在量子力学中, 算子代数上的线性映射称为超算符^[1], 在量子信息中扮演着重要的角色^[2].比如, 量子系统中的离散物理变换, 包括量子计算和量子噪声, 均是由完全正的保迹映射(称为量子运算)表示.另一个例子是部分转置, 虽然部分转置是正映射，但不是完全正映射, 它是纠缠理论中判断量子态是否纠缠的重要内容^[3-6].

众所周知, 一个量子系统是由一个Hilbert空间描述, 量子系统中的线性算子都是有界算子^[7].设${\Bbb X}$和${\Bbb Y}$为两个有限维的Hilbert空间, $B({\Bbb X}, {\Bbb Y})$为从${\Bbb X}$到${\Bbb Y}$上的所有有界线性算子之集, 特别地, 记$B({\Bbb X})=B({\Bbb X}, {\Bbb X})$.定义$B({\Bbb X}, {\Bbb Y})$上的内积为

$ \langle A|B\rangle={\rm Tr}(A^{\dagger}B), \forall A, B\in B({\Bbb X}, {\Bbb Y}), $

则$B({\Bbb X}, {\Bbb Y})$为一个复内积空间.线性映射$\Phi: B({\Bbb X})\rightarrow B({\Bbb Y})$称为${\Bbb X}$到${\Bbb Y}$上的一个超线性算符, 简称超算符.记$B(B({\Bbb X}), B({\Bbb Y}))$为从$B({\Bbb X})$到$B({\Bbb Y})$上的所有超算符之集.

在$B(B({\Bbb X}), B({\Bbb Y}))$上定义映射的加法和数乘:

(ⅰ) 加法: 设$\Phi_{1}, \Phi_{2}\in B(B({\Bbb X}), B({\Bbb Y}))$, 则$\Phi_{1}+\Phi_{2}$定义为

$ (\Phi_{1}+\Phi_{2})(A)=\Phi_{1}(A)+\Phi_{2}(A), \forall A\in B({\Bbb X}). $

(ⅱ) 数乘: 设$\Phi\in B(B({\Bbb X}), B({\Bbb Y}))$, $\alpha \in {\Bbb C}$, 则$\alpha\Phi$定义为

$ (\alpha\Phi)(A)=\alpha(\Phi(A)), \forall A\in B({\Bbb X}). $

则$B(B({\Bbb X}), B({\Bbb Y}))$构成了复数域${\Bbb C}$上的一个线性空间.

设超算符$\Phi\in B(B({\Bbb X}), B({\Bbb Y}))$, 对于任意$A_{1}\in B({\Bbb Y}), A_{2}\in B({\Bbb X})$, 定义算子$\Phi^{\dagger}$为

$ \langle\Phi^{\dagger}(A_{1})|A_{2}\rangle=\langle A_{1}|\Phi(A_{2})\rangle, $

则$\Phi^{\dagger}$存在且唯一, 并且$\Phi^{\dagger}\in B(B({\Bbb Y}), B({\Bbb X}))$, 称$\Phi^{\dagger}$为超线性算符$\Phi$的伴随.

在复合量子系统中, 类似算子张量积^[2]的定义, 也可以定义超算符的张量积.任给一组超算符$\Phi_{i}:B({\Bbb X}_{i})\rightarrow B({\Bbb Y}_{i}), i=1, 2, \cdots, n$, 则他们的张量积

$ \Phi_{1}\otimes\cdots\otimes\Phi_{n}:B({\Bbb X}_{1}\otimes\cdots\otimes{\Bbb X}_{n})\rightarrow B({\Bbb Y}_{1}\otimes\cdots\otimes{\Bbb Y}_{n}) $

定义为

$ (\Phi_{1}\otimes\cdots\otimes\Phi_{n})(A_{1}\otimes\cdots\otimes A_{n})=\Phi_{1}(A_{1})\otimes\cdots\otimes\Phi_{n}(A_{n}), \forall A_{i}\in B({\Bbb X}_{i}), i=1, 2, \cdots, n. $

在量子计算和量子信息中, 量子运算有一个强大的数学表示—算子和表示, 也称为Kraus表示, 在计算和理论中都有重要的作用^[9].虽然每一个量子运算都有Kraus表示, 但表示是不唯一的.事实上, 任何超算符都有Choi-Jamiolkowski表示, 并且文献[10-11]证明了Choi-Jamiolkowski表示的一些等价性质.另外, 对于每一个超算符而言, 还存在其他的表示形式, 具体参见文献[12-13].

