Quantum Repetition Code

The repetition code can be used to protect quantum information in the presence of a restricted error model. Let the physical system consist of three qubits. Errors act by independently applying, to each qubit, the flip operator $\sigma_x$ with probability $.25$. The classical code can be made into a quantum code by the superposition principle. Encoding one qubit is accomplished by

$$\alpha\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\... ...mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}.$$ (12)

The associated quantum code is the range of the encoding, that is, the two-dimensional subspace spanned by the encoded states $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{{\color{brown}\mathfrak{0}... ...{0}\mathfrak{0}}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ and $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{{\color{brown}\mathfrak{1}... ...{1}\mathfrak{1}}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$.

As in the classical case, decoding is accomplished by majority logic. However, it must be implemented carefully to avoid destroying quantum coherence in the stored information. One way to do that is to use only unitary operations to transfer the stored information to the output qubit. Fig. 5 shows a quantum network that accompishes this task.

FIG. 5: Quantum network for majority logic decoding into the output qubit $\mathsf {3}$. The effect of the quantum network on the basis states is shown. The top half shows the states with majority ${\color{brown}\mathfrak{0}}$. The decoded qubit is separated in the last step. The conventions for illustrating quantum networks are explained in [16].

As shown, the decoding network establishes an identification between the three physical qubits and a pair of subsystems consisting of two qubits representing the syndrome subsystem and one qubit for the information-carrying subsystem. On the left side of the correspondence, the information-carrying subsystem is not identifiable with any one (or two) of the physical qubits. Nevertheless it exists there through the identification.

To obtain a network for encoding, we reverse the decoding network and initialize qubits $\mathsf {2},\mathsf {3}$ in the state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{{\color{brown}\mathfrak{0}\mathfrak{0}}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$. Because of the initialization, the Toffoli gate becomes unnecessary. The complete system with a typical error is shown in Fig. 6.

FIG. 6: Encoding and decoding networks for the quantum repetition code with a typical error. The error that occurred can be determined from the state of the syndrome subsystem, which consists of the top two qubits. The encoding is shown as the reverse of the decoding, starting with an initialized syndrome subsystem. When the decoding is reversed to obtain the encoding, there is an initial Toffoli gate (shown in gray). Because of the initialization, this gate has no effect and is therefore omitted in an implementation.

As in the case of the classical repetition code, we can protect against cumulative errors without explicitly decoding and then re-encoding, which would cause a temporary loss of protection. Instead, one can find a means for directly resetting the syndrome subsystem to $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{{\color{brown}\mathfrak{0}\mathfrak{0}}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ (thus returning the information to the code) before the errors happen again. After resetting in this way, the errors in the correctable set have no effect on the encoded information because they act only on the syndrome subsystem.

Part of the task of designing error-correcting systems is to determine how well the system performs. An important performance measure is the probability of error. In quantum systems, the probability of error is intuitively interpreted as the maximum probability with which we can see a result different from the expected one in any measurement. Specifically, to determine the error, one compares the output $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_o}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ of the system to the input $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$. An upper bound is obtained if the output is written as a combination of the input state and an ``error'' state. For quantum information, combinations are linear combinations (that is, superpositions). Thus $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_o}\mbox{$\rangle\hspa... ...}{{\color{red}e}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ (see Fig. 7).The probability of error is bounded by ${\color{red}\epsilon}=\vert\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}... ...or{red}e}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\vert^2$ (which we call an ``error estimate''). In general, there are many different ways of writing the output as a combination of an acceptable state and an error term. One attempts to choose the combination that minimizes the error estimate. This choice yields the number ${\color{red}\epsilon}$, for which $1-{\color{red}\epsilon}$ is called the ``fidelity''. A fidelity of $1$ means that the output is the same (up to a phase factor) as the input.

FIG. 7: Representation of an error estimate. Any decomposition of the output state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_o}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ into a ``good'' state $\gamma\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ and an (unnormalized) error term $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{{\color{red}e}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ gives an estimate ${\color{red}\epsilon}=\vert\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}... ...or{red}e}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\vert^2$. For pure states, the optimum estimate is obtained when the error term is orthogonal to the input state. To obtain an error estimate for mixtures, one can use any representation of the state as a probabilistic combination of pure states and calculate the probabilistic sum of the pure state errors.

To illustrate error analysis, we calculate the error for the repetition code example for the two initial states $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ and ${1\over\sqrt{2}}\left(\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mat... ...thfrak{1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\right)$.

$$\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$} \stackrel{\mbox{\small encode}}{\longrightarrow} \mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{{\color{brow... ...{0}\mathfrak{0}}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$$

$${\color{red}\longrightarrow} \left\{ \begin{array}{rl} .75^3:&\mbox{$\vert\hspace*{-3pt}\vert\... ...box{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$} \end{array}\right.$$

$$\stackrel{\mbox{\small decode}}{\longrightarrow} \left\{ \begin{array}{rl} .4219:&\mbox{$\vert\hspace*{-3pt}\vert\... ...box{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$} \end{array}\right.$$ (13)

The final state is a mixture consisting of four correctly decoded components and four incorrectly decoded ones. The probability of each state in the mixture is shown before the colon. The incorrectly decoded information is orthogonal to the encoded information, and its probability is $0.1563$, an improvement over the one-qubit error-probability of $0.25$. The second state behaves quite differently:

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

$${\color{red}\longrightarrow} \left\{ \begin{array}{rl} \vdots&\\ .25^2*.75:&{1\over\sqrt{2}}\... ...{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\right) ,\\ \vdots& \end{array}\right.$$

$$\stackrel{\mbox{\small decode}}{\longrightarrow} \left\{ \begin{array}{rl} \vdots&\\ .0469:&{1\over\sqrt{2}} \mbo... ...*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}\right),\\ \vdots& \end{array}\right.$$ (14)

Not all error events have been shown, but in each case it can be seen that the state is decoded correctly, so the error is $0$. This shows that the error probability can depend significantly on the initial state. To remove this dependence and give a state independent error quantity, one can use the ``worst-case'', the ``average'' or the ``entanglement'' error. See Sect. 4.2.