# Preservers for the Tsallis Entropy of Convex Combinations of Density Operators

Authors: Xiaochun Fang; Yihui Lao
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

LetH 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.


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## Body

## 1. Introduction

In the mathematical framework of information theory, density operators (quantum states) are positive operators with trace 1 on complex separable Hilbert spaceH, and the set of all density operators is denoted by S(H) on H. S(H) 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 P1(H). We first introduce the definition of majorization given in [1]. Let l1(R+) be the set of all summable nonnegative real sequences and l11(R+) be the subset of l1(R+) consisting of all summable nonnegative real sequences whose sum is 1. Let x→,y→∈l1(R+) with x→=(x1,x2,…,xn,…), y→=(y1,y2,…,yn,…); then there exist (essentially unique) bijections π,ω:{1,2,…,n,…}→{1,2,…,n,…} such that xπ(1)≥xπ(2)≥⋯≥xπ(n)≥⋯ and yω(1)≥yω2≥⋯≥yω(n)≥⋯. We call x→≺y→ if ∑i=1kxπ(i)≤∑i=1kyω(i) for each k=1,2,…,n,…, and ∑i=1∞xπ(i)=∑i=1∞yω(i). We will denote x→↓=(xπ(1),xπ(2),…,xπ(n),…) and y→↓=(yω(1),yω(2),…,yω(n),…). Moreover, we have similar definitions and notations for x→,y→∈Rn with x→=(x1,x2,…,xn), y→=(y1,y2,…,yn).Letρ,σ∈S(H) with ρ=∑i=1∞xiPi and σ=∑i=1∞yiQi, where Pi,Qi∈P1(H); we denote ρ≺σ if (x1,x2,…,xn,…)=x→≺y→=(y1,y2,…,yn,…). For ρ∈S(H) and r∈R\{1}, Tsallis entropy Sr(ρ) [2] is defined as follows: (1)Srρ=1-r-1tr⁡ρr-1. In this paper we will always assume that r>0 and r≠1, i.e., r∈R+\{1}. If r>1, then tr⁡(ρr)≤1, for ρr≤ρ. Otherwise, Sr(ρ) can be infinite. Clearly Sr(ρ)≥0, and Sr(ρ)=0 if and only if ρ is a pure state. It is well-known that Sr(ρ) has concavity and subadditivity (see [3]). Using the spectral decomposition theorem, it is easy to see that limr→1Sr(ρ)=S(ρ), where Sρ=-tr⁡ρln⁡ρ 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 r in Sr(ρ) 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 S(H) is now important in quantum information. In [7], it is proved that, for a map ϕ:S(H)→S(H), ϕ(μρ+(1-μ)σ)=μϕ(ρ)+(1-μ)ϕ(σ)(∀ρ,σ∈SH,∀μ∈[0,1]), if and only if there exists a unitary or antiunitary operator U acting on H such that ϕ(ρ)=UρU∗ for all ρ∈S(H). 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 ρ,σ∈S(H) and dim⁡H<∞, then ρ≺σ⇔ρ=ϕ(σ), where ϕ is some mixed unitary quantum operation, i.e., ϕ(ρ)=∑i=1kλiUiρUi∗(∀ρ∈S(H)), where k<∞, λi>0, ∑i=1kλi=1, and all Ui are unitary operators acting on H. 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 S(ϕ(ρ))=S(ρ)(∀ρ∈S(H)) if and only if there exists a unitary operator U such that ϕ(ρ)=UρU†, where ϕ(ρ) is a bistochastic quantum operation, i.e., ϕ(ρ)=∑iEiρEi†(∀ρ∈S(H)), and {Ei} 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 (h,ϕ)-entropy H(h,ϕ)(ρ), where (h,ϕ) is called a pair of entropic functionals, and obtained that, for two quantum states ρ,  σ acting on finite dimensional Hilbert space H with ρ≺σ, H(h,ϕ)(ρ)=H(h,ϕ)(σ) holds for any pair (h,ϕ) of entropic functionals if and only if there exists a unitary and antiunitary operator U acting on H such that ρ=UσU† (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 Section2, 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.
LetH be a complex separable Hilbert space, ρ,σ∈S(H) such that ρ≺σ. Then the following are equivalent:(1)
Sr(ρ)=Sr(σ) holds for any r∈R+\{1}.(2)
Sr(ρ)=Sr(σ) holds for some r∈R+\{1} such that Sr(ρ)+Sr(σ)<∞.(3)
There exists a unitary operatorU acting on H such that ρ=UσU∗.It should be noted that, for anyr>1, Sr(ρ)+Sr(σ)<∞.In addition, Theorem1 can be revised in the case of finite dimensional Hilbert space without the conditions majorization ρ≺σ. We present it as Theorem 2 and give a proof by use of similar method in [12].Theorem 2.