本文结构安排如下:首先, 回顾超算符的自然表示和Choi-Jamiolkowski表示的相关概念和讨论了其相关性质; 其次, 建立了同一个超算符的这两种表示之间的相互转化关系, 并给出了一些重要例子.

设$\{|i\rangle\}\subset{\Bbb X}$和$\{|m\rangle\}\subset{\Bbb Y}$分别为${\Bbb X}$和${\Bbb Y}$上的正规正交基, $\forall A\in B({\Bbb X}, {\Bbb Y})$, 有

$ \begin{eqnarray} A=\sum\limits_{i, m} \langle m|A|i\rangle|m\rangle\langle i|, \end{eqnarray} $

(2.1)

相应地, 则$A$的复共轭, 转置和伴随分别表示为

$ \overline{A}=\sum\limits_{i, m} \overline{\langle m|A|i\rangle}|m\rangle\langle i|, \ \ A^{T}=\sum\limits_{i, m} \langle m|A|i\rangle|i\rangle\langle m|, \ \ A^{\dagger}=\sum\limits_{i, m} \overline{\langle m|A|i\rangle}|i\rangle\langle m|. $

空间$B({\Bbb X}, {\Bbb Y})$与${\Bbb Y}\otimes {\Bbb X}$之间存在一个对应关系, 且由映射${\rm vec}: B({\Bbb X}, {\Bbb Y})\rightarrow {\Bbb Y}\otimes {\Bbb X}$, 即

$ \begin{eqnarray} {\rm vec}(A)=\sum\limits_{i, m} \langle m|A|i\rangle|m\rangle|i\rangle \end{eqnarray} $

(2.2)

给出^[14].

注2.1  vec是一个线性双射且在内积意义下是一个同构映射, 即

$ \langle A_{1}|A_{2}\rangle=\langle{\rm vec}(A_{1})|{\rm vec}(A_{2})\rangle, \forall A_1, A_2\in B({\Bbb X}, {\Bbb Y}). $

并且(2.2) 式等价于

$ {\rm vec}(A)=(A\otimes I_{{\Bbb X}}){\rm vec}(I_{{\Bbb X}})=(I_{{\Bbb Y}}\otimes A^{T}){\rm vec}(I_{{\Bbb Y}}), $

其中$I_{{\Bbb H}}$是$B({\Bbb H})$上的恒等算子, ${\rm vec}(I_{{\Bbb H}})\in {\Bbb H}\otimes {\Bbb H}$是最大纠缠态, 即${\rm vec}(I_{{\Bbb H}})=\sum_{k}|k\rangle|k\rangle, \{|k\rangle\}$为${\Bbb H}$上的正规正交基, $A^{T}\in B({\Bbb Y}, {\Bbb X})$为$A$的转置.进一步, 容易验证映射${\rm vec}$满足下面三个性质:

(ⅰ) $(A_{1}\otimes A_{2}){\rm vec}(X)={\rm vec}(A_{1}XA_{2}^{T}), $$\forall A_{1}\in B({\Bbb X}_{1}, {\Bbb Y}_{1}), A_{2}\in B({\Bbb X}_2, {\Bbb Y}_2), $$X\in B({\Bbb X}_2, {\Bbb X}_1)$;

(ⅱ) ${\rm Tr}_{{\Bbb X}}({\rm vec}(A_{1}){\rm vec}(A_{2}^{\dagger}))=A_{1}A_{2}^{\dagger}, $$\rm{T}{{\rm{r}}_{\mathbb{Y}}}(\rm{vec}({{\mathit{A}}_{1}})\rm{vec}(\mathit{A}_{2}^{\dagger }))={{(\mathit{A}_{2}^{\dagger }{{\mathit{A}}_{1}})}^{\mathit{T}}}, $$\forall A_{1}, A_{2} \in B({\Bbb X, \Bbb Y})$;

(ⅲ) ${\rm vec}(uv^{\dagger})=u\otimes \overline{v}$, $\forall u\in{\Bbb X}$, $\forall v\in{\Bbb Y}$.特别地, ${\rm vec}(u)=u$, ${\rm vec}(v^{\dagger})=\overline{v}$.

