\documentclass[11pt]{beamer} \usetheme{Madrid} \usecolortheme{beaver} \usefonttheme{serif} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage[T1]{fontenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{mathrsfs} \usepackage{mathtools} \usepackage{graphicx} \usepackage{tabularx} \usepackage{colortbl} \usepackage{tikz} \DeclareMathOperator{\sen}{sen} \DeclareMathOperator{\tg}{tg} \DeclareMathOperator{\obdiam}{ObsDiam} \DeclareMathOperator{\diameter}{diam} \setbeamertemplate{caption}[numbered] \author[Zheyuan Wu]{Zheyuan Wu} \title{Measure concentration in complex projective space and quantum entanglement} \newcommand{\email}{w.zheyuan@wustl.edu} \setbeamertemplate{navigation symbols}{} \setbeamertemplate{footline} { \leavevmode% \hbox{% \begin{beamercolorbox}[wd=.15\paperwidth,ht=2.25ex,dp=1ex,center]{author in head/foot}% \usebeamerfont{author in head/foot}\insertshortauthor \end{beamercolorbox}% \begin{beamercolorbox}[wd=.6\paperwidth,ht=2.25ex,dp=1ex,center]{title in head/foot}% \usebeamerfont{title in head/foot}\insertsectionhead \end{beamercolorbox}% \begin{beamercolorbox}[wd=.15\paperwidth,ht=2.25ex,dp=1ex,center]{date in head/foot}% \usebeamerfont{author in head/foot}\insertshortdate \end{beamercolorbox}% \begin{beamercolorbox}[wd=.1\paperwidth,ht=2.25ex,dp=1ex,center]{institute in head/foot}% \insertframenumber{} / \inserttotalframenumber\hspace*{1ex} \end{beamercolorbox}}% \vskip0pt% } \setbeamercolor{block title}{fg=white, bg=red!50!black!60} \setbeamercolor{block body}{fg=black, bg=red!5} \setbeamercolor{item}{fg=red!60!black} \setbeamercolor{section number projected}{fg=white, bg=red!70!black} \institute[]{Washington University in St. Louis} \date{\today} \begin{document} \begin{frame} \titlepage \end{frame} \begin{frame}{Table of Contents} \hypersetup{linkcolor=black} \tableofcontents \end{frame} \section{Motivation} \begin{frame}{Light polarization and non-commutative probability} \begin{figure} \includegraphics[width=0.6\textwidth]{../latex/images/Filter_figure.png} \end{figure} \begin{itemize} \item Light passing through a polarizer becomes polarized in the direction of that filter. \item If two filters are placed with relative angle $\alpha$, the transmitted intensity decreases as $\alpha$ increases. \item In particular, the transmitted intensity vanishes when $\alpha=\pi/2$. \end{itemize} \end{frame} \begin{frame}{Polarization experiment} \vspace{0.5em} Now consider three filters $F_1,F_2,F_3$ with directions $$ \alpha_1,\alpha_2,\alpha_3. $$ Testing them pairwise suggests introducing three $0$--$1$ random variables $$ P_1,P_2,P_3, $$ where $P_i=1$ means that the photon passes filter $F_i$. \vspace{0.5em} If these were classical random variables on one probability space, they would satisfy a Bell-type inequality. \end{frame} \begin{frame}{A classical Bell-type inequality} \begin{block}{Bell-type inequality} For any classical random variables $P_1,P_2,P_3\in\{0,1\}$, $$ \operatorname{Prob}(P_1=1,P_3=0) \leq \operatorname{Prob}(P_1=1,P_2=0) + \operatorname{Prob}(P_2=1,P_3=0). $$ \end{block} \vspace{0.5em} \begin{proof} The event $\{P_1=1,P_3=0\}$ splits into two disjoint cases according to whether $P_2=0$ or $P_2=1$: $$ \{P_1=1,P_3=0\} = \{P_1=1,P_2=0,P_3=0\} \sqcup \{P_1=1,P_2=1,P_3=0\}. $$ Therefore, $$ \begin{aligned} \operatorname{Prob}(P_1=1,P_3=0) &= \operatorname{Prob}(P_1=1,P_2=0,P_3=0) \\ &\quad+ \operatorname{Prob}(P_1=1,P_2=1,P_3=0) \\ &\leq \operatorname{Prob}(P_1=1,P_2=0) + \operatorname{Prob}(P_2=1,P_3=0). \end{aligned} $$ \end{proof} \end{frame} \begin{frame}{Experimental