- Email: [email protected]

Optimum unambiguous discrimination between subsets of quantum states Daowen Qiu a,b a Department of Computer Science, Zhongshan University, Guangzhou 510275, People’s Republic of China b State Key Laboratory of Intelligent Technology and Systems, Department of Computer Science and Technology, Tsinghua University,

Beijing 100084, People’s Republic of China Received 25 November 2002; received in revised form 19 January 2003; accepted 28 January 2003 Communicated by P.R. Holland

Abstract Given n non-orthogonal quantum states |ψ1 , |ψ2 , . . . , |ψn having respective a priori probabilities, we propose a scheme for unambiguously distinguishing between subsets {|ψ1 , |ψ2 , . . . , |ψk } and the remainder {|ψk+1 , |ψk+2 , . . . , |ψn }. Particularly, when k = 1 and n = 3, we calculate the optimum probability for this discrimination, which is independent of the method of Lagrange multiplier used in the existing literature. Correspondingly, the optimal measurement for yielding the optimal solution is constructed and, some examples are included. Both the IDP limit and the Jaeger and Shimony’s bound are the special cases of our solution. 2003 Elsevier Science B.V. All rights reserved. PACS: 03.67.-a; 03.65.Ta Keywords: Unambiguous state discrimination; Quantum measurement

1. Introduction In quantum information processing, one of the intrinsic attributes is that non-orthogonal quantum states cannot be infallibly distinguished, which is especially interesting for its potential applications to cryptography [1]. However, as is known, it is possible for discriminating between non-orthogonal quantum states via probabilistic or approximate fashions and, various discrimination programs have been emerging. E-mail addresses: [email protected], [email protected] (D. Qiu).

Among those proposed and developed strategies for identifying between non-orthogonal quantum states, unambiguous state discrimination has likewise intrigued lots of authors, and produced plentiful achievements (see, for example [2] and the references therein). Unambiguous discrimination originated from the works of Ivnovic [3], Dieks [4] and Peres [5], who initially distinguished two non-orthogonal states |φ and |ψ with the same prior probabilities p and q, i.e., p = q = 1/2, and derived the maximum probability of success called IDP limit as 1 − |φ|ψ|. Subsequently, Jaeger and Shimony [6] generalized it to the case of unequal a priori probabilities and obtained the

0375-9601/03/$ – see front matter 2003 Elsevier Science B.V. All rights reserved. doi:10.1016/S0375-9601(03)00193-2

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√ results as 1 − 2 pq |φ|ψ| or p(1 − |φ|ψ|) in case p q. Remarkably, Barnett and Andersson [7] calculated these results of late, by exploiting the no-signal condition, independent of any quantum measurement transformation. Actually, in recent years, unambiguous state discrimination has undergone intriguing extensions and further development. In reality, various measurement schemes have been suggested; and some bounds on the probabilities for unambiguous discrimination have been derived. Peres and Terno [8] discussed in detail the problem of optimal distinction of three states having arbitrary a priori probabilities. Chefles [9] showed that a set {|ψi } of states is amenable to unambiguous state discrimination, if and only if they are linearly independent. The optimal unambiguous discrimination among linearly independent symmetric states was investigated by Chefles and Barnett [10]. But indeed the optimal solution is unknown yet. However, deriving some upper bounds on the success probability for unambiguous discrimination among states is always possible [11,12], and the recent derived bound is expressed as: 2 n pi pj ψi |ψj 1− (1) n−1 i =j

for any quantum states |ψi , i = 1, 2, . . . , n with respective a priori probabilities pi . Also, it is worth mentioning that many of the theoretically discovered optimal schemes for discrimination have been experimentally realized [13–15]. A variant of discriminating among states is the distinction between subsets of states, say {|ψ1 , |ψ2 , . . . , |ψk } and {|ψk+1 , |ψk+2 , . . . , |ψn }. In reality, we may meet such a problem on state discrimination. Assume a quantum system is described by one of states |ψi with a priori probabilities ηi (i = 1, 2, . . . , n), and, furthermore, we also know this state must belong to either {|ψ1 , |ψ2 , . . . , |ψk } or {|ψk+1 , |ψk+2 , . . . , |ψn }. Then we need search a procedure to identify which subset the state lives in. Like state discrimination, there are usually alternative strategies for processing subset discrimination, that is, ambiguity [16] and unambiguity [17]. In [17], the authors considered the case of unambiguously discriminating between {|ψ1 } and {|ψ2 , |ψ3 } with respective a priori probabilities η1 , η2 , η3 ; they derived that if |ψ1 is lin-

