The Exercise

Alice and Bob share $n$ Bell pairs $|\beta_{00}\rangle = (|00\rangle + |11\rangle)/\sqrt{2}$. Alice holds the first qubit of each pair and Bob the second, so each of them holds a register of $n$ qubits. In the computational basis their joint state is a sum over bit strings, each string in Alice’s register paired with the same string in Bob’s, all with the same amplitude.^✦ The exercise is to show that the state keeps this form in a basis built from a binary linear code $C_2$, whose states superpose the codewords shifted by one bit string, with signs set by another: each such state of Alice’s appears paired with the same state of Bob’s, again with equal amplitudes.

A Register of Qubits

A quantum system $S$ is described by its state space, a finite dimensional complex Hilbert space^✦, and every such space is the state space of a quantum system.^✦ A pure state $|\psi\rangle$ of $S$ is a vector of unit norm in the state space of $S$, determined physically up to a global phase.^✦

For a finite group $G$, the group system $\mathbb{C}[G]$ is the quantum system whose state space is the space of complex functions on $G$ with its standard inner product.^✦ The bit register of $n$ qubits, one for each element of a finite index set, is the group system of the additive group $\mathbb{Z}_2^n$ of bit strings.^✦ The computational basis $\{|w\rangle\}$ of the bit register of $n$ qubits is the standard basis of its state space, indexed by the bit strings $w$: $|w\rangle$ is the function on $\mathbb{Z}_2^n$ equal to $1$ at $w$ and to $0$ elsewhere.^✦

Coset States of a Code

A binary linear code $C_2$ on $n$ bits is a subspace of $\mathbb{Z}_2^n$, whose elements are its codewords. The dual code $C_2^\perp$ of a binary linear code $C_2$ on $n$ bits is the subspace of $\mathbb{Z}_2^n$ consisting of the strings $z$ with $z \cdot y = 0$ for every codeword $y$ of $C_2$, where $z \cdot y = \sum_i z_i y_i$ modulo $2$.^✦

For bit strings $u$, $v$ and $x$, the twisted coset state of $C_2$ is the state of the register $$|x + C_2\rangle_{u,v} = \frac{1}{\sqrt{|C_2|}} \sum_{y \in C_2} (-1)^{u \cdot y}\, |x + y + v\rangle .$$(1)^✦

For bit strings $a$ and $z$, the phase coset state $|a + C_2\rangle_z$ of $C_2$ is the twisted coset state of the register with $x = a$, $u = z$ and $v = 0$.^✦ These states are labelled by the pairs $p = (a + C_2,\ z + C_2^\perp)$ of a coset of $C_2$ in $\mathbb{Z}_2^n$ and a coset of $C_2^\perp$.^✦ With a representative $a$ fixed in each coset of $C_2$ and a representative $z$ in each coset of $C_2^\perp$, the state of the pair $p$ is $|\chi_p\rangle = |a + C_2\rangle_z$.^✦

The Claim

The $n$ pairs live on two copies of the register, the first qubit of every pair in the first copy and the second qubit in the second. The library’s vector of the pairs is unnormalised, $$|\Phi\rangle = \bigl(|00\rangle + |11\rangle\bigr)^{\otimes n} = 2^{n/2}\,|\beta_{00}\rangle^{\otimes n} .$$(2)^✦

What Is Checked

The claim is stated over the vector $|\Phi\rangle$ of (2) on two copies of the bit register, expanded in (3) over the register’s computational basis and in (4) over the phase coset states $|\chi_p\rangle$ of $C_2$, labelled by pairs of cosets of $C_2$ and $C_2^\perp$.

The library proves that the computational basis is orthonormal, and that every twisted coset state (1), the phase coset states among them, has unit norm and so is a pure state of the register.