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N量子比特系统的纠缠判据

谭维翰 ,  赵超樱 ,  郭奇志

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N量子比特系统的纠缠判据

谭维翰, 赵超樱, 郭奇志
物理学报, 2023, 72(1): 010301.
doi: 10.7498/aps.72.20221524

Entanglement criterion of N qubit system

Tan Wei-Han, Zhao Chao-Ying, Guo Qi-Zhi
Acta Phys. Sin., 2023, 72(1): 010301.
doi: 10.7498/aps.72.20221524
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  • 收稿日期:  2022-07-27
  • 修回日期:  2022-09-14
  • 上网日期:  2022-12-29
  • 刊出日期:  2023-01-05
物理学报. 2023, 72(1): 010301.
doi: 10.7498/aps.72.20221524.

N量子比特系统的纠缠判据

  • 1. 上海大学物理系, 上海　200444
  • 2. 杭州电子科技大学物理系, 杭州　310018
  • 3. 山西大学光电研究所, 量子光学与光量子器件国家重点实验室, 太原　030006
    • 通信作者: 赵超樱, zchy49@163.com
    基金项目: 教育部量子光学重点实验室(批准号: KF202004, KF202205)资助的课题.
  • 收稿日期:  2022-07-27
  • 修回日期:  2022-09-14
  • 上网日期:  2022-12-29
  • 刊出日期:  2023-01-05

摘要: 在前文(2019 Int. J. Mod. Phys. B 33 1950197; 2020 Int. J. Mod. Phys. B 34 2050022)中, 我们提出了一种判断2量子比特系统纠缠的方法, 2量子比特系统可分的充分必要条件是相关系数为正且主密度矩阵可分, 否则系统纠缠. 在本文中, 通过数值计算与讨论, 先将2量子比特系统纠缠判据的方法推广到3量子比特系统中去; 接着, 继续将3量子比特系统推广到N量子比特系统中去. 这是一个复杂而有趣的问题.

English Abstract

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    1.   引　言

    • 自从上世纪三十年代Schrödinger提出量子纠缠态的概念以来, 有关量子纠缠态的研究就从未间断过. 量子纠缠态在量子密钥分配、量子密集编码、量子隐形传态、量子纠错码和量子计算等诸多领域有着重要的应用. 但是, 迄今为止, 仍有许多理论问题没有得到解决, 这些都制约着实验工作的进展. 量子纠缠理论的发展将为量子信息技术打开广阔的应用前景, 其研究是当前国际上量子信息论和量子光学等学科关注的前沿课题之一.

      近年来, 纠缠态在量子计算、量子纠错等方面具有很大的应用前景[1-3], 但是关于纠缠本质等的基本问题还没有解决, 一个非常引人注目的量子现象是复合量子系统的纠缠[4-6]. Peres的PPT判据和Horodecki的约化判据. Peres和Horodecki等最早提出经PPT后系统特征值是否全为正来判别该系统为可分或纠缠. Peres判据, 从物理本质上来看就是对两体中的任一单体做部分时间反演操作. 对于2×2维和2×3维系统, 它是充要条件; 对于其他情况, 它是必要非充分条件. Horodecki等采用约化判据对2×4维和3×3维系统做了详细的研究. 最近又有关于$ 2 \times N $维和$ N \times N $维系统的研究报道, 计算相当繁琐. Werner[7]基于对局部隐变量模型的分析和翻转算子均值的正定性给出了一个判据. Horodecki等[8,9]得到了一个关于所谓熵的不等式形式的判据. Peres[10]判据基于部分转置后密度算符的正定性. 由两个子系统组成的量子系统是可分的, 当且仅当密度矩阵可以写成两个子系统的密度矩阵, 其中, 权重满足为正条件时, 系统是可分的, 否则系统是纠缠的. Peres-Horodecki准则是充要条件[11-16].

      由于非对角矩阵元的存在, 求密度矩阵$ {\boldsymbol{\rho}} $的解析解比获得$ {\boldsymbol{\rho}} $可分的充要条件要困难得多. 这是一个更为复杂和有趣的基础物理问题. 2019年, 我们课题组采用标准矩阵法研究了任意2×2复合系统密度矩阵$ {\boldsymbol{\rho}} $的解析解. 该解可以表示成主密度矩阵和可分离密度矩阵${{\boldsymbol{\rho}} _1}\text{—}{{\boldsymbol{\rho}} _4}$的和, 如果这些几率是全正的, 则密度矩阵$ {\boldsymbol{\rho}} $可分, 并且分离后的密度矩阵也一并求出来了. 如果这些几率不是全正的, 则复合系统就是纠缠的. 可以很容易就验证出几个已知2×2系统的不可分判据与PPT判据一致[17]. 2020年, 我们研究任意3×3复合系统密度矩阵$ {\boldsymbol{\rho}} $的解析解. 将密度矩阵$ {\boldsymbol{\rho}} $分解为主密度矩阵$ {{\boldsymbol{\rho}} _{\rm{p}}} $和六个约化密度矩阵$ {{\boldsymbol{\rho}} _{\text{1}}}\text{—}{{\boldsymbol{\rho}} _{\text{6}}} $的总和, 给出判断两个三量子态(0, 1, 2)系统纠缠的充要条件, 并得出几率是全正的解[18]. 在这篇文章中, 只讨论二量子态(0, 1), 即通常说的量子比特. 我们试图将前文[17]用方法推广到N量子比特系统. 第一步是3量子比特系统; 第二步到N量子比特系统. 本文的安排是: 第2节对2量子比特系统纠缠判据的简介. 第3节将上述解决方案推广到N量子比特系统. 第4节是关于2量子比特和3量子比特系统的数值计算.

    2.   量子比特系统纠缠态解析解简介

    • 对于一个单量子比特系统, 上能级$ \left| 1 \right\rangle $, 下能级$ \left| 0 \right\rangle $. 2量子比特的基向量$ \left| {00} \right\rangle $, $ \left| {01} \right\rangle $, $ \left| {10} \right\rangle $, $ \left| {11} \right\rangle $是所有纠缠态的基. 一般的, 假设密度矩阵元素是实数, 2量子比特系统的密度矩阵$ {\rho _{\rm II}} $具有如下的形式:

      $\begin{split} &{{\boldsymbol{\rho}} _{{\rm{II}}}} = \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 00}&{01}&{10}&{11} \end{array}} \\ {\left( {\begin{array}{*{20}{c}} {{a_0}}&\alpha &\beta &\gamma \\ \alpha &{{a_1}}&\delta &{ \lambda } \\ \beta &\delta &{{a_2} }&{ \mu } \\ \gamma &\lambda &{ \mu }&{{a_3}} \end{array}} \right)} \end{array}\text{, }~~ {{\boldsymbol{\rho}} _{{\rm{II}}}} = {p_{\rm{p}}}{{\boldsymbol{\rho}} _{\rm{p}}} + {\tilde p_1}{{\boldsymbol{\rho}} _1} + {\tilde p_2}{{\boldsymbol{\rho}} _2} + {\tilde p_3}{{\boldsymbol{\rho}} _3} + {\tilde p_4}{{\boldsymbol{\rho}} _4} \text{, }\\ &{{\boldsymbol{\rho}} _{\rm{p}}} = \frac{{1 - 2({x_1} + {x_2}) - x}}{2}{{\boldsymbol{S}}_1}+ \frac{{1 - 2({x_1} + {x_2}) + x}}{2}{{\boldsymbol{S}}_2} + {x_1}{{\boldsymbol{S}}_3} + {x_2}{{\boldsymbol{T}}_3} + ({x_1} + {x_2}){{\boldsymbol{S}}_4} \text{, } \end{split}$

      其中主密度矩阵$ {{\boldsymbol{\rho}} _{\rm{p}}} $分解为5个可分密度$ {{\boldsymbol{S}}_1} $, $ {{\boldsymbol{S}}_2} $, $ {{\boldsymbol{S}}_3} $, $ {{\boldsymbol{T}}_3} $, $ {{\boldsymbol{S}}_4} $其系数为正时, $ {{\boldsymbol{\rho}} _{\rm{p}}} $是可分的, 否则$ {{\boldsymbol{\rho}} _{\rm{p}}} $是纠缠的. 当$ {{\boldsymbol{\rho}} _{\rm{p}}} - {{\boldsymbol{\rho}} _4} $是可分离密度矩阵, 而且$ {p_p} \geqslant 0 $, $ {\tilde p_1} \geqslant 0 $, $ {\tilde p_2} \geqslant 0 $, $ {\tilde p_3} \geqslant 0 $, $ {\tilde p_4} \geqslant 0 $, 则密度矩阵$ {{\boldsymbol{\rho}} _{{\rm{II}}}} $是可分离的. 否则, 它就是纠缠的. (关于分离的密度矩阵的形式见附录A).

