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presentation/ZheyuanWu_HonorThesis_Presentation.pdf
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presentation/ZheyuanWu_HonorThesis_Presentation.tex
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@@ -18,8 +18,6 @@
\usepackage{tikz}
\usepackage{braket}
\DeclareMathOperator{\sen}{sen}
\DeclareMathOperator{\tg}{tg}
\DeclareMathOperator{\obdiam}{ObsDiam}
\DeclareMathOperator{\diameter}{diam}
@@ -157,112 +155,7 @@
\end{block}
\end{frame}
\begin{frame}{Backgrounds: Motivation of Tensor product}
Recall from the traditional notation of product space of two vector spaces $V$ and $W$, that is, $V\times W$, is the set of all ordered pairs $(\ket{v},\ket{w})$ where $\ket{v}\in V$ and $\ket{w}\in W$.
The space has dimension $\dim V+\dim W$.
We want to define a vector space with the notation of multiplication of two vectors from different vector spaces.
That is
$$
(\ket{v_1}+\ket{v_2})\otimes \ket{w}=(\ket{v_1}\otimes \ket{w})+(\ket{v_2}\otimes \ket{w})
$$
$$
\ket{v}\otimes (\ket{w_1}+\ket{w_2})=(\ket{v}\otimes \ket{w_1})+(\ket{v}\otimes \ket{w_2})
$$
and enables scalar multiplication by
$$
\lambda (\ket{v}\otimes \ket{w})=(\lambda \ket{v})\otimes \ket{w}=\ket{v}\otimes (\lambda \ket{w})
$$
And we wish to build a way to associate the basis of $V$ and $W$ with the basis of $V\otimes W$. That makes the tensor product a vector space with dimension $\dim V\times \dim W$.
\end{frame}
\begin{frame}{Backgrounds: Tensor product of vectors}
\begin{block}{Definition of Bilinear functional}
A bilinear functional is a bilinear function $\beta:V\times W\to \mathbb{F}$ satisfying the condition that $\ket{v}\to \beta(\ket{v},\ket{w})$ is a linear functional for all $\ket{w}\in W$ and $\ket{w}\to \beta(\ket{v},\ket{w})$ is a linear functional for all $\ket{v}\in V$.
\end{block}
The vector space of all bilinear functionals is denoted by $\mathcal{B}(V, W)$.
\begin{block}{Definition of Tensor product of vectors}
Let $V, W$ be two vector spaces.
Let $V'$ and $W'$ be the dual spaces of $V$ and $W$, respectively, that is $V'=\{\psi:V\to \mathbb{F}\}$ and $W'=\{\phi:W\to \mathbb{F}\}$, $\psi, \phi$ are linear functionals.
The \textbf{tensor product of vectors} $v\in V$ and $w\in W$ is the bilinear functional defined by $\forall (\psi,\phi)\in V'\times W'$ given by the notation
$$
(v\otimes w)(\psi,\phi)=\psi(v)\phi(w)
$$
\end{block}
\end{frame}
\begin{frame}{Backgrounds: Tensor product of vector spaces}
\begin{block}{Definition of Tensor product of vector spaces}
The tensor product of two vector spaces $V$ and $W$ is the vector space $\mathcal{B}(V',W')$
Notice that the basis of such vector space is the linear combination of the basis of $V'$ and $W'$, that is, if $\{e_i\}$ is the basis of $V'$ and $\{f_j\}$ is the basis of $W'$, then $\{e_i\otimes f_j\}$ is the basis of $\mathcal{B}(V', W')$.
Since $\{e_i\}$ and $\{f_j\}$ are bases of $V'$ and $W'$, respectively, then we can always find a set of linear functionals $\{\phi_i\}$ and $\{\psi_j\}$ such that $\phi_i(e_j)=\delta_{ij}$ and $\psi_j(f_i)=\delta_{ij}$. (Here $\delta_{ij}=1$ if $i=j$ and $0$ otherwise.)
$$
V\otimes W=\left\{\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w): \phi_i\in V', \psi_j\in W'\right\}
$$
\end{block}
Note that $\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w)$ is a bilinear functional that maps $V'\times W'$ to $\mathbb{F}$.
\end{frame}
\begin{frame}{Backgrounds: Trace}
\label{defn:trace}
\begin{block}{Trace}
Let $T$ be a linear operator on $\mathscr{H}$, $(e_1,e_2,\cdots,e_n)$ be a basis of $\mathscr{H}$ and $(\epsilon_1,\epsilon_2,\cdots,\epsilon_n)$ be a basis of dual space $\mathscr{H}^*$. Then the trace of $T$ is defined by
$$
\operatorname{Tr}(T)=\sum_{i=1}^n \epsilon_i(T(e_i))=\sum_{i=1}^n \langle e_i,T(e_i)\rangle
$$
\end{block}
This is equivalent to the sum of the diagonal elements of $T$.
\vspace{1em}
Q: How we generalize the trace to a subsystem of a larger, entangled quantum system $A\otimes B$?
