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presentation/backgrounds.tex

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+
+\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}