Quantum Code for a Cyclic System

The shift operators introduced earlier act as permutations of the seven states of the cyclic system. They can therefore be extended to unitary operators on a seven-state cyclic quantum system with logical basis $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{{\color{brown}0}}\mbox{$\r... ...{\color{brown}6}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$. The error model introduced earlier makes sense here without modification, as does the encoding. The subsystem identification now takes the six-dimensional subspace spanned by $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{{\color{brown}0}}\mbox{$\r... ...{\color{brown}5}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ to a pair consisting of a three-state system with basis $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{{\color{brown}-1}}\mbox{$\... ...{\color{brown}1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ and a qubit. The identification of Eq. 6 extends linearly to a unitary subsystem identification. The procedure for decoding is modified as follows: First, a measurement is performed to determine whether the state is in the six-dimensional subspace or not. If it is, the identification is used to extract the qubit. Here is an outline of what happens when the state ${1\over\sqrt{2}}\left(\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mat... ...thfrak{1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\right)$ is encoded:

$${1\over\sqrt{2}}\left(\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3p... ...thfrak{1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\right) \stackrel{\mbox{\small encode}}{\longrightarrow} {1\over\sqrt{2}}\left(\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3p... ...{brown}4}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\right)$$

$${\color{red}\longrightarrow} \left\{ \begin{array}{rl} \vdots&\\ .05641 e^{-4}:&{1\over\sqrt{... ...{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\right) ,\\ \vdots& \end{array}\right.$$

$$\stackrel{\mbox{\small detect}}{\longrightarrow} \left\{ \begin{array}{rl} \vdots&\\ .001:\left\{ \begin{array}{r... ...}\rangle\hspace*{-4.3pt}\rangle$} \end{array}\right. \vdots& \end{array}\right.$$

$$\stackrel{\mbox{\small decode}}{\longrightarrow} \left\{ \begin{array}{rl} \vdots&\\ .0005:& \mbox{fail}\\ .0005... ...e\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\\ \vdots& \end{array}\right.$$

$$= \left\{ \begin{array}{rl} \vdots&\\ .0005:& \mbox{fail}\\ .0005... ...}\rangle\hspace*{-4.3pt}\rangle$}\right)} \right)\\ \vdots& \end{array}\right.$$ (15)

A ``good'' state was separated from the output in the case that is shown. The leftover error term has probability amplitude $.0005*((1/2)^2+(1/2)^2)=.00025$, which contributes to the total error (not shown).