The Exercise

The Steane code encodes one qubit in seven. Its stabilizer has six generators, Pauli strings on the seven qubits, and brought to a standard form, which here takes a relabelling of the qubits, they give the logical operators of the code directly. The exercise asks for the logical $X$ that the standard form gives, an operator fixed by two columns of three bits, called $E$ and $C$, read off the generators, and asks that it act on the code as a logical $X$ should, exchanging the two logical states.

Seven Qubits

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ A pure state $\psi$ of $S$ is a unit vector in the state space of $S$.^✦ A unitary $U$ of $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = 1$, and it carries a pure state $\psi$ of $S$ to the pure state $U\psi$.^✦

The register of $n$ qubits is the system whose state space is $\mathbb{C}[\mathbb{Z}_2^n]$, the space of complex functions on the bit strings $w$ of length $n$^✦, with one computational basis vector $|w\rangle$ for each $w$; these vectors form an orthonormal basis.^✦ For a bit string $v$ of length $n$, the bit flip string $X^v = X^{v_1} \otimes \cdots \otimes X^{v_n}$, with $X$ the Pauli operator of one qubit that exchanges $|0\rangle$ and $|1\rangle$, is the unitary of the register that flips the qubits where $v$ has a $1$, sending $|w\rangle$ to $|w + v\rangle$.^✦

The Steane Code

The $[7,4,3]$ Hamming code is defined by its parity check matrix $H$, the binary matrix with three rows and seven columns whose $j$th column is the binary numeral of $j$, most significant bit on top: $$H = \begin{pmatrix} 0&0&0&1&1&1&1 \\ 0&1&1&0&0&1&1 \\ 1&0&1&0&1&0&1 \end{pmatrix}.$$(1) The Hamming code $C$ is the linear code of the words $x \in \mathbb{Z}_2^7$ with $Hx = 0$, and the dual Hamming code $C_2$ is its dual code $C^\perp$, a $[7,3]$ code with eight codewords.

For a binary linear code $D$ of length $n$ and a word $x \in \mathbb{Z}_2^n$, the coset state $|x + D\rangle$ is the pure state of the register of $n$ qubits $$|x + D\rangle = \frac{1}{\sqrt{|D|}} \sum_{y \in D} |x + y\rangle .$$(2)^✦ The logical states of the Steane code are the coset states of $C_2$ at the zero word and at the word of all ones: $$|0_L\rangle = \frac{1}{\sqrt 8} \sum_{y \in C_2} |y\rangle , \qquad |1_L\rangle = \frac{1}{\sqrt 8} \sum_{y \in C_2} |1111111 + y\rangle .$$(3)

The Standard Form

A single qubit Pauli has two check bits, its $X$ bit and its $Z$ bit: $(0,0)$ for $I$, $(1,0)$ for $X$, $(1,1)$ for $Y$ and $(0,1)$ for $Z$.^✦ The check row $r(g) = (x(g) \,|\, z(g))$ of a Pauli string $g$ on $n$ qubits is the pair of binary vectors of length $n$ whose $k$th entries are the $X$ bit and the $Z$ bit of the Pauli that $g$ places on qubit $k$.

In standard form the six generators of the Steane code are $$\begin{aligned} g_1 &= X\,I\,I\,I\,X\,X\,X, & g_4 &= Z\,I\,Z\,Z\,I\,I\,Z, \\ g_2 &= I\,X\,I\,X\,I\,X\,X, & g_5 &= I\,Z\,Z\,I\,Z\,I\,Z, \\ g_3 &= I\,I\,X\,X\,X\,X\,I, & g_6 &= Z\,Z\,Z\,I\,I\,Z\,I, \end{aligned}$$(4)^✦ with the $X$ parts of $g_1, g_2, g_3$ forming the identity on the first three qubits and the $Z$ parts of $g_4, g_5, g_6$ the identity on the next three. The column $C$ of the standard form holds the $Z$ bits of $g_1, g_2, g_3$ on the seventh qubit, and the column $E$ those of $g_4, g_5, g_6$: $$C = \bigl(z(g_1)_7,\, z(g_2)_7,\, z(g_3)_7\bigr)^T = (0, 0, 0)^T,$$(5)^✦ $$E = \bigl(z(g_4)_7,\, z(g_5)_7,\, z(g_6)_7\bigr)^T = (1, 1, 0)^T.$$(6)^✦ For two columns $E$ and $C$ of three bits, the standard form prescribes for the logical $X$ the check row $$\bar r(E, C) = \bigl(0, 0, 0, E_1, E_2, E_3, 1 \,\big|\, C_1, C_2, C_3, 0, 0, 0, 0\bigr).$$(7)^✦ The logical $X$ of the Steane code in standard form is the Pauli string $$\bar X = I\,I\,I\,X\,X\,I\,X = X_4 X_5 X_7 .$$(8)^✦

The standard form numbers the qubits differently from the Hamming code. The relabelling is the permutation $$\sigma = (6\;7)\,(3\;4)\,(1\;4),$$(9)^✦ composed from the right, which sends $1, 2, \dots, 7$ to $3, 2, 4, 1, 5, 7, 6$: qubit $k$ in the ordering of the Hamming code is qubit $\sigma(k)$ of the standard form. In that ordering $\bar X$ becomes the Pauli string $\bar X'$ that places on each qubit $k$ the Pauli that $\bar X$ places on qubit $\sigma(k)$: $$\bar X' = I\,I\,X\,I\,X\,X\,I = X_3 X_5 X_6 .$$(10)^✦ The support word $v$ of $\bar X'$ is the $X$ part of its check row, $$v = x(\bar X') = 0010110,$$(11) and as an operator on the register of seven qubits $\bar X'$ is the bit flip string $X^v$.^✦

The Claim

What Is Checked

The claim is stated over the six generators (4), the columns $C$ and $E$ read off them and the check row of the Pauli string $X_4 X_5 X_7$; its action on the code is stated over the bit flip string $X^v$ of the support word (11) and over the logical states (3).