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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?

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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).


A couple of limitations to this analysis:

  • From Eq. (1) we could equally have proceeded with the CNOT acting in the opposite direction, so this classical definition of the channel is not unique.
  • 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.
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