# Umambiguous discrimination using POVM with highest discriminate probability

I was studying Nielsen&Chuang's textbook (about page 92), and come up with a question that I cannot solve it.

Given one of the two state $$|\psi_1\rangle=|0\rangle$$ and $$|\psi_2\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$$ (both initial state has equal prior probability $$\frac{1}{2}$$). If I want to apply the umambiguous discrimination: Construct POVM $$\{\Pi_i\}$$ where $$i\in\{1,2,\text{inconclusive}\}$$, and outcome $$1$$ when given $$|\psi_1\rangle$$, $$2$$ when given $$|\psi_2\rangle$$ and $$\text{inconclusive}$$ if it is unluckily to determine the case.

How large can the success probability $$\sum_{i=1}^2\frac{1}{2}P(\text{outcome}~ i\mid\text{given}~|\psi_i\rangle)$$ be? Is it possible to attan as large as $$\frac{1}{4}$$? How to concretely construct one?

• – glS
May 23 at 18:23

## 1 Answer

I'll write $$\rho_1 = |\psi_1\rangle \langle \psi_1|$$ and $$\rho_2 = |\psi_2\rangle \langle \psi_2|$$. We want the discrimination to be unambiguous so we want, $$\mathrm{tr}[\rho_1 \Pi_2] = 0 = \mathrm{tr}[\rho_2 \Pi_1].$$ That is, when we get outcome $$i\in \{1,2\}$$ we know that we received $$\rho_i$$ as the other state has a zero probability of obtaining that outcome.

Well, we can just choose our POVM elements to be orthogonal to the other state, i.e. consider the projectors $$\hat{\Pi}_1 = |-\rangle \langle - | \qquad \text{and} \qquad \hat{\Pi}_2 = |1\rangle \langle 1|.$$ The only issue we have now is that if we choose $$\hat{\Pi}_3 = I - \hat{\Pi}_1 - \hat{\Pi}_2$$ then $$\hat{\Pi}_3 \not\geq 0$$. To fix this we can weight our projectors with $$\alpha,\beta \in [0,1]$$ $$\Pi_1 = \alpha \hat{\Pi}_1 \\ \Pi_2 = \beta \hat{\Pi}_2 \\ \Pi_3 = I - \Pi_1 -\Pi_2$$ which is a POVM if $$\alpha + \beta + \sqrt{\alpha^2 + \beta^2} \leq 2$$.

With such a POVM we get $$\mathrm{tr}[\rho_1 \Pi_1] = \frac{\alpha}{2} \qquad \text{and} \qquad \mathrm{tr}[\rho_2 \Pi_2] = \frac{\beta}{2}.$$ Putting it all together the maximum success probability is \begin{aligned} \max_{\alpha,\beta \in [0,1]}& \quad \frac{\alpha + \beta}{4} \\ \mathrm{s.t.}& \quad \alpha + \beta + \sqrt{\alpha^2 + \beta^2} \leq 2 \end{aligned} which should be maximized at $$\alpha = \beta = 2-\sqrt{2}$$. Giving a maximum success probability of $$\frac{2-\sqrt{2}}{4}$$.

• Many thanks!!! Very clear! May 23 at 14:36