This is a sequel to Motivation for the definition of k-distillability

Geometrical interpretation from the definition of 1-distillability

  • The eigenstate $|\psi\rangle$ of the partially transposed $1$-distillable states will have Schmidt rank at most 2, i.e.,

$$|\psi\rangle\langle\psi|=\sum_i\lambda_i^2|ii\rangle\langle ii|, \textit{ where } \sum_i\lambda_i^2=1\tag{17}$$

  • The constraint $\sum_i\lambda_i^2=1$ gives rise to a geometric structure in arbitrary $N$ dimensions.



  1. In the definition of $k$-distillability (cf. here) we were talking about bipartite density matrices $\rho$ in $H_A\otimes H_B$. In what sense is $|\psi\rangle$ an "eigenstate" of a partially transposed $1$-distillable state? Is for $1$-distillable $\rho$'s the (non-normalized) state $|\psi\rangle \in H$, such that $\langle \psi|\sigma|\psi\rangle < 0$ necessarily an eigenstate of $\rho$? Also, can we prove that $\langle \psi|\sigma|\psi\rangle < 0$ for any eigenstate of $\rho$?

  2. I do not see how the fact that "the eigenstate $|\psi\rangle$ of the partially transposed $1$-distillable states will have a Schmidt rank at most 2" is encapsulated within the statement "$|\psi\rangle\langle\psi|=\sum_i\lambda_i^2|ii\rangle\langle ii|$ where $\sum_i\lambda_i^2=1$".

    As far as I understand, the Schmidt decomposition of a pure state $|\Psi\rangle$ of a composite system AB, considering an orthonormal basis $|i_A\rangle$ for system A and $|i_B\rangle$ for system B, is $$|\Psi\rangle = \sum_i \lambda_i|i_A\rangle|i_B\rangle,$$ where $\lambda_i$ are non-negative real numbers satisfying $\sum_i\lambda_i^2=1$ known as Schmidt co-efficients. Now the number of non-zero $\lambda_i$'s in the Schmidt decomposition is called Schmidt rank or Schmidt number. So I don't quite understand the geometric constraint they're talking about; if the Schmidt rank is at most 2, then we'd be restricted to only two cases i.e. $\lambda_1^2=1$ and $\lambda_1^2+\lambda_2^2 = 1$...which aren't so interesting. Am I missing something?

  • $\begingroup$ I agree with your statement in question 2. I don't see it either. (This is often the problem with following slides rather than papers. Is this a typo? Is this an error in a non-refereed document?) $\endgroup$
    – DaftWullie
    Jul 20 '20 at 7:52
  • $\begingroup$ actually, it doesn't look like a typo because that structure is being used throughout the rest of the slides. It does not look right to me. $\endgroup$
    – DaftWullie
    Jul 20 '20 at 8:26

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