early independent from either |ψ2 or |ψ3 , then the optimum failure probability is as P=

3

ηi qi ,

(2)

i=1

where η2 η3 2 2 q = 1 η1 |O12 | + η1 |O13 | , 2 q2 = η2 |O212 | η3 , 2 η1 |O12 | + η1 |O13 | |O13 |2 , q 3 = η2 2 η3 2 η1 |O12 |

(3)

+ η |O13 | 1

with Oij = ψi |ψj ; and these solutions are subject to the constraints that |O12 O13 | ηη21 |O12 |2 + ηη31 |O13 |2 1, (4) qi 1, i = 1, 2, 3. Otherwise, if some inequalities in Eq. (4) are violated, then these violated inequalities are rethought of as equalities, and the solution to qi (i = 1, 2, 3) expressed by Eq. (3) is correspondingly changed and, consequently the other expressions for P will be accordingly derived. That is to say, the optimum solution for P may be considered by dividing into several cases, but the details were omitted in [17]. Particularly, the optimum solution in [17] is calculated via the method of Lagrange multiplier and, the calculation procedure is considerably complex. Especially, when the number of all distinguished states is more than three, the derivation will be very tedious. So, in this Letter, our goal is to present another way to calculate the optimum solution for discriminating between subsets of states, notably, which is independent of Lagrange multipliers. Our results are consistent with those in [17] described by Eqs. (2)–(4), but more explicit than the above expressions. The organization of this Letter is as follows. In Section 2, we discuss the optimum scheme for discriminating between subsets of states. Then, in Section 3 we calculate and specify the optimum solution for discriminating between {|ψ1 } and {|ψ2 , |ψ3 } with respective a priori probabilities η1 , η2 , η3 . The optimum measurement is derived in Section 4, in which some examples are described. Finally, some remarks are made in Section 5.

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2. Optimum strategy for discriminating between subsets

success and of failure are, respectively, as follows.

Given n non-orthogonal quantum states |ψ1 , |ψ2 , . . . , |ψn having respective a priori probabilities, η1 , η2 , . . . , ηn , assume that both Alice and Bob know the n states and the corresponding probabilities chosen. Now Alice chooses a state |ψt with probability ηt and gives it to Bob. Moreover, suppose both Alice and Bob know that this chosen state belongs to either subset S1 = {|ψ1 , |ψ2 , . . . , |ψk } or subset S2 = {|ψk+1 , |ψk+2 , . . . , |ψn }, but Bob do not know exactly in which one it is. So then Bob’s task is to identify it. From a viewpoint of von Neumann measurement, i.e., projective measurement [18], if ψi |ψj = 0 for each i ∈ {1, 2, . . . , k} and each j ∈ {k + 1, k + 2, . . . , n}, then one can reliably distinguish subset S1 from subset S2 with the projectors PS1 and PS2 onto subspaces spanned by S1 and S2 , respectively, since for each |ψk ∈ S1 , ψk |PS1 |ψk = 1, and 0 otherwise; and a similar result for PS2 likewise holds. In general, for non-orthogonal states |ψ1 , |ψ2 , . . . , |ψn , a fundamental scheme for unambiguously discriminating between the above sets S1 and S2 is by designing some general measurement operators Mi with i = 0, 1, 2, that is, the usual bounded linear operators Mi acting on the space spanned by {|ψ1 , |ψ2 , . . . , |ψn }, but subject to the constraints of {Ei } having a POVM formalism where Ei = Mi† Mi , more exactly, satisfying the completeness equation [18]