设$\{|i\rangle\}\subset{\Bbb X}$和$\{|m\rangle\}\subset{\Bbb Y}$分别为${\Bbb X}$和${\Bbb Y}$上的正规正交基, $A\in B({\Bbb X})$且$A=\sum_{i, j} \langle i|A|j\rangle|i\rangle\langle j|$, 超算符$\Phi: B({\Bbb X})\rightarrow B({\Bbb Y})$作用在$A$上为$\Phi(A)=\sum_{i, j} \langle i|A|j\rangle\Phi(|i\rangle\langle j|).$则存在线性映射$N_{\Phi}\in B({\Bbb X}\otimes{\Bbb X}, {\Bbb Y}\otimes{\Bbb Y})$满足

$ N_{\Phi}{\rm vec}(A)={\rm vec}(\Phi(A)), $

这时称$N_{\Phi}$为$\Phi$的自然表示^[14].由于

$ \begin{eqnarray*} {\rm vec}(\Phi(A))&=&\sum\limits_{i, j} \langle i|A|j\rangle{\rm vec}\left(\Phi(|i\rangle\langle j|)\right)\\ &=&\sum\limits_{i, j} \langle i|A|j\rangle{\rm vec}\Big(\sum\limits_{m, n}\langle m |\Phi(|i\rangle\langle j|)|n\rangle|m\rangle\langle n|\Big)\\ &=&\sum\limits_{i, j, m, n} \langle i|A|j\rangle\langle m |\Phi(|i\rangle\langle j|)|n\rangle|m\rangle|n\rangle, \end{eqnarray*} $

可知

$ \begin{eqnarray} N_{\Phi}=\sum\limits_{i, j, m, n} \langle m |\Phi(|i\rangle\langle j|)|n\rangle|m\rangle|n\rangle\langle i|\langle j| =\sum\limits_{i, j, m, n} \langle m |\Phi(|i\rangle\langle j|)|n\rangle|m\rangle\langle i|\otimes|n\rangle\langle j|. \end{eqnarray} $

(2.3)

任给一个超算符, 不仅有自然表示, 还存在另一种重要表示.设$\{|i\rangle\}\subset{\Bbb X}$为${\Bbb X}$上的一组正规正交基, $\Phi\in B(B({\Bbb X}), B({\Bbb Y}))$, 定义映射$J_{\Phi}:B(B({\Bbb X}), B({\Bbb Y}))\rightarrow B({\Bbb Y}\otimes{\Bbb X})$为

$ \begin{eqnarray} J_{\Phi}=\sum\limits_{i, j}\Phi(|i\rangle\langle j|)\otimes |i\rangle\langle j|, \end{eqnarray} $

(2.4)

则称映射$J_{\Phi}$为$\Phi$的Choi-Jamiolkowski表示^[14].

注2.2  (2.3) 式也可以写成

$ \begin{eqnarray*} J_{\Phi}=(\Phi\otimes I_{B({\Bbb X})})({\rm vec}(I_{{\Bbb X}}){\rm vec}(I_{{\Bbb X}})^{\dagger}). \end{eqnarray*} $

Choi-Jamiolkowski表示有个重要的特点是映射$\Phi$可以由算子$J_{\Phi}$通过等式

$ \Phi(A)={\rm tr}_{{\Bbb X}}(J_{\Phi}\cdot(I_{{\Bbb Y}}\otimes A^{_{T}})) $

来恢复.超算符$\Phi$和其Choi-Jamiolkowski表示之间建立的双射是一个同构, 并且满足下面的性质:

(ⅰ) $\Phi$是保迹映射当且仅当${\rm tr}_{{\Bbb Y}}(J_{\Phi})=I_{{\Bbb X}}$;

(ⅱ) $\Phi$是保自伴映射当且仅当$J_{\Phi}$是自伴的;

(ⅲ) $\Phi$是完全正的当且仅当$J_{\Phi}$是半正定的.

