On characterizations of bistochastic Kadison-Schwarz operators on $M_2(\mathbb{C})$

0.22 MB·

by  Farrukh Mukhamedov

ON CHARACTERIZATIONS OF BISTOCHASTIC KADISON-SCHWARZ
OPERATORS ON M (C)
2
FARRUKHMUKHAMEDOVAND HASANAKIN
6 Abstract. InthispaperwedescribebistochasticKadison-SchawrzoperatorsactingonM2(C).
1 Such a description allows us to find positive, but not Kadison-Schwarz operators. Moreover,
0 bymeansofthatcharacterizationweconstructKadison-Schawrzoperators,whicharenotcom-
2 pletely positive.

n
Mathematics Subject Classification: 47L07; 46L30; 47C15; 15A48; 81P68 60J99.
a
J Key words: bistochastic mapping, Kadison-Schwarz operator, complete positive.

5

]
A
F
. 1. Introduction
h
at Itisknownthatentanglement isoneoftheessentialfeatures ofquantumphysicsandisfunda-
m mental in modern quantum technologies [34]. One of the central problems in the entanglement
[ theory is the discrimination between separable and entangled states. There are several tools

which can be used for this purpose. There are many papers devoted to find a given state is
1
v separable (see [16]). The most general approach to characterize quantum entanglement uses a
9 notionof anentanglement witness[17,7,44]. Oneofthebigadvantages ofentanglement witness
1
is that they provide an economic method of detection which does not need the full information
7
about the quantum state (see for recent review [10]). Interestingly, the entanglement witnesses
0
0 are deeply connected to a theory of positive maps in operator algebras [6, 9, 15]. Therefore, it
1. would interesting to find some conditions for the positivity of given mappings. In this direction
0 there are several papers [4, 6, 8, 9, 15, 21, 40, 41]. Therefore, it would interesting to find some
6
conditionsforthepositivityofgivenmappings(see[21]-[25]). Intheliteraturethemosttractable
1
maps, the completely positive mapping, have proved to be of great importance in the study of
:
v
quantum system (see [11, 35, 36, 37, 42]). It is therefore of interest to study conditions stronger
i
X thanpositivity, butweaker thancompletepositivity. Suchaconditioniscalled Kadison-Schwarz
r (KS) property. Note that KS-operators no need to be completely positive. In [39] relations be-
a
tween n-positivity of a map φ and the KS property of certain map is established (see also [2]).
Someergodicpropertiesof theKadison-Schwarzmapswereinvestigated in[20,14,38]. Unfortu-
nately, like completely positive maps, the description of Kadison-Schwarz maps is not provided.
Very recently, one of the authors of this paper in [28] has described bistochastic KS-operators
from M (C) to itself. But, in general, the problem still remains open.
2
In [13] it was proposed to study positive operators P from a von Neumann algebra M to its
tensor square M ⊗M (we refer a reader to [12, 33] for recent review on quadratic operators).
It turns out that this kind of mappings have some applications to quantum information theory.
One of such an application is to detect entangled states. For example, let P be a block positive,
thenastateφonthealgebraM⊗M isseparable,thenthestateP∗φispositive. Ifφisentangled,
then P∗φ may not be positive. This observation leads to more investigation of operators from
M to M ⊗ M. In general, description of this kind of mappings was fully not studied yet.
Some positivity conditions were found in [21, 24]. In [30, 27] it was considered trace preserving
mappings from M (C) to M (C)⊗M (C), and each such kind of mappings can be written as
2 2 2
a sum of two ”linear” and ”nonlinear” operators (see (2.7)-(2.9)). In [29] mappings of the form
1
2 FARRUKHMUKHAMEDOVANDHASANAKIN
(2.8)havebeenstudied. Namely,somesufficientconditionsforpositivity(resp. Kadison-Schwarz
property) of the mentioned mappings were found.
In the present paper we are going to describe or characterize operators of the form (2.9). To
do it, we first in Section 3 we provide a characterization of KS-operators form M (C) to M (C)
2 2
which improves the main result of [28]. In section 4, we give a sufficient condition for a class of
bistochastic mappings from M (C) to M (C)⊗M (C) to be KS-operator. Note that this class
2 2 2
of operators are totally different from the operators studied in [29]. Such a description allows
us to find positive, but not Kadison-Schwarz operators. Moreover, by means of that conditions
one can construct KS-operators, which are not completely positive. Note that some parts of this
section have been announced in [31]. Moreover, our results allow to produce higher dimensional
examples of positive, but completely positive maps. The proposed approach can be extended
to a more general setting rather that M (C), and will produce non trivial examples of positive
2
mappings.
