M_{n}\) be a linear map. We say that Φ is a positive map if Φ(X) is a positive definite matrix in \(\mathbb M_{n}\) for every positive definite matrix \(X\in \mathbb M_{m}\). Be reminded that for \(k\in \mathbb N,\) there are the following identifications: \(\mathbb M_{k}(\mathbb M_{m})=\mathbb M_{k}\otimes \mathbb M_{m}=\mathbb M_{km}\), where \(\mathbb M_{k}(\mathbb M_{m})\) denotes the C*-algebra of k × k-matrices (Xij) with \(X_{ij}\in \mathbb M_{m}\). Define a map \({\Phi }_{k}:\mathbb M_{k}(\mathbb M_{m})\to \mathbb M_{k}(\mathbb M_{n})\) by Φk(Xij) = (Φ(Xij)). We say that the map Φ is k-positive if Φk is a positive map. If Φ is k-positive for every \(k\in \mathbb N,\) then it is called a completely positive map.

The transposition map \(\theta _{n}:\mathbb M_{n}\to \mathbb M_{n}:(x_{ij})\mapsto (x_{ji})\) is known to be not completely positive map (even not 2-positive) [22]. We say that a map \({\Phi }:\mathbb M_{n}\to \mathbb M_{n}\) is k-copositive (respectively completely copositive) if the composition 𝜃n ∘Φ is k-positive (respectively completely positive). A map Φ is said to be decomposable if it can be expressed as a sum Φ = Φ1 + Φ2, where Φ1 is a completely positive map while Φ2 is a completely copositive one. Otherwise, Φ is called indecomposable. For further details, we refer the reader to the book of Størmer [22].

The systematic study of positive maps on C*-algebras was started by a pioneering work of Erling Størmer in the 1960s of the last century [20]. Although the definitions are easy and the above notions seem to be rather elementary, the problem of description of all positive maps is still unsolved even in the case of low dimensional matrix algebras. In particular, still, there is no effective characterization of decomposable maps. In recent years, the importance of the theory of positive maps drastically increased because of its applications in mathematical physics, especially in rapidly emerging theory of quantum information [6, 10, 13].

It was shown by Størmer [20] and Woronowicz [25] that every positive map \({\Phi }:\mathbb M_{2}\to \mathbb M_{2}\) (Størmer) or \({\Phi }:\mathbb M_{2}\to \mathbb M_{3}\) (Woronowicz) is decomposable. The first example of an indecomposable positive map was given by Choi [3, 4]. It is a map \({\Phi }:\mathbb M_{3}\to \mathbb M_{3}\) given by
$$ {\Phi}(X)=\left( \begin{array}{ccc} x_{11}+x_{33} & -x_{12} & -x_{13}\\ -x_{21} & x_{22}+x_{11} & -x_{23}\\ -x_{31} & -x_{32} & x_{33}+x_{22} \end{array}\right). $$
(1.1)
Although there is no systematic prescription of indecomposable maps, there are numerous examples scattered across literature [5, 7, 8, 9, 14, 15, 16, 17, 18, 19, 23, 24].
An interesting approach to generalization of (1.1) was presented in [1]. For any triple of nonnegative numbers a,b,c define a map \({\Phi }_{[a,b,c]}:\mathbb M_{3}\to \mathbb M_{3}\) by
$${\Phi}_{[a,b,c]}(X)=\left( \begin{array}{ccc} ax_{11}+bx_{22}+cx_{33} & -x_{12} & -x_{13}\\ -x_{21} & cx_{11}+ax_{22}+bx_{33} & -x_{23}\\ -x_{31} & -x_{32} & bx_{11}+cx_{22}+ax_{33} \end{array}\right), $$
with xij being the matrix elements of \(X\in \mathbb M_{3}(\mathbb C)\). One proves

Theorem 1.1

([1]) The map Φ[a,b,c] is positive but not completely positive if and only if
  1. (1)

    0 ≤ a < 2,

     
  2. (2)

    a + b + c ≥ 2,

     
  3. (3)

    if a ≤ 1,then bc ≥ (1 − a)2.

     
Moreover, being positive it is indecomposable if and only if4bc < (2 − a)2.
Slightly different class was considered by Kye [12]
$${\Phi}_{[a;c_{1},c_{2},c_{3}]}[X]=\left( \begin{array}{ccc} ax_{11}+c_{1}x_{33} & -x_{12} & -x_{13}\\ -x_{21} & c_{2}x_{11}+ax_{22} & -x_{23}\\ -x_{31} & -x_{32} & c_{3}x_{22}+ax_{33} \end{array}\right), $$
with a,c1,c2,c3 ≥ 0. He proved the following

Theorem 1.2

A map \({\Phi }_{[a;c_{1},c_{2},c_{3}]}\) is positive but not completely positive if and only if (1) 1 ≤ a < 2,(2) c1c2c3 ≥ (2 − a)3.Moreover, \({\Phi }_{[a;c_{1},c_{2},c_{3}]}\) is indecomposable and atomic.

Interestingly, Osaka shown

Theorem 1.3

([17]) A map \({\Phi }_{[1;c_{1},c_{2},c_{3}]}\) is extremal if c1c2c3 = 1.

