The Exercise

A system can be in a pure state $|\psi\rangle$ or in a state $\sigma$ that may be mixed, and there are two ways to say how close the two are: their fidelity $F$ and their trace distance $D$. The exercise asks for a bound between the two measures when one of the states is pure: one minus the square of the fidelity is at most the trace distance.^✦

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 $S$; pure and mixed states alike are states in this sense.^✦

A pure state $\psi$ of $S$ is a vector $|\psi\rangle$ of unit norm in the state space of $S$, $\lVert \psi \rVert = 1$.^✦ A pure state $\psi$ of $S$ determines the state $|\psi\rangle\langle\psi|$ of $S$, the rank one projector onto $|\psi\rangle$.^✦

Fidelity and Trace Distance

The fidelity $F(\rho,\sigma)$ of two states $\rho$ and $\sigma$ of $S$ is the nonnegative real number equal to the trace of the square root of $\sqrt{\rho}\,\sigma\sqrt{\rho}$, where the square root of a positive operator is taken by applying the square root to each of its eigenvalues. $$F(\rho,\sigma) = \operatorname{tr}\sqrt{\sqrt{\rho}\,\sigma\sqrt{\rho}}.$$(1)^✦

The trace distance between two states $\rho$ and $\sigma$ of $S$ is half the trace norm of their difference, $$D(\rho,\sigma) = \tfrac{1}{2}\operatorname{tr}\lvert \rho - \sigma \rvert = \tfrac{1}{2}\sum_i \lvert \lambda_i \rvert,$$(2)^✦ where the $\lambda_i$ are the eigenvalues of $\rho - \sigma$.

The Claim

For a pure state $\psi$ and a state $\sigma$ of the same system, write $F(\psi,\sigma)$ and $D(\psi,\sigma)$ for the fidelity (1) and the trace distance (2) of $|\psi\rangle\langle\psi|$ and $\sigma$.

What Is Checked

The claim is stated over the state $|\psi\rangle\langle\psi|$ of a pure state $\psi$ and an arbitrary state $\sigma$ of the same system, compared by the fidelity (1) and the trace distance (2).