The Exercise

Two qubits are in one of the four Bell states $|\beta_{xy}\rangle$, labelled by two bits $x$ and $y$. Alice holds the first qubit and measures an observable $E$ on it, leaving the second untouched, so that the observable of the pair is $E \otimes I$, with $I$ the identity on the second qubit. The exercise is to show that the expectation value $\langle \beta_{xy} | E \otimes I | \beta_{xy} \rangle$ does not depend on $x$ and $y$.^✦

A Quantum System

A quantum system $S$ is 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 the state spaces of $S$ and $T$.^✦ The qubit is the two level system whose state space is $\mathbb{C}^2$ itself, with its standard inner product, and the two qubit system is the composite of the qubit with itself.

A pure state $\psi$ of a system $S$ is a unit vector $|\psi\rangle$ 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 $\mathbb{C}^2$.^✦ For pure states $\psi$ of $S$ and $\varphi$ of $T$, the product $\psi \otimes \varphi$ is the pure state of $S \otimes T$ whose vector is $|\psi\rangle \otimes |\varphi\rangle$.^✦ We write $|ab\rangle$ for the product $|a\rangle \otimes |b\rangle$ of two basis states.

The Bell States

For bits $x$ and $y$, the Bell vector $|\beta_{xy}\rangle$ is the vector of the two qubit system $$|\beta_{xy}\rangle = \frac{1}{\sqrt{2}} \bigl( |0\,y\rangle + (-1)^{x}\, |1\,\bar{y}\rangle \bigr),$$(1)^✦ where $\bar{y} = y \oplus 1$ is the other bit. The Bell state $\beta_{xy}$ is the pure state of the two qubit system whose vector is $|\beta_{xy}\rangle$.^✦

Observables and Expectation Values

An observable $A$ of a system $S$ is a self adjoint operator on the state space of $S$, $A^\dagger = A$.^✦ The identity observable $I$ of $S$ is the observable whose operator is the identity on the state space of $S$.^✦ For an observable $A$ of $S$ and an observable $B$ of $T$, the observable $A \otimes B$ of the composite system $S \otimes T$ is the one whose operator is the tensor product of the operators of $A$ and $B$.^✦ The expectation value of an observable $A$ of $S$ in a pure state $\psi$ of $S$ is the real number $$\langle A \rangle_\psi = \langle \psi | A | \psi \rangle .$$(2)^✦

The Claim

What Is Checked

The claim (3) is stated over the Bell states built on the vectors (1) and over the expectation value (2) of the observable $E \otimes I$, formed from $E$ and the identity observable of the second qubit.

About these constructions the library proves that every Bell vector has unit norm, so that $\beta_{xy}$ is a pure state, and that the vector of a product $\psi \otimes \varphi$ of two pure states has unit norm, the norm of a product vector being the product of the norms. It also proves that the tensor product of two self adjoint operators is self adjoint, which makes $A \otimes B$ an observable.