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两耦合谐振子纠缠熵计算.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{geometry}
+\geometry{a4paper, margin=1in}
+
+% 物理宏包测试
+\usepackage{physics}
+\usepackage{siunitx}
+
+\usepackage{enumitem}  % 控制列表样式
+
+% 中文支持（使用 XeLaTeX 编译）
+\usepackage{xeCJK}
+\setCJKmainfont{Noto Serif CJK SC}  % 确保系统有这个字体，或改为 "Noto Sans CJK SC"
+
+\title{小课题}
+
+\date{\today}
+
+\begin{document}
+
+\maketitle
+
+% 定义两个层级格式
+\setlist[enumerate,1]{label=\textbf{\arabic*}, leftmargin=*, itemsep=1em}
+\setlist[enumerate,2]{label=\arabic{enumi}.\arabic*, leftmargin=2em, itemsep=0.5em}
+
+% 使用
+\begin{enumerate}
+    \item 计算系统真空态
+    
+    将原哈密顿量$H=a^{+}a+b^{+}b+\lambda(a^{+}b^{+}+ab)$作对角化处理,引入如下线性变换（为保证产生湮灭算符的对易关系，同时满足$u^{2}-v^{2}=1$）
+    $$
+    \begin{cases} 
+    \alpha = ua + vb^{\dagger} \\ 
+    \beta = ub + va^{\dagger} 
+    \end{cases}
+    $$
+    反解后可得出
+    $$\begin{cases}
+    b=u\beta-v\alpha^{\dagger} \\
+    a=u\alpha-v\beta^{\dagger} & 
+    \end{cases}
+    $$
+    代入原哈密顿量可得出
+    $$\begin{aligned}
+    H & =(u\alpha^{\dagger}-v\beta)(u\alpha-v\beta^{\dagger})+(u\beta^{\dagger}-v\alpha)(u\beta-v\alpha^{\dagger}) \\
+     & +\lambda[(u\alpha^{\dagger}-v\beta)(u\beta^{\dagger}-v\alpha)+(u\alpha-v\beta^{\dagger})(u\beta-v\alpha^{\dagger})] \\
+     & =(u^{2}-\lambda uv)\alpha^{\dagger}\alpha+(v^{2}-\lambda uv)\alpha\alpha^{\dagger}+(u^{2}-\lambda uv)\beta^{\dagger}\beta+(v^{2}-\lambda uv)\beta\beta^{\dagger} \\
+     & +[\lambda(u^{2}+v^{2})-2uv]\alpha^{\dagger}\beta^{\dagger}+[\lambda(u^{2}+v^{2})-2uv]\alpha\beta
+    \end{aligned}$$
+    令$\lambda(u^{2}+v^{2})-2uv=0,u=\cosh r,v=\sinh r$
+
+    可得
+    $$\lambda=\tanh2r$$
+    利用恒等式可解得
+    $$\cosh(2r)=\frac{1}{\sqrt{1-\lambda^{2}}},\sinh(2r)=\frac{\lambda}{\sqrt{1-\lambda^{2}}}$$
+    再利用产生湮灭算符对易关系$[\alpha,\alpha^{+}]=1,[\beta,\beta^{+}]=1$
+
+    原哈密顿量可继续化简为如下
+    $$\begin{aligned}
+    H & =(u^{2}-\lambda w)(\alpha^{\dagger}\alpha+\beta^{\dagger}\beta)+(v^{2}-\lambda uv)(\alpha^{\dagger}\alpha+1+\beta^{\dagger}\beta+1) \\
+     & =\sqrt{1-\lambda^{2}}(\alpha^{\dagger}\alpha+\beta^{\dagger}\beta)+(\sqrt{1-\lambda^{2}}-1)
+    \end{aligned}$$
+
+    设真空态为$|0\rangle=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}C_{mn}|m\rangle_A|n\rangle_B$，由湮灭算符作用到真空态上为0有如下
+    $$\begin{aligned}
