3
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.