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In this paper, we show that any 2D measurement dim B. We consider ﬁnite-dimensional systems; NA H can be realized by one-way LOCC no matter how many and NB are ﬁnite. Assume that Alice and Bob share one POVM operators it has. Our result answers the above of a known collection of L quantum states represented L question: A global measurement is not needed for a 2D by density operators ρl l=1 on and want to optimally measurement in ﬁnite-dimensional systems, regardless of discriminate between{ them} in aH certain optimality crite- the optimality criterion used. rion. Our main result is that any 2D measurement (in It is worth noting that the problem of realizing a mea- ﬁnite-dimensional systems) can be realized by one-way surement by one-way LOCC is closely related to real- LOCC (see Corollary 3), which indicates that any op- izing a quantum receiver. Realization of an optimal or timal discrimination of ρl can be realized by one-way suboptimal receiver for optical states using linear opti- LOCC. { } cal feedback (or feedforward) and photon counting has We can easily extend our result to multipartite systems been widely studied both theoretically and experimen- in a way similar to [2]. Here, let us imagine a tripar- tally [21–30]. This type of receiver performs an individ- tite system: Alice, Bob, and Charlie share two linearly ual measurement on each temporal or spatial slot. A independent quantum states ψ and φ , which can be measurement can be decomposed into such individual represented by | i | i measurements if it can be realized by one-way LOCC; ′ ′ thus, our result indicates that any 2D measurement can ψ = pn A qn BC , be decomposed into individual measurements, at least | i n | i | i X in ﬁnite-dimensional systems. It is often important to ′ ′ φ = pn A rn BC , (2) investigate whether a measurement can be realized by | i n | i | i one-way LOCC to check whether it can be implemented X using only feasible resources when the whole system is instead of (1). In (2), Bob and Charlie are ﬁrst grouped spatially or temporally separated. as one party. Asuume that our main result, i.e., Corol- In Section II, we present some necessary preliminaries, lary 3, holds in a bipartite system; then, we can show that where we show that any 2D measurement can be real- any measurement on any tripartite 2D Hilbert space can ized by one-way LOCC if any measurement with ﬁnite also be realized by one-way LOCC. Indeed, Alice per- rank-one POVM operators on any 2D bipartite Hilbert forms a measurement on her system according to the space in which Alice’s subspace is two-dimensional can bipartite one-way LOCC protocol that we will propose be realized by one-way LOCC. In Section III, we recall in this paper and tells the result to Bob and Charlie. the idea of Walgate et al. [2], which provides a method Then, Bob and Charlie can again use the same protocol. for realizing a 2D projective measurement by one-way This argument can easily be extended to any multipar- LOCC. In Section IV, we consider realizing a 2D non- tite system, and thus, in the rest of paper, we consider projective measurement by one-way LOCC. We show only bipartite systems. that, by extending Walgate et al.’s idea, any measure- First, we show that our main reulst can be reduced ment with ﬁnite rank-one POVM operators on any 2D to a simpler one. For example, from [20], any quantum bipartite Hilbert space in which Alice’s subspace is two- measurement with a continuous set of outcomes (includ- dimensional can be realized by one-way LOCC (Proposi- ing the discrete outcomes) on a ﬁnite-dimensional Hilbert tions 6 and 8; also Theorem 2). We conclude the paper space is equivalent to a continuous random choice of mea- in Section V. surements with ﬁnite outcomes. Thus, it suﬃces to show that any 2D measurement with ﬁnite outcomes can be re- alized by one-way LOCC. We show the following lemma: II. PRELIMINARIES

