What is the "complementary map" of a channel with given Kraus decomposition?

Question

I have a quantum map described by the following Kraus operators

$$A_0 = c_0 \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}, \qquad A_1 = c_1 \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix},$$

such that $c_0^2 + c_1^2 = 1$. I want to know what is a complementary map and how to construct the same for the above-mentioned channel?

Edit 1: Checked for some literature. Here is the definition of the complementary map equations 37 and 38.

Answer 1

Let's start by finding a complementary channel for any channel given by a Kraus representation $$ \Phi(X) = \sum_{k=1}^n A_k X A_k^{\dagger}. $$ To make the necessary equations clear, let us assume that the channel has the form $\Phi:\mathrm{L}(\mathcal{X})\rightarrow \mathrm{L}(\mathcal{Y})$ for finite-dimensional Hilbert spaces $\mathcal{X}$ and $\mathcal{Y}$. Let us also define $\mathcal{Z} = \mathbb{C}^n$; the complementary channel we will define will take the form $\Psi:\mathrm{L}(\mathcal{X})\rightarrow \mathrm{L}(\mathcal{Z})$. (For the channel in the question itself, we will have $\mathcal{X}$, $\mathcal{Y}$, and $\mathcal{Z}$ all equal to $\mathbb{C}^2$, but it helps nevertheless to assign different names to these spaces.)

Define an operator $$ A = \sum_{k=1}^n A_k \otimes | k\rangle, $$ which is a linear operator mapping $\mathcal{X}$ to $\mathcal{Y}\otimes\mathcal{Z}$. This gives us a Stinespring representation $$ \Phi(X) = \operatorname{Tr}_{\mathcal{Z}} \bigl( A X A^{\dagger}\bigr). $$ The channel $$ \Psi(X) = \operatorname{Tr}_{\mathcal{Y}} \bigl( A X A^{\dagger}\bigr) $$ is therefore complementary to $\Phi$. We can simplify this expression by observing that $$ A X A^{\dagger} = \sum_{j=1}^n \sum_{k=1}^n A_j X A_k^{\dagger} \otimes | j \rangle \langle k |, $$ so that $$ \Psi(X) = \sum_{j=1}^n \sum_{k=1}^n \operatorname{Tr}\bigl(A_j X A_k^{\dagger}\bigr) | j \rangle \langle k |. $$ There's not too much more we can do with this, except perhaps to use the cyclic property of the trace to obtain the expression $$ \Psi(X) = \sum_{j=1}^n \sum_{k=1}^n \operatorname{Tr}\bigl(A_k^{\dagger} A_j X\bigr) | j \rangle \langle k |. $$

Now let's plug in the specific operators from the question to obtain $$ \Psi(X) = c_0^2 \operatorname{Tr}(X) | 0 \rangle \langle 0 | + c_1^2 \operatorname{Tr}(X) | 1 \rangle \langle 1 | + c_0 c_1 \operatorname{Tr}(\sigma_z X) | 0 \rangle \langle 1 | + c_0 c_1 \operatorname{Tr}(\sigma_z X) | 1 \rangle \langle 0 |. $$ Here $\sigma_z$ denotes the Pauli-Z operator, which we get because $A_0^{\dagger} A_1 = A_1^{\dagger}A_0 = c_0 c_1 \sigma_z$. (I am assuming $c_0$ and $c_1$ are real numbers.) The expression may look a bit nicer in matrix form: $$ \Psi\begin{pmatrix} \alpha & \beta\\ \gamma & \delta\end{pmatrix} = \begin{pmatrix} c_0^2(\alpha + \delta) & c_0 c_1 (\alpha - \delta)\\ c_0 c_1 (\alpha - \delta) & c_1^2 (\alpha + \delta) \end{pmatrix}. $$

Finally, the question asks for Kraus operators of $\Psi$, which we can get by computing the Choi operator of $\Psi$. In general, this is the operator $$ J(\Psi) = \sum_{j=1}^n\sum_{k=1}^n \Psi(|j\rangle\langle k|) \otimes |j\rangle\langle k|, $$ and in this particular case we obtain $$ J(\Psi) = \begin{pmatrix} c_0^2 & 0 & c_0 c_1 & 0\\ 0 & c_0^2 & 0 & -c_0 c_1 \\ c_0 c_1 & 0 & c_1^2 & 0\\ 0 & -c_0 c_1 & 0 & c_1^2 \end{pmatrix}. $$ This operator has rank 2, which means just 2 Kraus operators suffice. We can get them through a spectral decomposition of $J(\Psi)$. Specifically, we have $$ J(\Psi) = \begin{pmatrix} c_0\\ 0\\ c_1\\ 0 \end{pmatrix} \begin{pmatrix} c_0 & 0 & c_1 & 0 \end{pmatrix} + \begin{pmatrix} 0\\ c_0\\ 0\\ -c_1 \end{pmatrix} \begin{pmatrix} 0 & c_0 & 0 & -c_1 \end{pmatrix}, $$ and by "folding up" these vectors we get Kraus operators: $$ \Psi(X) = B_0 X B_0^{\dagger} + B_1 X B_1^{\dagger} $$ where $$ B_0 = \begin{pmatrix} c_0 & 0\\ c_1 & 0 \end{pmatrix} \quad\text{and}\quad B_1 = \begin{pmatrix} 0 & c_0 \\ 0 & -c_1 \end{pmatrix}. $$

Answer 2

I'll discuss a slightly different approach to obtain the same result given in the other answer.

You are given a completely positive quantum map in Kraus representation: some $\Phi:\mathrm L(\mathcal X)\to\mathrm L(\mathcal Y)$ with $$\Phi(X) = \sum_a A_a X A_a^\dagger, \qquad \forall X\in\mathrm L(\mathcal X).$$ Here $A_a\in\mathrm L(\mathcal X,\mathcal Y)$ are the Kraus operators of the map. Observe that you can always write a channel in its Stinespring representation as $\Phi(X)=\operatorname{Tr}_{\mathcal Z}(VXV^\dagger)$ for some isometry $V\in\mathrm U(\mathcal X,\mathcal Y\otimes\mathcal Z)$. Kraus and Stinespring dilations are tightly related via $A_a=(I\otimes\langle a|)V$ with $|a\rangle$ an orthonormal system for $\mathcal Z$ (more precisely, I should have said that given a channel corresponding to the isometry $V$, the set of operators $A_a$ thus defined gives a valid Kraus decomposition for $\Phi$, and vice versa given a Kraus decomposition the operator thus defined is an isometry). Note that different choices of orthonormal system $\{|a\rangle\}_a$ result in different Kraus decomposition, but same overall map.

In summary, we have $A_a=(I\otimes\langle a|)V$ for some isometry $V$. The complementary map of $\Phi$ is then the one whose Kraus operators are obtained "projecting on the other side" of $V$. Meaning $\tilde A_a\equiv(\langle a|\otimes I)V$. I would note that there might be some subtleties with this associated to the different dimensionality of $\mathcal Y$ and $\mathcal Z$, which should however disappear redefining things so that $\mathcal Y\simeq\mathcal X$ and choosing $\mathcal Z$ to have smallest required dimension (i.e. $\dim(\mathcal Z)=\operatorname{rank}(J(\Phi))$).

The intuition is pretty straightforward: if you understood $\Phi$ as evolution through some $V$ followed by ignoring part of the output, the complementary channel is the one corresponding to the same evolution through $V$, but instead ignoring the other (complementary!) part of the output.