The Exercise

The five qubit code stores one qubit in five. The exercise is to prepare its encoded $|0\rangle$ , written $|0_L\rangle$ , by measurement. Five Pauli strings on the five qubits, the four generators of the code and the logical $\bar Z$ , are measured one after another, each with an ancilla qubit of its own that starts in $|0\rangle$ and is read at the end of its circuit.

Systems and Gates

A quantum system $S$ is a physical object whose state space is a finite dimensional complex Hilbert space^✦, and the composite system $S \otimes T$ of two systems has as state space the tensor product of theirs.^✦ The qudit with $d$ levels is the system with state space $\mathbb{C}^d$ .^✦ The qubit is the qudit with two levels, and a register of $n$ qubits is the qudit with $2^n$ levels. The circuit here acts on an ancilla qubit and a register of five qubits, the composite system of the qubit with the qudit of $32$ levels.

A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$ , determined physically up to a global phase^✦; the basis states $|0\rangle$ and $|1\rangle$ of the qubit are the standard basis vectors of $\mathbb{C}^2$ .^✦ For pure states $|\psi\rangle$ of $S$ and $|\varphi\rangle$ of $T$ , the product $|\psi\rangle \otimes |\varphi\rangle$ is the pure state of $S \otimes T$ whose vector is the tensor product of the two.^✦ A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$ ^✦, and a pure state $|\psi\rangle$ determines the state $|\psi\rangle\langle\psi|$ , the projector onto its vector.^✦

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 $|\psi\rangle$ to the pure state $U|\psi\rangle$ .^✦ Two unitaries $U$ and $V$ of $S$ compose to the unitary $UV$ , first $V$ and then $U$ .^✦ The product unitary $U \otimes V$ of a unitary $U$ of $S$ and a unitary $V$ of $T$ is the unitary of $S \otimes T$ whose operator is the tensor product of theirs^✦; taking for $V$ the identity evolution $1$ of $T$ gives $U \otimes 1$ , the gate $U$ on the first factor alone.^✦ The Hadamard $H$ is the unitary of the qubit with matrix $\tfrac{1}{\sqrt 2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ .^✦ For a unitary $U$ of $S$ , the controlled gate $C(U)$ is the unitary of the composite of a control qubit with $S$ given by $C(U) = |0\rangle\langle 0| \otimes 1 + |1\rangle\langle 1| \otimes U.$ (1)^✦

Pauli Strings

The Pauli matrices are the four complex $2 \times 2$ matrices $\sigma_0 = I, \quad \sigma_1 = X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$ (2)^✦ For a choice $s = (s_1, \dots, s_n)$ of one Pauli matrix for each of $n$ qubits, the Pauli string $P_s = \sigma_{s_1} \otimes \cdots \otimes \sigma_{s_n}$ is the operator on the register whose matrix, with rows and columns labelled by bit strings $x$ and $y$ , or by the binary numbers they spell on the $2^n$ levels, has the entries $(P_s)_{xy} = \prod_{k=1}^{n} (\sigma_{s_k})_{x_k y_k}.$ (3)^✦ A string is written as a word, $XZZXI$ for $X \otimes Z \otimes Z \otimes X \otimes I$ .

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 $A|\psi\rangle = \mu|\psi\rangle$ .^✦ For a real coefficient $c$ , the Pauli string observable $c\,P_s$ is the observable on the register that acts as $c$ times the Pauli string, a Hermitian matrix.^✦ With $c = 1$ it squares to the identity, $P_s^2 = 1.$ (4)^✦

The Preparation Circuit

For a unitary $U$ of a system $S$ , the measurement circuit of $U$ is the unitary $(H \otimes 1)\, C(U)\, (H \otimes 1)$ of the composite of a control qubit with $S$ .^✦ The syndrome measuring block of a Pauli string $s$ is the measurement circuit of $P_s$ , a unitary of an ancilla qubit and the register.^✦

The four generators of the five qubit code and its logical $\bar Z$ are the Pauli strings $g_1 = XZZXI, \quad g_2 = IXZZX, \quad g_3 = XIXZZ, \quad g_4 = ZXIXZ, \quad g_5 = \bar Z = ZZZZZ,$ (5)^✦ each generator the one before it shifted cyclically one place to the right.

The preparation circuit is the sequence of five blocks $B_1, \dots, B_5$ applied one after another, where for $i = 1, \dots, 5$ the block $B_i$ is the syndrome measuring block of $g_i$ , $B_i = (H \otimes 1)\, C(P_{g_i})\, (H \otimes 1).$ (6)^✦

An encoded $|0\rangle$ of the five qubit code is a pure state $|\psi\rangle$ of the register of five qubits that is an eigenstate with eigenvalue $1$ of each of the five operators $P_{g_i}$ : $P_{g_i}\, |\psi\rangle = |\psi\rangle, \qquad i = 1, \dots, 5.$ (7)^✦

The Claim

Theorem 1. Let $i$ be one of $1, \dots, 5$ , let $b$ be a bit, and let $|\psi\rangle$ be a pure state of the register that is an eigenstate of $P_{g_i}$ with eigenvalue $(-1)^b$ . Then the $i$ th block, applied to the ancilla in $|0\rangle$ and the register in $|\psi\rangle$ , gives $B_i \bigl(|0\rangle \otimes |\psi\rangle\bigr) = |b\rangle \otimes |\psi\rangle.$ (8)^✦ Let $|\psi\rangle$ be an encoded $|0\rangle$ . Then for every block $B_i \bigl(|0\rangle \otimes |\psi\rangle\bigr) = |0\rangle \otimes |\psi\rangle, \qquad i = 1, \dots, 5.$ (9)^✦ There is an encoded $|0\rangle$ : a pure state $|0_L\rangle$ of the register with $P_{g_1} |0_L\rangle = P_{g_2} |0_L\rangle = \cdots = P_{g_5} |0_L\rangle = |0_L\rangle.$ (10)^✦ If $|\psi\rangle$ and $|\varphi\rangle$ are both encoded $|0\rangle$ states, they are the same state: $|\psi\rangle\langle\psi| = |\varphi\rangle\langle\varphi|.$ (11)^✦

What Is Checked

The claim is stated over the five blocks $B_i$ of (6), with their Hadamard gates on the ancilla and their controlled gates $C(P_{g_i})$ , and over pure states of the register: the eigenstates of the observables $P_{g_i}$ , the encoded $|0\rangle$ of (7), their products with the basis states of the ancilla, and the states $|\psi\rangle\langle\psi|$ they determine.