The Exercise

The controlled Z gate acts on two qubits, a control and a target. When the control is in $|0\rangle$ the target is left alone; when it is in $|1\rangle$ the target undergoes the phase flip $Z$. The exercise is to show that the two qubits play the same part: the gate with the second qubit as control and the first as target is the same gate as the one with the first qubit as control.

A Quantum System

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

The qubit $Q$ is the two level quantum system with state space $\mathbb{C}^2$.^✦ 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 $\mathbb{C}^2$, and $|0\rangle$ and $|1\rangle$ are the computational basis states.^✦ Two qubits form the composite system $Q \otimes Q$.

Unitary Gates

A unitary $U$ of a quantum system $S$ is a closed evolution of $S$: an operator on the state space of $S$ with $$U^\dagger U = U U^\dagger = 1 .$$(1)^✦ For two unitaries $U$ and $V$ of the same system $S$, the unitary $U$ after $V$ is the composite $UV$, which applies $V$ first and then $U$.^✦

The Pauli $Z$, the phase flip, is the unitary of the qubit whose matrix in the computational basis is $$Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} .$$(2)^✦

For a unitary $U$ of a system $S$, the controlled gate $C(U)$ is the unitary of the composite system $Q \otimes S$ of a control qubit and the target $S$ whose operator is $$C(U) = |0\rangle\langle 0| \otimes 1 + |1\rangle\langle 1| \otimes U ,$$(3)^✦ with $|0\rangle\langle 0|$ and $|1\rangle\langle 1|$ the projectors onto the basis states of the control and $1$ the identity on the state space of $S$. The controlled Z gate is $CZ = C(Z)$, with $Z$ the Pauli matrix (2), a unitary of $Q \otimes Q$.^✦

For a system $S$, the swap $\mathrm{SWAP}$ is the unitary of $S \otimes S$ that exchanges the two factors, so that for pure states $|\varphi\rangle$ and $|\chi\rangle$ of $S$: $$\mathrm{SWAP}\,\bigl(|\varphi\rangle \otimes |\chi\rangle\bigr) = |\chi\rangle \otimes |\varphi\rangle .$$(4)^✦ On $Q \otimes Q$ it exchanges the two qubits.

The Claim

Exchanging the roles of the two qubits amounts to swapping them before the gate and swapping them back after it: the control then sits on the second qubit and the phase flip acts on the first.

What Is Checked

The claim (5) is stated over two unitaries of the two qubit system $Q \otimes Q$, the controlled Z gate $C(Z)$ of (3) and the swap (4), and over their product $\mathrm{SWAP}\,CZ\,\mathrm{SWAP}$, a composite of unitaries of $Q \otimes Q$.

The library proves that the operator (3) satisfies (1) for every unitary $U$ of every system $S$, so that $C(U)$ is a unitary, and that $U$ after $V$ is a unitary whenever $U$ and $V$ are, being a product of unitaries.