The Exercise

A qudit with $d$ levels is prepared in a state $\rho$ and sent through a depolarizing channel, which pushes every state towards the maximally mixed one. How mixed a state is shows in the traces of its powers: $\mathrm{tr}(\rho^2)$ equals one for a pure state and $1/d$ for the maximally mixed state. The exercise asks to show that for every integer $k \geq 1$ the depolarizing channel never increases $\mathrm{tr}(\rho^k)$ .^✦

A Quantum System

A quantum system $S$ is given 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 $S$ .^✦ For a natural number $k$ , the $k$ th trace power of a state $\rho$ of $S$ is the real number $\mathrm{tr}(\rho^k)$ , the trace of the $k$ th power of the operator $\rho$ .^✦

The maximally mixed state of a system $S$ whose state space is not the zero space is the state $I/d$ of $S$ , the identity operator divided by the dimension $d$ of the state space.^✦

The qudit with $d$ levels is the quantum system whose state space is $\mathbb{C}^d$ with its standard inner product.^✦

The Claim

The depolarizing channel $\mathcal{E}$ on the qudit with $d$ levels, with a parameter $p$ in $0 \leq p \leq 1$ , replaces a state by the maximally mixed state with probability $p$ and leaves it as it was otherwise.^✦ It takes a state $\rho$ of the qudit to the state $\mathcal{E}(\rho) = (1-p)\,\rho + p\,\frac{I}{d}.$ (1)^✦

Theorem 1. Let $d \geq 1$ be an integer, let $p$ be a real number with $0 \leq p \leq 1$ , and let $\mathcal{E}$ be the depolarizing channel (1) with parameter $p$ on the qudit with $d$ levels. Then for every state $\rho$ of the qudit and every integer $k \geq 1$ , $\mathrm{tr}\bigl(\mathcal{E}(\rho)^k\bigr) \leq \mathrm{tr}(\rho^k).$ (2)^✦

What Is Checked

The claim is stated over the depolarizing channel (1) acting on the states of the qudit with $d$ levels, and over the trace powers of its input and its output.

About these constructions the library proves that $I/d$ is positive with unit trace, so that the maximally mixed state is a state. It proves as well that the output $\mathcal{E}(\rho)$ is again a state of the qudit, being a convex combination of the two states $\rho$ and $I/d$ .