设$\Phi\in B(B({\Bbb X}))$是一个超算符, 则$N_{\Phi}\in B({\Bbb X}\otimes{\Bbb X}), J_{\Phi}\in B({\Bbb X}\otimes{\Bbb X})$.一般地, $N_{\Phi}\neq J_{\Phi}$, 但它们之间存在一个重要的关系.

首先, 定义一个超算符$T:B({\Bbb X}\otimes{\Bbb X})\rightarrow B({\Bbb X}\otimes{\Bbb X})$为

$ \begin{eqnarray} T(S)=\sum\limits_{i, j}(I_{{\Bbb X}}\otimes E_{ij})S(E_{ij}\otimes I_{{\Bbb X}}), \forall S \in B({\Bbb X}\otimes{\Bbb X}), \end{eqnarray} $

(3.1)

其中$E_{ij}=|i\rangle\langle j|, \{|i\rangle\}\subset{\Bbb X}$为${\Bbb X}$上的一组正规正交基.设$A, B, C, D\in B({\Bbb X}\otimes{\Bbb X}), $则

$ T\left(\left[ \begin{array}{cc} A ~& B \\ C ~& D \\ \end{array} \right]\right)=\left(\begin{array}{cc} {\rm vec}(A)^T \\ {\rm vec}(B)^T \\ {\rm vec}(C)^T \\ {\rm vec}(D)^T \\ \end{array} \right). $

不难计算，超算符$T$满足下面的结论:

(ⅰ) $T$是一个线性双射, 即

$ T(\alpha S_1+\beta S_2)=\alpha T(S_1)+\beta T(S_2), \forall \alpha, \beta \in {\rm C}, \forall S_1, S_2 \in B({\Bbb X}\otimes{\Bbb X}). $

(ⅱ) $T^{2}=I$且$T^{-1}=T$, 因为$T^{2}(S)=\sum_{i}(I_{{\Bbb X}}\otimes E_{ii})S(E_{ii}\otimes I_{{\Bbb X}})=S$.

(ⅲ) ${\rm vec}(T(X\otimes Y))={\rm vec}(X)\otimes{\rm vec}(Y), $$ \forall X, Y\in B({\Bbb X})$, 因为

$ \begin{eqnarray*} T(X\otimes Y)&=&\sum\limits_{i, j}(I_{{\Bbb X}}\otimes E_{ij})(X\otimes Y)(E_{ij}\otimes I_{{\Bbb X}})\\ &=&\sum\limits_{i, j}XE_{ij}\otimes E_{ij}Y\\ &=&\sum\limits_{i, j}(X|i\rangle\otimes|i\rangle)(|j\rangle\otimes Y^{T}|j\rangle)^{T}\\ &=&{\rm vec} \bigg(\sum\limits_{i}X|i\rangle\langle i|\bigg)\bigg({\rm vec} \bigg(\sum\limits_{j}|j\rangle\langle j|Y\bigg)\bigg)^{T}\\ &=&{\rm vec}(X)\big({\rm vec}(Y)\big)^{T}. \end{eqnarray*} $

