update

presentation/ZheyuanWu_HonorThesis_Presentation.tex

@@ -18,8 +18,6 @@
 \usepackage{tikz}
 \usepackage{braket}
 
-\DeclareMathOperator{\sen}{sen}
-\DeclareMathOperator{\tg}{tg}
 \DeclareMathOperator{\obdiam}{ObsDiam}
 \DeclareMathOperator{\diameter}{diam}
 
@@ -157,112 +155,7 @@
 \end{block}
 \end{frame}
 
-\begin{frame}{Backgrounds: Motivation of Tensor product}
+\input{./backgrounds.tex}
-
-Recall from the traditional notation of product space of two vector spaces $V$ and $W$, that is, $V\times W$, is the set of all ordered pairs $(\ket{v},\ket{w})$ where $\ket{v}\in V$ and $\ket{w}\in W$.
-
-The space has dimension $\dim V+\dim W$.
-
-We want to define a vector space with the notation of multiplication of two vectors from different vector spaces.
-
-That is
-
-$$
-    (\ket{v_1}+\ket{v_2})\otimes \ket{w}=(\ket{v_1}\otimes \ket{w})+(\ket{v_2}\otimes \ket{w})
-$$
-$$
-    \ket{v}\otimes (\ket{w_1}+\ket{w_2})=(\ket{v}\otimes \ket{w_1})+(\ket{v}\otimes \ket{w_2})
-$$
-
-and enables scalar multiplication by
-
-$$
-    \lambda (\ket{v}\otimes \ket{w})=(\lambda \ket{v})\otimes \ket{w}=\ket{v}\otimes (\lambda \ket{w})
-$$
-
-And we wish to build a way to associate the basis of $V$ and $W$ with the basis of $V\otimes W$. That makes the tensor product a vector space with dimension $\dim V\times \dim W$.
-
-\end{frame}
-
-\begin{frame}{Backgrounds: Tensor product of vectors}
-
-    \begin{block}{Definition of Bilinear functional}
-    A bilinear functional is a bilinear function $\beta:V\times W\to \mathbb{F}$ satisfying the condition that $\ket{v}\to \beta(\ket{v},\ket{w})$ is a linear functional for all $\ket{w}\in W$ and $\ket{w}\to \beta(\ket{v},\ket{w})$ is a linear functional for all $\ket{v}\in V$.
-
-    \end{block}
-
-    The vector space of all bilinear functionals is denoted by $\mathcal{B}(V, W)$.
-    \begin{block}{Definition of Tensor product of vectors}
-        Let $V, W$ be two vector spaces.
-
-        Let $V'$ and $W'$ be the dual spaces of $V$ and $W$, respectively, that is $V'=\{\psi:V\to \mathbb{F}\}$ and $W'=\{\phi:W\to \mathbb{F}\}$, $\psi, \phi$ are linear functionals.
-
-        The \textbf{tensor product of vectors} $v\in V$ and $w\in W$ is the bilinear functional defined by $\forall (\psi,\phi)\in V'\times W'$ given by the notation
-
-        $$
-            (v\otimes w)(\psi,\phi)=\psi(v)\phi(w)
-        $$
-    \end{block}
-\end{frame}
-
-\begin{frame}{Backgrounds: Tensor product of vector spaces}
-    \begin{block}{Definition of Tensor product of vector spaces}
-    The tensor product of two vector spaces $V$ and $W$ is the vector space $\mathcal{B}(V',W')$
-
-    Notice that the basis of such vector space is the linear combination of the basis of $V'$ and $W'$, that is, if $\{e_i\}$ is the basis of $V'$ and $\{f_j\}$ is the basis of $W'$, then $\{e_i\otimes f_j\}$ is the basis of $\mathcal{B}(V', W')$.
-
-    Since $\{e_i\}$ and $\{f_j\}$ are bases of $V'$ and $W'$, respectively, then we can always find a set of linear functionals $\{\phi_i\}$ and $\{\psi_j\}$ such that $\phi_i(e_j)=\delta_{ij}$ and $\psi_j(f_i)=\delta_{ij}$. (Here $\delta_{ij}=1$ if $i=j$ and $0$ otherwise.)
-
-    $$
-        V\otimes W=\left\{\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w): \phi_i\in V', \psi_j\in W'\right\}
-    $$
-    \end{block}
-
-Note that $\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w)$ is a bilinear functional that maps $V'\times W'$ to $\mathbb{F}$.
-\end{frame}
-
-\begin{frame}{Backgrounds: Trace}
-    
-\label{defn:trace}
-
-\begin{block}{Trace}
-Let $T$ be a linear operator on $\mathscr{H}$, $(e_1,e_2,\cdots,e_n)$ be a basis of $\mathscr{H}$ and $(\epsilon_1,\epsilon_2,\cdots,\epsilon_n)$ be a basis of dual space $\mathscr{H}^*$. Then the trace of $T$ is defined by
-
-$$
-\operatorname{Tr}(T)=\sum_{i=1}^n \epsilon_i(T(e_i))=\sum_{i=1}^n \langle e_i,T(e_i)\rangle
-$$
-
-\end{block}
-
-This is equivalent to the sum of the diagonal elements of $T$.
-
-\vspace{1em}
-Q: How we generalize the trace to a subsystem of a larger, entangled quantum system $A\otimes B$?
-\end{frame}
-
-\begin{frame}{Backgrounds: Partial trace}
-
-\begin{block}{Definition of Partial trace}
-
-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.
-
-An operator $T$ on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$ can be written as 
-
-$$
-T=\sum_{i=1}^n a_i A_i\otimes B_i
-$$
-
-where $A_i$ is a linear operator on $\mathscr{A}$ and $B_i$ is a linear operator on $\mathscr{B}$.
-
-The $\mathscr{B}$-partial trace of $T$ ($\operatorname{Tr}_{\mathscr{B}}(T):\mathcal{L}(\mathscr{A}\otimes \mathscr{B})\to \mathcal{L}(\mathscr{A})$) is the linear operator on $\mathscr{A}$ defined by
-
-$$
-\operatorname{Tr}_{\mathscr{B}}(T)=\sum_{i=1}^n a_i \operatorname{Tr}(B_i) A_i
-$$
-
-\end{block}
-
-\end{frame}
 
