672 lines
20 KiB
TeX
672 lines
20 KiB
TeX
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\author[Zheyuan Wu]{Zheyuan Wu}
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\title{Measure concentration in complex projective space and quantum entanglement}
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\newcommand{\email}{w.zheyuan@wustl.edu}
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\institute[]{Washington University in St. Louis}
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\date{\today}
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\begin{document}
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\begin{frame}
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\titlepage
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\end{frame}
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\begin{frame}{Table of Contents}
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\hypersetup{linkcolor=black}
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\tableofcontents
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\end{frame}
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\section{Motivation}
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\begin{frame}{Light polarization and non-commutative probability}
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\begin{figure}
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\includegraphics[width=0.6\textwidth]{../latex/images/Filter_figure.png}
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\end{figure}
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\begin{itemize}
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\item Light passing through a polarizer becomes polarized in the direction of that filter.
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\item If two filters are placed with relative angle $\alpha$, the transmitted intensity decreases as $\alpha$ increases.
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\item In particular, the transmitted intensity vanishes when $\alpha=\pi/2$.
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\end{itemize}
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\end{frame}
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\begin{frame}{Polarization experiment}
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\vspace{0.5em}
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Now consider three filters $F_1,F_2,F_3$ with directions
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$$
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\alpha_1,\alpha_2,\alpha_3.
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$$
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Testing them pairwise suggests introducing three $0$--$1$ random variables
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$$
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P_1,P_2,P_3,
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$$
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where $P_i=1$ means that the photon passes filter $F_i$.
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\vspace{0.5em}
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If these were classical random variables on one probability space, they would satisfy a Bell-type inequality.
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\end{frame}
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\begin{frame}{A classical Bell-type inequality}
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\begin{block}{Bell-type inequality}
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For any classical random variables $P_1,P_2,P_3\in\{0,1\}$,
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$$
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\operatorname{Prob}(P_1=1,P_3=0)
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\leq
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\operatorname{Prob}(P_1=1,P_2=0)
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+
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\operatorname{Prob}(P_2=1,P_3=0).
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$$
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\end{block}
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\vspace{0.5em}
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\begin{proof}
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The event $\{P_1=1,P_3=0\}$ splits into two disjoint cases according to whether $P_2=0$ or $P_2=1$:
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$$
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\{P_1=1,P_3=0\}
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=
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\{P_1=1,P_2=0,P_3=0\}
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\sqcup
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\{P_1=1,P_2=1,P_3=0\}.
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$$
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Therefore,
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$$
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\begin{aligned}
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\operatorname{Prob}(P_1=1,P_3=0)
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&=
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\operatorname{Prob}(P_1=1,P_2=0,P_3=0) \\
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&\quad+
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\operatorname{Prob}(P_1=1,P_2=1,P_3=0) \\
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&\leq
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\operatorname{Prob}(P_1=1,P_2=0)
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+
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\operatorname{Prob}(P_2=1,P_3=0).
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\end{aligned}
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$$
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\end{proof}
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\end{frame}
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\begin{frame}{Experimental law}
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For unpolarized incoming light, the \textbf{observed transition law} for a pair of filters is
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$$
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\operatorname{Prob}(P_i=1,P_j=0)
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=
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\operatorname{Prob}(P_i=1)-\operatorname{Prob}(P_i=1,P_j=1).
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$$
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Using the polarization law,
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$$
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\operatorname{Prob}(P_i=1)=\frac12,
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\qquad
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\operatorname{Prob}(P_i=1,P_j=1)=\frac12\cos^2(\alpha_i-\alpha_j),
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$$
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hence
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$$
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\operatorname{Prob}(P_i=1,P_j=0)
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=
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\frac12-\frac12\cos^2(\alpha_i-\alpha_j)
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=
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\frac12\sin^2(\alpha_i-\alpha_j).
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$$
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\vspace{0.5em}
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So the experimentally observed probabilities depend only on the angle difference $\alpha_i-\alpha_j$.