      $ {{\boldsymbol{\rho}} _{{\rm{II}}}} $的对角矩阵元$ {a_0} - {a_3} $满足如下关系 (见附录A):

      $ \begin{split} &{p_{\rm{p}}}\frac{{1 - x}}{4} + {p_1}\frac{{1 + x}}{2} + {p_1}\frac{{1 + x}}{2} = {a_0}\text{, }\\ &{p_{\rm{p}}}\frac{{1 + x}}{4} + {p_1}\frac{{1 - x}}{2} + {p_1}\frac{{1 - x}}{2} = {a_1}\text{, }\\ &{p_{\rm{p}}}\frac{{1 + x}}{4} + {p_2}\frac{{1 - x}}{2} + {p_2}\frac{{1 - x}}{2} = {a_2}\text{, }\\ &{p_{\rm{p}}}\frac{{1 - x}}{4} + {p_2}\frac{{1 + x}}{2} + {p_2}\frac{{1 + x}}{2} = {a_3}\text{, }\\ &{{\boldsymbol{\rho}} _{{\rm{II}}}} = {p_{\rm{p}}}{{\boldsymbol{\rho}} _{\rm{p}}} + {p_1}{{\boldsymbol{\rho}} _1} + {p_2}{{\boldsymbol{\rho}} _2}. \end{split}$

      四个参数${p_{\rm{p}}}$, $x$, ${p_1}$, ${p_2}$刚好对应(2)式的四个式子.

    3.   N量子比特系统纠缠判据

      3.1.   3量子比特系统纠缠判据

    • 应用上方法研究了3量子比特的纠缠态问题. 这个方法包括写出基函数, 由此构成的可分离的密度矩阵, 进一步便是3量子系统主矩阵(详细见附录B). 先按附录A导出(2)式的办法导出下面3量子系统的相应方程(3). 这过程很复杂, 很难被推广到N量子比特系统. 但有趣且有用的是我们发现3量子系统主矩阵可表示为2量子系统的对角矩阵(见附录(B2)式). 这就为下面推广到N量子比特系统成为可能:

      $\begin{split} &{p_{\rm{p}}}\frac{{1 - x}}{8} + {p_1}\frac{{1 + x}}{4} + {p_1}\frac{{1 + x}}{4} = {a_0} \text{, } \\ &{p_{\rm{p}}}\frac{{1 + x}}{8} + {p_1}\frac{{1 - x}}{4} + {p_1}\frac{{1 - x}}{4} = {a_1}\text{, } \\ &{p_{\rm{p}}}\frac{{1 + x}}{8} + {p_2}\frac{{1 - x}}{4} + {p_2}\frac{{1 - x}}{4} = {a_2} \text{, } \\ &{p_{\rm{p}}}\frac{{1 - x}}{8} + {p_2}\frac{{1 + x}}{4} + {p_2}\frac{{1 + x}}{4} = {a_3} \text{, } \\ &{p_{\rm{p}}}\frac{{1 - y}}{8} + {p_3}\frac{{1 + y}}{4} + {p_3}\frac{{1 + y}}{4} = {a_4} \text{, } \\ &{p_{\rm{p}}}\frac{{1 + y}}{8} + {p_3}\frac{{1 - y}}{4} + {p_3}\frac{{1 - y}}{4} = {a_5} \text{, } \\ & {p_{\rm{p}}}\frac{{1 + y}}{8} + {p_4}\frac{{1 - y}}{4} + {p_4}\frac{{1 - y}}{4} = {a_6} \text{, }\\ &{p_{\rm{p}}}\frac{{1 - y}}{8} + {p_4}\frac{{1 + y}}{4} + {p_4}\frac{{1 + y}}{4} = {a_7} \text{, } \\ & {a_0} + {a_1} + {a_2} + {a_3} = \mu \text{, } {a_4} + {a_5} + {a_6} + {a_7} = 1 - \mu \text{, } \\ & {a_0} + {a_1} + {a_2} + {a_3} + {a_4} +{a_5} + {a_6} + {a_7} = 1 \text{, }\\ &{{\boldsymbol{\rho}} _{{\rm{III}}}} = {p_{\rm{p}}}{{\boldsymbol{\rho}} _{3{\rm{p}}}} + {p_1}{{\boldsymbol{\rho}} _1} + {p_2}{{\boldsymbol{\rho}} _2} + {p_3}{{\boldsymbol{\rho}} _3} + {p_4}{{\boldsymbol{\rho}} _4} .\\[-12pt] \end{split}$

      在给定对角元${a_0}\text{—}{a_7}$后(3)式中$ {p_{\rm{p}}} $, $ \mu $, $ x $, $ y $, $ {p_1} $, $ {p_2} $, $ {p_3} $, $ {p_4} $变量数恰等于方程数.

      • 3.2.   N量子比特系统纠缠判据

    • 一般来说, N量子比特系统的密度矩阵可以由矩阵$ {{\boldsymbol{\rho}} _n} = {{\boldsymbol{A}}_N} $表示出来, 其秩为$ N = {2^n} $. 对于3量子比特系统, 矩阵的秩为$ N = {2^3} = 8 $. 假设密度矩阵元素是实数, 3量子比特系统有如下形式[19]:

      其中

      $ \begin{split}&{{\boldsymbol{\rho}} _{{\rm{III}}}} = \left( {\begin{array}{*{20}{c}} {{a_0}}&\alpha &\beta &\gamma &{{b_0}}&\eta &{ \xi }&\zeta \\ {{\alpha ^*}}&{{a_1}}&\delta &\lambda &\nu &{{b_1}}&\vartheta &\sigma \\ {{\beta ^*}}&{{\delta ^*}}&{{a_2}}&\mu &\theta &\tau &{{b_2}}&\upsilon \\ {{\gamma ^*}}&{{\lambda ^*}}&{{\mu ^*}}&{{a_3}}&\chi &\omega &\kappa &{{b_3}} \\ {b_0^*}&{{\nu ^*}}&{{\theta ^*}}&{{\chi ^*}}&{{a_4}}&o&\pi &\rho \\ {{\eta ^*}}&{b_1^*}&{ {\tau ^*}}&{ {\omega ^*}}&{{o^*}}&{{a_5}}&\varphi &\phi \\ {{\xi ^*}}&{{\vartheta ^*}}&{b_2^*}&\kappa &{{\pi ^*}}&{{\varphi ^*}}&{{a_6}}&\varepsilon \\ {{\zeta ^*}}&{{\sigma ^*}}&{{\upsilon ^*}}&{b_3^*}&{\rho { ^*}}&{ \phi { ^*}}&{{\varepsilon ^*}}&{{a_7}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\boldsymbol{A}}&{\boldsymbol{D}} \\ {{{\tilde {\boldsymbol{D}}}^*}}&{\boldsymbol{B}} \end{array}} \right) \text{, } \\ &{\boldsymbol{A}} = \left( \begin{array}{*{20}{c}} {a_0}& \alpha & \beta & \gamma \\ {\alpha ^*}& {a_1}& \delta &\lambda \\ {\beta ^*}& \delta ^*&{a_2}& \mu \\ {\gamma ^*}&{\lambda ^*}& {\mu ^*}& {a_3} \\ \end{array} \right) \text{, } ~~{\boldsymbol{D}} = \left( \begin{array}{*{20}{c}} {c_0} & \eta & \xi & \zeta \\ \nu & {c_1} & \vartheta & \sigma \\ \theta & \tau & {c_2} & \upsilon \\ \chi & \omega & \kappa & {c_3} \end{array} \right) \text{, }~~ {\boldsymbol{B}} = \left( \begin{array}{*{20}{c}} {a_4} & o & \pi & \rho \\ {o^*} & {a_5} & \varphi & \phi \\ {\pi ^*} & {\varphi ^*} & {a_6} & \varepsilon \\ \rho { ^*} & \phi { ^*} & \varepsilon { ^*} & {a_7} \end{array} \right) .\end{split} $