P=

2

Mi† Mi = I,

(5)

i=0

where I denotes identity operator. Moreover, it is restricted that

(1) † ψk |M1 M1 |ψk = rk , |ψk ∈ S1 , (6) 0, |ψk ∈ S2 , and

(2) ψk |M2† M2 |ψk = rk , |ψk ∈ S2 , 0, |ψk ∈ S1 , (i)

(7) (i)

for some real numbers rk 0, i = 1, 2, where rk (i = 1, 2; k = 1, 2, . . . , n) represent the success probabilities, whereas ψk |M0† M0 |ψk (k = 1, 2, . . . , n) denote the failure ones. So, the average probabilities of

k

191

ηi ψi |M1† M1 |ψi

i=1 n

+

ηj ψj |M2† M2 |ψj ,

(8)

j =k+1

Q=

n

ηi ψi |M0† M0 |ψi .

(9)

i=1

Obviously, from Eqs. (5), (8) and (9) it follows that P + Q = 1 always holds. In what follows, we aim to provide a scheme for calculating the optimal values of P and Q, that is, the maximum P denoted by Popt or minimum Q by Qopt , subject to the constraints of Eqs. (5)–(7). Before processing the calculation, a natural query raised is when sets S1 and S2 above can be unambiguously discriminated with non-zero success probabilities, which is answered by the following lemmas. Lemma 1. For quantum states |ψi with respective non-zero a priori probabilities ηi (i = 1, 2, . . . , n), subsets S1 and S2 are as above, then the following three statements are equivalent: (i) S1 and S2 can be unambiguously distinguished with non-zero success probability, that is to say, there exists a general measurement {Mm } (i = 0, 1, 2) satisfying Eqs. (5)–(7) such that P = 0. (ii) Hilbert spaces H1 and H2 spanned by S1 and S2 , respectively, are unequal. (iii) There is at least a vector |ψi0 in S1 linearly independent from S2 , i.e., |ψi0 cannot be linearly represented by those vectors in S2 , or there exists at least a vector |ψj0 in S2 incapable of being linearly represented by those vectors in S1 . > 0 and We may further ask if we require that rk(1) 1 rk(1) > 0 for at least one k and one k , that is to say, the 1 2 2 probabilities for identifying the state in both subsets are non-zero, then how about the results? Indeed, we have Lemma 2. For quantum states |ψi with respective non-zero a priori probabilities ηi (i = 1, 2, . . . , n),

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subsets S1 and S2 are as above, then the following three statements are equivalent: (i) There exists a general measurement {Mm } (i = (1) 0, 1, 2) satisfying Eqs. (5)–(7) such that rk1 > 0

> 0 for at least one k1 and one k2 . and rk(1) 2 (ii) Hilbert spaces H1 and H2 spanned by S1 and S2 , respectively, satisfy that H1 \H2 = ∅ and H2 \H1 = ∅. (iii) There is at least a vector |ψi0 in S1 such that |ψi0 cannot be linearly represented by those vectors in S2 , and there exists at least a vector |ψj0 in S2 incapable of being linearly represented by those vectors in S1 . The proofs for the above lemmas are put in Appendix A. In this Letter, we focus on calculating the optimal solution for the case of distinguishing between {|ψ1 } and {|ψ2 , |ψ3 }. With Lemma 1, we obtain directly that, for quantum states |ψ1 , |ψ2 , |ψ3 , P = 0 if and only if |ψi = ci |φ

(10)

for some complex numbers ci (i = 1, 2, 3) and unit vector |φ. So, based on the above discussion, the optimum measurement {Mm } (m = 0, 1, 2) must entail that (i) {M1 |ψ1 } and {M2 |ψ2 , M2 |ψ3 } can be perfectly distinguished; and (ii) the probability of discriminating between {M0 |ψ1 } and {M0 |ψ2 , M0 |ψ3 } is zero. But the measurement {Mi } satisfying Eqs. (5)–(7) and the just presented two conditions (i) and (ii) may not be the optimal one. So, we aim to calculate the optimum solution not by first constructing the optimal measurement, but through deriving the minimum value Qopt under the action of these conditional measurements. Equivalently, as an optimum measurement, besides Eqs. (5)–(7), {Mm } (m = 0, 1, 2) must additionally satisfy ψj |M2† M1 |ψ1 = 0,

