updates

content/Math401/Math401_P1_1.md

@@ -1,4 +1,4 @@
-# Math 401, Paper 1, Side note 1: Quantum information theory and Measure concentration
+# Math 401 Paper 1, Side note 1: Quantum information theory and Measure concentration
 
 ## Typicality
 
@@ -56,7 +56,7 @@ Practically speaking:
 
 Recall that the bipartite state of a quantum system is a linear operator on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$, where $\mathscr{A}$ and $\mathscr{B}$ are finite-dimensional Hilbert spaces.
 
-#### Definition of partial trace
+#### Definition of partial trace for arbitrary linear operators
 
 Let $T$ be a linear operator on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$, where $\mathscr{A}$ and $\mathscr{B}$ are finite-dimensional Hilbert spaces.
 
@@ -74,16 +74,71 @@ $$
 \operatorname{Tr}_{\mathscr{B}}(T)=\sum_{i=1}^n a_i \operatorname{Tr}(B_i) A_i
 $$
 
+#### Partial trace for density operators
+
+Let $\rho$ be a density operator in $\mathscr{H}_1\otimes\mathscr{H}_2$, the partial trace of $\rho$ over $\mathscr{H}_2$ is the density operator in $\mathscr{H}_1$ (reduced density operator for the subsystem $\mathscr{H}_1$) given by:
+
+$$
+\rho_1\coloneqq\operatorname{Tr}_2(\rho)
+$$
+
+<details>
+<summary>Examples</summary>
+
+Let $\rho=\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)$ be a density operator on $\mathscr{H}=\mathbb{C}^2\otimes \mathbb{C}^2$. 
+
+Expand the expression of $\rho$ in the basis of $\mathbb{C}^2\otimes\mathbb{C}^2$ using linear combination of basis vectors:
+
+$$
+\rho=\frac{1}{2}(|01\rangle\langle 01|+|01\rangle\langle 10|+|10\rangle\langle 01|+|10\rangle\langle 10|)
+$$
+
+Note $\operatorname{Tr}_2(|ab\rangle\langle cd|)=|a\rangle\langle c|\cdot \langle b|d\rangle$.
+
+Then the reduced density operator of the subsystem $\mathbb{C}^2$ in first qubit is, note the $\langle 0|0\rangle=\langle 1|1\rangle=1$ and $\langle 0|1\rangle=\langle 1|0\rangle=0$:
+
+$$
+\begin{aligned}
+\rho_1&=\operatorname{Tr}_2(\rho)\\
+&=\frac{1}{2}(\langle 1|1\rangle |0\rangle\langle 0|+\langle 0|1\rangle |0\rangle\langle 1|+\langle 1|0\rangle |1\rangle\langle 0|+\langle 0|0\rangle |1\rangle\langle 1|)\\
+&=\frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|)\\
+&=\frac{1}{2}I
+\end{aligned}
+$$
+
+is a mixed state.
+
+</details>
+
 ### Purification
 
-Let $\rho$ be any [state](https://notenextra.trance-0.com/Math401/Math401_T6#pure-states) (may not be pure) on the finite dimensional Hilbert space $\mathscr{H}$. then there exists a unit vector $w\in \mathscr{H}\otimes \mathscr{H}$ such that $\rho=\operatorname{Tr}+2(|w\rangle\langle w|)$ is a pure state.
+Let $\rho$ be any [state](https://notenextra.trance-0.com/Math401/Math401_T6#pure-states) (may not be pure) on the finite dimensional Hilbert space $\mathscr{H}$. then there exists a unit vector $w\in \mathscr{H}\otimes \mathscr{H}$ such that $\rho=\operatorname{Tr}_2(|w\rangle\langle w|)$ is a pure state.
 
 <details>
 <summary>Proof</summary>
 
+Let $(u_1,u_2,\cdots,u_n)$ be an orthonormal basis of $\mathscr{H}$ consisting of eigenvectors of $\rho$ for the eigenvalues $p_1,p_2,\cdots,p_n$. As $\rho$ is a states, $p_i\geq 0$ for all $i$ and $\sum_{i=1}^n p_i=1$.
 
+We can write $\rho$ as
 
+$$
+\rho=\sum_{i=1}^n p_i |u_i\rangle\langle u_i|
+$$
+
+Let $w=\sum_{i=1}^n \sqrt{p_i} u_i\otimes u_i$, note that $w$ is a unit vector (pure state). Then
+
+$$
+\begin{aligned}
+\operatorname{Tr}_2(|w\rangle\langle w|)&=\operatorname{Tr}_2(\sum_{i=1}^n \sum_{j=1}^n \sqrt{p_ip_j} |u_i\otimes u_i\rangle \langle u_j\otimes u_j|)\\
+&=\sum_{i=1}^n \sum_{j=1}^n \sqrt{p_ip_j} \operatorname{Tr}_2(|u_i\otimes u_i\rangle \langle u_j\otimes u_j|)\\
+&=\sum_{i=1}^n \sum_{j=1}^n \sqrt{p_ip_j} \langle u_i|u_j\rangle |u_i\rangle\langle u_i|\\
+&=\sum_{i=1}^n \sum_{j=1}^n \sqrt{p_ip_j} \delta_{ij} |u_i\rangle\langle u_i|\\
+&=\sum_{i=1}^n p_i |u_i\rangle\langle u_i|\\
+&=\rho
+\end{aligned}
+$$
+
+is a pure state.
+
+QED
 </details>
-
-## MM space
-