+    \alpha| 0 \rangle &= (u a + v b^{\dagger}) \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} C_{mn} |m\rangle_A |n\rangle_B \\
+    &= \sum_{m,n} u C_{mn} \sqrt{m} |m-1\rangle_A |n\rangle_B + \sum_{m,n} v C_{mn} \sqrt{n+1} |m\rangle_A |n+1\rangle_B \\
+    &= \sum_{m,n} u C_{m+1,n} \sqrt{m+1} |m\rangle_A |n\rangle_B + \sum_{m,n} v C_{m,n-1} \sqrt{n} |m\rangle_A |n\rangle_B \\
+    &= \sum_{m,n} \left[ u C_{m+1,n} \sqrt{m+1} + v C_{m,n-1} \sqrt{n} \right] |m\rangle_A |n\rangle_B = 0
+    \end{aligned}$$
+    可得
+    $$u\sqrt{m+1}C_{m+1,n}=-v\sqrt{n}C_{m,n-1}$$
+    同理
+    $$\begin{aligned}
+    \beta|0\rangle &= (u b + v a^+) \sum_{m,n} C_{mn} |m\rangle_A |n\rangle_B \\
+    &= \sum_{m,n} u C_{mn} \sqrt{n} |m\rangle_A |n-1\rangle_B + \sum_{m,n} v C_{mn} \sqrt{m+1} |m+1\rangle_A |n\rangle_B \\
+    &= \sum_{m,n} u C_{m,n+1} \sqrt{n+1} |m\rangle_A |n\rangle_B + \sum_{m,n} v C_{m-1,n} \sqrt{m} |m\rangle_A |n\rangle_B \\
+    &= \sum_{m,n} \left[ u C_{m,n+1} \sqrt{n+1} + v C_{m-1,n} \sqrt{m} \right] |m\rangle_A |n\rangle_B = 0 
+    \end{aligned}$$
+    有
+    $$u\sqrt{n+1}C_{m,n+1}=-v\sqrt{m}C_{m-1,n}$$
+    由上述两个递推关系可知，只有当m=n时才能满足条件，所以真空态应该具有这样的形式$|0\rangle=\sum_{n=0}^{\infty}C_n|n\rangle_A|n\rangle_B$，在递推关系中令m=n-1,
+    可得系数的递推关系$C_{n}=(-\frac{v}{u})^{n}C_{0}$。由真空态归一化有
+    $$\sum_{n=0}^{\infty}|C_{n}|^{2}=\sum_{n=0}^{\infty}|(-\frac{v}{u})^{n}C_{0}|^{2}=|C_{0}|^{2}\sum_{n=0}^{\infty}(\frac{v^{2}}{u^{2}})^{n}=|C_{0}|^{2}\frac{u^{2}}{u^{2}-v^{2}}=1$$
+    计算得出$C_{0}=\frac{1}{u}$，代入真空态形式中有
+    \begin{align*}
+    |0\rangle &= \sum_{n=0}^{\infty} C_n |n\rangle_A |n\rangle_B
+    = \sum_{n=0}^{\infty} \left(-\frac{v}{u}\right)^n \frac{1}{u} |n\rangle_A |n\rangle_B \\
+    &= \frac{1}{\cosh r} \sum_{n=0}^{\infty} (-\tanh r)^n |n\rangle_A |n\rangle_B
+    \end{align*}
+
+    \item 证明真空态是纠缠态
+    
+    由1已得到真空态已是标准Schmidt形式
+    $$|0\rangle=\frac{1}{\cosh r}\sum_{n=0}^{\infty}(-\tanh r)^n|n\rangle_A\otimes|n\rangle_B$$
+    利用反证法证明，假设真空态是纯态，则应该能写成$|\phi\rangle_A=\sum_ma_m|m\rangle_A\text{ 和 }|\chi\rangle_B=\sum_nb_n|n\rangle_B$的直积形式，如下
+    $$\sum_{m,n}a_mb_n|m\rangle_A|n\rangle_B=\sum_{n=0}^\infty\frac{(-\tanh r)^n}{\cosh r}|n\rangle_A|n\rangle_B$$
+    对比得到系数应该满足$a_mb_n=c_n\delta_{mn}\text{,其中 }c_n=\frac{(-\tanh r)^n}{\cosh r}$，这意味着$\text{当 }m\neq n\text{ 时,}a_mb_n=0$，以及$\text{对所有 }m=n\mathrm{~,}a_nb_n=c_n\neq0$。
