Unambiguous discrimination using POVM with highest discriminate probability

Question

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 unambiguous 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 attain values as large as $\frac{1}{4}$? How to concretely construct one?

Answer 1

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 $\frac{2-\sqrt{2}}{4}$.