 \begin{frame}{Information theory in classical systems}
 
@@ -288,7 +181,7 @@ For a density matrix $\rho$,
 $$
 S(\rho)=-\operatorname{Tr}(\rho\log_2\rho).
 $$
-This measures the intrinsic mixedness of the quantum state and is basis-independent.
+This measures the intrinsic uncertainty of the quantum state and is basis-independent.
 \end{block}
 
 \begin{block}{Entanglement entropy}
@@ -513,9 +406,9 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
 
 \begin{frame}{Levy Concentration}
 
-    \begin{block}{Definition of Lipschitz function}
+    % \begin{block}{Definition of Lipschitz function}
-         A function $f:X\to Y$, where $X,Y$ are metric spaces, is $L$-Lipschitz if there exists a constant $L$ such that $|f(x)-f(y)|\leq L|y-x|$ for all $x,y\in S^n$.
+    %      A function $f:X\to Y$, where $X,Y$ are metric spaces, is $L$-Lipschitz if there exists a constant $L$ such that $|f(x)-f(y)|\leq L|y-x|$ for all $x,y\in S^n$.
-    \end{block}
+    % \end{block}
     \begin{block}{Levy's lemma}
         If $f:S^n\to \mathbb{R}$ is $1$-Lipschitz, then there exists a $a_0$ such that for $\epsilon>0$,
         $$
@@ -569,13 +462,10 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
         $$
     \end{block}
 
-    \begin{block}{Lipschitz estimate}
+    \begin{block}{Lipschitz estimate for $H(\psi_A)$}
-        $$
+        The Lipschitz constant for the function $
-        \|H(\psi_A)\|_{\mathrm{Lip}}
+        H(\psi_A)$ should be upper bounded by $
-        \leq
+        \sqrt{8}\,\log_2(d_A)$, for $d_A\geq 3$.
-        \sqrt{8}\,\log_2(d_A),
-        \qquad d_A\geq 3.
-        $$
     \end{block}
 
