如何理解施密特(Schmidt)分解定理

• 原文: https://zhuanlan.zhihu.com/p/86137107
• 发布日期: 2019-10-13
• 分类: 量子力学基础与量子信息

复合空间的施密特分解是一个比较不直观的过程, 虽然证明起来并不复杂, 但初学者可能会想不通怎么高维度的空间能用低于维度数的基矢分解, 这是本文讨论的一个重点.

初学者可能还会问到: "为什么看证明过程好像随便求和就能自动产生正交的矢量呢? 难道施密特分解这个操作没有运算量吗? "这也是本文讨论的一个重点.

如果教材用的是西安电子科技大学出版的那本量子信息的话可能会对证明过程产生如下疑问:

(2.156)式后的一句话中先指出 ${\left|C_m\right|}^2=\sum _n{\left|C_{mn}\right|}^2$ , 又说总可以选取系数 $C_m$ 使态归一. 这个态指的是什么态? 是 $C_m|u_m^2⟩$ 还是 $|u_m^2⟩$ ？ $C_m$究竟是 ${\left|C_m\right|}^2=\sum _n{\left|C_{mn}\right|}^2$ 确定的值还是仅仅是一个归一化系数? 如果 $C_m$ 是选取的一个归一化系数, 那仅顾2系统的去归一化会不会打乱1系统的状态信息? 在(2.160)式处又指出可以求得 {C}_{m}=\sqrt{\lambda }_{m}{e}^{i{\alpha }_{m} 这和前面确定的 $C_m$ 会不会有冲突?

这又是本文讨论的一个重点, 前面我会给出一个更直观一些的证明, 然后在最后会附上该教科书书上的证明方式, 并非照搬而是会在修改一些用词同时通过引用块的形式在中间解答这些疑惑. 当然没看过这本书的就不用管这部分了.

定理的证明如果用矩阵的奇异值分解定理的话会简单许多, 可以自行了解这里暂不做介绍.

Index :

I. 施密特(Schmidt)分解定理的内容

II. 施密特(Schmidt)分解定理的证明

III. 怎么高维度的空间的矢量能用低于维度数的基矢分解

IV. 这个分解过程真的没有运算量吗? 正交性是如何产生的

V. 如何理解西电出版的教科书上的证明

I. 施密特(Schmidt)分解定理的内容:

将一个复合系统分成 $A,B$ 两个子空间, 则复合系统的态矢可以用这两个子空间的基矢展开.

即 \left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle \left| {B}_{n} \right\rangle } .

其中 $\left\{|A_m⟩\right\},\left\{|B_n⟩\right\}$ 分别为 $A,B$ 空间的任意一组正交归一基.

而施密特分解定理指出有更简单的一般形式:

即 $|\psi ⟩=\sum _n{\sqrt{\lambda }}_n|a_n⟩|b_n⟩$.

$\left\{|a_n⟩\right\},\left\{|b_n⟩\right\}$ 分别为 $A,B$ 空间的密度算符 ${\rho }_A$ 与 ${\rho }_B$ 的本征矢, 都对应本征值 $\left\{{\lambda }_n\right\}$. 需要指出的是 $A,B$ 空间维度不必相同, 而求和指标 $n$ 的最大值为$A,B$ 中较小空间的维数.

这种形式被称作复合系统的施密特极化形式, 这个操作被称作施密特分解.

II. 施密特(Schmidt)分解定理的证明:

复合系统 $S$ 分成 $A,B$ 两个子系统, 记$\left\{|A_m⟩\right\},\left\{|B_n⟩\right\}$ 分别为 $A,B$ 空间的任意一组正交归一基. 则 $S$ 空间中任意纯态可以表示为 \left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle \left| {B}_{n} \right\rangle }. 下面不妨设 $B$ 空间的维度比较小.

设 $B$ 系统约化密度算符 ${\rho }_B$ 的本征方程为 ${\rho }_B|b_n⟩={\lambda }_n|b_n⟩$.

若记 \sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle }=\left| {a}'}_{n} \right\rangle, 则复合系统纯态可做如下分解:\left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle }\left| {b}_{n} \right\rangle =\sum\limits_{n}{\left( \sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle } \right)}\left| {b}_{n} \right\rangle =\sum\limits_{n}{\left| {a}'}_{n} \right\rangle }\left| {b}_{n} \right\rangle

