Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\chi_A:\mathcal{B}(\mathbb{C}^n)\rightarrow\mathcal{B}(\mathbb{C}^n)$ be a completely positive (cp) map defined as $\chi_A(x)=AxA^*$, where $A\in\mathcal{B}(\mathbb{C}^n)$. Clearly any cp map $\Psi$, on this space, is a finite sum of such maps (Krauss form). Let $\phi:\mathcal{B}(\mathbb{C}^n)\rightarrow \mathcal{B}(\mathbb{C}^n)$ is a positive map (need not be cp). Let $C_\phi=\sum_{i,j}e_{i,j}\otimes\phi(e_{i,j})$, is the Choi matrix (using Choi–Jamiołkowski isomorphism). Using these notations, I am asking a few questions.

For any arbitrary $\phi$ and an arbitrary $A$, can we make the following statement. \begin{equation} C_{\phi\circ\chi_A}=\sum_i(A_i\otimes B_i)C_\phi(A_i\otimes B_i)^*, \end{equation} for some $\lbrace A_i\rbrace$ and $\lbrace B_i\rbrace$.

If $\phi_n:M_n(\mathcal{B}(\mathbb{C}^n))\rightarrow M_n(\mathcal{B}(\mathbb{C}^n))$ be the canonical extension, then can we conclude that, $(\phi\circ\chi_A)_n=\sum_i\chi_{A_i\otimes B_i}\circ\phi_n$.

I believe that the above statements are true (at least the first one), but I could not prove it (probably by making some stupid mistakes). Advanced thanks for all helps and suggestions.

share|improve this question
    
OK. I got it. It follows from the change in basis. We just need to write the Choi matrix with respect to this new basis, which gives the corresponding transformation. For description, see "Tensor powers of 2-positive maps"- by Stormer –  RSG Sep 24 '12 at 6:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.