Let us be reminded that for a linear map \({\Phi }:\mathbb M_{m}\to \mathbb M_{n},\) one can define its Choi matrix [2]. It is an element \(C_{{\Phi }}\in \mathbb M_{m}(\mathbb M_{n})\) defined by
$$ C_{{\Phi}}=({\Phi}(E_{ij})), $$
(1.2)
where {Eij : 1 ≤ i,jn} is a system of matrix units in \(\mathbb M_{m}\). The mapping Φ↦CΦ is the so called Choi-Jamiołkowski isomorphism between \(L(\mathbb M_{m},\mathbb M_{n})\) and \(\mathbb M_{m}(\mathbb M_{n})\) [11]. One has

Theorem 1.4

([2]) A mapΦ is completely positive if and only if its Choi matrixCΦ is positive definite.

It turns out also that it is possible to characterize positivity and decomposability of Φ in terms of the matrix CΦ. To this end, let us recall that CΦ can be considered as an element of the tensor product \(\mathbb M_{m}\otimes \mathbb M_{n}\). An element ρMmMn is called a PPT matrix if both ρ and \((\text {id}_{\mathbb M_{m}}\otimes \theta _{n})(\rho )\) are positive definite.

Theorem 1.5

([10, 13, 21]) Let Φ be a linear map.
  1. (i)
    Φ is positive if and only if
    $$\left\langle\xi\otimes\eta,C_{{\Phi}}\xi\otimes\eta\right\rangle\geq 0 $$
    for every vectors \(\xi \in \mathbb C^{m}\), \(\eta \in \mathbb C^{n}\).
     
  2. (ii)
    Φ is decomposable if and only if
    $$\text{Tr}(\rho C_{{\Phi}})\geq 0 $$
    for every PPT matrix \(\rho \in \mathbb M_{m}\otimes \mathbb M_{n}\).
     

The property (i) in the above theorem is called block-positivity [13].

2 Choi-Like Maps

In this paper, we consider the following generalization of the maps described in Introduction. Let
$$ A=\left( \begin{array}{ccc} a_{1} & b_{1} & c_{1}\\ c_{2} & a_{2} & b_{2}\\ b_{3} & c_{3} & a_{3} \end{array}\right), $$
(2.1)
with ai,bj,ck ≥ 0, and define \({\Phi }_{A}:\mathbb M_{3}\to \mathbb M_{3}\) as follows
$$ {\Phi}_{A}(X)=\left( \begin{array}{ccc} a_{1}x_{11}+b_{1}x_{22}+c_{1}x_{33} & -x_{12} & -x_{13}\\ -x_{21} & c_{2}x_{11}+a_{2}x_{22}+b_{2}x_{33} & -x_{23}\\ -x_{31} & -x_{32} & b_{3}x_{11}+c_{3}x_{22}+a_{3}x_{33} \end{array}\right). $$
(2.2)
Our aim is to describe some conditions for positivity of ΦA as well as for decomposability.

Firstly, let us make the following observation.

Proposition 2.1

A mapΦA is completely positive if and only if
$$ \left( \begin{array}{ccc} a_{1} & -1 & -1\\ -1 & a_{2} & -1\\ -1 & -1 & a_{3} \end{array}\right) $$
(2.3)
is positive definite matrix.

Proof

Observe that the Choi matrix of the map ΦA is of the form
where we used dots instead of zeros. Now, we apply Theorem 1.4. Since bj,ck are nonnegative, positive definiteness of \(C_{{\Phi }_{A}}\) is equivalent to positive definiteness of the submatrix (2.3). □

Now, let us discuss some necessary conditions for positivity.

Theorem 2.2

A necessary condition for positivity of ΦA is positivity of \({\Phi }_{[\overline {a},\overline {b},\overline {c}]}\), where
$$\begin{array}{@{}rcl@{}} \overline{a}=\frac{1}{3}(a_{1}+a_{2}+a_{3}),\quad \overline{b}=\frac{1}{3}(b_{1}+b_{2}+b_{3}),\quad \overline{c}=\frac{1}{3}(c_{1}+c_{2}+c_{3}). \end{array} $$
(2.4)

Proof

Let ΦA be a positive map and consider a linear map
$$\overline{{\Phi}}_{A}(X)=\frac{1}{3}\left( {\Phi}_{A}(X)+S{\Phi}_{A}(S^{*}XS)S^{*}+S^{*}{\Phi}_{A}(SXS^{*})S\right), $$
where S denotes a unitary shift Sei = ei+ 1. It is clear that \(\overline {{\Phi }}_{A}\) is positive. Moreover, \(\overline {{\Phi }}_{A}={\Phi }[\overline {a},\overline {b},\overline {c}]\) with \(\overline {a},\overline {b},\overline {c}\) defined in (2.4). □
Let us introduce the following notation
$$\begin{array}{@{}rcl@{}} a_{*}=(a_{1}a_{2}a_{3})^{1/3},\quad b_{*}=(b_{1}b_{2}b_{3})^{1/3},\quad c_{*}=(c_{1}c_{2}c_{3})^{1/3}. \end{array} $$
To find sufficient conditions, consider a positive map Φ[a,b,c] and define
$${\Phi}_{[a,b,c]}^{V}(X):=V^{-1/2}{\Phi}_{[a,b,c]}\left( V^{1/2}XV^{1/2}\right)V^{-1/2}, $$
where \(V={\sum }_{i = 1}^{3}p_{i}E_{ii}\) with pi > 0. It is clear that \({\Phi }_{[a,b,c]}^{V}\) is positive. Moreover, one easily finds that \({\Phi }_{[a,b,c]}^{V}={\Phi }_{A}\) with
$$ a_{i}=a_{*},\quad b_{i}=\frac{p_{i + 1}}{p_{i}}b,\quad c_{i}=\frac{p_{i + 2}}{p_{i}}c,\quad i = 1,2,3, $$
(2.5)
that is, A = V − 1A[a,b,c]V. Note that b = b, c = c, and bici+ 1 = bc. Hence, one arrives at the following