law} For unpolarized incoming light, the \textbf{observed transition law} for a pair of filters is $$ \operatorname{Prob}(P_i=1,P_j=0) = \operatorname{Prob}(P_i=1)-\operatorname{Prob}(P_i=1,P_j=1). $$ Using the polarization law, $$ \operatorname{Prob}(P_i=1)=\frac12, \qquad \operatorname{Prob}(P_i=1,P_j=1)=\frac12\cos^2(\alpha_i-\alpha_j), $$ hence $$ \operatorname{Prob}(P_i=1,P_j=0) = \frac12-\frac12\cos^2(\alpha_i-\alpha_j) = \frac12\sin^2(\alpha_i-\alpha_j). $$ \vspace{0.5em} So the experimentally observed probabilities depend only on the angle difference $\alpha_i-\alpha_j$. \end{frame} \begin{frame}{Violation of the classical inequality} Substituting the experimental law into the classical inequality gives $$ \frac12\sin^2(\alpha_1-\alpha_3) \leq \frac12\sin^2(\alpha_1-\alpha_2) + \frac12\sin^2(\alpha_2-\alpha_3). $$ Choose $$ \alpha_1=0,\qquad \alpha_2=\frac{\pi}{6},\qquad \alpha_3=\frac{\pi}{3}. $$ Then $$ \begin{aligned} \frac12\sin^2\!\left(-\frac{\pi}{3}\right) &\leq \frac12\sin^2\!\left(-\frac{\pi}{6}\right) + \frac12\sin^2\!\left(-\frac{\pi}{6}\right) \\ \frac38 &\leq \frac18+\frac18 \\ \frac38 &\leq \frac14, \end{aligned} $$ which is false. \vspace{0.5em} Therefore the pairwise polarization data cannot come from one classical probability model with random variables $P_1,P_2,P_3$. \end{frame} \begin{frame}{The quantum model of polarization} The correct model uses a Hilbert space rather than classical events. \begin{itemize} \item A pure polarization state is a vector $$ \psi=\alpha|0\rangle+\beta|1\rangle \in \mathbb{C}^2. $$ \item A filter at angle $\alpha$ is represented by the orthogonal projection $$ P_\alpha= \begin{pmatrix} \cos^2\alpha & \cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha & \sin^2\alpha \end{pmatrix}. $$ \item For a pure state $\psi$, the probability of passing the filter is $$ \langle P_\alpha\psi,\psi\rangle. $$ \end{itemize} \vspace{0.4em} The key point is that sequential measurements are described by \emph{ordered products} of projections, and these need not commute. \end{frame} \begin{frame}{Recovering the observed law from the operator model} Assume the incoming light is unpolarized, so its state is the density matrix $$ \rho=\frac12 I. $$ The probability of passing the first filter $P_{\alpha_i}$ is $$ \operatorname{Prob}(P_i=1) = \operatorname{tr}(\rho P_{\alpha_i}) = \frac12\operatorname{tr}(P_{\alpha_i}) = \frac12. $$ If the photon passes the first filter, the post-measurement state is $$ \rho_i = \frac{P_{\alpha_i}\rho P_{\alpha_i}}{\operatorname{tr}(\rho P_{\alpha_i})} = P_{\alpha_i}. $$ $$ P_\alpha= \begin{pmatrix} \cos^2\alpha & \cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha & \sin^2\alpha \end{pmatrix}. $$ Therefore $$ \operatorname{Prob}(P_j=1\mid P_i=1) = \operatorname{tr}(\rho_i P_{\alpha_j}) = \operatorname{tr}(P_{\alpha_i}P_{\alpha_j}) = \cos^2(\alpha_i-\alpha_j). $$ \end{frame} \begin{frame}{Recovering the observed law from the operator model (cont.)} $$ \begin{aligned} \operatorname{Prob}(P_i=1,P_j=0) &= \operatorname{Prob}(P_i=1) \bigl(1-\operatorname{Prob}(P_j=1\mid P_i=1)\bigr) \\ &= \frac12\bigl(1-\cos^2(\alpha_i-\alpha_j)\bigr) \\ &= \frac12\sin^2(\alpha_i-\alpha_j). \end{aligned} $$ This matches the experiment exactly. \end{frame} \begin{frame}{Conclusion} \begin{itemize} \item The classical model predicts a Bell-type inequality for three $0$--$1$ random variables. \item The polarization experiment violates that inequality. \item The resolution is that the quantities measured are \emph{sequential probabilities}, not joint probabilities of classical random variables. \item In quantum probability, events are modeled by projections on a Hilbert space, and measurement order matters. \end{itemize} \vspace{0.6em} This is one of the basic motivations for passing from classical probability to non-commutative probability. \end{frame} \section{Concentration on Spheres and quantum states} \begin{frame}{Quantum states: pure vs.