We ﬁrst consider a bipartite system. Let ψ and φ Lemma 1 If any measurement with ﬁnite rank-one | i | i be two linearly independent quantum states shared by POVM operators on a 2D (2,N)-space (N is ﬁnite in- Alice and Bob. We can write, in general form, teger) can be realized by one-way LOCC, then any 2D measurement (in ﬁnite-dimensional systems) can be re- ψ = pn qn , A B alized by one-way LOCC. | i n | i | i X φ = pn A rn B , (1) Proof Assume that any measurement with ﬁnite rank- | i n | i | i X one POVM operators on a 2D (2,N)-space (denoted by where pn are quantum states of Alice, and qn 2) can be realized by one-way LOCC. {| iA} {| iB} H and rn are quantum states of Bob. pn , First, we show that any measurement on can be re- {| iB} {| iA} H2 qn , and rn are generally unnormalized and alized by one-way LOCC. From [20], any quantum mea- {| iB} {| iB} non-orthogonal. Let A = span( pn ) and B = surement, even if with a continuous set of outcomes, on H {| iA} H span( qn B , rn B ). Also, let be a 2D Hilbert 2 can always be realized as a random choice of extremal space{| spannedi } {| byi ψ} and φ . WeH denote such as measurementsH on , where an extremal measurement is | i | i H H2 a 2D (NA,NB)-space, where NA = dim A and NB = an extremal point of the set of all possible POVMs, which H 3 is a convex set. Moreover, from [19], an extremal mea- from each other and share a pair of particles in a state of surement on must be made of ﬁnite rank-one POVM either S or T . They want to perfectly discriminate H2 | i | 0i operators, apart from the trivial POVM Π1 = IH2 (IH2 between the orthogonal states S and T0 by one-way is the identity operator on ). The trivial{ POVM} is LOCC. This problem is identical| toi the problem| i of realiz- H2 obviously realized by one-way LOCC; thus, we consider ing the projective measurement S S , T0 T0 on only a nontrivial POVM. by one-way LOCC. If Alice simply{| i performs h | | i ah measure-|} H Next, we show that a 2D measurement Πm (on a 2D ment in the ONB + , , then Bob cannot dis- { } {| iA |−iA} (NA,NB)-space) can be realized by one-way LOCC. Let criminate between S and T ; for example, if the out- | i | 0i A and B be Alice’s and Bob’s Hilbert spaces, respec- come of Alice’s measurement is + , then Bob’s state is H H | iA tively. The case of NA 2 is trivial; assume that NA > 2. transformed into B, regardless of whether they share ≤ |−i Suppose without loss of generality that NA is even; other- S or T . Thus, Alice needs to use a proper ONB. S | i | 0i | i wise, expand Alice’s system into a (NA + 1)-dimensional and T0 are rewritten as Hilbert space. Alice’s system can be represented on the | i 0 1 + 1 0 tensor product of two- and (NA/2)-dimensional Hilbert S = − | iA | iB | iA | iB , spaces, denoted as A and A , respectively. Since | i √2 H 1 H 2 A B = A1 ( A2 B ) and dim A1 = 2, 0 0 1 1 H ⊗ H H ⊗ H ⊗ H H A B A B Πm can be realized by one-way LOCC between A T0 = | i | i − | i | i , (4) { } H 1 | i √2 and A B . Thus, it suﬃces to show that a measure- H 2 ⊗ H ment on a 2D subspace of A2 B can be realized by where 0 = ( + + )/√2, 1 = ( + H ⊗ H {| iα | iα |−iα | iα | iα − one-way LOCC. By repeating this procedure, the prob- α)/√2 (α A, B ) is the ONB in α. Alice may lem of realizing Πm by one-way LOCC is reduced to the |−i } ∈{ } H { } just perform a measurement in the ONB 0 A , 1 A and problem of realizing a measurement on a 2D (2,N)-space. tell the result to Bob, and he can then{| ﬁndi out| i which} Therefore, by the assumption, Πm can be realized by state they share by discriminating between 0 and 1 . { } B B one-way LOCC. From [2], for any 2D (2,N)-space, | ,i any ONB| i π , π⊥ in can be represented as theH following form In this paper, we will prove the following theorem: in{| Alice’si | i} properH ONB 0 , 1 : {| iA | iA} Theorem 2 Any measurement with ﬁnite rank-one π = 0 A η0 B + 1 A η1 B , POVM operators on a 2D (2,N)-space can be realized | i | i | i | i | i π⊥ = 0 ν + 1 ν , (5) by one-way LOCC. | i | iA | 0iB | iA | 1iB where ηk and νk are orthogonal for each k 0, 1 From Lemma 1 and Theorem 2, we can easily obtain the | iB | iB ∈{ } following corollary (proof omitted): but not necessarily normalized. We can see that (4) is a special form of (5). Similar to the above example, the projective measurement π π , π⊥ π⊥ can be real- Corollary 3 Any 2D measurement (in ﬁnite- ized by one-way LOCC if{| Alicei h | measures| i h her|} side of the dimensional systems) can be realized by one-way system in the ONB 0 , 1 and Bob discriminates LOCC. {| iA | iA} between ηk and νk . | iB | iB

III. REALIZATION OF 2D PROJECTIVE MEASUREMENT BY ONE-WAY LOCC IV. REALIZATION OF ANY 2D MEASUREMENT BY ONE-WAY LOCC In this section, using an example, we recall the idea of Now, we consider realizing a non-projective measure- Walgate et al. [2], which provides a way to realize a 2D M ment Πm m with ﬁnite rank-one POVM operators on projective measurement by one-way LOCC. Let ψ = S { } =1 | i | i a 2D (2,N)-space by one-way LOCC. Let us represent and φ = T0 , where H | i | i Π1 as + A B A + B S = | i |−i − |−i | i , Π1 = γ1 π π , (6) | i √2 | i h | ⊥ + A B + A + B with 0 <γ1 1 and π π = 1. Let π be a nor- T0 = | i |−i |−i | i , (3) malized vector≤ perpendicularh | i to π |so thati ∈ H π , π⊥ | i √2 | i {| i | i} is an ONB in . We choose an ONB 0 A, 1 A in A H ⊥ {| i | i } H and + α , α (α A, B ) is an orthonormal basis such that π and π are expressed in the form of (5). {| i |−i } ∈{ } (ONB) in α. In this example, = span( S , T ) (k) | i | i (0) (1) H H | i | 0i ⊆ Let B = span( ηk B , νk B); then, B = B B A B holds. We can easily see that S and T0 obviouslyH holds. Also,| i let| Pibe the orthogonalH H projection∪ H areH orthogonal.⊗ H If + and are the| spin-upi | andi | iα |−iα operator onto and IB be the identity operator on B. spin-down states of a spin-1/2 particle, then S and T0 Let H H are, respectively, singlet and triplet states of| twoi parti-| i cles. Suppose that Alice and Bob are spatially separated ηk = ηk ηk , νk = νk νk , k 0, 1 . (7) h | iB h | iB ∈{ } 4

From (5), we have In particular, setting cm = c2 for any m 3 in Lemma 5 gives the following proposition (proof in≥ Appendix B). η + η = π π = 1, 0 1 h | i ν + ν = π⊥ π⊥ = 1. (8) 0 1 h | i Proposition 6 Any measurement with ﬁnite rank-one Thus, we can assume without loss of generality (by suit- POVM operators on a 2D (2,N)-space can be realized ably permuting 0 and 1 ) that η ν . by one-way LOCC if | iA | iA 0 ≥ 0 γ η (1 γ )ν . (13) 1 ≥ 0 − − 1 0 A. Simple suﬃcient condition for realization by Note that if Πm is a projective measurement, then one-way LOCC { } since γ1 = 1 holds, (13) always