The Exercise

A unitary $U$ acts on a system $S$, and $|u\rangle$ is an eigenstate of $U$ with eigenvalue $e^{2\pi i\varphi}$ for a real phase $\varphi$. A control qubit is prepared in $|0\rangle$ beside the target in $|u\rangle$. A Hadamard gate acts on the control, then $U$ acts on the target if the control is $|1\rangle$, then a second Hadamard gate acts on the control, and the control is measured in the computational basis. The outcome $0$ occurs with probability $\cos^2(\pi\varphi)$, and with $U^k$ in place of $U$ it occurs with probability $\cos^2(\pi k\varphi)$.

Systems and States

A quantum system $S$ is a physical object whose state space is a finite dimensional complex Hilbert space.^✦ Any such space is the state space of a system, and the qubit is the two level system whose state space is $\mathbb{C}^2$. 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.^✦

A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$.^✦ The basis states $|0\rangle$ and $|1\rangle$ of the qubit, the computational basis, are the pure states whose vectors are the standard basis vectors $e_0$ and $e_1$ of $\mathbb{C}^2$.^✦ For pure states $|\psi\rangle$ of $S$ and $|\chi\rangle$ of $T$, the product $|\psi\rangle \otimes |\chi\rangle$ is the pure state of $S \otimes T$ whose vector is the tensor product of the two vectors.^✦

A state $\rho$ of $S$, also called a density operator, is a positive operator of unit trace on the state space of $S$.^✦ A pure state $|\psi\rangle$ of $S$ determines the state $|\psi\rangle\langle\psi|$ of $S$, the projector onto its vector.^✦

Gates and the Circuit

A unitary $U$ of $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = 1$.^✦ It carries a pure state $|\psi\rangle$ of $S$ to the pure state $U|\psi\rangle$.^✦ A pure state $|\psi\rangle$ of $S$ is an eigenstate of $U$ with eigenvalue the complex number $\mu$ when $U|\psi\rangle = \mu|\psi\rangle$.^✦

The identity evolution of $S$ is the unitary $1$.^✦ For two unitaries $U$ and $V$ of $S$, $U$ after $V$ is the unitary $UV$ of $S$, first $V$ then $U$, and for a natural number $k$ the power $U^k$ applies $U$ $k$ times in succession.^✦ The product unitary $U \otimes V$ of a unitary $U$ of $S$ and a unitary $V$ of $T$ is the unitary of $S \otimes T$ whose operator is the tensor product of the two operators. Acting with a unitary $U$ of $S$ on the first factor of $S \otimes T$ is the gate $U \otimes 1$, the product unitary of $U$ and the identity evolution of $T$.^✦

The Hadamard gate $H$ is the unitary of the qubit whose matrix in the computational basis is $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$$(1)^✦ For a unitary $U$ of $S$, the controlled gate $C(U)$ is the unitary of the composite of the qubit, the control, and $S$, the target, with the operator $$C(U) = |0\rangle\langle 0| \otimes 1 + |1\rangle\langle 1| \otimes U .$$(2) For a unitary $U$ of $S$, the circuit $W(U)$ of the exercise is the unitary of the composite of the qubit and $S$ given by $$W(U) = (H \otimes 1)\, C(U)\, (H \otimes 1)$$(3)

Measuring the Control

A measurement $M$ on $S$ with a finite set of outcomes $\iota$ is a family of operators $M_i$ on the state space of $S$, one for each outcome $i \in \iota$, satisfying the completeness relation $$\sum_i M_i^\dagger M_i = 1 .$$(4) The effects of the measurement are the positive operators $E_i = M_i^\dagger M_i$; together they form its positive operator valued measure (POVM).

In a state $\rho$ of $S$, a measurement $M$ on $S$ gives the outcome $i$ with probability, by the Born rule, $$p(i) = \operatorname{Tr}(E_i \rho) = \operatorname{Tr}(M_i \rho M_i^\dagger) .$$(5)^✦

The projective measurement of $S$ in an orthonormal basis $\{|b_i\rangle\}$ of its state space is the measurement whose operator for the outcome $i$ is the projector $|b_i\rangle\langle b_i|$.^✦ For a measurement with operators $M_i$ on $S$ and a further system $T$, the measurement on the first factor of $S \otimes T$ has the same outcomes and the operators $M_i \otimes 1$.^✦ The measurement of the control is the projective measurement of the qubit in the computational basis, performed on the first factor of the composite of the qubit and $S$: its operators are $|0\rangle\langle 0| \otimes 1$ and $|1\rangle\langle 1| \otimes 1$.^✦

The Claim

What Is Checked

The claim is stated over the circuit (3), built from the Hadamard gate (1) on the first factor and the controlled gate (2), acting on the product state $|0\rangle \otimes |u\rangle$, and over the Born probability (5) of the measurement of the control in the state that the circuit produces.

About these constructions the library proves that the operator (2) is unitary, so that $C(U)$ is a gate, and that the Hadamard matrix, composites of unitaries and their tensor products are unitary, so that $W(U)$ is a unitary of the composite. It proves that the projectors onto an orthonormal basis, and with them the operators $|i\rangle\langle i| \otimes 1$, for $i = 0, 1$, of the measurement of the control, satisfy the completeness relation (4), and that $|\psi\rangle\langle\psi|$ is a density operator.