Theorem 2.3

Assume that there exista,b,c and V such thatΦ[a,b,c] is a positive map and all coefficients of the matrix
$$M=A-V^{-1}A_{[a,b,c]}V, $$
are non-negative numbers. Then ΦA is a positive map.

Corollary 2.4

Assume that there exist a,b,c ≥ 0 such that
  1. (1)

    Φ[a,b,c] defines a positive map,

     
  2. (2)

    min{a1,a2,a3}≥ a,bb,cc,

     
  3. (3)

    bici+ 1bc for i = 1,2,3.

     
Then ΦA is a positive map.

Proof

We have b1b2b3b3 and c1c2c3c3. We construct a positive map with \(b^{\prime }_{k}\) and \(c^{\prime }_{k}\) such that \(b_{k} \geq b^{\prime }_{k}\) and \(c_{k} \geq c^{\prime }_{k}\). Let us take p1,p2,p3 such that
$$b_{1} = b^{\prime}_{1} := \frac{p_{2}}{p_{1}}b, \quad b_{2} = b^{\prime}_{2} := \frac{p_{3}}{p_{2}} b, \quad b_{3} \geq b^{\prime}_{3} := \frac{p_{1}}{p_{3}} b.$$
Clearly, \(b_{1} b_{2} b_{3} \geq b^{\prime }_{1} b^{\prime }_{2} b^{\prime }_{3} = b^{3}\). Now, b1c2bc and hence, let us take \(c_{2} \geq c^{\prime }_{2} := \frac {p_{1}}{p_{2}} c\) and similarly for c3 and c1.

Now, the map with \(b^{\prime }_{k}\) and \(c^{\prime }_{k}\) is positive due to (2.5). Hence, the map with bk and ck is positive as well. □

Remark 2.5

If b1b2b3 = 0, then a sufficient condition reduces to (1) a1,a2,a3a, (2) c1c2c3 ≥ (2 − a)3, with 1 ≤ a < 2. Interestingly, if a1 = a2 = a3 = a, then a ≥ 1 and c1c2c3 ≥ (2 − a)3 are necessary and sufficient [12]. Indeed, if ΦA is positive and c1c2c3 = c3 let us define V by taking
$$c_{1}=\frac{p_{3}}{p_{1}}c,\quad c_{2}=\frac{p_{1}}{p_{2}}c,\quad c_{3}=\frac{p_{2}}{p_{3}}c. $$
Note the map Φ[a,b,c] has to be positive and b = 0, hence a + c ≥ 2. Therefore, \(a+\sqrt [3]{c_{1}c_{2}c_{3}}\geq 2\) which implies c1c2c3 ≥ (2 − a)3.

We end this section with the following remark.

Remark 2.6

Let us observe that condition (3) in Theorem 1.1 can equivalently be written as
$$a+\sqrt{bc}\geq 1. $$
On the other hand, the condition (2) of Theorem 1.2 can be rewritten as
$$a+c_{*}\geq 2. $$
Let us also remind that for a matrix
$$A=\left( \begin{array}{cc}a_{1} & b_{1} \\b_{2} & a_{2} \end{array}\right)$$
one can define a map \({\Phi }_{A}:\mathbb M_{2}\to \mathbb M_{2}\) similarly to (2.2). It was shown by Kossakowski (see [6, Example 7.1]) that ΦA is positive if and only if ai ≥ 0, bi ≥ 0 and
$$\sqrt{a_{1}a_{2}}+\sqrt{b_{1}b_{2}}\geq1. $$
Having all these observations in mind, one can conjecture that positivity of the map (2.2) is related to the following set of conditions:
$$\begin{array}{@{}rcl@{}} & \sqrt{a_{i}a_{i + 1}}+\sqrt{b_{i}c_{i + 1}}\geq 1,& \end{array} $$
(2.6)
$$\begin{array}{@{}rcl@{}} & a_{*}+b_{*}+c_{*}\geq 2. & \end{array} $$
(2.7)
In next sections, we shall discuss these conditions.

3 Condition (2.6)

Now, we consider further generalization. Let A=\(\left (a_{ij}\right )_{i,j = 1}^{n}\) be a matrix with nonnegative coefficients. Let us define a map \({\Phi }_{A}:\mathbb M_{n}\to \mathbb M_{n}\) by
$$ {\Phi}_{A}\left( X\right)={\Delta}_{A}\left( X\right)-X, $$
(3.1)
where
$${\Delta}_{A}\left( X\right)=\text{diag}\left( {\sum}_{j = 1}^{n} a^{\prime}_{1j}x_{jj},{\sum}_{j = 1}^{n} a^{\prime}_{2j}x_{jj},\ldots,{\sum}_{j = 1}^{n} a^{\prime}_{nj}x_{jj}\right). $$
In the above formula \(a_{ij}^{\prime }=a_{ij}+\delta _{ij}\), where δij stands for the Kronecker delta.
Observe that positivity of ΦA is equivalent to the inequality 〈ηA (ζζ)η〉 ≥ 〈η,ζζη〉 for every \(\zeta ,\eta \in \mathbb C^{n}\). One can easily show that the above is equivalent to
$${\sum}_{i,j = 1}^{n} a_{ij}^{\prime} {p_{i}^{2}}{q_{j}^{2}}\geq\left( {\sum}_{i = 1}^{n} p_{i}q_{i}\right)^{2},\qquad p_{i},q_{i}\geq0. $$

In the following main result of this section, we discuss the role of the condition (2.6).

