updates

chapters/chap1.tex

@@ -182,9 +182,9 @@ This agrees with the experimentally observed transmission probabilities, but it
 	$\eta$-Lipschitz function
 
     Let $(X,\operatorname{dist}_X)$ and $(Y,\operatorname{dist}_Y)$ be two metric spaces. A function $f:X\to Y$ is said to be $\eta$-Lipschitz if there exists a constant $L\in \mathbb{R}$ such that
-    \[
+    $$
     \operatorname{dist}_Y(f(x),f(y))\leq L\operatorname{dist}_X(x,y)
-    \]
+    $$
     for all $x,y\in X$. And $\eta=\|f\|_{\operatorname{Lip}}=\inf_{L\in \mathbb{R}}L$.
 \end{defn}
 
@@ -202,9 +202,9 @@ This is a stronger condition than continuity, every Lipschitz function is contin
 
     Suppose $\sigma^n(\cdot)$ is the normalized volume measure on the sphere $S^n(1)$, then for any closed subset $\Omega\subset S^n(1)$, we take a metric ball $B_\Omega$ of $S^n(1)$ with $\sigma^n(B_\Omega)=\sigma^n(\Omega)$. Then we have
 
-    \[
+    $$
     \sigma^n(U_r(\Omega))\geq \sigma^n(U_r(B_\Omega))
-    \]
+    $$
 
     where $U_r(A)=\{x\in X:d(x,A)< r\}$
 \end{lemma}
@@ -224,20 +224,20 @@ To prove the lemma, we need to have a good understanding of the Riemannian geome
     An arbitrary 1-Lipschitz function $f:S^n\to \mathbb{R}$ concentrates near a single value $a_0\in \mathbb{R}$ as strongly as the distance function does.
 
     That is,
-    \[
+    $$
     \mu\{x\in S^n: |f(x)-a_0|\geq\epsilon\} < \kappa_n(\epsilon)\leq 2\exp\left(-\frac{(n-1)\epsilon^2}{2}\right)
-    \]
+    $$
     where 
-    \[
+    $$
     \kappa_n(\epsilon)=\frac{\int_\epsilon^{\frac{\pi}{2}}\cos^{n-1}(t)dt}{\int_0^{\frac{\pi}{2}}\cos^{n-1}(t)dt}
-    \]
+    $$
     $a_0$ is the \textbf{Levy mean} of function $f$, that is, the level set $f^{-1}:\mathbb{R}\to S^n$ divides the sphere into equal halves, characterized by the following equality:
-    \[
+    $$
     \mu(f^{-1}(-\infty,a_0])\geq \frac{1}{2} \text{ and } \mu(f^{-1}[a_0,\infty))\geq \frac{1}{2}
-    \]
+    $$
 \end{theorem}
 
-We will prove the theorem via the Maxwell-Boltzmann distribution law.~\cite{shioya2014metricmeasuregeometry}
+We will prove the theorem via the Maxwell-Boltzmann distribution law in this section for simplicity. ~\cite{shioya2014metricmeasuregeometry} The theorem will be discussed later in more general cases.
 
 \begin{defn}
 	\label{defn:Gaussian_measure}
@@ -271,15 +271,15 @@ If $X\sim \operatorname{Unif}(S^n(\sqrt{n}))$, then for any fixed unit vector $x
     Maxwell-Boltzmann distribution law:
     
     For any natural number $k$,
-    \[
+    $$
     \frac{d(\pi_{n,k})_*\sigma^n(x)}{dx}\to \frac{d\gamma^k(x)}{dx}
-    \]
+    $$
     where $(\pi_{n,k})_*\sigma^n$ is the push-forward measure of $\sigma^n$ by $\pi_{n,k}$.
     
