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\author[Zheyuan Wu]{Zheyuan Wu}
\title{Measure concentration in complex projective space and quantum entanglement}
\newcommand{\email}{w.zheyuan@wustl.edu}

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\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}
