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Simon map in a specific basis is defined as $$ \left[ {\begin{array}{ccc} A & B & C \\ D & E & F \\ G & H & I \\ \end{array} } \right] \rightarrow \left[ {\begin{array}{ccc} A +E & -B & -C \\ -D & E+I& -F \\ -G & -H & I+A \\ \end{array} } \right] $$ This looks similar to the reduction map ${(\rho \rightarrow tr(\rho)I -\rho )}$ with a minor difference which can be easily observed. I believe that the Simon map can be broken into a reduction map composed with some other map. However , despite many attempts I am unable to get a good decomposition . I would like someone to help me with positivity of the Simon map.

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This might be positive (I don't know how to check), but the computer tells me that it is not completely positive. This map acting on one share of a maximally entangled state yields something which is not positive. So there will not be Kraus operators. In fact, the reduction map doesn't seem to be completely positive either. –  Dan Stahlke Apr 2 at 12:56
    
And this Simon map seems to be the reduction map composed with a map that permutes the diagonal entries. But this latter map is not even positive (much less completely positive). –  Dan Stahlke Apr 2 at 12:57
    
@DanStahlke I would appreciate if you could share the counter-example (For complete positivity). –  Kishor Bharti Apr 2 at 13:25
    
The counterexample is the maximally entangled state, $\left|\psi\right> = \sum_{i=1}^3 \left|i\right> \otimes \left|i\right>$. Either the Simon map or the reduction map, tensored with the identity map, yields a non-positive state when acting on $\left|\psi\right>$. Note that by Choi's theorem, $(\Phi \ot I)(\left|\psi\right>\left<\psi\right|) \succeq 0$, where $\left|\psi\right>$ is the maximally entangled state, is a necessary and sufficient condition for complete positivity of map $\Phi$. –  Dan Stahlke Apr 2 at 15:24
    
Specifically, if $\Phi$ is the reduction map and $\left|\psi\right> = \sum_{i=1}^3 \left|i\right> \otimes \left|i\right>$ then $(\Phi \otimes I)(\left|\psi\right>\left<\psi\right|) = I \otimes I - \left|\psi\right>\left<\psi\right|$ which has a negative eigenvalue for the eigenvector $\left|\psi\right>\left<\psi\right|$. –  Dan Stahlke Apr 2 at 15:33

1 Answer 1

I may be misinterpreting the question (I have no background in quantum computing), but assuming you are using the definition of complete positivity in which $\Phi\geq0$ means $\Phi(\rho)\geq0$ for all $\rho\geq0$, then the statement that the map $$\Phi\left(\left[ {\begin{array}{ccc} A & B & C \\ D & E & F \\ G & H & I \\ \end{array} } \right]\right) =\left[ {\begin{array}{ccc} A +B & -B & -C \\ -D & E+F& -F \\ -G & -H & I+A \\ \end{array} } \right]$$ is completely positive is false.

A counterexample is the Hermitian matrix $$\rho=\left( \begin{array}{ccc} 0.924104\, +0. i & 0.577485\, +0.527832 i & -0.669071-0.336161 i \\ 0.577485\, -0.527832 i & 2.58896\, +0. i & -0.292335-0.540232 i \\ -0.669071+0.336161 i & -0.292335+0.540232 i & 2.98363\, +0. i \\ \end{array} \right)$$ which has $\text{Eigenvalues}\left(\Phi(\rho)\right)$ of {4.29122 - 0.0114877 I, 2.48666 - 0.255346 I, 0.928064 + 0.254433 I}, but $\text{Eigenvalues}\left(\rho\right)$ of {3.71306, 2.3659, 0.417722}.

Meanwhile, the reduction map $\Phi(\rho)=\text{tr}(\rho)I-\rho$ is trivially a positive map, due to the fact that if $\rho\geq0$, then the eigenvalues of $\Phi(\rho)$ are $$\lambda_1+\lambda_2+\lambda_3-\{\lambda_1\,\lambda_2,\lambda_3\}=\{\lambda_2+\lambda_3, \lambda_1+\lambda_3,\lambda_1+\lambda_3\}\geq0.$$

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Thanks for the answer !! However I am sorry for putting typo in my question which has been corrected. –  Kishor Bharti Apr 1 at 18:35

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