      由式(4)可知, 对角矩阵元$ {\boldsymbol{A}} $对应前面(1)式的2量子比特系统. 对角矩阵元$ {\boldsymbol{B}} $也可以采用与前面(1)式相同的方法获得解析解. 除此之外, 还有非对角矩阵元$ {\boldsymbol{D}} $和$ {\tilde {\boldsymbol{D}}^{\text{*}}} $的存在, 使得3量子比特的纠缠判据变得更为复杂. 为此就要使$ {{\boldsymbol{\rho}} _{{\rm{III}}}} $对角化(下面虽仍就$ {{\boldsymbol{\rho}} _{{\rm{III}}}} $进行讨论, 但所得结果很明显可推广到$ {{\boldsymbol{\rho}} _n} = {{\boldsymbol{A}}_N} $). 采用如下方法. 根据求本征函数和本征值的方法[19], $ {{\boldsymbol{C}}_{\boldsymbol{D}}} = [{d_0}, \cdots , {d_7}] $, 其中$ {d_0}, \cdots , {d_7} $是本征函数, ${b_0}, \cdots , - b _0$是$ \left( {\begin{array}{*{20}{c}} 0&{\boldsymbol{D}} \\ {{{\tilde {\boldsymbol{D}}}^*}}&0 \end{array}} \right) $的本征值.

      $ C_D^{ - 1}\left( {\begin{array}{*{20}{c}} 0 & D \\ {{{\tilde D}^*}} & 0 \end{array}} \right){C_D} = C_D^{ - 1}\left( {\begin{array}{*{20}{c}} 0 & 0 & 0 & 0 & {{c_0}} & { \eta } & { \xi } & \zeta \\ 0 & 0 & 0 & 0 & \nu & {{c_1}} & \vartheta & \sigma \\ 0 & 0 & 0 & 0 & \theta & { \tau } & {{c_2}} & \upsilon \\ 0 & 0 & 0 & 0 & \chi & \omega & \kappa & {{c_3}} \\ {c_0^*} & {\nu { ^*}} & {{\theta ^*}} & { {\chi ^*}} & 0 & 0 & 0 & 0 \\ {{\eta ^*}} & {c_1^*} & {{\tau ^*}} & {{\omega ^*}} & 0 & 0 & 0 & 0 \\ {{\xi ^*}} & {{\vartheta ^*}} & {c_2^*} & {{\kappa ^*}} & 0 & 0 & 0 & 0 \\ {{\zeta ^*}} & {{\sigma ^*}} & {{\upsilon ^*}} & {c_3^*} & 0 & 0 & 0 & 0 \end{array}} \right){C_D} = \left( {\begin{array}{*{20}{c}} {{b_0}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & {{b_1}} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & {{b_2}} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {{b_3}} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & { - {b_3}} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & { - {b_2}} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & { - {b_1}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & { - {b_0}} \end{array}} \right) . $

      我们得出一个分裂定理:

      $ \begin{split}{\boldsymbol{C}}_{\boldsymbol{D}}^{ - 1}{{\boldsymbol{\rho}} _{{\rm{III}}}}{{\boldsymbol{C}}_{\boldsymbol{D}}} =\ & {\boldsymbol{C}}_{\boldsymbol{D}}^{ - 1}\left( {\begin{array}{*{20}{c}} {\boldsymbol{A}}&0 \\ 0&{\boldsymbol{B}} \end{array}} \right){{\boldsymbol{C}}_{\boldsymbol{D}}} + \left( {\begin{array}{*{20}{c}} {{b_0}}&0&0&0 \\ 0&\cdot &0&0 \\ 0&0&\cdot &0 \\ 0&0&0&{ - {b_0}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\tilde {\boldsymbol{A}}}&0 \\ 0&{\tilde {\boldsymbol{B}}} \end{array}} \right) \text{, } \end{split}$

      $\begin{split}& \tilde {\boldsymbol{A}} = {\boldsymbol{C}}_{\boldsymbol{A}}^{ - 1}{\boldsymbol{A}}{{\boldsymbol{C}}_{\boldsymbol{A}}} + \left( {\begin{array}{*{20}{c}} {{b_0}}&0&0&0 \\ 0&{{b_1}}&0&0 \\ 0&0&{{b_2}}&0 \\ 0&0&0&{{b_3}} \end{array}} \right) \text{, } ~~~~\tilde {\boldsymbol{B}} = {\boldsymbol{C}}_{\boldsymbol{B}}^{ - 1}{\boldsymbol{B}}{{\boldsymbol{C}}_{\boldsymbol{B}}} + \left( {\begin{array}{*{20}{c}} { - {\kern 1pt} {b_3}}&0&0&0 \\ 0&{ - {\kern 1pt} {b_2}}&0&0 \\ 0&0&{ - {\kern 1pt} {b_1}}&0 \\ 0&0&0&{ - {\kern 1pt} {b_0}} \end{array}} \right) \text{, } \end{split}$

      其中

      $ {{\boldsymbol{C}}_{\boldsymbol{A}}} = ({d_0},{d_1},{d_2},{d_3}) \text{, }~~ {{\boldsymbol{C}}_{\boldsymbol{B}}} = ({d_4},{d_5}{\kern 1pt} ,{d_6},{d_7}) . $

      (7)式显示对角密度矩阵$ {{\boldsymbol{\rho}} _{{\rm{II}}{\boldsymbol{A}}}} = \tilde {\boldsymbol{A}} $, $ {{\boldsymbol{\rho}} _{{\rm{II}}{\boldsymbol{B}}}} = \tilde {\boldsymbol{B}} $, 其秩为$ N = 4 $. 密度矩阵$ {\boldsymbol{C}}_{\boldsymbol{D}}^{ - 1}{{\boldsymbol{\rho}} _{{\rm{III}}}}{{\boldsymbol{C}}_{\boldsymbol{D}}} $的秩为$ N = 8 $. $ {\boldsymbol{C}}_{\boldsymbol{D}}^{ - 1}{{\boldsymbol{\rho}} _{{\rm{III}}}}{{\boldsymbol{C}}_{\boldsymbol{D}}} $可分裂成(6)式所表述的对角矩阵$ {{\boldsymbol{\rho}} _{{\rm{II}}{\boldsymbol{A}}}} = \tilde {\boldsymbol{A}} $和$ {{\boldsymbol{\rho}} _{{\rm{II}}{\boldsymbol{B}}}} = \tilde {\boldsymbol{B}} $. 同样一个N量子比特系统可以通过这种分裂成为一个(N–1)量子比特系统. 这个分裂过程可以描述为$ {{\boldsymbol{\rho}} }_{n}\to {{\boldsymbol{\rho}} }_{n-1, {\boldsymbol{A}}}, {{\boldsymbol{\rho}} }_{n-1, {\boldsymbol{B}}} $, 其中$ {{\boldsymbol{\rho}} _n} $是$ {2^n} $秩的N量子比特系统的密度矩阵, $ {{\boldsymbol{\rho}} }_{n-1, A}, {{\boldsymbol{\rho}} }_{n-1, B} $是$ {2^{n - 1}} $秩的(N–1)量子比特系统的密度矩阵. 继续此过程, $ {\boldsymbol{\rho}} _n \to {\boldsymbol{\rho}} { _{n - 1}} \to {\boldsymbol{\rho}} _{n - 2} \to \cdots \to {\boldsymbol{\rho}} _{{\rm{II}}} $. 下图1所示的$ {2^4} $秩的整个分裂过程可以描述为$ {{\boldsymbol{\rho}} }_{4}\to {{\boldsymbol{\rho}} }_{{\rm{III}}, {\boldsymbol{A}}}, {{\boldsymbol{\rho}} }_{{\rm{III}}, {\boldsymbol{B}}}\to {{\boldsymbol{\rho}} }_{{\rm{II}}, {\boldsymbol{AA}}}, {{\boldsymbol{\rho}} }_{{\rm{II}}, {\boldsymbol{AB}}}; {{\boldsymbol{\rho}} }_{{\rm{II}}, {\boldsymbol{BA}}}, {{\boldsymbol{\rho}} }_{{\rm{II}}, {\boldsymbol{BB}}} $.

