Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

migrated from Apr 2 '14 at 13:14

This question came from our site for active researchers, academics and students of physics.

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 '14 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 '14 at 12:57
@DanStahlke I would appreciate if you could share the counter-example (For complete positivity). – Kishor Bharti Apr 2 '14 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 '14 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 '14 at 15:33

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 $$\lambda_1+\lambda_2+\lambda_3-\{\lambda_1\,\lambda_2,\lambda_3\}=\{\lambda_2+\lambda_3, \lambda_1+\lambda_3,\lambda_1+\lambda_3\}\geq0.$$

share|cite|improve this answer
Thanks for the answer !! However I am sorry for putting typo in my question which has been corrected. – Kishor Bharti Apr 1 '14 at 18:35

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.