The Exercise

A qubit passes through a device that applies one of the four operations $I$, $X$, $Y$ and $Z$, the identity and the three Pauli matrices of (1), each with probability $1/4$, and keeps no record of which. A state $\rho$ goes in, and the equal mixture of $\rho$, $X\rho X$, $Y\rho Y$ and $Z\rho Z$ comes out. The exercise is to show that the output is the maximally mixed state $I/2$, whatever the input.

Pauli and Bloch Matrices

The three Pauli matrices are the complex $2\times 2$ matrices $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(1)

For a real vector $\vec r = (r_1, r_2, r_3)$ in three dimensions, $\vec r \cdot \vec\sigma$ is the complex $2\times 2$ matrix $$\vec r \cdot \vec\sigma = r_1 X + r_2 Y + r_3 Z.$$(2)^✦

The Bloch matrix $\rho_{\vec r}$ of a real vector $\vec r$ in three dimensions is $$\rho_{\vec r} = \frac{1}{2}\bigl(I + \vec r \cdot \vec\sigma\bigr).$$(3)^✦ When $|\vec r| \le 1$ it is a density matrix, the state of a qubit with Bloch vector $\vec r$, and every state of a qubit is one of these.

The Pauli Average

The Pauli average $E(A)$ of a complex $2\times 2$ matrix $A$ is the matrix $$E(A) = \frac{1}{4}\bigl(A + XAX + YAY + ZAZ\bigr).$$(4)^✦ On a density matrix $\rho$, $E(\rho)$ is the output of the device of the exercise.

The Claim

For $|\vec r| \le 1$ this is the exercise.

What Is Checked

The claim is stated over the Pauli average (4), built from the Pauli matrices (1), and over the Bloch matrix (3) of an arbitrary real vector $\vec r$, which uses the matrix $\vec r \cdot \vec\sigma$ of (2).

About the Pauli average, the library proves that it fixes the identity and sends each Pauli matrix to zero: $$E(I) = I,$$(6)^✦ $$E(X) = 0,$$(7)^✦ $$E(Y) = 0,$$(8)^✦ $$E(Z) = 0.$$(9)^✦