The Exercise

Two qubits are prepared in the Bell state $$|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}.$$(1) One measures the Pauli observable $X$ on the first qubit and the Pauli observable $Z$ on the second, and records the product of the two outcomes, each of which is $\pm 1$. This is a measurement of the observable $X \otimes Z$ of the pair. The exercise is to show that its average over many runs, the expectation value of $X \otimes Z$ in $|\Phi^+\rangle$, is zero.

Two Qubits

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ The qubit is the quantum system whose state space is $\mathbb{C}^2$, the two level system. 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 pair of the exercise is the composite of two qubits, with state space $\mathbb{C}^2 \otimes \mathbb{C}^2$.

A pure state $|\psi\rangle$ of a quantum system $S$ is a vector of unit norm in the state space of $S$, $\lVert \psi \rVert = 1$.^✦ 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 $\mathbb{C}^2$; these are the computational basis states $|0\rangle$ and $|1\rangle$.^✦ Writing $|ij\rangle$ for $|i\rangle \otimes |j\rangle$, the Bell vector is the unnormalised vector $|00\rangle + |11\rangle$ in the state space of two qubits.^✦ The Bell state $|\Phi^+\rangle$ is the pure state of two qubits whose vector is the Bell vector scaled by $1/\sqrt{2}$, the state (1).^✦

Observables

An observable $A$ of a quantum system $S$ is a self adjoint operator on the state space of $S$, $A^\dagger = A$; its real spectrum is the set of values a measurement can return.^✦ The tensor product $A \otimes B$ of an observable $A$ of $S$ and an observable $B$ of $T$ is the observable of $S \otimes T$ whose operator is the tensor product of the operators of $A$ and $B$.^✦

The Pauli matrices are $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},$$(2) the components of the Pauli vector $\vec{\sigma} = (X, Y, Z)$. For a real vector $\vec{n} = (n_x, n_y, n_z)$ in three dimensions, $\vec{n} \cdot \vec{\sigma}$ is the complex $2 \times 2$ matrix $$\vec{n} \cdot \vec{\sigma} = n_x X + n_y Y + n_z Z,$$(3) and the Pauli observable $\vec{n} \cdot \vec{\sigma}$ is the observable of the qubit whose operator is this matrix acting on $\mathbb{C}^2$. The observables $X$ and $Z$ of the qubit are the Pauli observables for the unit vectors $\hat{n} = (1, 0, 0)$ and $\hat{n} = (0, 0, 1)$, with the matrices $X$ and $Z$ of (2) as operators.

The expectation value of an observable $A$ of $S$ in a pure state $|\psi\rangle$ of $S$ is $$\langle A \rangle_\psi = \langle \psi | A | \psi \rangle,$$(4)^✦ a real number because $A$ is self adjoint.

The Claim

What Is Checked

The claim is stated over the Bell state $|\Phi^+\rangle$, a pure state of the composite of two qubits, and over the observable $X \otimes Z$, the tensor product of the Pauli observables $X$ and $Z$, with the expectation value (4).

About these constructions the library proves that the Bell vector $|00\rangle + |11\rangle$ has norm $\sqrt{2}$, so that $|\Phi^+\rangle$ has unit norm. It proves that the matrix $\vec{n} \cdot \vec{\sigma}$ is self adjoint for every real $\vec{n}$, which makes each Pauli observable an observable, and that a tensor product of self adjoint operators is self adjoint, which makes $X \otimes Z$ an observable of the two qubits.