The Exercise

The Steane code encodes one qubit in seven. Its codewords are the states of the seven qubits left unchanged by six commuting Pauli operators, the generators of the code. Suppose a codeword suffers an error on at most one qubit: an $X$, a $Y$ or a $Z$ on one of the seven, or nothing at all, which makes 22 possible errors. Measuring the six generators then gives six outcomes $\pm 1$ that depend on the error alone, and written as bits they form the error syndrome. The exercise is to say which recovery operation each syndrome calls for, and to show that it undoes the error behind it.

Pauli Strings

In the library every operator of the exercise is a Pauli string on the seven qubits. A Pauli string $g$ on $n$ qubits assigns to each qubit $k = 1, \dots, n$ one of the four Paulis $g_k \in \{I, X, Y, Z\}$, and stands for the operator $g_1 \otimes \dots \otimes g_n$ up to an overall phase. We write $X_k$ for the string with $X$ on qubit $k$ and $I$ elsewhere, and likewise $Y_k$ and $Z_k$; a product such as $X_1 X_5$ carries $X$ on both qubits, and $I$ alone is the string with $I$ on every qubit.

The $X$ bit $x(a)$ and the $Z$ bit $z(a)$ of a single qubit Pauli $a$ are the two bits of its binary symplectic representation, which mark an $X$ factor and a $Z$ factor in $a$: $$x(I) = x(Z) = 0, \quad x(X) = x(Y) = 1, \qquad z(I) = z(X) = 0, \quad z(Y) = z(Z) = 1.$$(1) The symplectic pairing of two single qubit Paulis $a$ and $b$ is the number $\omega(a, b) = x(a)\, z(b) + z(a)\, x(b)$.^✦ The symplectic form of two Pauli strings $g$ and $h$ on $n$ qubits is the sum of the pairings of their factors, $$\omega(g, h) = \sum_{k=1}^{n} \omega(g_k, h_k),$$(2)^✦ and its parity is $0$ when $g$ and $h$ commute and $1$ when they anticommute.

The product $a \odot b$ of two single qubit Paulis $a$ and $b$ is the Pauli proportional to the operator $ab$: $$a \odot b = b \odot a, \quad I \odot a = a, \quad a \odot a = I, \quad X \odot Y = Z, \quad Y \odot Z = X, \quad Z \odot X = Y.$$(3)^✦ The product $g \odot h$ of two Pauli strings on $n$ qubits is taken qubit by qubit, $(g \odot h)_k = g_k \odot h_k$, and is the operator product $gh$ up to phase.^✦

Syndrome and Recovery

The standard form generators $g_1, \dots, g_6$ of the Steane code are six Pauli strings on the seven qubits, three of $X$ type and three of $Z$ type: $$\begin{aligned} g_1 &= X_1 X_5 X_6 X_7, & g_2 &= X_2 X_4 X_6 X_7, & g_3 &= X_3 X_4 X_5 X_6,\\ g_4 &= Z_1 Z_3 Z_4 Z_7, & g_5 &= Z_2 Z_3 Z_5 Z_7, & g_6 &= Z_1 Z_2 Z_3 Z_6. \end{aligned}$$(4)^✦ The syndrome $\beta(E)$ of a Pauli string $E$ on the seven qubits is the string of six bits whose bit $a$ is the parity of the symplectic form of $g_a$ with $E$: $$\beta_a(E) = \omega(g_a, E) \bmod 2, \qquad a = 1, \dots, 6.$$(5)^✦

The 22 errors $E_0, \dots, E_{21}$ are the Pauli strings on the seven qubits with at most one factor other than $I$, in the order $$E_0 = I, \qquad E_k = X_k, \quad E_{7+k} = Y_k, \quad E_{14+k} = Z_k \qquad (k = 1, \dots, 7).$$(6)^✦ The recovery $R(s)$ for a string $s$ of six bits is the Pauli string on the seven qubits read off the list (6): $$R(s) = \begin{cases} E_j, & j \text{ the least index with } \beta(E_j) = s,\\ I, & \text{no } E_j \text{ has syndrome } s. \end{cases}$$(7)^✦

The Claim

What Is Checked

The claim is stated over the generators (4), the syndrome (5) they assign to a Pauli string through the symplectic form (2), the list of errors (6), the recovery table (7) and the product $\odot$ of Pauli strings.

By (8), no two of the 22 errors share a syndrome.