2. Preliminaries
In this section we recall some definitions and notations.
Let M (C) be the algebra of n×n matrices over the complex field C. Recall that a linear
n
mapping Φ :M (C) → M (C) is called
n m
(i) positive if Φ(x) ≥ 0 whenever x ≥ 0;
(ii) unital if Φ(1I)= 1I;
(iii) trace preserving if τ(Φ(x)) = τ(x), where τ is the normalized trace;
(iv) bistochastic if Φ is unital and trace preserving;
(v) n-positiveifthemappingΦ : M (A) → M (B)definedbyΦ (a )= (Φ(a ))ispositive.
n n n n ij ij
Here M (A) denotes the algebra of n×n matrices with A-valued entries;
n
(vi) completely positive if it is n-positive for all n ∈N;
(vii) Kadison-Schwarz operator (KS-operator), if one has
∗ ∗
(2.1) Φ(x) Φ(x) ≤ Φ(x x) for all x ∈ A.
It is clear that any KS-operator is positive. Note that every unital 2- positive map is KS-
operator, and a famous result of Kadison states that any positive unital map satisfies the in-
equality (2.1) for all self-adjoint elements x ∈ A.
By KS(M ,M ) we denote the set of all KS-operators mapping from M (C) to M (C).
n m n m
Theorem 2.1. [28] The following assertions hold true:
(i) Let Φ,Ψ ∈ KS(M ,M ), then for any λ ∈ [0,1] the mapping Γ = λΦ+(1−λ)Ψ belongs
n m
to KS(M ,M ). This means KS(M ,M ) is convex;
n m n m
(ii) Let U,V be unitaries in M (C) and M (C), respectively, then for any Φ ∈ KS(M ,M )
n m n m
∗ ∗
the mapping Ψ (x) = UΦ(VxV )U belongs to KS(M ,M ).
U,V n m
ByM (C)⊗M (C)wemean tensorproductofM (C)intoitself. Wenotethat suchaproduct
2 2 2
can be considered as an algebra of 4×4 matrices M (C) over C. By S(M (C)) we denote the
4 2
set of all states (i.e. linear positive functionals which take value 1 at 1I) defined on M (C).
2
Recall that a linear operator ∆ : M (C) → M (C)⊗M (C) is said to be quantum quadratic
2 2 2
operator (q.q.o.) if it is unital and positive.
A state h∈ S(M (C)) is called a Haar state for a q.q.o. ∆ if for every x ∈ M (C) one has
2 2
(2.2) (h⊗id)◦∆(x) = (id⊗h)◦∆(x)= h(x)1I.
Remark 2.2. LetU : M (C)⊗M (C)→ M (C)⊗M (C)bealinearoperatorsuchthatU(x⊗y) =
2 2 2 2
y⊗x for all x,y ∈ M (C). If a q.q.o. ∆ satisfies U∆ =∆, then ∆ is called a quantum quadratic
2
stochastic operator or symmetric q.q.o. Recent reviews on this kind of operators can be found
in [12, 33]).
KADISON-SCHWARTZ OPERATORS 3
Recall [5] that the identity and Pauli matrices {1I,σ ,σ ,σ } form a basis for M (C), where
1 2 3 2
0 1 0 −i 1 0
σ = σ = σ = .
1 1 0 2 i 0 3 0 −1
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
In this basis every matrix x ∈ M (C) can be written as x = w 1I + wσ with w ∈ C,
2 0 0
w = (w ,w ,w )∈ C3, here wσ =w σ +w σ +w σ .
1 2 3 1 1 2 2 3 3
Lemma 2.3. [40] The following assertions hold true:
(a) x is self-adjoint iff w ,w are reals;
0
(b) Tr(x) = 1 iff w = 0.5, here Tr is the trace of a matrix x;
0
(c) x > 0 iff kwk ≤ w , where kwk = |w |2+|w |2+|w |2.
0 1 2 3
Note that any state ϕ∈ S(M (C)) can pbe represented by
2
(2.3) ϕ(w 1I+wσ) = w +hw,fi,
0 0
where f = (f ,f ,f ) ∈ R3 with kfk ≤ 1. Here as before h·,·i stands for the scalar product in
1 2 3
C3. Therefore, in the sequel we will identify a state ϕ with a vector f ∈ R3.
