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Quantum - Holevo's theorem

​χ(pi,∣vi⟩)=H(∑ipi∣vi⟩⟨vi∣\chi (p_i, |v_i \rangle) = H( \sum_i p_i | v_i \rangle \langle v_i|χ(pi​,∣vi​⟩)=H(∑i​pi​∣vi​⟩⟨vi​∣
 
Allowed to send p1,p2,p3,...,pk.p_1, p_2, p_3, ..., p_k.p1​,p2​,p3​,...,pk​.
 
Capacity is maxpi,piχ=maxpi,piχ−∑ipiH(pi)\frac{max}{p_i, p_i} \chi = \frac{max}{p_i, p_i} \chi - \sum_i p_i H(p_i)pi​,pi​max​χ=pi​,pi​max​χ−∑i​pi​H(pi​)
 
block length: n
capacity: C
pick 2n(c−ϵ)2^{n(c - \epsilon)}2n(c−ϵ)
 random codewords with letter chosen with probability maximizing H(B)−H(B∣A)H(B) - H(B | A)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→∞,∈→0n \rightarrow \infty ,  \in \rightarrow 0n→∞,∈→0
 .
How about quantum case?
Upper bound:
Alice sends pip_ipi​
 to Bob
Will show that for single state decoding, Shannon information provided by any measurement of Bob's <\lt<
 Holevo's information χ\chiχ
 .
Alice's record of pip_ipi​
​
​
 
​
​
​♡♡\def\myHearts#1#2{\color{#1}{\heartsuit}\kern-0pt\color{#2}{\heartsuit}} \myHearts{red} {blue}♡♡
 ∣0⟩| 0\rangle∣0⟩
 
 Brackets
\left\langle \frac{1}{2} \middle| 1 \right\rangle
​⟨12|1⟩\left\langle \frac{1}{2} \middle| 1 \right\rangle⟨21​∣∣​1⟩
 
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]
​
 
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.
​
 
​
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}
​Q(2,i)╱∣╲Q(2)Q(i2)Q(i)╲∣╱Q\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}Q(2​)​╱╲​Q(2​,i)∣Q(i2​)∣Q​╲╱​Q(i)​
 
Equations align
\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}
​ı^ȷ^a^bab^\hat\imath \\  \hat\jmath	\\  \hat ab \\ \hat{ab}^^​a^bab^
 
Matrices
Type 1
​
\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
}
​
 
​
Limits
\sum_{k=1}^n a_k
​∑k=1nak\sum_{k=1}^n a_k∑k=1n​ak​
 
References
 A complete list is available at https://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm​
Quantum operations
In general, what are the legal transformations ρ→ε(ρ)\rho \rightarrow \varepsilon (\rho)ρ→ε(ρ)
 ?
Def: Operation ε\varepsilonε
 is a valid quantum op iff
A1  Tr(ε(ρ))=1Tr(\varepsilon(\rho)) = 1Tr(ε(ρ))=1
 
A2 ε\varepsilonε
 is convex and linear. ε(∑kPkρk)=∑kPkε(ρk)\varepsilon (\sum_k P_k \rho_k) = \sum_k P_k \varepsilon (\rho_k)ε(∑k​Pk​ρk​)=∑k​Pk​ε(ρk​)
 
A3 ε\varepsilonε
is completely positive.
    a. if ρ≥0\rho \geq 0ρ≥0
 then ε(ρ)≥0\varepsilon(\rho) \geq 0ε(ρ)≥0
 
    b. (IRεQ)(ρRQ)≥0∀ρAB≥0(I_R \varepsilon_Q) (\rho_{RQ}) \geq 0 \forall \rho_{AB} \geq 0(IR​εQ​)(ρRQ​)≥0∀ρAB​≥0
 
Why A3b?
consider 
 
​
 