Advances in Mathematical Physics
Volume 2018 (2018), Article ID 5296085, 10 pages
https://doi.org/10.1155/2018/5296085

Research Article

Preservers for the Tsallis Entropy of Convex Combinations of Density Operators

Xiaochun Fang¹ and Yihui Lao1,2

¹Department of Mathematics, Tongji University, Shanghai 200092, China
²Department of Mathematics and Computer Science, Guangxi Normal College for Nationalities, Chongzuo, Guangxi 532200, China

Correspondence should be addressed to Yihui Lao; yihuilao@yeah.net

Received 16 January 2018; Revised 21 April 2018; Accepted 29 April 2018; Published 4 June 2018

Academic Editor: Pavel Kurasov

Copyright © 2018 Xiaochun Fang and Yihui Lao. 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.

Let be a complex separable Hilbert space; we first characterize the unitary equivalence of two density operators by use of Tsallis entropy and then obtain the form of a surjective map on density operators preserving Tsallis entropy of convex combinations.

1. Introduction

In the mathematical framework of information theory, density operators (quantum states) are positive operators with trace 1 on complex separable Hilbert space , and the set of all density operators is denoted by on . is a compact convex set. A pure state is a rank one density operator, i.e., a rank one projection, and the set of all pure states is denoted by . We first introduce the definition of majorization given in [1]. Let be the set of all summable nonnegative real sequences and be the subset of consisting of all summable nonnegative real sequences whose sum is 1. Let with , ; then there exist (essentially unique) bijections such that and . We call if for each , and . We will denote and . Moreover, we have similar definitions and notations for with , .

Let with and , where ; we denote if . For and , Tsallis entropy [2] is defined as follows:

In this paper we will always assume that and , i.e., . If , then , for . Otherwise, can be infinite. Clearly , and if and only if is a pure state. It is well-known that has concavity and subadditivity (see [3]). Using the spectral decomposition theorem, it is easy to see that , where is the von Neumann entropy. Therefore Tsallis entropy can be viewed as a generalization of the von Neumann entropy. Moreover Tsallis entropy is available for long-range interaction or fractal-type structure physical systems where the von Neumann entropy is not suitable and can be applied to thermostatistical formalism [4] or to image thresholding segmentation [5]. The parameter in provides flexibility and universality in image processing.

It is well-known that quantum operation is a completely positive linear map [1], and some entanglement witnesses appear to be some special positive maps [6]. Therefore characterizing some maps on is now important in quantum information. In [7], it is proved that, for a map , , if and only if there exists a unitary or antiunitary operator acting on such that for all . The map of this form is widely studied in the context of relative entropy by Monlár et al. in a series of articles [8–11]. In the quantum context, Uhlmann proved that if and , then , where is some mixed unitary quantum operation, i.e., , where , , , and all are unitary operators acting on . Uhlmann’s theorem is widely used to study the role of majorization in quantum mechanics.

In the past decades, more attention has been paid to von Neumann entropy than Tsallis entropy. In [12, 13], He et al. characterized the unitary equivalence of maps on quantum states by use of the von Neumann entropy and obtained the form of maps on quantum states preserving the von Neumann entropy in finite dimensional Hilbert space. In [14], Li et al. proved that if and only if there exists a unitary operator such that , where is a bistochastic quantum operation, i.e., , and is a set of matrices known as operation elements. Moreover, in [15], Li et al. extended the result to infinite dimensional separable Hilbert space case. Recently, in [16], Bosyk et al. introduced a more general entropy, called the quantum -entropy , where is called a pair of entropic functionals, and obtained that, for two quantum states ,   acting on finite dimensional Hilbert space with , holds for any pair of entropic functionals if and only if there exists a unitary and antiunitary operator acting on such that (see Definition 2 and Proposition  1 in [16]). In this paper, we would solve similar questions in the context of Tsallis entropy in complex separable Hilbert space.

The organization of this paper is as follows. In Section 2, we will characterize the unitary equivalence of two density operators by use of Tsallis entropy. In Section 3, the form of a surjective map on density operators preserving Tsallis entropy of convex combinations is obtained.

2. Tsallis Entropy Equivalence

The main result in this section is as follows.

Theorem 1. Let be a complex separable Hilbert space, such that . Then the following are equivalent:

(1) holds for any .(2) holds for some such that .(3)There exists a unitary operator acting on such that .

It should be noted that, for any , .