+    由$a_nb_n\neq0$知$a_n\neq0$且$b_n\neq0$对所有$n$。但若取$m=0,n=1$ ,则$a_0b_1=0$ ,$a_0\neq0,b_1\neq0$矛盾。因此假设不成立，真空态应为纠缠态。
+
+
+
+
+
+
+    \item 计算基态纠缠熵
+    \begin{enumerate}
+        \item 子系统A的约化密度矩阵
+        
+        由定义可得$\rho_{A}=\mathrm{tr}_{B}\rho=\sum_{m=0}^{\infty}{}_{B}\langle m|0\rangle\langle0|m\rangle_{B}$，其中
+        \begin{align*}
+        {}_B\langle m|0\rangle &= {}_B\langle m|\frac{1}{\cosh r}\sum_{n=0}^{\infty}(-\tanh r)^n|n\rangle_A|n\rangle_B \\
+        &= \frac{1}{\cosh r}\sum_{n=0}^{\infty}(-\tanh r)^n|n\rangle_A \, {}_B\langle m|n\rangle_B \\
+        &= \frac{(-\tanh r)^m}{\cosh r}|m\rangle_A
+        \end{align*}
+        则容易得到
+        $$\rho_{A}=\sum_{m=0}^{\infty}\frac{\tanh^{2m}r}{\cosh^{2}r}|m\rangle_{A}{}_A\langle m|$$
+
+        \item 子系统B的约化密度矩阵
+        
+        与A的计算同理
+        $$\rho_B=\mathrm{tr}_A\rho=\sum_{m=0}^{\infty}{}_A\langle m|0\rangle\langle 0|m\rangle_A=\sum_{m=0}^{\infty}\frac{\tanh^2m}{\cosh^2\gamma}|m\rangle_{B}{}_B\langle m|$$
+
+        \item 子系统A的纠缠熵
+        
+        由定义知$S_{A}=-\mathrm{tr}_{A}(\rho_{A}\log\rho_{A})$
+
+        记
+        $$\begin{aligned}
+        \rho_{A} &= \sum_{m=0}^{\infty} \frac{\tanh^{2m} r}{\cosh^{2} r} |m\rangle_{AA} \langle m| \\
+        &= \sum_{m=0}^{\infty} P_{m} |m\rangle_{AA} \langle m|
+        \end{aligned}$$
+
+        其中
+        $$\begin{aligned}
+        \rho_{A}\log\rho_{A}&=\left(\sum_{m=0}^{\infty}P_{m}|m\rangle_{A}{}_A\langle m|\right)\left(\sum_{n=0}^{\infty}\log p_{n}|n\rangle_{A}{}_A\langle n|\right) \\
+        &=\sum_{m,n}P_{m}\log P_{n}|m\rangle_{A}{}_A\langle n| \\
+        &=\sum_{m,n}P_{m}\log P_{n}\delta_{mn}|m\rangle_{A}{}_A\langle m| \\
+        &=\sum_{m}P_{m}\log P_{m}|m\rangle_{A}{}_A\langle m|
+        \end{aligned}$$
+        于是A的纠缠熵可化简为
+        $$\begin{aligned}
+        S_{A} &= - \sum_{k} {}_A \langle k | \rho_{A} \log \rho_{A} | k \rangle_{A} \\
+        &= - \sum_{k} \sum_{m} P_{m} \log P_{m} {}_A\langle k | m \rangle_{A} {}_A\langle m | k \rangle_{A} \\
+        &= - \sum_{m} P_{m} \log P_{m}
+        \end{aligned}$$
+        令$x=\tanh^{2}r$，可计算得出
+        $$\begin{aligned}
+        S_A &= - \sum_m  (1-x) x^m \log \left[ (1-x) x^m \right] \\
+        &= - \sum_m (1-x) x^m \left[ \log (1-x) + m \log x \right] \\
+        &= - \sum_m (1-x) x^m \log (1-x) - \sum_m (1-x) x^m m \log x \\