In what follows by τ we denote a normalized trace, i.e. τ(x) = 1 Tr(x), x ∈ M (C),
2 2
Let ∆ : M (C) → M (C)⊗M (C) be a q.q.o. We write the operator ∆ in terms of a basis in
2 2 2
M (C)⊗M (C) formed by the Pauli matrices. Namely,
2 2
(2.4) ∆1I = 1I⊗1I;
3 3 3
(1) (2)
(2.5) ∆(σ ) = b (1I⊗1I)+ b (1I⊗σ )+ b (σ ⊗1I)+ b (σ ⊗σ ),
i i ij j ij j ml,i m l
j=1 j=1 m,l=1
X X X
where i= 1,2,3.
In general, a description of positive operators is one of the main problems of quantum infor-
mation. In the literature most tractable maps are positive and trace-preserving ones, since such
maps arise naturally in quantum information theory (see [18, 19, 34, 40]). Therefore, in the
sequel we shall restrict ourselves to the trace preserving q.q.o. Hence, from (2.4),(2.5) one finds
3
(2.6) ∆(x) = w 1I⊗1I+B(1)w·σ⊗1I+1I⊗B(2)w·σ+ hb ,wiσ ⊗σ ,
0 ml m l
m,l=1
X
where x = w +wσ, b = (b ,b ,b ), and B(k) = (b(k))3 , k = 1,2.
0 ml ml,1 ml,2 ml,3 ij i,j=1
In general, to find some conditions for ∆ to be KS-operator, is a tricky job. Therefore, one
can rewrite (2.6) as follows
(2.7) ∆(x) = λ∆ (x)+(1−λ)∆ (x),
1 2
where
3
1
(2.8) ∆ (x) = w 1I⊗1I+ hb ,wiσ ⊗σ ,
1 0 ml m l
λ
m,l=1
X
1
(2.9) ∆ (x) = w 1I⊗1I+ B(1)w·σ⊗1I+1I⊗B(2)w·σ .
2 0
1−λ
(cid:18) (cid:19)
In [29, 32] we have studied q.q.o. of the form (2.8). It is found necessary conditions for (2.8)
kind of operators to be KS-operator. But operators of the form (2.9) has not been studied yet.
Therefore, main aim of this paper to find some conditions on operators (2.9) to be Kadison-
Schwarz. ThenusingTheorem 2.1 and ourfindingswith the results of [32], we can findsufficient
conditions for (2.7) to be KS-operator.
4 FARRUKHMUKHAMEDOVANDHASANAKIN
3. Kadison-Schwarz operators from M (C) to M (C)
2 2
To investigate operators of the form (2.9) (see section 4) we need some preliminary facts from
[28]. In this section we collect some of them, and improve a main result of [28].
It is known that every Φ : M (C) → M (C) linear mapping can also be represented in this
2 2
1 0
basis by a unique 4×4 matrix F. It is trace preserving if and only if F = where T
t T
(cid:18) (cid:19)
is a 3×3 matrix and 0 and t are row and column vectors, respectively, so that
(3.1) Φ(w 1I+w·σ) = w 1I+(w t+Tw)·σ.
0 0 0
WhenΦisalsopositive thenitmapsthesubspaceofself-adjoint matrices of M (C)intoitself,
2
which implies that T is real. A linear mapping Φ is unital if and only if t = 0. So, in this case
we have
(3.2) Φ(w 1I+w·σ) = w 1I+(Tw)·σ.
0 0
Hence, any bistochastic mapping Φ : M (C) → M (C) has a form (3.2). In [28] it has been
2 2
given a characterization bistochastic KS-operators, i.e. the following
Theorem 3.1. [28] Any bistochastic mapping Φ : M (C) → M (C) is KS-operator if and only
2 2
if one has
(3.3) kTwk ≤ kwk, Tw = Tw
(3.4) T[w,w]− Tw,Tw ≤ kwk2−kTwk2
(cid:13) (cid:13)
for all w ∈ C3. (cid:13)(cid:13) (cid:2) (cid:3)(cid:13)(cid:13)
(cid:13) (cid:13)
Let Φ bea bistochastic KS-operator on M (C), then it can berepresented by (3.2). Following
2
[18]let usdecomposethematrix T asfollows T = RS, hereR isarotation andS isaself-adjoint
matrix (see [18]). Define a mapping Φ as follows
S
(3.5) Φ (w 1I+w·σ) = w 1I+(Sw)·σ.
S 0 0
Every rotation is implemented by a unitary matrix in M (C), therefore there is a unitary U ∈
2
M (C) such that
2
(3.6) Φ(x)= UΦ (x)U∗, x ∈ M (C).