holds. M As an example, we consider Πm m (M 3) on { } =1 ≥ In this subsection, we consider the case in which = span( S , T0 ), where Πm = 0 (i.e., γ1 > 0). (0) M (0) H | i | i 6 there exist Bob’s measurements Φm m on B and For example, a measurement minimizing the average er- { } =1 H (1) M (1) ror probability for the M quantum states αm S + Φm m on B such that for any m with 1 m M, { } =1 H ≤ ≤ M 2 2 { | i Πm is expressed by βm T0 m=1 with αm + βm = 1 can often be written as this| i} form. π and| |π⊥ |can| be written as (0) (1) | i | i Πm = P 0 0 A Φm + 1 1 A Φm P. (9) | i h | ⊗ | i h | ⊗ π = x S + y T0 , |⊥i | ∗i | ∗i In this case, Πm is realized by one-way LOCC when π = y S + x T0 , (14) Alice measures{ her} side of the system in the ONB | i − | i | i 2 2 0 A , 1 A , as shown in the following lemma. with some complex values x and y with x + y = 1, {| i | i } where ∗ denotes the complex conjugate.| Indeed,| | we| can M ⊥ Lemma 4 Any measurement Πm m=1 with rank-one easily verify that π , π is an ONB in . Substitut- {| i | i} H ⊥ POVM operators on a 2D (2,N{ )-space} can be realized ing (4) into (14), we can represent π and π in the | i | i by one-way LOCC if there exist Bob’s measurements form of (5) as (0) M (1) M Φm m and Φm m that satisfy (9). { } =1 { } =1 y 0 B x 1 B x 0 B y 1 B π = 0 A | i − | i + 1 A | i − | i , Proof We consider the following one-way LOCC mea- | i | i √2 | i √2 ∗ ∗ ∗ ∗ surement: Alice measures her side in the ONB ⊥ x 0 B y 1 B y 0 B x 1 B π = 0 A | i − | i 1 A | i − | i . 0 A , 1 A and reports the result k 0, 1 to Bob, | i | i √2 − | i √2 {| i | i } ∈ { } and he then performs a corresponding measurement (15) (k) M Φm m . They regard Bob’s result m as the mea- { } =1 surement outcome. This measurement can obviously be From (15), η0 = ν0 = 1/2 holds, and thus (13) always M holds regardless of γ , x, and y. Therefore, from Propo- expressed by the POVM Ωm m=1, where 1 { } sition 6, Πm can be realized by one-way LOCC. (0) (1) { } Ωm = 0 0 A Φm + 1 1 A Φm . (10) Unfortunately, (13) does not always hold. For example, | i h | ⊗ | i h | ⊗ consider the measurement Πm = πm πm 3 on = From (9), Πm = P ΩmP holds for any m with 1 m m=1 ≤ ≤ span( S , T ) with { | i h |} H M, which means that Πm can be realized by one-way | i | +i LOCC. { } 2 1 1 π = T , π = T + S , We can derive a necessary and suﬃcient condition | 1i 3 | +i | 2i − 6 | +i 2 | i (0) M r r r that there exist Bob’s measurements Φm m=1 and 1 1 (1) M { } π = T S , (16) Φm that satisfy (9) as given in the following 3 + m=1 | i −r6 | i− r2 | i {lemma} (proof in Appendix A). where T+ = + A + B. After some algebra, we have M | i | i | i η0 = 1, ν0 = 1/2, and γ1 = 2/3, and thus (13) does Lemma 5 Let Πm m=1 be a 2D measurement with rank-one POVM{ operators} on a 2D (2,N)-space. A not hold. Actually, in this case, (13) can be satisﬁed by permuting Π1 and Π2. However, there exist 2D measure- necessary and suﬃcient condition that there exist Bob’s M (0) M (1) M ments Πm m=1 such that (13) does not hold for any measurements Φm m=1 and Φm m=1 satisfying (9) is { } M { } { } permutation of the POVM operators. that cm m exists such that { } =1 0 cm 1, 1 m M, M≤ ≤ ≤ ≤ B. Complete proof of Theorem 2 cmΠm = Z0, (11) m=1 X From Proposition 6, all we have to do now to prove where Theorem 2 is to show that a measurement Πm can be realized by one-way LOCC when (13) does not{ hold.