2

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!

Thinkpad
  • 415
  • Not my area of expertise, but I have limited hope that there's much to say. In vectorizing the density matrices, you've necessarily lost a lot of information about them. I'm not aware of any canonical way to vectorize a matrix, so I doubt properties like CP or TP will carry over in a meaningful way. – Charles Hudgins Aug 09 '22 at 04:54
  • 2
    @Thinkpad Usually, if a quantum channel is regarded as an $N^2 \times N^2$ matrix, then that matrix is the Choi representation of that map rather than the matrix of that map relative to the vectorized matrix. In terms of the Choi representation, there are nice characterizations of CPTP. – Ben Grossmann Aug 09 '22 at 10:50
  • 1
    @Thinkpad See in particular Choi's theoerm on completely positive maps. The trace preserving condition amounts to checking the partial trace of the representation matrix $C_{\Phi}$ – Ben Grossmann Aug 09 '22 at 10:57

1 Answers1

2
  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 $e_+\equiv \sum_j e_j\otimes e_j\equiv \sum_j |j,j\rangle$.

glS
  • 7,963