An optimal discrimination of two mixed qubit states with a fixed rate of inconclusive results

Abstract

In this paper we consider the optimal discrimination of two mixed qubit states for a measurement that allows a fixed rate of inconclusive results. Our strategy is to transform the problem of two qubit states into a minimum error discrimination for three qubit states by adding a specific quantum state $\rho _{0}$ and a prior probability $q_{0}$, which behaves as an inconclusive degree. First, we introduce the beginning and the end of practical interval of inconclusive result, $q_{0}^{(0)}$ and $q_{0}^{(1)}$, which are key ingredients in investigating our problem. Then we obtain the analytic form of them. Next, we show that our problem can be classified into two cases $q_{0}=q_{0}^{(0)}$ (or $q_{0}=q_{0}^{(1)}$) and $q_{0}^{(0)}<q_{0}<q_{0}^{(1)}$. In fact, by maximum confidences of two qubit states and non-diagonal element of $\rho _{0}$, our problem is completely understood. We provide an analytic solution of our problem when $q_{0}=q_{0}^{(0)}$ (or $q_{0}=q_{0}^{(1)}$). However, when $q_{0}^{(0)}<q_{0}<q_{0}^{(1)}$, we rather supply the numerical method to find the solution, because of the complex relation between inconclusive degree and corresponding failure probability. Finally we confirm our results using previously known examples.

Appendices

A Proofs of Lemmas in Sect. 2

Proof of Lemma 1

Suppose that when $\{q_{i},\rho _{i}\}_{i=1}^{N}$ is given, POVM $\{M_{i}\}_{i=0}^{N}$ can cause $P_\mathrm{I}=Q$ and $\bar{P}_\mathrm{cor}=\bar{P}_\mathrm{cor}^\mathrm{opt}(q)$. It follows that the POVM can make $P_\mathrm{cor}(Q)=\bar{P}_\mathrm{cor}^\mathrm{opt}(q)-qQ$. If there exists a POVM that can build $P_\mathrm{I}=Q$ and $P_\mathrm{cor}(Q)=P>\bar{P}_\mathrm{cor}^\mathrm{opt}(q)-qQ$, it can also construct $\bar{P}_\mathrm{cor}(q)=P+qQ$. However since $P+qQ$ is larger than $\bar{P}_\mathrm{cor}^\mathrm{opt}(q)$, this is contradictory. Therefore $\{M_{i}\}_{i=0}^{N}$ should produce $P_\mathrm{cor}^\mathrm{opt}(Q)$, which means $P_\mathrm{cor}^\mathrm{opt}(Q)=\bar{P}_\mathrm{cor}^\mathrm{opt}(q)-qQ$. $\square$

Proof of Lemma 2

Assume that when $\{q_{i},\rho _{i}\}_{i=1}^{N}$ is given, the POVM $\{M_{i}\}_{i=0}^{N}$ ($\{M_{i}'\}_{i=0}^{N}$) can produce $P_\mathrm{I}=Q$ and $\bar{P}_\mathrm{cor}=\bar{P}_\mathrm{cor}^\mathrm{opt}(q)$ ($P_\mathrm{I}=Q'$ and $\bar{P}_\mathrm{cor}=\bar{P}_\mathrm{cor}^\mathrm{opt}(q')$). If $q=q'$ and $Q<Q'$, the POVM $\{M_{i}''\}_{i=0}^{N}$ composed of $M_{i}''=pM_{i}+(1-p)M_{i}'$ ($0\le p \le 1$) will build $\bar{P}_\mathrm{cor}=\bar{P}_\mathrm{cor}^\mathrm{opt}(q)$ and $P_\mathrm{I}=pQ+(1-p)Q'$. Therefore $P_\mathrm{I}(q)$ becomes a convex set.

Now suppose that $q<q'$. $\{M_{i}\}_{i=0}^{N}$ constructs $\bar{P}_\mathrm{cor}=(q'-q)Q+\bar{P}_\mathrm{cor}^\mathrm{opt}(q)$ when $q_{0}=q'$, and the value should be equal to or less than $\bar{P}_\mathrm{cor}^\mathrm{opt}(q')=(q'-q)Q'+\bar{P}$, where $\bar{P}$ is $\bar{P}_\mathrm{cor}$ corresponding to $\{M_{i}'\}_{i=0}^{N}$ when $q_{0}=q$. This means that $(Q'-Q)\ge (\bar{P}_\mathrm{cor}^\mathrm{opt}(q)-\bar{P})/(q'-q)$. Therefore we have $Q\le Q'$. This means that $P_\mathrm{I}(q)\le P_\mathrm{I}(q')$. $\square$

Proof of Lemma 3

When $q_{0}<1/N$, we get $\bar{\tau }_{0}^\mathrm{opt}=(1/N-q_{0})I_{d}+(1/N)\sum _{i=1}^{N}\bar{\tau }_{i}^\mathrm{opt}$ by (ii) of (6). If we multiply $\bar{M}_{0}^\mathrm{opt}$ to both sides of the equation and take the trace of the result, we obtain $(1/N-q_{0})\mathrm{tr}[\bar{M}_{0}^\mathrm{opt}]\le 0$ by (iii) and the positivity of $\bar{\tau }_{i}^\mathrm{opt}$ ($\forall i$). From the assumption on $q_{0}$, $\bar{M}_{0}^\mathrm{opt}$ should be zero and we find $P_\mathrm{I}(q)=0(\forall q<1/N)$. Therefore using Lemma 2, we have $q_{0}^{(0)}\ge 1/N$.