} We Z = P ( 0 0 IB )P. (12) 0 | i h |A ⊗ here consider making Alice’s subsystem interact properly 5 with her auxiliary system. Let S be Alice’s 2D auxiliary Using Lemma 7, we can show the following proposition. H system and s0 , s1 be an ONB in S. Also, let {| i | i} H Proposition 8 (A)= U( s s A)U †, Any measurement with ﬁnite rank-one L | 0i h 0|⊗ POVM operators on a 2D (2,N)-space can be realized U = USA IB, (17) ⊗ by one-way LOCC if with an operator A on , where USA is a unitary operator H γ1 < η0 (1 γ1)ν0. (22) on S A. Also, let P˜ = (P ),ρ ˜l = (ρl), and − − H ⊗ H L L L ˜ = span( ρ˜l m ). We can easily see that P˜ is the M =1 Proof m H { } ˜ Let Π m=1 be a measurement with rank-one orthogonal projection operator onto . POVM operators{ } on a 2D (2,N)-space . Assume that We consider the following one-wayH LOCC measure- H Πm satisﬁes (22). From Lemma 7, it suﬃces to show ment: Alice prepares the auxiliary system in a state s { } | 0i that a unitary operator USA on S A exists such that and transforms ρl intoρ ˜l = (ρl) using USA. Then, Al- ˜ (0) HM ⊗H ˜ (1) M L M there exist measurements Φm m=1 and Φm m=1 that ice and Bob perform a measurement Π˜ m m , where { } { } { } =1 can be realized by one-way LOCC and satisfy (20). ˜ m m m Π = (Π ). Since Π is on , it follows that SA M L { } H First, we show a unitary operator U and measure- ˜ ˜ M M Πm m=1 is a 2D measurement on . From (17), for ˜ (0) ˜ (1) { } H ments Φm m=1 and Φm m=1 that satisfy (20). Also, any l with 1 l L and m with 1 m M, we have { ˜ (1)} { } ≤ ≤ ≤ ≤ we show Φ1 = 0. We choose USA such that Tr(˜ρlΠ˜ m) = Tr(( s0 s0 ρl)( s0 s0 Πm)) | i h |⊗ | i h |⊗ USA s0 0 A = (sin θ s0 + cos θ s1 ) 0 A , = Tr(ρlΠm), (18) | i | i | i | i | i USA s0 1 A = s1 1 A (23) which means that the measurement Π˜ m for ρ˜l is in- | i | i | i | i { } { } trinsically equivalent to the measurement Πm for ρl . for some real number θ. Such USA is not uniquely deter- { } { } Thus, to show that Πm can be realized by one-way mined; we can choose any USA satisfying (23). Let { } LOCC, it suﬃces to ﬁnd USA such that Π˜ m can be re- { } Z˜0 = P˜( s0 s0 IAB )P,˜ (24) alized by one-way LOCC. Note that since Π˜ m = (Πm), | i h |⊗ L for any m with 1 m M, Πm can be expressed by where IAB is the identity operator on A B . Using ≤ ≤ H ⊗ H † ˜ Lemma 5 with replacing A by S , B by A B, Πm = s0 U ΠmU s0 . (19) H H MH H ⊗ H h | | i and 0 A by s0 , we ﬁnd that if c˜m m=1 exists such that We consider the case in which there exist measure- | i | i { } (0) M (1) M ments Φ˜ m m and Φ˜ m m such that, for any m 0 c˜m 1, 1 m M, { } =1 { } =1 ≤ ≤ ≤ ≤ with 1 m M, Π˜ m is expressed by M ≤ ≤ c˜mΠ˜ m = Z˜ , (25) ˜ ˜ ˜ (0) ˜ (1) ˜ 0 Πm = P s0 s0 Φm + s1 s1 Φm P. (20) m=1 | i h |⊗ | i h |⊗ X (0) (1) The following lemma states that Πm can be realized then there exist POVMs Φ˜ m and Φ˜ m such that (20) (0) { (1)} { } { } by one-way LOCC if Φ˜ m and Φ˜ m can be realized holds. We can show that there exists c˜m such that (25) by one-way LOCC. { } { } ˜ (1) { } and Φ1 = 0 hold if M Lemma 7 Any measurement Πm m with rank-one 2 γ1 { } =1 sin θ = (26) POVM operators on a 2D (2,N)-space can be realized by η0 (1 γ1)ν0 − − one-way LOCC if a unitary operator USA on S A ˜ (0)HM ⊗ H holds (see Appendix C). Note that from (22), the right- exists such that there exist measurements Φm m=1 and hand side of (26) does not exeed 1, and thus there exists ˜ (1) M { } Φm m=1 that can be realized by one-way LOCC and θ satisfying (26). { } satisfy (20), where Π˜ m = (Πm). ˜ (0) L Next, we show that such measurements Φm and (1) { } Φ˜ m can be realized by one-way LOCC. Let k be the Proof As described above, if Π˜ m can be realized by { } { } outcome of the measurement