The Exercise

Alice and Bob share $n$ Bell pairs $|\beta_{00}\rangle = (|00\rangle + |11\rangle)/\sqrt{2}$; Alice holds one qubit of each pair and Bob the other. Without the normalization $2^{-n/2}$, their joint state is $$|\Phi\rangle = \sum_{x \in \{0,1\}^n} |x\rangle_A \, |x\rangle_B ,$$(1)^✦ the sum over the bit strings $x$ of length $n$, with Alice’s $k$th qubit paired with Bob’s $k$th.

They distill entanglement with a stabilizer code on $n$ qubits. Alice projects her qubits onto the code space, and Bob projects his onto the complex conjugate of that space. Noise may first have struck Alice’s qubits with a Pauli error $E$, which she undoes with a correction $C$ chosen from the measured syndrome before projecting. The exercise asks which vector the two projections leave, with and without the corrected error, and what its norm is.

The Pauli Group

The Pauli matrices are the four complex two by two matrices $\sigma_0 = I$, $\sigma_1 = X$, $\sigma_2 = Y$ and $\sigma_3 = Z$, with rows and columns indexed by $\{0,1\}$ and $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(2)^✦

For a string $a = (a_1, \dots, a_n)$ of indices $a_k \in \{0,1,2,3\}$, one for each of $n$ qubits, the Pauli string $\sigma_a = \sigma_{a_1} \otimes \cdots \otimes \sigma_{a_n}$ is the operator on the $n$ qubits whose matrix, with rows and columns labelled by bit strings $x$ and $y$ of length $n$, has the entries $$(\sigma_a)_{xy} = \prod_{k=1}^{n} (\sigma_{a_k})_{x_k y_k} .$$(3)^✦ The support of the string $a$ is the set of qubits $k$ with $a_k \neq 0$.

The Pauli group $G_n$ on $n$ qubits consists of the operators $i^s \sigma_a$: each element $p$ is given by its phase power $s$, an integer modulo $4$, and its Pauli string $a$.^✦ The phase $i^s$ is the imaginary unit raised to the representative of $s$ in $\{0,1,2,3\}$, one of the four fourth roots of unity.^✦ The matrix of $p$ is the complex $2^n \times 2^n$ matrix $i^s \sigma_a$, and below an element stands for its matrix.^✦ The element $-I$ of $G_n$ is the identity string with phase power $2$.^✦ The weight of an element $p$ of $G_n$ is the number of qubits in the support of its string.^✦

The Code

The stabilizer sum $Q$ of a subgroup $S$ of $G_n$ is the sum $\sum_{s \in S} s$ of the matrices of its elements.^✦ The code projector $P_S$ of $S$ is the stabilizer sum divided by the number $|S|$ of elements of $S$, $$P_S = \frac{Q}{|S|} = \frac{1}{|S|} \sum_{s \in S} s ,$$(4)^✦ the projector onto the code space $V_S$ of the stabilizer code with stabilizer $S$. The complex conjugate $P_S^{*}$ of the code projector $P_S$ is the matrix whose entries are the complex conjugates of those of $P_S$; it is Bob’s projection.

The normalizer $N(S)$ of $S$ is the set of elements $g$ of $G_n$ with $g S g^{-1} = S$. The distance $d(S)$ of the stabilizer code with stabilizer $S$ is the least weight of an element of $N(S)$ that is not equal to $i^r s$ for any element $s$ of $S$ and integer $r$; when there is no such element, $d(S)$ is infinite.^✦

The Distilled State

The distilled vector $|\psi_S\rangle$ of a subgroup $S$ of $G_n$ is the shared state (1) after Alice’s projection onto the code space: $$|\psi_S\rangle = (P_S \otimes I) \, |\Phi\rangle .$$(5)^✦

The Claim

What Is Checked

The claim is stated over the distilled vector $|\psi_S\rangle$ of (5), built from the code projector $P_S$ of (4), together with the matrices of the elements of $G_n$, their weights and the distance $d(S)$. The distilled vector $|\psi_S\rangle$ is the unnormalized maximally entangled state on the code space $V_S$, that of $m$ encoded Bell pairs $|\beta_{00}\rangle^{\otimes m}$ without the normalization.