S 2
Ontheotherhand,everyself-adjointoperatorS canbediagonalizedbysomeunitaryoperator,
i.e. there is a unitary V ∈ M (C) such that S = VD V∗, where
2 λ1,λ2,λ3
λ 0 0
1
(3.7) D = 0 λ 0 ,
λ1,λ2,λ3  2 
0 0 λ
3
where λ ,λ ,λ ∈ R.  
1 2 3
Consequently, the mapping Φ can be represented by
(3.8) Φ(x) =U˜Φ (x)U˜∗, x ∈M (C)
Dλ1,λ2,λ3 2
for some unitary U˜. Due to Theorem 2.1 the mapping Φ is also KS-operator. Hence,
Dλ1,λ2,λ3
all bistochastic KS-operators can be characterized by Φ and unitaries. In what follows,
Dλ1,λ2,λ3
for the sake of shortness by Φ we denote the mapping Φ . It is clear to observe
(λ1,λ2,λ3) Dλ1,λ2,λ3
from (3.3) that |λ |≤ 1,k = 1,2,3.
k
In [40] it has been given a characterization of completely positivity of Φ .
(λ1,λ2,λ3)
Using Theorem 2.1 we are going to characterize KS-operators of the form Φ .
(λ1,λ2,λ3)
KADISON-SCHWARTZ OPERATORS 5
Theorem 3.2. Φ is a KS-operator if and only if the following inequalities are satisfied:
(λ1,λ2,λ3)
(3.9) (1+λ2)(3+λ2+λ2−λ2)≤ 4(1+λ λ λ );
1 2 3 1 1 2 3
(3.10) (1+λ2)(3+λ2+λ2−λ2)≤ 4(1+λ λ λ );
2 1 3 2 1 2 3
(3.11) (1+λ2)(3+λ2+λ2−λ2)≤ 4(1+λ λ λ ).
3 1 2 3 1 2 3
where λ ,λ ,λ ∈ [−1,1].
1 2 3
Proof. ’only if’ part. Using simple calculation from (3.4) of Theorem 3.1 with T = D we
λ1,λ2,λ3
obtain
A|w w −w w |2 + B|w w −w w |2
2 3 3 2 1 3 3 1
(3.12) + C|w w −w w |2 ≤ α|w |2+β|w |2+γ|w |2 2,
1 2 2 1 1 2 3
where w = (w ,w ,w )∈ C3 and (cid:0) (cid:1)
1 2 3
(3.13) α= |1−λ2|, β = |1−λ2|, γ = |1−λ2|
1 2 3
(3.14) A= |λ −λ λ |2, B = |λ −λ λ |2, C = |λ −λ λ |2.
1 2 3 2 1 3 3 1 2
Due to the inequality |2ℑ(uv)| ≤|u|2 +|v|2, one has
(3.15) |w w −w w |2 = |2ℑ(w w )|2 ≤ |w |4+2|w |2|w |2+|w |4 (i 6= j)
i j j i i j i i j j
Note that this inequality is reachable by appropriate choosing of values w and w .
i j
Hence, we estimate LHS of (3.12) by
A(|w |4+2|w |2|w |2+|w |4)+B(|w |4+2|w |2|w |2+|w |4)+C(|w |4+2|w |2|w |2+|w |4)
2 2 3 3 1 1 3 3 1 1 2 2
Consequently, from (3.12) we derive the following one
|w |4(α2−B −C)+|w |4(β2−A−C)+|w |4(γ2−A−B)
1 2 3
(3.16) +2|w |2|w |2(αβ −C)+2|w |2|w |2(αγ −B)+2|w |2|w |2(βγ −A) ≥ 0
1 2 1 3 2 3
for all (w ,w ,w ) ∈C3. It is easy to see that (3.16) is satisfied if one has
1 2 3
α2 ≥ B +C, β2 ≥ A+C, γ2 ≥ A+B,
αβ ≥ C, αγ ≥B, βγ ≥ A.
Substituting above denotations (3.13),(3.14) to the last inequalities, and doing simple calcu-
lation one derives
(3.17) (1+λ2)(3+λ2+λ2−λ2)≤ 4(1+λ λ λ );
1 2 3 1 1 2 3
(3.18) (1+λ2)(3+λ2+λ2−λ2)≤ 4(1+λ λ λ );
2 1 3 2 1 2 3
(3.19) (1+λ2)(3+λ2+λ2−λ2)≤ 4(1+λ λ λ );
3 1 2 3 1 2 3
(3.20) λ2+λ2+λ2 ≤ 1+2λ λ λ .
1 2 3 1 2 3
where λ ,λ ,λ ∈ [−1,1].
