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It seems that Kraus operators cannot change a pure state into a mixed one (wrong). For any pure state can be written as $|\psi\rangle\langle\psi|$, so after the Kraus operators. It becomes $$\sum_l\Pi_l|\psi\rangle\langle\psi|\Pi_l^\dagger = |\phi\rangle\langle\phi|.$$

But does there exist some Kraus operators that can change the mixed state $\rho$ into a pure state?

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    $\begingroup$ They can change a pure state to a mixed state. Given a pure state $|+\rangle \langle +|$, and Kraus operators $N_{1}=|0\rangle \langle 0|$ and $N_{2}=|1\rangle \langle 1|$, you get $\sum_{i}N_{i}|+\rangle \langle +|N_{i}^{\dagger}=N_{1}|+\rangle \langle +|N_{1}^{\dagger}+N_{2}|+\rangle \langle +|N_{2}^{\dagger}=\frac{1}{2}|0\rangle \langle 0|+\frac{1}{2}|1\rangle \langle 1|$, which is a mixed state. $\endgroup$ Commented Jun 17, 2021 at 11:38
  • $\begingroup$ Given the Kraus operators describe the reduced dynamics of a non-local operation, even if the initial state is pure, after the evolution, it may be correlated, in which case the pure state in one subsystem will no longer be describable by information only available in that subsystem. $\endgroup$ Commented Jun 17, 2021 at 11:41
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    $\begingroup$ Note that any channel has a Kraus decomposition, so you're question is equivalent to asking if quantum channels can map pure to mixed and vice versa. To which the answer is yes. $\endgroup$
    – Rammus
    Commented Jun 17, 2021 at 12:16
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    $\begingroup$ Let me take the opportunity to blatantly self-advertise my list of canonical examples of quantum channels. $\endgroup$ Commented Jun 17, 2021 at 15:44

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More generally, given any two states, you can always find some channel sending one into the other. Consider for example replacement maps, which have the form $$\Phi_Y(X) = \operatorname{Tr}(X) Y.$$ Given any pair of states $\rho$ and $\sigma$, the channel $\Phi_\sigma$ will send $\rho$ (as well as any other state) into $\sigma$. The (or a) set of Kraus operators for $\Phi_Y$ is $\{\sqrt{y_\alpha}\lvert y_\alpha\rangle\!\langle \beta|\}_{\alpha,\beta}$, where $|y_\alpha\rangle$ form an orthonormal basis of eigenvectors for $Y$ (assuming $Y$ to be normal), $y_\alpha$ are the corresponding eigenvalues, and $|\beta\rangle$ is an arbitrary orthonormal basis (for the relevant space).

See also @Norbert's answer on physics.SE for other examples of quantum channels on which to test hypotheses.

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    $\begingroup$ I'd say "a channel can send any state into any other state" is somewhat misleading. I mean, your example sends every state to any state. Already if I give you two states which I want to be sent to any other two states, it is getting tricky. $\endgroup$ Commented Jun 17, 2021 at 15:45
  • $\begingroup$ @NorbertSchuch uhm, yes that was poor phrasing. I changed the sentence $\endgroup$
    – glS
    Commented Jun 17, 2021 at 15:48
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To add to the already great answers let me point out that if the initial mixed state $\rho$ has full rank, then there is only one type of channel which can send that state to a pure state $|\phi\rangle\langle\phi|$: the replacement/reset channel $X\mapsto{\rm tr}(X)|\phi\rangle\langle\phi|$. The reason for this is that all positive maps $\Phi$ "increase kernels" in the sense that given any $X,Y\geq 0$ with ${\rm ker}(X)\subseteq{\rm ker}(Y)$ one always has ${\rm ker}(\Phi(X))\subseteq{\rm ker}(\Phi(Y))$ (equivalently: positive maps increase ranges) as is shown, e.g., in Lemma 1.1 of this paper or Proposition 3.2 of one of my old papers.

Either way this kernel property implies that if a channel $\Phi$ maps $\rho>0$ to $|\phi\rangle\langle\phi|$, then for all states $\omega$ — because $\{0\}={\rm ker}(\rho)\subseteq{\rm ker}(\omega)$ — it holds that $|\phi\rangle^\perp={\rm ker}(\Phi(\rho))={\rm ker}(\Phi(\omega))$, i.e. $\Phi(\omega)=|\phi\rangle\langle\phi|$. By linearity this shows that $\Phi(X)={\rm tr}(X)|\phi\rangle\langle\phi|$ for all $X$, as claimed. More generally, the above kernel property places a restriction on all channels that map some non-pure state to a pure state. A reference for this more special result is, e.g., Lemma 5.6 in this paper

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    $\begingroup$ nice. I think this argument is also related (though not identical) to the one given in this answer: quantumcomputing.stackexchange.com/a/26379/55 $\endgroup$
    – glS
    Commented Apr 9 at 8:08
  • $\begingroup$ Right, the argument you linked is based on rank-1 projectors being the extreme points of the positive cone whereas the "increasing kernel" property is based on the fact that ${\rm ker}(X)\subseteq{\rm ker}(Y)$ for any $X,Y\geq 0$ is equivalent to the existence of $\lambda\geq 0$ such that $\lambda X-Y\geq 0$. So as you pointed out the methods do differ but they are similar in spirit. $\endgroup$ Commented Apr 9 at 8:21
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$\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right) $ and $\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right) $ can change one mixed state into a pure state.

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    $\begingroup$ As you've noted, this is indeed the case. Generally, a good example of operators that achieve what you want are non-orthogonal projectors which project all states to the same one. In the case of your answer is $|0\rangle\langle 0|$ and $|0\rangle \langle 1|$ $\endgroup$ Commented Jun 17, 2021 at 11:33

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