LetH be a complex Hilbert space with dim⁡H=n<+∞, ρ,σ∈S(H), 0<s≤1, and I be the identity on H. Then the following are equivalent.
(1)(2)Srtρ+1-tItr⁡I=Srtσ+1-tItr⁡Iholds for any r∈R+\{1} and any t∈[0,1].
(2) There existsr∈R+\{1} such that (3)Srtρ+1-tItr⁡I=Srtσ+1-tItr⁡Iholds for any t∈(0,s).
(3) There exists a unitary operatorU acting on H such that ρ=UσU∗.Proof.
( 1 ) ⇒ ( 2 ). It is obvious.
( 2 ) ⇒ ( 3 ). Let r∈R+\{1}. Since ρ,σ∈S(H), we may let ρ=∑i=1kxiPi, σ=∑i=1lyiQi, where 1≤k,l≤n, Pi,Qi∈P1(H), 0<xi,yi≤1, and ∑i=1kxi=∑i=1lyi=1. Hence we get (4)∑i=1ktxi+1-tnr+n-k1-tnr=tr⁡tρ+1-tInr=tr⁡tσ+1-tInr=∑i=1ltyi+1-tnr+n-l1-tnr.
Taking the Taylor series of(txi+(1-t)/n)r and (1-t)r at t=0, we have(5)txi+1-tnr=1nr+1+nxi-1tr=1nr1+∑p=1∞rr-1⋯r-p+1p!nxi-1ptp=1nr+∑p=1∞rr-1⋯r-p+1nr-pp!xi-1nptp,(6)1-tr=1+∑p=1∞rr-1⋯r-p+1p!-1ptp.
Thus we get(7)∑i=1ktxi+1-tnr+n-k1-tnr=∑i=1ktxi+1-tnr+n-knr1-tr=1nr-1+∑p=1∞rr-1⋯r-p+1nr-pp!∑i=1kxi-1np+-1pn-knptp=1nr-1+∑p=1∞rr-1⋯r-p+1nr-pp!∑i=1k∑q=0p-1-1qCpqnqxip-q+-1pnp-1tp,where Cpq=p!/q!(p-q)!. Similarly, we can have the expansion of formula ∑i=1l(tyi+(1-t)/n)r+(n-l)((1-t)/n)r.(8)∑i=1ltyi+1-tnr+n-l1-tnr=1nr-1+∑p=1∞rr-1⋯r-p+1nr-pp!∑i=1l∑q=0p-1-1qCpqnqyip-q+-1pnp-1tp.
Thus we have(9)∑i=1kxid-∑i=1lyid=0,d∈N+.Without the loss of generality, let x1≥x2≥⋯≥xk>0, y1≥y2≥⋯≥yl>0. Dividing both sides of equality (9) by y1d and x1d, respectively, we have (10)x1y1d+x2y1d+⋯+xky1d=1+y2y1d+⋯+yly1d,(11)1+x2x1d+⋯+xkx1d=y1x1d+y2x1d+⋯+ylx1d.If x1>y1 and letting d→+∞, then the left side of equality (10) is infinite while its right side is less than l, which is a contradiction; therefore x1≤y1. Similarly, we have x1≥y1 by use of the equality (11), and so x1=y1. Repeating the process, we get k=l and xi=yi(i=1,2,…,k). Hence ρ,  σ have the same spectrum. This implies that ρ,  σ are unitary equivalent.
( 3 ) ⇒ ( 1 ). Let r∈R+\{1} and t∈[0,1]. Denote Aρ=tρ+(1-t)(I/n) and Aσ=tσ+(1-t)(I/n). Since both Aρ and Aσ are positive operators and ρ=UσU∗, we have Aρr=(UAσU∗)r=UAσrU∗. Thus Sr(Aρ)=(tr⁡(Aρr)-1)/(1-r)=Sr(Aσ).To prove the main Theorem1, we need some preparation.Firstly, the following lemma is a direct corollary of [1, Theorem 8.0.1] as the functions f(x)=xr(r>1) and f(x)=-xr(0<r<1) are convex.Lemma 3.