When $C=\max _{i}C_{i}$, $\bar{M}_{i}=\delta _{i0}\rho _{0}$ and $\bar{\tau }_{i}=CI_{d}-\bar{\rho }_{i}$ satisfy the optimality condition (6) of $q_{0}=C$, and $1\in P_\mathrm{I}(C)$. Therefore we get $q_{0}^{(1)}\le C$ by Lemma 2. $\square$

B Proofs of Lemmas in Sect. 3

Proof of Lemma 4

When $q_{0}=C_{2}$, since $\{\bar{M}_{i}=\delta _{i0}\rho _{0}\}_{i=0}^{2}$ and $\{\bar{\tau }_{i}^{\star }=C_{2}I_{2}-\bar{\rho }_{i}\}_{i=0}^{2}$ satisfy the KKT optimality condition (6), $\bar{P}_\mathrm{cor}^\mathrm{opt}(C_{2})=C_{2}$. This means that $\bar{\tau }_{i}^\mathrm{opt}=\bar{\tau }_{i}^{\star }(\forall i)$. Since the rank of $\bar{\tau }_{2}^\mathrm{opt}$ should be one by $C_{1}+C_{2}>1$, we have $\bar{M}_{2}^\mathrm{opt}=\beta |\nu _{2}\rangle \!\langle \nu _{2}|$ from (iii). The form of $\bar{M}_{0}^\mathrm{opt}$ and $\bar{M}_{1}^\mathrm{opt}$ can be classified into the cases of $C_{1}=C_{2}$ and $C_{1}<C_{2}$.

If $C_{1}=C_{2}$, the rank of $\bar{\tau }_{1}^\mathrm{opt}$ becomes 1, and we get $\bar{M}_{1}^\mathrm{opt}=\alpha |\nu _{1}\rangle \!\langle \nu _{1}|$ from (iii). Furthermore (i) indicates that $\bar{M}_{0}^\mathrm{opt}=\rho _{0}-\alpha |\nu _{1}\rangle \!\langle \nu _{1}|-\beta |\nu _{2}\rangle \!\langle \nu _{2}|$, $0\le \alpha \le \rho _{11}$, $0\le \beta \le \rho _{22}$, and $(\rho _{11}-\alpha )(\rho _{22}-\beta )\ge |\rho _{12}|^{2}$. Since $\mathrm{tr}[\bar{M}_{0}^\mathrm{opt}]=1-\alpha -\beta$, the maximum can be found at $\alpha =\beta =0$. $\alpha$ and $\beta$ corresponding to its minimum can be different. When $\rho _{11}<|\rho _{12}|\le \rho _{22}$, we have $\alpha =0$ and $\beta =1-Q_{1}$. When $\rho _{22}<|\rho _{12}|\le \rho _{11}$, we obtain $\alpha =1-Q_{2}$ and $\beta =0$. When $|\rho _{12}|\le \rho _{11},\rho _{22}$, we have $\alpha =\rho _{11}-|\rho _{12}|$ and $\beta =\rho _{22}-|\rho _{12}|$. Therefore $P_\mathrm{I}(C_{2})$ becomes (16).

If $C_{1}<C_{2}$, the rank of $\bar{\tau }_{1}^\mathrm{opt}$ becomes 2, and (iii) implies that $\bar{M}_{1}^\mathrm{opt}=0$. Then (i) means $\bar{M}_{0}^\mathrm{opt}=\rho _{0}-\beta |\nu _{2}\rangle \!\langle \nu _{2}|$ and $0\le \beta \le 1-Q_{1}$. Since $\mathrm{tr}[\bar{M}_{0}^\mathrm{opt}]=1-\beta$, the minimum(maximum) can be found at $\beta = 1-Q_{1}$ ($\beta =0$). Therefore we get $P_\mathrm{I}(C_{2})=[Q_{1},1]$. $\square$

Proof of Lemma 5

When $C_{1}\le 1/2<C_{2}$, since $\{\bar{M}_{i}=\delta _{i2}\rho _{0}\}_{i=0}^{2}$ and the following $\{\bar{\tau }_{i}^{\star }\}_{i=0}^{2}$ satisfies the optimality condition (6) of $q_{0}=1-C_{1}$, $\bar{P}_\mathrm{cor}^\mathrm{opt}(1-C_{1})=q_{2}$.

$$\begin{aligned} \bar{\tau }_{0}^{\star }\!=\!(C_{1}\!+\!C_{2}\!-\!1)|\nu _{2}\rangle \!\langle \nu _{2}|,\quad \bar{\tau }_{1}^{\star }=(1-2C_{1})|\nu _{1}\rangle \!\langle \nu _{1}|+(2C_{2}-1)|\nu _{2}\rangle \!\langle \nu _{2}|,\quad \bar{\tau }_{2}^{\star }=0.\nonumber \\ \end{aligned}$$

(54)

This implies that $\bar{\tau }_{i}^\mathrm{opt}=\bar{\tau }_{i}^{\star }(\forall i)$ when $q_{0}=1-C_{1}$. Since the rank of $\bar{\tau }_{0}^\mathrm{opt}$ is 1 given that $C_{1}+C_{2}>1$, (iii) implies $\bar{M}_{0}^\mathrm{opt}=\alpha |\nu _{1}\rangle \!\langle \nu _{1}|$. However $\bar{M}_{1}^\mathrm{opt}$ and $\bar{M}_{2}^\mathrm{opt}$ are classified into cases where $C_{1}<1/2$ and $C_{1}=1/2$.

