The Exercise

Two states $\rho$ and $\sigma$ of a quantum system pass through the same quantum channel $\mathcal{E}$. The exercise is to show that the angle between the outputs $\mathcal{E}(\rho)$ and $\mathcal{E}(\sigma)$ is never larger than the angle between $\rho$ and $\sigma$.^✦

A Quantum System

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$.^✦

Fidelity and Angle

The fidelity of two states $\rho$ and $\sigma$ of $S$ is the nonnegative real number $$F(\rho,\sigma) = \operatorname{tr}\sqrt{\sqrt{\rho}\,\sigma\sqrt{\rho}},$$(1)^✦ where the trace is over the state space of $S$ and every square root is the positive square root of a positive operator, taken eigenvalue by eigenvalue.

The angle between two states $\rho$ and $\sigma$ of $S$ is the arccosine of their fidelity, $$A(\rho,\sigma) = \arccos F(\rho,\sigma).$$(2)^✦

Quantum Channels

A quantum channel $\mathcal{E}$ on $S$ is a map taking states of $S$ to states of $S$ that agrees on states with a completely positive, trace preserving linear map $\Phi$ on the operators of the state space of $S$: $$\mathcal{E}(\rho) = \Phi(\rho) \quad \text{for every state } \rho, \qquad \operatorname{tr}\Phi(X) = \operatorname{tr} X \quad \text{for every operator } X.$$(3)^✦

The Claim

What Is Checked

The claim is stated over a quantum channel on $S$ in the sense of (3) and over the angle (2) between two states of $S$, built from their fidelity (1). What the library proves about these constructions is the inequality (4) for every channel on $S$ and every pair of states of $S$, pure and mixed alike.^✦