Letx→,y→∈Rn such that xi,yi≥0 and x→≺y→. Then ∑i=1nxir≥∑i=1nyir(0<r<1), and ∑i=1nxir≤∑i=1nyir(r>1).Theorem 4.
Letx→,y→∈l1(R+) such that x→≺y→; then ∑i=1∞xir≥∑i=1∞yir(0<r<1), and ∑i=1∞xir≤∑i=1∞yir(r>1).Proof.
Suppose thatα=(α1,α2,…,αn,…), where αi≥0 and ∑i=1∞αi<∞. For arbitrary n, we have (12)α1,α2,…,αn≺∑i=1nαi,0,0,…,0. Then Lemma 3 implies that ∑i=1nαir≥(∑i=1nαi)r(0<r<1) and ∑i=1nαir≤(∑i=1nαi)r(r>1). Let n→∞, by the continuity of the function f(x)=xr, we get (13)∑i=1∞αir≥∑i=1∞αir0<r<1,∑i=1∞αir≤∑i=1∞αirr>1.For any m>1, we have that (14)x→≺y→≺y1,y2,…,ym,∑i=m+1∞yi,0,0,….
It is clear that we may assume∑i=1∞xi=∑i=1∞yi≠0. Since ∑i=1∞xi<∞, there exists a positive integer N≥m such that (15)∑i=N+1∞xi<min⁡y1,y2,…,ym,∑i=m+1∞yi-0.Therefore (16)x1,x2,…,xN,∑i=N+1∞xi≺y1,y2,…,ym,∑i=m+1∞yi,0,…,0. Then Lemma 3 implies that (17)∑i=1Nxir+∑i=N+1∞xir≥∑i=1myir+∑i=m+1∞yir≥∑i=1myir0<r<1,∑i=1Nxir+∑i=N+1∞xir≤∑i=1myir+∑i=m+1∞yirr>1. By inequalities (13), we have (18)∑i=1∞xir≥∑i=1myir0<r<1,∑i=1∞xir≤∑i=1myir+∑i=m+1∞yirr>1.
Letm→∞; we obtained the conclusion ∑i=1∞xir≥∑i=1∞yir(0<r<1), ∑i=1∞xir≤∑i=1∞yir(r>1).Proposition 5.
Letx→,y→∈Rn such that xi,yi≥0. If x→≺y→ and ∑i=1nxir=∑i=1nyir for some r∈R+\{1}, then x→↓=y→↓.Proof.
For0<λ<1, denote (19)z→=λx1+1-λy1,λx2+1-λy2,…,λxn+1-λyn.It is obvious that x→≺z→≺y→. By Lemma 3 and ∑i=1nxir=∑i=1nyir, we have ∑i=1n(λxi+(1-λ)yi)r=∑i=1nxir. Taking the second derivative with respect to λ on both sides of this equality, we obtain (20)∑i=1nrr-1λxi+1-λyir-2xi-yi2=0. It is then easy to see that xi=yi(1≤i≤n); i.e., x→↓=y→↓.Now we may generalize Proposition5 to the infinite dimension case.Theorem 6.
Letx→,y→∈l1(R+) such that x→≺y→ and ∑i=1∞xir=∑i=1∞yir<∞ for some r∈R+\{1}; then x→↓=y→↓.Proof.
Without the loss of generality, letx→=x→↓=(x1,x2,…,xn,…) and y→=y→↓=(y1,y2,…,yn,…). We first claim that there is n1≥1 such that x1=yn1. In fact, if it is not right, then x1≠yn for any n≥1. Since x→≺y→, we may assume that x→≠0, 0≠x1≤y1. Then there exists k≥1 such that x1∈(yk+1,yk), and x1=tyk+1+(1-t)yk for some t∈(0,1). For s∈(0,t), let (21)z→=y1,y2,…,yk-1,syk+1+1-syk,1-syk+1+syk,yk+2⋯.It is obvious that x→≺z→≺y→. By Theorem 4 and the condition ∑i=1∞xir=∑i=1∞yir<∞, we have (22)∑i=1∞xir=∑i=1k-1yir+syk+1+1-sykr+1-syk+1+sykr+∑i=k+2∞yir=∑i=1∞yir<∞.Therefore, for s∈(0,t), (23)syk+1+1-sykr+1-syk+1+sykr=ykr+yk+1r.Taking the second derivative with respect to s on both sides of this equality above, we obtain (24)rr-1syk+1+1-sykr-2+1-syk+1+sykr-2yk-yk+12=0.Therefore yk=yk+1, which is a contradiction.
