Theorem (Kraus Theorem): A map $\mathcal{N}_{A\rightarrow B}$ from a finite-dimensional Hilbert space $\mathcal{H}_A$ to a finite-dimensional space $\mathcal{H}_B$ is:

• Convex linear:

$\mathcal{N}_{A\rightarrow B}(\sum_i p_i\rho_i)=\sum_ip_i\mathcal{N}_{A\rightarrow B}(\rho_i)$

• Completely positive:

$\mathcal{N}_{A\rightarrow B}(A)$ is positive for any positive (semi-) definite density operator $A$.

Furthermore, if we introduce an extra system $R$ of arbitrary dimensionality, it must be true that $(\mathcal{I}\otimes \mathcal{N}_{A\rightarrow B})(A)$ is positive (semi-) definite for any positive (semi-) definite operator $A$ on the combined system $RA$, where $\mathcal{I}$ is the identity map on $\mathcal{H}_R$.

Remark: for

$\forall A = \begin{pmatrix} a_{11}&...&a_{1k}\\ \vdots&\ddots&\vdots\\ a_{k1}&...& a_{kk} \end{pmatrix}\in\mathcal{H}_{RA}, a_{ij}\in\mathcal{H}_A\\ (\mathcal{I}\otimes\mathcal{N}_{A\rightarrow B})(A)=\begin{pmatrix} \mathcal{N}_{A\rightarrow B}(a_{11})&...&\mathcal{N}_{A\rightarrow B}(a_{1k})\\ \vdots &\ddots &\vdots\\ \mathcal{N}_{A\rightarrow B}(a_{k1})&...&\mathcal{N}_{A\rightarrow B}(a_{kk}) \end{pmatrix}\in\mathcal{H}_{RB}$

• Trace preserving:

$0\leq tr(\mathcal{N}_{A\rightarrow B}(\rho))\leq 1$

IF AND ONLY IF the map has a Kraus decomposition as follows:

$\mathcal{N}_{A\rightarrow B}(\rho)=\sum_{l=0}^{d-1}V_l\rho V_l^\dagger$

satisfying $\sum_{l=0}^{d-1}V_l^\dagger V_l=I$ and $d\leq dim(\mathcal{H}_A)dim(\mathcal{H}_B)$.

Proof:

• If the map has the form:

$\mathcal{N}_{A\rightarrow B}(\rho)=\sum_{l=0}^{d-1}V_l\rho V_l^\dagger$

• Convex linear:

\begin{aligned} \mathcal{N}_{A\rightarrow B}(\sum_ip_i\rho_i)&=\sum_{l=0}^{d-1}V_l\sum_{i}p_i\rho_i V_l^\dagger\\ &=\sum_ip_i\sum_{l=0}^{d-1}V_l\rho_i V_l^\dagger\\ &=\sum_ip_i\mathcal{N}_{A\rightarrow B}(\rho_i) \end{aligned}

• Complete positive:

For any valid density operator $\rho\in\mathcal{H}_A$, it’s positive (semi-) definite. Then:

\begin{aligned} \forall |\psi\rangle\in\mathcal{H}_{RB},\\ \langle\psi|(\mathcal{I}\otimes \mathcal{N}_{A\rightarrow B})(\rho)|\psi\rangle &=\langle\psi|\sum_{l=0}^{d-1}(I\otimes V_l)\rho(I\otimes V_l^\dagger)|\psi\rangle\\ &=\langle\psi|\sum_{l=0}^{d-1}(I\otimes V_l)\rho(I\otimes V_l)^\dagger|\psi\rangle\\ &=\sum_{l=0}^{d-1}[\langle\psi|(I\otimes V_l)]\cdot\rho\cdot[(I\otimes V_l)^\dagger|\psi\rangle]\\ &=\sum_{l=0}^{d-1}\geq 0\\ &\geq 0 \end{aligned}

It’s because $\rho$ is positive (semi-) definite, so for $|\varphi\rangle=(I\otimes V_l)^\dagger|\psi\rangle$, $\langle\varphi|\rho|\varphi\rangle\geq 0$.

• Trace preserving:

\begin{aligned} tr(\mathcal{N}_{A\rightarrow B}(\rho))&=tr(\sum_{l=0}^{d-1}V_l\rho V_l^\dagger)\\ &=tr(\rho\sum_{l=0}^{d-1}V_l^\dagger V_l)\\ &=tr(\rho) \end{aligned}

where we use $tr(AB)=tr(BA)$.

• On the other hand, if the map is linear, completely positive, trace preserving, we prove that the map has Kraus decomposition.