Theorem 3.1

LetΦA be given by (3.1). If the mapΦA is positive, then
$$ \sqrt{a_{ii}a_{jj}}+\sqrt{a_{ij}a_{ji}}\geq1,\quad i\neq j $$
(3.2)

Proof

Let ij be fixed. Observe that
$$\begin{array}{@{}rcl@{}} &&{\sum\limits}_{kl}a_{kl}^{\prime} {p_{k}^{2}}{q_{l}^{2}}-\left( {\sum\limits}_{k}p_{k}q_{k}\right)^{2} \end{array} $$
(3.3)
$$\begin{array}{@{}rcl@{}} & = &{\sum\limits}_{kl}a_{kl}^{\prime} {p_{k}^{2}}{q_{l}^{2}}-{\sum\limits}_{kl}p_{k}p_{l}q_{k}q_{l} \end{array} $$
(3.4)
$$\begin{array}{@{}rcl@{}} &=&{\sum\limits}_{k} a_{kk} {p_{k}^{2}}{q_{k}^{2}}+{\sum}_{k<l}\left( a_{kl}{p_{k}^{2}}{q_{l}^{2}}+a_{lk}{p_{l}^{2}}{q_{k}^{2}}\right)-2{\sum\limits}_{k<l}p_{k}p_{l}q_{k}q_{l} \end{array} $$
(3.5)
$$\begin{array}{@{}rcl@{}} & = & {\sum\limits}_{k} a_{kk} {p_{k}^{2}}{q_{k}^{2}} + {\sum}_{k<l}(\sqrt{a_{kl}}p_{k}q_{l}-\sqrt{a_{lk}}p_{l}q_{k})^{2}+ 2 {\sum\limits}_{k<l}(\sqrt{a_{kl}a_{lk}}-1)p_{k}p_{l}q_{k}q_{l}. \end{array} $$
(3.6)
Consider the case pk = 0 and qk = 0 for every k∉{i,j}. Then, the above expression is equal to
$$\begin{array}{@{}rcl@{}} &&a_{ii}{p_{i}^{2}}{q_{i}^{2}} + a_{jj}{p_{j}^{2}}{q_{j}^{2}} +(\sqrt{a_{ij}}p_{i}q_{j}-\sqrt{a_{ji}}p_{j}q_{i})^{2}+ 2(\sqrt{a_{ij}a_{ji}}-1)p_{i}p_{j}q_{i}q_{j}\\ &=& (\sqrt{a_{ii}}p_{i}q_{i}-\sqrt{a_{jj}}p_{j}q_{j})^{2} + (\sqrt{a_{ij}}p_{i}q_{j}-\sqrt{a_{ji}}p_{j}q_{i})^{2} \end{array} $$
(3.7)
$$\begin{array}{@{}rcl@{}} &&+ 2\left( \sqrt{a_{ii}a_{jj}}+\sqrt{a_{ij}a_{ji}}-1\right)p_{i}p_{j}q_{i}q_{j}. \end{array} $$
(3.8)
Now, take
$$p_{i}=q_{j}= 1,\quad p_{j}=\left( \frac{a_{ij}a_{ii}}{a_{ji}a_{jj}}\right)^{1/4}, \quad q_{i}=\left( \frac{a_{ij}a_{jj}}{a_{ji}a_{ii}}\right)^{1/4}. $$
Then, the expression from lines (3.7) and (3.8) reduces to
$$2\left( \sqrt{a_{ii}a_{jj}}+\sqrt{a_{ij}a_{ji}}-1\right)\left( \frac{a_{ij}}{a_{ji}}\right)^{1/2}.$$
It follows from the assumption that the expression (3.3) is nonnegative for every choice of pk, qk. Thus, we arrive at the inequality
$$\sqrt{a_{ii}a_{jj}}+\sqrt{a_{ij}a_{ji}}-1\geq 0$$
which is equivalent to (3.2). □

Having in mind, Theorem 1.1 one cannot expect that a converse theorem is also true. However, it turns out that a slight modification of (3.2) gives a sufficient condition for positivity and even decomposability.

Theorem 3.2

If
$$ \frac{1}{n-1}\sqrt{a_{ii}a_{jj}}+\sqrt{a_{ij}a_{ji}}\geq1,\quad i\neq j $$
(3.9)
then the map ΦA is positive and decomposable.

