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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 278606, 9 pages
http://dx.doi.org/10.1155/2013/278606
Research Article

On Kadison-Schwarz Type Quantum Quadratic Operators on

Department of Computational & Theoretical Sciences, Faculty of Science, International Islamic University Malaysia, P.O. Box 141, 25710 Kuantan, Pahang, Malaysia

Received 3 January 2013; Revised 13 March 2013; Accepted 27 March 2013

Academic Editor: Natig M. Atakishiyev

Copyright © 2013 Farrukh Mukhamedov and Abduaziz Abduganiev. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We study the description of Kadison-Schwarz type quantum quadratic operators (q.q.o.) acting from into . Note that such kind of operators is a generalization of quantum convolution. By means of such a description we provide an example of q.q.o. which is not a Kadison-Schwartz operator. Moreover, we study dynamics of an associated nonlinear (i.e., quadratic) operators acting on the state space of .

1. Introduction

It is known that one of the main problems of quantum information is the characterization of positive and completely positive maps on -algebras. There are many papers devoted to this problem (see, e.g., [14]). In the literature the completely positive maps have proved to be of great importance in the structure theory of -algebras. However, general positive (order-preserving) linear maps are very intractable [2, 5]. It is therefore of interest to study conditions stronger than positivity, but weaker than complete positivity. Such a condition is called Kadison-Schwarz property, that is, a map satisfies the Kadison-Schwarz property if holds for every . Note that every unital completely positive map satisfies this inequality, and a famous result of Kadison states that any positive unital map satisfies the inequality for self-adjoint elements . In [6] relations between -positivity of a map and the Kadison-Schwarz property of certain map is established. Certain relations between complete positivity, positive, and the Kadison-Schwarz property have been considered in [79]. Some spectral and ergodic properties of Kadison-Schwarz maps were investigated in [1012].

In [13] we have studied quantum quadratic operators (q.q.o.), that is, maps from into , with the Kadison-Schwarz property. Some necessary conditions for the trace-preserving quadratic operators are found to be the Kadison-Schwarz ones. Since trace-preserving maps arise naturally in quantum information theory (see, e.g., [14]) and other situations in which one wishes to restrict attention to a quantum system that should properly be considered a subsystem of a larger system with which it interacts. Note that in [15, 16] quantum quadratic operators acting on a von Neumann algebra were defined and studied. Certain ergodic properties of such operators were studied in [17, 18] (see for review [19]). In the present paper we continue our investigation; that is, we are going to study further properties of q.q.o. with Kadison-Schwarz property. We will provide an example of q.q.o. which is not a Kadison-Schwarz operator and study its dynamics. We should stress that q.q.o. is a generalization of quantum convolution (see [20]). Some dynamical properties of quantum convolutions were investigated in [21].

Note that a description of bistochastic Kadison-Schwarz mappings from into has been provided in [22].

2. Preliminaries

In what follows, by we denote an algebra of matrices over complex filed . By we mean tensor product of into itself. We note that such a product can be considered as an algebra of matrices over . In the sequel means an identity matrix, that is, . By we denote the set of all states (i.e., linear positive functionals which take value 1 at ) defined on .

Definition 1. A linear operator is said to be (a)a quantum quadratic operator (q.q.o.) if it satisfies the following conditions: (i)unital, that is, ;(ii) is positive, that is, whenever ; (b)a Kadison-Schwarz operator (KS) if it satisfies

One can see that if is unital and KS operator, then it is a q.q.o. A state is called a Haar state for a q.q.o. if for every one has

Remark 2. Note that if a quantum convolution on becomes a -homomorphic map with a condition then a pair is called a compact quantum group [20]. It is known [20] that for any given compact quantum group there exists a unique Haar state w.r.t. .

Remark 3. Let be a linear operator such that for all . If a q.q.o. satisfies , then is called a quantum quadratic stochastic operator. Such a kind of operators was studied and investigated in [17].
Each q.q.o. defines a conjugate operator by
One can define an operator by which is called a quadratic operator (q.c.). Thanks to conditions (a) (i), (ii) of Definition 1 the operator maps to .

3. Quantum Quadratic Operators with Kadison-Schwarz Property on

In this section we are going to describe quantum quadratic operators on and find necessary conditions for such operators to satisfy the Kadison-Schwarz property.

Recall [23] that the identity and Pauli matrices form a basis for , where

In this basis every matrix can be written as with , , here .

Lemma 4 (see [3]). The following assertions hold true: (a) is self-adjoint if and only if are reals; (b) if and only if ; here is the trace of a matrix ; (c) if and only if , where .