Now letx1=yn1, and x′→=(x2,…,xn,…), y′→=(y1,y2,…,yn1-1,yn1+1,…); then x′→ and y′→ have the same properties as x→ and y→. By the claim above again, there is n2≥1 such that x2=yn2 and n2≠n1. Repeating this process, we have that, for any xk, there is nk≥1 such that xk=ynk and nk≠nl if k≠l. Since x→,y→∈l1(R+) and ∑i=1∞xi=∑i=1∞yi<∞, we have x→↓=y→↓.Proof of Theorem1.
(1)⇒  (2). It is obvious.
(2)⇒  (3). Since ρ,σ∈S(H), we may let ρ=∑i=1∞xiPi, σ=∑i=1∞yiQi, where Pi,Qi∈P1(H) and 0≤xi,yi≤1 such that ∑i=1∞Pi=∑i=1∞Qi=idH and ∑i=1∞xi=∑i=1∞yi=1. Let x→=(x1,x2,…), y→=(y1,y2,…). Noting that ρ≺σ, then we have x→≺y→. Since Sr(ρ)=Sr(σ)<∞, then ∑i=1∞xir=∑i=1∞yir<∞. Therefore by Theorem 6, we have x→↓=y→↓. Thus there exists a unitary operator U acting on H such that ρ=UσU∗.
(3)⇒  (1). Since both ρ and σ are positive operators and ρ=UσU∗, then, for r∈R+\{1}, we have (25)Srρ=1-r-1tr⁡ρr-1=1-r-1tr⁡UσU∗r-1=1-r-1tr⁡UσrU∗-1=Srσ.


## 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.
LetH be a complex separable Hilbert space and ϕ:S(H)→S(H) be a surjective map such that, for any ρ∈S(H), ρ and ϕ(ρ) are comparable, i.e., ρ≺ϕ(ρ) or ϕ(ρ)≺ρ. Then the following statements are equivalent:
(1)(26)Srtρ+1-tσ=Srtϕρ+1-tϕσholds for any ρ,σ∈S(H), any r∈R+\{1}, and any t∈[0,1].
(2) There existsr>1 such that Sr(ρ)=Sr(ϕ(ρ))(∀ρ∈S(H)) and that for any ρ,σ∈S(H) there exists t∈(0,1) with (27)Srtρ+1-tσ=Srtϕρ+1-tϕσ.
(3) There exists a unitary or antiunitary operatorU on H such that ϕ(ρ)=UρU∗ for all ρ∈S(H).Remark 8.
In this proposition, scalart∈[0,1] is defined by the arbitrariness of ρ and σ. 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 t, if Sr(tρ+(1-t)σ)=Sr(tϕ(ρ)+(1-t)ϕ(σ)), for any ρ,σ∈S(H), then there exists a unitary or antiunitary operator U acting on H such that ϕ(ρ)=UρU∗ for all ρ∈S(H).To prove Proposition7, we need some preparation.Lemma 9.
LetH be a complex separable Hilbert space, A be a positive compact operator on H, and x1≥x2≥⋯≥xn≥⋯ be all the eigenvalues of A; then (28)xn=minK⊆H,dim⁡K=n-1⁡suph∈K⊥,h=1⁡Ah,h.Proof.
LetK⊆H with dim⁡(K)=n-1, and let the unit vector ei be the mutually orthogonal eigenvector of A with respect to xi. Then ei=ki⊕hi, where ki∈K and hi∈K⊥. Since dim⁡(K)=n-1 there are α1,α2,…,αn such that ∑i=1nαiki=0 and ∑i=1nαi2=1. Then ∑i=1nαiei=∑i=1nαihi∈(K⊥∩Hn)\{0}, where Hn=span{e1,e2,…,en}. Then (29)A∑i=1nαiei,∑i=1nαiei=∑i=1nxiαiei,∑i=1nαiei≥xn. On the other hand it is easy to see that xn=suph∈Hn-1⊥,h=1Ah,h, and this completes the proof.Corollary 10.
LetH be a complex separable Hilbert space, A and B be positive compact operators on H such that A≥B. If x1≥x2≥⋯≥xn≥⋯ and y1≥y2≥⋯≥yn≥⋯ are all the eigenvalues of A and B, respectively, then xi≥yi for i=1,2,….Lemma 11.