in the ONB s0 , s1 . If one-way LOCC, then Πm can also be realized by one- {| i | i} { } k = 0, then, from (23), the state of A is always pro- way LOCC. We consider the following one-way LOCC H(0) M ˜ m measurement for ρ˜l (denoted by Ω˜ m m ): Alice ﬁrst jected onto 0 A, which indicates that Φ can be written { } { } =1 | i performs a measurement on S in the ONB s , s . in the form H {| 0i | 1i} Let k 0, 1 be its result. Alice and Bob then perform (0) ˜ m ∈{ } (k) Φm = 0 0 A Ψ , (27) a measurement Φ˜ m and regard its result as the result | i h | ⊗ { } M of Ω˜ m . Ω˜ m is obviously expressed by where Ψm m=1 is a POVM on Bob’s side of the system. { } Thus,{ in this} case, Bob may simply perform the mea- (0) (1) Ω˜ m = s s Φ˜ m + s s Φ˜ m . (21) | 0i h 0|⊗ | 1i h 1|⊗ surement Ψm . If k = 1, then Alice and Bob have to { } ˜ (1) M ˜ (1) From (20), Π˜ m = P˜Ω˜ mP˜ holds for any m with 1 m perform the 2D measurement Φm m=1. Since Φ1 =0 ≤ ≤ (1) { } M, which means that Π˜ m can be realized by one-way holds, Φ˜ m has less than M non-zero POVM operators. { } { } LOCC. Therefore, the problem of realizing Πm with M POVM { } 6

START with ﬁnite rank-one POVM operators on a 2D (2,N)- space. 2: repeat 3: A or S A or S Compute γ1, ν0, and η0 from (6) and (7). 4: Compute 0 and 1 such that (5) holds. | iA | iA Alice S S 5: if (13) holds then (2) (2) 6: Alice performs a measurement in the ONB 0 , 1 and reports her result k 0, 1 to Bob. {| iA | iA} ∈{ } A A A B? B? k k (2) (2) (2) 7: ( ) M ( ) Bob performs a measurement Φm m=1 (Φm is ob- tained from (A12)). { } 8: else B B B B B Bob … 9: Compute USA such that (23) holds (θ is obtained (M) (M) (M−1) (M−1) (2) from (26)). 10: Alice prepares the auxiliary system in a state s and | 0i FIG. 1. A schematic diagram for realizing a measurement transforms ρl intoρ ˜l = (ρl). with ﬁnite rank-one POVM operators on a 2D (2, N)-space 11: Alice performs a measurementL in the ONB H by one-way LOCC. Diamonds represent decisions. Rectan- (k) s0 , s1 (denote its result as k). gles represent measurements on HS , HA, or HB (k ∈ {0, 1}). 12: if{| ki=0| theni} Each measurement on HS is performed after Alice’s state in- 13: M Bob performs a measurement Ψm m=1 satisfying teracts with her auxiliary system. Values in the brackets show (27). { } the number of measurement outcomes. 14: else (1) M M 15: Regardρ ˜l and Φ˜ m m as ρl and Πm m , re- { } =1 { } =1 operators by one-way LOCC is reduced to the problem spectively. 16: end if of realizing Φ˜ (1) with M ′ POVM operators by one-way 1 17: end if LOCC, where{ M}′ < M. Therefore, by iteratively per- 18: until Bob performs a measurement. ˜ (1) forming the procedure stated in this paper, Φm can 19: Output: the outcome of Bob’s measurement. be realized by one-way LOCC, since any 2D{ measure-} ment with less than three non-zero POVM operators can obviously be realized by one-way LOCC [2]. V. CONCLUSION

Proof of Theorem 2 Obvious from Propositions 6 In conclusion, we have proved that any 2D measure- and 8. ment in ﬁnite-dimensional multipartite systems can be realized by one-way LOCC. This implies that multipar- tite quantum states on a 2D Hilbert space can always C. Schematic diagram for realizing 2D be optimally distinguished by one-way LOCC no matter measurement which optimality criterion is applied. This also means that in a 2D case, any entangled information of quan- A schematic diagram of our measurement process in tum states obtained by a global measurement can also the case of a measurement with ﬁnite rank-one POVM be