The Exercise

The five qubit code is a stabilizer code: its code space is the space of states of five qubits left fixed by every element of a group $S$ of Pauli operators, the stabilizer, which is generated by four tensor products of Pauli matrices. An error acts on at most one qubit, as a Pauli $X$, $Y$ or $Z$ there; with the case of no error, that makes sixteen errors. The exercise is to show that the code corrects all of them: one recovery operation, applied without knowing which error occurred, returns every encoded state.

The Pauli Group

The Pauli matrices are the four complex $2 \times 2$ matrices $$\sigma_0 = I, \quad \sigma_1 = X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(1)^✦

For a choice $a$ of one Pauli matrix $\sigma_{a_k}$ for each qubit $k$ of $n$ qubits, the Pauli string $\sigma_a$ is the operator on the $n$ qubits $$\sigma_a = \sigma_{a_1} \otimes \sigma_{a_2} \otimes \cdots \otimes \sigma_{a_n},$$(2)^✦ a $2^n \times 2^n$ matrix with rows and columns labelled by the bit strings in $\{0,1\}^n$.

The Pauli group $G_n$ on $n$ qubits consists of the operators $$p = i^s \sigma_a,$$(3)^✦ for a phase power $s$, an integer modulo $4$, and a Pauli string $a$ on $n$ qubits.

The stabilizer sum $Q$ of a finite subgroup $S$ of $G_n$ is the sum of the matrices of all its elements^✦, and the code projector $P_S$ of $S$ is their average: $$Q = \sum_{g \in S} g, \qquad P_S = \frac{1}{|S|}\, Q .$$(4)^✦ When $S$ is the stabilizer of a code, $P_S$ is the projector onto its code space $V_S$.

The Five Qubit Code

The generators of the stabilizer of the five qubit code are built from the four Pauli strings of Figure 10.12, each written by the letters of its matrices, qubit $1$ first: $$a^{(1)} = XZZXI, \quad a^{(2)} = IXZZX, \quad a^{(3)} = XIXZZ, \quad a^{(4)} = ZXIXZ.$$(5) The generator $g_k$ is the element of $G_5$ with phase power $0$ and string $a^{(k)}$, $$g_k = \sigma_{a^{(k)}}, \qquad k = 1, 2, 3, 4,$$(6)^✦ and the stabilizer $S$ of the code is the subgroup of $G_5$ the four generate, $$S = \langle g_1, g_2, g_3, g_4 \rangle .$$(7)^✦ The code projector $P_S$ of the stabilizer $S$ is the $32 \times 32$ matrix (4) for this $S$.

The errors on the five qubits are the sixteen elements $E_0, \dots, E_{15}$ of $G_5$, all of phase power $0$: the identity, and on each qubit $j$ the three Pauli matrices acting on that qubit alone, $$E_0 = I, \qquad E_{3j-2} = X_j, \quad E_{3j-1} = Y_j, \quad E_{3j} = Z_j, \qquad j = 1, \dots, 5,$$(8)^✦ where $X_j$ is the Pauli string with $X$ on qubit $j$ and $I$ on the other four, and likewise $Y_j$ and $Z_j$.

The Claim

A recovery $\mathcal{R}$ on the five qubits is a quantum operation, $\mathcal{R}(\rho) = \sum_o R_o \rho R_o^\dagger$, given by finitely many operation elements $R_o$, operators on the $32$ dimensional state space of the five qubits. The recovery $\mathcal{R}$ corrects an error $E$ on the code space $V_S$ when, applied after $E$, it returns every encoded state.

What Is Checked

The claim is stated over the code projector (4) of the stabilizer (7), with its generators (6) taken from the strings (5), and over the matrices (3) of the errors (8).