I was trying to understand the fact that non-orthogonal quantum states cannot be reliably distinguished and I came across this link.

The explanation finishes with the result that the probability of successfully identifying a state between two non-orthogonal states is $Pr(success) ⩽ (1/2)(1 + sin(\alpha))$

This another reference discuss the fact in a more intuitively manner and then says that the probability of correctly inferring the state, when they are two non-orthogonal states with some angle $\theta$ between then is exactly the same as the first link. See the image below

enter image description here

It says that this result is "captured by the Helstrom-Holevo bound". From what I remember the Holevo bound is that it is a mathematical expression that gives an upper bound on the amount of accessible information that can be obtained from a quantum state. My questions are:

  1. This result $ 1 + sin(\alpha) $ comes from the use of the Holevo bound? If so.... how? I cannot see it
  2. The Holevo bound comes from the fact that non-orthogonal quantum states cannot be distinguished or it is a proof of this fact?

The only proof - or sketch of a proof - that I remember of this fact comes from Nielsen and Chuang but I never really liked - or understood - the proof, so if a good reference that proves it or discuss it in details could be given here I would really appreciate.

  • 1
    $\begingroup$ to clarify: you are asking for a proof of the bound for the optimal discrimination probability? $\endgroup$
    – glS
    Apr 24, 2023 at 20:59
  • 3
    $\begingroup$ Theorem 3.4 of Watrous gives a nice proof of the HH bound $\endgroup$
    – Rammus
    Apr 24, 2023 at 21:38
  • $\begingroup$ @glS that's right. I mean, the second reference from my post just use the result abd the first one I'm not sure if is formal enough. I understand what the author did but I cannot see how it connects to the HH bound $\endgroup$
    – Dimitri
    Apr 25, 2023 at 2:14
  • $\begingroup$ Or bow can I use the HH bound to derive this 1 + sin(alpha) $\endgroup$
    – Dimitri
    Apr 25, 2023 at 2:15
  • 1
    $\begingroup$ The Helstrom-Holevo bound and the Holevo bound are two different things, and you don't need the Holevo bound to prove the Helstrom-Holevo bound. The Helstrom-Holevo bound is the simpler one to prove by far, and I hope the proof in my book that Rammus linked will be helpful. Sometimes it's just called the Helstrom bound or Helstrom's theorem, so you can also try searching for those terms. $\endgroup$ Apr 25, 2023 at 12:08


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