     Levy concentration plus these two estimates produces the exponential entropy tail bound.
@@ -605,7 +495,7 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
         \right)=1-\Theta(e^{-c d_B})$, a random pure state is almost maximally entangled $\log_2(d_A)-\frac{1}{2\ln(2)}\frac{d_A}{d_B}=\log_2(d_A)-\Theta(\frac{1}{d_B})$.
 \end{frame}
 
-\begin{frame}{A natual question from the observables}
+\begin{frame}{A natural question from the observables}
     
     \textbf{What does the hayden--leung--winter theorem generalize the behavior of the lipschitz function $S^n\to \mathbb{R}$ and the lipschitz function $\mathbb{C}P^n\to \mathbb{R}$ as $n\to \infty$?}
 \end{frame}
@@ -614,13 +504,10 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
 
 \begin{frame}{Observable diameter: the inner definition}
     \begin{block}{Partial diameter on $\mathbb{R}$}
-        Let $\nu$ (nu) be a Borel probability measure on $\mathbb{R}$ and let $\alpha \in (0,1]$.
+        Let $\mu$ be a Borel probability measure on $\mathbb{R}$ and let $\alpha \in (0,1]$.
-        The \textbf{partial diameter} of $\nu$ at mass level $\alpha$ is
+        The \textbf{partial diameter} of $\mu$ at mass level $\alpha$ is
         $$
-        \diameter(\nu;\alpha):=
+        \diameter(\mu;\alpha):=\inf_{A\subset \mathbb{R}}\left\{ \diameter(A): \mu(A)\geq \alpha \right\}.
-        \{\diameter(A):A \subseteq \mathcal{B}(\mathbb{R}),
-             \nu(A)\ge \alpha
-        \},
         $$
         where
         $$
@@ -630,8 +517,8 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
 
     \vspace{0.4em}
     \begin{itemize}
-        \item This asks for the shortest interval-like region containing at least $\alpha$ of the total mass.
+        \item The partial diameter asks for: what is the shortest interval I need to capture at least $\alpha$ of the mass (measure)?
-        \item So $\diameter(\nu;1-\kappa)$ measures how tightly we can capture \emph{most} of the distribution, allowing us to discard a set of mass at most $\kappa$.
+        \item If $1$ is the total measure of the space, $\diameter(\mu;1-\kappa)$, measures how tightly we can capture \emph{most} of the distribution, allowing us to discard a set of mass at most $\kappa$.
     \end{itemize}
 
