The Exercise

A unitary $U$ on a quantum system $S$ that is also Hermitian, $U^\dagger = U$, is at once a gate and an observable, and each of its eigenvalues is $+1$ or $-1$. The exercise measures this observable with the help of one qubit. The qubit starts in $|0\rangle$ beside a vector $|x\rangle$ of $S$; a Hadamard gate acts on the qubit, then the controlled gate $C(U)$ with the qubit as control and $S$ as target, then a second Hadamard gate on the qubit. Afterwards the qubit holds the outcome: beside its $|0\rangle$ stands the part of $|x\rangle$ in the $+1$ eigenspace of $U$, beside its $|1\rangle$ the part in the $-1$ eigenspace.

The library states the exercise in two theorems. One is about the measurement of any observable whose square is the identity and the states its outcomes leave; the other runs the circuit for a Hermitian unitary $U$ on the input $|0\rangle \otimes |x\rangle$.

A Quantum System

A quantum system $S$ is given by its state space, a finite dimensional complex Hilbert space. The qubit $Q$ 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 the state spaces of $S$ and $T$; the circuit acts on $Q \otimes S$, a control qubit beside the target $S$.^✦

A pure state $\psi$ of $S$ is a unit vector $|\psi\rangle$ in the state space of $S$.^✦ The basis states $|0\rangle$ and $|1\rangle$ of the qubit are the pure states whose vectors are the standard basis vectors $e_0$ and $e_1$ of $\mathbb{C}^2$.^✦ For a nonzero vector $|v\rangle$ in the state space of $S$, the normalized pure state of $|v\rangle$ is the pure state of $S$ whose vector is $|v\rangle$ divided by its norm.^✦ A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$, and a pure state $\psi$ gives the state $|\psi\rangle\langle\psi|$, the projector onto its vector.

Gates and the Circuit

A unitary $U$ of $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = 1$, and the identity evolution of $S$ is the unitary $1$. For two unitaries $U$ and $V$ of $S$, the composite $UV$ is the unitary of $S$ that applies first $V$, then $U$.^✦ Unitaries $U$ of $S$ and $V$ of $T$ form the product unitary $U \otimes V$ of $S \otimes T$, whose operator is the tensor product of the two operators; a gate $U$ on the first factor of $S \otimes T$ is the product $U \otimes 1$ with the identity evolution of $T$.

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 $Q \otimes S$ with operator $$C(U) = |0\rangle\langle 0| \otimes 1 + |1\rangle\langle 1| \otimes U,$$(2) The measuring circuit of $U$ is the unitary of $Q \otimes S$ $$(H \otimes 1)\, C(U)\, (H \otimes 1).$$(3)

Observables and Their Measurement

An observable $A$ on $S$ is a Hermitian operator on the state space of $S$, $A^\dagger = A$.^✦ A pure state $\psi$ of $S$ is an eigenstate of $A$ with eigenvalue $\mu$, a real number, when $A|\psi\rangle = \mu|\psi\rangle$.^✦ A Hermitian unitary $U$ of $S$, regarded as an observable, is the observable of $S$ whose operator is $U$ itself.^✦

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

The projectors $P_+$ and $P_-$ of an observable $A$ of $S$ are the operators $$P_+ = \tfrac{1}{2}(1 + A), \qquad P_- = \tfrac{1}{2}(1 - A),$$(7) which project onto the $+1$ and $-1$ eigenspaces of $A$ when $A^2 = 1$. For such an observable, the sign measurement of $A$ is the measurement on $S$ with two outcomes, $0$ and $1$, and operators $M_0 = P_+$ and $M_1 = P_-$: the projective measurement of $A$, the outcome $0$ standing for the eigenvalue $+1$ and the outcome $1$ for $-1$.^✦

For a pure state $\psi$ of $S$ with $P_+|\psi\rangle \neq 0$, the state $\psi_+$ after the outcome $+1$ in the measurement of $A$ is the normalized pure state of $P_+|\psi\rangle$.^✦ For an observable $A$ of $S$ with $A^2 = 1$ and a pure state $\psi$ of $S$ with $P_-|\psi\rangle \neq 0$, the state $\psi_-$ of $\psi$ after the outcome $-1$ in the sign measurement of $A$ is the pure state of $S$ with vector $$|\psi_-\rangle = \frac{P_-|\psi\rangle}{\lVert P_-|\psi\rangle \rVert}.$$(8)^✦

The Claim

A Hermitian unitary $U$, regarded as an observable, squares to the identity, $U^2 = 1$, so it has a sign measurement, with operators $M_0 = \tfrac{1}{2}(1 + U)$ and $M_1 = \tfrac{1}{2}(1 - U)$.

What Is Checked

The first theorem is stated over the sign measurement of an observable with $A^2 = 1$, its projectors (7), the states (8) its outcomes leave, and the general rule (6) for the state after an outcome, taken on $|\psi\rangle\langle\psi|$. The second is stated over the measuring circuit (3), with the Hadamard gate (1) and the controlled gate (2).

About these constructions the library proves that the Hadamard matrix and the controlled gate are unitary, and that composites and tensor products of unitaries are again unitary, so the circuit is a unitary of $Q \otimes S$. It proves as well that the projector onto a unit vector, and the state after an outcome of nonzero probability, are density operators.