How to find the operator-sum representation of the depolarizing channel?

Question

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?)

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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?

Answer 1

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.

Answer 2

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.