The Exercise

Let $\mathcal{E}$ be a quantum channel on a system that brings any two different states strictly closer in trace distance, so that its two outputs are always harder to tell apart than the two inputs. Show that $\mathcal{E}$ has a unique fixed point: exactly one state that the channel leaves unchanged.^✦

A Quantum System

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

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

Strictly Contractive Channels

A quantum channel $\mathcal{E}$ on $S$ is a map on the states of $S$ given by a completely positive, trace preserving linear map $\Phi$ on the operators on the state space of $S$: $$\mathcal{E}(\rho) = \Phi(\rho) \ \text{for every state } \rho, \qquad \operatorname{tr}\Phi(A) = \operatorname{tr}A \ \text{for every operator } A .$$(2)^✦

A map $\mathcal{E}$ on the states of $S$ is strictly contractive when, for any two different states, the trace distance between their images is smaller than the distance between them: $$D\bigl(\mathcal{E}(\rho),\mathcal{E}(\sigma)\bigr) < D(\rho,\sigma) \quad \text{for all states } \rho \neq \sigma \text{ of } S .$$(3)^✦

The Claim

What Is Checked

The claim is stated over the states of $S$ with the trace distance (1) between them, and over a map $\mathcal{E}$ on those states that is a channel in the sense of (2) and strictly contractive in the sense of (3).