The Exercise

Box 6.1 of the book draws a circuit on two qubits inside a dotted box, built from Pauli $X$ and Hadamard gates around a controlled NOT (CNOT). The exercise is to show that, up to a global phase, the circuit performs the conditional phase shift $$2\lvert 00 \rangle\langle 00 \rvert - I ,$$(1) which leaves $\lvert 00 \rangle$ alone and changes the sign of every vector orthogonal to it.

Two Qubits

A quantum system $S$ is a physical system given by its state space, a finite dimensional complex Hilbert space.^✦ The composite system $S \otimes T$ of two systems $S$ and $T$ is the system whose state space is the tensor product of their state spaces.^✦ The qubit is the quantum system whose state space is $\mathbb{C}^2$, and two qubits form the composite $\mathrm{qubit} \otimes \mathrm{qubit}$.

A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$; pure and mixed states alike are states in this sense.^✦ A pure state $\psi$ of $S$ is a unit vector in the state space of $S$.^✦ The basis states $\lvert 0 \rangle$ and $\lvert 1 \rangle$ of the qubit are the pure states whose vectors are the standard basis vectors $e_0$ and $e_1$ of $\mathbb{C}^2$.^✦ Given pure states $\psi$ of $S$ and $\varphi$ of $T$, their product $\psi \otimes \varphi$ is the pure state of $S \otimes T$ whose vector is the tensor product of the two vectors.^✦ The basis state $\lvert 00 \rangle$ of two qubits is the product $\lvert 0 \rangle \otimes \lvert 0 \rangle$.^✦

Gates

A unitary $U$ of a system $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = I$.^✦ A unitary $U$ of $S$ evolves a state $\rho$ of $S$ into the state $$\rho \;\longmapsto\; U \rho U^\dagger .$$(2)^✦

Unitaries combine in two ways. For unitaries $U$ and $V$ of the same system, $U$ after $V$ is the unitary $UV$ whose operator is the composite of the two, first $V$ and then $U$.^✦ 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 the two operators.^✦ The identity evolution of $S$ is the unitary $I$ of $S$.^✦ Given a unitary $V$ of $T$ and a further system $S$, the gate $V$ on the last wires is the product unitary $I \otimes V$ of $S \otimes T$.^✦

The Hadamard gate $H$ and the Pauli $X$ are the unitaries of the qubit whose operators act on $\mathbb{C}^2$ by the matrices $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$(3) For a unitary $U$ of a system $S$, the controlled gate $C(U)$ is the unitary of $\mathrm{qubit} \otimes S$, a control qubit and the target $S$, whose operator is $$C(U) = \lvert 0 \rangle\langle 0 \rvert \otimes I + \lvert 1 \rangle\langle 1 \rvert \otimes U .$$(4)^✦ The CNOT gate is the controlled gate $C(X)$ of two qubits, which flips the second qubit when the first is $\lvert 1 \rangle$.^✦ The conditional phase shift is the unitary $U_{00}$ of two qubits whose operator is (1).^✦

The Circuit

The dotted box circuit of Box 6.1 is the unitary of two qubits $$U_{\mathrm{box}} = (X \otimes X)\,(I \otimes H)\,\mathrm{CNOT}\,(I \otimes H)\,(X \otimes X),$$(5)^✦ in which the rightmost factor acts first.

The Claim

A global phase $e^{i\theta}$ on a unitary $U$ meets $e^{-i\theta}$ on $U^\dagger$ in (2) and cancels, so two unitaries compared by the states they produce are compared up to a global phase, as the exercise asks.

What Is Checked

The claim is stated over the states of two qubits and two unitaries of that system, the circuit (5) and the conditional phase shift (1), each acting on states by (2).

The library proves that the matrices (3) are unitary, so that $H$ and $X$ are unitaries of the qubit, and that $C(U)$ in (4) is unitary for every unitary $U$ of any system, which makes the CNOT a unitary. Composites $UV$ and tensor products $U \otimes V$ of unitaries are again unitaries, so $U_{\mathrm{box}}$ is one. The operator $2\lvert 00 \rangle\langle 00 \rvert - I$ is self adjoint and squares to $I$, hence is unitary. A product of pure states is a unit vector, since the norm of a product vector is the product of the norms, and conjugating a density operator by a unitary gives a density operator, so (2) is again a state.