What does CPTP mean to the $N^2 \times N^2$ matrix form of quantum channels?

Question

This question might be basic knowledge, but I found nothing on the internet. So any reference would also be highly appreciated!

I have the quantum channel regarded as an $N^2 \times N^2$ matrix (when the density matrices are regarded as $N^2$-length vectors). The question is, how are the properties of a quantum channel related to the corresponding matrix representation? Specifically, what does CP (completely positive) or TP (trace-preserving) mean to the $N^2 \times N^2$ matrix form of an arbitrary quantum channel?

Thank you!

Answer 1

1. Given a map $\Phi\in \operatorname{Lin}(\operatorname{Lin}(\mathbb{C}^N))$, $\Phi:\operatorname{Lin}(\mathbb{C}^N)\to\operatorname{Lin}(\mathbb{C}^N)$, it seems like you are asking about the properties of its natural representation (see e.g. Watrous' book for more on this terminology). This is the linear map $K(\Phi)\in\operatorname{Lin}(\mathbb{C}^N\otimes\mathbb{C}^N)$ such that $$K(\Phi)\operatorname{vec}(\rho)=\operatorname{vec}(\Phi(\rho)),$$ for any linear operator $\rho\in\operatorname{Lin}(\mathbb{C}^N)$, and with $\operatorname{vec}:\operatorname{Lin}(\mathbb{C}^N,\mathbb{C}^M)\to \mathbb{C}^M\otimes\mathbb{C}^N$ denoting the vectorisation operation: $\operatorname{vec}(A)_{ij}\equiv A_{ij}$.
2. We know that a map $\Phi$ is CPTP iff its Choi operator $J(\Phi)$ is positive semidefinite and satisfies $\operatorname{Tr}_1(J(\Phi))=I$. The Choi is defined here as $$J(\Phi) \equiv \sum_{ij} \Phi(e_i e_j^\dagger)\otimes e_i e_j^\dagger.$$ Note that this is also a linear map $\mathbb{C}^N\otimes\mathbb{C}^N\to\mathbb{C}^N\otimes\mathbb{C}^N$ in this case (though it has different dimensions than $K(\Phi)$ when $\Phi$ has input and output of different dimensions).
3. The relation between $K(\Phi)$ and $J(\Phi)$ can be written as $$J(\Phi)_{ij,k\ell} \equiv \langle i,j|J(\Phi)|k,\ell\rangle = \langle i,k|K(\Phi)|j,\ell\rangle \equiv K(\Phi)_{ik,j\ell}.\tag1$$

In light of the above, what you can say is that given your $N^2\times N^2$ matrix, which amounts to $K(\Phi)$, the map $\Phi$ being CPTP corresponds to $K(\Phi)$ being positive semidefinite, after transposing some of the indices as per (1) above. I'm not aware of a more direct way to express such condition directly on $K(\Phi)$. See also this answer.

On the other hand, the trace-preserving conditions is manageable: $\Phi$ being trace-preserving amounts to $e_+^\dagger K(\Phi)=e_+$, where $e_+\equiv \sum_j e_j\otimes e_j\equiv \sum_j |j,j\rangle$.