In addition, Theorem 1 can be revised in the case of finite dimensional Hilbert space without the conditions majorization . We present it as Theorem 2 and give a proof by use of similar method in [12].

Theorem 2. Let be a complex Hilbert space with , , , and be the identity on . Then the following are equivalent.
(1)

holds for any and any .
(2) There exists such that

holds for any .
(3) There exists a unitary operator acting on such that .

Proof. . It is obvious.
. Let . Since , we may let , , where , , , and . Hence we get

Taking the Taylor series of and at , we have

Thus we get

where . Similarly, we can have the expansion of formula

Thus we have

Without the loss of generality, let , . Dividing both sides of equality (9) by and , respectively, we have

If and letting , then the left side of equality (10) is infinite while its right side is less than , which is a contradiction; therefore . Similarly, we have by use of the equality (11), and so . Repeating the process, we get and . Hence ,   have the same spectrum. This implies that ,   are unitary equivalent.
. Let and . Denote and . Since both and are positive operators and , we have . Thus .

To prove the main Theorem 1, we need some preparation.

Firstly, the following lemma is a direct corollary of [1, Theorem 8.0.1] as the functions and are convex.

Lemma 3. Let such that and . Then , and .

Theorem 4. Let such that ; then , and .

Proof. Suppose that , where and . For arbitrary , we have

Then Lemma 3 implies that and . Let , by the continuity of the function , we get

For any , we have that

It is clear that we may assume . Since , there exists a positive integer such that

Therefore

Then Lemma 3 implies that

By inequalities (13), we have

Let ; we obtained the conclusion , .

Proposition 5. Let such that . If and for some , then .

Proof. For , denote

It is obvious that . By Lemma 3 and , we have . Taking the second derivative with respect to on both sides of this equality, we obtain

It is then easy to see that ; i.e., .

Now we may generalize Proposition 5 to the infinite dimension case.

Theorem 6. Let such that and for some ; then .

Proof. Without the loss of generality, let and . We first claim that there is such that . In fact, if it is not right, then for any . Since , we may assume that , . Then there exists such that , and for some . For , let

It is obvious that . By Theorem 4 and the condition , we have

Therefore, for ,

Taking the second derivative with respect to on both sides of this equality above, we obtain

Therefore , which is a contradiction.
Now let , and , ; then and have the same properties as and . By the claim above again, there is such that and . Repeating this process, we have that, for any , there is such that and if . Since and , we have .

Proof of Theorem 1. (1)    (2). It is obvious.
(2)    (3). Since , we may let , , where and such that and . Let , . Noting that , then we have . Since , then . Therefore by Theorem 6, we have . Thus there exists a unitary operator acting on such that .
(3)    (1). Since both and are positive operators and , then, for , we have

3. The Characterization of Mappings That Preserve Tsallis Entropy

In this section, we shall determine the structure of surjective maps on density operators preserving Tsallis entropy of convex combinations. The following is our result.

Proposition 7. Let be a complex separable Hilbert space and be a surjective map such that, for any , and are comparable, i.e., or . Then the following statements are equivalent:
(1)

holds for any , any , and any .
(2) There exists such that and that for any there exists with

(3) There exists a unitary or antiunitary operator on such that for all .

Remark 8. In this proposition, scalar is defined by the arbitrariness of and . We advanced this proposition into infinite dimension by use of comparable condition. It should be pointed out that a similar conclusion is discussed in [17]. Their result is as follows: for a given , if , for any , then there exists a unitary or antiunitary operator acting on such that for all .

To prove Proposition 7, we need some preparation.

Lemma 9. Let be a complex separable Hilbert space, be a positive compact operator on , and be all the eigenvalues of ; then

Proof. Let with , and let the unit vector be the mutually orthogonal eigenvector of with respect to . Then , where and . Since there are such that and . Then , where . Then

On the other hand it is easy to see that , and this completes the proof.

Corollary 10. Let be a complex separable Hilbert space, and be positive compact operators on such that . If and are all the eigenvalues of and , respectively, then for .

Lemma 11. Let be a complex separable Hilbert space, , and be a unitary operator acting on . If there exist and such that holds for any , then .