\ mixed} \begin{itemize} \item A finite-dimensional quantum system is modeled by a complex Hilbert space (a complete inner product space) $$ \mathcal H \cong \mathbb C^{n+1}. $$ \item A \textbf{pure state} is represented by a unit vector $$ \psi \in \mathcal H, \qquad \|\psi\|=1. $$ \item A \textbf{mixed state} is represented by a density matrix $$ \rho \geq 0, \qquad \operatorname{tr}(\rho)=1. $$ \item Pure states describe maximal information; mixed states describe probabilistic mixtures or partial information. \end{itemize} \vspace{0.4em} \begin{block}{Key distinction} Pure states form a curved geometric space; mixed states form a convex set inside the space of matrices. \end{block} \end{frame} \begin{frame}{Why pure states are not vectors} \begin{itemize} \item Two nonzero vectors that differ by a nonzero complex scalar represent the same physical state: $$ \psi \sim \lambda \psi, \qquad \lambda \in \mathbb C^\times. $$ \item In particular, multiplying by a phase $e^{i\theta}$ does not change any physical predictions. \item Therefore the physical pure state is not a single vector, but the \emph{complex line} spanned by that vector. \end{itemize} \vspace{0.4em} Hence the space of pure states is $$ \mathbb P(\mathcal H) = (\mathcal H \setminus \{0\})/\mathbb C^\times. $$ After choosing a basis $\mathcal H \cong \mathbb C^{n+1}$, this becomes $$ \mathbb P(\mathcal H) \cong \mathbb C P^n. $$ \end{frame} \begin{frame}{Relation with the sphere} \begin{itemize} \item Every nonzero vector can be normalized, so each pure state has a representative on the unit sphere $$ S^{2n+1} \subset \mathbb C^{n+1}. $$ \item Two unit vectors represent the same pure state exactly when they differ by a phase: $$ z \sim e^{i\theta} z. $$ \item Therefore $$ \mathbb C P^n = S^{2n+1}/S^1. $$ \end{itemize} \vspace{0.4em} The quotient map $$ p:S^{2n+1}\to \mathbb C P^n, \qquad p(z)=[z]=\{\lambda z : \lambda \in \mathbb C^\times\}, $$ is the \textbf{Hopf fibration}. \end{frame} \begin{frame}{How the metric descends to $\mathbb C P^n$} \begin{itemize} \item The sphere $S^{2n+1}$ inherits the round metric from the Euclidean metric on $$ \mathbb C^{n+1} \cong \mathbb R^{2n+2}. $$ \item The fibers of the Hopf map are circles $$ p^{-1}([z]) = \{e^{i\theta}z : \theta \in \mathbb R\}. $$ \item Tangent vectors split into: \begin{itemize} \item \textbf{vertical directions}: tangent to the $S^1$-fiber, \item \textbf{horizontal directions}: orthogonal complement to the fiber. \end{itemize} \item The differential $dp$ identifies horizontal vectors on the sphere with tangent vectors on $\mathbb C P^n$. \end{itemize} \vspace{0.4em} This allows the round metric on $S^{2n+1}$ to define a metric on $\mathbb C P^n$. \end{frame} \begin{frame}{The induced metric: Fubini--Study metric} \begin{itemize} \item The metric on $\mathbb C P^n$ obtained from the Hopf quotient is the \textbf{Fubini--Study metric}. \item So the geometric picture is: $$ S^{2n+1} \xrightarrow{\text{Hopf fibration}} \mathbb C P^n $$ $$ \text{round metric} \rightsquigarrow \text{Fubini--Study metric}. $$ \item The normalized Riemannian volume measure induced by this metric gives the natural