     In other words,
-    \[
+    $$
     (\pi_{n,k})_*\sigma^n\to \gamma^k\text{ weakly as }n\to \infty
-    \]
+    $$
 \end{lemma}
 
 \begin{proof}
@@ -288,12 +288,12 @@ If $X\sim \operatorname{Unif}(S^n(\sqrt{n}))$, then for any fixed unit vector $x
     Observe that $\pi_{n,k}^{-1}(x),x\in \mathbb{R}^k$ is isometric to $S^{n-k}(\sqrt{n-\|x\|^2})$, that is, for any $x\in \mathbb{R}^k$, $\pi_{n,k}^{-1}(x)$ is a sphere with radius $\sqrt{n-\|x\|^2}$ (by the definition of $\pi_{n,k}$).
     
     So,
-    \[
+    $$
     \begin{aligned}
     \frac{d(\pi_{n,k})_*\sigma^n(x)}{dx}&=\frac{\operatorname{vol}_{n-k}(\pi_{n,k}^{-1}(x))}{\operatorname{vol}_k(S^n(\sqrt{n}))}\\
     &=\frac{(n-\|x\|^2)^{\frac{n-k}{2}}}{\int_{\|x\|\leq \sqrt{n}}(n-\|x\|^2)^{\frac{n-k}{2}}dx}\\
     \end{aligned}
-    \]
+    $$
     as $n\to \infty$.
     
     Note that $\lim_{n\to \infty}(1-\frac{a}{n})^n=e^{-a}$ for any $a>0$.
@@ -301,13 +301,13 @@ If $X\sim \operatorname{Unif}(S^n(\sqrt{n}))$, then for any fixed unit vector $x
     $(n-\|x\|^2)^{\frac{n-k}{2}}=\left(n(1-\frac{\|x\|^2}{n})\right)^{\frac{n-k}{2}}\to n^{\frac{n-k}{2}}\exp(-\frac{\|x\|^2}{2})$
     
     So
-    \[
+    $$
     \begin{aligned}
     \frac{(n-\|x\|^2)^{\frac{n-k}{2}}}{\int_{\|x\|\leq \sqrt{n}}(n-\|x\|^2)^{\frac{n-k}{2}}dx}&=\frac{e^{-\frac{\|x\|^2}{2}}}{\int_{x\in \mathbb{R}^k}e^{-\frac{\|x\|^2}{2}}dx}\\
     &=\frac{1}{(2\pi)^{\frac{k}{2}}}e^{-\frac{\|x\|^2}{2}}\\
     &=\frac{d\gamma^k(x)}{dx}
     \end{aligned}
-    \]
+    $$
 \end{proof}
 
 Now we can prove Levy's concentration theorem, the proof is from~\cite{shioya2014metricmeasuregeometry}.
@@ -335,25 +335,25 @@ Now we can prove Levy's concentration theorem, the proof is from~\cite{shioya201
 
     So the push-forward measure of $(f_{n_i})_*\sigma^{n_i}$ of $[x'-\epsilon_1,x'+\epsilon_2]$ is
 
-    \[
+    $$
     \begin{aligned}
     (f_{n_i})_*\sigma^{n_i}[x'-\epsilon_1,x'+\epsilon_2]&=\sigma^{n_i}(x'-\epsilon_1\leq f_{n_i}\leq x'+\epsilon_2)\\
     &\geq \sigma^{n_i}(U_{\epsilon_1}(\Omega_+)\cap U_{\epsilon_2}(\Omega_-))\\
     &=\sigma^{n_i}(U_{\epsilon_1}(\Omega_+))+\sigma^{n_i}(U_{\epsilon_2}(\Omega_-))-1\\
     \end{aligned}
-    \]
+    $$
 
     By the lemma~\ref{lemma:isoperimetric_inequality_on_sphere}, we have
 
-    \[
+    $$
     \sigma^{n_i}(U_{\epsilon_1}(\Omega_+))\geq \sigma^{n_i}(U_{\epsilon_1}(B_{\Omega_+}))\quad \text{and} \quad \sigma^{n_i}(U_{\epsilon_2}(\Omega_-))\geq \sigma^{n_i}(U_{\epsilon_2}(B_{\Omega_-}))
-    \]
+    $$
 
