# Deriving the depolarizing channel

Consider a circuit built as follows: take two ancilla states and an operator $$U$$ made of a series of controlled gates which act on a pure state $$\rho$$ as follows:

• $$X$$ if the ancilla is in $$|00\rangle$$;
• $$Y$$ if the ancilla is in $$|01\rangle$$;
• $$Z$$ if the ancilla is in $$|10\rangle$$;
• $$\mathbb{I}$$ if the ancilla is in $$|11\rangle$$.

Prepare the ancilla in the state $$|\psi\rangle_{A}=\alpha|00\rangle+\beta|01\rangle+\gamma|10\rangle+\delta|11\rangle.$$I want to find $$\rho'$$ using the operator-sum formalism. Using the notation $$C^2_O$$ for a double-controlled gate, we have $$\begin{equation} E_{ij}=\langle ij|C_Z^2C_Y^2C_X^2(\alpha|00\rangle+\beta|01\rangle+\gamma|10\rangle+\delta|11\rangle) \\ =\langle ij|00\rangle\alpha X+\langle ij|01\rangle\beta Y\langle ij|10\rangle\gamma Z+\langle ij|11\rangle\delta \mathbb{I}, \end{equation}$$ so that, for example, $$E_{00}=\alpha X$$, $$E_{00}^\dagger=\alpha^*X$$. Therefore $$\begin{equation} \rho'=\sum_{i,j\in\{0,1\}}E_{ij}\rho E_{ij}^\dagger=|\alpha|^2X\rho X+|\beta|^2 Y\rho Y+|\gamma|^2Z\rho Z+|\delta|^2\rho. \end{equation}$$ (as a side note: what is the standard notation to indicate a doubly controlled quantum gate?)

This model can be used to study several noisy channels. In particular, $$|\alpha|^2=|\beta|^2=|\gamma|^2=p/3$$ gives the so-called depolarizing channel: $$\rho'=\frac{p}{3}(X\rho X+Y\rho Y+Z \rho Z)+(1-p)\rho.$$ To see this, I should be able to write the first term as $$p\mathbb{I}/2$$. The problem is, I can't. I haven't tried anything fancy, just writing $$\rho=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ and doing the calculations explicitly. What's the best way to show it?

Consider the map $$\mathcal{E}: L_H \to L_H$$ given by

$$\mathcal{E}(\rho) = \frac{1}{4} (\rho + X\rho X + Y\rho Y + Z\rho Z)$$

where $$L_H$$ denotes the four-dimensional real vector space of $$2\times 2$$ Hermitian matrices with complex entries.

It is straightforward to calculate that $$\mathcal{E}(I) = I$$ and $$\mathcal{E}(X) = \mathcal{E}(Y) = \mathcal{E}(Z) = 0$$. However, $$I, X, Y, Z$$ form a basis of $$L_H$$, so any $$\rho \in L_H$$ can be written as $$\rho = aI + bX + cY + dZ$$ and then $$\mathcal{E}(\rho) = aI$$. If $$\rho$$ is a density matrix then $$a = \frac{1}{2}$$ and $$\mathcal{E}(\rho) = I/2$$.

We can use this result to extract the term proportional to the identity in

\begin{align} \rho'&=\frac{p}{3}(X\rho X+Y\rho Y+Z \rho Z)+(1-p)\rho \\ &= \frac{p}{3}(\rho + X\rho X+Y\rho Y+Z \rho Z) + \left(1 - \frac{4p}{3}\right)\rho \\ & = \frac{4p}{3} \frac{I}{2} + \left(1 - \frac{4p}{3}\right)\rho \\ & = \lambda \frac{I}{2} + (1 - \lambda)\rho \end{align}

where $$p \in [0, 1]$$ is the Pauli error probability and $$\lambda = \frac{4p}{3} \in [0, \frac{4}{3}]$$ is the depolarization parameter. These two parameters are different because the maximally mixed state $$I/2$$ and $$(X\rho X + Y\rho Y + Z\rho Z)/3$$ are different states.

If you expand $$\rho$$ in terms of its Bloch vector, you get \begin{align} \rho = \frac{1}{2}(I + r_x X + r_y Y + r_z Z) \end{align} Plug this into your expression, and work out all of the terms like $$XPX, YPY, ZPZ$$ where $$P \in {I,X,Y,Z}$$, using the standard Pauli identities (such as those listed on slide 14 of this powerpoint). This should yield the desired result.