给上式举一个 $3$ 维 $A$ 空间与 $2$ 维 $B$ 空间的直观例子: \begin{align} & \left| \psi \right\rangle =\sum\limits_{m=1}^{3}{\sum\limits_{n=1}^{2}{c}_{mn}\left| {A}_{m} \right\rangle \left| {b}_{n} \right\rangle } \\ & \ \ \ \ \ ={c}_{11}\left| {A}_{1} \right\rangle \left| {b}_{1} \right\rangle +{c}_{12}\left| {A}_{1} \right\rangle \left| {b}_{2} \right\rangle +{c}_{21}\left| {A}_{2} \right\rangle \left| {b}_{1} \right\rangle +{c}_{22}\left| {A}_{2} \right\rangle \left| {b}_{2} \right\rangle +{c}_{31}\left| {A}_{3} \right\rangle \left| {b}_{1} \right\rangle +{c}_{32}\left| {A}_{3} \right\rangle \left| {b}_{2} \right\rangle \\ & \ \ \ \ \ =\left( {c}_{11}\left| {A}_{1} \right\rangle +{c}_{21}\left| {A}_{2} \right\rangle +{c}_{31}\left| {A}_{3} \right\rangle \right)\left| {b}_{1} \right\rangle +\left( {c}_{12}\left| {A}_{1} \right\rangle +{c}_{22}\left| {A}_{2} \right\rangle +{c}_{32}\left| {A}_{3} \right\rangle \right)\left| {b}_{2} \right\rangle \\ & \ \ \ \ \ =\left| {a}'}_{1} \right\rangle \left| {b}_{1} \right\rangle +\left| {a}'}_{2} \right\rangle \left| {b}_{2} \right\rangle \\ & \ \ \ \ \ =\sum\limits_{n=1}^{2}{\left| {a}'}_{n} \right\rangle \left| {b}_{n} \right\rangle } \\ \end{align}

那么现在得到的式子 \left| \psi \right\rangle =\sum\limits_{n}{\left| {a}'}_{n} \right\rangle }\left| {b}_{n} \right\rangle 中的 \left\{ \left| {a}'}_{n} \right\rangle \right\} 是不是一个正交集呢?

\left| {a}'}_{n} \right\rangle 的表达式已由等式 \sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle }=\left| {a}'}_{n} \right\rangle 给出.

现求内积验证是否正交:

\left\langle {a}'}_{n}'} | {a}'}_{n} \right\rangle =\sum\limits_{m}'}{c_{m}'{n}'}^{*}\left\langle {A}_{m}'} \right|}\sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle }=\sum\limits_{m}{c_{m{n}'}^{*}{c}_{mn}

其中 $c_{mn}$ 的表达式为: \left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle }\left| {b}_{n} \right\rangle \Rightarrow {c}_{mn}=\left\langle {A}_{m} \right|\left\langle {b}_{n} | \psi \right\rangle .

将 $c_{mn}$ 的表达式代入内积表达式:

\begin{align} & \left\langle {a}'}_{n}'} | {a}'}_{n} \right\rangle =\sum\limits_{m}{c_{m{n}'}^{*}{c}_{mn}=\sum\limits_{m}{\left\langle \psi | {A}_{m} \right\rangle \left| {b}_{n}'} \right\rangle \left\langle {A}_{m} \right|\left\langle {b}_{n} | \psi \right\rangle } \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left\langle {b}_{n} \right|\left( \sum\limits_{m}{\left\langle {A}_{m} | \psi \right\rangle \left\langle \psi | {A}_{m} \right\rangle } \right)\left| {b}_{n}'} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left\langle {b}_{n} \right|t{r}_{A}\left( {\rho }_{AB} \right)\left| {b}_{n}'} \right\rangle =\left\langle {b}_{n} \right|{\rho }_{B}\left| {b}_{n}'} \right\rangle ={\lambda }_{n}{\delta }_{n{n}'} \\ \end{align}

别忘了前面假设的 ${\rho }_B|b_n⟩={\lambda }_n|b_n⟩$ , 即 $|b_n⟩$ 是约化密度算符 ${\rho }_B$ 的本征矢.

至此我们发现光就是通过等式 \sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle }=\left| {a}'}_{n} \right\rangle 定义的\left\{ \left| {a}'}_{n} \right\rangle \right\} 就是一个正交集.

且不难发现 \left| {a}'}_{n} \right\rangle 模长 \left| \left| {a}'}_{n} \right\rangle \right|=\sqrt{\lambda }_{n} . 那么由 \left| {a}_{n} \right\rangle =\frac{\left| {a}'}_{n} \right\rangle }{\sqrt{\lambda }_{n} 构成的集合 $\left\{|a_n⟩\right\}$ 就是正交归一集.

综上所述:

$|\psi ⟩=\sum _n{\sqrt{\lambda }}_n|a_n⟩|b_n⟩$ , 其中 $\left\{|a_n⟩\right\},\left\{|b_n⟩\right\}$ 均为正交归一基, 且有 ${\rho }_B|b_n⟩={\lambda }_n|b_n⟩$.

接下来就是要证明 ${\lambda }_n$ 同时也是 $A$ 系统约化密度算符 ${\rho }_A$ 的本征矢:

$A$ 系统约化密度算符 ${\rho }_A=t{r}_B\left({\rho }_{AB}\right)=\sum _n⟨b_n|\psi ⟩⟨\psi |b_n⟩$ 其中 $⟨b_n|\psi ⟩=⟨b_n|\sum _m{\sqrt{\lambda }}_m|a_m⟩|b_m⟩={\sqrt{\lambda }}_n|a_n⟩$ . 所以显然有 {\rho }_{A}=t{r}_{B}\left( {\rho }_{AB} \right)=\sum\limits_{n}{\lambda }_{n}\left| {a}_{n} \right\rangle \left\langle {a}_{n} \right|} , 也即存在关系 ${\rho }_A|a_n⟩={\lambda }_n|a_n⟩$ .