(ⅳ)令$K=\sum_{i, j}E_{ij}\otimes E_{ji}$, 则$\forall X, Y\in B({\Bbb X})$,

$ \ \ \ \ \ X=\sum\limits_{i, j}\langle i|X|j\rangle|i\rangle\langle j|=\sum\limits_{i, j}|i\rangle\langle i|X|j\rangle\langle j|=\sum\limits_{i, j}E_{ii}XE_{jj}, \\ \ \ \ \ \ X^{T}=\sum\limits_{i, j}\langle j|X|i\rangle|i\rangle\langle j|=\sum\limits_{i, j}|i\rangle\langle j|X|i\rangle\langle j|=\sum\limits_{i, j}E_{ij}XE_{ij}, \\ K{\rm vec}(X)=\sum\limits_{i, j}E_{ij}\otimes E_{ji}{\rm vec}(X)=\sum\limits_{i, j}{\rm vec}(E_{ij}X E_{ij}) ={\rm vec}(X^{T}), \\ \ \ \ K(X\otimes Y)K=\bigg(\sum\limits_{i, j}E_{ij}\otimes E_{ji}\bigg)(X\otimes Y) \bigg(\sum\limits_{m, n}E_{mn}\otimes E_{nm}\bigg)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{i, j, m, n}E_{ij}XE_{mn}\otimes E_{ji}YE_{nm}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{i, n}\langle i|Y|n\rangle E_{in}\otimes \sum\limits_{j, m}\langle j|X|m\rangle E_{jm}=Y\otimes X, \\ (I\otimes K \otimes I){\rm vec}(X\otimes Y)= \sum\limits_{i, j}{\rm vec}\left[(I\otimes E_{ij})(X\otimes Y)(E_{ij}\otimes I)\right]\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\rm vec}\bigg(\sum\limits_{i, j}XE_{ij}\otimes E_{ij}Y\bigg)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\rm vec}(T(X\otimes Y)). $

定理3.1  设$N_{\Phi}$与$J_{\Phi}$分别为$\Phi\in B(B({\Bbb X}))$的自然表示和${\rm Choi}$-${\rm Jamiolkowski}$表示, 则

$ \begin{eqnarray} N_{\Phi}=T(J_{\Phi}). \end{eqnarray} $

(3.2)

证  根据$N_{\Phi}$与$J_{\Phi}$的形式如等式(2.3) 和(2.4) 所示, 计算(3.2) 式的右边, 得

$ \begin{eqnarray*} T(J_{\Phi})&=&\sum\limits_{k, l}(I_{{\Bbb X}}\otimes E_{kl}) \bigg(\sum\limits_{ij}\Phi(E_{ij})\otimes E_{ij}\bigg)(E_{kl}\otimes I_{{\Bbb X}})\\ &=&\sum\limits_{i, j, k}(\Phi(E_{ij})E_{ki}\otimes E_{kj})\\ &=&\sum\limits_{i, j, k}\bigg(\sum\limits_{m, n}\langle E_{mn}, \Phi(E_{ij})\rangle E_{mn}\bigg)E_{ki}\otimes E_{kj}\\ &=&\sum\limits_{i, j, k, m}\langle E_{mk}, \Phi(E_{ij})\rangle E_{mi}\otimes E_{kj}\\ &=&N_{\Phi}. \end{eqnarray*} $

因而, 等式(3.2) 成立.

注3.1  一般情况下, 设$\{|i\rangle\}\subset{\Bbb X}$和$\{|m\rangle\}\subset{\Bbb Y}$分别为${\Bbb X}$和${\Bbb Y}$上的正规正交基.定义线性映射

$ \widetilde{T}:B({\Bbb Y})\otimes B({\Bbb X})\rightarrow B({\Bbb X}, {\Bbb Y})\otimes B({\Bbb X}, {\Bbb Y}), $

作用在基底上为

$ \widetilde{T}(|m\rangle\langle n|\otimes|i\rangle\langle j|)=|m\rangle\langle i|\otimes|n\rangle\langle j|, $

那么$\forall \Phi\in B(B({\Bbb X}), B({\Bbb Y}))$, 有

$ \begin{eqnarray*} \widetilde{T}(J_{\Phi})&=&\widetilde{T} \bigg(\sum\limits_{i, j}\Phi(|i\rangle\langle j|)\otimes |i\rangle\langle j|\bigg)\\ &=&\widetilde{T}\bigg(\sum\limits_{i, j, m, n}\langle m|\Phi(|i\rangle\langle j|)|n\rangle|m\rangle\langle n|\otimes |i\rangle\langle j|\bigg)\\ &=&\sum\limits_{i, j, m, n}\langle m|\Phi(|i\rangle\langle j|)|n\rangle\widetilde{T}(|m\rangle\langle n|\otimes |i\rangle\langle j|)\\ &=&\sum\limits_{i, j, m, n}\langle m|\Phi(|i\rangle\langle j|)|n\rangle|m\rangle\langle i|\otimes |n\rangle\langle j|\\ &=&N_{\Phi}. \end{eqnarray*} $

