# Complementary channel of binary sum channel

This isn't strictly a quantum question but the idea of complementary channels is the following: Take any channel $$N_{A\rightarrow B}$$. Take it's Stinespring dilation (which is an isometry) $$V_{A\rightarrow BE}$$. Now trace out the $$B$$ system to obtain $$N^c_{A\rightarrow E}$$, which the the complementary channel to $$N_{A\rightarrow B}$$.

This should also apply to all classical channels. So for the case of $$N_{XY\rightarrow Z}$$, where $$Z= X\oplus Y$$, how does one define the Stinespring and complementary channels? Note that this a multiple-access channel and it's purely classical. Also note that one classical way to make $$X\oplus Y$$ reversible is to make a copy of $$X$$. But how should one formally do this in the quantum framework?

Here's one idea: For alphabets $$\mathcal{X},\mathcal{Y},\mathcal{Z}=\{0,1\}$$ and random variables $$X, Y,Z$$ taking values in these alphabets respectively, we can take advantage of the relationship $$X \oplus Y = \text{CNOT}_{\mathcal{X}\rightarrow \mathcal{Y}}(X, Y) = \text{CNOT}_{\mathcal{Y}\rightarrow \mathcal{X}}(X, Y), \tag{1}$$ where the subscript $$\mathcal{A}\rightarrow \mathcal{B}$$ denotes the variable in $$\mathcal{A}$$ as control bit acting on the variable in $$\mathcal{B}$$ as target bit, and in each case the output is read from the target register (I'm using the same notation for alphabets and "registers" containing variables). So, define a pmf $$p_X: \mathcal{X}\rightarrow \mathbb{R}$$ where $$p_X(i) = \text{Pr}(X=i)$$ and define $$p_Y, p_Z$$ similarly. Then states describing the variables $$X$$ and $$Y$$ are \begin{align} \rho := \text{diag}(p_X(0), p_X(1)), \tag{2} \\ \sigma := \text{diag}(p_Y(0), p_Y(1)).\tag{3} \end{align} We will overload the classical notation for a CNOT to define a unitary $$\text{CNOT}_{\mathcal{X}\rightarrow \mathcal{Y}}: \mathcal{X}\otimes \mathcal{Y} \rightarrow \mathcal{X'} \otimes \mathcal{Z}$$ acting in the obvious way on states in $$\mathcal{X}\otimes \mathcal{Y}$$. Then one way to define the binary sum channel in a way that reproduces the desired classical behavior is with $$\Phi: \mathcal{X}\otimes \mathcal{Y} \rightarrow \mathcal{Z}$$ given by $$\begin{equation} \Phi(\rho\otimes \sigma) = \text{Tr}_\mathcal{X'} \left(\text{CNOT}_{\mathcal{X}\rightarrow \mathcal{Y}} (\rho \otimes \sigma)\right). \tag{4} \end{equation}$$ Since $$\text{CNOT}$$ is unitary we already have our channel in Stinespring form (i.e. any unitary $$U$$ acting on a state $$\rho$$ may be represented in Stinespring form as $$\Psi(\rho) = \text{Tr}_Z (U\rho U^\dagger)$$ for $$Z=\mathbb{C}$$), then the complementary channel becomes $$\begin{equation} \Phi^c(\rho\otimes \sigma) = \text{Tr}_\mathcal{Z} \left(\text{CNOT}_{\mathcal{X}\rightarrow \mathcal{Y}} (\rho \otimes \sigma)\right). \tag{5} \end{equation}$$ In the special case where we're interested in classical input states like Eqs. (2)-(3), we get the expected outcomes \begin{align} \Phi(\rho\otimes \sigma) &\rightarrow \text{Tr}_\mathcal{X'} \tag{6a-d}\left(\text{CNOT}_{\mathcal{X}\rightarrow \mathcal{Y}} \left[\text{diag}(p_X(0)p_Y(0),p_X(0)p_Y(1), p_X(1)p_Y(0), p_X(1)p_Y(1))\right]\right) \\ &= \text{Tr}_\mathcal{X'} \left( \text{diag}(p_X(0)p_Y(0),p_X(0)p_Y(1), p_X(1)p_Y(1), p_X(1)p_Y(0)) \right) \\&=\text{diag}\left(p_X(0)p_Y(0) + p_X(1)p_Y(1),p_X(0)p_Y(1) + p_X(1)p_Y(0)\right) \\&\sim X\oplus Y \\ \Phi^c(\rho\otimes \sigma) &\rightarrow \text{Tr}_\mathcal{Z} \left( \text{diag}(p_X(0)p_Y(0),p_X(0)p_Y(1), p_X(1)p_Y(1), p_X(1)p_Y(0)) \right)\tag{7a-d} \\&= \text{diag}(p_X(0)p_Y(0) + p_X(0)p_Y(1), p_X(1)p_Y(1) + p_X(1)p_Y(0)) \\&=\text{diag}(p_X(0) , p_X(1)) \\&\sim X, \end{align} where $$\sim$$ here just means "describes the distribution of these classical random variables." Eq. (6d) follows because the distribution of $$Z$$ is induced by the distributions of $$X,Y$$ according to \begin{align} \text{Pr}(Z = 0) &= \text{Pr}(X\oplus Y=0) \\&=\text{Pr}(X=0 \text{ and } Y=0) + \text{Pr}(X=1 \text{ and }Y=1) \\&= p_X(0) p_Y(0) + p_X(1) p_Y(1), \end{align} which we can read off of the first entry of the state describing $$Z$$ in Eq. (6c).
• Defining $$\Phi$$ according to its behavior on classical states leaves a significant degree of freedom for ways to define $$\Phi$$, so this choice of $$\Phi^c$$ is far from unique.