综上所述:

将一个复合系统分成 $A,B$ 两个子空间, 则复合系统的态矢可以用这两个子空间的基矢展开.

这个展开具有一般形式: $|\psi ⟩=\sum _n{\sqrt{\lambda }}_n|a_n⟩|b_n⟩$

其中 $\left\{|a_n⟩\right\},\left\{|b_n⟩\right\}$ 分别为 $A,B$ 空间的密度算符 ${\rho }_A$ 与 ${\rho }_B$ 的本征矢, 都对应本征值 $\left\{{\lambda }_n\right\}$.

这也就是施密特分解定理.

III. 怎么高维度的空间的矢量能用低于维度数的基矢分解：

Exp : $3$ 维 $A$ 系统 $\left\{|A_1⟩,|A_2⟩,|A_3⟩\right\}$ 与 $1$ 维 $B$ 系统 $\left\{|B⟩\right\}$ , 这是一个比较极端的例子.

先设 $B$ 系统的本征方程 ${\rho }_B|b⟩=\lambda |b⟩$ , 但 $B$ 空间说实话就只能存在一种态...

复合系统任何一个态矢都可以表达成 $|\psi ⟩=c_1|A_1⟩|b⟩+c_2|A_2⟩|b⟩+c_3|A_3⟩|b⟩$

再具体一点:

\left| \psi \right\rangle =\frac{1}{\sqrt{7}\left| {A}_{1} \right\rangle \left| b \right\rangle +\frac{\sqrt{2}{\sqrt{7}\left| {A}_{2} \right\rangle \left| b \right\rangle +\frac{2}{\sqrt{7}\left| {A}_{3} \right\rangle \left| b \right\rangle =\left( \frac{1}{\sqrt{7}\left| {A}_{1} \right\rangle +\frac{\sqrt{2}{\sqrt{7}\left| {A}_{2} \right\rangle +\frac{2}{\sqrt{7}\left| {A}_{3} \right\rangle \right)\left| b \right\rangle

如同前面说的, 记 \left| a \right\rangle =\frac{1}{\sqrt{7}\left| {A}_{1} \right\rangle +\frac{\sqrt{2}{\sqrt{7}\left| {A}_{2} \right\rangle +\frac{2}{\sqrt{7}\left| {A}_{3} \right\rangle

而这本身就是归一的了, 从前面的推到来看就是说 ${\rho }_A,{\rho }_B$ 的共同本征值是 $1$ , 这个很容易验证.

就是说 \left| \psi \right\rangle =\frac{1}{\sqrt{7}\left| {A}_{1} \right\rangle \left| b \right\rangle +\frac{\sqrt{2}{\sqrt{7}\left| {A}_{2} \right\rangle \left| b \right\rangle +\frac{2}{\sqrt{7}\left| {A}_{3} \right\rangle \left| b \right\rangle 分解为 $|\psi ⟩=|a⟩|b⟩$ .

在这个具体例子中这一切是显然的, $B$ 系统当然一定处于 $|b⟩$ 态；至于 $A$ 系统, 纵使你空间有一万维我们也能把你任意处于的那个态作为你密度算符的本征态. 在这里是 $3$ 维的情况, 那厄米算符 ${\rho }_A$ 的本征矢应该是完备的啊, 起码三个吧？这是不难理解的, 密度算符也是算符, 而算符的基矢是任意选取的, 这里只是选取了特别合适的一组正交基矢 $\{|a⟩,|a_2⟩,|a_3⟩\}$ 使得密度算符表达为: ${\rho }_A=|a⟩⟨a|+0|a_2⟩⟨a_2|+0|a_3⟩⟨a_3|$ .

IV. 这个分解过程真的没有运算量吗? 正交性是如何产生的:

从证明过程来看我们似乎什么都没做, 就是合并了一下同类项就得到了正交的 \left\{ \left| {a}'}_{1} \right\rangle \right\} 是不是太神奇了点? 施密特正交化的过程真的没有运算量吗? 让我们来观察一个例子:

Exp : $3$ 维 $A$ 系统 $\left\{|A_1⟩,|A_2⟩,|A_3⟩\right\}\$与$2$维$B$系统$\left{ \left| {B}{1} \right\rangle ,\left| {B} \right\rangle \right}$ , 看正交性如何产生.