When $C_{1}<1/2$, since the rank of $\bar{\tau }_{1}^\mathrm{opt}$ becomes 2, (iii) gives $\bar{M}_{1}^\mathrm{opt}=0$ and (i) means $\bar{M}_{2}^\mathrm{opt}=\rho _{0}-\alpha |\nu _{1}\rangle \!\langle \nu _{1}|$ and $0\le \alpha \le 1-Q_{2}$. Then, since $\mathrm{tr}[\bar{M}_{0}^\mathrm{opt}]=\alpha$ shows a minimum at $\alpha =0$ and a maximum at $\alpha =1-Q_{2}$, we have $P_\mathrm{I}(1-C_{1})=[0,1-Q_{2}]$. However, when $C_{1}=1/2$, the rank of $\bar{\tau }_{1}^\mathrm{opt}$ is 1 and (iii) implies $\bar{M}_{1}^\mathrm{opt}=\beta |\nu _{1}\rangle \!\langle \nu _{1}|$. Therefore (i) means $\bar{M}_{2}^\mathrm{opt}=\rho _{0}-(\alpha +\beta )|\nu _{1}\rangle \!\langle \nu _{1}|$ and we have $\alpha ,\beta \ge 0$, and $\alpha +\beta \le 1-Q_{2}$. $\mathrm{tr}[\bar{M}_{0}^\mathrm{opt}]=\alpha$ has a minimum(maximum) at $\alpha =0$ ($\alpha =1-Q_{2}$ and $\beta =0$). Therefore, we obtain $P_\mathrm{I}(1-C_{1})=[0,1-Q_{2}]$. $\square$

Proof of Lemma 6

When $1/2<C_{1}\le C_{2}$,$\rho _{12}=0$, since the following $\{\bar{M}_{i}\}_{i=0}^{2}$, $\{\bar{\tau }_{i}^{\star }\}_{i=0}^{2}$ satisfies the optimality condition (6) of $q_{0}=C_{1}$, $\bar{P}_\mathrm{cor}^\mathrm{opt}(C_{1})=\rho _{11}C_{1}+\rho _{22}C_{2}$.

$$\begin{aligned} \begin{array}{lllllllll} \bar{M}_{0}&{}=&{}0,\, &{}\bar{M}_{1}&{}=&{}\rho _{11}|\nu _{1}\rangle \!\langle \nu _{1}|,\, &{}\bar{M}_{2}&{}=&{}\rho _{22}|\nu _{2}\rangle \!\langle \nu _{2}|,\\ \bar{\tau }_{0}^{\star }&{}=&{}(C_{2}-C_{1})|\nu _{2}\rangle \!\langle \nu _{2}|,\, &{}\bar{\tau }_{1}^{\star }&{}=&{}(2C_{2}-1)|\nu _{2}\rangle \!\langle \nu _{2}|,\, &{}\bar{\tau }_{2}^{\star }&{}=&{}(2C_{1}-1)|\nu _{1}\rangle \!\langle \nu _{1}|. \end{array}\nonumber \\ \end{aligned}$$

(55)

This means that $\bar{\tau }_{i}^\mathrm{opt}=\bar{\tau }_{i}^{\star }(\forall i)$ when $q_{0}=C_{1}$. Since, if $C_{1}<C_{2}$, the rank of $\bar{\tau }_{i}^\mathrm{opt}$ is one, (iii) tells that $\bar{M}_{0}^\mathrm{opt}$ and $\bar{M}_{1}^\mathrm{opt}$ are proportional to $|\nu _{1}\rangle \!\langle \nu _{1}|$, and $\bar{M}_{2}^\mathrm{opt}$ is proportional to $|\nu _{2}\rangle \!\langle \nu _{2}|$. By (i), $\bar{M}_{i}^\mathrm{opt}$ can be expressed as (25). However, since $C_{1}=C_{2}$ implies $\bar{\tau }_{0}^\mathrm{opt}=0$, $\bar{M}_{i}^\mathrm{opt}$ becomes (26). Therefore $P_\mathrm{I}(C_{1})$ can be written as $[0, \rho _{11}+\rho _{22}\delta _{C_{1},C_{2}}]$. $\square$

