Theorem on projective measurement and 100% distinguishable states

[provolus]

Hi everybody! I'm studying this paper "Unambiguous discrimination among quantum operations" http://pra.aps.org/abstract/PRA/v73/i4/e042301 and they state that

It is well known that a set of
quantum states can be perfectly distinguished if and only if
they are orthogonal to each other.

Ok, it's well known, but then I took a review of the notes to my course of quantum information and i found this not proven proposition:

consider two states \rho_1 and \rho_2 (density operators) of a finite quantum system: if exists a projector \Pi_1 such that trace(\rho_1\Pi_1)=1 and trace(\rho_2\Pi_1)=0 then the two states are disntinguishable with certainty if and only if supp(\rho_1) is orthogonal to supp(\rho_2)

Well, I'm not able to arrange a sounded proof of this simple proposition: i guess i have to choose a base and decompose spectrally the operator but I'm not able to go on. could someone be so kind to show me how to do or at least to suggest me a book or a paper where i could find a proof?

Thank!

 

[eljose79]

Hi there! Thank you for sharing your thoughts and questions on this paper. It's great to see that you are actively engaging with the material and trying to understand it more deeply.

First of all, I want to clarify that the statement in the paper "Unambiguous discrimination among quantum operations" is referring specifically to quantum operations, not just quantum states. In other words, the paper is discussing the ability to distinguish between different operations that can be applied to a quantum system, rather than just distinguishing between different states.

With that being said, your proposition is correct and can be proven using spectral decomposition. Here's a brief outline of the proof:

1. Assume that there exists a projector Pi_1 such that trace(rho_1 * Pi_1) = 1 and trace(rho_2 * Pi_1) = 0.

2. Using spectral decomposition, we can write rho_1 = sum_i p_i * |psi_i><psi_i| and rho_2 = sum_i q_i * |phi_i><phi_i|, where p_i and q_i are the eigenvalues of rho_1 and rho_2, respectively, and |psi_i> and |phi_i> are the corresponding eigenvectors.

3. Since trace(rho_1 * Pi_1) = 1, we know that Pi_1 must have at least one eigenvector |psi_j> with eigenvalue 1, and similarly, for trace(rho_2 * Pi_1) = 0, Pi_1 must have at least one eigenvector |phi_k> with eigenvalue 0.

4. Now, if supp(rho_1) is not orthogonal to supp(rho_2), then there must exist an eigenvector |psi_j> of Pi_1 that is also an eigenvector of rho_2 with eigenvalue q_k > 0. This implies that trace(rho_2 * Pi_1) > 0, which contradicts our initial assumption that trace(rho_2 * Pi_1) = 0.

5. Therefore, we can conclude that if supp(rho_1) is not orthogonal to supp(rho_2), then there does not exist a projector Pi_1 that satisfies the conditions in the proposition.

I hope this helps to clarify the proof for you. If you would like to explore this topic further, I would suggest looking into the book "Quantum