Let $d_A\equiv dim(\mathcal{H}_A),d_B\equiv dim(\mathcal{H}_B)$ and $R$ is an auxiliary system with the same orthonormal basis with $A$. Let $|\Gamma\rangle_{RA}$ denote the following unnormalized maximally entangled vector:

$|\Gamma\rangle_{RA}\equiv\sum_{i=0}^{d_A-1}|i\rangle_R\otimes |i\rangle_A$

then we have

$(\mathcal{I}\otimes\mathcal{N}_{A\rightarrow B})(|\Gamma\rangle\langle\Gamma|_{RA})=\sum_{i,j=0}^{d_A-1}|i\rangle\langle j|_R\otimes\mathcal{N}_{A\rightarrow B}(|i\rangle\langle j|_A)$

Since $|\Gamma\rangle\langle \Gamma|_{RA}$ is positive semi-definite, since $\mathcal{N}_{A\rightarrow B}$ is completely positive, the RHS is positive semi-definite, too. So it can be diagonalized:

$\sum_{i,j=0}^{d_A-1}|i\rangle\langle j|_R\otimes\mathcal{N}_{A\rightarrow B}(|i\rangle\langle j|_A)=\sum_{l=0}^{d-1}|\phi_l\rangle\langle\phi_l|_{RB}$

where $d\leq d_Ad_B$, $d_Ad_B$ is the dimension of system $RB$, and $|\phi_l\rangle$ does not necessarily have to be orthonormal.

we expand the $|\phi_l\rangle$ in terms of an orthonormal basis of $RB$:

$|\phi\rangle_{RB}=\sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1}\alpha_{ij}|i\rangle_R\otimes|j\rangle_B$

let $\mathcal{V}_{A\rightarrow B}$ denote the following linear operator:

$\mathcal{V}_{A\rightarrow B}\equiv\sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1}\alpha_{i,j}|j\rangle_B\langle i|_A$

Then we see:

\begin{aligned} (\mathcal{I}\otimes\mathcal{V}_{A\rightarrow B})|\Gamma\rangle_{RA}&=(I\otimes\sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1}\alpha_{i,j}|j\rangle_B\langle i|_A)(\sum_{k=0}^{d_A-1}|k\rangle_R\otimes|k\rangle_A)\\ &=\sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1}\sum_{k=0}^{d_A-1}\alpha_{i,j}|k\rangle_R\otimes|j\rangle_B\langle i|k\rangle_A\\ &=\sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1}\alpha_{i,j}|i\rangle_R\otimes|j\rangle_B\\ &=|\phi\rangle_{RB} \end{aligned}

So we see that for all vector $|\phi\rangle_{RB}$, we can find a linear operator $\mathcal{V}_{A\rightarrow B}$ such that $(\mathcal{I}\otimes\mathcal{V}_{A\rightarrow B})|\Gamma\rangle_{RA}=|\phi\rangle_{RB}$.

And we have:

\begin{aligned} (\langle i|_R\otimes I)|\phi\rangle_{RB}&=(\langle i|_R\otimes I)\cdot(\mathcal{I}\otimes\mathcal{V}_{A\rightarrow B})|\Gamma\rangle_{RA}\\ &=(\langle i|_R\otimes I)(\mathcal{I}\otimes\mathcal{V}_{A\rightarrow B})\sum_{j=0}^{d_A-1}|j\rangle_R\otimes|j\rangle_A\\ &=\sum_{j=0}^{d_A-1}\langle i|j\rangle_R\otimes \mathcal{V}_{A\rightarrow B}|j\rangle_A\\ &=\mathcal{V}_{A\rightarrow B}|i\rangle_A \end{aligned}

Finally we have:

\begin{aligned} \mathcal{N}_{A\rightarrow B}(|i\rangle\langle j|_A)&=(\langle i|_R\otimes I)\cdot \sum_{p,q=0}^{d_A-1}|p\rangle\langle q|_R\otimes\mathcal{N}_{A\rightarrow B}(|p\rangle\langle q|_A)\cdot(|j\rangle_R\otimes I)\\ &=(\langle i|_R\otimes I)\cdot\sum_{l=0}^{d-1}|\phi_l\rangle\langle\phi_l|_{RB}\cdot(|j\rangle_R\otimes I)\\ &=\sum_{l=0}^{d-1}(\langle i|_R\otimes I)|\phi_l\rangle\cdot\langle\phi_l|(|j\rangle_R\otimes I)\\ &=\sum_{l=0}^{d-1}(\mathcal{V}_l)_{A\rightarrow B}|i\rangle\langle j|_A(\mathcal{V}_l)_{A\rightarrow B}^\dagger \end{aligned}

We still need to prove that $\sum_{l}\mathcal{V}_l^\dagger\mathcal{V}_l=I$. Since $\mathcal{N}_{A\rightarrow B}$ is trace preserving, then:

$tr(\mathcal{N}_{A\rightarrow B}(|i\rangle\langle j|_A))=tr(|i\rangle\langle j|_A)=\delta_{ij}$

Consider:

\begin{aligned} \delta_{ij}&=tr(\mathcal{N}_{A\rightarrow B}(|i\rangle\langle j|_A))\\ &=tr(\sum_{l}\mathcal{V}_l|i\rangle\langle j|_A\mathcal{V}_l^\dagger)\\ &=tr(\sum_l\mathcal{V}_l^\dagger\mathcal{V}_l|i\rangle\langle j|_A)\\ &=\langle j|_A\sum_l\mathcal{V}_l^\dagger\mathcal{V}_l|i\rangle_A\\ \end{aligned}

That is, $\sum_l\mathcal{V}_l^\dagger\mathcal{V}_l=I$.

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