updated section 6 and corrected the final-2022

contents/final-2022/._wordcount_selection.tex

@@ -1,26 +1 @@
-根据 $J = 0$, 而 $-|J|\leq M\leq |J|$, 夹逼定理得到 $M= 0$. 而磁量子数守恒, 所以 $J_{1,z} + J_{2,z} = J_{z} = 0$. 已知 C-G 系数可以用于将 $|J,M;j_{1},j_{2}\rangle$ 以基矢 $|j_{1},m_{1};j_{2},m_{2}\rangle$ 展开, 代入上述讨论结果有
-
-\begin{align*}
-  |0,0;j,j\rangle = \sum_{m_,-m}^{-j\leq m\leq j}C_{j,j,m,-m}^{0,0}|j,m;j,-m\rangle
-\end{align*}
-
-概率即为 $P(m_{1}=m,m_{2}=-m) = |C_{j,j,m,-m}^{0,0}|^{2}$. 那么问题就来到如何计算这个特殊的 C-G 系数. 根据 C-G 系数的递推定义, 可以得到其解析表达式
-
-\begin{align*}
-  &\langle j_{1},m_{1};j_{2},m_{2}|J,M;j_{1},j_{2}\rangle \\
-  &=\sqrt{\frac{(2J+1)(J+j_{1}-j_{2})!(J-j_{1}+j_{2})!(j_{1}+j_{2}-J)!}{(j_{1}+j_{2}+J+1)!}}\\
-  &\times \sqrt{(J+M)!(J-M)!(j_{1}+m_{1})!(j_{1}-m_{1})!(j_{2}+m_{2})!(j_{2}-m_{2})!}\\
-  &\times \sum_{k_{\text{min}}}^{k_{\text{max}}}\frac{(-1)^{k}}{k!(j_{1}+j_{2}-J-k)!(j_{1}-m_{1}-k)!(j_{2}+m_{2}-k)!(J-M-k)!}\\
-  &\times \frac{1}{(J-j_{2}+m_{1}+k)!(J-j_{1}-m_{2}+k)!}\\
-  k_{\text{min}} &= \text{max}\{0,j_{2}-m_{1}-J,j_{1}+m_{2}-J\},\quad
-  k_{\text{max}} = \text{min}\{j_{1}+j_{2}-J, j_{1}-m_{1},j_{2}+m_{2}\}
-\end{align*}
-
-所以代入 $j_{1}=j_{2}=j$, $m_{1}=-m_{2}=m$, 即有 $\begin{aligned}
-  C_{j,m,j,-m}^{0,0} = \frac{(-1)^{j-m}}{\sqrt{2j+1}}
-\end{aligned}$, 显然因为平方消去了可能存在的负号, 使得 
-$|j,m;j,-m\rangle,\quad\forall m\in \{-j,-j+1,\cdots,j-1,j\}$ 等概率, 所以得到
-
-\begin{align*}
-  \boxed{P(m_{1}=m,m_{2}=-m) = \frac{1}{2j+1}}
-\end{align*}
+\left[(\langle i|_{A})\rho_{A}(|i\rangle_{A})\right]

contents/final-2022/single-choice.tex

@@ -165,13 +165,13 @@ \section{单项选择}
       0 & 1
     \end{pmatrix}
   \end{align*}
-  计算 $\rho_{A}$ 的纠缠熵:
+  由于 $\rho_{A}$ 已经是对角阵, 所以对角线上元素即为特征值 $\lambda_{A,i}$. 计算 $\rho_{A}$ 的纠缠熵:
   \begin{align*}
     S(A) &= 
     -\text{Tr}[\rho_{A}\ln{\rho_{A}}] 
     = -\sum_{i}^{\uparrow,\downarrow} 
-    (\langle i|_{A})\rho_{A}(|i\rangle_{A})
-    \ln{\left[(\langle i|_{A})\rho_{A}(|i\rangle_{A})\right]}\\
+    \lambda_{A,i}
+    \ln{\lambda_{A,i}}\\
     &= -\left(\frac{1}{2}\ln{\frac{1}{2}} + \frac{1}{2}\ln{\frac{1}{2}}\right) = \boxed{\ln{2} = 1\text{ bit}}
   \end{align*}}}
 

