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Measurement operators for those that one creates for unambiguous state discrimination, $\Pi_0,~\Pi_1,~\Pi_?$, are such that $\Pi_0 + \Pi_1 + \Pi_? = \mathbb{1}$ and the probabilities for a measurement click are $p_i = \langle E_i | \Pi_i | E_i \rangle$.

From what I gather, the following table sums up the possibilities:

Possible outcomes from measuring two possible state

Let's say we use $\Pi_1$ to measure either $E_0$ or $E_1$. From what I read, a click tells us $E_1$ while no-click tells us nothing since both $E_0$ and $E_1$ can trigger non-clicks. That doesn't seem correct. Let's say $p_1$ is close to 1. Then a no-click is very unlikely to be from $E_1$ and we can say with some probability that the state was $E_0$ (unequal a priori probabilities would make this even more apparent). This wouldn't be an unambiguous determination but it wouldn't be a complete guess either. Am I on the right track? I hadn't seen any references bring this point up.

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2 Answers 2

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I think you're essentially asking what's the probability of the state being $E_0$ if you get the "?" outcome. So let's work this out: $$\operatorname{Pr}(E_0|\Pi_?) = \frac{\operatorname{Pr}(\Pi_?|E_0)\operatorname{Pr}(E_0)}{\operatorname{Pr}(\Pi_?)} = \frac{\operatorname{Pr}(\Pi_?|E_0)\operatorname{Pr}(E_0)}{\operatorname{Pr}(\Pi_?|E_0)\operatorname{Pr}(E_0)+\operatorname{Pr}(\Pi_?|E_1)\operatorname{Pr}(E_1)}.$$ Assume unbiased priors: $\operatorname{Pr}(E_0)=\operatorname{Pr}(E_1)=1/2$. You're asking about the scenario where $\operatorname{Pr}(\Pi_1|E_1)\equiv p_1\simeq 1$, which then also implies $\operatorname{Pr}(\Pi_?|E_1)= 1-p_1\simeq 0$. It would seem that plugging this in the above formula would give $\operatorname{Pr}(E_0|\Pi_?)\simeq 1$, and thus yes, in such a scenario, you'd conclude that the $\Pi_?$ outcome actually tells you quite a bit about the input state. However, the issue in this reasoning is that an optimal discrimination strategy such that $\operatorname{Pr}(\Pi_1|E_1)\simeq 1$ corresponds to a scenario where the states to discriminate are orthogonal, and then also $\operatorname{Pr}(\Pi_0|E_0)\simeq 1$, and $\Pi_?$ is altogether observed with vanishing probability.

Another way to see this is that having $\Pi_?$ give you meaningful information violates the premise of the measurement being optimal for unambiguous state discrimination. If the $\Pi_?$ outcome tells you any amount of information about the input state, then there must be a better way to measure that makes use of this information.

In fact, the optimal unambiguous state discrimination strategy never gives you such a scenario. For unbiased priors, the success probabilities for the optimal strategy are equal: $p_0=p_1$ in your notation. More generally, for priors $\eta_i$, they satisfy $1-p_0=\sqrt{\eta_1/\eta_0}\cos\Theta$ and $1-p_1=\sqrt{\eta_0/\eta_1}\cos\Theta$ (using the notation of the pdf notes you linked previously). Notice that $\operatorname{Pr}(\Pi_?|E_i)=1-p_i$, and thus the formula above for $\operatorname{Pr}(E_0|\Pi_?)$ works out to $$\operatorname{Pr}(E_0|\Pi_?) = \frac{\eta_0 \sqrt{\eta_1/\eta_0}}{\eta_0\sqrt{\eta_1/\eta_0} + \eta_1\sqrt{\eta_0/\eta_1} } = \frac12,$$ which is exactly what we should expect from the optimal strategy: the $\Pi_?$ outcome gives no information whatsoever.

