The Exercise

A qubit in a state $\rho$ passes through a noisy channel that leaves it alone or applies one of the Pauli operators $X$, $Y$ and $Z$, each of the four with probability one quarter. This is the completely depolarizing channel $\mathcal{E}$. The exercise is to show that $\mathcal{E}$ is a quantum channel^✦ and that it erases its input: whatever state enters, the qubit leaves in the maximally mixed state $I/2$, half the identity.^✦ The library states both facts, with the channel given by four operation elements, the operators $E_k$ for which $\mathcal{E}(\rho)$ is the sum of the terms $E_k \rho E_k^\dagger$.

The Qubit

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

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 $S$ with a nonzero state space is the state $I/d$ of $S$, the identity divided by the dimension $d$ of that space; for the qubit it is $I/2$.^✦

An observable $A$ of $S$ is a self adjoint operator on the state space of $S$, $A^\dagger = A$.^✦

The Pauli matrices are $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},$$(1) and for a real vector $\vec{n} = (n_1, n_2, n_3)$ in three dimensions the matrix $$\vec{n} \cdot \vec{\sigma} = n_1 X + n_2 Y + n_3 Z$$(2)^✦ is Hermitian. The Pauli observable $\vec{n} \cdot \vec{\sigma}$ is the observable of the qubit with this matrix as its operator on $\mathbb{C}^2$.^✦ The observables $X$, $Y$ and $Z$ are the Pauli observables $\hat{n} \cdot \vec{\sigma}$ along the coordinate axes, $\hat{n} = (1,0,0)$, $(0,1,0)$ and $(0,0,1)$, whose operators are the matrices (1).

The Channel

The Pauli family of the qubit is the four operators $\sigma_0 = I$, $\sigma_1 = X$, $\sigma_2 = Y$ and $\sigma_3 = Z$: the identity and the operators of the three Pauli observables.^✦ The completely depolarizing channel $\mathcal{E}$ of the qubit is the map on its states with operation elements $E_k = \sigma_k/2$, $k = 0, \dots, 3$: $$\mathcal{E}(\rho) = \sum_{k=0}^{3} E_k \rho E_k^\dagger = \tfrac{1}{4}\left(\rho + X\rho X + Y\rho Y + Z\rho Z\right).$$(3)^✦

A quantum channel $f$ of a system $S$ is a map on the states of $S$ that is the restriction to states of a linear map $\Phi$ on the operators on the state space of $S$ which is trace preserving and completely positive, meaning that $\Phi \otimes \mathrm{id}_R$ takes positive operators to positive operators for every auxiliary system $R$: $$\operatorname{tr} \Phi(A) = \operatorname{tr} A \ \text{for every operator } A, \qquad \Phi(\rho) = f(\rho) \ \text{for every state } \rho .$$(4)

The Claim

What Is Checked

The claim is stated over the states of the qubit, the completely depolarizing channel (3) with the Pauli family scaled by one half as its operation elements, the maximally mixed state of the qubit, and the definition (4) of a quantum channel.

The library proves that these operation elements satisfy the completeness relation $\sum_k E_k^\dagger E_k = I$, so that $\mathcal{E}(\rho)$ is a state of the qubit for every state $\rho$, and that $I/d$ is positive with unit trace, so that the maximally mixed state is a state.