I am trying to prove that the following superoperators are quantum channels, that is completely positive and trace-perserving linear maps

1 $\Psi[M]=WMW^\dagger$ where $W$ is an isometry

2 $\Psi[M_A]=M_A\otimes\sigma_B$ where $\sigma_B$ is a state

3 partial trace: $\Psi[M_{AB}]=Tr_B[M_{AB}]$

4 $\Psi[M]=\sum_{x,y}p(y|x)\langle x|M|x\rangle|y\rangle\langle y|$ with $p(y|x)$ a conditional probability distribution

I have already verified the trace-preserving property but I have no idea how to deal with the completely positive definition . How does one do that?

My defintion is:

Definition: A superoperator $\Phi$ in $L(L(H_A),L(H_B)) $is completely positive if $(\Phi \otimes \mathcal{I}_R)(M_A) \geq 0$ is true for all $\mathcal{H}_R$, given that $M_A\geq0\,.$

  • $\begingroup$ Check the relevant properties of the Choi matrix or show it has a Kraus form, then you know it is a quantum channel. $\endgroup$
    – Rammus
    Commented Mar 8 at 14:07

1 Answer 1


The standard method is the apply your superoperator to one half of a maximally entangled state on $\mathbb{C}^d\otimes\mathbb{C}^d$ (as a density matrix) where your operator takes inputs from the space $\mathbb{C}^d$. The outcome is a state called the Choi map. Its eigenvalues are non-negative if and only if the superoperator is completely positive.

Take case (2) as an example. if $|B\rangle$ is your Bell state, then $$ I\otimes\Psi(|B\rangle\langle B|)=|B\rangle\langle B| \otimes \sigma_B. $$ This is clearly a valid density matrix, so the eigenvalues are all between 0 and 1. Hence, the operation is completely positive.

  • $\begingroup$ Could you show (2) as an example? $\endgroup$
    – darkside
    Commented Mar 8 at 14:36
  • 1
    $\begingroup$ Shoudn't it be "Choi state" rather than "Choi map"? $\endgroup$ Commented Mar 8 at 19:11
  • $\begingroup$ Possibly. I certainly concede it's not a map. But it also might not be a state... $\endgroup$
    – DaftWullie
    Commented Mar 12 at 12:25
  • $\begingroup$ In the unnormalized case---i.e. when it's not a state so "Choi state" is not appropriate---it's usually just called "Choi matrix" $\endgroup$ Commented Apr 30 at 8:23

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