The Exercise

Two qubits in states $|q_A\rangle$ and $|q_B\rangle$ are to come out of a controlled NOT (CNOT) gate as two encoded qubits, each held in a quantum system $C$. The gate is applied beforehand, to a resource of four qubits numbered $1$ to $4$: qubits $1,2$ and qubits $3,4$ start as two Bell pairs $|\Phi^+\rangle$, and a CNOT acts on qubits $2$ and $3$ with $2$ as the control. Then $|q_A\rangle$ is measured together with qubit $1$ in the Bell basis, and $|q_B\rangle$ together with qubit $4$. Qubits $2$ and $3$ are left in $\mathrm{CNOT}\,|q_A\rangle|q_B\rangle$ up to a Pauli operator on each, fixed by the four bits of the two outcomes. The same holds when qubits $2$ and $3$ are held encoded, by a linear map $E$ from the qubit into $C$. The exercise asks for the identity behind the protocol: the state of the six qubits before the measurements, written in the Bell bases of the two measured pairs.

Qubits and Gates

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ A pure state $\psi$ of $S$ is a unit vector of the state space of $S$, $\lVert\psi\rVert = 1$, determined physically up to a global phase.^✦ The composite system $S\otimes T$ of two systems $S$ and $T$ is the system whose state space is the tensor product of theirs^✦, and the product $\psi\otimes\varphi$ of pure states of $S$ and $T$ is the pure state of $S\otimes T$ whose vector is the tensor product of their vectors.^✦ The qubit $Q$ is the quantum system with state space $\mathbb{C}^2$^✦; its basis states $|0\rangle$ and $|1\rangle$ are the pure states with the standard basis vectors $e_0$ and $e_1$ as vectors.^✦

A unitary $U$ of a system $S$, a closed evolution of $S$, is an operator on the state space of $S$ with $U^\dagger U = UU^\dagger = 1$^✦; it takes a pure state $\psi$ of $S$ to the pure state $U\psi$.^✦ For unitaries $U$ and $V$ of $S$, $U$ after $V$ is the unitary $UV$, first $V$ then $U$^✦, and the identity evolution is the unitary $1$.^✦ 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 theirs.^✦ Its special cases $U\otimes 1$ and $1\otimes V$ are $U$ acting on the first system alone and $V$ acting on the second system alone.

The Pauli gates $X$ and $Z$ are the unitaries of the qubit whose matrices in the basis $|0\rangle$, $|1\rangle$ are $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(1) For bits $x$ and $y$, the Pauli byproduct $X^yZ^x$ is the unitary of the qubit that applies $Z$ if $x = 1$ and then $X$ if $y = 1$, the identity standing in for either one whose bit is $0$.^✦

For a unitary $U$ of a system $S$, the controlled gate $C(U)$ is the unitary of $Q\otimes S$, a control qubit and the target $S$, with operator $$C(U) = |0\rangle\langle 0|\otimes 1 + |1\rangle\langle 1|\otimes U.$$(2)^✦ The CNOT gate is the unitary $C(X)$ of two qubits.

For bits $x$ and $y$, the Bell vector $|\beta_{xy}\rangle$ is the vector of two qubits $$|\beta_{xy}\rangle = \frac{1}{\sqrt{2}}\bigl(|0\rangle|y\rangle + (-1)^x|1\rangle|\bar y\rangle\bigr), \qquad \bar y = y + 1 \bmod 2.$$(3)^✦ The Bell state $|\Phi^+\rangle$ is the pure state of two qubits whose vector is the unnormalised Bell vector $|00\rangle + |11\rangle$ divided by its norm $\sqrt{2}$.^✦

The Resource

An identification $e$ of a quantum system $S$ with a system $T$ is a linear isometric bijection from the state space of $S$ onto that of $T$^✦; it relabels a pure state $\psi$ of $S$ as the pure state $e\psi$ of $T$.^✦ Identifications compose^✦ and have inverses^✦, the identity map of a state space is one^✦, and identifications $e$ of $S$ with $S'$ and $f$ of $T$ with $T'$ give their product $e\otimes f$, an identification of $S\otimes T$ with $S'\otimes T'$.^✦ The elementary ones regroup factors: the regrouping $\alpha$ takes $u\otimes(v\otimes w)$ to $(u\otimes v)\otimes w$^✦, the swap takes $u\otimes v$ to $v\otimes u$^✦, and the interchange takes $(a\otimes b)\otimes(c\otimes d)$ to $(a\otimes c)\otimes(b\otimes d)$.^✦

From these are built the regrouping $R$ of a register of four qubits, which brings its middle qubits together, and the regrouping $P$ of the six qubits of the protocol, the two data qubits followed by the four qubits of the resource, which puts each data qubit beside the resource qubit it is measured with, the two measured pairs first and the two output qubits last: $$R\bigl((a\otimes b)\otimes(c\otimes d)\bigr) = a\otimes\bigl((b\otimes c)\otimes d\bigr),$$(4)^✦ $$P\Bigl((a\otimes b)\otimes\bigl((c\otimes d)\otimes(e\otimes f)\bigr)\Bigr) = \bigl((a\otimes c)\otimes(b\otimes f)\bigr)\otimes(d\otimes e).$$(5)^✦ The first is the inverse of $\alpha$ followed by $1\otimes\alpha$^✦, the second two interchanges with regroupings and a swap between them.^✦

The CNOT on the middle qubits is the unitary $1\otimes(\mathrm{CNOT}\otimes 1)$ of the regrouped register $Q\otimes((Q\otimes Q)\otimes Q)$, the CNOT on qubits $2$ and $3$ and the identity on qubits $1$ and $4$.^✦ The CNOT teleportation resource $|\chi\rangle$ is the pure state of four qubits obtained from two Bell pairs by regrouping with $R$, applying the CNOT on the middle qubits and regrouping back: $$|\chi\rangle = R^{-1}\bigl(1\otimes(\mathrm{CNOT}\otimes 1)\bigr)R\, \bigl(|\Phi^+\rangle\otimes|\Phi^+\rangle\bigr).$$(6)^✦

The Claim

What Is Checked

The claim is stated over the resource $|\chi\rangle$ of (6) and the regrouping $P$ of (5) on the left, and over the Bell vectors (3), the Pauli byproducts and the CNOT gate on the right, for any linear map $E$ into any quantum system $C$ and any two vectors of the qubit.

The library proves that the matrices (1) are unitary and that the controlled gate (2) is unitary for every unitary $U$.^✦ Products and composites of unitaries are shown to be unitaries, which makes the CNOT on the middle qubits and every Pauli byproduct unitaries too. Finally, $|\Phi^+\rangle$ has unit norm, and a product, a relabelling or a unitary image of pure states is again a pure state, so that $|\chi\rangle$ is a pure state of four qubits.