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Let $A$ and $B$ be finite dimensional Hilbert spaces, and let $\mathcal{L}(A \to B)$ be the space of linear operators from $A$ to $B$. Say that a subspace $K \subseteq \mathcal{L}(A \to B)$ is a span of Kraus operators if there are operators $\{K_i\}$ such that $\sum_i K_i^\dagger K_i = I$ and $K = \textrm{span}_i\{K_i\}$.

Equivalently, $K$ is a span of Kraus operators if there is an ancillary Hilbert space $C$ and an isometry $J:A \to B \otimes C$ such that $K = \textrm{span}_{\left| \psi \right> \in C} \{ (I \otimes \left<\psi\right|) J \}$.

Are there any non-trivial (and hopefully simple) necessary and sufficient conditions for an operator subspace $K$ to be a span of Kraus operators? The only necessary condition I can come up with is that $I \in K^\dagger K := \textrm{span}\{ x^\dagger y : x \in K, y \in K \}$, but this is not sufficient.

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migrated from Sep 30 '13 at 22:55

This question came from our site for professional mathematicians.

Out of curiosity, why did you ask for migration from MO? It feels so much more like an MO question... – Emilio Pisanty Oct 3 '13 at 16:26
That's why I asked there originally. But this has to do with quantum information, a topic which spans the math, cstheory, and physics sites. Although many QI folk subscribe to all three sites, physics seems to have a larger population. The question itself is math, but QI people have the intuition on trace preserving maps and Kraus operators. – Dan Stahlke Oct 3 '13 at 20:43
Well, good luck here, then, though I suspect you could have been better of in cstheory, since they might have more of the more abstract chops needed for this. It's a valuable question for all three (four?) sites anyway, I think, so I'd recommend judicious cross-posting instead of a hopping migration. – Emilio Pisanty Oct 3 '13 at 20:50
up vote 0 down vote accepted

Consider a space $\textrm{span}_j\{L_j\}$ with the $\{L_j\}$ linearly independent. This is a span of Kraus operators if it can be written as $\textrm{span}_i\{K_i\}$ with the $K_i$ being Kraus operators. This is possible if and only if there is a set of Kraus operators $\{K_i\}$ such that $K_i = \sum_j F_{ij} L_j$ for some matrix $F$ with full support (i.e. $F^\dagger F$ invertible). Applying the closure condition for Kraus operators, this becomes $\sum_i K_i^\dagger K_i = \sum_{ijj'} (F_{ij'} L_{j'})^\dagger (F_{ij} L_j) = I$, equivalently $\sum_{jj'} (F^\dagger F)_{j'j} L_{j'}^\dagger L_j = I$. So one is looking for a positive definite matrix $M$ such that $\sum_{jj'} M_{j'j} L_{j'}^\dagger L_j = I$. The problem is then reduced to finding a positive definite operator satisfying an affine constraint. Note that this is equivalent to finding a positive definite operator which is perpendicular to the space $\textrm{span}\{ \sum_{jj'} \textrm{Tr}(X L_{j'}^\dagger L_j)^* \left|j'\right>\left<j\right| : \textrm{Tr}(X)=0 \}$.

If one is instead interested in whether $\textrm{span}\{ L_j \}$ contains a span of Kraus operators, then look for nonzero positive semidefinite operators $M$ rather than positive definite operators.

Unfortunately I do not know the computational complexity of finding a positive (semi)definite operator lying within a subspace. However, an SDP solver should find such a PSD operator (or a certificate for its nonexistence) for non-pathological cases.

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