Note that any state can be represented by where with . Here as before stands for the scalar product in . Therefore, in the sequel we will identify a state with a vector .

In what follows by we denote a normalized trace, that is, , .

Let be a q.q.o. with a Haar state . Then one has which means that is an invariant state for .

Let us write the operator in terms of a basis in formed by the Pauli matrices, namely, where ().

One can prove the following.

Theorem 5 (see [13, Proposition 3.2]). Let be a q.q.o. with a Haar state , then it has the following form: where ,  , .

Let us turn to the positivity of . Given vector put Define a matrix .

By we denote a norm of the matrix associated with Euclidean norm in . Put and denote

Proposition 6 (see [13, Proposition 3.3]). Let be a q.q.o. with a Haar state , then .

Let be a liner operator with a Haar state . Then due to Theorem 5   has the form (10). Take arbitrary states and let be the corresponding vectors (see (7)). Then one finds that

Thanks to Lemma 4 the functional is a state if and only if the vector satisfies .

So, we have the following.

Proposition 7 (see [13, Proposition 4.1]). Let be a liner operator with a Haar state . Then for any if and only if the following holds:

From the proof of Proposition 6 and the last proposition we conclude that holds if and only if (16) is satisfied.

Remark 8. Note that characterizations of positive maps defined on were considered in [24] (see also [25]). Characterization of completely positive mappings from into itself with invariant state was established in [3] (see also [26]).

Next we would like to recall (see [13]) some conditions for q.q.o. to be the Kadison-Schwarz ones.

Let be a linear operator with a Haar state ; then it has the form (10). Now we are going to find some conditions to the coefficients when is a Kadison-Schwarz operator. Given and state , let us denote where . Here and in what follows stands for the usual cross-product in . Note that here the numbers are skew symmetric, that is, . By we will denote mapping to defined by .

Denote where (see (11)).

Theorem 9 (see [13, Theorem 3.6]). Let be a Kadison-Schwarz operator with a Haar state ; then it has the form (10) and the coefficients satisfy the following conditions: for all . Here as before ; , and , , and are defined in (19), (17), and (18), respectively.

Remark 10. The provided characterization with [2, 3] allows us to construct examples of positive or Kadison-Schwarz operators which are not completely positive (see next section).

Now we are going to give a general characterization of KS operators. Let us first give some notations. For a given mapping , by we denote the vector , and by we mean the following: where . Note that the last equality (22), due to the linearity of , can also be written as .

Theorem 11. Let be a unital -preserving linear mapping. Then is a KS operator if and only if one has for all with .

Proof. Let be an arbitrary element, that is, . Then . Therefore Consequently, we have From (26) and (27) one gets So, the positivity of the last equality implies that Now dividing both sides by we get the required inequality. Hence, this completes the proof.

4. An Example of Q.Q.O. Which Is Not Kadison-Schwarz One

In this section we are going to study dynamics of (57) for a special class of quadratic operators. Such class operators are associated with the following matrix given by and .

Via (10) we define a liner operator , for which is a Haar state. In the sequel we would like to find some conditions to which ensures positivity of .

It is easy that for given one can find a form of as follows. where, as before, .

Theorem 12. A linear operator given by (31) is a q.q.o. if and only if .

Proof. Let be a positive element from . Let us show positivity of the matrix . To do it, we rewrite (31) as follows: ; here
where positivity of yields that are real numbers. In what follows, without loss of generality, we may assume that , and therefore . It is known that positivity of is equivalent to positivity of the eigenvalues of .
Let us first examine eigenvalues of . Simple algebra shows us that all eigenvalues of can be written as follows: Now examine maximum and minimum values of the functions on the ball .
One can see that
Note that the functions , can reach values at .
Now let us rewrite and as follows:
One can see that where . Therefore, the functions ,   reach their maximum and minimum on the sphere (i.e., ). Hence, denoting from (37) and (36) we introduce the following functions: where .
One can find that the critical values of are , and the critical value of is . Consequently, extremal values of and on are the following: Therefore, from (37) and (38) we conclude that
It is known that for the spectrum of one has Therefore, So, if then one can see for all ,   . This implies that the matrix is positive for all with .
Now assume that is positive. Then is positive whenever is positive. This means that for all (). From (34) and (41) we conclude that . This completes the proof.

Theorem 13. Let then the corresponding q.q.o. is not KS operator.

