# Completely positive quantum channel

Let $$\mathcal{E} \to CP(A \to B)$$ (completely positive linear map) be a trace non-increasing $$CP$$ map. Show that any operator sum representation $$\{M_x\}_{x=1}^{m}$$ of $$\mathcal{E}$$ satisfes $$\sum_{x=1}^{m}M^{*}_xM_x \leq I^A$$. Show that the marginal of the Choi matrix $$J^{AB}{\mathcal{E}}$$ satisfies $$J^A{\mathcal{E}}:= Tr_B[J^{AB}_{\mathcal{E}}] \leq I^A$$.

The proof the second part is as follows, but I dont know how to prove first part. Let$$\mathcal{E}:\mathcal{B}(\mathcal{H}^A)\to \mathcal{B}(\mathcal{H}^B)$$ be a linear map. Clearly if $$\mathcal{E}$$ is completely positive then by definition $$J^{AB}{\mathcal{E}}:=\text{id} \otimes \mathcal{E}(\phi^{AA^{\prime}}_{+})\geq0$$.

Suppose now that $$J^{AB}_{\mathcal{E}} \geq 0$$. Let $$k \in \mathbb{N}$$, and $$|\psi^{RA}\rangle \in \mathbb{C}^k \otimes \mathbb{C}^d$$, where R is a k-dimensional (reference) system. $$|\psi^{RA}\rangle = M_{\psi} \otimes I^A |\phi^{A A^{\prime}}_{+}\rangle$$. Where $$M_{\psi}:\mathcal{H}^{A^{\prime}} \to \mathcal{H}^R$$is a linear operator. We therefore have $$\text{id}_k \otimes \mathcal{E} |\psi^{RA}\rangle\langle\psi^{RA}| = (\text{id}_k \otimes \mathcal{E}) ( M_{\psi} \otimes I^A ) |\phi^{A A^{\prime}}_{+}\rangle \langle\phi^{A A^{\prime}}_{+}| ( M_{\psi}^{*} \otimes I^A )$$ $$= ( M_{\psi} \otimes I^A ) J^{AB}_{\mathcal{E}} ( M_{\psi}^{*} \otimes I^A )$$ Finally any operator $$\rho^{RA}\geq 0$$ can be diagonalized as $$\rho^{RA}= \sum_{x=1}^m |\psi_x^{RA}\rangle \langle\psi_x^{RA}|$$, where $$|\psi_x^{RA}\rangle \in \mathbb{C}^k \otimes \mathbb{C}^d$$ are some (possibly unnormalized) pure states. Since $$|\psi^{RA}\rangle$$ above was arbitrary, we conclude that $$(\text{id}_k \otimes \mathcal{E}) \rho^{RA} = \sum_{x=1}^m (\text{id}_k \otimes \mathcal{E}) (|\phi_x^{RA}\rangle \langle\phi_x^{RA}|) \geq 0$$

We are given $$\mathcal{E}_{A\rightarrow B}$$ to be some completely positive trace nonincreasing map with Kraus operators $$\{M_i\}$$ so that $$\mathcal{E}(\rho) = \sum_i M_i\rho M_i^\dagger$$.
For any $$\rho$$, it holds that \begin{align} \langle I_A, \rho\rangle &= \text{tr}(\rho)\\ &\geq \text{tr}(\mathcal{E}(\rho)) \\ &= \langle I_B, \mathcal{E}(\rho)\rangle \\ &= \langle I_B, \sum_i M_i \rho M_i^\dagger\rangle \\ &= \langle \sum_i M_i^\dagger M_i, \rho\rangle \end{align}
Thus, $$\langle I_A - \sum_i M_i^\dagger M_i, \rho\rangle \geq 0$$ for any $$\rho$$ and hence, $$\sum_i M^\dagger_i M_i \leq I_A$$.
For the second question, let us choose an orthonormal basis $$\{\vert i\rangle\}$$ for $$\mathcal{H}_A$$. Note that $$J(\mathcal{E}) = \sum_{i,j} \vert i\rangle\langle j\vert_A\otimes\mathcal{E}(\vert i\rangle\langle j\vert)_B$$. Since $$\mathcal{E}$$ is trace nonincreasing, it holds that $$\text{tr}(\mathcal{E}(\vert i\rangle\langle j\vert))\leq \text{tr}(\vert i\rangle\langle j\vert) = \delta_{ij}$$ and hence
$$\text{tr}_B\left(J(\mathcal{E})\right) = \sum_{i,j} \vert i\rangle\langle j\vert_A\otimes\text{tr}\left(\mathcal{E}(\vert i\rangle\langle j\vert)\right)\leq \sum_{i,j}\vert i\rangle\langle j\vert \delta_{ij} = I_A$$