# Is the map $\rho\rightarrow Tr(\sigma\rho)$ completely positive?

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

$$N:H\rightarrow\mathbb{C}$$

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

• 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 $Tr(\sigma\rho)=2$. Jun 16 at 12:54
• @narip that's of course only true if $\sigma$ has trace one. Jun 16 at 13:01
• $\sigma$ can't be equal to $\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$ if $\sigma$ is to be a valid density matrix Jun 16 at 13:02
• @GaussStrife but the assumption was only psd ... so narip is correct Jun 16 at 13:02
• @MarkusHeinrich correct. My bad. 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 $$A_i = \langle i | \sqrt{\sigma} : H\rightarrow\mathbb{C}$$.

• These wouldn't satisfy the completeness relation though, correct? $\sum_{i}A_{i}^\dagger A_{i}=\sigma$ Jun 16 at 13:39
• @GaussStrife Yes, the completeness relation is equivalent to the preservation of trace. Jun 16 at 13:47
• 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. 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.

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