Let $\sigma$ be a fixed positive semidefinite matrix (edit: need unit trace too as pointed out if we want trace nonincreasing). Is the map


given by $N(\rho) = Tr(\sigma\rho)$ completely positive? Note that it is positive and trace nonincreasing. If yes, what are its Kraus operators?

  • 1
    $\begingroup$ Why trace nonincreasing? If $\sigma= \begin{matrix} 2 & 0\\ 0 & 2\\ \end{matrix} \tag{1}$ and $\rho=\begin{matrix} 1 & 0\\ 0 & 0\\ \end{matrix} \tag{2}$ Then $Tr(\sigma\rho)=2$. $\endgroup$
    – narip
    Jun 16 at 12:54
  • $\begingroup$ @narip that's of course only true if $\sigma$ has trace one. $\endgroup$ Jun 16 at 13:01
  • $\begingroup$ $\sigma$ can't be equal to $\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$ if $\sigma$ is to be a valid density matrix $\endgroup$ Jun 16 at 13:02
  • $\begingroup$ @GaussStrife but the assumption was only psd ... so narip is correct $\endgroup$ Jun 16 at 13:02
  • $\begingroup$ @MarkusHeinrich correct. My bad. $\endgroup$ Jun 16 at 13:05

It's the composition of 2 maps: $x \rightarrow \sqrt{\sigma}x\sqrt{\sigma}$ and $x \rightarrow \text{Tr}(x)$. Both are completely positive.

The first map is already in the Kraus decomposition form.
For the second map we can take the decomposition $\text{Tr}(x) = \sum_i \langle i |x|i\rangle$.

So, the Kraus operators for the whole map are $A_i = \langle i | \sqrt{\sigma} : H\rightarrow\mathbb{C}$.

  • $\begingroup$ These wouldn't satisfy the completeness relation though, correct? $\sum_{i}A_{i}^\dagger A_{i}=\sigma$ $\endgroup$ Jun 16 at 13:39
  • $\begingroup$ @GaussStrife Yes, the completeness relation is equivalent to the preservation of trace. $\endgroup$
    – Danylo Y
    Jun 16 at 13:47
  • $\begingroup$ ah I see. I made the assumption that the map they were referring to was a Quantum Map, which it clearly can't be due to violation of the completness relation. $\endgroup$ Jun 16 at 13:54

Yes, it is. Let's show this by computing the Choi state of $N$: $$ \mathcal{J}(N) = \sum_{i,j} N(|i\rangle\langle j|) \otimes |i\rangle\langle j| = \sum_{i,j} \langle j|\sigma|i\rangle |i\rangle\langle j| = \sigma^T. $$ Since the transposition map ${}^T$ is positive (btw not completely positive), $\sigma^T$ is a positive semi-definite operator and thus $N$ is CP.

  • $\begingroup$ Thank you - I had to pick one answer but yours was also super helpful and a nice way to see it $\endgroup$
    – polynotexp
    Jun 17 at 4:51

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