contents/homework/._wordcount_selection.tex

@@ -1 +1 @@
-n_{i\uparrow}\langle
+gray

contents/lecture-note/._wordcount_selection.tex

@@ -1 +1 @@
-        0\sigma^{2} & 1\sigma^{2} \\ 1\sigma^{2} & 0\sigma^{2}
+\rho_{A}^{ii}

contents/lecture-note/section6.tex

@@ -59,11 +59,123 @@ \subsubsection{双量子比特算符}
 
 \subsubsection{双量子比特模型}
 
+双量子比特 Heisenberg 模型哈密顿量为 $\begin{aligned}
+    H = \frac{J}{4}\vec{\sigma}_{A}\cdot\vec{\sigma}_{B}
+\end{aligned}$, 将其写作矩阵形式:
+\begin{align*}
+    H = \frac{J}{4}(\sigma^{11}+\sigma^{22}+\sigma^{33}) = \frac{J}{4} \begin{pmatrix}
+        1 &  & & \\
+         & -1 & 2 & \\
+         & 2 & -1 & \\
+            &  &  & 1
+    \end{pmatrix}
+\end{align*}
+
+将其对角化以计算本征值, 并找到本征值对应的本征态, 结果为
+\begin{align*}
+    E_{s} &= -\frac{3}{4}J,\quad |s\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle),\quad \text{自旋单态}\\
+    E_{t} &= \frac{J}{4},\quad \left\{\begin{aligned}
+        |t_{1}\rangle &= \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle)\\
+        |t_{2}\rangle &= \frac{1}{\sqrt{2}}(|\uparrow\uparrow\rangle + |\downarrow\downarrow\rangle)\\
+        |t_{3}\rangle &= \frac{1}{\sqrt{2}}(|\uparrow\uparrow\rangle - |\downarrow\downarrow\rangle)
+    \end{aligned}\right.,\quad \text{自旋三重态}
+\end{align*}
+
 \subsubsection{自旋单态}
 