LetH be a complex separable Hilbert space, ρ∈S(H), and U be a unitary operator acting on H. If there exist r∈R+\{1} and t∈(0,1) such that Sr(tρ+(1-t)P)=Sr(tUρU∗+(1-t)P)<∞ holds for any P∈P1(H), then ρ=UρU∗.Proof.
Letx1≥x2≥⋯≥xn≥⋯ be all the eigenvalues of ρ. More precisely, we assume x1=⋯=xn1>xn1+1=⋯=xn2>⋯>xnk-1+1=⋯=xnk>⋯. Let Hn be the eigenspace of ρ with respect to xn and Qn be the projection on Hn for any n≥1.
Assume that the unit vectore1 is the eigenvector of ρ with respect to x1, i.e., ρe1=x1e1. Let P=e1⊗e1; then ρP=Pρ, and all the eigenvalues of tρ+(1-t)P are tx1+1-t≥tx2≥⋯≥txn≥⋯ for t∈(0,1). Let k=U∗e1; then U∗PU=k⊗k. Now fix r and t as given in the theorem; by Corollary 10 there are z1≥0, z2≥0,…, zn≥0,… such that ∑i=1∞zi=1-t, and all the eigenvalues of tρ+(1-t)k⊗k are tx1+z1≥tx2+z2≥⋯≥txn+zn≥⋯. Let x→=(tx1+1-t,tx2,…,txn,…), y→=(tx1+z1,tx2+z2,…,txn+zn,…), then y→≺x→. Since Sr(tρ+(1-t)P)=Sr(tUρU∗+(1-t)P)=Sr(tρ+(1-t)k⊗k)<∞, by Theorem 6 we have y→=x→. It follows that all the eigenvalues of tρ+(1-t)k⊗k are tx1+1-t≥tx2≥⋯≥txn≥⋯. We claim that k∈H1. In fact, otherwise, k=k1⊕k2 with k1∈H1, k2⊥H1, and k2≠0. Since tx1+1-t is the eigenvalue of tρ+(1-t)k⊗k, there is unit vector h∈H such that(30)tx1+1-t=tρ+1-tk⊗kh,h=tρh,h+1-tk⊗kh,h=tx1h12+tρh2,h2+1-th,k2≤tx1h12+txn2h22+1-th1,k1+h2,k22≤tx1-tx1-xn2h22+1-th1k1+h2k22≤tx1-tx1-xn2h22+1-th12+h22k12+k22=tx1-tx1-xn2h22+1-t, where h=h1⊕h2 with h1∈H1 and h2⊥H1. Since k2≠0, the equality above holds only if h2≠0 and so (31)tx1+1-t≤tx1-tx1-xn2h22+1-t<tx1+1-t,which is a contradiction. This completes the proof of k=U∗e1∈H1. By the arbitrariness of e1∈H1 and the fact that dim⁡(H1)<∞ we have U∗=U-1 is a bijection on H1 and so UQ1=Q1UQ1, U∗Q1=Q1U∗Q1. Therefore UQ1=Q1U. Let P=en2⊗en2, where en2 is a unit eigenvector of ρ with respect to xn2. Since U(1-Q1)=(1-Q1)U and P≤1-Q1, tx1=tx2=⋯=txn1 are the eigenvalues of both tρ+(1-t)U∗PU and tρ+(1-t)P. Then we get UQ2=Q2U by similar discussion as above with (1-Q1)ρ(1-Q1) instead of ρ. Repeating this process, we have UQn=QnU for any n, and so Uρ=ρU; i.e., UρU∗=ρ.Lemma 12.
LetH be a complex Hilbert space with dim⁡H=n<∞ and ρ∈S(H). Then ρ=I/n if and only if Sr(ρ)=(n1-r-1)/(1-r), where r∈R+\{1} and I is the identity on H.Proof.
Ifρ=I/n, then Sr(ρ)=(tr⁡((I/n)r)-1)/(1-r)=(n1-r-1)/(1-r) for any r∈R+\{1}. Let ρ∈S(H); then ρ=∑i=1kxiPi, where Pi∈P1(H), 1≤k≤n, 0≤xi≤1, and ∑i=1kxi=1. Assume that Sr(ρ)=(n1-r-1)/(1-r) for some r>1. By the convexity of the function y=xr, we get (32)tr⁡ρr=∑i=1kxir=k·1k∑i=1kxir≥k∑i=1k1kxir=k1-r.
Equality holds if and only ifxi=1/k(i=1,2,…,k). So (33)n1-r-11-r=Srρ=tr⁡ρr-11-r≤k1-r-11-r≤n1-r-11-rr>1.Then k=n, tr⁡(ρr)=k1-r=n1-r. Therefore xi=1/n(i=1,2,…,k); i.e., ρ=I/n.