先设 $B$ 系统的本征方程 ${\rho }_B|b⟩=\lambda |b⟩$.

\begin{align} & \left| \psi \right\rangle =\frac{1}{\sqrt{17}\left| {A}_{1} \right\rangle \left| {b}_{1} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{2} \right\rangle \left| {b}_{1} \right\rangle +\frac{\sqrt{3}{\sqrt{17}\left| {A}_{3} \right\rangle \left| {b}_{1} \right\rangle +\frac{2}{\sqrt{17}\left| {A}_{1} \right\rangle \left| {b}_{2} \right\rangle +\frac{\sqrt{5}{\sqrt{17}\left| {A}_{2} \right\rangle \left| {b}_{2} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{3} \right\rangle \left| {b}_{2} \right\rangle \\ & \ \ \ \ \ =\left( \frac{1}{\sqrt{17}\left| {A}_{1} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{2} \right\rangle +\frac{\sqrt{3}{\sqrt{17}\left| {A}_{3} \right\rangle \right)\left| {b}_{1} \right\rangle +\left( \frac{2}{\sqrt{17}\left| {A}_{1} \right\rangle +\frac{\sqrt{5}{\sqrt{17}\left| {A}_{2} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{3} \right\rangle \right)\left| {b}_{2} \right\rangle \\ \end{align}

记

\left| {a}'}_{1} \right\rangle =\frac{1}{\sqrt{17}\left| {A}_{1} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{2} \right\rangle +\frac{\sqrt{3}{\sqrt{17}\left| {A}_{3} \right\rangle \ \ \ \ \ \left| {a}'}_{2} \right\rangle =\frac{2}{\sqrt{17}\left| {A}_{1} \right\rangle +\frac{\sqrt{5}{\sqrt{17}\left| {A}_{2} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{3} \right\rangle

则得到 \left| \psi \right\rangle =\left| {a}'}_{1} \right\rangle \left| {b}_{1} \right\rangle +\left| {a}'}_{2} \right\rangle \left| {b}_{2} \right\rangle

现在求内积来证明这样得到的 \left\{ \left| {a}'}_{1} \right\rangle \right\} 是正交的:

\left\langle {a}'}_{1} | {a}'}_{2} \right\rangle =\left( \frac{1}{\sqrt{17}\left\langle {A}_{1} \right|+\frac{\sqrt{2}{\sqrt{17}\left\langle {A}_{2} \right|+\frac{\sqrt{3}{\sqrt{17}\left\langle {A}_{\text{3} \right| \right)\left( \frac{2}{\sqrt{17}\left| {A}_{1} \right\rangle +\frac{\sqrt{5}{\sqrt{17}\left| {A}_{2} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{3} \right\rangle \right)=\frac{2+\sqrt{10}+\sqrt{6}{17}

不正交??? 这是为什么呢? 前面证明定理本身的过程中是这样的:

现求内积 \left\langle {a}'}_{n}'} | {a}'}_{n} \right\rangle =\sum\limits_{m}'}{c_{m}'{n}'}^{*}\left\langle {A}_{m}'} \right|}\sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle }=\sum\limits_{m}{c_{m{n}'}^{*}{c}_{mn} 求 $c_{mn}$ 的表达式: \left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle }\left| {b}_{n} \right\rangle \Rightarrow {c}_{mn}=\left\langle {A}_{m} \right|\left\langle {b}_{n} | \psi \right\rangle , 再代入上式: \begin{align} & \left\langle {a}'}_{n}'} | {a}'}_{n} \right\rangle =\sum\limits_{m}{c_{m{n}'}^{*}{c}_{mn}=\sum\limits_{m}{\left\langle \psi | {A}_{m} \right\rangle \left| {b}_{n}'} \right\rangle \left\langle {A}_{m} \right|\left\langle {b}_{n} | \psi \right\rangle } \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left\langle {b}_{n} \right|\left( \sum\limits_{m}{\left\langle {A}_{m} | \psi \right\rangle \left\langle \psi | {A}_{m} \right\rangle } \right)\left| {b}_{n}'} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left\langle {b}_{n} \right|t{r}_{A}\left( {\rho }_{AB} \right)\left| {b}_{n}'} \right\rangle =\left\langle {b}_{n} \right|{\rho }_{B}\left| {b}_{n}'} \right\rangle ={\lambda }_{n}{\delta }_{n{n}'} \\ \end{align}

这里\left\langle {a}'}_{1} | {a}'}_{2} \right\rangle =\frac{1}{\sqrt{17}\frac{2}{\sqrt{17}+\frac{\sqrt{2}{\sqrt{17}\frac{\sqrt{5}{\sqrt{17}+\frac{\sqrt{3}{\sqrt{17}\frac{\sqrt{2}{\sqrt{17} 和前面的 \left\langle {a}'}_{n}'}|{a}'}_{n} \right\rangle =\sum\limits_{m}{c_{m{n}'}^{*}{c}_{mn}是一致的.