      图  1  纠缠态的“N-AB-tree结构”

      Figure 1.  24-AB-tree to four 2 qubit string.

      3量子比特密度矩阵的解析解可以由两个2量子比特系统密度矩阵表示. n量子比特密度矩阵的解析解可以由两个n–1量子比特系统密度矩阵表示. 即密度矩阵${{\boldsymbol{\rho}} _n}$的解析求解转化为$ {{\boldsymbol{\rho}} }_{n-1, A}, {{\boldsymbol{\rho}} }_{n-1, B} $. 计算过程只需将(7)式的本征函数改写为$ {{\boldsymbol{C}}_{\boldsymbol{D}}} = [{d_0}, \cdots , {d_N}] $, 本征值改写为$ ({b_0}, \cdots {b_{N/2}}, {b_{ - N/2}} \cdots b{_0}) $, 相应的$ {{\boldsymbol{C}}_{\boldsymbol{A}}} = ({d_0}, \cdots, {d_{N/2}}) $, $ {{\boldsymbol{C}}_{\boldsymbol{B}}} = ({d_{N/2 + 1}}, \cdots , {d_N}) $.

    4.   2量子比特和3量子比特系统主纠缠态${p_p}$的数值计算

    • 对于2量子比特与3量子比特系统主纠缠态${p_p}$的计算分别由图2和图3—图5给出. (2)式右端的系数分别表示为 ${a_0} + {a_1} + {a_2} + {a_3} = {r^2}$, ${a_0} = 0.5\times {(r{\rm{cos}}[\theta ])^2} + \varDelta$, ${a_3} = 0.5{(r{\rm{cos}}[\theta ])^2} - \varDelta$, ${a_1} = (r{\text{sin}}[\theta ]\times {\text{cos}}[\omega ])^2$, ${a_2} = {(r{\text{sin}} [\theta ]{\text{sin}}[\omega ])^2}$, 参照(1)式密度矩阵${{\boldsymbol{\rho}} _{{\rm{II}}}}$, ${a_0}$, ${a_1}$, ${a_2}$, ${a_3}$分别表示对角矩阵元, 考虑到参量${a_0}$, ${a_3}$分别为两个二能级原子分别同时取基态或激发态的几率. 故${a_0} - {a_3} = 2\Delta $就代表集居数的差. 这个差与驱动场的失谐有关$\Delta \omega + {\rm{i}}\gamma = ({\omega _{21}} - \omega ) + {\rm{i}}({1 / {{T_1}}} + {\gamma _c})$, ${T_1}$即纵弛豫时间, ${\gamma _c}$即碰撞系数[20].

      图  2  2量子比特系统主密度矩阵系数${\kern 1 pt} {p_p}$随$\{ \theta , 0, 0.2\} $, $\{ w, - \pi , \pi \} $变化的三维曲线图, $r = 1$　(a) $\varDelta = 0.002$; (b) $\varDelta = 0.00103$; (c) $\varDelta = 0.0002$; (d) $\varDelta = 0.0001$

      Figure 2.  Principal entangled state${p_p}$of 2 qubit system with the parameters$\{ \theta , 0, 0.2\} $, $\{ w, - \pi , \pi \} $, $r = 1$: (a)$\varDelta = 0.002$; (b)$\varDelta = $$ 0.00103$; (c)$\varDelta = 0.0002$; (d)$\varDelta = 0.0001$.

      图  3  3量子比特系统主密度矩阵系数${\kern 1 pt} {p_p}$随$\{ \theta , 0, 0.2\} $, $\{ w, - \pi , \pi \} $变化的三维曲线图　(a) ${r_a} = 0.867$, $\varDelta = 0.0001$; (b) ${r_b} = 0.547$, $\varDelta = 0.001$, $\xi + \eta = r_a^2 + r_b^2 = 1$

      Figure 3.  Principal entangle state${p_p}$of 3 qubit system with the parameters$\{ \theta , 0, 0.2\} $, $\{ w, - \pi , \pi \} $: (a) ${r_a} = 0.867$, $\varDelta = 0.0001$; (b) ${r_b} = $$ 0.547$, $\varDelta = 0.001$, $\xi + \eta = r_a^2 + r_b^2 = 1$.

      图  5  3量子比特系统主密度矩阵系数${\kern 1 pt} {p_p}$随$\{ \theta , 0, 0.2\} $, $\{ w, - \pi , \pi \} $变化的三维曲线图, ${r_a} = {r_b} = 0.707$　(a) $\varDelta = 0.0001$; (b) $\varDelta = $$ 0.0002$; (c) $\varDelta = 0.00103$

      Figure 5.  Principal entangled state${p_p}$of 3 qubit system with the parameters$\{ \theta , 0, 0.2\} $, $\{ w, - \pi , \pi \} $, ${r_a} = {r_b} = 0.707$; (a) $\varDelta = 0.0001$; (b) $\varDelta = 0.0002$; (c) $\varDelta = 0.00103$.

      我们注意到上述表式中的参数$r$. 对于${{\boldsymbol{\rho}} _{{\rm{II}}}}$, $r = 1$, 对于分裂后${{\boldsymbol{\rho}} _{{\rm{III}}}} = \left( {\begin{array}{*{20}{c}} {{{\boldsymbol{\rho}} _{{\rm{II}}{\boldsymbol{A}}}}}&0 \\ 0&{{{\boldsymbol{\rho}} _{{\rm{II}}{\boldsymbol{B}}}}} \end{array}} \right)$, 其中${{\boldsymbol{\rho}} _{{\rm{II}}{\boldsymbol{A}}}}$, ${{\boldsymbol{\rho}} _{{\rm{II}}{\boldsymbol{B}}}}$的对角矩阵元的和即权重分别为$\xi = r_a^2$, $\eta = r_b^2$, $\xi + \eta = r_a^2 + r_b^2 = 1$.

      1) 2量子比特系统

      2) 3量子比特系统

      图  4  3量子比特系统主密度矩阵系数${\kern 1 pt} {p_p}$随$\{ \theta , 0, 0.2\} $, $\{ w, - \pi , \pi \} $变化的三维曲线图　(a) ${r_a} = 0.724$, $\varDelta = 0.0001$; (b) ${r_b} = 0.652$, $\varDelta = 0.001$, $\xi + \eta = r_a^2 + r_b^2 = 1$

      Figure 4.  Principal entangled state${p_p}$of 3 qubit system with the parameters$\{ \theta , 0, 0.2\} $, $\{ w, - \pi , \pi \} $: (a) ${r_a} = 0.724$, $\varDelta = 0.0001$; (b) ${r_b} = 0.652$, $\varDelta = 0.001$, $\xi + \eta = r_a^2 + r_b^2 = 1$.

      在图2和图3—图5描述了2量子比特和3量子比特系统的主密度矩阵系数${p_p}$的数值计算结果. 一般的, 存在两种类型的纠缠: 一个是浅纠缠, 处于初始阶段, 最大值从10到100; 另一个是深度纠缠, 处于末了阶段, 最大值达到1000. 拥有大量的深纠缠的2量子比特系统. 所谓深浅只是对纠缠程度一个唯象的描述, 并没严格定义. 纠缠对失谐参数的的变化很敏感. 图6描绘出了图2(b)曲线的轮廓. 图7描绘出了图5(c)曲线的轮廓.

