IsCP: Difference between revisions

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|desc=Determines whether or not a [[superoperator]] is [[completely positive]]
|desc=Determines whether or not a [[superoperator]] is [[completely positive]]
|rel=[[IsHermPreserving]]
|rel=[[IsHermPreserving]]
|cat=[[List of functions#Superoperators|Superoperators]]
|upd=January 4, 2013
|upd=January 4, 2013
|v=1.01}}
|v=0.50}}
<tt>'''IsCP'''</tt> is a [[List of functions|function]] that determines whether or not a given [[superoperator]] is [[completely positive]].
<tt>'''IsCP'''</tt> is a [[List of functions|function]] that determines whether or not a given [[superoperator]] is [[completely positive]].


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==Examples==
==Examples==
The following code verifies that the map $\Phi$ defined by $\Phi(X) = X - UXU^*$ is not completely positive, where $U = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}$.
The following code verifies that the map $\Phi$ defined by $\Phi(X) = X - UXU^*$ is not completely positive, where $U = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}$.
<pre>
<syntaxhighlight>
>> U = [1 1;-1 1]/sqrt(2);
>> U = [1 1;-1 1]/sqrt(2);
>> Phi = {eye(2),eye(2); U,-U};
>> Phi = {eye(2),eye(2); U,-U};
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     0
     0
</pre>
</syntaxhighlight>
 
{{SourceCode|name=IsCP}}

Latest revision as of 15:28, 29 September 2014

IsCP
Determines whether or not a superoperator is completely positive

Other toolboxes required none
Related functions IsHermPreserving
Function category Superoperators

IsCP is a function that determines whether or not a given superoperator is completely positive.

Syntax

  • CP = IsCP(PHI)
  • CP = IsCP(PHI,TOL)

Argument descriptions

  • PHI: A superoperator. Should be provided as either a Choi matrix, or as a cell with either 1 or 2 columns (see the tutorial page for more details about specifying superoperators within QETLAB).
  • TOL (optional, default eps^(3/4)): The numerical tolerance used when determining complete positivity.

Examples

The following code verifies that the map $\Phi$ defined by $\Phi(X) = X - UXU^*$ is not completely positive, where $U = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}$.

>> U = [1 1;-1 1]/sqrt(2);
>> Phi = {eye(2),eye(2); U,-U};
>> IsCP(Phi)

ans =

     0

Source code

Click here to view this function's source code on github.