The Exercise

A device acts on a qubit by picking one of the unitaries $U_1, \dots, U_n$ at random, $U_k$ with probability $p_k$, and applying it. Averaged over the choice, a state $\rho$ goes to $\sum_k p_k U_k \rho U_k^\dagger$.

The exercise runs the other way. A channel on a qubit that leaves the maximally mixed state $I/2$ unchanged is always of this kind, a unitary drawn at random. On a qutrit, a three level system, the converse fails: some channel leaves $I/3$ unchanged and is no mixture of unitaries.^✦

Qubits and Qutrits

Both halves concern maps that take the states of a system to its states.

A quantum system $S$ is a physical object whose state space is a finite dimensional complex Hilbert space.^✦ The qudit with $d$ levels is the quantum system whose state space is $\mathbb{C}^d$ with its standard inner product.^✦ The qubit is the two level system, with state space $\mathbb{C}^2$, and the qutrit is the qudit with three levels.

A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$.^✦ The maximally mixed state of a system $S$ whose state space is nonzero, of dimension $d$, is the state $I/d$, where $I$ is the identity operator.^✦ A unitary $U$ of $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = I$.^✦ Under $U$ a state $\rho$ of $S$ evolves to $U \rho U^\dagger$.^✦

Channels

A quantum channel $\mathcal{E}$ on $S$ is a map taking states of $S$ to states of $S$ that comes from a completely positive linear map $\Phi$ on the operators of the state space of $S$, one that preserves the trace of every operator: $$\mathcal{E}(\rho) = \Phi(\rho) \ \text{for every state } \rho, \qquad \operatorname{tr} \Phi(A) = \operatorname{tr} A \ \text{for every operator } A.$$(1)^✦

A map $\mathcal{E}$ taking states of $S$ to states of $S$ is unital if it fixes the maximally mixed state of $S$, $\mathcal{E}(I/d) = I/d$.^✦

A random unitary channel $C$ on $S$ is a finite family of unitaries $U_1, \dots, U_n$ of $S$ together with probabilities $p_1, \dots, p_n$, nonnegative and summing to one.^✦ It acts on a state $\rho$ of $S$ by $$C(\rho) = \sum_{k=1}^{n} p_k \, U_k \rho U_k^\dagger .$$(2)^✦ A map $\mathcal{E}$ taking states of $S$ to states of $S$ is random unitary if some random unitary channel $C$ on $S$ agrees with it on every state, $C(\rho) = \mathcal{E}(\rho)$.^✦

The Claim

What Is Checked

The claims (3) and (4) are stated for the quantum channels (1) that fix the maximally mixed state, against the random unitary channels (2).

The library proves that a unitary carries a state to a state, $U \rho U^\dagger$ being again positive of unit trace, and that a random unitary channel does too, its output (2) being a mixture of such states with weights $p_k$.