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Zheyuan Wu
2026-03-29 18:08:48 -05:00
parent dd10a1969b
commit 2949c3e5b6
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52
latex/chapters/chap1.tex
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@@ -121,6 +121,8 @@ $$
Other revised experiments (e.g., Aspect's experiment, calcium entangled photon experiment) are also conducted and the inequality is still violated.
\subsection{The true model of light polarization}
The contradiction above marks the point where classical probability stops being adequate. To continue, the sample-space picture must be replaced by states in a Hilbert space and by projections representing measurements. This operator model keeps the experimental probabilities but no longer forces incompatible measurements into a single classical joint distribution.
The full description of the light polarization is given below:
@@ -160,8 +162,8 @@ The probability that a photon passes the first filter $P_{\alpha_i}$ is given by
$$
\operatorname{Prob}(P_i=1)
=\operatorname{tr}(\rho P_{\alpha_i})
=\frac{1}{2} \operatorname{tr}(P_{\alpha_i})
=\operatorname{Tr}(\rho P_{\alpha_i})
=\frac{1}{2} \operatorname{Tr}(P_{\alpha_i})
=\frac{1}{2}
$$
@@ -170,7 +172,7 @@ If the photon passes the first filter, the post-measurement state is given by th
$$
\rho \longmapsto
\rho_i
=\frac{P_{\alpha_i}\rho P_{\alpha_i}}{\operatorname{tr}(\rho P_{\alpha_i})}
=\frac{P_{\alpha_i}\rho P_{\alpha_i}}{\operatorname{Tr}(\rho P_{\alpha_i})}
= P_{\alpha_i}.
$$
@@ -178,7 +180,7 @@ The probability that the photon then passes the second filter is
$$
\operatorname{Prob}(P_j=1 \mid P_i=1)
=\operatorname{tr}(P_{\alpha_i} P_{\alpha_j})
=\operatorname{Tr}(P_{\alpha_i} P_{\alpha_j})
=\cos^2(\alpha_i-\alpha_j).
$$
@@ -198,6 +200,8 @@ This agrees with the experimentally observed transmission probabilities, but it
\section{Concentration of measure phenomenon}
The operator model explains why entanglement is a meaningful observable, but it does not yet explain why large random systems are typically highly entangled. That is the role of concentration of measure. The next section moves from quantum motivation back to geometry and probability, where high-dimensional spheres already exhibit the same kind of rigidity that later reappears in the entropy of random bipartite states.
\begin{defn}
$\eta$-Lipschitz function
@@ -251,9 +255,9 @@ To prove the lemma, we need to have a good understanding of the Riemannian geome
$$
\kappa_n(\epsilon)=\frac{\int_\epsilon^{\frac{\pi}{2}}\cos^{n-1}(t)dt}{\int_0^{\frac{\pi}{2}}\cos^{n-1}(t)dt}
$$
$a_0$ is the \textbf{Levy mean} of function $f$, that is, the level set $f^{-1}:\mathbb{R}\to S^n$ divides the sphere into equal halves, characterized by the following equality:
$a_0$ is a \textbf{median} of $f$, characterized by the following inequalities:
$$
\mu(f^{-1}(-\infty,a_0])\geq \frac{1}{2} \text{ and } \mu(f^{-1}[a_0,\infty))\geq \frac{1}{2}
\mu(f^{-1}((-\infty,a_0]))\geq \frac{1}{2} \text{ and } \mu(f^{-1}([a_0,\infty)))\geq \frac{1}{2}
$$
\end{theorem}
@@ -275,7 +279,7 @@ We will prove the theorem via the Maxwell-Boltzmann distribution law in this sec
Basically, you can consider the Gaussian measure as the normalized Lebesgue measure on $\mathbb{R}^k$ with standard deviation $1$.
It also has another name, the Projective limit theorem.~\cite{romanvershyni}
It also has another name, the Poincar\'e limit theorem.~\cite{romanvershyni}
If $X\sim \operatorname{Unif}(S^n(\sqrt{n}))$, then for any fixed unit vector $x$ we have $\langle X,x\rangle\to N(0,1)$ in distribution as $n\to \infty$.