     By the lemma~\ref{lemma:Maxwell-Boltzmann_distribution_law}, we have
 
-    \[
+    $$
     \sigma^{n_i}(U_{\epsilon_1}(\Omega_+))+\sigma^{n_i}(U_{\epsilon_2}(\Omega_-))\to \gamma^1[x'-\epsilon_1,x'+\epsilon_2]+\gamma^1[x-\epsilon_1,x+\epsilon_2]
-    \]
+    $$
 
     Therefore,
 
@@ -392,12 +392,12 @@ To surpass the Holevo bound, we need to use the entanglement of quantum states.
         
         The Bell states are the following four states:
         
-        \[
+        $$
         |\Phi^+\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle),\quad |\Phi^-\rangle=\frac{1}{\sqrt{2}}(|00\rangle-|11\rangle)
-        \]
+        $$
-        \[
+        $$
         |\Psi^+\rangle=\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle),\quad |\Psi^-\rangle=\frac{1}{\sqrt{2}}(|01\rangle-|10\rangle)
-        \]
+        $$
         These are a basis of the 2-qubit Hilbert space.
 \end{defn}
 
@@ -505,7 +505,73 @@ Then we have bound for Lipschitz constant $\eta$ of the map $S(\varphi_A): \math
 \end{lemma}
 