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  • $\begingroup$ by "get the $\Pi_?$ outcome", does that mean you get a click when the $\Pi_?$ measurement operator is used? $\endgroup$
    – BeauGeste
    Jul 17, 2023 at 15:16
  • $\begingroup$ "using the $\Pi_?$ measurement operator" doesn't mean much. The measurement is the POVM $\{\Pi_0,\Pi_1,\Pi_?\}$. The possible outcomes are 0,1,? (or equivalently, $\Pi_0,\Pi_1,\Pi_?$). Each element of the POVM is an operator telling you the probability of getting that outcome, via the usual rule $p_i(\rho)=\operatorname{tr}(\Pi_i \rho)$. The whole thinking in terms of "clicks" is a bit misguided as well here: every time you measure a state, you get a "click" (that is, an outcome) in one of the three outcomes. You don't decide a priori which measurement operator you're "using" $\endgroup$
    – glS
    Jul 17, 2023 at 16:26
  • $\begingroup$ Can you prove that $p_0 = p_1$ in an easier way than solving the entire problem? Or, more interestingly, can you prove that the probability of getting ? is the same for all states when you have more than two states? $\endgroup$ Jul 18, 2023 at 10:33
  • $\begingroup$ do you mean if one can prove, in the unbiased priors scenario, that you must have $p_0=p_1$ without actually going through the full calculation? I'm not sure tbh. Given the first equation here, for unbiased priors it indeed seem that you must have $p_0=p_1$ in order to get ${\rm Pr}(E_i|\Pi_?)=1/2$ for all $i$, simply because $p_i/(p_0+p_1)=1/2$ implies $p_0=p_1$. The idea that the optimal measurement must have ${\rm Pr}(E_i|\Pi_?)=1/2$ is still a bit informal though. I can't see how this could be any other way, but I'm not sure what's a proper way to prove it $\endgroup$
    – glS
    Jul 18, 2023 at 10:44
  • $\begingroup$ @gIS ok, that's the crux. I have thinking of measurements using a single $\Pi$ and not all three. I do not really get what it means to use all three but perhaps that's the subject of another question. Thanks. I do appreciate your patience. $\endgroup$
    – BeauGeste
    Jul 19, 2023 at 6:14
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For two states this doesn't work, as @glS pointed out: the probability of getting an inconclusive outcome is equal to $|\langle \psi_0 | \psi_1 \rangle| $ for both $|\psi_0\rangle$ and $|\psi_1\rangle$.

However, for three states it does work. Consider that we have again $|\psi_0\rangle$ and $|\psi_1\rangle$ embedded into some two-dimensional subspace, and add a new state $|\psi_2\rangle$ that is orthogonal to both of them. Now the optimal unambiguous state discrimination is $E_0,E_1,E_2,E_?$, with $E_0$ and $E_1$ the same operators as before, and $E_2 = |\psi_2\rangle\langle \psi_2|$. The probability of an inconclusive measurement is still $|\langle \psi_0 | \psi_1 \rangle| $ for both $|\psi_0\rangle$ and $|\psi_1\rangle$, but it is equal to zero for $|\psi_2\rangle$. Therefore, if you get an inconclusive measurement you know you have either $|\psi_0\rangle$ and $|\psi_1\rangle$.

This extreme example is just to make the calculation easy, in general even if one of the states is not orthogonal you still learn something from the inconclusive measurement.

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  • $\begingroup$ is nontrivial unambiguous state discrimination actually possible for more than two states in a two-dimensional space? You need $\Pi_0$ orthogonal to $E_1,E_2$, $\Pi_1$ orthogonal to $E_0, E_2$, and $\Pi_2$ orthogonal to $E_0,E_1$. In two dimensions, unless some of the states to discriminate are identical (in which case there clearly isn't a nontrivial solution), how can you have this? There's no state orthogonal to $E_0$ and $E_1$ if $E_0\neq E_1$ (and I'm referring to orthogonality between the underlying pure states) $\endgroup$
    – glS
    Jul 18, 2023 at 12:35
  • $\begingroup$ No, you need all states to be linearly independent, so the dimension must be at least equal to the number of states. Note that I wrote that $|\psi_0\rangle$ and $|\psi_1\rangle$ are in a two-dimensional subspace, and $|\psi_2\rangle$ is orthogonal to them, so the minimal dimension for my example is 3. $\endgroup$ Jul 18, 2023 at 12:50
  • $\begingroup$ ah, yes, I misread that sentence as saying the overall space was two-dimensional, sorry. I wonder how this generalises though. If no pair of the 3 states is orthogonal, will the statement hold? Feels like it won't, as otherwise it wouldn't match with your limiting example. I guess in such scenarios the "?" outcome is not really a "I don't know" answer, but rather should be thought of as an outcome which "discriminates" between subspaces rather than individual states. Like in your example, where it separates the span of two states from the third. $\endgroup$
    – glS
    Jul 18, 2023 at 14:51
  • $\begingroup$ in other words, in a 3-states scenario, should one really consider POVMs of the form $\Pi_i,i=1,2,3$ with an added $\Pi_?$? Maybe it's more appropriate to consider more outcomes, where you use $\Pi_1,\Pi_2,\Pi_3$ to optimally extract what you can to discriminate individual states, but then consider discrimination of pairs of states vs the third, which gives 3 additional outcome, and then a final outcome that is the "true" $\Pi_?$ outcome, which really tells you nothing about the input state. Of course, these are different notions of optimality, so it's a slightly different problem setting $\endgroup$
    – glS
    Jul 18, 2023 at 14:55
  • $\begingroup$ That's an interesting problem, you can definitely extract more information than "for sure this state". It does fall into a nasty combinatorial explosion, though, for $n=4$ it's already unwieldy. I don't see any reason for this "true" $\Pi_?$ to give you no information about the state, though. It cannot give deterministic information, by construction, but it might be biased towards some state. Or can you prove that it becomes completely unbiased? $\endgroup$ Jul 18, 2023 at 16:00

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