Proof of Lemma 7

When $1/2<C_{1}\le C_{2}$ and $\rho _{12}\ne 0$, Lemma 2 and Corollary 1 reveal that $q_{0}^{(0)}<C_{2}$. By (ii) of optimality condition (6) and the nonnegativity of $\bar{\tau }_{i}(\forall i)$, $\bar{\tau }_{0}^\mathrm{opt}=0$ includes $q_{0}\ge C_{2}$, and $\bar{\tau }_{1(2)}^\mathrm{opt}=0$ contains $C_{2(1)}\le 1/2$. This implies that $\bar{\tau }_{i}^\mathrm{opt}\ne 0(\forall i)$ if $q_{0}<q_{0}^{(1)}$. Then $\bar{M}_{0}^\mathrm{opt}=0$ means that $\bar{M}_{1}^\mathrm{opt},\bar{M}_{2}^\mathrm{opt}\ne 0$ because $\bar{M}_{0}^\mathrm{opt}=\bar{M}_{1}^\mathrm{opt}=0$ implies $\bar{\tau }_{2}^\mathrm{opt}=0$ and $\bar{M}_{0}^\mathrm{opt}=\bar{M}_{2}^\mathrm{opt}=0$ contains $\bar{\tau }_{1}^\mathrm{opt}=0$. When $\bar{M}_{0}^\mathrm{opt}=0$, in order to obtain the explicit form of $\{\bar{M}_{i}^\mathrm{opt},\bar{\tau }_{i}^\mathrm{opt}\}_{i=1}^{2}$, we use the optimality condition (8). Since $\bar{M}_{0}=0$ includes $p_{0}=0$, it has no effect on $r_{0}$ and ${{\varvec{w}}}_{0}$. $\bar{M}_{i},\bar{\tau }_{i}\ne 0(i=1,2)$ implies $p_{i},r_{i}\ne 0 (i=1,2)$, and by (iii) we have $\Vert {{\varvec{w}}}_{i}\Vert _{2}=1$, ${{\varvec{u}}}_{i}=-{{\varvec{w}}}_{i}(i=1,2)$. Since $r_{1}+r_{2}=l$ and $r_{2}-r_{1}=q_{1}-q_{2}$ should be satisfied by (ii), $\{p_{i}^\mathrm{opt},{{\varvec{u}}}_{i}^\mathrm{opt}\}_{i=1}^{2}$ and $\{r_{i}^\mathrm{opt},{{\varvec{w}}}_{i}^\mathrm{opt}\}_{i=1}^{2}$ can be obtained as follows:

$$\begin{aligned} p_{1}^\mathrm{opt}= & {} p_{2}^\mathrm{opt}=\frac{1}{2},\quad r_{1}^\mathrm{opt}=\frac{1+l}{2}-q_{1},\quad r_{2}^\mathrm{opt}=\frac{1+l}{2}-q_{2},\nonumber \\ {{\varvec{u}}}_{1}^\mathrm{opt}= & {} {{\varvec{w}}}_{2}^\mathrm{opt}=\frac{q_{1}{{\varvec{v}}}_{1}-q_{2}{{\varvec{v}}}_{2}}{\Vert q_{1}{{\varvec{v}}}_{1}-q_{2}{{\varvec{v}}}_{2}\Vert _{2}},\quad {{\varvec{u}}}_{2}^\mathrm{opt}={{\varvec{w}}}_{1}^\mathrm{opt}=\frac{q_{2}{{\varvec{v}}}_{2}-q_{1}{{\varvec{v}}}_{1}}{\Vert q_{1}{{\varvec{v}}}_{1}-q_{2}{{\varvec{v}}}_{2}\Vert _{2}}. \end{aligned}$$

(56)

From these, we find $\bar{P}_\mathrm{cor}^\mathrm{opt}(q)=(1+l)/2(\forall q\le q_{0}^{(0)})$, and can decide the explicit form of $\{{M}_{i}^\mathrm{opt}\}_{i=0}^{2}$ and $\{\tau _{i}^\mathrm{opt}\}_{i=1}^{2}$. However $r_{0}^\mathrm{opt}$ and ${{\varvec{w}}}_{0}^\mathrm{opt}$ are not decided yet. These are affected only by (ii). The triangle made of $\{q_{i}{{\varvec{v}}}_{i}\}_{i=0}^{2}$ lies in the plane with the origin, and the triangle consisting of $\{-r_{i}^\mathrm{opt}{{\varvec{w}}}_{i}^\mathrm{opt}\}_{i=0}^{2}$ should be located in the same plane. Since the two triangles are congruent, then as $\Vert r_{0}^\mathrm{opt}{{\varvec{w}}}_{0}^\mathrm{opt}\Vert _{2}$ grows larger $\Vert q_{0}{{\varvec{v}}}_{0}\Vert _{2}$ becomes larger. Since $q_{0}+r_{0}^\mathrm{opt}$ is fixed as $(1+l)/2$, when $\Vert {{\varvec{w}}}_{0}^\mathrm{opt}\Vert _{2}$ reaches the maximum(that is, when $\Vert {{\varvec{w}}}_{0}^\mathrm{opt}\Vert _{2}=1$), $q_{0}$ reaches the maximum. Therefore the determinant of $\bar{\tau }_{0}^\mathrm{opt}$ is 0 when $q_{0}=q_{0}^{(0)}$. From (ii), we have $(\chi _{1}-q_{0}^{(0)})(\chi _{2}-q_{0}^{(0)})=|\gamma _{12}|^{2}$. Though there are two roots of this equation, the nonnegativity of $\bar{\tau }_{0}^\mathrm{opt}$ implies that $q_{0}^{(0)}\le \min \{\chi _{1},\chi _{2}\}$, and the analytic form of $q_{0}^{(0)}$ can be obtained as $\chi$ of (29). The optimal POVM of $q_{0}=\chi$ is unique since $r_{i}^\mathrm{opt}\ne 0(\forall i)$ and $\{q_{i}{{\varvec{v}}}_{i}\}_{i=0}^{2}$ forms a triangle; see Appendix D. This means that ${P}_\mathrm{I}(\chi )=0$. $\square$