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\end{frame}
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\begin{frame}{Violation of the classical inequality}
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Substituting the experimental law into the classical inequality gives
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$$
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\frac12\sin^2(\alpha_1-\alpha_3)
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\leq
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\frac12\sin^2(\alpha_1-\alpha_2)
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+
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\frac12\sin^2(\alpha_2-\alpha_3).
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$$
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Choose
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$$
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\alpha_1=0,\qquad
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\alpha_2=\frac{\pi}{6},\qquad
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\alpha_3=\frac{\pi}{3}.
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$$
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Then
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$$
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\begin{aligned}
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\frac12\sin^2\!\left(-\frac{\pi}{3}\right)
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&\leq
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\frac12\sin^2\!\left(-\frac{\pi}{6}\right)
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+
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\frac12\sin^2\!\left(-\frac{\pi}{6}\right) \\
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\frac38 &\leq \frac18+\frac18 \\
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\frac38 &\leq \frac14,
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\end{aligned}
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$$
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which is false.
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\vspace{0.5em}
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Therefore the pairwise polarization data cannot come from one classical probability model with random variables $P_1,P_2,P_3$.
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\end{frame}
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\begin{frame}{The quantum model of polarization}
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The correct model uses a Hilbert space rather than classical events.
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\begin{itemize}
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\item A pure polarization state is a vector
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$$
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\psi=\alpha|0\rangle+\beta|1\rangle \in \mathbb{C}^2.
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$$
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\item A filter at angle $\alpha$ is represented by the orthogonal projection
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$$
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P_\alpha=
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\begin{pmatrix}
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\cos^2\alpha & \cos\alpha\sin\alpha \\
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\cos\alpha\sin\alpha & \sin^2\alpha
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\end{pmatrix}.
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$$
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\item For a pure state $\psi$, the probability of passing the filter is
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$$
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\langle P_\alpha\psi,\psi\rangle.
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$$
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\end{itemize}
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\vspace{0.4em}
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The key point is that sequential measurements are described by \emph{ordered products} of projections, and these need not commute.
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\end{frame}
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\begin{frame}{Recovering the observed law from the operator model}
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Assume the incoming light is unpolarized, so its state is the density matrix
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$$
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\rho=\frac12 I.
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$$
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The probability of passing the first filter $P_{\alpha_i}$ is
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$$
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\operatorname{Prob}(P_i=1)
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=
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\operatorname{tr}(\rho P_{\alpha_i})
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=
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\frac12\operatorname{tr}(P_{\alpha_i})
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=
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\frac12.
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$$
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If the photon passes the first filter, the post-measurement state is
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$$
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\rho_i
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=
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\frac{P_{\alpha_i}\rho P_{\alpha_i}}{\operatorname{tr}(\rho P_{\alpha_i})}
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=
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P_{\alpha_i}.
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$$
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$$
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P_\alpha=
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\begin{pmatrix}
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\cos^2\alpha & \cos\alpha\sin\alpha \\
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\cos\alpha\sin\alpha & \sin^2\alpha
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\end{pmatrix}.
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$$
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Therefore
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$$
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\operatorname{Prob}(P_j=1\mid P_i=1)
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=
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\operatorname{tr}(\rho_i P_{\alpha_j})
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=
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\operatorname{tr}(P_{\alpha_i}P_{\alpha_j})
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=
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\cos^2(\alpha_i-\alpha_j).
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$$
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\end{frame}
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\begin{frame}{Recovering the observed law from the operator model (cont.)}
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$$
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\begin{aligned}
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\operatorname{Prob}(P_i=1,P_j=0)
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&=
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\operatorname{Prob}(P_i=1)
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\bigl(1-\operatorname{Prob}(P_j=1\mid P_i=1)\bigr) \\
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&=
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\frac12\bigl(1-\cos^2(\alpha_i-\alpha_j)\bigr) \\
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&=
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\frac12\sin^2(\alpha_i-\alpha_j).
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\end{aligned}
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$$
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This matches the experiment exactly.
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\end{frame}
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\begin{frame}{Conclusion}
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\begin{itemize}
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\item The classical model predicts a Bell-type inequality for three $0$--$1$ random variables.