      图  6  图2(b)曲线的轮廓.

      Figure 6.  The profile of Fig. 2(b). ${p_{\rm{p}}}~~vs~~(\varDelta = 0.00102 + $$ x \times {10^{ - 5}})$.

      图  7  图5(c)曲线的轮廓.

      Figure 7.  The profile of Fig. 5(c). ${p_{\rm{p}}}~~vs~~(\varDelta = 0.00102 + $$ x \times {10^{ - 5}})$.

      图6和图7的结果显示曲线的轮廓依赖于纠缠度. 在初始阶段, 纠缠度从0逐渐变到100. 随后, 快速变化到–600 (见图6)和–1400 (见图7). 最后, 突然返回0. 从上面的关系可以获知:

      1) 越大粒子量子比特系统会产生更深的纠缠度和更窄的轮廓带宽.

      2) 曲线的轮廓与线性色散理论的线型相似[20]. 使用拉普拉斯变换可以得到纠缠度将随时间$ \tau $衰减为$ \propto {{\rm{e}}^{ - \gamma \tau }}$, $ \gamma = {1 \mathord{\left/ {\vphantom {1 {{T_1}}}} \right. } {{T_1}}} + {\gamma _c} $为色散曲线的带宽. 产生大纠缠的物理条件是上下能级的集居数差即$2\varDelta$要求比较小, 但不要求反转.

    5.   结　论

    • 1) 在前文的基础上, 我们用同样方法求得3量子比特系统的主密度矩阵, 并发现其结构恰是两个2量子比特系统的主密度矩阵的对角矩阵. 我们称此为A-B分裂. 并在这个基础上, 应用数学归纳法求得n量子比特系统的主密度矩阵解, 我们将其称之为“${2^N}$-AB-树”.

      2) 通过数值计算2量子比特和3量子比特揭示出存在2种类型的纠缠: 浅纠缠和深纠缠, 曲线的轮廓具有典型的洛伦兹线型[20]. 经过对线型的拉普拉斯变换可得出: 1) 纠缠态的衰变时间反比于线型的宽度; 2) 线型的宽度又反比于$ {2^n} $, n为参与纠缠的粒子数. 换句话说, 参与的粒子数n越大, 线型的宽度越乍纠缠态存在时间越长.

    附录A　关于(2)式的计算及分离态的矩阵形式

    • 1) (2)式的计算

      (1) 式对角化元素相等, 有如下的方程$: $

      $\begin{split} &{p_p}\frac{{1 - x}}{4} + ({\tilde p_1} + {\tilde p_2})\frac{{1 + x}}{4} = {a_0} \text{, }\\ &{p_p}\frac{{1 + x}}{4} + ({\tilde p_1} + {\tilde p_4})\frac{{1 - x}}{4} = {a_1} \text{, } \\ &{p_p}\frac{{1 + x}}{4} + ({\tilde p_2} + {\tilde p_3})\frac{{1 - x}}{4} = {a_2} \text{, } \\ &{p_p}\frac{{1 - x}}{4} + ({\tilde p_3} + {\tilde p_4})\frac{{1 + x}}{4} = {a_3} \text{, } \end{split} \tag{A1}$

      令

      $ \begin{split}&{\tilde p_1} = {p_1} - {\delta / 2} \text{, } ~~{\tilde p_2} = {p_1} + {\delta / 2} \text{, }\\ &{\tilde p_3} = {p_2} + {\delta / 2} \text{, } ~~{\tilde p_4} = {p_2} - {\delta / 2} .\end{split}\tag{A2} $

      将(A2)式代入(A1)式, 有

      $\begin{split} &{p_p}\frac{{1 - x}}{4} + ({p_1} + {p_1})\frac{{1 + x}}{4} = {a_0} \text{, } \\ &{p_p}\frac{{1 - x}}{4} + ({p_2} + {p_2})\frac{{1 + x}}{4} = {a_3} \text{, } \\ &{p_p}\frac{{1 + x}}{4} + ({p_1} + {p_1})\frac{{1 - x}}{4} + ({p_2} - {p_1} - \delta )\frac{{1 - x}}{4} = {a_1} \text{, } \\ &{p_p}\frac{{1 + x}}{4} + ({p_2} + {p_2})\frac{{1 - x}}{4} - ({p_2} - {p_1} - \delta )\frac{{1 - x}}{4} = {a_2} . \end{split}\tag{A3}$

      若下面的方程成立:

      $ ({p_2} - {p_1} - \delta )\frac{{1 - x}}{4} = 0 .\tag{A4} $

      根据(A1)式—(A4)式, 可以得到(2)式:

      $ \begin{split} &{p_p}\frac{{1 - x}}{4} + ({p_1} + {p_1})\frac{{1 + x}}{4} = {a_0} \text{, }\\ &{p_p}\frac{{1 + x}}{4} + ({p_1} + {p_1})\frac{{1 - x}}{4} = {a_1} \text{, }\\ &{p_p}\frac{{1 + x}}{4} + ({p_2} + {p_2})\frac{{1 - x}}{4} = {a_2} \text{, } \\ &{p_p}\frac{{1 - x}}{4} + ({p_2} + {p_2})\frac{{1 + x}}{4} = {a_3} .\end{split} \tag{A5}$

      (A4)式包括两个解, 其中一个是特解:

      $ 1 - x = 0 \text{, }~ \delta = 0 \text{, }~ {\tilde p_1} = {p_1} \text{, }~ {\tilde p_2} = {p_1} \text{, } ~ {\tilde p_3} = {p_2} \text{, }~ {\tilde p_4} = {p_2} . \tag{A6}$

      另一个是通解:

      $ \begin{split} & 1 - x \ne 0 \text{, } ~~ \delta = {p_2} - {p_1} \text{, }~~ {\tilde p_1} = {p_1} - \frac{\delta }{2} = \frac{{3{p_1} - {p_2}}}{2} \text{, } \\ & {\tilde p_2} = {p_1} + \frac{\delta }{2} = \frac{{{p_1} + {p_2}}}{2} \text{, }~~ {\tilde p_3} = {p_2} + \frac{\delta }{2} = \frac{{3{p_2} - {p_1}}}{2} \text{, } \\ & {\tilde p_4} = {p_2} - \frac{\delta }{2} = \frac{{{p_2} + {p_1}}}{2} .\end{split} \tag{A7}$

      根据(A6)式, 若$ {p_1} > 0 $, $ {p_2} > 0 $, 则$ {\tilde p_1} - {\tilde p_4} > 0 $. 然而, 根据(A7)式, 除了$ {p_1} > 0 $, $ {p_2} > 0 $. 必须施加一个更加严格的要求$ 3 > {{{p_2}} \mathord{\left/ {\vphantom {{{p_2}} {{p_1}}}} \right. } {{p_1}}} > {1 \mathord{\left/ {\vphantom {1 3}} \right. } 3} $, 则必要条件$ {\tilde p_1} - {\tilde p_4} > 0 $必须满足.