Proof

Let us observe that
$$\begin{array}{@{}rcl@{}} {\sum\limits}_{i} a_{ii} {p_{i}^{2}}{q_{i}^{2}} &\,=\,&\frac{1}{n-1} {\sum\limits}_{i<j}\left( a_{ii}{p_{i}^{2}}{q_{i}^{2}}+a_{jj}{p_{j}^{2}}{q_{j}^{2}} \right) \\ &\,=\,&{\sum\limits}_{i<j}\left( \sqrt{\frac{a_{ii}}{n\,-\,1}}p_{i}q_{i}\,-\,\sqrt{\frac{a_{jj}}{n\,-\,1}}p_{j}q_{j}\right)^{2}\,+\,2{\sum\limits}_{i<j}\frac{\sqrt{a_{ii}a_{jj}}}{n\,-\,1}p_{i}p_{j}q_{i}q_{j}. \end{array} $$
(3.10)
According to the lines (3.3)–(3.6) and (3.10), one has
$$\begin{array}{@{}rcl@{}} &&{\sum\limits}_{ij}a_{ij}^{\prime} {p_{i}^{2}}{q_{j}^{2}}-\left( {\sum}_{i}p_{i}q_{i}\right)^{2}\\ & = & {\sum\limits}_{i}a_{ii}{p_{i}^{2}}{q_{i}^{2}} + {\sum}_{i<j}(\sqrt{a_{ij}}p_{i}q_{j}-\sqrt{a_{ji}}p_{j}q_{i})^{2}+ 2 {\sum\limits}_{i<j}(\sqrt{a_{ij}a_{ji}}-1)p_{i}p_{j}q_{i}q_{j}\\ &=& {\sum\limits}_{i<j}\left( \sqrt{\frac{a_{ii}}{n-1}}p_{i}q_{i}-\sqrt{\frac{a_{jj}}{n-1}}p_{j}q_{j}\right)^{2} + {\sum}_{i<j}(\sqrt{a_{ij}}p_{i}q_{j}-\sqrt{a_{ji}}p_{j}q_{i})^{2} \\ &&+ 2{\sum\limits}_{i<j}\left( \frac{\sqrt{a_{ii}a_{jj}}}{n-1}+\sqrt{a_{ij}a_{ji}}-1\right)p_{i}p_{j}q_{i}q_{j}. \end{array} $$
It follows from (3.9) that the above expression is nonnegative for every nonnegative pi and qi, so ΦA is positive.
Now, we will show that ΦA is decomposable. Let \(\rho \in \mathbb M_{n}(\mathbb M_{n})\) be a PPT matrix. We have ρ = (ρij), where \(\rho _{ij}\in \mathbb M_{n}\). For ij, let rij be the (i,j)-coefficient of the matrix ρij. For any i,j, let βij denote the j-th diagonal term in the matrix ρii. Thus, the matrix ρ looks like
while
where ρΓ = (𝜃n(ρij)) is the partially transposed ρ, and stars stand for any numbers. Since ρ and ρΓ are positive definite, βij ≥ 0, \(r_{ji}=\overline {r_{ij}}\), and
$$ \beta_{ii}\beta_{jj}\geq |r_{ij}|^{2},\qquad \beta_{ij}\beta_{ji}\geq |r_{ij}|^{2},\qquad i\neq j. $$
(3.11)
Now, observe that
$$\begin{array}{@{}rcl@{}} \text{Tr}(\rho C_{{\Phi}_{A}})&\,=\,&\sum\limits_{i,j = 1}^{n} a_{ij}\beta_{ij}-2\sum\limits_{1\leq i<j\leq n}\text{Re}(r_{ij}) \\ & \!\geq\! & \sum\limits_{i = 1}^{n} a_{ii}\beta_{ii} + \sum\limits_{1\leq i<j\leq n}(a_{ij}\beta_{ij}+a_{ji}\beta_{ji})-2\sum\limits_{1\leq i<j\leq n}|r_{ij}|\\ &\,=\,& \frac{1}{n\,-\,1}\sum\limits_{1\leq i<j\leq n}(a_{ii}\beta_{ii}+a_{jj}\beta_{jj})\,+\,\sum\limits_{1\leq i<j\leq n}(a_{ij}\beta_{ij}+a_{ji}\beta_{ji})\,-\,2\sum\limits_{1\leq i<j\leq n}|r_{ij}| \\ &\!\geq\! & 2\sum\limits_{i<j}\left( \frac{1}{n\,-\,1}\sqrt{a_{ii}a_{jj}\beta_{ii}\beta_{jj}}+\sqrt{a_{ij}a_{ji}\beta_{ij}\beta_{ji}}-|r_{ij}|\right) \end{array} $$
(3.12)
$$\begin{array}{@{}rcl@{}} &\!\geq\! & 2{\sum}_{i<j}|r_{ij}|\left( \frac{1}{n-1}\sqrt{a_{ii}a_{jj}}+\sqrt{a_{ij}a_{ji}}-1\right)\geq 0. \end{array} $$
(3.13)
The expression (3.12) was obtained by arithmetic-geometric mean inequality, while (3.13) is due to inequalities (3.11). Since ρ is an arbitrary PPT matrix, we conclude that ΦA is decomposable (cf. Theorem 1.5 (ii)). □

For n = 2, we immediately get the following corollary (cf. Remark 2.6).

Corollary 3.3

For n = 2, a map ϕA is positive if and only if
$$\sqrt{a_{11}a_{22}}+\sqrt{a_{12}a_{21}}\geq 1.$$

Remark 3.4

For n = 3, the above theorem assures that the condition (2.6) is necessary for positivity of ΦA.

