The Exercise

Three qubits are in one of the two states $(|000\rangle + |111\rangle)/\sqrt{2}$ and $(|000\rangle - |111\rangle)/\sqrt{2}$. A phase flip $Z$ strikes one of the three qubits, and which one is not known. Show that applying $Z$ to every qubit, the operator $Z_1 Z_2 Z_3$, returns the qubits to the state they started in, whichever qubit was flipped.^✦

A Quantum System

A quantum system $S$ is given by its state space, a finite dimensional complex Hilbert space.^✦ A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$, $\lVert \psi \rVert = 1$.^✦ For a nonzero vector $v$ in the state space of $S$, the normalized pure state of $v$ is the pure state of $S$ with vector $v / \lVert v \rVert$.^✦

The composite system $S \otimes T$ of two quantum systems $S$ and $T$ is the quantum system whose state space is the tensor product of the state spaces of $S$ and $T$.^✦ 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 vectors.^✦

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 of $S$ whose operator is the identity $1$.^✦ For unitaries $U$ of $S$ and $V$ of $T$, the product unitary $U \otimes V$ is the unitary of $S \otimes T$ whose operator is the tensor product of the operators of $U$ and $V$.^✦

Three Qubits

The qubit is the quantum system whose state space is $\mathbb{C}^2$.^✦ For a bit $i$, its basis state $|i\rangle$ is the pure state whose vector is the standard basis vector $e_i$ of $\mathbb{C}^2$.^✦ The three qubit register is the composite system of three qubits.^✦ For bits $q_1, q_2, q_3$ its basis state is the product $|q_1 q_2 q_3\rangle = |q_1\rangle \otimes |q_2\rangle \otimes |q_3\rangle$, and these eight states form its computational basis.^✦

The Pauli matrix $Z$ is the two by two complex matrix $$Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},$$(1)^✦ and the Pauli $Z$ gate, the phase flip, is the unitary of the qubit whose operator is the matrix (1) acting on $\mathbb{C}^2$.^✦ The phase flip $Z_i$ on qubit $i$ of the register is the product unitary with the Pauli $Z$ gate on qubit $i$ and the identity evolution on the other two, $$Z_1 = Z \otimes 1 \otimes 1, \qquad Z_2 = 1 \otimes Z \otimes 1, \qquad Z_3 = 1 \otimes 1 \otimes Z,$$(2)^✦ and the recovery operator of the register is the product unitary of the register with the Pauli $Z$ gate on each of the three qubits, $$Z_1 Z_2 Z_3 = Z \otimes Z \otimes Z.$$(3)

For a bit $\sigma$, the block vector of $\sigma$ is the vector $|000\rangle + (-1)^\sigma |111\rangle$ in the state space of the register^✦, and the block state $|c_\sigma\rangle$ is the normalized pure state of the block vector, $$|c_\sigma\rangle = \frac{|000\rangle + (-1)^\sigma |111\rangle}{\sqrt{2}}.$$(4)^✦ The values $\sigma = 0$ and $\sigma = 1$ give the two states of the exercise.

The Claim

What Is Checked

The claim is stated over the block states (4) of the three qubit register, over the three phase flips (2) and the recovery operator (3), which are product unitaries of the Pauli $Z$ gate and the identity evolution, and over the action of a unitary on a pure state.

The library shows that the block vector $|000\rangle + (-1)^\sigma |111\rangle$ is nonzero for both values of $\sigma$, so the block state (4) is a pure state of the register.^✦ The matrix (1) is unitary, and the tensor product of two unitaries is unitary, so the flips and the recovery operator are unitaries of the register. The norm of a product vector is the product of the norms, so a product of pure states is a pure state; a unitary preserves norms, so it carries a pure state to a pure state.