那为什么前面发展到了 ${\lambda }_n{\delta }_{n{n}^{\prime }}$ 而这个例子却得到了一个实数呢? 我们来一步步分析看看:

\begin{align} & \left\langle {a}'}_{1} | {a}'}_{2} \right\rangle =\frac{1}{\sqrt{17}\frac{2}{\sqrt{17}+\frac{\sqrt{2}{\sqrt{17}\frac{\sqrt{5}{\sqrt{17}+\frac{\sqrt{3}{\sqrt{17}\frac{\sqrt{2}{\sqrt{17} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left\langle \psi | {A}_{1} \right\rangle \left| {b}_{1} \right\rangle \left\langle {A}_{1} \right|\left\langle {b}_{2} | \psi \right\rangle +\left\langle \psi | {A}_{2} \right\rangle \left| {b}_{1} \right\rangle \left\langle {A}_{2} \right|\left\langle {b}_{2} | \psi \right\rangle +\left\langle \psi | {A}_{3} \right\rangle \left| {b}_{1} \right\rangle \left\langle {A}_{3} \right|\left\langle {b}_{2} | \psi \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left\langle {b}_{2} \right|\left( \left\langle {A}_{1} | \psi \right\rangle \left\langle \psi | {A}_{1} \right\rangle \right)\left| {b}_{1} \right\rangle +\left\langle {b}_{2} \right|\left( \left\langle {A}_{2} | \psi \right\rangle \left\langle \psi | {A}_{2} \right\rangle \right)\left| {b}_{1} \right\rangle +\left\langle {b}_{2} \right|\left( \left\langle {A}_{3} | \psi \right\rangle \left\langle \psi | {A}_{3} \right\rangle \right)\left| {b}_{1} \right\rangle \\ \end{align} 将 \left| \psi \right\rangle =\left| {a}'}_{1} \right\rangle \left| {b}_{1} \right\rangle +\left| {a}'}_{2} \right\rangle \left| {b}_{2} \right\rangle 与 \left\{ \begin{align} & \left| {a}'}_{1} \right\rangle =\frac{1}{\sqrt{17}\left| {A}_{1} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{2} \right\rangle +\frac{\sqrt{3}{\sqrt{17}\left| {A}_{3} \right\rangle \\ & \left| {a}'}_{2} \right\rangle =\frac{2}{\sqrt{17}\left| {A}_{1} \right\rangle +\frac{\sqrt{5}{\sqrt{17}\left| {A}_{2} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{3} \right\rangle \\ \end{align} \right. 代入上式得到: \begin{align} & \left\langle {a}'}_{1} | {a}'}_{2} \right\rangle =\left\langle {b}_{2} \right|\left( \frac{1}{\sqrt{17}\left| {b}_{1} \right\rangle +\frac{2}{\sqrt{17}\left| {b}_{2} \right\rangle \right)\left( \frac{1}{\sqrt{17}\left\langle {b}_{1} \right|+\frac{2}{\sqrt{17}\left\langle {b}_{2} \right| \right)\left| {b}_{1} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\left\langle {b}_{2} \right|\left( \frac{\sqrt{2}{\sqrt{17}\left| {b}_{1} \right\rangle +\frac{\sqrt{5}{\sqrt{17}\left| {b}_{2} \right\rangle \right)\left( \frac{\sqrt{2}{\sqrt{17}\left\langle {b}_{1} \right|+\frac{\sqrt{5}{\sqrt{17}\left\langle {b}_{2} \right| \right)\left| {b}_{1} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\left\langle {b}_{2} \right|\left( \frac{\sqrt{3}{\sqrt{17}\left| {b}_{1} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {b}_{2} \right\rangle \right)\left( \frac{\sqrt{3}{\sqrt{17}\left\langle {b}_{1} \right|+\frac{\sqrt{2}{\sqrt{17}\left\langle {b}_{2} \right| \right)\left| {b}_{1} \right\rangle \\ \end{align} 这个结果对应着前面的 \left\langle {a}'}_{n}'}|{a}'}_{n} \right\rangle =\langle {b}_{n}|{\rho }_{B}\left| {b}_{n}'} \right\rangle ={\lambda }_{n}{\delta }_{n{n}'}但我们这里 ${\rho }_{B}\left| {b}_{1} \right\rangle =\left[ \begin{align} & \ \ \ \ \left( \frac{1}{\sqrt{17}\left| {b}_{1} \right\rangle +\frac{2}{\sqrt{17}\left| {b}_{2} \right\rangle \right)\left( \frac{1}{\sqrt{17}\langle {b}_{1}|+\frac{2}{\sqrt{17}\langle {b}_{2}| \right) \\ & +\left( \frac{\sqrt{2}{\sqrt{17}\left| {b}_{1} \right\rangle +\frac{\sqrt{5}{\sqrt{17}\left| {b}_{2} \right\rangle \right)\left( \frac{\sqrt{2}{\sqrt{17}\langle {b}_{1}|+\frac{\sqrt{5}{\sqrt{17}\langle {b}_{2}| \right) \\ & +\left( \frac{\sqrt{3}{\sqrt{17}\left| {b}_{1} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {b}_{2} \right\rangle \right)\left( \frac{\sqrt{3}{\sqrt{17}\langle {b}_{1}|+\frac{\sqrt{2}{\sqrt{17}\langle {b}_{2}| \right) \\ \end{align} \right]\left| {b}_{1} \right\rangle$ $\begin{align} & \Rightarrow {\rho }_{B}\left| {b}_{1} \right\rangle =\left[ \frac{6}{17}\left| {b}_{1} \right\rangle \langle {b}_{1}|+\frac{2+\sqrt{10}+\sqrt{6}{17}\left| {b}_{1} \right\rangle \langle {b}_{2}|+\frac{2+\sqrt{10}+\sqrt{6}{17}\left| {b}_{2} \right\rangle \langle {b}_{1}|+\frac{11}{17}\left| {b}_{2} \right\rangle \langle {b}_{2}| \right]\left| {b}_{1} \right\rangle \\ & \Rightarrow {\rho }_{B}\left| {b}_{1} \right\rangle =\frac{6}{17}\left| {b}_{1} \right\rangle ++\frac{2+\sqrt{10}+\sqrt{6}{17}\left| {b}_{2} \right\rangle \ne {\lambda }_{1}\left| {b}_{1} \right\rangle \\ \end{align}$ 这说明什么呢? 这说明前面的 $\left\{|b_n⟩\right\}$ 根本就不是 ${\rho }_B$ 的本征矢! 你不能说他是他就是了. 不过在一般性的证明的过程中我们总可以假设已经找到 ${\rho }_B$ 的本征矢$\left\{|b_n⟩\right\}$ 然后把 $|\psi ⟩$ 表示为 \left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle }\left| {b}_{n} \right\rangle .

