# Is the process matrix of a quantum channel always positive?

Consider the following Kraus decomposition of a quantum channel: $$\mathcal E(\rho) = \sum_i A_i \rho A_i^\dagger.$$ By choosing a basis for space of operators like $$\{E_i\}$$ where these bases are orthonormal by trace norm, one can find such representation for this channel: $$\mathcal E(\rho) = \sum_{m,n} \chi_{mn}E_m \rho E_n^\dagger.$$ I'm trying to prove whether the matrix $$\chi$$ defined this way is positive. I know that it's Hermitian, but I cannot definitely conclude that it's positive.

Can anyone help me?

• Please use latex for formulas! Also, "orthonormal by trace norm" makes no sense, the trace norm does not induce a scalar product. Do you mean the Hilbert-Schmidt scalar product? Commented Jan 9, 2021 at 9:55
• Hint: Expand each $A_i$ in the basis $E_n$, and then derive an explicit expression for $\chi$. What do you get? Commented Jan 9, 2021 at 9:57

## 1 Answer

Decomposing the Kraus operators in terms of the basis $$\{E_i\}$$ we write $$A_i = \sum_j c_{ij} E_j$$. Then $$\mathcal E(\rho) = \sum_{jk}\Big(\underbrace{\sum_i c_{ij}\bar c_{ik}}_{\equiv \chi_{jk}}\Big) E_j \rho E_k^\dagger.$$

It follows that $$\chi$$, as a matrix, has the form $$\chi = CC^\dagger$$ with $$C_{ij}\equiv c_{ji}$$. For any $$C$$, the operator $$CC^\dagger$$ is positive semidefinite, hence the conclusion.

Equivalently, you may notice that the Choi representation of $$\mathcal E$$, which is defined as $$J(\mathcal E)=(\mathcal E\otimes I)(d|+\rangle\!\langle +|)$$ with $$d$$ the dimension of the underlying space and $$|+\rangle=\frac{1}{\sqrt d}\sum_i |i,i\rangle$$ the maximally entangled state, can be written in terms of the Kraus operators as $$J(\mathcal E) = \sum_i \mathrm{vec}(A_i)\mathrm{vec}(A_i)^\dagger,$$ where $$\mathrm{vec}(A_i)$$ denotes the vectorisation of $$A_i$$. Decomposing $$A_i$$ as before, we get $$J(\mathcal E) = \sum_{jk} \Big(\sum_i c_{ij} \bar c_{ik}\Big) \mathrm{vec}(E_j)\mathrm{vec}(E_k)^\dagger.$$ We can write this equivalently as $$\chi_{jk}=\langle\!\langle E_j|J(\mathcal E)|E_k\rangle\!\rangle$$, where I'm using the notation $$|E_j\rangle\!\rangle\equiv\mathrm{vec}(E_j)$$ and $$\langle\!\langle E_j|\equiv\mathrm{vec}(E_j)^\dagger$$. In other words, $$\chi$$ is the Choi represented in the basis of (vectorisations of) the operators $$\{E_j\}_j$$. The Choi is positive semidefinite iff $$\mathcal E$$ is a channel, and thus so must be $$\chi$$.