例3.1  设$\Phi_H$为一个乘法算子, $\forall \rho \in D({\Bbb C}^2)$, $\Phi_H(\rho)=H\rho$, 其中$H\in B({\Bbb C}^2)$, 则

$ {\rm vec}(\Phi_H(\rho))={\rm vec}(H\rho)=(H\otimes I){\rm vec}(\rho)=N_{\Phi_H}{\rm vec}(\rho). $

证  令$H=\left[\begin{array}{cc} h_1~&h_2 \\ h_3 ~& h_4 \\ \end{array} \right]$.因为

$ \begin{eqnarray*} R_{\Phi_H}=H\otimes I=\left[ \begin{array}{cccc} h_1&0&h_2&~0 \\ 0&~h_1~&0&~h_2 \\ h_3&0&h_4&~0 \\ 0&h_3&0&~h_4 \\ \end{array} \right] \end{eqnarray*} $

和

$ \begin{eqnarray*} J_{\Phi_H}&=&(\Phi_H\otimes I)(|00\rangle+|00\rangle)(\langle00|+\langle11|)\\ &=&H|0\rangle\langle0|\otimes|0\rangle\langle0|+H|0\rangle\langle1|\otimes|0\rangle\langle1| +H|1\rangle\langle0|\otimes|1\rangle\langle0| +H|1\rangle\langle1|\otimes|1\rangle\langle1|\\ &=&\left[\begin{array}{cccc} h_1&~~0~~&0 &~~ h_1 \\ h_2&0&0&~~h_2 \\ h_3&0&0&~~h_3 \\ h_4&0&0&~~h_4 \\ \end{array} \right]. \end{eqnarray*} $

所以$T(J_{\Phi_H})=R_{\Phi_H}$.

例3.2  设$\Phi$为一个酉量子信道, $\Phi(\rho)=U\rho U^{\dagger}, \forall \rho \in D({\Bbb C}^2), $其中$U$为酉算子, 则

$ N_{\Phi}=U\otimes \overline{U}, \ \ J_{\Phi}={\rm vec}(U){\rm vec}(U)^{\dagger}. $

证  令$U=\left[\begin{array}{cc} u_{11} ~& u_{12} \\ u_{21} ~& u_{22} \\ \end{array} \right]$.因为

$ U\otimes \overline{U}=\left[ \begin{array}{cc} u_{11} ~& u_{12} \\ u_{21} ~& u_{22} \\ \end{array} \right]\otimes \left[ \begin{array}{cc} \bar{u}_{11} ~& \bar{u}_{12} \\ \bar{u}_{21} ~& \bar{u}_{22} \\ \end{array} \right]=\left[ \begin{array}{cccc} |u_{11}|^{2}&~u_{11}\bar{u}_{12}~&u_{12}\bar{u}_{11} &~ |u_{12}|^{2} \\ u_{11}\bar{u}_{21}&u_{11}\bar{u}_{22}&u_{12}\bar{u}_{21} &~ u_{12}\bar{u}_{22} \\ u_{21}\bar{u}_{11}&u_{21}\bar{u}_{12}&u_{22}\bar{u}_{11}&~u_{22}\bar{u}_{12} \\ |u_{21}|^{2}&u_{21}\bar{u}_{22}&u_{22}\bar{u}_{21} &~ |u_{22}|^{2} \end{array} \right] $

