Making an ambiguous and unambiguous state determinations together

Question

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:

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.

Answer 1

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.

Answer 2

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.