1 2 3
Now we would like to show that (3.20) is an extra condition, i.e. the inequality (3.20) always
satisfies when (3.17), (3.18) and (3.19) are true. Suppose that
(3.21) λ2+λ2+λ2 = 1+2λ λ λ
1 2 3 1 2 3
is true. We will show that the elements of the surface do not satisfy the inequalities (3.17),
(3.18) and (3.19) except for (0,0,0), (±1,±1,±1). Using simple algebra from (3.17), (3.18) and
6 FARRUKHMUKHAMEDOVANDHASANAKIN
(3.19) with (3.21) we obtain the followings
(1−λ2)(λ2−λ λ λ ) ≤ 0;
1 1 1 2 3
(1−λ2)(λ2−λ λ λ ) ≤ 0;
2 2 1 2 3
(1−λ2)(λ2−λ λ λ ) ≤ 0,
3 3 1 2 3
where λ ,λ ,λ ∈ [−1,1]. Due to our assumption λ 6= ±1,λ 6= ±1,λ 6= ±1 from the last
1 2 3 1 2 3
inequalities we infer that
(3.22) λ (λ −λ λ ) ≤ 0
1 1 2 3
(3.23) λ (λ −λ λ ) ≤ 0
2 2 1 3
(3.24) λ (λ −λ λ ) ≤ 0,
3 3 1 2
where λ ,λ ,λ ∈ (−1,1). Let λ > 0, then one gets λ ≤ λ λ . It implies λ > 0,λ > 0 or
1 2 3 1 1 2 3 2 3
λ < 0,λ < 0. Now assume λ > 0,λ > 0, then from (3.23) and (3.24) one gets
2 3 2 3
λ ≤ λ λ , λ ≤ λ λ .
2 1 3 3 1 2
From λ ≤ λ λ and λ ≤ λ λ one has λ ≤ λ λ2. This means 1 ≤ λ2. This contradicts to our
1 2 3 2 1 3 2 2 3 3
assumption.
Now let λ > 0,λ < 0 and λ < 0, then from (3.23) and (3.24) one finds
1 2 3
(3.25) λ ≥ λ λ , λ ≥λ λ .
2 1 3 3 1 2
From (3.25) one finds λ ≥ λ2λ . This implies that λ2 ≥ 1. It is again a contradiction. In case
3 1 3 1
λ < 0, using the similar argument we will get again contradiction. This implies the required
1
assertion.
’if’ part. Let (3.9)-(3.11) be satisfied. Then it implies that (3.20) is always true. This means
(3.16) is satisfied. This yields (3.12), hence Φ is a KS-operator. This completes the
(λ1,λ2,λ3)
proof. (cid:3)
Note that the proved theorem provided necessary and sufficient conditions for the mapping
Φ tobeKS-operator. In[28]itwas proved only sufficientconditions tobeKS-operators.
(λ1,λ2,λ3)
Therefore,thelasttheoremessentiallyimprovesamainresultof[28]. Moreover, thelasttheorem
allows us to construct lots of KS-operators, which are not completely positive.
4. A class of Kadison-Schwarz operators from M (C) to M (C)⊗M (C)
2 2 2
In this section we are going to provide description of operators of the form (2.9). First we
need the following auxiliary
Lemma 4.1. Let x = w 1I⊗1I+w·σ⊗1I+1I⊗r·σ. Then the following statements hold true:
0
(i) x is self-adjoint if and only if w ∈ R and w,r ∈ R3;
0
(ii) x is positive if and only if w > 0 and kwk+krk ≤ w .
0 0
Proof. (i). One can see that
∗
x = w 1I⊗1I+w·σ⊗1I+1I⊗r·σ
0
So, self adjointness x implies w = w , w = w, r = r.
0 0
(ii). Let x be self-adjoint. Then from the definition of Pauli matrices one finds
w +w +r w −iw r −ir 0
0 3 3 1 2 1 2
w +iw w −w +r 0 r −ir
x =  1 2 0 3 3 1 2 
r +ir 0 w +w −r w −iw
1 2 0 3 3 1 2
 0 r1+ir2 w1+iw2 w0−w3−r3 
 
 
KADISON-SCHWARTZ OPERATORS 7
It is easy to calculate that eigenvalues of last matrix are the followings
λ = w −krk+kwk, λ = w −krk−kwk,
1 0 2 0
λ = w +krk+kwk, λ = w +krk−kwk
3 0 4 0
So, we can conclude that x is positive if and only if the smallest eigenvalue is positive. This
means w −krk−kwk ≥ 0, which completes the proof. (cid:3)
0
Now we rewrite operator (2.9) as T : M (C)→ M (C)⊗M (C) given by
2 2 2
(4.1) T(w 1I+w·σ) = w 1I⊗1I+Aw·σ⊗1I+1I⊗Cw·σ
0 0
where A,C are linear operators on C3.