+\begin{align*}
+    |s\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle) = \frac{1}{\sqrt{2}}\begin{pmatrix}
+        0 \\ 1 \\ -1 \\ 0
+    \end{pmatrix},\vec{\sigma}_{A} = (\sigma^{10},\sigma^{20},\sigma^{30}) = \left(\begin{pmatrix}
+         &  & 1 &  \\  &  &  & 1\\ 1 &  &  &  \\  & 1 &  & 
+    \end{pmatrix}, \begin{pmatrix}
+         &  & -i &  \\  &  &  & -i\\ i &  &  &  \\  & i &  & 
+    \end{pmatrix}, \begin{pmatrix}
+        1 &  &  &  \\  & 1 &  & \\  &  & -1 &  \\  &  &  & -1
+    \end{pmatrix}
+    \right)
+\end{align*}
+
 \subsubsection{纠缠熵}
+双量子比特态 $|\psi\rangle$ 中量子比特 $A$ 的纠缠熵: $\begin{aligned}
+    S(A) = -\text{Tr}[\rho_{A}\ln{\rho_{A}}]
+\end{aligned}$. 其中约化密度矩阵 $\begin{aligned}
+    \rho_{A} = \text{Tr}_{B}|\psi\rangle\langle\psi|
+\end{aligned}$
+
+更广义的 Renyi 纠缠熵 $\begin{aligned}
+    S^{(n)}(A) = \frac{1}{1-n}\ln{\text{Tr}\rho_{A}^{n}}
+\end{aligned}$. 
+
+接下来介绍如何部分求迹. 
+
+\begin{enumerate}
+    \item 自旋单态 $\begin{aligned}
+        |\psi\rangle = |s\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)
+    \end{aligned}$. 其总密度矩阵为
+    \begin{align*}
+        \rho &= |s\rangle\langle s| = \frac{1}{2}\begin{pmatrix}
+            0 \\ 1 \\ -1 \\ 0
+        \end{pmatrix}\begin{pmatrix}
+            0 & 1 & -1 & 0
+        \end{pmatrix} = \frac{1}{2}\begin{pmatrix}
+             & & & \\
+                & 1 & -1 & \\
+                & -1 & 1 & \\
+                & & &
+        \end{pmatrix}\\
+        \rho_{A} &= \text{Tr}_{B}\rho = \frac{1}{2}\begin{pmatrix}
+            \text{Tr}\begin{pmatrix}
+                & \\ & 1
+            \end{pmatrix} & \text{Tr}\begin{pmatrix}
+                & \\ -1 &
+            \end{pmatrix} \\ \text{Tr}\begin{pmatrix}
+                 & -1 \\ & 
+            \end{pmatrix} & \text{Tr}\begin{pmatrix}
+                1 & \\ &
+            \end{pmatrix}
+        \end{pmatrix} = \frac{1}{2}\begin{pmatrix}
+            1 & 0 \\ 0 & 1
+        \end{pmatrix},\quad \lambda_{A}^{1} = \frac{1}{2}, \quad \lambda_{A}^{2} = \frac{1}{2}\\
+        S(A) &= -\text{Tr}\rho_{A}\ln{\rho_{A}} = - \sum_{i}\lambda_{A}^{i}\ln{\lambda_{A}^{i}} = -\left(\frac{1}{2}\ln{\frac{1}{2}} + \frac{1}{2}\ln{\frac{1}{2}}\right) = \ln{2}
+    \end{align*}
+    \item 乘积态 $\begin{aligned}
+        |\psi\rangle = \frac{1}{2}(|\uparrow\uparrow\rangle + |\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle + |\downarrow\downarrow\rangle)
+    \end{aligned}$. 总密度矩阵为
+    \begin{align*}
+        \rho &= |\psi\rangle\langle\psi| = \frac{1}{4}\begin{pmatrix}
+            1 \\ 1 \\ 1 \\ 1
+        \end{pmatrix}\begin{pmatrix}
+            1 & 1 & 1 & 1
+        \end{pmatrix} = \frac{1}{4}\begin{pmatrix}
+            1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1
+        \end{pmatrix}\\
+        \rho_{A} &= \text{Tr}_{B}\rho = \frac{1}{4}\begin{pmatrix}
+            \text{Tr}\begin{pmatrix}
+                1 & 1 \\ 1 & 1
+            \end{pmatrix} &  \text{Tr}\begin{pmatrix}
+                1 & 1 \\ 1 & 1
+            \end{pmatrix}\\
+            \text{Tr}\begin{pmatrix}
+                1 & 1 \\ 1 & 1
+            \end{pmatrix} &  \text{Tr}\begin{pmatrix}
+                1 & 1 \\ 1 & 1
+            \end{pmatrix}
+        \end{pmatrix} = \frac{1}{4}\begin{pmatrix}
+            2 & 2 \\ 2 & 2
+        \end{pmatrix} = \frac{1}{2}\begin{pmatrix}
+            1 & 1 \\ 1 & 1
+        \end{pmatrix},\quad \lambda_{A}^{1} = 0,\quad \lambda_{A}^{2} = 1\\
+        S(A) &= -\text{Tr}\rho_{A}\ln{\rho_{A}} = -\sum_{i}\lambda_{A}^{i}\ln{\lambda_{A}^{i}} = - (0\ln{0} + 1\ln{1}) = 0.
+    \end{align*}
+\end{enumerate}
 
 \subsubsection{互信息}
+双量子比特体系, A 和 B 之间的互信息为
+
+\begin{align*}
+    I(A:B) = S(A) + S(B) - S(A\cup B)
+\end{align*}
 
 \subsubsection{EPR 佯谬和 Bell 不等式}