所以当题目给定一个待分解的矢量 $|\psi ⟩$ 比如说:

\left| \psi \right\rangle =\frac{1}{\sqrt{17}\left| {A}_{1} \right\rangle \left| {B}_{1} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{2} \right\rangle \left| {B}_{1} \right\rangle +\frac{\sqrt{3}{\sqrt{17}\left| {A}_{3} \right\rangle \left| {B}_{1} \right\rangle +\frac{2}{\sqrt{17}\left| {A}_{1} \right\rangle \left| {B}_{2} \right\rangle +\frac{\sqrt{5}{\sqrt{17}\left| {A}_{2} \right\rangle \left| {B}_{2} \right\rangle +\frac{\sqrt{2}{\sqrt{17}\left| {A}_{3} \right\rangle \left| {B}_{2} \right\rangle 的时候

我们要做的第一件事就是找到 ${\rho }_B$ 的本征矢$\left\{|b_n⟩\right\}$

再将其改写为\left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle }\left| {b}_{n} \right\rangle的形式.

而这就是施密特分解的运算量所在之处.

下面我们仍然以此为例来分解一下试试?

{\rho }_{B}=\frac{6}{17}\left| {B}_{1} \right\rangle \langle {B}_{1}|+\frac{2+\sqrt{10}+\sqrt{6}{17}\left| {B}_{1} \right\rangle \langle {B}_{2}|+\frac{2+\sqrt{10}+\sqrt{6}{17}\left| {B}_{2} \right\rangle \langle {B}_{1}|+\frac{11}{17}\left| {B}_{2} \right\rangle \left\langle {B}_{2} \right|

本征值分别为 \left\{ \begin{align} & {\lambda }_{1}=\frac{1}{34}\left( 17+\sqrt{105+16\sqrt{6}+16\sqrt{10}+16\sqrt{15} \right)\approx 0.97128 \\ & {\lambda }_{2}=\frac{1}{34}\left( 17-\sqrt{105+16\sqrt{6}+16\sqrt{10}+16\sqrt{15} \right)\approx 0.02872 \\ \end{align} \right.

注意这里 ${\lambda }_1+{\lambda }_2=1$ 这非常棒!

对应的本征矢分别为(约等号后进行了归一化操作):

第一次忘了归一化, 全都重来了一遍, 这是血的教训.

\left\{ \begin{align} & \left| {b}_{1} \right\rangle =-\frac{5-\sqrt{105+16\sqrt{6}+16\sqrt{10}+16\sqrt{15}{2\left( 2+\sqrt{6}+\sqrt{10} \right)}\left| {B}_{1} \right\rangle +\left| {B}_{2} \right\rangle \\ & \ \ \ \ \ \approx 0.58650\left| {B}_{1} \right\rangle +0.80995\left| {B}_{2} \right\rangle \\ & \left| {b}_{2} \right\rangle =-\frac{5+\sqrt{105+16\sqrt{6}+16\sqrt{10}+16\sqrt{15}{2\left( 2+\sqrt{6}+\sqrt{10} \right)}\left| {B}_{1} \right\rangle +\left| {B}_{2} \right\rangle \\ & \ \ \ \ \ \approx -0.80995\left| {B}_{1} \right\rangle +0.58650\left| {B}_{2} \right\rangle \\ \end{align} \right.

\Rightarrow \left\{ \begin{align} & \left| {B}_{1} \right\rangle =0.58650\left| {b}_{1} \right\rangle -0.80995\left| {b}_{2} \right\rangle \\ & \left| {B}_{2} \right\rangle =0.80995\left| {b}_{1} \right\rangle +0.58650\left| {b}_{2} \right\rangle \\ \end{align} \right.