(11)

for j = 2, 3; and M0 |ψi = ci |φ,

(12)

for some complex numbers ci (i = 1, 2, 3) and unit vector |φ with |ci |2 1, (i = 1, 2, 3). That is to say,

an optimal measurement {Mm } has to satisfy Eqs. (5)– (7), (11) and (12), but the converse may not be true. However, we can still calculate the optimum solution to discrimination via using all those possible optimal measurements, that is, choosing an optimal value in all those solutions resulting from those conditional measurements. Therefore, the optimal failure probability Qopt is the minimum of those Q deriving from the measurements satisfying Eqs. (5)–(7), (11) and (12), which can be achieved by adopting an algebraic and analytical method of calculation.

3. Optimum solution In this section, we calculate the optimal probability for discriminating S1 = {|ψ1 } from S2 = {|ψ2 , |ψ3 } under Eqs. (5)–(7), (11) and (12). Scenario I. If |ψ1 ∈ span S2 , then, by Eq. (6) we know that for any measurement satisfying Eqs. (5)– (7), (11) and (12), M1 |ψ1 = 0, that is to say, the success probability for identifying |ψ1 is zero and, therefore, |c1 |2 = 1. With Eqs. (5)–(7) and (12), we have 2 |cj |2 = ψ1 |ψj , j = 2, 3, (13) and hence, we obtain the optimal average probability as Qopt = η1 |c1 |2 +

3

ηj |cj |2

j =2

= η1 +

3

2 ηj ψ1 |ψj .

(14)

j =2

Remark 1. In this situation, Eqs. (3) correspond to q1 = 1, q2 = |ψ1 |ψ2 |2 , and q3 = |ψ1 |ψ3 |2 . Therefore, Eq. (14) agrees with Eq. (2). / span S2 , then we set Scenario II. If |ψ1 ∈ ψ1 |ψj = max ψ1 |ψj : j = 2, 3 0

(15)

for some 2 j0 3. Denote x = |cj0 |2 . Then, with Eqs. (5)–(7) and (12) we have 2 |c1 |2 = |ψ1 |ψj0 | , x (16) 2 |cj |2 = |ψ1 |ψj | x, 2 j 3. |ψ |ψ |2 1

j0

D. Qiu / Physics Letters A 309 (2003) 189–197

So we have

Case 2. If

Q = Q(x)

3

|ψ1 |ψj0 |2 |ψ1 |ψj |2 + = η1 ηj x. x |ψ1 |ψj0 |2 3

(17)

j =2

j =2

3 |ψ1 |ψj0 |2 |ψ1 |ψj |2 = η1 + ηj xopt xopt |ψ1 |ψj0 |2 j =2 3 2 = 2 η1 ηj ψ1 |ψj ,

(18)

j =2

dQ dx

= 0 we can obtain

η1 |ψ1 |ψj0 |4 2 = 3 . xopt 2 j =2 ηj |ψ1 |ψj |

(19)

Since |ci |2 1 for 1 i 2, with Eqs. (16) and (19) we have ψ1 |ψj 2 x 1. (20) 0 Next we consider three cases: Case 1. If η1

3

2 ηj ψ1 |ψj ,

(21)

where xopt =

2 ηj ψ1 |ψj .

η1 |ψ1 |ψj0 |4

3 . 2 j=2 ηj |ψ1 |ψj |

j =2

which is also consistent with Eq. (26). (22)

By inequality (20), inequality (22) implies that in this case, x in Eq. (17) cannot achieve its optimal value. With Eq. (18) we know dQ dx x=x0 > 0 when x0 belongs to the domain defined by inequality (20). So, Q(x) increases in this domain and, therefore, Q = Q(x) gets its minimum value when x = |ψ1 |ψj0 |2 , that is, 2 Qopt = Q ψi |ψj0 = η1 +

(26)

Remark 3. In Case 2, inequality (24) implies that all the inequalities in (4) hold. So, from Eq. (3) it follows that 3 3 2 ηi qi = 2 η1 ηj ψ1 |ψj , P= i=1

3

(25)

j =2

j =2

then by Eq. (19), we have 2 xopt ψ1 |ψj0 .