probability measure on pure states. \end{itemize} \vspace{0.5em} \begin{block}{Proof roadmap} To prove this carefully, one usually shows: \begin{enumerate} \item $p:S^{2n+1}\to \mathbb C P^n$ is a smooth surjective submersion, \item the vertical space is the tangent space to the $S^1$-orbit, \item horizontal lifts are well defined, \item the quotient metric is exactly the Fubini--Study metric. \end{enumerate} \end{block} \end{frame} \begin{frame}{Maxwell-Boltzmann Distribution Law} \begin{columns}[T] \column{0.58\textwidth} Consider the orthogonal projection $$ \pi_{n,k}:S^n(\sqrt{n})\to \mathbb{R}^k. $$ Its push-forward measure converges to the standard Gaussian: $$ (\pi_{n,k})_*\sigma^n\to \gamma^k. $$ \vspace{0.5em} This explains why Gaussian behavior emerges from high-dimensional spheres and supports the proof strategy for Levy concentration. \column{0.42\textwidth} \begin{figure} \includegraphics[width=\textwidth]{../latex/images/maxwell.png} \end{figure} \end{columns} \end{frame} \begin{frame}{Levy Concentration} \begin{block}{Levy's theorem} If $f:S^n\to \mathbb{R}$ is $1$-Lipschitz, then there exists a median $a_0$ such that $$ \mu\{x\in S^n:|f(x)-a_0|\geq \epsilon\} \leq 2\exp\left(-\frac{(n-1)\epsilon^2}{2}\right). $$ \end{block} \begin{itemize} \item In high dimension, most Lipschitz observables are almost constant. \item This is the geometric mechanism behind generic entanglement. \end{itemize} \end{frame} \section{Main Result} \begin{frame}{Generic Entanglement Theorem} \begin{block}{Hayden--Leung--Winter} Let $\psi\in \mathcal{P}(A\otimes B)$ be Haar-random and define $$ \beta=\frac{1}{\ln(2)}\frac{d_A}{d_B}. $$ For $d_B\geq d_A\geq 3$, $$ \operatorname{Pr}[H(\psi_A)<\log_2(d_A)-\alpha-\beta] \leq \exp\left( -\frac{1}{8\pi^2\ln(2)} \frac{(d_Ad_B-1)\alpha^2}{(\log_2 d_A)^2} \right). $$ \end{block} With overwhelming probability, a random pure state is almost maximally entangled. \end{frame} \begin{frame}{How the Entropy Observable Fits In} \begin{figure} \centering \begin{tikzpicture}[node distance=30mm, thick, main/.style={draw, draw=white}, towards/.style={->}, towards_imp/.style={->,red}, mutual/.style={<->} ] \node[main] (cp) {$\mathbb{C}P^{d_A d_B-1}$}; \node[main] (pa) [left of=cp] {$\mathcal{P}(A\otimes B)$}; \node[main] (sa) [below of=pa] {$\mathcal{S}(A)$}; \node[main] (rng) [right of=sa] {$[0,\log_2 d_A]$}; \draw[mutual] (cp) -- (pa); \draw[towards] (pa) -- node[left] {$\operatorname{Tr}_B$} (sa); \draw[towards_imp] (pa) -- node[above right] {$\psi\mapsto H(\psi_A)$} (rng); \draw[towards] (sa) -- node[above] {$H$} (rng); \end{tikzpicture} \end{figure} \begin{itemize} \item The red arrow is the observable to which concentration is applied. \item The projective description is natural because global phase does not change the physical state. \end{itemize} \end{frame} \begin{frame}{Ingredients Behind the Tail Bound} \begin{block}{Page-type lower bound} $$ \mathbb{E}[H(\psi_A)] \geq \log_2(d_A)-\frac{1}{2\ln(2)}\frac{d_A}{d_B}. $$ \end{block} \begin{block}{Lipschitz estimate} $$ \|H(\psi_A)\|_{\mathrm{Lip}} \leq \sqrt{8}\,\log_2(d_A), \qquad d_A\geq 3. $$ \end{block} Levy concentration plus these two estimates produces the exponential entropy tail bound. \end{frame} \section{Geometry of State Space} \begin{frame}{Observable diameter: the inner definition} \begin{block}{Partial diameter on $\mathbb{R}$} Let $\nu$ 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 $$ \diameter(\nu;\alpha):= \{\diameter(A):A \subseteq \mathcal{B}(\mathbb{R}), \nu(A)\ge \alpha \}, $$ where $$ \diameter(A):=\sup_{x,y\in