We first find conditions when T is positive. This is given by the following
Theorem 4.2. The mapping T given by (4.1) is positive if and only if
kAwk+kCwk ≤ 1,
for all w ∈ R3 with kwk = 1.
Proof. Letx = w 1I+w·σbepositive, i.e. w > 0,kwk ≤ w . Withoutlostofgenerality wemay
0 0 0
assume w = 1. Now Lemma 4.1 yields that T(x) is positive if and only if kAwk+kCwk ≤ 1.
0
This competes the proof. (cid:3)
Corollary 4.3. Let A = C then T is positive if and only if kAk ≤ 1.
2
Now let us turn to the Kadison-Schwarz property.
Define the following mappings
(4.2) Φ(x) = w 1I+2Aw·σ
0
(4.3) Ψ(x)= w 1I+2Cw·σ
0
Then one finds
1
(4.4) T(x) = Φ(x)⊗1I+1I⊗Ψ(x) .
2
(cid:18) (cid:19)
Theorem 4.4. Let T be a mapping given by (4.4). If one has
(4.5) kwk2−2kAwk2 −2kCwk2 ≥ 0
(4.6)kA[w,w]−2[Aw,Aw]k+kC[w,w]−2[Cw,Cw]k ≤ kwk2−2kAwk2 −2kCwk2
Then T is a Kadison-Schwarz operator.
Proof. From (4.4) one finds that
1
∗ ∗ ∗ ∗
T(x x)−T(x) T(x) = Φ(x x)−Φ(x) Φ(x) ⊗1I
2
(cid:18)
(cid:0) (cid:1)
∗ ∗
+1I⊗ Ψ(x x)−Ψ(x) Ψ(x)
(cid:19)
1 (cid:0) ∗(cid:1)
(4.7) + 1I⊗Ψ(x)−Φ(x)⊗1I 1I⊗Ψ(x)−Φ(x)⊗1I .
4
(cid:18) (cid:19) (cid:18) (cid:19)
Now taking into account the following formula
x∗x = |w |2+kwk2 1I+ w w+w w−i w,w ·σ
0 0 0
(cid:0) (cid:1) (cid:0) (cid:2) (cid:3)(cid:1)
8 FARRUKHMUKHAMEDOVANDHASANAKIN
from (4.2) and (4.3) we have
Φ(x∗x)−Φ(x)∗Φ(x) = kwk2−k2Awk2 1I−2i A[w,w]−2[Aw,Aw] σ,
Ψ(x∗x)−Ψ(x)∗Ψ(x)= kwk2−k2Cwk2 1I−2i C[w,w]−2[Cw,Cw] σ.
(cid:0) (cid:1) (cid:0) (cid:1)
Therefore, one gets
(cid:0) (cid:1) (cid:0) (cid:1)
∗ ∗ ∗ ∗
Φ(x x)−Φ(x) Φ(x) ⊗1I+1I⊗ Ψ(x x)−Ψ(x) Ψ(x)
(cid:18) (cid:19) (cid:18) (cid:19)
= kwk2 −4kAwk2 1I−2i A[w,w]−2[Aw,Aw] σ ⊗1I
(cid:18) (cid:19)
(cid:0) (cid:1) (cid:0) (cid:1)
+1I⊗ kwk2 −4kCwk2 1I−2i C[w,w]−2[Cw,Cw] σ
(cid:18) (cid:19)
= 2kwk2(cid:0)−4kAwk2 −4kC(cid:1)wk2 1I(cid:0)⊗1I (cid:1)
−(cid:0) 2i A[w,w]−2[Aw,Aw] σ(cid:1)⊗1I−1I⊗2i C[w,w]−2[Cw,Cw] σ
According to Lemma 4.1 we conclude that the last expression is positive if and only if (4.5) and
(cid:0) (cid:1) (cid:0) (cid:1)
(4.6) are satisfied. Consequently, from (4.7) we infer that under the last conditions the mapping
T is a KS operator. This completes the proof. (cid:3)
We should stress that the conditions (4.5),(4.6) are sufficient to be KS-operator.
Corollary 4.5. If the mappings Φ and Ψ are KS operators, then T is also KS operator.
The proof immediately follows from (4.7).
Remark 4.6. We have to stress that if T is KS operator, then the mappings Φ and Ψ no need
to be KS.
