# What is a complementary map?

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.

• Please do not drastically edit your question after you've already received and accepted an answer based on a previous version. If you need to add something mention it in "Edit 3". – Sanchayan Dutta Apr 10 at 12:13
• The question is too broad and independent of the form of c's. Moreover, it is better to use map rather than channel. – Tobias Fritzn Apr 10 at 13:05
• Ideally, that should have been taken care of before this was answered, not after. – Sanchayan Dutta Apr 10 at 13:10
• Yeah, but it was due to lack of information at that time. I am sure these edits will make it easy for people to search. – Tobias Fritzn Apr 10 at 13:12

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 $$B_0 = \begin{pmatrix} c_0 & 0\\ c_1 & 0 \end{pmatrix} \quad\text{and}\quad B_1 = \begin{pmatrix} 0 & c_0 \\ 0 & -c_1 \end{pmatrix}.$$

• Thank you so much, @John Watrous. Just a small request. How to obtain the Choi operator? – Tobias Fritzn Mar 28 at 18:13
• The answer now defines the Choi operator. – John Watrous Mar 28 at 18:54
• Hello, @John Watrous. Before Choi operator, everything was nice 2-dimensional. What does the Choi operator actually ("physically") do? – Tobias Fritzn Mar 29 at 5:38
• To be precise, if $\Psi$ maps a 2D state to a 2D state, why is it that in order to get the Kraus operators, one has to perform a "bipartite" kind of operation? – Tobias Fritzn Mar 29 at 7:02
• The Choi operator of a channel is useful in multiple ways, including the fact that it provides a mechanical way to compute Kraus operators, which is how it was used in this answer. If you would like to know more about Choi operators, let me suggest that you ask that as a separate question, and I am sure you will get some informative answers. – John Watrous Mar 29 at 12:40