@@ -348,7 +352,7 @@ Now we can prove Levy's concentration theorem, the proof is from~\cite{shioya201
It is sufficient to show that,
$$
U_{\epsilon_1}(\Omega_+)\cup U_{\epsilon_2}(\Omega_-)\subset \{x'-\epsilon_1\leq f_{n_i}\leq x'+\epsilon_2\}
U_{\epsilon_1}(\Omega_+)\cap U_{\epsilon_2}(\Omega_-)\subset \{x'-\epsilon_1\leq f_{n_i}\leq x'+\epsilon_2\}
$$
By 1-Lipschitz continuity of $f_{n_i}$, we have for all $\zeta\in U_{\epsilon_1}(\Omega_+)$, there is a point $\xi\in \Omega_+$ such that $d(\zeta,\xi)\leq \epsilon_1$. So $U_{\epsilon_1}(\Omega_+)\subset \{f_{n_i}\geq x'-\epsilon_1\}$. With the same argument, we have $U_{\epsilon_2}(\Omega_-)\subset \{f_{n_i}\leq x+\epsilon_2\}$.
@@ -359,7 +363,7 @@ Now we can prove Levy's concentration theorem, the proof is from~\cite{shioya201
\begin{aligned}
(f_{n_i})_*\sigma^{n_i}[x'-\epsilon_1,x'+\epsilon_2]&=\sigma^{n_i}(x'-\epsilon_1\leq f_{n_i}\leq x'+\epsilon_2)\\
&\geq \sigma^{n_i}(U_{\epsilon_1}(\Omega_+)\cap U_{\epsilon_2}(\Omega_-))\\
&=\sigma^{n_i}(U_{\epsilon_1}(\Omega_+))+\sigma^{n_i}(U_{\epsilon_2}(\Omega_-))-1\\
&\geq \sigma^{n_i}(U_{\epsilon_1}(\Omega_+))+\sigma^{n_i}(U_{\epsilon_2}(\Omega_-))-1\\
\end{aligned}
$$
@@ -380,7 +384,7 @@ Now we can prove Levy's concentration theorem, the proof is from~\cite{shioya201
$$
\begin{aligned}
\sigma_\infty[x'-\epsilon_1,x'+\epsilon_2]&\geq \liminf_{i\to \infty}(f_{n_i})_*\sigma^{n_i}[x'-\epsilon_1,x'+\epsilon_2]\\
&\geq \gamma^1[x'-\epsilon_1,\infty)\cap \gamma^1(-\infty,x+\epsilon_2]-1\\
&\geq \gamma^1[x'-\epsilon_1,\infty)+\gamma^1(-\infty,x+\epsilon_2]-1\\
&=\gamma^1[x-\epsilon_1,x+\epsilon_2]
\end{aligned}
$$
@@ -391,6 +395,8 @@ The full proof of Levy's concentration theorem requires more digestion for cases
\section{The application of the concentration of measure phenomenon in non-commutative probability theory}
Having established concentration for Lipschitz observables on high-dimensional spheres, we can now return to quantum information. The remaining step is to identify a physically meaningful observable on pure states whose geometry is controlled well enough for Levy-type bounds to apply. In this thesis that observable is entanglement entropy, viewed after partial trace.
In quantum communication, we can pass classical bits by sending quantum states. However, by the indistinguishability (Proposition~\ref{prop:indistinguishability}) of quantum states, we cannot send an infinite number of classical bits over a single qubit. There exists a bound for zero-error classical communication rate over a quantum channel.
\begin{theorem}
@@ -424,6 +430,8 @@ To surpass the Holevo bound, we need to use the entanglement of quantum states.
\subsection{Superdense coding and entanglement}
Superdense coding is the operational reason entanglement matters in this chapter. It shows that entangled states are not merely algebraically interesting: they change communication capacity. Once that point is clear, the natural probabilistic question is whether highly entangled states are rare or typical, which leads directly to Hayden's concentration theorem.
The description of the superdense coding can be found in~\cite{gupta2015functionalanalysisquantuminformation} and~\cite{Hayden}.
Suppose $A$ and $B$ share a Bell state (or other maximally entangled state) $|\Phi^+\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$, where $A$ holds the first part and $B$ holds the second part.
@@ -453,6 +461,8 @@ Additionally, no information can be gained by measuring a pair of entangled qubi
\subsection{Hayden's concentration of measure phenomenon}
The geometric and communication-theoretic threads now meet. Random pure states live on a projective state space, partial trace sends them to mixed states, and entropy turns those mixed states into real numbers. Hayden's theorem is precisely the statement that this entropy observable is concentrated when the ambient dimension is large.