Proof. Let be all the eigenvalues of . More precisely, we assume . Let be the eigenspace of with respect to and be the projection on for any .
Assume that the unit vector is the eigenvector of with respect to , i.e., . Let ; then , and all the eigenvalues of are for . Let ; then . Now fix and as given in the theorem; by Corollary 10 there are , , such that , and all the eigenvalues of are . Let , , then . Since , by Theorem 6 we have . It follows that all the eigenvalues of are . We claim that . In fact, otherwise, with , , and . Since is the eigenvalue of , there is unit vector such that

where with and . Since , the equality above holds only if and so

which is a contradiction. This completes the proof of . By the arbitrariness of and the fact that we have is a bijection on and so , . Therefore . Let , where is a unit eigenvector of with respect to . Since and , are the eigenvalues of both and . Then we get by similar discussion as above with instead of . Repeating this process, we have for any , and so ; i.e., .

Lemma 12. Let be a complex Hilbert space with and . Then if and only if , where and is the identity on .

Proof. If , then for any . Let ; then , where , , , and . Assume that for some . By the convexity of the function , we get

Equality holds if and only if . So

Then , . Therefore ; i.e., .
In the case for some , the discussion is similar.

Lemma 13 (see [3] Lemma 1). Let be a complex separable Hilbert space, and, for , let be the subset of consisting of all states with . Then is convex and the Tsallis entropy is strictly concave on ; i.e., for , , where . Moreover equality holds if and only if .
It should be noted that if or , .

Lemma 14 (see [3], Lemma 1). With the notations as above, for , ,

If the involved quantities are finite, then the equality holds if and only if .

Proposition 15. Let be a complex Hilbert space with and be a surjective map. Then the following statements are equivalent:
(1)

holds for any , any , and any .
(2) There exists such that

holds for any and any .
(3) There exists a unitary or antiunitary operator on such that for all .

Proof. and are obvious.
: we divided it into two cases: and .
Case 1  . Let . Take in equality (36); we have .
Then is injective and so bijective. In fact, if with , then, for ,

Thus we have by Lemma 13.
Claim. preserves orthogonality in both directions; i.e., for all .
In fact, if , by Lemma 14 we have

Since , by equality (36) we get

By Lemma 13 again, we get . Similarly, we can show . This completes the proof of the claim.
Since

is a bijection from onto .
By Corollary  1 in [18], there exists a unitary or antiunitary operator on such that for any .
Let for any ; it is obvious that . Since , by Lemma 12 we have , where is the identity on . Then .
To prove the Theorem, it is enough to prove for each fixed . By equality (36), we have

By Theorem 2, there exists a unitary operator (depending on and ) on such that ; i.e., . By equality (36) again, we get, for any ,

By Lemma 11, we have . Thus ; i.e., .
Case 2  . If , let us identify with the subset of density matrices

Note that satisfies if and only if the corresponding matrix is a rank one projection.
By equality (36), we still have for all and some . Since still preserves pure states in both directions, then there exists a unitary operator such that

Let for all ; then

For any , let

Taking in equality (36), then for any and some , we have

Since, for two qubit states , , we always have or , noting equality (47) and by Proposition 5, we know that the following two states have the same eigenvalues and eigenpolynomials.

Then we get their eigenpolynomials

It follows that and , so there exists a function such that for all .
Now, for arbitrary , let

and then

By equality (36) again, we have

It follows that

In equality (53), if , we have . Noting , so . It means that is a constant for

If , in equality (53), we have either

or

For any

equality (55) means while equality (56) results in . Next we shall show that either equality (55) holds for all states or equality (56) holds for all states . Assume on the contrary that there exist ,   such that the equality (55) holds for and equality (56) holds for . Let , it follows from equality (53) that

This is a contradiction. As a whole, if equality (55) holds for states , we have

So , where

If equality (56) holds for all states , we have

So , where is the transpose of the matrix and

Let , the proof is completed.

Proof of Proposition 7. and are obvious.
. Since , all the involved quantities are finite. Then by the same discussion as the proof of Proposition 15 in Case 1, we have , , and that is a bijective map which preserves orthogonality in both directions. By Corollary  1 in [18], there exists a unitary or antiunitary operator on such that for any .
Let for any ; it is obvious that . Since or , we have or . Now fix a . Since , by Theorem 1, there exists a unitary operator (depending on and ) on such that ; i.e., . By equality (27), we have, for any ,

By Lemma 11, we know ; then .

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The authors would like to thank Doctor Xin Li and Doctor Junqi Yang. This work is supported by National Natural Science Foundation of China [Grant no. 11371279] and the Fundamental Research Funds for the Central Universities [Grant no. 22120170035].

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