      2) 分离态的矩阵形式

      $\begin{split}& \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\quad\qquad 00}&{11} \end{array}} \\ {{{\boldsymbol{S}}_1} = \dfrac{1}{2}\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)} \end{array} , ~~\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\quad\qquad 01}&{10} \end{array}} \\ {{{\boldsymbol{S}}_2} = \dfrac{1}{2}\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)} \end{array} , ~~ \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\quad\qquad 00}&{01}&{10}&{11} \end{array}} \\ {{{\boldsymbol{S}}_3} = \dfrac{1}{4}\left( {\begin{array}{*{20}{c}} 1&0&0&{ - 1} \\ 0&1&1&0 \\ 0&1&1&0 \\ { - 1}&0&0&1 \end{array}} \right)} \end{array} , ~~\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\quad\qquad 00}&{01}&{10}&{11} \end{array}} \\ {{{\boldsymbol{S}}_4} = \dfrac{1}{4}\left( {\begin{array}{*{20}{c}} 1&0&0&{ - 1} \\ 0&1&{ - 1}&0 \\ 0&{ - 1}&1&0 \\ { - 1}&0&0&1 \end{array}} \right)} \end{array} , ~~\\ & \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\quad\qquad 00}&{01}&{10}&{01} \end{array}} \\ {{{\boldsymbol{T}}_3} = \dfrac{1}{4}\left( {\begin{array}{*{20}{c}} 1&0&0&1 \\ 0&1&{ - 1}&0 \\ 0&{ - 1}&1&0 \\ 1&0&0&1 \end{array}} \right)} \end{array} ,~~ {{\boldsymbol{\rho}} _1} = \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{\kern 1pt} {\kern 1pt} 00}&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 01} \end{array}} \\ {\left( {\begin{array}{*{20}{c}} {\dfrac{{1 + x}}{2}}&{\dfrac{{{y_1}}}{2}} \\ {\dfrac{{{y_1}}}{2}}&{\dfrac{{1 - x}}{2}} \end{array}} \right)} \end{array}{\kern 1pt} , ~~{\kern 1pt} {\kern 1pt} {{\boldsymbol{\rho}} _2} = \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {00}&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 10} \end{array}} \\ {\left( {\begin{array}{*{20}{c}} {\dfrac{{1 + x}}{2}}&{\dfrac{{{y_2}}}{2}} \\ {\dfrac{{{y_2}}}{2}}&{\dfrac{{1 - x}}{2}} \end{array}} \right)} \end{array} , ~~\\ & {{\boldsymbol{\rho }}_3} = \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {10}&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 11} \end{array}} \\ {\left( {\begin{array}{*{20}{c}} {\dfrac{{1 - x}}{2}}&{\dfrac{{{z_2}}}{2}} \\ {\dfrac{{{z_2}}}{2}}&{\dfrac{{1 + x}}{2}} \end{array}} \right)} \end{array} , ~~ {{\boldsymbol{\rho }}_4} = \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{\kern 1pt} {\kern 1pt} {\kern 1pt} 01}&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 11} \end{array}} \\ {\left( {\begin{array}{*{20}{c}} {\dfrac{{1 - x}}{2}}&{\dfrac{{{z_1}}}{2}} \\ {\dfrac{{{z_1}}}{2}}&{\dfrac{{1 + x}}{2}} \end{array}} \right)} \end{array} , ~~ \begin{gathered} \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 00}&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 01}&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 10}&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 11} \end{array}} \\ {{{\boldsymbol{\rho}} _{\rm{p}}} = \left( {\begin{array}{*{20}{c}} {({{1 - x}})/{4}}&0&0&{ - {{{x_1}}}/{2}} \\ 0&{({{1 + x}})/{4}}&{ - {{{x_2}}}/{2}}&0 \\ 0&{ - {{{x_2}}}/{2}}&{({{1 + x}})/{4}}&0 \\ { - {{{x_1}}}/{2}}&0&0&{ ({{1 - x}})/{4}} \end{array}} \right)} \end{array} \\ \end{gathered} .\end{split} $

    附录B 3量子系统的基函数, 可分离矩阵及主矩阵

    • 1) 基函数 ($ {\omega ^3} = 1 $)

      $ \begin{split}&|{\varphi }_{1}\rangle \langle {\varphi }_{1}|=\frac{|000\rangle +|111\rangle }{\sqrt{2}}\frac{\langle 000|+\langle 111|}{\sqrt{2}}=\frac{1}{2}({{\boldsymbol{\rho}} }_{000}'{{\boldsymbol{\rho}} }_{000}''+{{\boldsymbol{\rho}} }_{111}'{{\boldsymbol{\rho}} }_{111}''+{{\boldsymbol{\rho}} }_{000}'{{\boldsymbol{\rho}} }_{111}''+{{\boldsymbol{\rho}} }_{111}'{{\boldsymbol{\rho}} }_{000}'')\text{, } \\ &|{\varphi }_{2}\rangle \langle {\varphi }_{2}|=\frac{|000\rangle +\omega |111\rangle }{\sqrt{2}}\frac{\langle 000|+{\omega }^{2}\langle 111|}{\sqrt{2}}=\frac{1}{2}({{\boldsymbol{\rho}} }_{000}'{{\boldsymbol{\rho}} }_{000}''+{{\boldsymbol{\rho}} }_{111}'{{\boldsymbol{\rho}} }_{111}''+{\omega }^{2}{{\boldsymbol{\rho}} }_{000}'{{\boldsymbol{\rho}} }_{111}''+\omega {{\boldsymbol{\rho}} }_{111}'{{\boldsymbol{\rho}} }_{000}'')\text{, } \\ &|{\varphi }_{3}\rangle \langle {\varphi }_{3}|=\frac{|000\rangle +{\omega }^{2}|111\rangle }{\sqrt{2}}\frac{\langle 000|+\omega \langle 111|}{\sqrt{2}}=\frac{1}{2}({{\boldsymbol{\rho}} }_{000}'{{\boldsymbol{\rho}} }_{000}''+{{\boldsymbol{\rho}} }_{111}'{{\boldsymbol{\rho}} }_{111}''+\omega {{\boldsymbol{\rho}} }_{000}'{{\boldsymbol{\rho}} }_{111}''+{\omega }^{2}{{\boldsymbol{\rho}} }_{111}'{{\boldsymbol{\rho}} }_{000}'')\text{, } \\ &|{\phi }_{1}\rangle \langle {\phi }_{1}|=\frac{|010\rangle +|101\rangle }{\sqrt{2}}\frac{\langle 010|+\langle 101|}{\sqrt{2}}=\frac{1}{2}({{\boldsymbol{\rho}} }_{010}'{{\boldsymbol{\rho}} }_{010}''+{{\boldsymbol{\rho}} }_{101}'{{\boldsymbol{\rho}} }_{101}''\pm {{\boldsymbol{\rho}} }_{010}'{{\boldsymbol{\rho}} }_{101}''\pm {{\boldsymbol{\rho}} }_{101}'{{\boldsymbol{\rho}} }_{010}'')\text{, } \\ &|{\phi }_{2}\rangle \langle {\phi }_{2}|=\frac{|010\rangle +\omega |101\rangle }{\sqrt{2}}\frac{\langle 010|+{\omega }^{2}\langle 101|}{\sqrt{2}}=\frac{1}{2}({{\boldsymbol{\rho}} }_{010}'{{\boldsymbol{\rho}} }_{010}''+{{\boldsymbol{\rho}} }_{101}'{{\boldsymbol{\rho}} }_{101}''+\omega {{\boldsymbol{\rho}} }_{010}'{{\boldsymbol{\rho}} }_{101}''+{\omega }^{2}{{\boldsymbol{\rho}} }_{101}'{{\boldsymbol{\rho}} }_{010}'')\text{, } \\ &|{\phi }_{3}\rangle \langle {\phi }_{3}|=\frac{|010\rangle +{\omega }^{2}|101\rangle }{\sqrt{2}}\frac{\langle 010|+\omega \langle 101|}{\sqrt{2}}=\frac{1}{2}({{\boldsymbol{\rho}} }_{010}'{{\boldsymbol{\rho}} }_{010}''+{{\boldsymbol{\rho}} }_{101}'{{\boldsymbol{\rho}} }_{101}''+{\omega }^{2}{{\boldsymbol{\rho}} }_{010}'{{\boldsymbol{\rho}} }_{101}''+\omega {{\boldsymbol{\rho}} }_{101}'{{\boldsymbol{\rho}} }_{010}'')\text{, } \\ &|{\psi }_{1}\rangle \langle {\psi }_{1}|=\frac{|001\rangle +|110\rangle }{\sqrt{2}}\frac{\langle 001|+\langle 110|}{\sqrt{2}}=\frac{1}{2}({{\boldsymbol{\rho}} }_{001}'{{\boldsymbol{\rho}} }_{001}''+{{\boldsymbol{\rho}} }_{110}'{{\boldsymbol{\rho}} }_{110}''+{{\boldsymbol{\rho}} }_{001}'{{\boldsymbol{\rho}} }_{110}''+{{\boldsymbol{\rho}} }_{110}'{{\boldsymbol{\rho}} }_{001}'')\text{, } \\ &|{\psi }_{2}\rangle \langle {\psi }_{2}|=\frac{|001\rangle +\omega |110\rangle }{\sqrt{2}}\frac{\langle 001|+{\omega }^{2}\langle 110|}{\sqrt{2}}=\frac{1}{2}{{\boldsymbol{\rho}} }_{001}'{{\boldsymbol{\rho}} }_{001}''+{{\boldsymbol{\rho}} }_{110}'{{\boldsymbol{\rho}} }_{110}''+{\omega }^{2}{{\boldsymbol{\rho}} }_{001}'{{\boldsymbol{\rho}} }_{110}''+\omega {{\boldsymbol{\rho}} }_{110}'{{\boldsymbol{\rho}} }_{001}''))\text{, } \\ &|{\psi }_{3}\rangle \langle {\psi }_{3}|=\frac{|001\rangle +{\omega }^{2}|110\rangle }{\sqrt{2}}\frac{\langle 001|+\omega \langle 110|}{\sqrt{2}}=\frac{1}{2}({{\boldsymbol{\rho}} }_{001}'{{\boldsymbol{\rho}} }_{001}''+{{\boldsymbol{\rho}} }_{110}'{{\boldsymbol{\rho}} }_{110}''+\omega {{\boldsymbol{\rho}} }_{001}'{{\boldsymbol{\rho}} }_{110}''+{\omega }^{2}{{\boldsymbol{\rho}} }_{110}'{{\boldsymbol{\rho}} }_{001}'')).\end{split} \tag{B1}$