4.1. Case: C = A. Now let usstudytheoperator T given by (4.1)whenC= A. Consequently
from (4.1) one finds
(4.8) T (w 1I+w·σ) = w 1I⊗1I+Aw·σ⊗1I+1I⊗Aw·σ.
A 0 0
From Theorem 4.4 we immediately have the following
Corollary 4.7. Let T be a mapping given by (4.8). If one has
A
kwk2 −4kAwk2 ≥ 0
(4.9) 2kA[w,w]−2[Aw,Aw]k ≤ kwk2 −4kAwk2.
Then T is a Kadison-Schwarz operator.
A
Now using the same argument as in section 3, we can write
(4.10) T (x) = U˜T (x)U˜∗, x ∈ M (C)
A Dλ1,λ2,λ3 2
for some unitary U˜. Due to Theorem 2.1 all bistochastic KS-operators can be characterized by
T and unitaries. In what follows, for the sake of shortness by T we denote the
Dλ1,λ2,λ3 (λ1,λ2,λ3)
mapping T .
Dλ1,λ2,λ3
Next we want to characterize KS operators of the form T .
(λ1,λ2,λ3)
Theorem 4.8. If
4(1+8λ λ λ ) ≥ (1+4λ2)(3+4λ2+4λ2−4λ2),
1 2 3 1 2 3 1
4(1+8λ λ λ ) ≥ (1+4λ2)(3+4λ2+4λ2−4λ2),
1 2 3 2 1 3 2
4(1+8λ λ λ ) ≥ (1+4λ2)(3+4λ2+4λ2−4λ2)
1 2 3 3 1 2 3
are satisfied, then T is a KS operator.
(λ1,λ2,λ3)
KADISON-SCHWARTZ OPERATORS 9
Proof. Taking A = D in (4.9), we obtain
λ1,λ2,λ3
4A |w w −w w |2+4A |w w −w w |2+4A |w w −w w |2
1 2 3 2 3 2 1 3 1 3 3 1 2 1 2
2
(4.11) ≤ B |w |2+B |w |2+B |w |2 ,
1 1 2 2 3 3
where w = (w ,w ,w(cid:16) )∈ C3 and (cid:17)
1 2 3
(4.12) A =|λ −2λ λ |2, A = |λ −2λ λ |2, A = |λ −2λ λ |2,
1 1 2 3 2 2 1 3 3 3 1 2
(4.13) B = (1−4λ2), B = (1−4λ2), B = (1−4λ2).
1 1 2 2 3 3
By (3.15) LHS of (4.11) can be evaluated as follows
4A |w |4+2|w |2|w |2+|w |4 + 4A |w |4+2|w |2|w |2+|w |4
1 2 2 3 3 2 1 1 3 3
(cid:16) (cid:17) (cid:16) (cid:17)
+ 4A |w |4+2|w |2|w |2+|w |4 .
3 1 1 2 2
(cid:16) (cid:17)
Therefore, from (4.11) one gets
B2−4A −4A |w |4+ B2−4A −4A |w |4+ B2−4A −4A |w |4
1 2 3 1 2 1 3 2 3 1 2 3
+(cid:16)2|w |2|w |2(B B(cid:17) −4A )(cid:16)+2|w |2|w |2(B B(cid:17) −4A (cid:16))+2|w |2|w |2(B(cid:17)B −4A )≥ 0
2 3 2 3 1 1 3 1 3 2 1 2 1 2 3
It is obvious that above inequality is satisfied if one has
B2 ≥4A +4A , B2 ≥ 4A +4A , B2 ≥ 4A +4A ,
1 2 3 2 1 3 3 1 2
B B ≥ 4A , B B ≥ 4A , B B ≥4A .
2 3 1 1 3 2 1 2 3
Substitutingabovedenotations(4.12),(4.13)tothelastinequalities,anddoingsomecalculations
one derives
(4.14) 4(1+8λ λ λ ) ≥ (1+4λ2)(3+4λ2+4λ2−4λ2),
1 2 3 1 2 3 1
(4.15) 4(1+8λ λ λ ) ≥ (1+4λ2)(3+4λ2+4λ2−4λ2),
1 2 3 2 1 3 2
(4.16) 4(1+8λ λ λ ) ≥ (1+4λ2)(3+4λ2+4λ2−4λ2),
1 2 3 3 1 2 3
(4.17) 1+16λ λ λ ≥ 4λ2+4λ2+4λ2,
1 2 3 1 2 3
where λ ,λ ,λ ∈ − 1, 1 .
1 2 3 2 2
Now using the shame argiument as in the proof of Theorem 3.2 one can show that (4.17) is an
extra condition. This completes the proof. (cid:3)
ItisinterestingtostudywhentheoperatorT iscompletepositive. Letuscharacterize
(λ1,λ2,λ3)
completely positivity of T .