In the caseSr(ρ)=(n1-r-1)/(1-r) for some 0<r<1, the discussion is similar.Lemma 13 (see [3] Lemma 1).
LetH be a complex separable Hilbert space, and, for r∈R+\{1}, let Sr(H) be the subset of S(H) consisting of all states ρ with Sr(ρ)<∞. Then Sr(H) is convex and the Tsallis entropy Sr(ρ) is strictly concave on Sr(H); i.e., for t∈(0,1), Sr(tρ+(1-t)σ)≥tSr(ρ)+(1-t)Sr(σ), where ρ,σ∈Sr(H). Moreover equality holds if and only if ρ=σ.
It should be noted that ifr>1 or dim⁡(H)<∞, Sr(H)=S(H).Lemma 14 (see [3], Lemma 1).
With the notations as above, for0<t<1, r∈R+\{1}, (34)Srtρ+1-tσ≤trSrρ+1-trSrσ+r-1-11-tr-1-tr. If the involved quantities are finite, then the equality holds if and only if ρσ=0.Proposition 15.
LetH be a complex Hilbert space with dim⁡H=n<∞ and ϕ:S(H)→S(H) be a surjective map. Then the following statements are equivalent:
(1)(35)Srtρ+1-tσ=Srtϕρ+1-tϕσholds for any ρ,σ∈S(H), any r∈R+\{1}, and any t∈[0,1].
(2) There existsr∈R+\{1} such that (36)Srtρ+1-tσ=Srtϕρ+1-tϕσholds for any ρ,σ∈S(H) and any t∈[0,1].
(3) There exists a unitary or antiunitary operatorV on H such that ϕ(ρ)=VρV∗ for all ρ∈S(H).Proof.
( 1 ) ⇒ ( 2 ) and (3)⇒(1) are obvious.
( 2 ) ⇒ ( 3 ): we divided it into two cases: 2<dim⁡H=n and dim⁡H=2.
Case 1 ( 2 < dim ⁡ H = n ) .Let r∈R+\{1}. Take t=1 in equality (36); we have Sr(ρ)=Sr(ϕ(ρ))(∀ρ∈S(H)).
Thenϕ is injective and so bijective. In fact, if ρ,σ∈S(H) with ϕ(ρ)=ϕ(σ), then, for t∈(0,1),(37)tSrρ+1-tSrσ=tSrϕρ+1-tSrϕσ=Srtϕρ+1-tϕσ=Srtρ+1-tσ. Thus we have ρ=σ by Lemma 13.
Claim. ϕ preserves orthogonality in both directions; i.e., ρσ=0⇔ϕ(ρ)ϕ(σ)=0 for all ρ,σ∈S(H).
In fact, ifρσ=0, by Lemma 14 we have (38)Srtρ+1-tσ=trSrρ+1-trSrσ+r-1-11-tr-1-tr. Since Sr(ρ)=Sr(ϕ(ρ)), by equality (36) we get(39)Srtρ+1-tσ=trSrρ+1-trSrσ+r-1-11-tr-1-tr=trSrϕρ+1-trSrϕσ+r-1-11-tr-1-tr=Srtϕρ+1-tϕσ.By Lemma 13 again, we get ϕ(ρ)ϕ(σ)=0. Similarly, we can show ϕ(ρ)ϕ(σ)=0⇒ρσ=0. This completes the proof of the claim.
Since(40)ρ∈P1H⟺0=Srρ⟺0=Srρ=Srϕρ⟺Srϕρ=0⟺ϕρ∈P1H,
ϕ P 1 ( H ) is a bijection from P1(H) onto P1(H).
By Corollary  1 in [18], there exists a unitary or antiunitary operator V on H such that ϕ(P)=VPV∗ for any P∈P1(H).
Letψ(ρ)=V∗ϕ(ρ)V for any ρ∈S(H); it is obvious that ψ(P)=V∗VPV∗V=P(∀P∈P1(H)). Since Sr(I/n)=Sr(ϕ(I/n)), by Lemma 12 we have ϕ(I/n)=I/n, where I is the identity on H. Then ψ(I/n)=V∗ϕ(I/n)V=V∗(I/n)V=I/n.