Proof of Lemma 8

When $C_{1}<C_{2}$ and $\rho _{12}=0$, if $q_{0}^{(0)}<q<q_{0}^{(1)}$, POVM, defined as (34), and $\{\tau _{i}^{\star }\}_{i=0}^{2}$ satisfies KKT optimality condition (6) to $q_{0}=q$:

$$\begin{aligned} \bar{\tau }_{0}^{\star }= & {} (C_{2}-q)|\nu _{2}\rangle \!\langle \nu _{2}|,\nonumber \\ \bar{\tau }_{1}^{\star }= & {} (q-C_{1})|\nu _{1}\rangle \!\langle \nu _{1}|+(2C_{2}-1)|\nu _{2}\rangle \!\langle \nu _{2}|,\nonumber \\ \bar{\tau }_{2}^{\star }= & {} (q-1+C_{1})|\nu _{1}\rangle \!\langle \nu _{1}|. \end{aligned}$$

(57)

This means that $\bar{\tau }_{i}^\mathrm{opt}=\bar{\tau }_{i}^{\star }(\forall i)$. Since the rank of $\bar{\tau }_{1}^\mathrm{opt}$ is two, (iii) implies $\bar{M}_{1}^\mathrm{opt}=0$. However, since the rank of $\bar{\tau }_{0}^\mathrm{opt}$ and $\bar{\tau }_{2}^\mathrm{opt}$ are one, $\bar{M}_{0}^\mathrm{opt}$ and $\bar{M}_{2}^\mathrm{opt}$ are proportional to $|\nu _{1}\rangle \!\langle \nu _{1}|$ and $|\nu _{2}\rangle \!\langle \nu _{2}|$, respectively. Therefore (i) means that $\bar{M}_{i}^\mathrm{opt}$ is unique as (34). $\square$

Proof of Lemma 9

First of all, let us consider the case of $C_{1}\le \frac{1}{2}<C_{2}$ and $\rho _{12}\ne 0$. In the region of $q_{0}^{(0)}<q_{0}<q_{0}^{(1)}$, since $\lambda _{1}$ of (44) is less than 0, we find $M_{1}^\mathrm{opt}=0$ or $M_{2}^\mathrm{opt}=0$ by Theorem 3. If $M_{2}^\mathrm{opt}=0$, since optimality condition (6) means $\bar{\tau }_{1}^\mathrm{opt}=\bar{\tau }_{0}^\mathrm{opt}+q_{0}I_{2}-\bar{\rho }_{1}$ and $\det (\bar{\tau }_{0}^\mathrm{opt})=\det (\bar{\tau }_{1}^\mathrm{opt})=0$, $t_{i}=\langle \nu _{i}|\bar{\tau }_{0}^\mathrm{opt}|\nu _{i}\rangle$ satisfies $t_{1}t_{2}=(t_{1}+q_{0}-C_{1})(t_{2}+q_{0}-1+C_{2})$. However, this result is contradictory because $(t_{1}+q_{0}-C_{1})(t_{2}+q_{0}-1+C_{2})$ is greater than $t_{1}t_{2}$ in the region of $(C_{1},1-C_{2}\le 1-C_{1}=)q_{0}^{(0)}<q_{0}<q_{0}^{(1)}$. Therefore we get $M_{1}^\mathrm{opt}=0$.

Next, let us consider the case of $\frac{1}{2}<C_{1}\le C_{2}$ and $\rho _{12}\ne 0$. Here $(q_{0}^{(0)},q_{0}^{(1)})$ is divided into two cases: $(\chi ,C_{1}]$ and $(C_{1},C_{2})$. In latter case, because of $\lambda _{1}< 0$, we can obtain $M_{1}^\mathrm{opt}=0$. $\square$

C Proof of Theorem 3

Proof

When $\rho _{xy}\ne 0$ and $q_{0}=q\in (q_{0}^{(0)},q_{0}^{(1)})$, the line intersecting ${{\varvec{v}}}_{1}$ and ${{\varvec{v}}}_{2}$ does not contain the origin, and $\{q_{i}{{\varvec{v}}}_{i}\}_{i=0}^{2}$ forms a triangle. $r_{0}^\mathrm{opt}=0$ implies that $\{M_{i}=\delta _{i0}I_{2}\}_{i=0}^{2}$ provide an optimal POVM, which includes $q_{0}^{(1)}\le q$. Since $r_{k}^\mathrm{opt}=0(k\in \{1,2\})$ indicates that $\{M_{i}=\delta _{ik}I_{2}\}_{i=0}^{2}$ yields the optimal POVM, this means $q\le q_{0}^{(0)}$. Therefore the element of $\{r_{i}^\mathrm{opt}\}_{i=0}^{2}$ are all nonzero. In this case, the optimal POVM is unique; see Appendix D. In addition, $M_{0}^\mathrm{opt}$ is nonzero, and at least one of $M_{1}^\mathrm{opt}$ and $M_{2}^\mathrm{opt}$ is nonzero.