A}|x-y|. $$ \end{block} \vspace{0.4em} \begin{itemize} \item This asks for the shortest interval-like region containing at least $\alpha$ of the total mass. \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 $\kappa$. \end{itemize} \end{frame} \begin{frame}{Observable diameter of a metric-measure space} \begin{block}{Definition} Let $X=(X,d_X,\mu_X)$ be a metric-measure space and let $\kappa>0$. The \textbf{observable diameter} of $X$ is $$ \obdiam_{\mathbb{R}}(X;-\kappa) := \sup_{f\in \operatorname{Lip}_1(X,\mathbb{R})} \diameter(f_*\mu_X;1-\kappa), $$ where $\operatorname{Lip}_1(X,\mathbb{R})$ is the set of all $1$-Lipschitz functions $f:X\to\mathbb{R}$, and $f_*\mu_X$ is the pushforward measure on $\mathbb{R}$. \end{block} \vspace{0.4em} \begin{itemize} \item Each $1$-Lipschitz function $f$ is viewed as an \textbf{observable} on $X$. \item The pushforward measure $f_*\mu_X$ is the distribution of the values of that observable. \item If $\obdiam_{\mathbb{R}}(X;-\kappa)$ is small, then \emph{every} $1$-Lipschitz observable is strongly concentrated. \end{itemize} \end{frame} \begin{frame}{A Geometric Consequence} In this thesis, entropy functions are used as concrete observables to estimate observable diameter, and the Hopf fibration helps transfer information between $S^{2n+1}$ and $\mathbb{C}P^n$. \vspace{0.4em} \begin{block}{Projective-space estimate} For $0<\kappa<1$, $$ \obdiam(\mathbb{C}P^n(1);-\kappa)\leq O(\sqrt{n}). $$ \end{block} \begin{itemize} \item First estimate observable diameter on spheres via Gaussian limits. \item Then use the Hopf map $S^{2n+1}(1)\to \mathbb{C}P^n$. \item This gives a geometric explanation for why many projective-space observables concentrate. \end{itemize} \end{frame} \section{Numerical Section} \begin{frame}{Entropy-Based Simulations} \begin{itemize} \item Sample Haar-random pure states in $\mathbb{C}^{d_A}\otimes\mathbb{C}^{d_B}$. \item Compute reduced density matrices and entanglement entropy. \item Measure shortest intervals containing mass $1-\kappa$ in the entropy distribution. \item Compare concentration across: \begin{itemize} \item real spheres, \item complex projective spaces, \item symmetric states via Majorana stellar representation. \end{itemize} \end{itemize} \end{frame} \begin{frame}{What the Data Suggests} \begin{columns}[T] \column{0.5\textwidth} \begin{figure} \includegraphics[width=\textwidth]{../latex/images/entropy_vs_dim.png} \end{figure} \centering Entropy vs.\ ambient dimension \column{0.5\textwidth} \begin{figure} \includegraphics[width=\textwidth]{../latex/images/entropy_vs_dA.png} \end{figure} \centering Entropy vs.\ subsystem dimension \end{columns} \vspace{0.6em} As dimension increases, the entropy distribution concentrates near the maximal value. \end{frame} \section{Conclusion} \begin{frame}{Conclusion and Outlook} \begin{itemize} \item Concentration of measure explains generic high entanglement in large bipartite systems. \item Complex projective space provides the natural geometric setting for pure quantum states. \item Observable diameter gives a way to phrase concentration geometrically. \item Ongoing directions: \begin{itemize} \item sharper estimates for $\mathbb{C}P^n$, \item deeper use of Fubini--Study geometry, \item Majorana stellar representation for symmetric states. \end{itemize} \end{itemize} \end{frame} \section{References} \begin{frame}[allowframebreaks]{References} \nocite{*} \bibliographystyle{apalike} \bibliography{references} \end{frame} \begin{frame} \begin{center} Q\&A \end{center} \end{frame} \end{document}