(24)

j =2

Qopt = Q(xopt )

3

by setting

3 2 |ψ1 |ψj |2 η ηj ψ1 |ψj , 1 4 |ψ1 |ψj0 |

then Q can achieve its optimal value, that is,

Due to −|ψ1 |ψj0 |2 |ψ1 |ψj |2 dQ = η1 + ηj , dx x2 |ψ1 |ψj0 |2

ηj

193

(23)

j =2

Remark 2. In Case 1, with inequality (4), inequality (21) implies that η2 η3 |O12 |2 + |O13 |2 = 1. η1 η1 So, in Eq. (3), q1 = 1, q2 = |ψ1 |ψ2 |2 , and q3 = |ψ1 |ψ3 |2 , which shows the consistence of Eq. (23) with Eq. (2).

Case 3. If 3 j =2

ηj

|ψ1 |ψj |2 η1 , |ψ1 |ψj0 |4

(27)

2 1, and so, x = 1 and thus then xopt

Qopt = Q(1) 3 2 |ψ1 |ψj |2 ηj . = η1 ψ1 |ψj0 + |ψ1 |ψj0 |2

(28)

j =2

Remark 4. With inequality (27), we can directly check that Eq. (28) agrees with Eq. (2). So, by utilizing general quantum measurement, together with the analytical method, we have precisely derived the optimal probability for unambiguously distinguishing subset S1 = {|ψ1 } from subset S2 = {|ψ2 , |ψ3 }, which is embodied in Eqs. (14), (23), (26) and (28) in terms of various cases, and, really

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D. Qiu / Physics Letters A 309 (2003) 189–197

consistent with those described by Eqs. (2)–(4) that were obtained in [17]. But, clearly, our conclusions are more evident than those in [17]. However, as we know, a likewise important problem is how to get these information as best as possible, that is to say, what are the corresponding optimal measurements to produce such these optimal solutions? This is the main topic of next section. To conclude this section, we consider two special cases. If η3 = 0, that is to say, the problem reduces to unambiguous discrimination between |ψ1 and |ψ2 , then Eqs. (23), (26) and (28), respectively reduce to Eqs. (29), (30) and (31) as follows: 2 Qopt = η1 + η2 ψ1 |ψ2 , (29) if 2 η1 η2 ψ1 |ψ2 ;

|Ψ =

and √ Qopt = 2 η1 η2 ψ1 |ψ2 ,

3

xi |χi ,

i=1

(30)

in case η2 ψ1 |ψ2 2 ; η η 1 2 |ψ1 |ψ2 |2 as well as 2 Qopt = η1 ψ1 |ψ2 + η2 ,

unambiguous subset discrimination. Particularly, we aim to construct some general optimal measurement to produce the optimal solution for unambiguous discrimination between subsets {|ψ1 } and {|ψ2 , |ψ3 }. Also, to demonstrate our method, some examples are described. According to Lemma 1 in Section 2, if H1 spanned by S1 equals to H2 spanned by S2 , then the success probability for discriminating S1 from S2 is zero, so we may take the corresponding measurement as {Mm } where M1 = M2 =√0 and M0 = I , or M1 = M2 = |ψψ| and M0 = I − |ψψ| for some unit vector |ψ orthogonal to H1 and H2 . In general, suppose H1 = H2 , and |ψi (i = 1, 2, 3) live in 3-dimensional space. Let |χ1 , |χ2 , |χ3 be an orthonormal basis of a Hilbert space containing |ψi (i = 1, 2, 3). Setting