In the case of $M_{x}^\mathrm{opt}\ne 0$, $M_{y}^\mathrm{opt}=0(\{x,y\}=\{1,2\})$, the index x turns out to be the index i in $\max _{i\in {1,2}}[q_{i}+\Vert q{{\varvec{v}}}_{0}-q_{i}{{\varvec{v}}}_{i}\Vert _{2}]$ because $\bar{P}_\mathrm{cor}^\mathrm{opt}(q)=\max _{i\in {1,2}}[q+q_{i}+\Vert q\rho _{0}-q_{i}\rho _{i}\Vert _{1}]/2$. The optimal POVM, by the optimality condition (8), can be expressed as (42).

In the case of $M_{i}^\mathrm{opt}\ne 0(\forall i)$, by the optimality condition (6), $\{\bar{M}_{i}^\mathrm{opt},\bar{\tau }_{i}^\mathrm{opt}\}_{i=0}^{2}$ can be found explicitly. From condition (ii), $\{\bar{\tau }_{i}^\mathrm{opt}\}_{i=0}^{2}$ are given as follows.

$$\begin{aligned} \bar{\tau }_{0}^\mathrm{opt}= & {} \tau _{11}|\nu _{1}\rangle \!\langle \nu _{1}|+\tau _{12}|\nu _{1}\rangle \!\langle \nu _{2}|+\tau _{21} |\nu _{2}\rangle \!\langle \nu _{1}|+\tau _{22}|\nu _{2}\rangle \!\langle \nu _{2}|,\nonumber \\ \bar{\tau }_{1}^\mathrm{opt}= & {} (\tau _{11}+q-C_{1})|\nu _{1}\rangle \!\langle \nu _{1}|+\tau _{12}|\nu _{1}\rangle \!\langle \nu _{2}|+\tau _{21} |\nu _{2}\rangle \!\langle \nu _{1}|\nonumber \\&+(\tau _{22}+q-1+C_{2})|\nu _{2}\rangle \!\langle \nu _{2}|,\nonumber \\ \bar{\tau }_{2}^\mathrm{opt}= & {} (\tau _{11}+q-1+C_{1})|\nu _{1}\rangle \!\langle \nu _{1}|+\tau _{12}|\nu _{1}\rangle \!\langle \nu _{2}|+\tau _{21} |\nu _{2}\rangle \!\langle \nu _{1}|\nonumber \\&+(\tau _{22}+q-C_{2})|\nu _{2}\rangle \!\langle \nu _{2}|.~~ \end{aligned}$$

(58)

By the complementary slackness condition (iii) of (6), the every rank of $\{\bar{M}_{i}^\mathrm{opt},\bar{\tau }_{i}^\mathrm{opt}\}_{i=0}^{2}$ is one. Therefore their determinants become 0, which means

$$\begin{aligned} |\tau _{12}|=\sqrt{\tau _{11}\tau _{22}}\quad \text{ and }\quad \bar{M}_{i}^\mathrm{opt}=\mathrm{tr}\left[ \bar{M}_{i}^\mathrm{opt}\right] \cdot \left[ I_{2}-\frac{\bar{\tau }_{i}^\mathrm{opt}}{\mathrm{tr}\left[ \bar{\tau }_{i}^\mathrm{opt}\right] }\right] \ \forall i. \end{aligned}$$

(59)

Then, we have $\tau _{11}=\lambda _{1}$ and $\tau _{22}=\lambda _{2}$. Since $\mathrm{tr}[\bar{M}_{i}^\mathrm{opt}]$ is the probability that $M_{i}^\mathrm{opt}$ may be detected, $\mathrm{tr}[\bar{M}_{0}^\mathrm{opt}]$ becomes $P_\mathrm{I}(q)$. The phase of $\tau _{12}$ and the form of $\mathrm{tr}[\bar{M}_{i}^\mathrm{opt}]$ can be obtained by condition (i). The completeness condition of the POVM is represented as

$$\begin{aligned} \frac{\mathrm{tr}\left[ \bar{M}_{0}^\mathrm{opt}\right] }{\mathrm{tr}\left[ \bar{\tau }_{0}^\mathrm{opt}\right] }\cdot \bar{\tau }_{0}^\mathrm{opt}+ \frac{\mathrm{tr}\left[ \bar{M}_{1}^\mathrm{opt}\right] }{\mathrm{tr}\left[ \bar{\tau }_{1}^\mathrm{opt}\right] }\cdot \bar{\tau }_{1}^\mathrm{opt}+ \frac{\mathrm{tr}\left[ \bar{M}_{2}^\mathrm{opt}\right] }{\mathrm{tr}[\bar{\tau }_{2}^\mathrm{opt}]}\cdot \bar{\tau }_{2}^\mathrm{opt} =I_{2}-\rho _{0}. \end{aligned}$$