+        &= - \log (1-x) - \frac{x \log x}{1-x}
+        \end{aligned}$$
+
+
+        \item 子系统B的纠缠熵
+        
+        B的纠缠熵计算过程与A同理
+        $$\begin{aligned}
+        S_{B} & =-tr_{B}(\rho_{B}\log \rho_{B}) \\
+        & =S_{A}
+        \end{aligned}$$
+
+
+    \end{enumerate}
+
+    \item 计算激发态纠缠熵和能量
+    
+    \begin{enumerate}
+        \item 哈密顿量形式为$H=\theta a^{\dagger}a$
+        
+        此时的酉算符为$U_{A}(\theta)=e^{iH}=e^{i\theta a^{\dagger}a}$，作用到真空态得激发态如下
+        $$\begin{aligned}
+        |\psi(\theta)\rangle &= U_A(\theta)|0\rangle = e^{i\theta a^\dagger a} \frac{1}{\cosh r} \sum_{n=0}^{\infty} (-\tanh r)^n |n\rangle_A |n\rangle_B \\
+        &= \frac{1}{\cosh r} \sum_{n=0}^{\infty} (-\tanh r)^n e^{i\theta n} |n\rangle_A |n\rangle_B \\
+        &= \sum_{n=0}^{\infty} \frac{(-\tanh r e^{i\theta})^n}{\cosh r} |n\rangle_A |n\rangle_B 
+        \end{aligned}$$
+        此时A的约化密度矩阵为
+        $$\begin{aligned}
+        \rho_{A} &= \mathrm{tr}_{B} \rho = \sum_{m=0}^{\infty}  {}_B\langle m| \psi(\theta) \rangle \langle \psi(\theta) | m \rangle_{B} \\
+        &= \sum_{m=0}^{\infty} \frac{\tanh^{2m} r}{\cosh^{2} r} |m\rangle_{AA} \langle m|
+        \end{aligned}$$
+        按照定义，A的纠缠熵应为
+        $$S_A(\theta)=-\mathrm{tr}_A(\rho_A\log\rho_A)=S_A$$
+
+        此激发态的能量为$E(\theta)=\langle\psi(\theta)|H|\psi(\theta)\rangle$，$H=a^{\dagger}a+b^{\dagger}b+\lambda(a^{\dagger}b^{\dagger}+ab)$,拆开为四项分别计算
+
+        第一项结果为$\langle\psi(\theta)|a^{\dagger}a|\psi(\theta)\rangle=\frac{x}{1-x}=\sinh^{2}r$
+        详细计算过程如下，其中
+        $$\begin{aligned}
+        a|\psi(\theta)\rangle &= \sum_{n=0}^{\infty} \frac{(-\tanh r\, e^{i\theta})^n}{\cosh r} |n\rangle_A |n\rangle_B \\
+        &= \sum_{n=0}^{\infty} \frac{(-\tanh r\, e^{i\theta})^n}{\cosh r} \sqrt{n} |n-1\rangle_A |n\rangle_B
+        \end{aligned}$$
+        则
+        $$\begin{aligned}
+        \langle \psi(\theta) | a^\dagger a | \psi(\theta) \rangle &= \sum_{m,n}^\infty \sqrt{mn} \frac{(-\tanh r\, e^{-i\theta})^m}{\cosh r} \frac{(-\tanh r\, e^{i\theta})^n}{\cosh r} {}_A\langle m-1|n-1\rangle_A {}_B\langle m|n\rangle_B \\
+        &= \sum_{n=0}^\infty n \frac{\tanh^{2n} r}{\cosh^2 r} \\
+        &= \sum_{n=0}^\infty n(1-x)\,x^n = \frac{x}{1-x} = \sinh^2 r
+        \end{aligned}$$
+
+        第二项计算与第一项同理$\langle\psi(\theta)|b^{\dagger}b|\psi(\theta)\rangle=\sinh^{2}r$
