The Exercise

The exercise concerns syndrome measurement for two codes, the five qubit code and Shor’s nine qubit code. Each code is given by a list of generators, tensor products of Pauli matrices on the code’s qubits, and the syndrome is read one generator at a time. An ancilla qubit is prepared in $|0\rangle$ beside the code’s qubits, and a circuit acts on both. If the state $|\psi\rangle$ of the code’s qubits is an eigenstate of the generator with eigenvalue $(-1)^b$, for a bit $b$, the circuit must leave the ancilla in $|b\rangle$ and the code’s qubits in $|\psi\rangle$. A measurement of the ancilla in the computational basis then gives the syndrome bit $b$.

The library states this for every generator of both codes, with one circuit built the same way from each generator.

A Quantum System

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ Every such space $\mathcal{H}$ is the state space of a system, the system modelled on $\mathcal{H}$.^✦ The qudit with $d$ levels is the system modelled on $\mathbb{C}^d$ with its standard inner product^✦; the qubit is the qudit with two levels^✦, and the qudit with $2^n$ levels is a register of $n$ qubits. The composite system $S \otimes T$ of two systems $S$ and $T$ is the system whose state space is the tensor product of their state spaces.^✦

A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$, $\lVert \psi \rVert = 1$.^✦ For pure states $|\psi\rangle$ of $S$ and $|\varphi\rangle$ of $T$, the product state $|\psi\rangle \otimes |\varphi\rangle$ is the pure state of $S \otimes T$ whose vector is the tensor product of the two vectors.^✦ The basis states $|0\rangle$ and $|1\rangle$ of the qubit are the pure states whose vectors are the standard basis vectors $e_0$ and $e_1$ of $\mathbb{C}^2$.^✦

A unitary $U$ of $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = 1$.^✦ It carries a pure state $|\psi\rangle$ of $S$ to the pure state $U|\psi\rangle$.^✦ The identity evolution of $S$ is the unitary $1$.^✦ For two unitaries $U$ and $V$ of $S$, the composite $UV$ is the unitary of $S$ that applies $V$ first and then $U$.^✦ For a unitary $U$ of $S$ and a unitary $V$ of $T$, the product unitary $U \otimes V$ is the unitary of $S \otimes T$ whose operator is the tensor product of the two operators^✦; with $V = 1$ it is the gate $U \otimes 1$, which acts with $U$ on the first factor alone.^✦

An observable $A$ on $S$ is a self adjoint operator on the state space of $S$, $A^\dagger = A$.^✦ A pure state $|\psi\rangle$ of $S$ is an eigenstate of $A$ with eigenvalue $\mu$, a real number, when $A|\psi\rangle = \mu|\psi\rangle$.^✦

Pauli Strings

The Pauli matrices are the four complex $2 \times 2$ matrices $$\sigma_0 = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad \sigma_1 = X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(1)^✦ For a choice $p = (p_1, \dots, p_n)$ of one label in $\{0, 1, 2, 3\}$ for each of $n$ qubits, the Pauli string $P = \sigma_{p_1} \otimes \cdots \otimes \sigma_{p_n}$ is the operator on the $n$ qubits whose matrix, with rows and columns labelled by strings $i$ and $j$ of $n$ bits, has the entries $$P_{ij} = \prod_{k=1}^{n} (\sigma_{p_k})_{i_k j_k}.$$(2)^✦ With $i$ and $j$ read as binary numbers from $0$ to $2^n - 1$, $P$ is a matrix on $\mathbb{C}^{2^n}$, the state space of the register of $n$ qubits.^✦ A Pauli string is written as a word, $XZZXI$ for $X \otimes Z \otimes Z \otimes X \otimes I$, or by the qubits on which it is not the identity, $Z_1 Z_2$ for $Z \otimes Z \otimes I \otimes \cdots \otimes I$.

A Pauli string is Hermitian. For a Pauli string $P$ on $n$ qubits and a real coefficient $c$, the observable $cP$ on the register of $n$ qubits is the operator whose matrix is $c$ times that of $P$; with $c = 1$ it is the observable $P$.^✦

The Syndrome Circuits

The Hadamard gate $H$ is the unitary of the qubit whose matrix is $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$$(3)^✦ For a unitary $U$ of a system $S$, the controlled gate $C(U)$ is the unitary of the composite system of a control qubit and the target $S$ whose operator is $$C(U) = |0\rangle\langle 0| \otimes 1 + |1\rangle\langle 1| \otimes U.$$(4)^✦

The measurement circuit $(H \otimes 1)\, C(U)\, (H \otimes 1)$ of $U$ is the unitary of the control qubit and $S$ that applies a Hadamard gate to the control alone, then $C(U)$, then a second Hadamard gate to the control alone.^✦ For a Pauli string $P$ on $n$ qubits, the syndrome circuit of $P$ is the measurement circuit of the observable $P$, taken as a unitary of the register: $$M_P = (H \otimes 1)\, C(P)\, (H \otimes 1).$$(5)^✦

The five qubit code has the four generators of Figure 10.12, each the one before it shifted cyclically one place to the right: $$g_1 = XZZXI, \quad g_2 = IXZZX, \quad g_3 = XIXZZ, \quad g_4 = ZXIXZ.$$(6)^✦ Its syndrome circuits are the four unitaries of a control qubit and a register of five qubits $$M^{(5)}_i = M_{g_i}, \qquad i = 1, \dots, 4.$$(7)^✦

Shor’s code has eight generators on nine qubits: six of $Z$ type, against bit flips within each block of three qubits, and two of $X$ type, against phase flips between the blocks, $$\begin{gathered} s_1 = Z_1 Z_2, \quad s_2 = Z_2 Z_3, \quad s_3 = Z_4 Z_5, \quad s_4 = Z_5 Z_6, \quad s_5 = Z_7 Z_8, \quad s_6 = Z_8 Z_9, \\ s_7 = X_1 X_2 X_3 X_4 X_5 X_6, \quad s_8 = X_4 X_5 X_6 X_7 X_8 X_9. \end{gathered}$$(8)^✦ On a control qubit and a register of nine qubits, the syndrome circuits of Shor’s code are $$M^{(9)}_j = M_{s_j}, \qquad j = 1, \dots, 8.$$(9)^✦

The Claim

What Is Checked

The two claims are stated over the syndrome circuits (7) and (9), each the circuit (5) of one generator, applied to the product state of the ancilla’s $|0\rangle$ with a pure state of the code’s register, and over the eigenstates of each generator as an observable.

The library proves that every Pauli string squares to the identity, $P^2 = 1$^✦, which makes the observable $P$ a unitary of the register, and that the controlled gate $C(U)$ of (4) is unitary for every unitary $U$.^✦