 \end{frame}

presentation/backgrounds.tex

@@ -0,0 +1,107 @@
+
+\begin{frame}{Backgrounds: Motivation of Tensor product}
+
+Recall from the traditional notation of product space of two vector spaces $V$ and $W$, that is, $V\times W$, is the set of all ordered pairs $(\ket{v},\ket{w})$ where $\ket{v}\in V$ and $\ket{w}\in W$.
+
+The space has dimension $\dim V+\dim W$.
+
+We want to define a vector space with the notation of multiplication of two vectors from different vector spaces.
+
+That is
+
+$$
+    (\ket{v_1}+\ket{v_2})\otimes \ket{w}=(\ket{v_1}\otimes \ket{w})+(\ket{v_2}\otimes \ket{w})
+$$
+$$
+    \ket{v}\otimes (\ket{w_1}+\ket{w_2})=(\ket{v}\otimes \ket{w_1})+(\ket{v}\otimes \ket{w_2})
+$$
+
+and enables scalar multiplication by
+
+$$
+    \lambda (\ket{v}\otimes \ket{w})=(\lambda \ket{v})\otimes \ket{w}=\ket{v}\otimes (\lambda \ket{w})
+$$
+
+And we wish to build a way to associate the basis of $V$ and $W$ with the basis of $V\otimes W$. That makes the tensor product a vector space with dimension $\dim V\times \dim W$.
+
+\end{frame}
+
+\begin{frame}{Backgrounds: Tensor product of vectors}
+
+    \begin{block}{Definition of Bilinear functional}
+    A bilinear functional is a bilinear function $\beta:V\times W\to \mathbb{F}$ satisfying the condition that $\ket{v}\to \beta(\ket{v},\ket{w})$ is a linear functional for all $\ket{w}\in W$ and $\ket{w}\to \beta(\ket{v},\ket{w})$ is a linear functional for all $\ket{v}\in V$.
+
+    \end{block}
+
+    The vector space of all bilinear functionals is denoted by $\mathcal{B}(V, W)$.
+    \begin{block}{Definition of Tensor product of vectors}
+        Let $V, W$ be two vector spaces.
+
+        Let $V'$ and $W'$ be the dual spaces of $V$ and $W$, respectively, that is $V'=\{\psi:V\to \mathbb{F}\}$ and $W'=\{\phi:W\to \mathbb{F}\}$, $\psi, \phi$ are linear functionals.
+
+        The \textbf{tensor product of vectors} $v\in V$ and $w\in W$ is the bilinear functional defined by $\forall (\psi,\phi)\in V'\times W'$ given by the notation
+
+        $$
+            (v\otimes w)(\psi,\phi)=\psi(v)\phi(w)
+        $$
+    \end{block}
+\end{frame}
+
+\begin{frame}{Backgrounds: Tensor product of vector spaces}
+    \begin{block}{Definition of Tensor product of vector spaces}
+    The tensor product of two vector spaces $V$ and $W$ is the vector space $\mathcal{B}(V',W')$
+
+    Notice that the basis of such vector space is the linear combination of the basis of $V'$ and $W'$, that is, if $\{e_i\}$ is the basis of $V'$ and $\{f_j\}$ is the basis of $W'$, then $\{e_i\otimes f_j\}$ is the basis of $\mathcal{B}(V', W')$.
+
+    Since $\{e_i\}$ and $\{f_j\}$ are bases of $V'$ and $W'$, respectively, then we can always find a set of linear functionals $\{\phi_i\}$ and $\{\psi_j\}$ such that $\phi_i(e_j)=\delta_{ij}$ and $\psi_j(f_i)=\delta_{ij}$. (Here $\delta_{ij}=1$ if $i=j$ and $0$ otherwise.)
+
+    $$
+        V\otimes W=\left\{\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w): \phi_i\in V', \psi_j\in W'\right\}
+    $$
+    \end{block}
+
+Note that $\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w)$ is a bilinear functional that maps $V'\times W'$ to $\mathbb{F}$.
+\end{frame}
+
+\begin{frame}{Backgrounds: Trace}
+    
+\label{defn:trace}
+
+\begin{block}{Trace}
+Let $T$ be a linear operator on $\mathscr{H}$, $(e_1,e_2,\cdots,e_n)$ be a basis of $\mathscr{H}$ and $(\epsilon_1,\epsilon_2,\cdots,\epsilon_n)$ be a basis of dual space $\mathscr{H}^*$. Then the trace of $T$ is defined by
+
+$$
+\operatorname{Tr}(T)=\sum_{i=1}^n \epsilon_i(T(e_i))=\sum_{i=1}^n \langle e_i,T(e_i)\rangle
+$$
+
+\end{block}
+
+This is equivalent to the sum of the diagonal elements of $T$.
+
+\vspace{1em}
+Q: How we generalize the trace to a subsystem of a larger, entangled quantum system $A\otimes B$?
+\end{frame}
+
+\begin{frame}{Backgrounds: Partial trace}
+
+\begin{block}{Definition of Partial trace}
+
+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.
+
+An operator $T$ on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$ can be written as 
+
+$$
+T=\sum_{i=1}^n a_i A_i\otimes B_i
+$$
+
+where $A_i$ is a linear operator on $\mathscr{A}$ and $B_i$ is a linear operator on $\mathscr{B}$.
+
+The $\mathscr{B}$-partial trace of $T$ ($\operatorname{Tr}_{\mathscr{B}}(T):\mathcal{L}(\mathscr{A}\otimes \mathscr{B})\to \mathcal{L}(\mathscr{A})$) is the linear operator on $\mathscr{A}$ defined by
+
+$$
+\operatorname{Tr}_{\mathscr{B}}(T)=\sum_{i=1}^n a_i \operatorname{Tr}(B_i) A_i
+$$
+
+\end{block}
+
+\end{frame}