1
$\begingroup$

Theorem 12.15 of Nielsen and Chuang's 10th anniversary edition is Nielsen's Theorem (1999). In particular, it says,

Theorem 12.15: A bipartite pure state $\mid \psi \rangle$ may be transformed to another pure state $\mid \phi \rangle$ by LOCC if and only if $\lambda_\psi \prec \lambda_\phi$.

Here, $\mid \psi \rangle$ is an entangled pure state in the composite system $AB$ and $\lambda_\psi$ denotes the vector of eigenvalues of $\rho_{\psi} = Tr_B(\mid \psi \rangle \langle \psi \mid)$ (which is the density matrix for the state of A's system).

In the proof of the converse, the authors assume that $\rho_{\psi}$ is invertible in order to define measurement operations $M_j = \sqrt{p_j \rho_\phi}U_j^{\dagger}\rho_{\psi}^{-1/2}$ for Alice's system. They say "this assumption is easily removed; see Exercise 12.20" but that exercise doesn't give any indication on how to remove this assumption.

Exercise 12.20: Show that the assumption that $\rho_{\psi}$ is invertible may be removed from the proof of the converse part of Theorem 12.15.

It is not clear to me how to define appropriate measurement operators if we do not have invertibility of $\rho_{\psi}$. My question is, how can this assumption be removed?

Below is a picture of the proof given in Nielsen-Chuang, to show exactly how invertibility of $\rho_{\psi}$ is used.

Theorem 12.15 Nielsen-Chuang


Cross-posted on qc.SE

$\endgroup$
2
  • 1
    $\begingroup$ I haven't looked at the proof by my guess is that you can work in the range of $\rho_\psi$ where the density matrix is actually invertible $\endgroup$
    – lcv
    Commented Dec 13, 2022 at 14:36
  • $\begingroup$ Have you checked the original paper? $\endgroup$ Commented Dec 13, 2022 at 17:05

1 Answer 1

2
$\begingroup$

As already commented by @lcv, if $\rho_\psi$ is not full range, you can initially restrict the construction of the POVM operators $M_i$ to the support of $\rho_\psi$ -- let's call those $\hat M_i$. Of course, this will not directly be a POVM on the full space, but you can clearly complete it to a POVM, by building the full $M_i$ as a direct sum of $\hat M_i$ on the support, and something trivial on its orthogonal complement. (What you choose is arbitrary, since this part of the POVM is never actually used for anything, as the state does not have support in that subspace.)

$\endgroup$

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.