# Quantum - Holevo's theorem $$\chi (p\_i, |v\_i \rangle) = H( \sum\_i p\_i | v\_i \rangle \langle v\_i|$$ Allowed to send $$p\_1, p\_2, p\_3, ..., p\_k.$$ Capacity is $$\frac{max}{p\_i, p\_i} \chi = \frac{max}{p\_i, p\_i} \chi - \sum\_i p\_i H(p\_i)$$ block length: n capacity: C pick $$2^{n(c - \epsilon)}$$ random codewords with letter chosen with probability maximizing $$H(B) - H(B | A)$$ . Steps: * Alice sends Bob codeword * Bob gets codeword with noise * Find the codeword most likely to have been the input words as $$n \rightarrow \infty , \in \rightarrow 0$$ . #### How about quantum case? Upper bound: * Alice sends $$p\_i$$ to Bob Will show that for single state decoding, Shannon information provided by any measurement of Bob's $$\lt$$ Holevo's information $$\chi$$ . Alice's record of $$p\_i$$ $$|i \rangle \longrightarrow p\_i \buildrel Bob \over \longrightarrow | b\_j\rangle$$ ### $$\def\myHearts#1#2{\color{#1}{\heartsuit}\kern-0pt\color{#2}{\heartsuit}} \myHearts{red} {blue}$$ $$| 0\rangle$$ ### Brackets ``` \left\langle \frac{1}{2} \middle| 1 \right\rangle ``` $$\left\langle \frac{1}{2} \middle| 1 \right\rangle$$ ### Fractions and brackets ``` \left( \begin{array}{cc} 2\tau & 7\phi-frac5{12} \\ 3\psi & \frac{\pi}8 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) \mbox{~and~} \left[ \begin{array}{cc|r} 3 & 4 & 5 \\ 1 & 3 & 729 \end{array} \right] ``` $$\left( \begin{array}{cc} 2\tau & 7\phi-frac5{12} \ 3\psi & \frac{\pi}8 \end{array} \right) \left( \begin{array}{c} x \ y \end{array} \right) \mbox{~~and~~} \left\[ \begin{array}{cc|r} 3 & 4 & 5 \ 1 & 3 & 729 \end{array} \right]$$ ### Function definition ``` f(z) = \left\{ \begin{array}{rcl} \overline{\overline{z^2}+\cos z} & \mbox{for} & |z|<3 \\ 0 & \mbox{for} & 3\leq|z|\leq5 \\ \sin\overline{z} & \mbox{for} & |z|>5 \end{array}\right. ``` $$f(z) = \left{ \begin{array}{rcl} \overline{\overline{z^2}+\cos z} & \mbox{for} & |z|<3 \ 0 & \mbox{for} & 3\leq|z|\leq5 \ \sin\overline{z} & \mbox{for} & |z|>5 \end{array}\right.$$ ### Diagrams ``` \begin{matrix} && \Bbb Q(\sqrt{2},i) & \\ &\huge\diagup & \huge| & \huge\diagdown \\ \Bbb Q(\sqrt{2}) & & \Bbb Q(i\sqrt{2})& & \Bbb Q(i)\\ &\huge\diagdown & \huge| & \huge\diagup \\ &&\Bbb Q \end{matrix} ``` $$\begin{matrix} && \Bbb Q(\sqrt{2},i) & \ &\huge\diagup & \huge| & \huge\diagdown \ \Bbb Q(\sqrt{2}) & & \Bbb Q(i\sqrt{2})& & \Bbb Q(i)\ &\huge\diagdown & \huge| & \huge\diagup \ &&\Bbb Q \end{matrix}$$ ### Equations align ``` \eqalign{ (a+b)^2 &= (a+b)(a+b) \\ &= a^2 + ab + ba + b^2 \\ &= a^2 + 2ab + b^2 } ``` $$\eqalign{ (a+b)^2 &= (a+b)(a+b) \ &= a^2 + ab + ba + b^2 \ &= a^2 + 2ab + b^2 }$$ ### Hat ``` \hat\imath \hat\jmath \hat ab \hat{ab} ``` $$\hat\imath \ \hat\jmath \ \hat ab \ \hat{ab}$$ ### Matrices #### Type 1 ``` \begin{matrix} \hline x_{11} & x_{12} \\ x_{21} & x_{22} \strut \\ \hdashline x_{31} & x_{32} \strut \end{matrix} ``` $$\begin{matrix} \hline x\_{11} & x\_{12} \ x\_{21} & x\_{22} \strut \ \hdashline x\_{31} & x\_{32} \strut \end{matrix}$$ #### Type 2 ``` A = \pmatrix{ a_{11} & a_{12} & \ldots & a_{1n} \cr a_{21} & a_{22} & \ldots & a_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr a_{m1} & a_{m2} & \ldots & a_{mn} \cr } ``` $$A = \pmatrix{ a\_{11} & a\_{12} & \ldots & a\_{1n} \cr a\_{21} & a\_{22} & \ldots & a\_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr a\_{m1} & a\_{m2} & \ldots & a\_{mn} \cr }$$ ### Limits ``` \sum_{k=1}^n a_k ``` $$\sum\_{k=1}^n a\_k$$ ### References A complete list is available at ## Quantum operations In general, what are the legal transformations $$\rho \rightarrow \varepsilon (\rho)$$ ? Def: Operation $$\varepsilon$$ is a valid quantum op iff A1 $$Tr(\varepsilon(\rho)) = 1$$ A2 $$\varepsilon$$ is convex and linear. $$\varepsilon (\sum\_k P\_k \rho\_k) = \sum\_k P\_k \varepsilon (\rho\_k)$$ A3 $$\varepsilon$$is completely positive. a. if $$\rho \geq 0$$ then $$\varepsilon(\rho) \geq 0$$ b. $$(I\_R \varepsilon\_Q) (\rho\_{RQ}) \geq 0 \forall \rho\_{AB} \geq 0$$ Why A3b? consider $$\varepsilon: \matrix{ a & b \cr c & d } \longrightarrow \pmatrix{ a & c \cr b & d }$$ $$\sum\_{jk} C\_{jk} | j k | \buildrel \varepsilon \over \rightarrow \sum\_{jk} C\_{jk} | k j |$$