(λ1,λ2,λ3)
Theorem 4.9. A map T is complete positive if and only if the followings inequalities
(λ1,λ2,λ3)
are satisfied
(1) |λ | < 1;
3 2
4λ2+4λ2+4λ2 ≤ 1+16λ λ λ ;
1 2 3 1 2 3
2
λ2+λ2+ λ2+λ2 −4λ λ λ +λ2 ≤ 1;
1 2 1 2 1 2 3 3 2
r
(2) λ = 1, λ ,(cid:16)λ ∈ −(cid:17)1,1
3 2 1 2 2 2
(3) λ = −1, λ = ±h1, λ =i ∓1
3 2 1 2 2 2
Proof. From[11]weknowthatthecompletepositivityofT isequivalenttothepositivity
(λ1,λ2,λ3)
of the following matrix
T (e ) T (e )
T = (λ1,λ2,λ3) 11 (λ1,λ2,λ3) 12 .
(λ1,λ2,λ3) T (e ) T (e )
(cid:18) (λ1,λ2,λ3) 21 (λ1,λ2,λ3) 22 (cid:19)
b
10 FARRUKHMUKHAMEDOVANDHASANAKIN
It is clear that
1+2λ 0 0 0
3
1 0 1 0 0
T(λ1,λ2,λ3)(e11) = 2 0 0 1 0 ,
 0 0 0 1−2λ3 
 
 0 λ +λ λ +λ 0
1 2 1 2
1 λ −λ 0 0 λ +λ
T(λ1,λ2,λ3)(e12) = 2 λ11−λ22 0 0 λ11+λ22 
 0 λ1−λ2 λ1−λ2 0 
 
 ∗
and T (e )= 1I⊗1I−T (e ), T (e ) = T (e ) .
(λ1,λ2,λ3) 22 (λ1,λ2,λ3) 11 (λ1,λ2,λ3) 21 (λ1,λ2,λ3) 12
(1). According to [3, Theorem 1.3.3] the matrix T is positive if and only if
(λ1,λ2,λ3)
(4.18) T (e )−T (e )T (e )−1T (e )≥ 0,
(λ1,λ2,λ3) 11 (λ1,λ2,λ3) 12 (λ1,λ2,λ3b) 22 (λ1,λ2,λ3) 21
where T (e ) and T (e ) are positive matrices.
(λ1,λ2,λ3) 11 (λ1,λ2,λ3) 22
It is easy to see that T (e ) and T (e ) are positive if and only if
(λ1,λ2,λ3) 11 (λ1,λ2,λ3) 22
1
(4.19) |λ |≤ .
3
2
One can calculate that (4.18) is equivalent to
α 0 0 α
1 4
0 1+α α 0
 3 3  ≥ 0
0 α 1+α 0
3 3
 α4 0 0 α2 
 
 
where
α = 1+2λ −2(λ +λ )2, α = 1−2λ −2(λ −λ )2,
1 3 1 2 2 3 1 2
(λ −λ )2 (λ +λ )2
α = 1 2 − 1 2 , α = −2 λ2−λ2 .
3 2λ −1 2λ +1 4 1 2
3 3
(cid:16) (cid:17)
It is known that the matrix is positive if and only if the eigenvalues are positive. The eigen-
values of the last matrix can be calculated as follows
4λ2+4λ2+4λ2−16λ λ λ −1
s = 1, s = 1 2 3 1 2 3 ,
1 2 4λ2−1
3
2
s = 1−2λ2−2λ2 +2 λ2+λ2 −4λ λ λ +λ2,
3 1 2 1 2 1 2 3 3
r
(cid:16) (cid:17)
2
s = 1−2λ2−2λ2 −2 λ2+λ2 −4λ λ λ +λ2.
4 1 2 1 2 1 2 3 3
r
(cid:16) (cid:17)
To check the their positivity, it is enough to have s ≥ 0 and s ≥ 0. These mean
2 4
1
(4.20) λ 6= ;
3
2
(4.21) 4λ2+4λ2+4λ2 ≤ 1+16λ λ λ ;
1 2 3 1 2 3
2 1
(4.22) λ2+λ2+ λ2+λ2 −4λ λ λ +λ2 ≤ ;
1 2 1 2 1 2 3 3 2
r
2 (cid:16) (cid:17)
(4.23) λ2+λ2 +λ2 ≥ 4λ λ λ .
1 2 3 1 2 3
(cid:16) (cid:17)