Difference between revisions of "IsCP"

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(Created page with "{{Function |name=IsCP |desc=Determines whether or not a superoperator is completely positive |req=ApplyMap<br />ChoiMatrix<br />iden<br />IsPSD<br />[[...")
 
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|name=IsCP
 
|name=IsCP
 
|desc=Determines whether or not a [[superoperator]] is [[completely positive]]
 
|desc=Determines whether or not a [[superoperator]] is [[completely positive]]
|req=[[ApplyMap]]<br />[[ChoiMatrix]]<br />[[iden]]<br />[[IsPSD]]<br />[[MaxEntangled]]<br />[[opt_args]]<br />[[PermuteSystems]]
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|rel=[[IsHermPreserving]]
|upd=January 3, 2013
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|upd=January 4, 2013
|v=1.00}}
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|v=1.01}}
<tt>'''IsCP'''</tt> is a [[List of functions|function]] that determines whether or not a given superoperator is [[completely positive]].
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<tt>'''IsCP'''</tt> is a [[List of functions|function]] that determines whether or not a given [[superoperator]] is [[completely positive]].
  
 
==Syntax==
 
==Syntax==
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==Examples==
 
==Examples==
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The following code verifies that the map $\Phi$ defined by $\Phi(X) = X - UXU^*$ is not completely positive, where $U = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}$.
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<pre>
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>> U = [1 1;-1 1]/sqrt(2);
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>> Phi = {eye(2),eye(2); U,-U};
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>> IsCP(Phi)
  
Add some examples here.
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ans =
 +
 
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    0
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</pre>

Revision as of 16:35, 21 January 2013

IsCP
Determines whether or not a superoperator is completely positive

Other toolboxes required none
Related functions IsHermPreserving

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