4 Condition (2.7)

Now, we are going to discuss necessity of condition (2.7). Let us start with the following

Theorem 4.1

Assume that the matrix (2.1) is of the form
$$ A=\left( \begin{array}{ccc} a_{1} & b & c\\ c & a_{2} & b\\ b & c & a_{3} \end{array}\right) $$
(4.1)
for some arbitrary, a1,a2,a3 ≥ 1 and b,c > 0. If the map ΦA given by (2.2) is positive, then
$$ (a_{1}a_{2}a_{3})^{1/3}+b+c\geq 2. $$
(4.2)

Proof

Observe that the Choi matrix of the map ΦA is of the form
It follows from positivity of ΦA that CA is block positive, i.e.,
$$ \langle \xi\otimes\eta, C_{A}\xi\otimes\eta\rangle\geq 0 $$
(4.3)
for every \(\xi ,\eta \in \mathbb C^{3}\).
Let
$$\xi=\eta=\left( \left( a_{1}^{-1}a_{2}^{\phantom{1}}a_{3}^{\phantom{1}}\right)^{\frac{1}{12}}, \left( a_{1}^{\phantom{1}}a_{2}^{-1}a_{3}^{\phantom{1}}\right)^{\frac{1}{12}}, \left( a_{1}^{\phantom{1}}a_{2}^{\phantom{1}}a_{3}^{-1}\right)^{\frac{1}{12}} \right)^{\mathrm{T}}. $$
Then,
$$\xi\otimes\eta= \left( \begin{array}{ccc|ccc|ccc} \left( a_{1}^{-1}a_{2}^{\phantom{1}}a_{3}^{\phantom{1}}\right)_{\phantom{1}}^{\frac{1}{6}}, & a_{3}^{\frac{1}{6}}, & a_{2}^{\frac{1}{6}} & a_{3}^{\frac{1}{6}}, & \left( a_{1}^{\phantom{1}}a_{2}^{-1}a_{3}^{\phantom{1}}\right)_{\phantom{1}}^{\frac{1}{6}}, & a_{1}^{\frac{1}{6}} & a_{2}^{\frac{1}{6}}, & a_{1}^{\frac{1}{6}}, & \left( a_{1}^{\phantom{1}}a_{2}^{\phantom{1}}a_{3}^{-1}\right)_{\phantom{1}}^{\frac{1}{6}} \end{array} \right)^{\mathrm{T}}. $$
Hence,
$$\begin{array}{@{}rcl@{}} \langle \xi\otimes\eta,C_{A}\xi\otimes \eta\rangle &=& a_{1}^{\frac{2}{3}}a_{2}^{\frac{1}{3}}a_{3}^{\frac{1}{3}}+ a_{1}^{\frac{1}{3}}a_{2}^{\frac{2}{3}}a_{3}^{\frac{1}{3}}+ a_{1}^{\frac{1}{3}}a_{2}^{\frac{1}{3}}a_{3}^{\frac{2}{3}}+ ba_{3}^{\frac{1}{3}}+ ba_{1}^{\frac{1}{3}}+ ba_{2}^{\frac{1}{3}}\\ &&+ca_{2}^{\frac{1}{3}}+ ca_{3}^{\frac{1}{3}}+ ca_{1}^{\frac{1}{3}} -2a_{3}^{\frac{1}{3}}-2a_{1}^{\frac{1}{3}}-2a_{2}^{\frac{1}{3}}\\ &=& \left( a_{1}^{\frac{1}{3}}+a_{2}^{\frac{1}{3}}+a_{3}^{\frac{1}{3}}\right) \left( \left( a_{1}a_{2}a_{3}\right)^{\frac{1}{3}}+b+c-2\right). \end{array} $$
Thus, inequality (4.3) leads to (4.2). □

Remark 4.2

The vector ξ in the above proof can be taken even simpler:
$$ \xi=\eta=\left( a_{1}^{-\frac{1}{6}},a_{2}^{-\frac{1}{6}},a_{3}^{-\frac{1}{6}}\right)^{\mathrm{T}}. $$
(4.4)

Remark 4.3

For the case n = 2, one can prove a similar result for
$$A=\left( \begin{array}{cc} a_{1} & b_{1} \\ b_{2} & a_{2} \end{array}\right),$$
where ai ≥ 1, bi > 0 are arbitrary. In this case, one takes
$$\xi=\left( a_{1}^{-\frac{1}{4}}b_{2}^{\frac{1}{4}}, a_{2}^{-\frac{1}{4}}b_{1}^{\frac{1}{4}}\right)^{\mathrm{T}},\quad \eta= \left( a_{1}^{-\frac{1}{4}}b_{2}^{-\frac{1}{4}}, a_{2}^{-\frac{1}{4}}b_{1}^{-\frac{1}{4}}\right)^{\mathrm{T}}. $$

Theorem 4.4

Assume that the matrix (2.1) is of the form
$$A=\left( \begin{array}{ccc} a_{1} & b_{1} & 0\\ 0 & a_{2} & b_{2}\\ b_{3} & 0 & a_{3} \end{array}\right) $$
for some arbitrary, a1,a2,a3 ≥ 1 and b1,b2,b3 > 0. If the map ΦA given by (2.2) is positive, then
$$(a_{1}a_{2}a_{3})^{1/3}+(b_{1}b_{2}b_{3})^{1/3}\geq 2. $$