(31)

when |ψ η|ψ2 |2 η1 , which really describe the failure 1 2 limits for the case of unambiguously distinguishing |ψ1 from |ψ2 by Jaeger and Shimony [6]. More specially, if η1 = η2 = 1/2, then, due to |ψ1 |ψ2 |2 1, the above solution Eq. (30) becomes exactly the well-known IDP limit [3–5] |ψ1 |ψ2 | of the failure probability of unambiguous discrimination between |ψ1 and |ψ2 .

and letting transformation M0 = |φΨ | for some unit vector |φ with xi to be determined, then in terms of the preceding conditions described by inequalities (21), (24) and (27), and the corresponding solutions expressed by Eqs. (23), (26) and (28), respectively, we can also derive respective M0 by determining the values of xi , and hence further calculate M1 and M2 such that {Mm } constitutes a measurement satisfying Eqs. (5)–(7), (11) and (12) as well as yielding such a solution. Next we deal with this problem in detail for unambiguous discrimination between subsets {|ψ1 } and {|ψ2 , |ψ3 }. Scenario I. If |ψ1 ∈ span S2 , then by recalling Eq. (14) we know Qopt = η1 +

3

2 ηj ψ1 |ψj .

j =2

4. Discussion on optimum measurement and examples For n non-orthogonal quantum states |ψ1 , |ψ2 , |ψ3 having respective a priori probabilities, we have derived the optimal probability for unambiguously distinguishing between S1 = {|ψ1 } and the remainder S2 = {|ψ2 , |ψ3 }. In this section, we try to provide a scheme for deducing a general measurement for

Let |ψ1 , |χ2 , |χ3 be an orthonormal basis of a Hilbert space containing S1 ∪ S2 . Then set M0 |Ψ = |φΨ | for some unit vector |φ, where |Ψ = with xi to be determined. In terms of Ψ |ψ1 2 = 1, Ψ |ψj 2 = ψ1 |ψj 2 , 2 j 3,

3

i=1 xi |χi

D. Qiu / Physics Letters A 309 (2003) 189–197

we may take |Ψ = |ψ1 and M = |φψ |, 0 1 M1 = 0, M2 = |χ2 χ2 | + |χ3 χ3 |.

(32)

Then M1 |ψj = 0 for 2 j 3, M2 |ψ1 = 0,

2 † m=0 Mm Mm = I ; as well as 3

ηi ψi |M0† M0 |ψi = η1 +

3

2 ηj ψ1 |ψj ,

j =2

i=1

η1 ψ1 |M1† M1 |ψ1 +

3

ηj ψj |M2† M2 |ψj

j =2

=

3

2 ηj 1 − ψ1 |ψj .

j =2

So the constructed measurement {Mm } (m = 0, 1, 2) satisfies exactly Eqs. (5)–(7), (11) and (12), as well as yields the above optimal solution. Example 1. Let |0, |1, |2 be an orthonormal basis of Hilbert space H. Suppose quantum states |ψ1 = √ √ |0, |ψ2 = |0−|1−|2 , and |ψ3 = |0+|1+|2 with the 3 3 equal a priori probabilities, i.e., η1 = η2 = η3 = 1/3. What are the optimal probability and the corresponding optimal measurement producing such a probability for unambiguously distinguishing between {|ψ1 } and {|ψ2 , |ψ3 }? First we note that |ψ1 ∈ span{|ψ2 , |ψ3 }, and ψ1 |ψ2 2 = ψ1 |ψ3 2 = 1 . 3 It follows from Eq. (14) that the optimal failure probability for distinguishing between {|ψ1 } and {|ψ2 , |ψ3 } is as Qopt = η1 +

3 j =2

2 5 ηj ψ1 |ψj = . 9

The optimal measurement is {Mm } where M0 = |00|, M1 = 0 and M2 = |11| + |22|. / span S2 , then we consider Scenario II. If |ψ1 ∈ three cases corresponding to the preceding discussion in Section 2.