Proof. It is enough to show the dissatisfaction of (21) at some values of () and .
Assume that ; then a little algebra shows that (21) reduces to the following one: where
Now choose as follows: Then calculations show that Hence, we find which means that (45) is not satisfied. Hence, is not a KS operator at .

Recall that a linear operator is completely positive if for any positive matrix the matrix is positive for all . Now we are interested when the operator is completely positive. It is known [1] that the complete positivity of is equivalent to the positivity of the following matrix: here () are the standard matrix units in .

From (31) one can calculate that where Hence, we find that where is the unit matrix in and So, the matrix is positive if and only if where .

One can easily calculate that . Therefore, we have the following.

Theorem 14. Let be given by (31). Then is completely positive if and only if .

5. Dynamics of

Let be a q.q.o. on . Let us consider the corresponding quadratic operator defined by , . From Theorem 5 one can see that the defined operator maps into itself if and only if or equivalently (16) holds. From (14) we find that Here, as before, .

So, (56) suggests that we consider the following nonlinear operator defined by where .

It is worth to mention that uniqueness of the fixed point (i.e., ) of the operator given by (57) was investigated in [13, Theorem 4.4].

In this section, we are going to study dynamics of the quadratic operator corresponding to (see (31)), which has the following form

Let us first find some condition on which ensures (16).

Lemma 15. Let be given by (58). Then maps into itself if and only if is satisfied.

Proof. “If” Part. Assume that maps into itself. Then (16) is satisfied. Take , . Then from (16) one finds that which yields .
“Only If” Part. Assume that . Take any . Then one finds that This completes the proof.

Remark 16. We stress that condition (16) is necessary for to be a positive operator. Namely, from Theorem 12 and Lemma 15 we conclude that if then the operator is not positive, while (16) is satisfied.

In what follows, to study dynamics of we assume . Recall that a vector is a fixed point of if . Clearly is a fixed point of . Let us find others. To do it, we need to solve the following equation:

We have the following.

Proposition 17. If   then has a unique fixed point in . If   then has the following fixed points: and in .

Proof. It is clear that is a fixed point of . If , for some then due to , one can see that the only solution of (61) belonging to is . Therefore, we assume that (). So, from (61) one finds Denoting From (62) it follows that
According to our assumption are nonzero, so from (64) one gets where and .
Dividing the second equality of (65) to the first one of (65) we find that which with yields Simplifying the last equality one gets This means that either or .
Assume that . Then from , one finds . Moreover, from the second equality of (65) we have . So, ; therefore, the solutions of the last one are . Hence, .
Now suppose that ; then . We note that , since . So, from the second equality of (65) we find So, which yields the solutions . Therefore, we obtain , and , .
Consequently, solutions of (65) are the following ones:
Now owing to (63) we need to solve the following equations: According to our assumption , we consider cases when .
Now let us start to consider several cases.
Case  1. Let , . Then from (71) one gets . So, from (61) we find , that is, . Now taking into account one gets . From the last inequality we have . Due to Lemma 15 the operator is well defined if and only if ; therefore, one gets . Hence, in this case a solution is .
Case  2. Let , . Then from (71) one finds . Substituting the last ones to (61) we get . Then, we have . Taking into account we find . This means ; due to Lemma 15 in this case the operator is not well defined; therefore, we conclude that there is no fixed point of belonging to .
Using the same argument for the rest of the cases we conclude the absence of solutions. This shows that if the operator has unique fixed point in . If , then has three fixed points belonging to . This completes the proof.

Now we are going to study dynamics of operator .

Theorem 18. Let be given by (58). Then the following assertions hold true: (i)if , then for any one has as . (ii)if , then for any with one has as .

Proof. Let us consider the following function . Then we have
This means
Due to from (73) one finds that which yields that the sequence is convergent. Next we would like to find the limit of .(i)First we assume that ; then from (73) we obtain This yields that as , for all .(ii)Now let . Then consider two distinct subcases.
Case A. Let and denote . Then one gets Hence, we have . This means . Hence, as .
Case B. Now take and assume that is not a fixed point. Therefore, we may assume that for some , otherwise from Proposition 17 one concludes that is a fixed point. Hence, from (58) one finds Similarly, one gets It is clear that (). According to our assumption () we conclude that one of is strictly less than ; this means . Therefore, from Case A, one gets that as .

Acknowledgments

The authors acknowledge the MOHE Grant FRGS11-022-0170. The first named author acknowledges the junior associate scheme of the Abdus Salam International Centre for Theoretical Physics, Trieste, Italy. The authors would like to thank an anonymous referee whose useful suggestions and comments improved the content of the paper.

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