The Exercise

The three qubit phase flip code stores one qubit in the two states $|{+}{+}{+}\rangle$ and $|{-}{-}{-}\rangle$ of three qubits. The errors here are not phase flips but projections: $P_k$ projects qubit $k$ onto $|0\rangle$, $Q_k$ projects it onto $|1\rangle$, and both leave the other two qubits alone. The exercise is to show that the code protects against the error set $\{I, P_1, Q_1, P_2, Q_2, P_3, Q_3\}$, in the sense of the quantum error correction conditions: for the projector $P$ onto the code and the errors $E_i$, the operators $P E_i^\dagger E_j P$ are multiples of $P$.

A Quantum System

A quantum system $S$ is a physical object whose state space is a finite dimensional complex Hilbert space.^✦ 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 $T$.^✦ A pure state $\psi$ of a quantum system $S$ is a vector of unit norm, $\lVert \psi \rVert = 1$, in the state space of $S$.^✦

The qubit is the quantum system whose state space is $\mathbb{C}^2$.^✦ For a bit $i$, the basis state $|i\rangle$ of the qubit is the pure state whose vector is the standard basis vector $e_i$ of $\mathbb{C}^2$; these are the computational basis states $|0\rangle$ and $|1\rangle$.^✦ The states $|+\rangle$ and $|-\rangle$ of the qubit are the pure states with the vectors $$|\pm\rangle = \begin{pmatrix} 1/\sqrt{2} \\ \pm 1/\sqrt{2} \end{pmatrix} = \frac{|0\rangle \pm |1\rangle}{\sqrt{2}},$$(1) the eigenvectors of the Pauli operator $X$ with eigenvalues $+1$ and $-1$.

The Phase Flip Code

Three qubits form the composite system with state space $\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$. On each qubit the code uses the pair of states $|+\rangle, |-\rangle$, in that order.^✦ For signs $\sigma_1, \sigma_2, \sigma_3$, each $+$ or $-$, the product state $|\sigma_1 \sigma_2 \sigma_3\rangle = |\sigma_1\rangle \otimes |\sigma_2\rangle \otimes |\sigma_3\rangle$ is a state of the three qubits.

The codeword $|a_L\rangle$ of the phase flip code, for a bit $a$, takes the $a$th state of the pair on every qubit: $$|0_L\rangle = |{+}{+}{+}\rangle, \qquad |1_L\rangle = |{-}{-}{-}\rangle.$$(2)^✦ The code projector $P$ is the sum of the projectors onto the two codewords, $$P = |0_L\rangle\langle 0_L| + |1_L\rangle\langle 1_L|.$$(3)^✦

The Errors

The projector $|0\rangle\langle 0|$ of the qubit onto its basis state $|0\rangle$ is the operator on $\mathbb{C}^2$ that sends a vector $\phi$ to $$|0\rangle\langle 0|\,\phi = \langle 0|\phi\rangle\,|0\rangle,$$(4)^✦ and the projector $|1\rangle\langle 1|$ onto $|1\rangle$ sends it to $$|1\rangle\langle 1|\,\phi = \langle 1|\phi\rangle\,|1\rangle.$$(5)^✦

The error set is a family of seven operators $E_0, \dots, E_6$ on the three qubits: the identity, and for each qubit $k$ the projectors $P_k$ and $Q_k$ that act on qubit $k$ as $|0\rangle\langle 0|$ and $|1\rangle\langle 1|$ and as the identity on the other two, $$\begin{aligned} E_0 &= I, \\ E_1 &= P_1 = |0\rangle\langle 0| \otimes I \otimes I, & E_2 &= Q_1 = |1\rangle\langle 1| \otimes I \otimes I, \\ E_3 &= P_2 = I \otimes |0\rangle\langle 0| \otimes I, & E_4 &= Q_2 = I \otimes |1\rangle\langle 1| \otimes I, \\ E_5 &= P_3 = I \otimes I \otimes |0\rangle\langle 0|, & E_6 &= Q_3 = I \otimes I \otimes |1\rangle\langle 1|. \end{aligned}$$(6)^✦

The Claim

An operator $P$ on the state space of a quantum system $S$ and a family of operators $E_i$ on the same space, indexed by a set $\iota$, satisfy the quantum error correction conditions when some Hermitian complex matrix $\alpha$ with rows and columns indexed by $\iota$ gives $P E_i^\dagger E_j P = \alpha_{ij} P$ for all $i$ and $j$.^✦

What Is Checked

The claim is stated over the composite system of three qubits, the code projector (3) built from the codewords (2), and the seven errors (6) built from the projectors (4) and (5) of a single qubit.

About these constructions the library proves that the vectors (1) and the standard basis vectors $e_0$, $e_1$ have unit norm, so that $|+\rangle$, $|-\rangle$, $|0\rangle$ and $|1\rangle$ are pure states of the qubit.