The Exercise

An operator $U$ on a quantum system that is both unitary and Hermitian is at once a gate and an observable, and as an observable its eigenvalues are $\pm 1$. The exercise measures the observable with a circuit that uses $U$ only as a gate. A control qubit is prepared in $|0\rangle$ beside the system, which is in a pure state $|\psi\rangle$. A Hadamard gate acts on the control, then the controlled gate that applies $U$ to the system when the control is $|1\rangle$, then a second Hadamard gate on the control; last, the control is measured in the computational basis.

The exercise is to show that this is a measurement of $U$ on $|\psi\rangle$. The outcome $0$ comes with the probability of the eigenvalue $+1$ and the outcome $1$ with that of $-1$. After the outcome the control is left in $|0\rangle$ or $|1\rangle$, and the system in $|\psi\rangle$ projected onto the matching eigenspace and normalized.

The library states this for a target system $S$ of any finite dimension, and compares the two measurements outcome by outcome: in the probability of each outcome and in the state it leaves.

Systems and States

A quantum system $S$ is a physical system whose state space is 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 theirs.^✦ The qubit $Q$ is the two level system with state space $\mathbb{C}^2$; its basis states $|0\rangle$ and $|1\rangle$ are the standard basis vectors of $\mathbb{C}^2$.

A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$.^✦ A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$, and a pure state $|\psi\rangle$ gives the state $|\psi\rangle\langle\psi|$, the projector onto it. For pure states $|\psi\rangle$ of $S$ and $|\varphi\rangle$ of $T$, the product $|\psi\rangle|\varphi\rangle = |\psi\rangle \otimes |\varphi\rangle$ is the pure state of $S \otimes T$ whose vector is the tensor product of the two.^✦

Gates

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 $U|\psi\rangle$. Given two unitaries $U$ and $V$ of $S$, the unitary $U$ after $V$ is the product $UV$, first $V$ and then $U$.^✦ The identity evolution of $S$ is the unitary $1$.^✦ 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, and a gate $U$ on the first wires of $S \otimes T$ is the unitary $U \otimes 1$, with the identity evolution on $T$.

The Hadamard gate $H$ is the unitary of the qubit whose matrix in the basis $|0\rangle, |1\rangle$ 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 $Q \otimes S$ $$C(U) = |0\rangle\langle 0| \otimes 1 + |1\rangle\langle 1| \otimes U.$$(2)^✦ The circuit of the exercise is the unitary of $Q \otimes S$ $$W = (H \otimes 1)\, C(U)\, (H \otimes 1).$$(3)^✦

Measurements

A measurement $m$ on $S$ with a finite set of outcomes is a family of operators $M_i$ on the state space of $S$, one for each outcome $i$, that satisfies the completeness relation $$\sum_i M_i^\dagger M_i = 1.$$(4)^✦ The positive operator valued measure (POVM) of $m$ has the elements $E_i = M_i^\dagger M_i$.^✦ Performed on a state $\rho$ of $S$, the measurement gives the outcome $i$ with the probability $$p_m(i \mid \rho) = \operatorname{Tr}(E_i \rho),$$(5) and when $p_m(i \mid \rho) \neq 0$ the state after the outcome is $$\rho_{m,i} = \frac{M_i \rho M_i^\dagger}{p_m(i \mid \rho)}.$$(6)

The measurement of $S$ in an orthonormal basis $\{|b_i\rangle\}$ is the measurement with the operators $M_i = |b_i\rangle\langle b_i|$.^✦ The measurement on the first wires of $S \otimes T$ made from a measurement on $S$ with operators $M_i$ is the measurement with the same outcomes and the operators $M_i \otimes 1$.^✦ The measurement of the control $m_{\mathrm{c}}$ is the measurement of $Q \otimes S$ in the basis $|0\rangle, |1\rangle$ of the qubit, made on the first wires, so that its operators are $|0\rangle\langle 0| \otimes 1$ and $|1\rangle\langle 1| \otimes 1$.^✦ Write $p_{\mathrm{c}}$ for its probabilities and $\rho_{\mathrm{c},i}$ for the states after its outcomes.

Observables

An observable $A$ on $S$ is a Hermitian operator on the state space of $S$, $A^\dagger = A$.^✦ Its two operators $$P_\pm = \tfrac{1}{2}(1 \pm A)$$(7) are, when $A^2 = 1$, the projectors onto its eigenspaces for $+1$ and $-1$. For such an $A$, the measurement of $A$ is the measurement on $S$ with the operators $M_0 = P_+$ and $M_1 = P_-$, the outcome $0$ standing for the eigenvalue $+1$ and the outcome $1$ for $-1$.^✦

The state $|\psi_+\rangle$ after the outcome $+1$ of $A$, measured on a pure state $|\psi\rangle$ of $S$ with $P_+|\psi\rangle \neq 0$, is the normalized pure state $$|\psi_+\rangle = \frac{P_+|\psi\rangle}{\lVert P_+|\psi\rangle \rVert}.$$(8)^✦ The state $|\psi_-\rangle$ after the outcome $-1$ is formed in the same way from $P_-$, when $P_-|\psi\rangle \neq 0$.^✦

A unitary $U$ of $S$ that is Hermitian is an observable with the same operator, and since $U^\dagger = U$ and $U^\dagger U = 1$ it squares to the identity, $$U^2 = 1.$$(9) Its measurement $m_U$ is the measurement of this observable. Write $p_U$ for its probabilities, $P_\pm$ for the projectors (7) with $A = U$, and $|\psi_\pm\rangle$ for the states it leaves on $|\psi\rangle$.

The Claim

What Is Checked

The claim is stated over the circuit $W$ of (3), made of the Hadamard gate (1) and the controlled gate (2), and over two measurements: that of the control, and that of $U$ with its states $|\psi_\pm\rangle$ of (8), both with the probabilities (5) and the states after an outcome (6).

About these constructions the library proves that $C(U)$ is unitary, $C(U)^\dagger C(U) = C(U) C(U)^\dagger = 1$, and that a Hermitian unitary squares to the identity, (9).