﻿ 保持量子态凸组合的Tsallis的映射
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 同济大学学报(自然科学版)  2019, Vol. 47 Issue (5): 720-722.  DOI: 10.11908/j.issn.0253-374x.2019.05.018 0

### 引用本文

LAO Yihui, YANG Junqi. Maps on Quantum States Preserving Tsallis Entropy of Convex Combinations[J]. Journal of Tongji University (Natural Science), 2019, 47(5): 720-722. DOI: 10.11908/j.issn.0253-374x.2019.05.018

### 文章历史

1. 同济大学 数学科学学院，上海 200092;
2. 广西民族师范学院 数学与计算机科学学院，广西 崇左 532200

Maps on Quantum States Preserving Tsallis Entropy of Convex Combinations
LAO Yihui 1,2, YANG Junqi 1
1. School of Mathematical Sciences, Tongji University, Shanghai 200092, China;
2. Mathematics and Computer Sciences, Guangxi Normal College for Nationalities, Chongzuo 532200, China
Abstract: Let Hm be the complex Hilbert space with dimension m, S(Hm$\otimes$Hn) be all the quantum states acting on complex bipartite Hilbert space Hm$\otimes$Hn and Ssep(Hm$\otimes$Hn) be the convex set of comparable quantum states. $\varphi : S\left(H_{m} \otimes H_{n}\right) \rightarrow \mathrm{S}\left(H_{m} \otimes H_{n}\right)$ be a surjective map and $\varphi\left(S_{\mathrm{sep}}\left(H_{m} \otimes H_{n}\right)=S_{\mathrm{sep}}\left(H_{m} \otimes H_{n}\right)\right)$ For some rR+\{1}, if φ satisfies Tsallis entropy $S^{r}(t \rho+(1-t) \sigma)=S^{r}(t \varphi(\rho)+(1-t) \varphi(\sigma))$ for any ρ, σS(Hm$\otimes$Hn) and for any t∈[0, 1], there exist unitary operators Um, Vn acting on Hm, Hn such that $\varphi(\rho)=\left(U_{m} \otimes V_{n}\right) \rho\left(U_{m} \otimes V_{n}\right)^{*}$ for any $\rho \in S_{\mathrm{sep}}\left(\mathrm{H}_{\mathrm{m}} \otimes \mathrm{H}_{\mathrm{n}}\right)$.
Key words: maps    Tsallis entropy    quantum states
1 问题的提出

 $H=H_{1} \otimes H_{2} \otimes \cdots \otimes H_{k}$

k=2时，系统就叫作双系统.若H1H2都是复有限维希尔伯特空间，那么对于ρS(H1$\otimes$H2)，如果$\rho = \sum\limits_{i = 1}^n {{p_i}} {\rho _i} \otimes {\sigma _i}$，其中ρiS(H1)、σiS(H2)、0≤pi≤1、$\sum\limits_{i = 1}^n {{p_i}} = 1$.则这时量子态ρ是可分的，否则就说量子态ρ是纠缠的.Ssep(H1$\otimes$H2)和Pursep(H1$\otimes$H2)分别表示双系统中所有可分量子态和可分量子纯态.一个积态ω$\otimes$τ是纯态当且仅当ωτ都是纯态.用符号A*表示矩阵A的共轭转置.

 $S^{r}(\rho)=\frac{{Tr}\left(\rho^{r}\right)-1}{1-r}$

2 定理的证明

 $\varphi : S\left(H_{m} \otimes H_{n}\right) \rightarrow S\left(H_{m} \otimes H_{n}\right)$

 $\varphi\left(S_{\mathrm{sep}}\left(H_{m} \otimes H_{n}\right)\right)=S_{\mathrm{sep}}\left(H_{m} \otimes H_{n}\right)$

 $S^{r}\left(t_{\rho}+(1-t)_{\sigma}\right)=S^{r}(t \varphi(\rho)+(1-t) \varphi(\sigma))$

 $S^{r}\left(t_{\rho}+(1-t) \sigma\right)=S^{r}(t \varphi(\rho)+(1-t) \varphi(\sigma))$

 $\varphi : S_{\mathrm{sep}}\left(H_{m} \otimes H_{n}\right) \rightarrow S_{\mathrm{sep}}\left(H_{m} \otimes H_{n}\right)$

(1) 存在可逆算子SB(Hm)、TB(Hn)，使得

 $\varphi(\rho)=\frac{(S \otimes T) \psi(\rho)(S \otimes T)^{*}}{{Tr}\left((S \otimes T) \psi(\rho)(S \otimes T)^{*}\right)}$

(2) 存在可逆算子SB(Hm, Hn)、TB(Hn, Hm)，使得

 $\varphi(\rho)=\frac{(S \otimes T) \psi(\theta(\rho))(S \otimes T)^{*}}{{Tr}\left((S \otimes T) \psi(\theta(\rho))(S \otimes T)^{*}\right)}$

 $I_{m n}=(A \otimes B)(A \otimes B)^{*}=\left(A A^{*}\right) \otimes\left(B B^{*}\right)$

AA*=(dij)，则Imn=(dijBB*).从而必须所有的dij(ij)都为零，且所有的dii(i=1, 2, …, m)都相等且非零，记其为α，于是αBB*=InAA*=αIm.类似地，考虑

 $I_{m n}=(A \otimes B)^{*}(A \otimes B)=\left(A^{*} A\right) \otimes\left(B^{*} B\right)$

 $\alpha=\left|a_{11}\right|^{2}+\left|a_{12}\right|^{2}+\cdots+\left|a_{1 n}\right|^{2}>0$

 $\varphi\left(S_{\mathrm{sep}}\left(H_{m} \otimes H_{n}\right)\right)=S_{\mathrm{sep}}\left(H_{m} \otimes H_{n}\right)$

 $\varphi(\rho)=V_{\rho} V^{*}=\frac{(S \otimes T) \psi(\rho)(S \otimes T)^{*}}{{Tr}\left((S \otimes T) \psi(\rho)(S \otimes T)^{*}\right)}$

 $\rho=\left(\frac{1}{m} I_{m}\right) \otimes\left(\frac{1}{n} I_{n}\right)=\frac{1}{m m} I_{m n}$

 $\frac{1}{m m} I_{m n}=\frac{(S \otimes T)(S \otimes T)^{*}}{{Tr}\left((S \otimes T)(S \otimes T)^{*}\right)}$

$\frac{1}{m n} {Tr}\left((S \otimes T)(S \otimes T)^{*}\right)=k$，则k>0(因为ST都是可逆的).记$B=\frac{1}{\sqrt{k}} S \otimes T$，则BB*=I，因此B是酉矩阵.由引理3知，存在λ>0，使得$\frac{\lambda}{\sqrt{k}} S$$\frac{1}{\lambda} T$同时为酉矩阵，分别记其为UV.则

 \begin{aligned} \varphi(\rho)=& \frac{(\sqrt{k} B) \rho\left(\sqrt{k} B^{*}\right)}{{Tr}\left(\sqrt{k} B_{\rho} \sqrt{k} B^{*}\right)}=\frac{B_{\rho} B^{*}}{{Tr}\left(B \rho B^{*}\right)}=\\ & B_{\rho} B^{*}=(U \otimes V)_{\rho}(U \otimes V)^{*} \end{aligned}

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