 \begin{proof}
-    Consider the Lipschitz constant of the function $g:A\otimes B\to \R$ defined as $g(\varphi)=H(M(\varphi_A))$, where $M:A\otimes B\to \mathcal{P}(A)$ is the complete von Neumann measurement and $H: \mathcal{P}(A)\otimes \mathcal{P}(B)\to \R$ is the Shannon entropy.
+    Consider the Lipschitz constant of the function $g:A\otimes B\to \R$ defined as $g(\varphi)=H(M(\varphi_A))$, where $M:A\otimes B\to \mathcal{P}(A)$ is any fixed complete von Neumann measurement and $H: \mathcal{P}(A)\otimes \mathcal{P}(B)\to \R$ is the Shannon entropy.
+
+    Let $\{\ket{e_j}_A\}$ be the orthonormal basis for $A$ and $\{\ket{f_k}_B\}$ be the orthonormal basis for $B$. Then we decompose the state as spectral form $\ket{\varphi}=\sum_{j=1}^{d_A}\sum_{k=1}^{d_B}\varphi_{jk}\ket{e_j}_A\ket{f_k}_B$.
+
+    By unitary invariance, suppose $M_j=\ket{e_j}\bra{e_j}_A$, and define
+
+    $$
+    p_j(\varphi)=\bra{e_j}\varphi_A \ket{e_j}=\sum_{k=1}^{d_B}|\varphi_{jk}|^2
+    $$
+
+    Then 
+    
+    $$
+    g(\varphi)=H(M(\varphi_A))=-\sum_{j=1}^{d_A}p_j(\varphi)\log_2(p_j(\varphi))
+    $$
+
+    Let $h(p)=-p\log_2(p)$, $h(p)=-\frac{p\ln p}{\ln 2}$, and $h'(p)=-\frac{\ln p+1}{\ln 2}$. Let $\varphi_{jk}=x_{jk}+i y_{jk}$, then $p_j(\varphi)=\sum_{k=1}^{d_B}(x_{jk}^2+y_{jk}^2)$, $\frac{\partial p_j}{\partial x_{jk}}=2x_{jk}$, $\frac{\partial p_j}{\partial y_{jk}}=2y_{jk}$.
+
+    Therefore 
+    
+    $$
+    \frac{\partial g}{\partial x_{jk}}=\frac{\partial g}{\partial p_j}\frac{\partial p_j}{x_{jk}}=-\frac{1+\ln p_j}{\ln 2}\cdot 2x_{jk}
+    \qquad 
+    \frac{\partial g}{\partial y_{jk}}=-\frac{1+\ln p_j}{\ln 2}\cdot 2y_{jk}
+    $$
+    
+    Then the lipschitz constant of $g$ is
+
+    $$
+    \begin{aligned}
+    \eta^2&=\sup_{\langle \varphi|\varphi\rangle \leq 1}\nabla g\cdot \nabla g\\
+    &=\sum_{j=1}^{d_A}\sum_{k=1}^{d_B}\left(\frac{\partial g}{\partial x_{jk}}\right)^2+\left(\frac{\partial g}{\partial y_{jk}}\right)^2\\
+    &=\sum_{j=1}^{d_A}\sum_{k=1}^{d_B}\frac{4(x_{jk}^2+y_{jk}^2)}{(\ln 2)^2}[1+\ln p_j(\varphi)]^2\\
+    &=\sum_{j=1}^{d_A}\sum_{k=1}^{d_B}\frac{4|\varphi_{jk}|^2}{(\ln 2)^2}[1+\ln p_j(\varphi)]^2\\
+    \end{aligned}
+    $$
+
+    Note that $\sum_{k=1}^{d_B}|\varphi_{jk}|^2=p_j(\varphi)$, $\nabla g\cdot \nabla g=\frac{4}{(\ln 2)^2}\sum_{j=1}^{d_A}p_j(\varphi)(1+\ln p_j(\varphi))^2$.
+
+    Since $0\leq p_j\leq 1$, we have $\ln p_j(\varphi)\leq 0$, hence $\sum_{j=0}^{d_A}p_j(\varphi)\ln p_j(\varphi)\leq 0$.
+    
+    $$
+    \begin{aligned}
+        \sum_{j=1}^{d_A}p_j(\varphi)(1+\ln p_j(\varphi))^2&=\sum_{j=1}^{d_A}p_j(\varphi)(1+2\ln p_j(\varphi)+(\ln p_j(\varphi))^2)\\
+        &=1+2\sum_{j=1}^{d_A} p_j(\varphi)\ln p_j(\varphi)+\sum_{j=1}^{d_A}p_j(\varphi)(\ln p_j(\varphi))^2\\
+        &\leq 1+\sum_{j=1}^{d_A}p_j(\varphi)(\ln p_j(\varphi))^2\\
+    \end{aligned}
+    $$
+
+    Thus,
+    $$
+    \begin{aligned}
+    \nabla g\cdot \nabla g&\leq \frac{4}{(\ln 2)^2}[1+\sum_{j=1}^{d_A}p_j(\varphi)(\ln p_j(\varphi))^2]\\
+    &\leq \frac{4}{(\ln 2)^2}[1+(\ln d_A)^2]\\
+    &\leq 8(\log_2 d_A)^2
+    \end{aligned}
+    $$
+
+    Proving $\sum_j^{d_A} p_j(\varphi)\ln p_j(\varphi)\leq (\ln d_A)^2$ for $d_A\geq 3$ takes some efforts and we will continue that later.
+
+    Consider any two unit vectors $\ket{\varphi}$ and $\ket{\psi}$, assume $S(\varphi_A)\leq S(\psi_A)$. If we choose the measurement $M$ to be along the eigenbasis of $\varphi_A$, $H(M(\varphi_A))=S(\varphi_A)$ and we have
+
+    $$
+    S(\psi_A)-S(\varphi_A)\leq H(M(\psi_A))-H(M(\varphi_A))\leq \eta\|\ket{\psi}-\ket{\varphi}\|
+    $$
+
+    Thus the lipschitz constant of $S(\varphi_A)$ is upper bounded by $\sqrt{8}\log_2(d_A)$.
 \end{proof}
 
 From Levy's lemma, we have

chapters/chap2.tex

@@ -13,6 +13,8 @@ In this section, we will explore how the results from Hayden's concentration of
 
 \section{Observable diameters}
 
+Recall from previous sections, an arbitrary 1-Lipschitz function $f:S^n\to \mathbb{R}$ concentrates near a single value $a_0\in \mathbb{R}$ as strongly as the distance function does.
+
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