(60)

$\rho _{12}$ and $\tau _{12}$ have the relation of $\rho _{12}/\tau _{12}=-\sum _{i=0}^{2}(\mathrm{tr}[\bar{M}_{i}^\mathrm{opt}]/\mathrm{tr}[\bar{\tau }_{i}^\mathrm{opt}])$. By $M_{i}^\mathrm{opt}\ne 0(\forall i)$ and the nonnegativity of POVM, the right-hand side of the equation is always negative, and we get $\rho _{12}/\tau _{12}=-|\rho _{12}|/|\tau _{12}|$. That is, $\tau _{12}=-(\rho _{12}/|\rho _{12}|)\sqrt{\lambda _{1}\lambda _{2}}$. And $\bar{P}_\mathrm{cor}^\mathrm{opt}(q)$ is found as (43). Then we have $\mathrm{tr}[\bar{M}_{i}^\mathrm{opt}]=\eta _{i}$ by the following relation:

$$\begin{aligned} \left( \begin{array}{c} \mathrm{tr}[\bar{M}_{0}^\mathrm{opt}]\\ \mathrm{tr}[\bar{M}_{1}^\mathrm{opt}]\\ \mathrm{tr}[\bar{M}_{2}^\mathrm{opt}]\\ \end{array}\right) = \left( \begin{array}{ccc} 1&{}\quad 1&{}\quad 1\\ \frac{1}{\mathrm{tr}\left[ \bar{\tau }_{0}^\mathrm{opt}\right] } &{} \quad \frac{1}{\mathrm{tr}\left[ \bar{\tau }_{1}^\mathrm{opt}\right] } &{}\quad \frac{1}{\mathrm{tr}\left[ \bar{\tau }_{2}^\mathrm{opt}\right] }\\ \frac{\lambda _{1}}{\mathrm{tr}\left[ \bar{\tau }_{0}^\mathrm{opt}\right] } &{}\quad \frac{\lambda _{1}+q-C_{1}}{\mathrm{tr}\left[ \bar{\tau }_{1}^\mathrm{opt}\right] } &{}\quad \frac{\lambda _{1}+q -1 +C_{1}}{\mathrm{tr}\left[ \bar{\tau }_{2}^\mathrm{opt}\right] } \end{array}\right) ^{-1} \left( \begin{array}{c} 1\\ \frac{|\rho _{12}|}{\sqrt{\lambda _{1}\lambda _{2}}}\\ \rho _{22}\\ \end{array}\right) . \end{aligned}$$

(61)

Therefore $\bar{M}_{i}^\mathrm{opt}$ is represented as (45). The result implies the following. If $\lambda _{i}\ge 0(\forall i)$ and $\eta _{i}>0(\forall i)$, we have $M_{i}^\mathrm{opt}\ne 0(\forall i)$. Otherwise, we find $M_{1}^\mathrm{opt}=0$ or $M_{2}^\mathrm{opt}=0$. $\square$

D Proof of uniqueness of optimal POVM

Here we prove the following fact: When $\{q_{i}{{\varvec{v}}}_{i}\}_{i=0}^{2}$ forms a triangle, if $r_{i}^\mathrm{opt}\ne 0 (\forall i)$, then the POVM $\{M_{i}=p_{i}(I_2+{{\varvec{u}}}_{i}\cdot {\varvec{\sigma }})\}_{i=0}^{2}$ fulfilling the optimality condition (8) is unique. For the proof, we use ${{\varvec{v}}}_{0}$ as an arbitrary Bloch vector extrinsic to ${{\varvec{v}}}_{1}$,${{\varvec{v}}}_{2}$. Since $M_{k}^\mathrm{opt}=I_2$ implies $r_{k}^\mathrm{opt}=0$, at least two of $\{M_{i}^\mathrm{opt}\}_{i=0}^{2}$ are nonzero.

First, we consider the case that there exists $\{p_{i}\ne 0,{{\varvec{u}}}_{i}\}_{i=0}^{2}$ and $\{r_{i}\ne 0,{{\varvec{w}}}_{i}\}_{i=0}^{2}$ fulfilling optimality condition (8). Without loss of generality, we can set $q_{0}\ge q_{1},q_{2}$. Then (iii) becomes ${{\varvec{u}}}_{i}\cdot {{\varvec{w}}}_{i}=-1(\forall i)$. This can be rewritten as $\Vert {{\varvec{u}}}_{i}\Vert _{2}=1$, ${{\varvec{w}}}_{i}=-{{\varvec{u}}}_{i}(\forall i)$, and (ii) is as follows: $r_{i}-r_{0}=e_{i}$, $\mathbf{R}\equiv q_{i}{{\varvec{v}}}_{i}-r_{i}{{\varvec{u}}}_{i}(i=0,1,2)$. $e_{i}$ is the difference between two prior probabilities $q_{0}$ and $q_{i}$. This condition means the following; $\{r_{i}{{\varvec{u}}}_{i}\}_{i=0}^{2}$ forms a triangle congruent to a triangle $\{q_{i}{{\varvec{v}}}_{i}\}_{i=0}^{2}$, and $\{r_{i}{{\varvec{u}}}_{i}\}_{i=0}^{2}$ coincides with $\{q_{i}{{\varvec{v}}}_{i}\}_{i=0}^{2}$ by parallel transport $\mathbf{R}$. Then (i) contains the following statement. $\mathbf{R}$ lies in the interior of the triangle $\{q_{i}{{\varvec{v}}}_{i}\}_{i=0}^{2}$, and the distance from this point to the vertex of the triangle $q_{i}{{\varvec{v}}}_{i}$ is $r_{i}$. The points fulfilling $r_{i}-r_{0}=e_{i}$ satisfy the following hyperbolic equation:

