The Exercise

Two states $\rho$ and $\sigma$ of one quantum system have a fidelity $F(\rho,\sigma)$, and the fidelity fixes an angle $A(\rho,\sigma)$ between them. The exercise asks to show that this angle lies between $0$ and $\pi/2$, and that it is zero exactly when $\rho$ and $\sigma$ are the same state.^✦

A Quantum System

A quantum system $S$ is a physical object 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 $S$.^✦

The Fidelity and the 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 square root of a positive operator is taken eigenvalue by eigenvalue.

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

The Claim

What Is Checked

The theorem is stated over two arbitrary states $\rho$ and $\sigma$ of a quantum system $S$ and over the angle (2) between them, built on their fidelity (1).