代入待分解的态, 整理得到:

\begin{align} & \left| \psi \right\rangle =\left[ \left( 0.58650\frac{1}{\sqrt{17}+0.80995\frac{2}{\sqrt{17} \right)\left| {A}_{1} \right\rangle \left| {b}_{1} \right\rangle +\left( 0.58650\frac{2}{\sqrt{17}-0.80995\frac{1}{\sqrt{17} \right)\left| {A}_{1} \right\rangle \left| {b}_{2} \right\rangle \right] \\ & \ \ \ \ \ \ +\left[ \left( 0.58650\frac{\sqrt{2}{\sqrt{17}+0.80995\frac{\sqrt{5}{\sqrt{17} \right)\left| {A}_{2} \right\rangle \left| {b}_{1} \right\rangle +\left( 0.58650\frac{\sqrt{5}{\sqrt{17}-0.80995\frac{\sqrt{2}{\sqrt{17} \right)\left| {A}_{2} \right\rangle \left| {b}_{2} \right\rangle \right] \\ & \ \ \ \ \ \ +\left[ \left( 0.58650\frac{\sqrt{3}{\sqrt{17}+0.80995\frac{\sqrt{2}{\sqrt{17} \right)\left| {A}_{3} \right\rangle \left| {b}_{1} \right\rangle +\left( 0.58650\frac{\sqrt{2}{\sqrt{17}-0.80995\frac{\sqrt{3}{\sqrt{17} \right)\left| {A}_{3} \right\rangle \left| {b}_{2} \right\rangle \right] \\ \end{align}

进一步化简: $|\psi ⟩\approx \left(0.53513|A_1⟩+0.64042|A_2⟩+0.52419|A_3⟩\right)|b_1⟩+\left(0.08805|A_1⟩+0.04026|A_2⟩-0.13908|A_3⟩\right)|b_2⟩$

记 \left\{ \begin{align} & \left| {a}'}_{1} \right\rangle =0.53513\left| {A}_{1} \right\rangle +0.64042\left| {A}_{2} \right\rangle +0.52419\left| {A}_{3} \right\rangle \\ & \left| {a}'}_{2} \right\rangle =0.08805\left| {A}_{1} \right\rangle +0.04026\left| {A}_{2} \right\rangle -0.13908\left| {A}_{3} \right\rangle \\ \end{align} \right.

验证正交性:

\begin{align} & \left\langle {a}'}_{1}|{a}'}_{2} \right\rangle =0.53513\times 0.08805+0.64042\times 0.04026-0.52419\times 0.13908 \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.0471181965+0.0257833092-0.0729043452 \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.0729015057-0.0729043452\approx -0.0000028\approx 0 \\ \end{align}

验证模长: \left\{ \begin{align} & {\left| \left| {a}'}_{1} \right\rangle \right|}^{2}={0.53513}^{2}+{0.64042}^{2}+{0.52419}^{2}=0.9712770494\approx {\lambda }_{1} \\ & {\left| \left| {a}'}_{2} \right\rangle \right|}^{2}={0.08805}^{2}+{0.04026}^{2}+{0.13908}^{2}=0.0287169165\approx {\lambda }_{2} \\ \end{align} \right.

所以真正奇妙的点是, 为何两个系统的密度算符拥有同样的本征值. 以及为何把 $B$ 系统进行本征普分解之后 $A$ 系统也跟着被分解了.

这是一个极其乐色的例子, 但我仍然莽出了一个漂亮的结果.

V. 如何理解西电出版的教科书上的证明:

为了方便于前面方法对比, 特意将与产生的疑惑无关的部分的形式整理的与前面方法相似.

复合系统 $S$ 分成 $A,B$ 两个子系统, 记$\left\{|A_m⟩\right\},\left\{|B_n⟩\right\}$ 分别为 $A,B$ 空间的任意一组正交归一基. 则 $S$ 空间中任意纯态可以表示为 \left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle \left| {B}_{n} \right\rangle }. 下面不妨设 $B$ 空间的维度比较小.

设 $B$ 系统约化密度算符 ${\rho }_B$ 本征方程为 ${\rho }_B|b_n⟩={\lambda }_n|b_n⟩$, 且记 \sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle }={a}_{n}\left| {a}_{n} \right\rangle.

\left| {a}'}_{n} \right\rangle =\sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle }={a}_{n}\left| {a}_{n} \right\rangle 中 $|a_n⟩$ 是一个单位矢量. 系数 $a_n$ 是总矢量 \left| {a}'}_{n} \right\rangle 的模长. 故书上说的归一化并非指的非恒等变形, 而是单纯的把一个矢量表达为模长乘以方向罢了.