Proof

The same idea as above. Observe that the Choi matrix of ΦA has the form
Consider the following vectors ξ and η (see (4.4)):
$$\xi=\left( \begin{array}{c} a_{1}^{-\frac{1}{6}}b_{1}^{\frac{1}{6}}b_{3}^{-\frac{1}{6}} \\ a_{2}^{-\frac{1}{6}}b_{2}^{\frac{1}{6}}b_{1}^{-\frac{1}{6}} \\ a_{3}^{-\frac{1}{6}}b_{3}^{\frac{1}{6}}b_{2}^{-\frac{1}{6}} \end{array}\right),\quad \eta=\left( \begin{array}{c} a_{1}^{-\frac{1}{6}}b_{1}^{-\frac{1}{6}}b_{3}^{\frac{1}{6}} \\ a_{2}^{-\frac{1}{6}}b_{2}^{-\frac{1}{6}}b_{1}^{\frac{1}{6}} \\ a_{3}^{-\frac{1}{6}}b_{3}^{-\frac{1}{6}}b_{2}^{\frac{1}{6}} \end{array}\right). $$
Then, the vector ξη is of the form
Like in the proof of Theorem 4.1, we calculate
$$\begin{array}{@{}rcl@{}} &&\langle \xi\otimes\eta,C_{A}\xi\otimes \eta\rangle\\ =&&a_{1}^{\frac{1}{3}}+a_{2}^{\frac{1}{3}}+ a_{3}^{\frac{1}{3}}+a_{1}^{-\frac{1}{3}}a_{3}^{-\frac{1}{3}}b_{1}^{\frac{1}{3}}b_{2}^{\frac{1}{3}}b_{3}^{\frac{1}{3}}+a_{1}^{-\frac{1}{3}}a_{2}^{-\frac{1}{3}}b_{1}^{\frac{1}{3}}b_{2}^{\frac{1}{3}}b_{3}^{\frac{1}{3}}+a_{2}^{-\frac{1}{3}}a_{3}^{-\frac{1}{3}}b_{1}^{\frac{1}{3}}b_{2}^{\frac{1}{3}}b_{3}^{\frac{1}{3}}\\ &&-2a_{1}^{-\frac{1}{3}}a_{2}^{-\frac{1}{3}}-2a_{1}^{-\frac{1}{3}}a_{3}^{-\frac{1}{3}}-2a_{2}^{-\frac{1}{3}}a_{3}^{-\frac{1}{3}}\\ =&&a_{1}^{\frac{1}{3}}+a_{2}^{\frac{1}{3}}+ a_{3}^{\frac{1}{3}} +a_{2}^{\frac{1}{3}}a_{*}^{-1}b_{*}+a_{3}^{\frac{1}{3}}a_{*}^{-1}b_{*} + a_{1}^{\frac{1}{3}}a_{*}^{-1}b_{*} -2a_{3}^{\frac{1}{3}}a_{*}^{-1}-2a_{2}^{\frac{1}{3}}a_{*}^{-1}-2a_{1}^{\frac{1}{3}}a_{*}^{-1} \\ =&&a_{*}^{-1}\left( a_{1}^{\frac{1}{3}}+a_{2}^{\frac{1}{3}}+ a_{3}^{\frac{1}{3}}\right)\left( a_{*}+b_{*}-2\right). \end{array} $$

5 Problem of Sufficiency

Let us consider the matrix A given by (4.1). We have shown that the following conditions (cf. (3.2), (4.2))
$$\begin{array}{@{}rcl@{}}&(a_{i}a_{i + 1})^{1/2}+(bc)^{1/2}\geq 1,\quad i = 1,2,3,&\\ &(a_{1}a_{2}a_{3})^{1/3}+b+c\geq 2& \end{array} $$
are necessary for positivity of ΦA. For a1 = a2 = a3, these condition are also sufficient [1]. Note that they are special cases of conditions (2.6) and (2.7).

However, these conditions are not sufficient for general case. Namely, we have the following no-go result.

Proposition 5.1

Let a1,a2,a3 > 0 and b ≥ 0 be such that
$$(a_{1}a_{2}a_{3})^{1/3}+b = 2.$$
Let A be the matrix in given by (4.1) with c = 0. Then, ΦA is positive if and only if a1 = a2 = a3.

