Pairwise and joint distinguishability of quantum states

[entanglement witness]

Hi!

I'm struggling with the following question:

Show that if n quantum states ρ₁, ..., ρ_n are pairwise perfectly distinguishable, they are also jointly perfectly distinguishable.

Perfect distinguishability means that there is a set of psd matrices $$\{E_{1}, ..., E_{n}\},\, \sum_{i} E_{i} = \mathbb{1}$$ s.t. $$tr\left( E_{i} \rho_{j} \right) = \delta_{ij}$$

I don't understand what exactly I have to show here. My assumption is that if for example $$(E_{1}^{(2)}, \mathbb{1} - E_{1}^{(2)})$$ distinguishes ρ₁ and ρ₂, $$(E_{1}^{(3)}, \mathbb{1} - E_1^{(3)})$$ ρ₁ and ρ₃ and so on, that $$E_{1}^{(2)} = ... = E_{1}^{(n)} =: E_{1}$$ (and likewise for all others), which would give the set I was looking for?

Now two states are perf. distinguishable if and only if $$\rho_{i} \rho_{j} = 0\, \forall i \neq j$$ i.e. the density matrices have orthogonal ranges. Since for psd matrices A, B $$tr(AB) = 0 \iff AB = 0$$ it follows that $$E_{1}^{(2)} \rho_{2} = ... = E_{1}^{(2)} \rho_{n} = 0$$ the same for $$E_{1}^{(3)} ... E_{1}^{(n)}$$ thus $$E_{1}^{(2)} = ... = E_{1}^{(n)}$$

But those are just my ideas. I hope somebody can explain how to tackle this correctly.

 

[mark726]

Hi there!

To show that the n quantum states ρ1, ..., ρn are jointly perfectly distinguishable, we need to show that there exists a set of psd matrices \{E_{1}, ..., E_{n}\},\, \sum_{i} E_{i} = \mathbb{1} s.t. tr\left( E_{i} \rho_{j} \right) = \delta_{ij} for all i, j.

Your idea is on the right track, but we need to show that the set of psd matrices \{E_{1}, ..., E_{n}\} is the same for all pairs of ρi and ρj, not just for ρ1 and ρ2.

To do this, we can use the fact that the n quantum states ρ1, ..., ρn are pairwise perfectly distinguishable. This means that for any two states ρi and ρj, there exists a psd matrix E_{ij} such that tr\left( E_{ij} \rho_{i} \right) = \delta_{ij} and tr\left( E_{ij} \rho_{j} \right) = \delta_{ij}. We can then define the set of psd matrices \{E_{1}, ..., E_{n}\} as follows:

E_{i} = \sum_{j \neq i} E_{ij}

This set of psd matrices satisfies the conditions we need, i.e. tr\left( E_{i} \rho_{j} \right) = \delta_{ij} for all i, j, and \sum_{i} E_{i} = \mathbb{1}, since for any i, tr\left( E_{i} \rho_{i} \right) = \delta_{ii} = 1 and for any i \neq j, tr\left( E_{i} \rho_{j} \right) = \delta_{ij} = 0.

Therefore, we have shown that if n quantum states ρ1, ..., ρn are pairwise perfectly distinguishable, they are also jointly perfectly distinguishable. I hope this helps clarify things for you! Let me know if you have any further questions.