\end{frame}
\begin{frame}{Backgrounds: Partial trace}
\begin{block}{Definition of Partial trace}
Let $T$ be a linear operator on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$, where $\mathscr{A}$ and $\mathscr{B}$ are finite-dimensional Hilbert spaces.
An operator $T$ on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$ can be written as
$$
T=\sum_{i=1}^n a_i A_i\otimes B_i
$$
where $A_i$ is a linear operator on $\mathscr{A}$ and $B_i$ is a linear operator on $\mathscr{B}$.
The $\mathscr{B}$-partial trace of $T$ ($\operatorname{Tr}_{\mathscr{B}}(T):\mathcal{L}(\mathscr{A}\otimes \mathscr{B})\to \mathcal{L}(\mathscr{A})$) is the linear operator on $\mathscr{A}$ defined by
$$
\operatorname{Tr}_{\mathscr{B}}(T)=\sum_{i=1}^n a_i \operatorname{Tr}(B_i) A_i
$$
\end{block}
\end{frame}
\input{./backgrounds.tex}
\begin{frame}{Information theory in classical systems}
@@ -288,7 +181,7 @@ For a density matrix $\rho$,
$$
S(\rho)=-\operatorname{Tr}(\rho\log_2\rho).
$$
This measures the intrinsic mixedness of the quantum state and is basis-independent.
This measures the intrinsic uncertainty of the quantum state and is basis-independent.
\end{block}
\begin{block}{Entanglement entropy}
@@ -513,9 +406,9 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
\begin{frame}{Levy Concentration}
\begin{block}{Definition of Lipschitz function}
A function $f:X\to Y$, where $X,Y$ are metric spaces, is $L$-Lipschitz if there exists a constant $L$ such that $|f(x)-f(y)|\leq L|y-x|$ for all $x,y\in S^n$.
\end{block}
% \begin{block}{Definition of Lipschitz function}
% A function $f:X\to Y$, where $X,Y$ are metric spaces, is $L$-Lipschitz if there exists a constant $L$ such that $|f(x)-f(y)|\leq L|y-x|$ for all $x,y\in S^n$.
% \end{block}
\begin{block}{Levy's lemma}
If $f:S^n\to \mathbb{R}$ is $1$-Lipschitz, then there exists a $a_0$ such that for $\epsilon>0$,
$$
@@ -569,13 +462,10 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
$$
\end{block}
\begin{block}{Lipschitz estimate}
$$
\|H(\psi_A)\|_{\mathrm{Lip}}
\leq
\sqrt{8}\,\log_2(d_A),
\qquad d_A\geq 3.
$$
\begin{block}{Lipschitz estimate for $H(\psi_A)$}
The Lipschitz constant for the function $
H(\psi_A)$ should be upper bounded by $
\sqrt{8}\,\log_2(d_A)$, for $d_A\geq 3$.
\end{block}
Levy concentration plus these two estimates produces the exponential entropy tail bound.
@@ -605,7 +495,7 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
\right)=1-\Theta(e^{-c d_B})$, a random pure state is almost maximally entangled $\log_2(d_A)-\frac{1}{2\ln(2)}\frac{d_A}{d_B}=\log_2(d_A)-\Theta(\frac{1}{d_B})$.
\end{frame}
\begin{frame}{A natual question from the observables}
\begin{frame}{A natural question from the observables}
\textbf{What does the hayden--leung--winter theorem generalize the behavior of the lipschitz function $S^n\to \mathbb{R}$ and the lipschitz function $\mathbb{C}P^n\to \mathbb{R}$ as $n\to \infty$?}
\end{frame}
@@ -614,13 +504,10 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
\begin{frame}{Observable diameter: the inner definition}
\begin{block}{Partial diameter on $\mathbb{R}$}
Let $\nu$ (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
Let $\mu$ be a Borel probability measure on $\mathbb{R}$ and let $\alpha \in (0,1]$.
The \textbf{partial diameter} of $\mu$ at mass level $\alpha$ is
$$
\diameter(\nu;\alpha):=
\{\diameter(A):A \subseteq \mathcal{B}(\mathbb{R}),
\nu(A)\ge \alpha
\},
\diameter(\mu;\alpha):=\inf_{A\subset \mathbb{R}}\left\{ \diameter(A): \mu(A)\geq \alpha \right\}.
$$
where
$$
@@ -630,8 +517,8 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
\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 $\kappa$.
\item The partial diameter asks for: what is the shortest interval I need to capture at least $\alpha$ of the mass (measure)?
\item If $1$ is the total measure of the space, $\diameter(\mu;1-\kappa)$, 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}

107
presentation/backgrounds.tex Normal file
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@@ -0,0 +1,107 @@
\begin{frame}{Backgrounds: Motivation of Tensor product}
Recall from the traditional notation of product space of two vector spaces $V$ and $W$, that is, $V\times W$, is the set of all ordered pairs $(\ket{v},\ket{w})$ where $\ket{v}\in V$ and $\ket{w}\in W$.
The space has dimension $\dim V+\dim W$.
We want to define a vector space with the notation of multiplication of two vectors from different vector spaces.