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\item The polarization experiment violates that inequality.
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\item The resolution is that the quantities measured are \emph{sequential probabilities}, not joint probabilities of classical random variables.
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\item In quantum probability, events are modeled by projections on a Hilbert space, and measurement order matters.
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\end{itemize}
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\vspace{0.6em}
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This is one of the basic motivations for passing from classical probability to non-commutative probability.
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\end{frame}
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\section{Concentration on Spheres and quantum states}
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\begin{frame}{Quantum states: pure vs.\ mixed}
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\begin{itemize}
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\item A finite-dimensional quantum system is modeled by a complex Hilbert space (a complete inner product space)
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$$
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\mathcal H \cong \mathbb C^{n+1}.
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$$
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\item A \textbf{pure state} is represented by a unit vector
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$$
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\psi \in \mathcal H, \qquad \|\psi\|=1.
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$$
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\item A \textbf{mixed state} is represented by a density matrix
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$$
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\rho \geq 0, \qquad \operatorname{tr}(\rho)=1.
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$$
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\item Pure states describe maximal information; mixed states describe probabilistic mixtures or partial information.
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\end{itemize}
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\vspace{0.4em}
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\begin{block}{Key distinction}
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Pure states form a curved geometric space; mixed states form a convex set inside the space of matrices.
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\end{block}
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\end{frame}
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\begin{frame}{Why pure states are not vectors}
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\begin{itemize}
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\item Two nonzero vectors that differ by a nonzero complex scalar represent the same physical state:
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$$
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\psi \sim \lambda \psi, \qquad \lambda \in \mathbb C^\times.
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$$
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\item In particular, multiplying by a phase $e^{i\theta}$ does not change any physical predictions.
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\item Therefore the physical pure state is not a single vector, but the \emph{complex line} spanned by that vector.
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\end{itemize}
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\vspace{0.4em}
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Hence the space of pure states is
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$$
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\mathbb P(\mathcal H)
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=
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(\mathcal H \setminus \{0\})/\mathbb C^\times.
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$$
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After choosing a basis $\mathcal H \cong \mathbb C^{n+1}$, this becomes
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$$
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\mathbb P(\mathcal H) \cong \mathbb C P^n.
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$$
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\end{frame}
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\begin{frame}{Relation with the sphere}
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\begin{itemize}
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\item Every nonzero vector can be normalized, so each pure state has a representative on the unit sphere
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$$
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S^{2n+1} \subset \mathbb C^{n+1}.
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$$
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\item Two unit vectors represent the same pure state exactly when they differ by a phase:
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$$
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z \sim e^{i\theta} z.
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$$
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\item Therefore
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$$
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\mathbb C P^n = S^{2n+1}/S^1.
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$$
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\end{itemize}
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\vspace{0.4em}
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The quotient map
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$$
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p:S^{2n+1}\to \mathbb C P^n, \qquad p(z)=[z]=\{\lambda z : \lambda \in \mathbb C^\times\},
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$$
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is the \textbf{Hopf fibration}.
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\end{frame}
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\begin{frame}{How the metric descends to $\mathbb C P^n$}
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\begin{itemize}
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\item The sphere $S^{2n+1}$ inherits the round metric from the Euclidean metric on
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$$
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\mathbb C^{n+1} \cong \mathbb R^{2n+2}.
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$$
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\item The fibers of the Hopf map are circles
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$$
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p^{-1}([z]) = \{e^{i\theta}z : \theta \in \mathbb R\}.
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$$
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\item Tangent vectors split into:
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\begin{itemize}
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\item \textbf{vertical directions}: tangent to the $S^1$-fiber,
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\item \textbf{horizontal directions}: orthogonal complement to the fiber.
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\end{itemize}
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\item The differential $dp$ identifies horizontal vectors on the sphere with tangent vectors on $\mathbb C P^n$.
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\end{itemize}
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\vspace{0.4em}
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This allows the round metric on $S^{2n+1}$ to define a metric on $\mathbb C P^n$.