195

Case 1. If inequality (21) holds, then it is similar to Eq. (32), by taking M = |φψ |, 0

1

M1 = 0, M2 = |χ2 χ2 | + |χ3 χ3 |,

(33)

where |ψ1 , |χ2 , |χ3 be an orthonormal basis of a Hilbert space containing |ψi (i = 1, 2, 3), and |φ a unit vector. Example 2. Let |0, |1, |2 be an orthonormal basis of Hilbert space H. Suppose quantum states |ψ1 = |0+|1 √ √ √ , |ψ2 = |1+|2 , and |ψ3 = |1−|2 with re2 2 2 spective prior probabilities, η1 = 1/9, η2 = η3 = 4/9. What are the optimal probability and the corresponding optimal measurement producing such a probability for unambiguously distinguishing between {|ψ1 } and {|ψ2 , |ψ3 }? First we note that |ψ1 ∈ / span{|ψ2 , |ψ3 }, and ψ1 |ψ2 2 = ψ1 |ψ3 2 = 1 . 4 It is direct to check that inequality (21) is satisfied. So, we have that Qopt = η1 +

3 j =2

2 1 ηj ψ1 |ψj = . 3

(34)

√ √ Because |0+|1 , |0−|1 , |2 constitute an orthonormal 2 2 basis of Hilbert space H containing the above |ψi (i = 1, 2, 3), so the corresponding optimal measurement {Mm } can be constructed as M0 = |φψ1 |, M1 = 0, and M2 = |χ2 χ2 | + |22|, where |χ2 = |0−|1 √ . 2

Case 2. If inequality (24) holds, then we first give an orthonormal basis as |χ1 , |χ2 , |χ3 for the space including |ψi (i = 1, 2, 3). Set

M0 = |φΨ | where |φ is a unit vector and, |Ψ = nl=1 xl |χl , with xl to be determined in terms of the following equations: ψ |ψ = ψ |Ψ Ψ |ψ , 2 j 3, 1 j 1 j 2 ηj † n ψ1 |M0 M0 |ψ1 = j =2 η1 ψ1 |ψj , √ (35) 2 |ψ |ψ | η1 , ψj |M0† M0 |ψj = n 1 j 2 j=2 ηj |ψ1 |ψj | 2 j 3.

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Then, M1 and M2 can be derived from the following equations: ψ1 |M1† M1 |ψ1 + ψ1 |M0† M0 |ψ1 = 1, (36) ψj |M1† M1 |ψj = 0, 2 j 3, and † † ψj |M2 M2 |ψj + ψj |M0 M0 |ψj = 1, 2 j 3, ψ1 |M2† M2 |ψ1 = 0.

(37)

We use an example to demonstrate this procedure of calculation. Example 3. Let |0, |1, |2 be an orthonormal basis of Hilbert space H. Suppose quantum states |ψ1 = |0+|1 √ √ √ , |ψ2 = |1+|2 , and |ψ3 = |1−|2 with the 2 2 2 equal a priori probabilities, i.e., η1 = η2 = η3 = 1/3. What are the optimal probability and the corresponding optimal measurement producing such a probability for unambiguously distinguishing between {|ψ1 } and {|ψ2 , |ψ3 }? / span{|ψ2 , |ψ3 }, and Since |ψ1 ∈ ψ1 |ψ2 2 = ψ1 |ψ3 2 = 1 , 4 inequality (24) holds and, with Eq. (26) the optimal failure probability is as √ 2 . Qopt = 3 Assume M0 = |0Ψ | where |Ψ = x0 |0 + y0 |1 + z0 |2. Then according to Eq. (35), we have y0 +z0 1 √ = 2, 2 y −z 0√2 0 = 12 , x +y 2 √2 (38) 0√ 0 = , 2 2 √ y0√−z0 2 = y0√+z0 2 = 2 . 2