+
+        第三项为
+        $$\begin{aligned}
+        \langle\psi(\theta)|a^{\dagger}b^{\dagger}|\psi(\theta)\rangle&=\sum_{m,n=0}^{\infty}\sqrt{m+1}\sqrt{n+1}\frac{(-\tanh r e^{i\theta})^n (-\tanh r e^{-i\theta})^{m+1}}{\cosh^2 r}{}_A{\langle m|n\rangle_{A}}{}_{B}{\langle m+1|n+1\rangle_B} \\
+        &=\sum_{n=0}^{\infty}(n+1)\frac{\tanh^{2n}r}{\cosh^2r}(-\tanh re^{-i\theta}) \\
+        &=(-\tanh re^{-i\theta})\sum_{n=0}^{\infty}(n+1)\frac{\tanh^{2n}r}{\cosh^2r} \\
+        &=\frac{-\tanh r e^{-i\theta}}{1 - \tanh^2 r}=-\sinh r\cosh re^{-i\theta}
+        \end{aligned}$$
+
+        第四项与第三项类似
+        $$\langle\psi(\theta)|ab|\psi(\theta)\rangle=-\sinh r\cosh re^{i\theta}$$
+        将上述四项合并后得到总的能量，并化简为如下结果
+        $$\begin{aligned}
+        E(\theta) & =2\sinh^{2}r+\lambda[-\sinh r\cosh r(e^{-i\theta}+e^{i\theta})] \\
+        & =2\sinh^{2}r-2\lambda\sinh r\cos\theta r\cos\theta \\
+        & =2\sinh^{2}r-\lambda\sinh(2r)\cos\theta=E_{0}+2\lambda\sinh(2r)\sin^{2}\frac{\theta}{2}
+        \end{aligned}$$
+
+        \item 哈密顿量形式为$H=\theta(a^\dagger b^\dagger + ab)$,已作厄米化处理
+        
+        此时的酉算符作用到真空态上得到激发态，并设激发态形式如下
+        $$\begin{aligned}
+        |\psi(\theta)\rangle &= U_{AB}(\theta)|0\rangle = e^{i\theta(a^\dagger b^\dagger + ab)}|0\rangle \\
+        &= \sum_{n=0}^{\infty} C_n(\theta) |n_A\rangle |n_B\rangle
+        \end{aligned}$$
+        接下来对上式左右两边同时对$\theta$求导可得到系数的微分方程，结合初始条件求得系数从而得到激发态的具体表达式
+        $$\begin{aligned}
+        \frac{d|\psi(\theta)\rangle}{d\theta} &= i(a^\dagger b^\dagger + ab)|\psi(\theta)\rangle \\
+        &= i\sum_{n=0}^{\infty} C_n(\theta)(a^\dagger b^\dagger + ab)|n\rangle_A |n\rangle_B \\
+        &= i\sum_{n=0}^{\infty} C_n(\theta)\left[(n+1)|n+1\rangle_A |n+1\rangle_B + n|n-1\rangle_A |n-1\rangle_B\right] \\
+        &= i\left[\sum_{n=0}^{\infty} C_n(\theta)(n+1)|n+1\rangle_A |n+1\rangle_B + \sum_{n=1}^{\infty} C_n(\theta)n|n-1\rangle_A |n-1\rangle_B\right] \\
+        &= i\left[\sum_{n=0}^{\infty} C_{n-1}(\theta)n|n\rangle_A |n\rangle_B + \sum_{n=0}^{\infty} C_{n+1}(\theta)(n+1)|n\rangle_A |n\rangle_B\right] 
+        \end{aligned}$$
+        由此得出系数$C_{n}$满足方程，结合初始条件如下
+        $$\frac{dC_{n}}{d\theta}=i[(n+1)C_{n+1}+nC_{n+1}],C_{n}(0)=\frac{(-\tanh r)^{n}}{\cosh r}$$
+        设$C_{n}(\theta)=A(\theta)[B(\theta)]^{n}$，相应初始条件为$A(0)=\frac{1}{\cosh r},B(0)=-\tanh r$