To prove the Theorem, it is enough to proveψ(ρ)=ρ for each fixed ρ∈S(H). By equality (36), we have(41)Srtρ+1-tIn=Srtϕρ+1-tϕIn=SrtVψρV∗+1-tVψInV∗=Srtψρ+1-tψIn=Srtψρ+1-tIn. By Theorem 2, there exists a unitary operator Uρ (depending on ρ and ψ) on H such that ψ(ρ)=UρρUρ∗; i.e., ρ=Uρ∗ψ(ρ)Uρ. By equality (36) again, we get, for any P∈P1(H), (42)Srtρ+1-tP=Srtϕρ+1-tϕP=SrtVψρV∗+1-tVψPV∗=Srtψρ+1-tψP=Srtψρ+1-tP=SrtUρρUρ∗+1-tP. By Lemma 11, we have ρ=UρρUρ∗. Thus ρ=UρρUρ∗=UρUρ∗ψ(ρ)UρUρ∗=ψ(ρ)=V∗ϕ(ρ)V; i.e., ϕ(ρ)=VρV∗.
Case 2 ( dim ⁡ H = 2 ). If dim⁡H=2, let us identify S(H) with the subset of 2×2 density matrices (43)SH=121+zx+iyx-iy1-z∣x,y,z∈R3:x2+y2+z2≤1. Note that (x,y,z) satisfies x2+y2+z2=1 if and only if the corresponding matrix is a rank one projection.
By equality (36), we still have Sr(ρ)=Sr(ϕ(ρ)) for all ρ∈S(H) and some r∈R+\{1}. Since ϕ still preserves pure states in both directions, then there exists a unitary operator V such that (44)ϕ1000=V1000V∗. Let ψ(ρ)=V∗ϕ(ρ)V for all ρ∈S(H); then (45)ψ1000=1000.
For anyρ∈S(H), let (46)ρ=1+z2x+iy2x-iy21-z2,ψρ=1+z′2x′+iy′2x′-iy′21-z′2.
Takingσ=1000 in equality (36), then for any t∈[0,1] and some r∈R+\{1}, we have (47)Srt1+z2x+iy2x-iy21-z2+1-t1000=Srt1+z′2x′+iy′2x′-iy′21-z′2+1-t1000.
Since, for two qubit states(x1,x2), (y1,y2), we always have (x1,x2)≺(y1,y2) or (y1,y2)≺(x1,x2), noting equality (47) and by Proposition 5, we know that the following two states have the same eigenvalues and eigenpolynomials. (48)t1+z2x+iy2x-iy21-z2+1-t1000,t1+z′2x′+iy′2x′-iy′21-z′2+1-t1000. Then we get their eigenpolynomials (49)λ2-λ-x2+y2+1-z24t2+1-z2t=λ00λ-t1+z2x+iy2x-iy21-z2+1-t1000=λ00λ-t1+z′2x′+iy′2x′-iy′21-z′2+1-t1000=λ2-λ-x′2+y′2+1-z′24t2+1-z′2t.
It follows thatz=z′ and x+iy=x′+iy′, so there exists a function θ:S(H)→[0,2π) such that eiθ(ρ)(x+iy)=x′+iy′ for all ρ∈S(H).
Now, for arbitraryρ,σ∈S(H), let(50)ρ=1+zρ2xρ+iyρ2xρ-iyρ21-zρ2,σ=1+zσ2xσ+iyσ2xσ-iyσ21-zσ2,and then(51)ψρ=1+zρ2xρ+iyρ2eiθρxρ-iyρ2e-iθρ1-zρ2,ψσ=1+zσ2xσ+iyσ2eiθσxσ-iyσ2e-iθσ1-zσ2.By equality (36) again, we have(52)λ2-λ-xρ-xσ2+yρ-yσ2+zρ-zσ2t2+2xσ2+2yσ2+2zσ2-2xρxσ-2yρyσ-2zρzσt+1-xσ2-yσ2-zσ2=λ2-λ-xρ2+yρ2+zρ2+xσ2+yσ2+zσ2-2xρxσcos⁡θρ-θσ-2yρyσcos⁡θρ-θσ-2xρyσsin⁡θρ-θσ+2xσyρsin⁡θρ-θσ-2zρzσt2+2xσ2+2yσ2+2zσ2-2xρxσcos⁡θρ-θσ-2yρyσcos⁡θρ-θσ-2zρzσ-2xρyσsin⁡θρ-θσ+2xσyρsin⁡θρ-θσt+1-xσ2-yσ2-zσ2.