2

4

By solving the above equations (38) we obtain (Notably it may be not unique) √ √ 4 x0 = 2 − 22 , √ y0 = 22 , z0 = 0. Similarly, set M1 = |1Ψ1 |, where |Ψ1 = x1 |0 + y1 |1 + z1 |2. Then by virtue of Eq. (36) we aim to

solve the equation: x1√+y1 2 + x0√+y0 2 = 1, 2

2

|y1 + z1 |2 = |y1 − z1 |2 = 0, √ 1/2 and obtain √ 1/2x1 = (2 − 2) , y1 = z1 = 0. So M1 = (2 − 2 ) |10|. By utilizing the analogous method, one can derive M2 = |2Ψ2 | where √ 1/2 √ 1/2 2 2 |Ψ2 = 1 − |0 − 1 − |1 + i|2. 2 2 We can directly check that the calculated {Mm } is the optimal measurement satisfying Eqs. (5)–(7), √ (11) and (12), and, producing the solution Qopt = 2/3. Case 3. If inequality (27) holds, then with the method exactly similar to that in Case 2, one can construct the corresponding measurement to yield the solution by Eq. (28). We here omit the details.

5. Conclusion In conclusion, for non-orthogonal quantum states |ψ1 , |ψ2 , . . . , |ψn , with the a priori probabilities ηi of states |ψi (i = 1, 2, . . . , n), we propose a scheme for unambiguously distinguishing between subsets {|ψ1 , |ψ2 , . . . , |ψk } and the remainder {|ψk+1 , |ψk+2 , . . . , |ψn }, and, a sufficient and necessary condition for such a discrimination having non-zero success probabilities is presented. In particular, we have calculated the optimum probability of unambiguously distinguishing between subsets {|ψ1 } and {|ψ2 , |ψ3 }, and, the derived results are consistent with those in [17], while the expressions for our conclusions are more specific. Remarkably, our method is different from that in [17] on unambiguous discrimination of |ψ1 from {|ψ2 , |ψ3 }, where the method of Lagrange multiplier was applied. Also, we have constructed the corresponding optimal measurement to realize the optimal solution and, some examples have been described. However, we have not calculated optimal probability for discrimination between {|ψ1 , |ψ2 , . . . , |ψk } and {|ψk+1 , |ψk+2 , . . . , |ψn } in general situation, especially in case k 2 and n 4. It is still open and worth further investigating. Meanwhile, we do not yet

D. Qiu / Physics Letters A 309 (2003) 189–197

know the concrete results for the optimum probability of unambiguous discrimination among n states. We take into account that the method utilized in this Letter may be applicable to this issue, and shall study it elsewhere.

Acknowledgements I am grateful to the referee for invaluable comments and suggestions that help me to improve the presentation of this Letter. This work was supported by the National Key Project for Basic Research, the Natural Science Foundation of Guangdong Province of China (Grant No. 020146) and the Young Foundation of Zhongshan University.

197

On the other hand, suppose (iii) holds, for example, there is |ψi0 ∈ S1 incapable of being linearly represented by S2 . Let |ψ0⊥ be a unit vector belonging to the orthogonal complement space H2⊥ of H2 spanned by S2 , such that ψ0⊥ |ψi0 = 0 (this is absolutely possible since |ψi0 is independent of S2 ). Then, set transformations M1 = |ψ0⊥ ψ0⊥ |, M2 = 0,

and M0 = I − M1† M1 . It is easy to check that measurement {Mm } satisfies Eqs. (5)–(7) and P = 0. This has completed the proof. ✷

The proof of Lemma 2 is exactly similar to Lemma 1, so we omit the details.

References Appendix A Proof of Lemma 1. The equivalence between (ii) and (iii) is clear, so we show that (i) is equivalent to (iii). If (i) holds, then from Eqs. (5) and (6) it follows that there is at least one non-zero term in the summation of Eq. (8), for example, ηi0 ψi0 |M1† M1 |ψi0 = 0 for some i0 ∈ {1, 2, . . . , k}. We claim |ψi0 cannot be linearly represented by S2 . Indeed, suppose |ψi0 =

n

cj |ψj ,

[14]

j =k+1

then with Eq. (5) we have M1 |ψi0 =

n

[1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]

cj M1 |ψj = 0,

j =k+1

which contradicts ψi0 |M1† M1 |ψi0 = 0. Therefore, the above claim is correct, and consequently, (iii) holds.

[15] [16] [17] [18]

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