+        由此可得出系数A,B满足的方程组
+        $$\frac{dC_{n}}{d\theta}=A^{\prime}B^{n}+AnB^{n-1}B^{\prime}=i[(n+1)AB^{n+1}+nAB^{n-1}]$$
+        $$\frac{A^{\prime}}{A}B+nB^{\prime}=i[(n+1)B^{2}+n]=i(B^{2}+1)n+B^{2}$$
+        $$\begin{cases}
+        B^{\prime}=i(B^{2}+1) \\
+        \frac{A^{\prime}}{A}=iB & 
+        \end{cases}$$
+        只需分别求出系数A,B就能得到系数$C_{n}$，其中B的求解如下
+        $$\begin{aligned}
+        B' &= i(B^2 + 1) \Rightarrow \arctan B = i\theta + C \\
+        B(\theta) &= \tan(i\theta + C) = \frac{\tan C + \tan i\theta}{1 - \tan C \tan i\theta} = \frac{\tan C + i \tan \theta}{1 - i \tan C \tan \theta} \\
+        B(0) &= \tan C = -\tanh r \\
+        B(\theta) &= \frac{-\tanh r + i \tan \theta}{1 + i \tanh r \tanh \theta}
+        \end{aligned}$$
+        $$\Rightarrow B(\theta)=\frac{-\tanh r+i\tanh\theta}{1+i\tanh r\tanh\theta}$$
+        其中A的求解如下
+        $$\begin{aligned}
+        \frac{A'}{A} &= iB = i \frac{-\tanh r + i \tanh \theta}{1 + i \tanh r \tanh \theta} = \frac{-i \tanh r - \tanh \theta}{1 + i \tanh r \tanh \theta} \\
+        \Rightarrow A(\theta) &= \frac{C}{\cosh \theta + i \tanh r \sinh \theta}, \quad A(0) = \frac{1}{\cosh r} = C \\
+        \Rightarrow A(\theta) &= \frac{1}{\cosh \theta \cosh r + i \sinh r \sinh \theta}
+        \end{aligned}$$
+        最后得出系数满足如下
+        $$\begin{aligned}
+        C_n(\theta)&=A(\theta)[B(\theta)]^n \\
+        &=\frac{1}{\cosh\theta\cosh r+i\sinh r\sinh\theta}\left[\frac{-\tanh r+i\tanh\theta}{1+i\tanh r\tanh\theta}\right]^n
+        \end{aligned}$$
+        接下来就可以计算子系统A的纠缠熵了,由约化密度矩阵带入纠缠熵定义得
+        $$\rho = |\psi(\theta) \rangle \langle \psi(\theta)|$$
+        $$\rho_A=\mathrm{tr}_B\rho=\sum_{m=0}^{\infty}{}_B\langle m|\psi(\theta)\rangle\langle\psi(\theta)|m\rangle_B=\sum_{m=0}^{\infty}|C_m(\theta)|^2|m\rangle_A{}_A\langle m|$$
+        $$\begin{aligned}
+        S_A(\theta) &= -\mathrm{tr}_A(\rho_A \log \rho_A) \\
+        &= -\sum_{k=0}^{\infty} {}_A\langle k|\rho_A \log \rho_A|k\rangle_A \\
+        &= -\sum_{k=0}^{\infty} {}_A\langle k|\left(\sum_{m=0}^{\infty} |C_m(\theta)|^2 \log |C_m(\theta)|^2 |m\rangle_A\langle m|\right)|k\rangle_A \\
+        &= -\sum_{m=0}^{\infty} |C_m(\theta)|^2 \log |C_m(\theta)|^2
+        \end{aligned}$$
+        其中