$$\begin{aligned} r_{0}=\frac{l_{i}^{2}-e_{i}^{2}}{2(l_{i}\cos \theta _{i}+e_{i})}. \end{aligned}$$

(62)

Above $l_{i}$ is the distance between two vectors $q_{0}{{\varvec{v}}}_{0}$ and $q_{i}{{\varvec{v}}}_{i}$, and $\theta _{i}$ is the angle between two sides $\{\mathbf{R},q_{0}{{\varvec{v}}}_{0}\}$ and $\{q_{0}{{\varvec{v}}}_{0},q_{i}{{\varvec{v}}}_{i}\}$. As $\theta _{i}$ increases, $r_{0}$ also increases, and inside the triangle $\{q_{i}{{\varvec{v}}}_{i}\}_{i=0}^{2}$ the position of $\mathbf{R}$ is unique. This means that the $\{p_{i},{{\varvec{u}}}_{i}\}_{i=0}^{2}$ are unique. Therefore, the optimal POVM in which every element is nonzero is unique. To make a distinction, we denote this POVM as $\{M_{i}'\}_{i=0}^{2}$. Suppose that there exists another POVM satifying the optimality condition and denote it as $\{M_{i}''\}_{i=0}^{2}$. Then the POVM consisting of $M_{i}=\epsilon M_{i}'+(1-\epsilon )M_{i}''(0<\epsilon <1)$ is optimal, and we have $M_{i}\ne 0(\forall i)$. This is contradictory, and therefore the optimal POVM is unique.

Next, we consider the case that there exist $\{p_{i},{{\varvec{u}}}_{i}\}_{i=0}^{2}$ and $\{r_{i}\ne 0,{{\varvec{w}}}_{i}\}_{i=0}^{2}$ fulfilling optimality condition (8) and one of $\{p_{i}\}_{i=0}^{2}$ is zero and the others are nonzero. Without loss of generality, we can set $p_{0}=0$. Then (iii) becomes ${{\varvec{u}}}_{i}\cdot {{\varvec{w}}}_{i}=-1(i=1,2)$. This can turn into $\Vert {{\varvec{u}}}_{i}\Vert _{2}=1$, ${{\varvec{w}}}_{i}=-{{\varvec{u}}}_{i}(i=1,2)$, and (ii) can be expressed in the following way: $r_{1}-r_{2}=q_{2}-q_{1}$, $\mathbf{R}\equiv q_{1}{{\varvec{v}}}_{1}-r_{1}{{\varvec{u}}}_{1}=q_{2}{{\varvec{v}}}_{2}-r_{2}{{\varvec{u}}}_{2}$. This condition implies that $\{r_{i}{{\varvec{u}}}_{i}\}_{i=1}^{2}$ coincides with the line segment $\{q_{i}{{\varvec{v}}}_{i}\}_{i=1}^{2}$ by parallel translation $\mathbf{R}$. (i) means that $\mathbf{R}$ lies in the interior of $\{q_{i}{{\varvec{v}}}_{i}\}_{i=1}^{2}$ and the distance from the point to $q_{i}{{\varvec{v}}}_{i}$ is $r_{i}$. That is, we have $r_{1}+r_{2}=l_{12}$. $l_{12}$ is the distance between two vectors $q_{1}{{\varvec{v}}}_{1}$ and $q_{2}{{\varvec{v}}}_{2}$. Then $r_{1}$ and $r_{2}$ satisfying $r_{1}-r_{2}=q_{2}-q_{1}$ are apparently unique. This implies that $\{p_{i},{{\varvec{u}}}_{i}\}_{i=1}^{2}$ are unique. Therefore the optimal POVM satisfying $M_{0}=0$,$M_{1}\ne 0$,$M_{2}\ne 0$ is unique. To differentiate from the other POVM, we represent this POVM as $\{M_{i}'\}_{i=0}^{2}$. We assume that there exists a POVM satisfying $M_{0}\ne 0$ and the optimality condition, and denote it as $\{M_{i}''\}_{i=0}^{2}$. Then the POVM consisting of $M_{i}=\epsilon M_{i}'+(1-\epsilon )M_{i}''$ ($0<\epsilon <1$) is optimal. The result is that POVM fulfilling $M_{i}\ne 0 (\forall i)$ and geometric optimality condition is not unique. This contradicts the previous result, and the optimal POVM is unique.

In conclusion, when $\{q_{i}{{\varvec{v}}}_{i}\}_{i=0}^{2}$ forms a triangle and $r_{i}^\mathrm{opt}\ne 0 (\forall i)$, the optimal POVM is unique. $\square$