It follows that(53)xρxσ+yρyσ1-cos⁡θρ-θσ=xρyσ-xσyρsin⁡θρ-θσ.In equality (53), if xρ,xσ≠0,yρ=yσ=0, we have cos⁡(θρ-θσ)=1. Noting θρ,θσ∈[0,2π), so θρ=θσ. It means that θρ is a constant θ0 for(54)ρ=1+zρ2xρ2xρ21-zρ2.
Ifxρ,yρ,xσ≠0, yσ=0 in equality (53), we have either(55)cos⁡θρ-θ0=1,sin⁡θρ-θ0=0 or (56)cos⁡θρ-θ0=xρ2-yρ2xρ2+yρ2,sin⁡θρ-θ0=-2xρyρxρ2+yρ2.For any(57)ρ=1+zρ2xρ+iyρ2xρ-iyρ21-zρ2∈SH,equality (55) means (xρ+iyρ)ei(θρ-θ0)=xρ+iyρ while equality (56) results in (xρ+iyρ)ei(θρ-θ0)=xρ-iyρ. Next we shall show that either equality (55) holds for all states ρ or equality (56) holds for all states ρ. Assume on the contrary that there exist ρ,  σ such that the equality (55) holds for xρ+iyρ and equality (56) holds for xσ+iyσ. Let yρ,yσ≠0, it follows from equality (53) that(58)2xρxσ-yρyσ=xρ+iyρxσ+iyσ+xρ-iyρxσ-iyσ=xρ+iyρeiθρ-θ0xσ-iyσe-iθσ-θ0+xρ-iyρe-iθρ-θ0xσ+iyσeiθσ-θ0=2xρxσ+yρyσcos⁡θρ-θσ+2xρyσ-xσyρsin⁡θρ-θσ=2xρxσ+yρyσ. This is a contradiction. As a whole, if equality (55) holds for states ρ∈S(H), we have (59)
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.So ψ(ρ)=WρW∗, where(60)W=eiθ0/200e-iθ0/2.
If equality (56) holds for all states ρ∈S(H), we have(61)ψρ=1+zρ2xρ+iyρ2eiθρxρ-iyρ2e-iθρ1-zρ2=1+zρ2xρ+iyρ2eiθρ-θ0eiθ0xρ-iyρ2e-iθρ-θ0e-iθ01-zρ2=1+zρ2xρ-iyρ2eiθ0xρ+iyρ2e-iθ01-zρ2=eiθ0/200e-iθ0/21+zρ2xρ-iyρ2xρ+iyρ21-zρ2e-iθ0/200eiθ0/2.
Soψ(ρ)=WρTW∗, where ρT is the transpose of the matrix ρ and(62)W=eiθ0/200e-iθ0/2.Let U=VW, the proof is completed.Proof of Proposition7.
( 1 ) ⇒ ( 2 ) and (3)⇒(1) are obvious.
( 2 ) ⇒ ( 3 ). Since r>1, all the involved quantities are finite. Then by the same discussion as the proof of Proposition 15 in Case 1, we have Sr(ρ)=Sr(ϕ(ρ)), ϕ(P1(H))=P1(H), and that ϕ is a bijective map which preserves orthogonality in both directions. By Corollary  1 in [18], there exists a unitary or antiunitary operator U on H such that ϕ(P)=UPU∗ for any P∈P1(H).
Letψ(ρ)=U∗ϕ(ρ)U for any ρ∈S(H); it is obvious that ψ(P)=U∗UPU∗U=P(∀P∈P1(H)). Since ρ≺ϕ(ρ) or ϕ(ρ)≺ρ, we have ρ≺ψ(ρ) or ψ(ρ)≺ρ. Now fix a ρ∈S(H). Since Sr(ρ)=Sr(ϕ(ρ))=Sr(Uψ(ρ)U∗)=Sr(ψ(ρ)), by Theorem 1, there exists a unitary operator Wρ (depending on ρ and ψ) on H such that ψ(ρ)=WρρWρ∗; i.e., ρ=Wρ∗ψ(ρ)Wρ. By equality (27), we have, for any P∈P1(H), (63)Srtρ+1-tP=Srtϕρ+1-tϕP=SrtUψρU∗+1-tUψPU∗=Srtψρ+1-tψP=Srtψρ+1-tP=SrtWρρWρ∗+1-tP. By Lemma 11, we know ρ=WρρWρ∗=WρWρ∗ψ(ρ)WρWρ∗=ψ(ρ); then ϕ(ρ)=Uψ(ρ)U∗=UρU∗.



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*Source: 5296085-2018-06-04.xml*