Sequential state discrimination with quantum correlation

Abstract

The sequential unambiguous state discrimination (SSD) of two states prepared in arbitrary prior probabilities is studied and compared with three strategies that allow classical communication. The deviation from equal probabilities contributes to the success in all the tasks considered. When one considers at least one of the parties succeeds, the protocol with probabilistic cloning is superior to others, which is not observed in the special case with equal prior probabilities. We also investigate the roles of quantum correlations in SSD and show that the procedure requires discords but rejects entanglement. The left and right discords correspond to the part of information extracted by the first observer and the part left to his successor, respectively. Their relative difference is extended by the imbalance of prior probabilities.

Appendices

Calculations for protocol (2) that allows classical communication

The optimization of success probability for both Bob and Charlie to succeed in identifying the state can be written as

$$\begin{aligned}&{\mathrm{maximize:}}\ P^{(2)}=\left[ P_1\left( 1-q_1^b\right) +P_2\left( 1-q_2^b\right) \right] \left[ P_1'\left( 1-q_1^c\right) +P_2'\left( 1-q_2^c\right) \right] , \nonumber \\\end{aligned}$$

(21)

$$\begin{aligned}&{\mathrm{subject\ to:}}\ P_i'=\frac{P_i\left( 1-q_i^b\right) }{P_1\left( 1-q_1^b\right) +P_2\left( 1-q_2^b\right) },\quad i=1,2,\quad q_1^cq_2^c=q_1^bq_2^b=s^2, \nonumber \\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \qquad q_1^b,q_2^b,q_1^c,q_2^c\in \left[ s^2,1\right] , \quad P_1\in \left( 0,1/2\right] .\nonumber \\ \end{aligned}$$

(22)

The values of $P_1'$ and $P_2'$ are derived as

$$\begin{aligned} \mathrm {(i):}\, \, \,&P_1'=P_1^c,\, P_2'=P_2^c,\,&{\mathrm{when}}\, \frac{s^2}{1+s^2}\le P_1\le \frac{1}{2}; \end{aligned}$$

(23a)

$$\begin{aligned} \mathrm {(ii):}\, \, \,&P_1'=0,\, P_2'=1,\,&{\mathrm{when}}\, 0<P_1<\frac{t^2}{1+t^2}. \end{aligned}$$

(23b)

The case (i) in Eq. (23a) is divided into two subcases: (ia) $\frac{s^2}{1+s^2}<P_1'\le \frac{1}{2}$ and (ib) $0\le P_1'\le \frac{s^2}{1+s^2}$ which correspond to the results in Eq. (9a), (9b), respectively. The corresponding critical values $P_{c1}$ in Eq. (9a) and (9b) can be acquired after solving the equation which satisfy the successive boundary condition.

For case (ii) in Eq. (23b), Bob gets optimized success probability for $q_1^b=1$. Then, for the next observer Charlie, the conditional probability is found to be 0 ($P_1'=0$) according to Eq. (22) and the state $|\psi _1\rangle$ is completely impossible to appear. Charlie can succeed in identifying the state with $100\%$ probability because he has learned that his state is actually $|\psi _2\rangle$. Thus, the results in Eq. (9c) are obtained.

Calculations for protocol (3) where probabilistic cloning occurs

Bob’s unitary cloning operation is given by [25]

$$\begin{aligned} U\left( |\Psi _i\rangle )|0\rangle \right) =\sqrt{\gamma _i}|\Psi _i\rangle |\Psi _i\rangle |\lambda _i\rangle +\sqrt{1-\gamma _i}|\beta \rangle |\beta \rangle |\lambda _0\rangle , \quad i=1,2, \end{aligned}$$

(24)

where $|0\rangle$ is a initialized state of the ancillas and $|\lambda _i\rangle$, $|\lambda _0\rangle$ are orthogonal states of the flag associated with successful cloning and failure cloning, respectively. $\gamma _i$ is the success probability of the cloning for the state $|\Psi _i\rangle$ and $|\beta \rangle$ is a genetic failure state.

Thus we can get an optimized successful cloning probability as

$$\begin{aligned}&{\mathrm{maximize:}}\ P^{\mathrm{cl}}=P_1\gamma _1+P_2\gamma _2 \end{aligned}$$

(25)

$$\begin{aligned}&{\mathrm{subject\ to:}}\quad s=\sqrt{\gamma _1\gamma _2}s^2\langle \lambda _1|\lambda _2\rangle +\sqrt{(1-\gamma _1)(1-\gamma _2)}. \end{aligned}$$

(26)

according to Eq. (24), where $|\lambda _1\rangle =|\lambda _2\rangle$ is required for optimal cloning [25].