+        $$|C_m(\theta)|^2=\frac{1}{\cosh^2r(\cosh^2\theta+\tanh^2r\sinh^2\theta)}\left[\frac{\tanh^2r+\tanh^2\theta}{1+\tanh^2r\tanh^2\theta}\right]^m$$
+        为计算方便，令
+        $$\frac{\tanh^{2}r+\tanh^{2}\theta}{1+\tanh^{2}r\tan^{2}\theta}=\tanh^{2}R=X$$
+        则容易得到
+        $$\begin{aligned}
+        1-\tanh^{2}R & =sech^{2}R \\
+        & =\frac{1+\tanh^{2}r\tanh^{2}\theta-\tanh^{2}r-\tanh^{2}\theta}{1+\tanh^{2}r\tanh^{2}\theta} \\
+        & =\frac{\cosh^{2}r\cosh^{2}\theta+\sinh^{2}r\sinh^{2}\theta-\sinh^{2}r\cosh^{2}\theta-\sinh^{2}\theta\cosh^{2}r}{\cosh^{2}r\cosh^{2}\theta+\sinh^{2}r\sinh^{2}\theta} \\
+        & =\frac{1}{\cosh^{2}r\cosh^{2}\theta+\sinh^{2}r\sinh^{2}\theta}
+        \end{aligned}$$
+        则
+        $$|C_{m}(\theta)|^{2}=\frac{1}{\cosh^{2}R}\tanh^{2m}R=(1-X)X^{m}$$
+        此时的纠缠熵与真空态形式类似
+        $$\begin{aligned}
+        S_{A}(\theta) & =-\sum_{m=0}^{\infty}(1-X)X^{m}\log[(1-X)X^{m}] \\
+        & =-\log(1-X)-\frac{X\log X}{1-X}
+        \end{aligned}$$
+
+        此时的激发态能量为
+        $$\langle \psi(\theta) | H | \psi(\theta) \rangle= \langle \psi(\theta) | a^\dagger a + b^\dagger b + \lambda (a^\dagger b^\dagger + a b) | \psi(\theta) \rangle$$
+        依然分为四项计算，第一项结果为
+        $$\begin{aligned}
+        \langle \psi(\theta) | a^\dagger a | \psi(\theta) \rangle
+        &= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} n |C_n(\theta)|^2 {}_A \langle n-1 | n-1 \rangle_A {}_B \langle n | n \rangle_B \\
+        &= \sum_{n=1}^{\infty} n |C_n(\theta)|^2 = \sum_{n=0}^{\infty} n |C_n(\theta)|^2 \\
+        &= \sum_{n=0}^{\infty} n (1-x) x^n = \sinh^2R
+        \end{aligned}$$
+        第二项计算与第一项同理
+        $$\langle\psi(\theta)|b^\dagger b|\psi(\theta)\rangle=\sinh^2x$$
+        第三项为
+        $$\begin{aligned}
+        \langle\psi(\theta)|a^{\dagger}b^{\dagger}|\psi(\theta)\rangle &=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sqrt{m+1}\sqrt{n+1}\ C_{m+1}^{*}(\theta)C_{n}(\theta) {}_A\langle m|n\rangle_A {}_B\langle m+1|m+1\rangle_B \\
+        &=\sum_{n=0}^{\infty}(n+1)C_{n+1}^{*}(\theta)C_{n}(\theta) \\
+        &=\sum_{n=0}^{\infty}(n+1)B^{*}(\theta)C_{n}^{*}(\theta)C_{n}(\theta) \\
+        &=B^{*}(\theta)\sum_{n=0}^{\infty}(n+1)(1-X)X^{n} \\
+        &=B^{*}(\theta)\frac{1}{1-X}=B^{*}(\theta)\cosh^2R
+        \end{aligned}$$
+        第四项与第三项类似
+        $$\langle\psi(\theta)|ab|\psi(\theta)\rangle=B(\theta)\cosh^2R$$