The application of the concentration of measure phenomenon in the superdense coding can be realized in random sampling the entangled qubits~\cite{Hayden}:
It is a theorem connecting the following mathematical structure:
@@ -468,8 +478,8 @@ It is a theorem connecting the following mathematical structure:
% define nodes
\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] {$S_A$};
\node[main] (rng) [right of=sa] {$[0,\infty)\subset \mathbb{R}$};
\node[main] (sa) [below of=pa] {$\mathcal{S}(A)$};
\node[main] (rng) [right of=sa] {$[0,\log_2(d_A)]\subset \mathbb{R}$};
% draw edges
\draw[mutual] (cp) -- (pa);
@@ -483,7 +493,7 @@ It is a theorem connecting the following mathematical structure:
\begin{itemize}
\item The red arrow is the concentration of measure effect. $f=H(\operatorname{Tr}_B(\psi))$.
\item $S_A$ denotes the mixed states on $A$.
\item $\mathcal{S}(A)$ denotes the mixed states on $A$.
\end{itemize}
To prove the concentration of measure phenomenon, we need to analyze the following elements involved in figure~\ref{fig:Hayden_concentration_of_measure_phenomenon}:
@@ -491,7 +501,7 @@ To prove the concentration of measure phenomenon, we need to analyze the followi
The existence and uniqueness of the Haar measure is a theorem in compact lie group theory. For this research topic, we will not prove it.
Due to time constrains of the projects, the following lemma is demonstrated but not investigated thoroughly through the research:
Due to time constraints of the project, the following lemma is demonstrated but not investigated thoroughly through the research:
\begin{lemma}
@@ -499,9 +509,9 @@ Due to time constrains of the projects, the following lemma is demonstrated but
Page's lemma for expected entropy of mixed states
Choose a random pure state $\sigma=|\psi\rangle\langle\psi|$ from $A'\otimes A$.
Choose a random pure state $\sigma=|\psi\rangle\langle\psi|$ from $A\otimes B$.
The expected value of the entropy of entanglement is known and satisfies a concentration inequality known as Page's formula~\cite{Pages_conjecture,Pages_conjecture_simple_proof,Bengtsson_Zyczkowski_2017}[15.72].
The expected value of the entropy of entanglement is known and is given by Page's formula~\cite{Pages_conjecture,Pages_conjecture_simple_proof,Bengtsson_Zyczkowski_2017}[15.72].
$$
\mathbb{E}[H(\psi_A)]=\frac{1}{\ln(2)}\left(\sum_{j=d_B+1}^{d_Ad_B}\frac{1}{j}-\frac{d_A-1}{2d_B}\right) \geq \log_2(d_A)-\frac{1}{2\ln(2)}\frac{d_A}{d_B}
@@ -546,12 +556,12 @@ Then we have bound for Lipschitz constant $\eta$ of the map $S(\varphi_A): \math
Therefore
$$
\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}
\frac{\partial g}{\partial x_{jk}}=\frac{\partial g}{\partial p_j}\frac{\partial p_j}{\partial 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}
$$
Then the lipschitz constant of $g$ is
Then the Lipschitz constant of $g$ is
$$
\begin{aligned}
@@ -564,7 +574,7 @@ Then we have bound for Lipschitz constant $\eta$ of the map $S(\varphi_A): \math
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 $\sum_{j=0}^{d_A}p_j(\varphi)\ln p_j(\varphi)\leq 0$.
Since $0\leq p_j\leq 1$, we have $\ln p_j(\varphi)\leq 0$, hence $\sum_{j=1}^{d_A}p_j(\varphi)\ln p_j(\varphi)\leq 0$.
$$
\begin{aligned}
@@ -583,7 +593,7 @@ Then we have bound for Lipschitz constant $\eta$ of the map $S(\varphi_A): \math
\end{aligned}
$$
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.
Proving $\sum_{j=1}^{d_A} p_j(\varphi)(\ln p_j(\varphi))^2\leq (\ln d_A)^2$ 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
@@ -591,7 +601,7 @@ Then we have bound for Lipschitz constant $\eta$ of the map $S(\varphi_A): \math
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 $\sqrt{8}\log_2(d_A)$.
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