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A completely positive map $\mathbb{W}$ is a map from $\mathbb{C}^{n\times n} \to \mathbb{C}^{n\times n}$ that can be written in terms of $n\times n$ matrices $K$ as

$$ \mathbb{W}(\rho) = \sum_{i=1}^N K_i \rho K^\dagger_i$$

Note that I do not require $\sum_i K_i^\dagger K_i = I$, so this is generically not a quantum channel, as it is generically not trace preserving (and hence generically doesn't take density matrices to density matrices).

Consider the set of eigenvalues of $\mathbb{W}$, and consider the eigenvalue $\lambda_m$ with the maximum absolute value.

Am I guaranteed that there exists at least one $\rho_m$ such that $\mathbb{W}(\rho_m) = \lambda_m \rho_m$ and $Tr[\rho_m] \neq 0$ are both satisfied?

That is, can I always find a not-traceless matrix in the leading-eigenvalue eigenspace of $\mathbb{W}$?


I actually suspect I might be able to ask for even more properties, like at least one positive semi-definite matrix in the leading-eigenvalue eigenspace of $\mathbb{W}(\rho)$. However, for my purposes, I'm most interested in the weakened statement above of a not-traceless matrix.

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  • $\begingroup$ Those are called completely positive maps. (I sometimes also call them quantum channels, though I think the mainstream use of the latter includes trace preserving.) $\endgroup$ Oct 24, 2023 at 8:37
  • $\begingroup$ @NorbertSchuch Thanks for the edits! $\endgroup$
    – user196574
    Oct 24, 2023 at 19:01

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The reference of choice for anything about quantum channels are arguably Michael Wolf's lecture notes Quantum Channels and Operations - Guided Tour.

There, Theorem 6.5 states that any positive map (and thus, in particular, any completely positive map) with spectral radius $\varrho$ has $\varrho$ as an eigenvalue, and there is a positive semi-definite $X$ which is an eigenvector with that eigenvalue.

(In essence, this comes from a quantum version of the Perron-Frobenius theorem, which says that the leading eigenvector of a stochastic matrix, i.e. a random process, is a probability distribution.)


One intuition of why this is the case might be taken from an iterated application of the map $X\mapsto \mathbb W(X)/|\lambda_m|$, which will converge to an eigenvector with eigenvalue $\lambda_m$ for any initial $X$. On the other hand, for a positive $X$, this map will only ever return positive semi-definite matrices. What remains open in this argument is to show that there is at least one $X\ge0$ which has non-zero overlap with a fixed point.

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  • $\begingroup$ Thanks, this is great to see, and I'll study those notes. As an aside, I have a thought on making the intuition a bit more rigorous: $\endgroup$
    – user196574
    Oct 24, 2023 at 18:30
  • $\begingroup$ Convergence to an eigenvector with maximal eigenvalue will happen only for initial $X$ that have overlap with an eigenvector with maximal eigenvalue. If the identity $I$ is such a vector, we get using positivity that the maximal-eigenvalue eigenvector it has overlap with is itself positive semi-definite. cont'd $\endgroup$
    – user196574
    Oct 24, 2023 at 18:32
  • $\begingroup$ If the identity $I$ is not such a vector, consider a vector $X$ that does have overlap with an eigenvector of maximal eigenvalue, and note $X+a I $ will converge to that eigenvector under iterated application of the scaled map for any $a$. In particular, we can take $a$ large and positive, and hence the maximal-eigenvalue eigenvector must itself be positive semi-definite. $\endgroup$
    – user196574
    Oct 24, 2023 at 18:32

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