则存在分解 \left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle }\left| {b}_{n} \right\rangle =\sum\limits_{n}{\left( \sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle } \right)}\left| {b}_{n} \right\rangle =\sum\limits_{n}{a}_{n}\left| {a}_{n} \right\rangle }\left| {b}_{n} \right\rangle

\left| \psi \right\rangle =\sum\limits_{n}{a}_{n}\left| {a}_{n} \right\rangle \left| {b}_{n} \right\rangle } 已经很像需要证明的结论了, 接下来就是研究 $\left\{|a_n⟩\right\}$ 是否互相正交：

$|a_n⟩$ 表达式已经给出: \sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle }={a}_{n}\left| {a}_{n} \right\rangle \Rightarrow \left| {a}_{n} \right\rangle =\sum\limits_{m}{\frac{c}_{mn}{a}_{n}\left| {A}_{m} \right\rangle }

求内积

\left\langle {a}_{n}'} | {a}_{n} \right\rangle =\sum\limits_{m}'}{\frac{c_{m}'{n}'}^{*}{a_{n}'}^{*}\left\langle {A}_{m}'} \right|}\sum\limits_{m}{\frac{c}_{mn}{a}_{n}\left| {A}_{m} \right\rangle }=\sum\limits_{m}',m}{\frac{c_{m}'{n}'}^{*}{c}_{mn}{a_{n}'}^{*}{a}_{n}\left\langle {A}_{m}'} | {A}_{m} \right\rangle =\sum\limits_{m}{\frac{c_{m{n}'}^{*}{c}_{mn}{a_{n}'}^{*}{a}_{n}

现在研究 $c_{mn}$, 其表达式不难得出:

\left| \psi \right\rangle =\sum\limits_{m,n}{c}_{mn}\left| {A}_{m} \right\rangle \left| {B}_{n} \right\rangle }\Rightarrow {c}_{mn}=\left\langle {A}_{m} \right|\left\langle {B}_{n} | \psi \right\rangle

\begin{align} & \left\langle {a}_{n}'} | {a}_{n} \right\rangle =\sum\limits_{m}{\frac{c_{m{n}'}^{*}{c}_{mn}{a_{n}'}^{*}{a}_{n}=\sum\limits_{m}{\frac{\left\langle \psi | {b}_{n}'} \right\rangle \left| {A}_{m} \right\rangle \left\langle {A}_{m} \right|\left\langle {b}_{n} | \psi \right\rangle }{a_{n}'}^{*}{a}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\sum\limits_{m}{\left\langle {A}_{m} \right|\left\langle {b}_{n} | \psi \right\rangle \left\langle \psi | {b}_{n}'} \right\rangle \left| {A}_{m} \right\rangle }{a_{n}'}^{*}{a}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\left\langle {b}_{n} \right|\left( \sum\limits_{m}{\left\langle {A}_{m} \right|{\rho }_{AB}\left| {A}_{m} \right\rangle } \right)\left| {b}_{n}'} \right\rangle }{a_{n}'}^{*}{a}_{n}=\frac{\left\langle {b}_{n} \right|{\rho }_{B}\left| {b}_{n}'} \right\rangle }{a_{n}'}^{*}{a}_{n}=\frac{\lambda }_{n}{\delta }_{n{n}'}{a_{n}'}^{*}{a}_{n} \\ \end{align}

显然 {a}_{n}=\sqrt{\lambda }_{n}{e}^{i{\varphi }_{n} (其中 ${\phi }_n$ 为实数)才能保证 $\left\{|a_n⟩\right\}$ 归一, 而相位可以放在后面态矢里.

这里需要指出的是并非 $a_n$ 被式子 {a}_{n}=\sqrt{\lambda }_{n}{e}^{i{\varphi }_{n} 确定了, 而是 ${\lambda }_n$ 被此式确定为 ${\left|a_n\right|}^2$ 而由 \sum\limits_{m}{c}_{mn}\left| {A}_{m} \right\rangle }={a}_{n}\left| {a}_{n} \right\rangle 知 $a_n$ 是被式子 ${\left|a_n\right|}^2=\sum _m{\left|c_{mn}\right|}^2$ 确定了模长.

既有 $|\psi ⟩=\sum _n{\sqrt{\lambda }}_n|a_n⟩|b_n⟩$ , 现在求 $A$ 的约化密度矩阵:

{\rho }_{A}=t{r}_{B}{\rho }_{AB}=\sum\limits_{n}{\left\langle {b}_{n} \right|{\rho }_{AB}\left| {b}_{n} \right\rangle }=\sum\limits_{n}{\left\langle {b}_{n} \right|\sum\limits_{m}',m}{\lambda }_{m}\left| {a}_{m} \right\rangle }\left| {b}_{m} \right\rangle \left\langle {a}_{m}'} \right|\left\langle {b}_{m}'} | {b}_{n} \right\rangle }=\sum\limits_{n}{\lambda }_{n}\left| {a}_{n} \right\rangle \left\langle {a}_{n} \right|}

显然存在关系 ${\rho }_A|a_n⟩={\lambda }_n|a_n⟩$ .

所以证明了存在分解 $|\psi ⟩=\sum _n{\sqrt{\lambda }}_n|a_n⟩|b_n⟩$.

其中 $\left\{|a_n⟩\right\},\left\{|b_n⟩\right\}$ 分别为 $A,B$ 空间的密度算符 ${\rho }_A$ 与 ${\rho }_B$ 的本征矢, 都对应本征值 $\left\{{\lambda }_n\right\}$.