Prove a linear map $\mathcal{N}$ is completely positive if its Choi operator is positive semi-definite

Question

I'm doing exercise 4.4.1 in Quantum information theory by Wilde. The exercise asks to prove that a linear map $\mathcal{N}_{A\to B}$ is completely positive if its Choi operator is a positive semi-definite operator.

By definition, to show that $\mathcal{N}_{A\to B}$ is completely positive semi-definite, we need to show $(id_R\otimes \mathcal{N}_{A\to B})X_{RA}\geq 0$ for any positive semi-definite operator $X_{RA}$, where $R$ is the reference system. By the hint given in the exercise, the positive semi-definite operator $X_{RA}$ can be decomposed by $X_{RA} = \sum_{l}|\psi_l\rangle \langle \psi_l|$.

Again by the hint and Equation (4.202)-(4.205), we can find a linear operator $V_{A\to A}$ such that $$(I_R \otimes V_{A \to A})|\Gamma\rangle_{RA}=|\psi\rangle$$ where $|\Gamma\rangle_{RA}$ is the maximally entangled state. Then we have $$(id_R\otimes \mathcal{N}_{A\to B})X_{RA}=(id_R\otimes \mathcal{N}_{A\to B})\sum_{l}(I_R \otimes V^l_{A \to A})|\Gamma\rangle_{RA} \langle \Gamma|_{RA}(I_R \otimes V^{l\dagger}_{A \to A})$$.

It seems like if I can find an argument to exchange the positions of $(id_R\otimes \mathcal{N}_{A\to B})$ and $(I_R \otimes V^l_{A \to A})$, I have the desired result, but this does not hold in general. Is there anything wrong or missing in my proof?

Answer 1

I feel like this should have already been discussed in the site, but I can't find the post, so here we are.

Let $J(\mathcal N)\in\mathrm{Lin}(\mathbb{C}^m\otimes\mathbb{C}^n)$ be the Choice of the quantum map $\mathcal N:\mathrm{Lin}(\mathbb{C}^n)\to\mathrm{Lin}(\mathbb{C}^m)$.

1. Observe that $J(\mathcal N)=uu^\dagger$ for some vector $u\in \mathbb{C}^m\otimes\mathbb{C}^n$ iff $\mathcal N(X) = AXA^\dagger$ with $u=\operatorname{vec}(A)$.
2. Observe that $J(\mathcal N)\ge0$ iff there's a set of orthogonal vectors $u_k$ such that $J(\mathcal N)=\sum_k u_k u_k^\dagger$. Or equivalently, there's a set of orthonormal vectors $\tilde u_k$ and positive reals $\lambda_k$ such that $$J(\mathcal N)=\sum_k \lambda_k \tilde u_k\tilde u_k^\dagger.$$
3. Thus $J(\mathcal N)\ge0$ iff $\mathcal N$ admits a Kraus decomposition, i.e. there's operators $A_k$ such that $$\mathcal N(X)=\sum_k A_k X A_k^\dagger.$$
4. Observe that a map $\mathcal N$ is completely positive iff it's completely positive on pure states, meaning $(\mathcal N\otimes\operatorname{Id}_p)(\mathbb{P}_\psi)\ge0$ for all pure states $|\psi\rangle\in\mathbb{C}^n\otimes\mathbb{C}^p$ and integers $p$, using the shorthand notation $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$.
5. Observe that a map $\mathcal N$ has a Kraus decomposition iff it has a Steinspring dilation, meaning there's an isometry $V:\mathbb{C}^n\to\mathbb{C}^m\otimes \mathbb{C}^q$ such that $\mathcal N(X)=\operatorname{tr}_2[VXV^\dagger]$ for all $X$.
6. Observe that a map $\mathcal N$ is completely positive iff it has a Kraus decomposition, and thus iff $J(\mathcal N)\ge0$. This is immediate from the Stinespring dilation form, because $$(\mathcal N\otimes \operatorname{Id}_p)\mathbb{P}_\psi = \operatorname{tr}_2[\mathbb{P}((V\otimes I)|\psi\rangle)]\ge0,$$ with the RHS clearly positive semidefinite because it's the partial trace applied to the pure state $(V\otimes I)|\psi\rangle$.

You can also skip some of these steps by observing that maps of the form $X\mapsto AXA^\dagger$ are always completely positive, and that sums of completely positive maps are completely positive.

An alternative route is to observe that a map is completely positive iff it's $n$-positive, and that $J(\mathcal N)\ge0$ is the same as saying $\mathcal N$ is $n$-positive.