和

$ \begin{eqnarray*} {\rm vec}(U){\rm vec}(U)^{\dagger}&=&\left( \begin{array}{c} u_{11} \\ u_{12} \\ u_{21} \\ u_{22} \\ \end{array} \right)\left( \begin{array}{cccc} \overline{u_{11}}& ~~\overline{u_{12}} ~~& \overline{u_{21}}&~~\overline{u_{22}} \\ \end{array} \right)\\ &=&\left[\begin{array}{cccc} |u_{11}|^{2}&~u_{11}\bar{u}_{12} ~& u_{11}\bar{u}_{21} &~ u_{11}\bar{u}_{22} \\ u_{12}\bar{u}_{11}&|u_{12}|^{2}&u_{12}\bar{u}_{21}&~u_{12}\bar{u}_{22} \\ u_{21}\bar{u}_{11}&u_{21}\bar{u}_{12} &|u_{21}|^{2}&~u_{21}\bar{u}_{22} \\ u_{22}\bar{u}_{11}&u_{22}\bar{u}_{12}&u_{22}\bar{u}_{21} &~ |u_{22}|^{2} \end{array} \right]. \end{eqnarray*} $

所以

$ T({\rm vec}(U){\rm vec}(U)^{\dagger})=U\otimes \overline{U}. $

证毕.

注3.2  设$\Phi$为${\rm 1}$比特量子系统中的酉量子信道, 则$N_{\Phi}\neq J_{\Phi}$.

证  假设存在一个酉量子信道$\Phi$满足$N_{\Phi}= J_{\Phi}$, 则$U\otimes \overline{U}={\rm vec}(U){\rm vec}(U)^{\dagger}$.令

$ U=\left[ \begin{array}{cc} u_{11} ~& u_{12} \\ u_{21} ~& u_{22} \\ \end{array} \right], $

则有

$ \left\{ \begin{array}{ll} u_{11}\bar{u}_{21}=u_{12}\bar{u}_{11}; \\ u_{11}\bar{u}_{22}=|u_{12}|^{2}; \\ u_{22}\bar{u}_{11}=|u_{21}|^{2}; \\ u_{21}\bar{u}_{22}=u_{22}\bar{u}_{12}. \\ \end{array} \right. $

由酉算子的性质知

$ \left\{ \begin{array}{ll} |u_{11}|^{2}+|u_{12}|^{2}=1, |u_{21}|^{2}+|u_{22}|^{2}=1, u_{11}\bar{u}_{21}=-u_{12}\bar{u}_{22}; \\ |u_{11}|^{2}+|u_{21}|^{2}=1, |u_{12}|^{2}+|u_{22}|^{2}=1, u_{11}\bar{u}_{12}=-u_{21}\bar{u}_{22}. \\ \end{array} \right. $

这个方程组无解, 那么$U\otimes \overline{U}\neq{\rm vec}(U){\rm vec}(U)^{\dagger}$, 因而, $N_{\Phi}\neq J_{\Phi}$.

例3.3  设超算符$\Phi$是一个量子运算, 其Kraus表示为

$ \begin{eqnarray} \Phi(X)=\sum\limits_{j=1}^{m}{E_j X E_j^\dagger}, \forall X\in B({\Bbb X}), \end{eqnarray} $

(3.3)

其中$E_j\in B({\Bbb X}, {\Bbb Y})$满足$\sum_{j=1}^{m}{E_j^\dagger E_j}=I_{{\Bbb X}}$, 且称为$\Phi$的Kraus算子.

进一步, 形如等式(3.3) 的量子运算$\Phi$, 其自然表示和Choi-Jamiolkowski表示分别为

$\begin{eqnarray*} N_{\Phi}=\sum\limits_{j=1}^{m}{E_j\otimes(\overline{E_j})}, \ \ J_{\Phi}=\sum\limits_{j=1}^{m}{{\rm vec}(E_j){\rm vec}(E_j)^{\dagger}}.\end{eqnarray*} $

根据算子$T$的性质(ⅱ)和(ⅲ), 有

$\begin{eqnarray*} T(J_{\Phi})=\sum\limits_{j=1}^{m}{T({\rm vec}(E_j){\rm vec}(E_j)^{\dagger})} =\sum\limits_{j=1}^{m}{T^{-1}({\rm vec}(E_j){\rm vec}(E_j)^{\dagger})} =\sum\limits_{j=1}^{m}{E_j\otimes \overline{E_j}}=N_{\Phi}.\end{eqnarray*} $