The Exercise

In the B92 protocol, Bennett’s scheme of 1992 for quantum key distribution, Alice sends Bob one qubit prepared according to a classical bit $a$ that she keeps: the state $|0\rangle$ if $a = 0$, the state $|+\rangle$ if $a = 1$. Bob holds a bit $a'$ of his own. He measures the Pauli operator $Z$ on the qubit if $a' = 0$ and the Pauli operator $X$ if $a' = 1$, and records $b = 0$ for the eigenvalue $+1$ and $b = 1$ for the eigenvalue $-1$. The exercise is to show that Bob obtains $b = 1$ with nonzero probability exactly when $a \neq a'$.^✦ Each $b = 1$ then tells him Alice’s bit, $a = 1 - a'$.

A Quantum System

A quantum system $S$ is described by its state space, a finite dimensional complex Hilbert space.^✦ A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$.^✦ A pure state $\psi$ of $S$ is a unit vector $|\psi\rangle$ in the state space of $S$^✦; it determines the state $|\psi\rangle\langle\psi|$ of $S$, the rank one projector onto $|\psi\rangle$.^✦

The qubit is the quantum system whose state space is $\mathbb{C}^2$^✦, with the computational basis kets $$|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$$(1)^✦ The pure state $|0\rangle$ of the qubit has the first of these kets as its vector^✦, and the pure state $|+\rangle$ has the vector $$|+\rangle = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} = \frac{|0\rangle + |1\rangle}{\sqrt{2}}.$$(2)^✦

Observables and Measurements

An observable $A$ on $S$ is a self adjoint operator on the state space of $S$, $A^\dagger = A$.^✦ The Pauli matrices $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$(3) make up the Pauli vector $\vec{\sigma} = (X, Y, Z)$. For a real vector $\vec{n} = (n_x, n_y, n_z)$, the Pauli observable $\vec{n} \cdot \vec{\sigma}$ is the observable on the qubit with the matrix $$\vec{n} \cdot \vec{\sigma} = n_x X + n_y Y + n_z Z.$$(4)^✦ The observables $X$ and $Z$ are the Pauli observables of the unit vectors $\hat{x} = (1, 0, 0)$ and $\hat{z} = (0, 0, 1)$.

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 $i$, satisfying the completeness relation $$\sum_i M_i^\dagger M_i = 1.$$(5)^✦ The effect of the outcome $i$ of a measurement $\{M_i\}$ on $S$ is the positive operator $E_i = M_i^\dagger M_i$ on the state space of $S$, and the effects $E_i$ together form the positive operator valued measure (POVM) of the measurement. The probability of the outcome $i$ when the measurement is performed on a state $\rho$ of $S$ is given by the Born rule, $$p(i) = \operatorname{Tr}(M_i \rho M_i^\dagger) = \operatorname{Tr}(E_i \rho).$$(6)^✦

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

The Protocol

Alice’s state $|\psi_a\rangle$ for a bit $a$ is the pure state of the qubit $$|\psi_0\rangle = |0\rangle, \qquad |\psi_1\rangle = |+\rangle.$$(8)^✦ Bob’s observable $B_{a'}$ for a bit $a'$ is the observable $Z$ on the qubit when $a' = 0$ and the observable $X$ when $a' = 1$.^✦ Bob’s measurement for the bit $a'$ is the sign measurement of $B_{a'}$.^✦ The probability $P(b = 1 \mid a, a')$ that Bob obtains $b = 1$ is the probability of the outcome $1$ of his measurement on the state $|\psi_a\rangle\langle\psi_a|$, $$P(b = 1 \mid a, a') = \operatorname{Tr}\bigl(P_- \, |\psi_a\rangle\langle\psi_a| \, P_-^\dagger\bigr), \qquad P_- = \tfrac{1}{2}(1 - B_{a'}).$$(9)^✦

The Claim

What Is Checked

The claim is stated over the probability $P(b = 1 \mid a, a')$ of (9), the Born probability of the eigenvalue $-1$ when the sign measurement of $Z$ or of $X$ is performed on the state $|0\rangle\langle 0|$ or $|+\rangle\langle +|$ of the qubit.

The library proves that $|0\rangle$ and $|+\rangle$ have unit norm, and that the projector onto a unit vector is a density operator, so that $|\psi_a\rangle\langle\psi_a|$ is a state of the qubit. It proves that $\vec{n} \cdot \vec{\sigma}$ is Hermitian, so that each Pauli observable is self adjoint, and that $X$ and $Z$ square to the identity. For an observable with $A^2 = 1$ it proves that $P_+$ and $P_-$ are self adjoint and idempotent with $P_+ + P_- = 1$, which gives the completeness relation (5) for the sign measurement; for any measurement, it proves that the effects $E_i$ are positive and sum to the identity.