That is
$$
(\ket{v_1}+\ket{v_2})\otimes \ket{w}=(\ket{v_1}\otimes \ket{w})+(\ket{v_2}\otimes \ket{w})
$$
$$
\ket{v}\otimes (\ket{w_1}+\ket{w_2})=(\ket{v}\otimes \ket{w_1})+(\ket{v}\otimes \ket{w_2})
$$
and enables scalar multiplication by
$$
\lambda (\ket{v}\otimes \ket{w})=(\lambda \ket{v})\otimes \ket{w}=\ket{v}\otimes (\lambda \ket{w})
$$
And we wish to build a way to associate the basis of $V$ and $W$ with the basis of $V\otimes W$. That makes the tensor product a vector space with dimension $\dim V\times \dim W$.
\end{frame}
\begin{frame}{Backgrounds: Tensor product of vectors}
\begin{block}{Definition of Bilinear functional}
A bilinear functional is a bilinear function $\beta:V\times W\to \mathbb{F}$ satisfying the condition that $\ket{v}\to \beta(\ket{v},\ket{w})$ is a linear functional for all $\ket{w}\in W$ and $\ket{w}\to \beta(\ket{v},\ket{w})$ is a linear functional for all $\ket{v}\in V$.
\end{block}
The vector space of all bilinear functionals is denoted by $\mathcal{B}(V, W)$.
\begin{block}{Definition of Tensor product of vectors}
Let $V, W$ be two vector spaces.
Let $V'$ and $W'$ be the dual spaces of $V$ and $W$, respectively, that is $V'=\{\psi:V\to \mathbb{F}\}$ and $W'=\{\phi:W\to \mathbb{F}\}$, $\psi, \phi$ are linear functionals.
The \textbf{tensor product of vectors} $v\in V$ and $w\in W$ is the bilinear functional defined by $\forall (\psi,\phi)\in V'\times W'$ given by the notation
$$
(v\otimes w)(\psi,\phi)=\psi(v)\phi(w)
$$
\end{block}
\end{frame}
\begin{frame}{Backgrounds: Tensor product of vector spaces}
\begin{block}{Definition of Tensor product of vector spaces}
The tensor product of two vector spaces $V$ and $W$ is the vector space $\mathcal{B}(V',W')$
Notice that the basis of such vector space is the linear combination of the basis of $V'$ and $W'$, that is, if $\{e_i\}$ is the basis of $V'$ and $\{f_j\}$ is the basis of $W'$, then $\{e_i\otimes f_j\}$ is the basis of $\mathcal{B}(V', W')$.
Since $\{e_i\}$ and $\{f_j\}$ are bases of $V'$ and $W'$, respectively, then we can always find a set of linear functionals $\{\phi_i\}$ and $\{\psi_j\}$ such that $\phi_i(e_j)=\delta_{ij}$ and $\psi_j(f_i)=\delta_{ij}$. (Here $\delta_{ij}=1$ if $i=j$ and $0$ otherwise.)
$$
V\otimes W=\left\{\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w): \phi_i\in V', \psi_j\in W'\right\}
$$
\end{block}
Note that $\sum_{i=1}^n \sum_{j=1}^m a_{ij} \phi_i(v)\psi_j(w)$ is a bilinear functional that maps $V'\times W'$ to $\mathbb{F}$.
\end{frame}
\begin{frame}{Backgrounds: Trace}
\label{defn:trace}
\begin{block}{Trace}
Let $T$ be a linear operator on $\mathscr{H}$, $(e_1,e_2,\cdots,e_n)$ be a basis of $\mathscr{H}$ and $(\epsilon_1,\epsilon_2,\cdots,\epsilon_n)$ be a basis of dual space $\mathscr{H}^*$. Then the trace of $T$ is defined by
$$
\operatorname{Tr}(T)=\sum_{i=1}^n \epsilon_i(T(e_i))=\sum_{i=1}^n \langle e_i,T(e_i)\rangle
$$
\end{block}
This is equivalent to the sum of the diagonal elements of $T$.
\vspace{1em}
Q: How we generalize the trace to a subsystem of a larger, entangled quantum system $A\otimes B$?
\end{frame}
\begin{frame}{Backgrounds: Partial trace}
\begin{block}{Definition of Partial trace}
Let $T$ be a linear operator on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$, where $\mathscr{A}$ and $\mathscr{B}$ are finite-dimensional Hilbert spaces.
An operator $T$ on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$ can be written as
$$
T=\sum_{i=1}^n a_i A_i\otimes B_i
$$
where $A_i$ is a linear operator on $\mathscr{A}$ and $B_i$ is a linear operator on $\mathscr{B}$.
The $\mathscr{B}$-partial trace of $T$ ($\operatorname{Tr}_{\mathscr{B}}(T):\mathcal{L}(\mathscr{A}\otimes \mathscr{B})\to \mathcal{L}(\mathscr{A})$) is the linear operator on $\mathscr{A}$ defined by
$$
\operatorname{Tr}_{\mathscr{B}}(T)=\sum_{i=1}^n a_i \operatorname{Tr}(B_i) A_i
$$
\end{block}
\end{frame}