If we set $\sin {\theta _i}=\sqrt{1-\gamma _i}$ ($i=1,2$) for $0\le \theta _i\le \pi /2$, the variables $x=\cos (\theta _1+\theta _2)$, $y=\cos (\theta _1-\theta _2)$ are further introduced. Eq. (26) is equivalent to $2s=(1+s^2)y-(1-s^2)x$. And then we find an intermediate parameter $\omega$ which satisfies

$$\begin{aligned} x=\frac{1-\left( 1+s^2\right) \omega }{s}, y=\frac{1-\left( 1-s^2\right) \omega }{s}. \end{aligned}$$

(27)

The range of the parameter $\omega$ is given in Eq. (31). It’s found that

$$\begin{aligned} \gamma _i=\frac{1}{2}\left[ 1+xy+(-1)^i\sqrt{\left( 1-x^2\right) \left( 1-y^2\right) }\right] . \end{aligned}$$

(28)

To seek the optimal value $P_{\max }^{\mathrm{cl}}$, the following equation should be satisfied $(P_{\max }^{\mathrm{cl}})'=\frac{dP_{\max }^{\mathrm{cl}}}{d\omega }=0$. This equation is equivalent to $P_1\gamma '_1+(1-P_1)\gamma '_2=0$, thus the following results are obtained

$$\begin{aligned} P_1=\frac{\gamma '_2}{\gamma '_2-\gamma '_1}, P_{\max }^{\mathrm{cl}}=\frac{\gamma '_2\gamma _1-\gamma '_1\gamma _2}{\gamma '_2-\gamma '_1}. \end{aligned}$$

(29)

where

$$\begin{aligned} \gamma '_i=\frac{d\gamma _i}{d\omega }=\frac{\sqrt{\gamma _i(1-\gamma _i)}}{s}\left[ -\frac{1+s^2}{\sqrt{1-x^2}}+(-1)^i\frac{1-s^2}{\sqrt{1-y^2}}\right] . \end{aligned}$$

(30)

And then, the conditional probabilities $P_i^{\mathrm{cl}}$ ($i=1,2$) of $|\Psi _i\rangle$ for the following two discriminations can be obtained as $P_i^{\mathrm{cl}}=\frac{P_i\gamma _i}{P_1\gamma _1+P_2\gamma _2}$. Hence, for the optimized successful cloning probability, $P_i$, $P_i^{\mathrm{cl}}$, $P_{\max }^{\mathrm{cl}}$, $P_{b,\max }^{\mathrm{cl}}$ and $P_{c,\max }^{\mathrm{cl}}$ are all obtained as parametric functions of $\omega$ with the range

$$\begin{aligned} \omega _1\le \omega \le \omega _2,\quad \omega _1=\frac{1}{1+s},\quad \omega _2=\frac{1}{1+s^2}, \end{aligned}$$

(31)

where $\omega _1$ and $\omega _2$ correspond to the cases for $P_1=P_2=\frac{1}{2}$ and $P_1=0$, respectively.

At last, the optimal success probability for both Bob and Charlie to identify the state is obtained as

$$\begin{aligned}&{\mathrm{maximize:}}\ P^{(3)}=P_{\max }^{\mathrm{cl}}\left[ P_1^{\mathrm{cl}}\left( 1-q_1^b\right) +P_2^{\mathrm{cl}}\left( 1-q_2^b\right) \right] \left[ P_1^{\mathrm{cl}}\left( 1-q_1^c\right) \right. \nonumber \\&\left. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +P_2^{\mathrm{cl}}\left( 1-q_2^c\right) \right] \end{aligned}$$

(32)

$$\begin{aligned}&{\mathrm{subject\ to:}}\ q_1^cq_2^c=q_1^bq_2^b=s^2, \quad q_1^b,q_2^b,q_1^c, q_2^c\in \left[ s^2,1\right] . \end{aligned}$$

(33)

Thus, we can acquire the results in Eq. (11) analytically.

Optimal probability for at least one of Bob and Charlie succeeding in identifying the states

It is obvious that the optimized probability $P_{\max }^*$ for one of their succeeding in discrimination for protocol (1) and (2) is equivalent to the results in Eq. (8). For SSD protocol, we can obtain the optimization as

$$\begin{aligned}&{\mathrm{maximize:}}\ P^{SSD*}=P_1\left( 1-q_1^bq_1^c\right) +P_2\left( 1-q_2^bq_2^c\right) \end{aligned}$$

(34)

$$\begin{aligned}&{\mathrm{subject\ to:}}\ P_1\in (0,1/2],\ q_1^bq_2^b=s^2/t^2,\ q_1^cq_2^c=t^2, \nonumber \\&\quad \quad \quad \qquad \qquad \qquad q_1^b,q_2^b\in \left[ s^2/t^2,1\right] ,\ q_1^c,q_2^c\in \left[ t^2,1\right] . \end{aligned}$$

(35)

Thus, this result is also equal to $P_{\max }^{(1)}$. For protocol (3), the maximal probability is derived as

$$\begin{aligned}&{\mathrm{maximize:}}\ P^{(3)*}=1-\left( P_1^{\mathrm{cl}}q_1^b+P_2^{\mathrm{cl}}q_2^b\right) \left( P_1^{\mathrm{cl}}q_1^c+P_2^{\mathrm{cl}}q_2^c\right) \end{aligned}$$

(36)

$$\begin{aligned}&{\mathrm{subject\ to:}}\ q_1^bq_2^b=q_1^cq_2^c=s^2,\ q_1^b, q_2^b, q_1^c, q_2^c\in \left[ s^2,1\right] ,\ P_1\in \left( 0,1/2\right] \nonumber \\ \end{aligned}$$

(37)

Thus, the result in Eq. (12) can be easily obtained.