Proof

Sufficiency part follows from Theorem 1.1. We will show necessity. Assume that ΦA is a positive map. Let
$$\xi=\left( a_{1}^{-\frac{1}{6}}a_{3}^{\frac{1}{6}},a_{2}^{-\frac{1}{6}}a_{1}^{\frac{1}{6}}, a_{3}^{-\frac{1}{6}}a_{2}^{\frac{1}{6}}\right)^{\mathrm{T}}.$$
Then,
$$\xi\xi^{*}=\left( \begin{array}{ccc} a_{1}^{-\frac{1}{3}}a_{3}^{\frac{1}{3}} & a_{2}^{-\frac{1}{6}}a_{3}^{\frac{1}{6}} & a_{1}^{-\frac{1}{6}}a_{2}^{\frac{1}{6}} \\ a_{2}^{-\frac{1}{6}}a_{3}^{\frac{1}{6}} & a_{2}^{-\frac{1}{3}}a_{1}^{\frac{1}{3}} & a_{3}^{-\frac{1}{6}}a_{1}^{\frac{1}{6}} \\ a_{1}^{-\frac{1}{6}}a_{2}^{\frac{1}{6}} & a_{3}^{-\frac{1}{6}}a_{1}^{\frac{1}{6}} & a_{3}^{-\frac{1}{3}}a_{2}^{\frac{1}{3}} \end{array}\right)$$
and
$${\Phi}_{A}(\xi\xi^{*})=\left( \begin{array}{ccc} a_{1}^{\frac{2}{3}}a_{3}^{\frac{1}{3}} + ba_{2}^{-\frac{1}{3}}a_{1}^{\frac{1}{3}} & -a_{2}^{-\frac{1}{6}}a_{3}^{\frac{1}{6}} & -a_{1}^{-\frac{1}{6}}a_{2}^{\frac{1}{6}} \\ -a_{2}^{-\frac{1}{6}}a_{3}^{\frac{1}{6}} & a_{2}^{\frac{2}{3}}a_{1}^{\frac{1}{3}} + ba_{3}^{-\frac{1}{3}}a_{2}^{\frac{1}{3}} & -a_{3}^{-\frac{1}{6}}a_{1}^{\frac{1}{6}} \\ -a_{1}^{-\frac{1}{6}}a_{2}^{\frac{1}{6}} & -a_{3}^{-\frac{1}{6}}a_{1}^{\frac{1}{6}} & a_{3}^{\frac{2}{3}}a_{2}^{\frac{1}{3}} + ba_{1}^{-\frac{1}{3}}a_{3}^{\frac{1}{3}} \end{array}\right). $$
Calculate the determinant of the above matrix
$$\begin{array}{@{}rcl@{}} &&\det {\Phi}_{A}(\xi\xi^{*})\\ &=& \left( a_{1}^{\frac{2}{3}}a_{3}^{\frac{1}{3}} + ba_{2}^{-\frac{1}{3}}a_{1}^{\frac{1}{3}}\right)\left( a_{2}^{\frac{2}{3}}a_{1}^{\frac{1}{3}} + ba_{3}^{-\frac{1}{3}}a_{2}^{\frac{1}{3}}\right)\left( a_{3}^{\frac{2}{3}}a_{2}^{\frac{1}{3}} + ba_{1}^{-\frac{1}{3}}a_{3}^{\frac{1}{3}}\right) - 2 \\ &&-a_{1}^{-\frac{1}{3}}a_{2}^{\frac{1}{3}}\left( a_{2}^{\frac{2}{3}}a_{1}^{\frac{1}{3}} + ba_{3}^{-\frac{1}{3}}a_{2}^{\frac{1}{3}}\right) -a_{3}^{-\frac{1}{3}}a_{1}^{\frac{1}{3}}\left( a_{1}^{\frac{2}{3}}a_{3}^{\frac{1}{3}} + ba_{2}^{-\frac{1}{3}}a_{1}^{\frac{1}{3}}\right)\\ && -a_{2}^{-\frac{1}{3}}a_{3}^{\frac{1}{3}}\left( a_{3}^{\frac{2}{3}}a_{2}^{\frac{1}{3}} + ba_{1}^{-\frac{1}{3}}a_{3}^{\frac{1}{3}}\right) \\ &=&a_{1}a_{2}a_{3} + 3ba_{1}^{\frac{2}{3}}a_{2}^{\frac{2}{3}}a_{3}^{\frac{2}{3}}+ 3b^{2}a_{1}^{\frac{1}{3}}a_{2}^{\frac{1}{3}}a_{3}^{\frac{1}{3}}+b^{3}-2 \\ &&-a_{2}-ba_{1}^{-\frac{1}{3}}a_{2}^{\frac{2}{3}}a_{3}^{-\frac{1}{3}} -a_{1}-ba_{1}^{\frac{2}{3}}a_{2}^{-\frac{1}{3}}a_{3}^{-\frac{1}{3}} -a_{3}-ba_{1}^{-\frac{1}{3}}a_{2}^{-\frac{1}{3}}a_{3}^{\frac{2}{3}} \\ &=&(a_{*}+b)^{3}-2-3\overline{a}-3b\overline{a}a_{*}^{-1}\\ &=& a_{*}^{-1}\left( 6a_{*}-3a_{*}\overline{a}-3b\overline{a}\right) = 6a_{*}^{-1}\left( a_{*}-\overline{a}\right). \end{array} $$
In the last line, the assumption that a + b = 2 was applied. It follows from positivity of ΦA that \(a_{*}\geq \overline {a}\). Hence, \(a_{*}=\overline {a}\) and consequently all ai are equal. □

Remark 5.2

The above considerations show that (2.6) and (2.7) do not form a set of sufficient conditions for positivity of the map ΦA in the case when ai are distinct numbers. At this stage, it is still an open problem how to generalize Theorems 1.1 and 1.2 for arbitrary numbers ai, bj, ck.

Notes

Funding Information

DC was supported by the Polish National Science Centre project 2015/19/B/ST1/03095, while MM was supported by Templeton Foundation Project “Quantum objectivity: between the whole and the parts”.

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Authors and Affiliations

  1. 1.Institute of PhysicsNicolaus Copernicus UniversityToruńPoland
  2. 2.Institute of Theoretical Physics and Astrophysics, Faculty of Mathematics, Physics and InformaticsUniversity of GdańskGdańskPoland

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