SEASON 3
Quantum - Holevo's theorem
χ(pi,vi)=H(ipivivi\chi (p_i, |v_i \rangle) = H( \sum_i p_i | v_i \rangle \langle v_i|
Allowed to send
p1,p2,p3,...,pk.p_1, p_2, p_3, ..., p_k.
Capacity is
maxpi,piχ=maxpi,piχipiH(pi)\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
2n(cϵ)2^{n(c - \epsilon)}
random codewords with letter chosen with probability maximizing
H(B)H(BA)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 0
    .

How about quantum case?

Upper bound:
  • Alice sends
    pip_i
    to Bob
Will show that for single state decoding, Shannon information provided by any measurement of Bob's
<\lt
Holevo's information
χ\chi
.
Alice's record of
pip_i

\def\myHearts#1#2{\color{#1}{\heartsuit}\kern-0pt\color{#2}{\heartsuit}} \myHearts{red} {blue}
0| 0\rangle

Brackets

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\left\langle \frac{1}{2} \middle| 1 \right\rangle
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12|1\left\langle \frac{1}{2} \middle| 1 \right\rangle

Fractions and brackets

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\left( \begin{array}{cc} 2\tau & 7\phi-frac5{12} \\
2
3\psi & \frac{\pi}8 \end{array} \right)
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\left( \begin{array}{c} x \\ y \end{array} \right)
4
\mbox{~and~} \left[ \begin{array}{cc|r}
5
3 & 4 & 5 \\ 1 & 3 & 729 \end{array} \right]
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Function definition

1
f(z) = \left\{ \begin{array}{rcl}
2
\overline{\overline{z^2}+\cos z} & \mbox{for}
3
& |z|<3 \\ 0 & \mbox{for} & 3\leq|z|\leq5 \\
4
\sin\overline{z} & \mbox{for} & |z|>5
5
\end{array}\right.
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Diagrams

1
\begin{matrix}
2
&& \Bbb Q(\sqrt{2},i) & \\
3
&\huge\diagup & \huge| & \huge\diagdown \\
4
\Bbb Q(\sqrt{2}) & & \Bbb Q(i\sqrt{2})& & \Bbb Q(i)\\
5
&\huge\diagdown & \huge| & \huge\diagup \\
6
&&\Bbb Q
7
\end{matrix}
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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}

Equations align

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\eqalign{
2
(a+b)^2 &= (a+b)(a+b) \\
3
&= a^2 + ab + ba + b^2 \\
4
&= a^2 + 2ab + b^2
5
}
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Hat

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\hat\imath
2
3
\hat\jmath
4
5
\hat ab
6
7
\hat{ab}
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ı^ȷ^a^bab^\hat\imath \\ \hat\jmath \\ \hat ab \\ \hat{ab}

Matrices

Type 1

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2
\begin{matrix}
3
\hline
4
x_{11} & x_{12} \\
5
x_{21} & x_{22} \strut \\
6
\hdashline
7
x_{31} & x_{32} \strut
8
\end{matrix}
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Type 2

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A = \pmatrix{
2
a_{11} & a_{12} & \ldots & a_{1n} \cr
3
a_{21} & a_{22} & \ldots & a_{2n} \cr
4
\vdots & \vdots & \ddots & \vdots \cr
5
a_{m1} & a_{m2} & \ldots & a_{mn} \cr
6
}
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Limits

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\sum_{k=1}^n a_k
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k=1nak\sum_{k=1}^n a_k

References

Quantum operations

In general, what are the legal transformations
ρε(ρ)\rho \rightarrow \varepsilon (\rho)
?
Def: Operation
ε\varepsilon
is a valid quantum op iff
A1
Tr(ε(ρ))=1Tr(\varepsilon(\rho)) = 1
A2
ε\varepsilon
is convex and linear.
ε(kPkρk)=kPkε(ρk)\varepsilon (\sum_k P_k \rho_k) = \sum_k P_k \varepsilon (\rho_k)
A3
ε\varepsilon
is completely positive.
a. if
ρ0\rho \geq 0
then
ε(ρ)0\varepsilon(\rho) \geq 0
b.
(IRεQ)(ρRQ)0ρAB0(I_R \varepsilon_Q) (\rho_{RQ}) \geq 0 \forall \rho_{AB} \geq 0
Why A3b?
consider