# Can Kraus operators change a mixed state into a pure state?

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?

• 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. Jun 17 at 11:38
• 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. Jun 17 at 11:41
• 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. Jun 17 at 12:16
• Let me take the opportunity to blatantly self-advertise my list of canonical examples of quantum channels. Jun 17 at 15:44

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).
$$\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right)$$ and $$\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right)$$ can change one mixed state into a pure state.
• 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 $|0\rangle \langle 1|$ Jun 17 at 11:33