+        总能量为
+        $$\begin{aligned}
+        \langle \psi(\theta) | H | \psi(\theta) \rangle &= 2 \sinh^2 R + \lambda (\cosh^2 r \cosh^2 \theta + \sinh^2 r \sinh^2 \theta)  \frac{-2 \tanh r}{\cosh^2 \theta + \tanh^2 r \sinh^2 \theta} \\
+        &= 2 \sinh^2 R - 2 \lambda \sinh r \cosh r \\
+        &= E_o - E_o + 2 \sinh^2 R - 2 \lambda \sinh r \cosh r \\
+        &= E_o + 2 \sinh^2 R - 2 \sinh^2 r
+        \end{aligned}$$
+
+
+    \end{enumerate}
+
+    \item 
+    基态表达式为
+    $$|0\rangle=\sum_{n=0}^{\infty}\frac{(-\tanh r)^n}{\cosh r}|n\rangle_A|n\rangle_B$$
+    其中
+    $$\begin{aligned}
+    | n \rangle _ { A } | n \rangle _ { B } &= \frac { a ^ {\dagger} } { \sqrt { n } } \frac { b ^ {\dagger} } { \sqrt { n } } | n - 1 \rangle _ { A } | n - 1 \rangle _ { B } \\
+    &= \frac { \left( a ^ {\dagger} \right) ^ { 2 } } { \sqrt { n ( n - 1 ) } } \frac { \left( b ^ {\dagger} \right) ^ { 2 } } { \sqrt { n ( n - 1 ) } } | n - 2 \rangle _ { A } | n - 2 \rangle _ { B } \\
+    &= \frac { \left( a ^ {\dagger} b ^ {\dagger} \right) ^ { n } } { n ! } | 0 \rangle _ { A } | 0 \rangle _ { B }
+    \end{aligned}$$
+    由此真空态可表示成指数形式
+    $$\begin{aligned}
+    |0\rangle &= \frac{1}{\cosh r} \sum_{n=0}^{\infty} \frac{(-\tanh r\, a^{\dagger} b^{\dagger})^n}{n!} |0\rangle_A |0\rangle_B \\
+    &= \frac{1}{\cosh r} e^{-\tanh r\, a^{\dagger} b^{\dagger}} |0\rangle_A |0\rangle_B
+    \end{aligned}$$
+    真空态时子系统A的纠缠熵为
+    $$\begin{aligned}
+    S_{A} &= -\log{(1-x)} - \frac{x\log{x}}{1-x} \\
+    &= -\log{(1-\tanh^{2}{r})} - \frac{\tanh^{2}{r}\log{\tanh^{2}{r}}}{1-\tanh^{2}{r}} \\
+    &= -\log{\operatorname{sech}^{2}{r}} - \sinh^{2}{r}\log\tanh^{2}{r} \\
+    &= -\log{\operatorname{sech}^{2}{r}} - \sinh^{2}{r}\left[\log{\sinh^{2}{r}} - \log{\cosh^{2}{r}}\right] \\
+    &= \sinh^{2}{r}\log{\cosh^{2}{r}} + \log{\cosh^{2}{r}} - \sinh^{2}{r}\log{\sinh^{2}{r}} \\
+    &= (\sinh^{2}{r}+1)\log{(\sinh^{2}{r}+1)} - \sinh^{2}{r}\log{\sinh^{2}{r}}
+    \end{aligned}$$
+    令$y=\sinh^{2}r=\frac{1}{2}(\frac{1}{\sqrt{1-\lambda^{2}}}-1)$
+    由此纠缠熵可化简为
+    $$S_{A}=(y+1)\log(y+1)-y\log y$$
+    由于在y>0的范围内$\frac{dS_{A}}{dy}=\ln(1+\frac{1}{y})>0$，因此$S_{A}$随y严格单调递增。
+    在$\lambda^2$<1范围内y随$\lambda^2$也单调递增，因此$S_{A}$随$\lambda^2$单调递增，所以当$\lambda^2$趋近于1时纠缠熵趋于无穷大。
+
+\end{enumerate}
+
+\end{document}