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\end{frame}
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\begin{frame}{The induced metric: Fubini--Study metric}
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\begin{itemize}
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\item The metric on $\mathbb C P^n$ obtained from the Hopf quotient is the
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\textbf{Fubini--Study metric}.
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\item So the geometric picture is:
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$$
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S^{2n+1}
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\xrightarrow{\text{Hopf fibration}}
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\mathbb C P^n
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$$
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$$
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\text{round metric}
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\rightsquigarrow
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\text{Fubini--Study metric}.
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$$
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\item The normalized Riemannian volume measure induced by this metric gives the natural probability measure on pure states.
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\end{itemize}
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\vspace{0.5em}
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\begin{block}{Proof roadmap}
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To prove this carefully, one usually shows:
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\begin{enumerate}
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\item $p:S^{2n+1}\to \mathbb C P^n$ is a smooth surjective submersion,
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\item the vertical space is the tangent space to the $S^1$-orbit,
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\item horizontal lifts are well defined,
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\item the quotient metric is exactly the Fubini--Study metric.
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\end{enumerate}
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\end{block}
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\end{frame}
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\begin{frame}{Maxwell-Boltzmann Distribution Law}
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\begin{columns}[T]
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\column{0.58\textwidth}
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Consider the orthogonal projection
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$$
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\pi_{n,k}:S^n(\sqrt{n})\to \mathbb{R}^k.
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$$
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Its push-forward measure converges to the standard Gaussian:
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$$
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(\pi_{n,k})_*\sigma^n\to \gamma^k.
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$$
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\vspace{0.5em}
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This explains why Gaussian behavior emerges from high-dimensional spheres and supports the proof strategy for Levy concentration.
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\column{0.42\textwidth}
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\begin{figure}
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\includegraphics[width=\textwidth]{../latex/images/maxwell.png}
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\end{figure}
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\end{columns}
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\end{frame}
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\begin{frame}{Levy Concentration}
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\begin{block}{Levy's theorem}
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If $f:S^n\to \mathbb{R}$ is $1$-Lipschitz, then there exists a median $a_0$ such that
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$$
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\mu\{x\in S^n:|f(x)-a_0|\geq \epsilon\}
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\leq
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2\exp\left(-\frac{(n-1)\epsilon^2}{2}\right).
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$$
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\end{block}
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\begin{itemize}
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\item In high dimension, most Lipschitz observables are almost constant.
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\item This is the geometric mechanism behind generic entanglement.
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\end{itemize}
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\end{frame}
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\section{Main Result}
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\begin{frame}{Generic Entanglement Theorem}
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\begin{block}{Hayden--Leung--Winter}
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Let $\psi\in \mathcal{P}(A\otimes B)$ be Haar-random and define
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$$
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\beta=\frac{1}{\ln(2)}\frac{d_A}{d_B}.
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$$
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For $d_B\geq d_A\geq 3$,
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$$
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\operatorname{Pr}[H(\psi_A)<\log_2(d_A)-\alpha-\beta]
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\leq
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\exp\left(
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-\frac{1}{8\pi^2\ln(2)}
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\frac{(d_Ad_B-1)\alpha^2}{(\log_2 d_A)^2}
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\right).
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$$
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\end{block}
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With overwhelming probability, a random pure state is almost maximally entangled.
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\end{frame}
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\begin{frame}{How the Entropy Observable Fits In}
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\begin{figure}
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\centering
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|
\begin{tikzpicture}[node distance=30mm, thick,
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main/.style={draw, draw=white},
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towards/.style={->},
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|
towards_imp/.style={->,red},
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mutual/.style={<->}
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]
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|
\node[main] (cp) {$\mathbb{C}P^{d_A d_B-1}$};
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|
\node[main] (pa) [left of=cp] {$\mathcal{P}(A\otimes B)$};
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|
\node[main] (sa) [below of=pa] {$\mathcal{S}(A)$};
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\node[main] (rng) [right of=sa] {$[0,\log_2 d_A]$};
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\draw[mutual] (cp) -- (pa);
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\draw[towards] (pa) -- node[left] {$\operatorname{Tr}_B$} (sa);
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\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}
|