Updated and edited question:

Let $N_{\delta}:P(\mathcal{H}_A)\rightarrow P(\mathcal{H}_B)$ be a completely positive trace nonincreasing map from the set of positive semidefinite operators in $\mathcal{H}_A$ to positive semidefinite operators in $\mathcal{H}_B$. For any $\rho\in P(\mathcal{H}_{A})$, we want $\sigma = N_{\delta}(\rho)$ such that all its positive eigenvalues are greater than $\delta$ and all remamining eigenvalues are zero.

Can such a map $N_\delta$ be constructed?

Based on a previous version of the question, the comment of @Rammus and the answer of @NorbertSchuch suggested using a projection operation $P$ into the eigenspace with eigenvalues larger than $\delta$. However, the required projector depends on the input state $\rho$, which I would like to avoid.

  • 1
    $\begingroup$ The TNICP map that projects onto the positive eigenspace of your density matrix should do the trick I think. $\endgroup$
    – Rammus
    Jun 9, 2020 at 21:38
  • $\begingroup$ With your edit, it makes no sense. The way you write it the map should neither depend on rho nor on $\delta$. Or is this what you really mean? Please write a formal statement (which means: a formula, with clear statements on what the map can depend.) -- For instance, a formal statement would read: "For every delta, there is a CP map E such that for all rho, XYZ holds." $\endgroup$ Jun 9, 2020 at 21:51
  • $\begingroup$ Much better! I guess what is missing is that you want that $\sigma=P\rho P^\dagger$, where $P$ is the projector onto the eigenvalues of $\rho$ larger than $\delta$. $\endgroup$ Jun 9, 2020 at 22:17
  • $\begingroup$ @Rammus and Norbert Schuch, thank you - your answers to both this and the old question were helpful! $\endgroup$ Jun 9, 2020 at 22:20

1 Answer 1


Answer to the new version:

No, such a map cannot exist. CP maps are linear, i.e., $$ \mathcal E(\lambda\rho) = \lambda \mathcal E(\rho)\ . $$ Thus, you can easily see that for $\lambda\ne 1$, this is inconsistent with your definition.

Answer to the original version:

Sure, just project onto the eigenspace which you want to keep: $$ \rho\mapsto P\rho P^\dagger $$ with $P$ the orthogonal projector onto the eigenspace.

This is in Kraus form, and thus CP.

  • $\begingroup$ Sorry, I should have clarified in the question but I meant a CP map that works for any input $\rho$, not just a specific one. That is, it always checks if any eigenvalues are smaller than $\delta$ and sets them to zero if this is so. Is that possible? $\endgroup$ Jun 9, 2020 at 21:38
  • $\begingroup$ @user1936752 Linearity should instantly imply that "no". (For instance, just apply the map to $10\rho$ instead of $\rho$!) But before I take the effort to write this in an answer please ask this in a clean and formal way in the question. $\endgroup$ Jun 9, 2020 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.