      2) 可分离矩阵 ${{{\rho}} _1}\text{—}{{{\rho}} _4}$及主矩阵$ {{{\rho}} _{\rm{p}}} $

      $ \begin{split} &\begin{array}{*{20}{r}} \begin{array}{*{20}{r}} 000 & 111\end{array}\\ {S}_{1}=\dfrac{1}{3}(|{\varphi }_{1}\rangle \langle {\varphi }_{1}|+|{\varphi }_{2}\rangle \langle {\varphi }_{2}|+|{\varphi }_{2}\rangle \langle {\varphi }_{3}|)=\dfrac{1}{2}({{\boldsymbol{\rho}} }_{000}'{{\boldsymbol{\rho}} }_{000}''+{{\boldsymbol{\rho}} }_{111}'{{\boldsymbol{\rho}} }_{111}'')=\dfrac{1}{2}\left(\begin{array}{ll}1&{}\\ {}&1\end{array}\right){, }\end{array} \\ &\begin{array}{*{20}{r}} \begin{array}{*{20}{r}}001& 110\end{array}\\ {S}_{2}=\dfrac{1}{3}(|{\phi }_{1}\rangle \langle {\phi }_{1}|+|{\phi }_{2}\rangle \langle {\phi }_{2}|+|{\phi }_{3}\rangle \langle {\phi }_{3}|)=\dfrac{1}{2}({{\boldsymbol{\rho}} }_{001}'{{\boldsymbol{\rho}} }_{001}''+{{\boldsymbol{\rho}} }_{110}'{{\boldsymbol{\rho}} }_{110}'')=\dfrac{1}{2}\left(\begin{array}{ll}1&{}\\ {} &1\end{array}\right){, }\end{array} \\ & \begin{array}{*{20}{r}}\begin{array}{*{20}{r}} 010& 101\end{array}\\ {S}_{3}=\dfrac{1}{3}(|{\psi }_{1}\rangle \langle {\psi }_{1}|+|{\psi }_{2}\rangle \langle {\psi }_{2}|+|{\psi }_{3}\rangle \langle {\psi }_{3}|)=\dfrac{1}{2}({{\boldsymbol{\rho}} }_{010}'{{\boldsymbol{\rho}} }_{010}''+{{\boldsymbol{\rho}} }_{101}'{{\boldsymbol{\rho}} }_{101}'')=\dfrac{1}{2}\left(\begin{array}{ll}1& {} \\ {}&1\end{array}\right){, }\end{array} \\ & \begin{array}{*{20}{r}}\begin{array}{*{20}{r}}100& 011\end{array}\\ {S}_{4}=\dfrac{1}{3}(|{\chi }_{1}\rangle \langle {\chi }_{1}|+|{\chi }_{2}\rangle \langle {\chi }_{2}|+|{\chi }_{3}\rangle \langle {\chi }_{3}|)=\dfrac{1}{2}({{\boldsymbol{\rho}} }_{100}'{{\boldsymbol{\rho}} }_{100}''+{{\boldsymbol{\rho}} }_{011}'{{\boldsymbol{\rho}} }_{011}'')=\dfrac{1}{2}\left(\begin{array}{ll}1&{}\\ {}& 1\end{array}\right){, }\end{array} \\ & \begin{array}{*{20}{r}}\begin{array}{*{20}{l}} 000 &\; 111 \end{array}\quad\\ {S}_{5}=\dfrac{1}{2}(|{\varphi }_{2}\rangle \langle {\varphi }_{2}|+|{\varphi }_{2}\rangle \langle {\varphi }_{3}|)=\dfrac{1}{2}({{\boldsymbol{\rho}} }_{000}'{{\boldsymbol{\rho}} }_{000}''+{{\boldsymbol{\rho}} }_{111}'{{\boldsymbol{\rho}} }_{111}''-{{\boldsymbol{\rho}} }_{000}'{{\boldsymbol{\rho}} }_{111}''-{{\boldsymbol{\rho}} }_{111}'{{\boldsymbol{\rho}} }_{000}'')=\dfrac{1}{2}\left(\begin{array}{ll}1&-1\\ -1&1\end{array}\right){, }\end{array} \\ & \begin{array}{*{20}{r}}\begin{array}{*{20}{l}} 001&\; 110\end{array}\\ {S}_{6}=\dfrac{1}{2}(|{\psi }_{2}\rangle \langle {\psi }_{2}|+|{\psi }_{3}\rangle \langle {\psi }_{3}|)=\dfrac{1}{2}({{\boldsymbol{\rho}} }_{001}'{{\boldsymbol{\rho}} }_{001}''+{{\boldsymbol{\rho}} }_{110}'{{\boldsymbol{\rho}} }_{110}''-{{\boldsymbol{\rho}} }_{110}'{{\boldsymbol{\rho}} }_{001}''-{{\boldsymbol{\rho}} }_{001}'{{\boldsymbol{\rho}} }_{110}'')=\dfrac{1}{2}\left(\begin{array}{ll}1&-1\\ -1&1\end{array}\right){, }\end{array} \\ & \begin{array}{*{20}{r}}\begin{array}{*{20}{l}} 010&\; 101\end{array}\\ {S}_{7}=\dfrac{1}{2}(|{\phi }_{2}\rangle \langle {\phi }_{2}|+|{\phi }_{3}\rangle \langle {\phi }_{3}|)=\dfrac{1}{2}({{\boldsymbol{\rho}} }_{010}'{{\boldsymbol{\rho}} }_{010}''+{{\boldsymbol{\rho}} }_{101}'{{\boldsymbol{\rho}} }_{101}''-{{\boldsymbol{\rho}} }_{101}'{{\boldsymbol{\rho}} }_{010}''-{{\boldsymbol{\rho}} }_{010}'{{\boldsymbol{\rho}} }_{101}'')=\dfrac{1}{2}\left(\begin{array}{ll}1&-1\\ -1&1\end{array}\right){, }\end{array} \\ &\begin{array}{*{20}{r}}\begin{array}{*{20}{l}} 011&\; 100\end{array}\\ {S}_{8}=(|{\chi }_{2}\rangle \langle {\chi }_{2}|+|{\chi }_{3}\rangle \langle {\chi }_{3}|)=\dfrac{1}{2}({{\boldsymbol{\rho}} }_{100}'{{\boldsymbol{\rho}} }_{100}''+{{\boldsymbol{\rho}} }_{011}'{{\boldsymbol{\rho}} }_{011}''-{{\boldsymbol{\rho}} }_{100}'{{\boldsymbol{\rho}} }_{011}''-{{\boldsymbol{\rho}} }_{011}'{{\boldsymbol{\rho}} }_{100}'')=\dfrac{1}{2}\left(\begin{array}{ll}1&-1\\ -1&1\end{array}\right){, }\end{array} \\ & {{\boldsymbol{\rho}} _{\rm{p}}} = \dfrac{{1 - 2({x_1} + {x_2}) - x}}{4}{{\boldsymbol{S}}_1} + \dfrac{{1 - 2({x_1} + {x_2}) + x}}{4}{{\boldsymbol{S}}_2} + \dfrac{{1 - 2({x_3} + {x_4}) - y}}{4}{{\boldsymbol{S}}_3} + \dfrac{{1 - 2({x_3} + {x_4}) + y}}{4}{{\boldsymbol{S}}_4} \\ &\qquad + \dfrac{{{x_1}}}{2}{{\boldsymbol{S}}_5} + \dfrac{{{x_2}}}{2}{{\boldsymbol{S}}_6}{{ + }} \dfrac{{{x_3}}}{2}{{\boldsymbol{S}}_7} + \dfrac{{{x_4}}}{2}{{\boldsymbol{S}}_8} {, } \\ &\begin{array}{*{20}{l}}\begin{array}{*{20}{l}} \qquad \qquad 000 & \quad 001 & \quad 010 & \quad 011 & \quad 100 & \quad 101 & \quad 110 & \quad 111\end{array}\\ \quad = \left( \begin{array}{*{20}{l}} \dfrac{{1 - x}}{8}& {}& {}& {}& {}& {}& {}& \dfrac{{ - {x_1}}}{4} \\ {}& \dfrac{{1 + x}}{8} & {}& {}& {}& {}& \dfrac{{ - {x_2}}}{4}& {} \\ {} & {} & \dfrac{{1 - y}}{8} & {}& {}& \dfrac{{ - {x_3}}}{4} & {} & {}\\ {} & {} & {} &\dfrac{{1 + y}}{8}& \dfrac{{ - {x_4}}}{4}& {}& {}& {}\\ {} & {} & {} & \dfrac{{ - {x_4}}}{4}&\dfrac{{1 + y}}{8}& {}& {}& {} \\ {} & {} & \dfrac{{ - {x_3}}}{4} & {}& {}& \dfrac{{1 - y}}{8} & {} & {} \\ {}& \dfrac{{ - {x_2}}}{4}& {}& {}& {}& {}& \dfrac{{ 1 + x}}{8}& {} \\ \dfrac{{ - {x_1}}}{4}& {}& {}& {}& {}& {}& {}& \dfrac{{1 - x}}{8} \end{array} \right) \end{array} \\ &\quad \begin{array}{*{20}{l}} \begin{array}{*{20}{l}}\qquad\qquad 000 &\quad 001 &\quad 110 &\quad 111 &\quad 010 & \quad 011 &\quad 100 &\quad 101\end{array} \\ = \left( \begin{array}{*{20}{l}} \dfrac{{1 - x}}{8} & {} & {} & \dfrac{{ - {x_1}}}{4} & {} & {} & {} & {}\\ {} & \dfrac{{1 + x}}{8} & \dfrac{{ - {x_2}}}{4} & {} & {} & {} & {} & {} \\ {} & \dfrac{{ - {x_2}}}{4} & \dfrac{{1 + x}}{8} & {} & {} & {} & {} & {} \\ \dfrac{{ - {x_1}}}{4} & {} & {} & \dfrac{{1 - x}}{8} & {} & {} & {} & {} \\ {} & {} & {} & {} & \dfrac{{1 - y}}{8} & {} & {} & \dfrac{{ - {x_3}}}{4}\\ {} & {} & {} & {} & {} & \dfrac{{1 + y}}{8} & \dfrac{{ - {x_4}}}{4} & {}\\ {} & {} & {} & {} & {} & \dfrac{{ - {x_4}}}{4} & \dfrac{{1 + y}}{8} & {} \\ {} & {} & {} & {} & \dfrac{{ - {x_3}}}{4} & {} & {} & \dfrac{{1 - y}}{8} \end{array} \right)\\ \end{array} .\end{split} \tag{B2}$

