Is the Holevo Quantity Preserved under Channel Applications?

  • Thread starter Thread starter rmp251
  • Start date Start date
  • Tags Tags
    Channel
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
rmp251
Messages
8
Reaction score
0
I'm trying to prove that the Holevo quantity does not increase when a channel is applied to the ensemble of states.

So, if

[itex]\Phi[/itex](ε) = { (p(a), [itex]\Phi[/itex](ρa)) : a[itex]\in[/itex][itex]\Gamma[/itex]},

then I want to prove that

[itex]\chi[/itex]([itex]\Phi[/itex](ε)) ≤ [itex]\chi[/itex](ε)

where [itex]\chi[/itex] refers to the Holevo quantity. I'm trying an approach similar to the proof for Holevo's theorem, but I can's say I totally understand that proof... but I don't think this should be too difficult. Please help!

Thank you!
 
Physics news on Phys.org
First, let us recall the definition of the Holevo quantity. The Holevo quantity of a given ensemble \Phi is defined as\chi(\Phi) = S(\sum_{a \in \Gamma} p_a \Phi(ρ_a)) - \sum_{a \in \Gamma} p_a S(\Phi(ρ_a))where S is the von Neumann entropy. Now, suppose that a channel \mathcal{N} is applied to the ensemble \Phi. By linearity of the channel, we have \Phi'(ε) = { (p(a), \mathcal{N}(\Phi(ρa)) : a\in\Gamma}, where \Phi' refers to the new ensemble after the channel has been applied. For this, the Holevo quantity of the new ensemble \Phi' is then \chi(\Phi') = S(\sum_{a \in \Gamma} p_a \mathcal{N}(\Phi(ρ_a))) - \sum_{a \in \Gamma} p_a S(\mathcal{N}(\Phi(ρ_a)))Since the von Neumann entropy is non-increasing under quantum operations, it follows that \chi(\Phi') ≤ S(\sum_{a \in \Gamma} p_a \Phi(ρ_a)) - \sum_{a \in \Gamma} p_a S(\Phi(ρ_a)) = \chi(\Phi)Therefore, the Holevo quantity does not increase when a channel is applied to an ensemble of states.