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Unambiguous state discrimination, as outlined here starting at p. 421, introduces an average success probability $P = 1 - Q^{POVM}$ where $Q^{POVM}$ is the average failure probability. However, I fail to see the value in these quantities when trying to distinguish two states.

Let's say I have two states. I make a single measurement and can unambiguously conclude what state I measured if it 'clicks'. Since these are POVMs, there will be times when there is no 'click' for the state I'm trying to measured - this would be an instance of failure. I do not see how the average failure probability is an useful metric though since it takes into account two measurements, not one, since $Q^{POVM} = \eta_1 q_1 + \eta_2 q_2$. But of course, I can't (or don't want to) make two measurements on the state.

To me, it seems like the more pertinent metric is $p_i = 1 - q_i$ as the probability of successfully determining the state to be $i$.

EDIT: Here's an example of what I'm having difficulty understanding.

Let's say we have a priori probabilities $\eta_0$ and $\eta_1$ for the two different possible states. If I use measurement operator $\Pi_1 $ I have a $p_1 = \langle \psi_1 | \Pi_1 | \psi_1 \rangle$ probability of getting a click from a $\psi_1$ state. If I have 100 states to measure, then $\eta_1 p_1$ fraction will be clicks where I can definitely say the state was $\psi_1$. For the remaining measurements I cannot say anything since there were no clicks.

In the link above, the average success probability is defined as $\eta_0 p_0 + \eta_1 p_1$. But above, it's clear (to me) that the success probability is $\eta_1 p_1$. This is the inconsistency that is troubling me.

BeauGeste
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  • not sure I understand your point. As you mention, when discussing the optimal average success probability, you generally get strategies that also have nonzero failure probability. What's the objection exactly here? If you want strategies that have zero failure probability, that's also possible (though these will generally have an overall lower average success probability). – glS Jul 15 '23 at 07:12
  • @gIS I just updated my question with some more detail on my sticking point. – BeauGeste Jul 16 '23 at 23:13
  • What you will generally find is that the optimal measurement choice has the same probability of failure for the two states you're measuring. That means that all your possible success metrics end up being the same thing. – DaftWullie Jul 17 '23 at 06:17
  • what's $\Pi_1$ in your edit? Are you talking about optimal discrimination of optimal unambiguous discrimination? Also, why do you say that the success probability is $\eta_1 p_1$? There's also the probability of the state being $\psi_0$ to take into account, hence the correct expression $\eta_0 p_0+\eta_1 p_1$. Of course, doing multiple measurement changes this probability, amplifying the probability of the discrimination being successful – glS Jul 17 '23 at 06:22
  • $\Pi_1 \propto |\psi_0^{\perp} \rangle \langle \psi_0^{\perp} |$ . In my edit, I am talking about generic discrimination (not being optimized necessarily). The reason I say $\eta_1 p_1$ is that $p_1$ is the probability of $\psi_1$ clicking. But only $\eta_1$ fraction of measurements will be on $\psi_1$ so total probability of clicking is $\eta_1 p_1$. – BeauGeste Jul 17 '23 at 15:05
  • yes, and for the remaining $\eta_0$ fraction, there's a success probability $\eta_0 p_0$. Hence the overall success probability being the sum of the two terms. I don't quite understand where's your doubt in this reasoning. As per the other post, "measuring $\Pi_1$" in this context is another way to say you find the outcome "1". But the actual measurement you're doing is always the POVM ${\Pi_0,\Pi_1,\Pi_?}$. So you have a probability success for both possible input states – glS Jul 17 '23 at 16:31

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