The Exercise

A qubit is prepared in one of four pure states $|X_0\rangle, \dots, |X_3\rangle$, which form a symmetric informationally complete (SIC) set, each chosen with probability $1/4$. A measurement is then made on the qubit to learn which state it is in. Let $X$ be the label $k$ of the prepared state and $Y$ the outcome of the measurement. The exercise is to show that, whatever the measurement, the mutual information $H(X:Y)$ is strictly less than $\log 2$, one bit.^✦ The Holevo bound gives $H(X:Y) \le \log 2$ for any ensemble of qubit states; for this one the inequality is strict.

A Quantum System

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$, determined physically up to a global phase.^✦ A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$^✦; a pure state $|\psi\rangle$ gives the state $|\psi\rangle\langle\psi|$, the projector onto $|\psi\rangle$.^✦

The superposition of orthogonal pure states $|\psi\rangle$ and $|\phi\rangle$ of $S$, $\langle\psi|\phi\rangle = 0$, with complex amplitudes $a$ and $b$, $|a|^2 + |b|^2 = 1$, is the pure state of $S$ with vector $a|\psi\rangle + b|\phi\rangle$.^✦

The qubit is the two level system whose state space is $\mathbb{C}^2$.^✦ The basis states $|0\rangle$ and $|1\rangle$ of the qubit are the pure states given by the standard basis vectors of $\mathbb{C}^2$.^✦ For amplitudes $a$ and $b$ with $|a|^2 + |b|^2 = 1$, the qubit state $a|0\rangle + b|1\rangle$ is the superposition of the two basis states with these amplitudes.^✦

The Four States

Let $\omega = e^{2\pi i/3}$ be the cube root of unity.^✦ The four amplitude vectors $(a_k, b_k)$ in $\mathbb{C}^2$, for $k = 0, 1, 2, 3$, are $$(a_0, b_0) = (1, 0), \qquad (a_k, b_k) = \Bigl(\frac{1}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}\,\omega^{k-1}\Bigr) \quad (k = 1, 2, 3).$$(1)^✦ The SIC state $|X_k\rangle$ is the qubit state with these amplitudes, $$|X_k\rangle = a_k|0\rangle + b_k|1\rangle, \qquad k = 0, 1, 2, 3.$$(2)^✦ The label $k$ is drawn with the uniform weights $$p_k = \frac{1}{4}, \qquad k = 0, 1, 2, 3.$$(3)^✦

A Measurement

A measurement on a system $S$ with a finite set of outcomes is a family of measurement operators $M_y$ on the state space of $S$, one for each outcome $y$, satisfying the completeness relation $\sum_y M_y^\dagger M_y = I$.^✦ The positive operator valued measure (POVM) of the measurement is the family of effects $E_y = M_y^\dagger M_y$, one for each outcome.^✦ Performed on a state $\rho$ of $S$, the measurement gives the outcome $y$ with the Born probability $$p(y \mid \rho) = \operatorname{Tr}\bigl(M_y \rho M_y^\dagger\bigr) = \operatorname{Tr}(E_y \rho).$$(4)^✦ Prepare the qubit in the state $|X_k\rangle\langle X_k|$ with $k$ drawn as in (3) and make the measurement on it; the label $X$ and the outcome $Y$ then have the joint distribution $$p(k, y) = p_k \, p\bigl(y \mid |X_k\rangle\langle X_k|\bigr).$$(5)

The Claim

The mutual information of $X$ and $Y$ is $H(X:Y) = H(X) + H(Y) - H(X,Y)$, where $H$ is the Shannon entropy of the joint distribution (5) or of one of its marginals.^✦

What Is Checked

The claim is stated over the four qubit states $|X_k\rangle$ of (2), built from the amplitudes (1) and prepared with the uniform weights (3), and over every measurement on the qubit with a finite set of outcomes, through its Born probabilities (4) on the states $|X_k\rangle\langle X_k|$.

The library proves that each amplitude vector has unit length, $|a_k|^2 + |b_k|^2 = 1$^✦, and that the superposition of two orthogonal pure states with such amplitudes is again a unit vector, so that each $|X_k\rangle$ is a pure state of the qubit. It also proves that the projector $|\psi\rangle\langle\psi|$ onto a unit vector is a density operator, and that the effects $E_y = M_y^\dagger M_y$ of a measurement are positive and sum to the identity.