The Exercise

Alice wants to send Bob the state of a qubit, qubit 0, and the two share a pair of qubits, 1 and 2, in the Bell state $(|00\rangle + |11\rangle)/\sqrt{2}$, qubit 2 on Bob’s side. Alice applies a controlled NOT gate from qubit 0 to qubit 1 and a Hadamard gate to qubit 0, measures qubits 0 and 1 in the computational basis with outcomes $m_0$ and $m_1$, and sends Bob the two bits. Bob corrects qubit 2 with Pauli operators chosen by the bits, and qubit 2 then holds the state Alice started with.

The exercise is to verify this with stabilizers: a state is followed through the circuit by the Pauli operators that fix it, where a Pauli operator $P$ fixes a vector $|\psi\rangle$ when $P|\psi\rangle = |\psi\rangle$. The inputs are eigenstates of the Pauli operators $X$ and $Z$: the state $|+\rangle$ for $b = 0$ or $|-\rangle$ for $b = 1$, which $X$ multiplies by $(-1)^b$, and the state $|a\rangle$ for $a = 0$ or $1$, which $Z$ multiplies by $(-1)^a$. Write $X_k$ and $Z_k$ for $X$ and $Z$ acting on qubit $k$, with the identity on the other qubits.

The library states what Bob’s correction makes of the measured state, for each of the two inputs.

Pauli Operators

The Pauli matrices $\sigma_0, \dots, \sigma_3$ are the four complex $2 \times 2$ matrices $\sigma_0 = I$, the identity, and $\sigma_1 = X$, $\sigma_2 = Y$, $\sigma_3 = Z$, which in the computational basis $|0\rangle, |1\rangle$ are $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(1)^✦

A Pauli string on $n$ qubits is the operator $\sigma_{g_0} \otimes \cdots \otimes \sigma_{g_{n-1}}$ that acts on each qubit $k$ with one Pauli matrix $\sigma_{g_k}$, chosen by an index $g_k \in \{0, 1, 2, 3\}$. Its matrix element between the bit strings $s = s_0 \cdots s_{n-1}$ and $t = t_0 \cdots t_{n-1}$ is the product of single qubit entries, $$\langle s |\, \sigma_{g_0} \otimes \cdots \otimes \sigma_{g_{n-1}} \,| t \rangle = \prod_{k=0}^{n-1} (\sigma_{g_k})_{s_k t_k}.$$(2)^✦ The operators $X_k$ and $Z_k$ on three qubits and their products, such as $Z_0X_2 = Z \otimes I \otimes X$ and $Z_1Z_2 = I \otimes Z \otimes Z$, are Pauli strings.

Bob’s correction for the outcomes $m_0$ and $m_1$ is the operator $C_{m_0m_1}$ on the three qubits that applies $X^{m_1}$ and then $Z^{m_0}$ to qubit 2: $$C_{m_0m_1} = Z_2^{m_0} X_2^{m_1}.$$(3)^✦

The Claim

What Is Checked

The claim is stated over six Pauli strings on three qubits, $Z_0$, $Z_1$, $Z_0X_2$, $Z_1Z_2$, $X_2$ and $Z_2$, each built by (2) from the Pauli matrices (1), and over Bob’s correction (3). About these operators the library proves the two parts of the theorem, (4) for the input in the $X$ basis and (5) for the input in the $Z$ basis.