      $ \begin{split}&{{\boldsymbol{\rho}} }_{1}={{\boldsymbol{\rho}} }_{0}^{\prime}\otimes {{\boldsymbol{\rho}} }_{{y}_{1}}^{\prime\prime}=\begin{array}{c}\begin{array}{cc}0& 1\end{array}\\ \left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\end{array}\otimes \begin{array}{c}\begin{array}{cc}00& 01\end{array}\\ \left(\begin{array}{cc}\dfrac{1-x}{2}& \dfrac{{z}_{1}}{2}\\ \dfrac{{z}_{1}}{2}& \dfrac{1+x}{2}\end{array}\right)\end{array}=\begin{array}{c}\begin{array}{cc}000& 001\end{array}\\ \left(\begin{array}{cc}\dfrac{1-x}{2}& \dfrac{{y}_{1}}{2}\\ \dfrac{{y}_{1}}{2}& \dfrac{1+x}{2}\end{array}\right)\end{array} \text{, } \\ &{{\boldsymbol{\rho}} }_{2}={{\boldsymbol{\rho}} }_{0}^{\prime}\otimes {{\boldsymbol{\rho}} }_{{y}_{1}}^{\prime\prime}=\begin{array}{c}\begin{array}{cc}0& 1\end{array}\\ \left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\end{array}\otimes \begin{array}{c}\begin{array}{cc}10& 11\end{array}\\ \left(\begin{array}{cc}\dfrac{1-y}{2}& \dfrac{{z}_{2}}{2}\\ \dfrac{{z}_{2}}{2}& \dfrac{1+y}{2}\end{array}\right)\end{array}=\begin{array}{c}\begin{array}{cc}010& 011\end{array}\\ \left(\begin{array}{cc}\dfrac{1-y}{2}& \dfrac{{y}_{1}}{2}\\ \dfrac{{y}_{1}}{2}& \dfrac{1+y}{2}\end{array}\right)\end{array}\text{, } \\ &{{\boldsymbol{\rho}} }_{3}={{\boldsymbol{\rho}} }_{0}^{\prime}\otimes {{\boldsymbol{\rho}} }_{{y}_{1}}^{\prime\prime}=\begin{array}{c}\begin{array}{cc}0& 1\end{array}\\ \left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)\end{array}\otimes \begin{array}{c}\begin{array}{cc}00& 01\end{array}\\ \left(\begin{array}{cc}\dfrac{1+y}{2}& \dfrac{{z}_{3}}{2}\\ \dfrac{{z}_{3}}{2}& \dfrac{1-y}{2}\end{array}\right)\end{array}=\begin{array}{c}\begin{array}{cc}100& 101\end{array}\\ \left(\begin{array}{cc}\dfrac{1+y}{2}& \dfrac{{y}_{1}}{2}\\ \dfrac{{y}_{1}}{2}& \dfrac{1-y}{2}\end{array}\right)\end{array}\text{, } \\ &{{\boldsymbol{\rho}} _4} = \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 0&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 1} \end{array}} \\ {\left( {\begin{array}{*{20}{c}} 0&0 \\ 0&1 \end{array}} \right)} \end{array} \otimes = \begin{array}{*{20}{c}} {10{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 11} \\ {\left( {\begin{array}{*{20}{c}} {\dfrac{{1 + x}}{2}}&{\dfrac{{{z_4}}}{2}} \\ {\dfrac{{{z_4}}}{2}}&{\dfrac{{1 - x}}{4}} \end{array} = } \right)} \end{array}\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{\kern 1pt} {\kern 1pt} 110}&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 111} \end{array}} \\ {\left( {\begin{array}{*{20}{c}} {\dfrac{{1 + x}}{2}}&{\dfrac{{{y_1}}}{2}} \\ {\dfrac{{{y_1}}}{2}}&{\dfrac{{1 - x}}{2}} \end{array}} \right)} \end{array}. \end{split}\tag{B3}$

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