obtained only by one-way LOCC, at least in ﬁnite- operators on a 2D (2,N)-space is sketched in Fig. 1. Alice dimensional systems. ﬁrst determines whether she performs a binary measure- ment on A or makes her system interact with her auxil- H iary system S followed by performing a binary measure- Appendix A: Proof of Lemma 5 H ment on S. The decision rule is given by (13). Then, in the formerH case, Alice tells the result k to Bob, and 1. Preparations (k) he performs a measurement on B . In the latter case, whether Alice or Bob performsH a measurement is deter- First, we deﬁne some operators. Let mined by the result of Alice’s measurement in the ONB ⊥ Sk = π ηk B + π νk B , k 0, 1 , (A1) s0 , s1 . Alice repeats the above sequence the neces- | i h | | i h | ∈{ } {| i | i} − − ⊥ sary number of times. This procedure stops after a ﬁnite Tk = ηk ηk B π + νk νk B π , k 0, 1 ,(A2) | i h | | i h | ∈{ } number of steps. Bob may perform a measurement only where x− is deﬁned as x−1 if x = 0 and zero otherwise. once at an appropriate time. 6 (k) Sk and Tk are operators from B to and from to The entire algorithm for realizing such a (k) H H H measurement is found in the following pseu- B , respectively. Let Pk = SkTk; then, from (A1) and H(A2), for any k 0, 1 , we have docode: ∈{ } M − − ⊥ ⊥ 1: Input: a quantum state ρl and a POVM Πm m Pk = ηkη π π + νkν π π . (A3) { } =1 k | i h | k | i h | 7

− − Since ηkηk and νkνk are 0 or 1, Pk is the orthogonal 3. Suﬃciency ⊥ projection operator onto span(ηk π ,νk π ) (note that | i | i if ηk = 0 and νk = 0, then Pk = P ). Also, for any Here, we prove the suﬃciency. Assume that there ex- 6 6 M k 0, 1 , we have ists cm m=1 satisfying (11). It is suﬃcient to show that ∈{ } { } (0) (1) − − (k) POVMs Φm and Φm exist such that (A10) holds. TkSk = η ηk ηk + ν νk νk = IB , (A4) { } { } k | i h |B k | i h |B Indeed, in this case (9) is obtained from the sum of the ﬁrst and second lines of (A10). Let where the second equality follows since ηk and νk are k | i | i ( ) (k) (k) † orthogonal vectors of B , and (7) holds. Moreover, for Φm = cm TkΠmT , 1 m M, k 0, 1 , H k ≤ ≤ ∈{ } any operator X on B and k 0, 1 , we have (0) H ∈{ } cm = cm, 1 m M, ≤ ≤ † (1) P ( k k A X)P = SkXSk. (A5) cm =1 cm, 1 m M. (A12) | i h | ⊗ − ≤ ≤ ⊥ ⊥ (k) (k) Indeed, from P = π π + π π , we have Φm is obviously a positive semideﬁnite operator on B . | i h | | i h | (0) (1) H ⊥ ⊥ We show that Φm and Φm are POVMs satisfying P ( k A IB ) = ( π π + π π )( k A IB ) { } { } | i ⊗ | i h | | i h | | i ⊗ (A10). Since Z0 + Z1 = P holds from (8) and (A9), = Sk, (A6) M m=1(1 cm)Πm = P Z0 = Z1 holds from (11), which gives − − where the second line follows from (5). Thus, since P k k A X = ( k A IB)X( k A IB ), (A5) holds. M | i h | ⊗ | i ⊗ h | ⊗ (k) We also deﬁne cm Πm = Zk, k 0, 1 . (A13) ∈{ } m=1 Zk = P ( k k A IB )P, k 0, 1 , (A7) X | i h | ⊗ ∈{ } Thus, from (A12), for any k 0, 1 , we have ∈{ } which includes the deﬁnition of Z0 in (12). We can easily M obtain SkIB = Sk from (A1); thus, from (A5), we have (k) † † † (k) Φm = TkZkTk = TkSkSkTk = IB , (A14) † † m=1 Zk = SkIB Sk = SkSk. (A8) X where the second and third equalities follow from (A8) Substituting (A1) into (A8) yields (0) (1) and (A4), respectively. Therefore, Φm and Φm are ⊥ ⊥ { } { } Zk = ηk π π + νk π π . (A9) POVMs. From (A5) and (A12), for any k 0, 1 , we | i h | | i h | have ∈ { }

(k) (k) † P k k A Φm P = SkΦm Sk 2. Necessity | i h | ⊗ (k) † † = cm SkTkΠmTk Sk Here, we prove the necessity. Since Πm is a rank-one (k) M = cm PkΠmPk. (A15) operator, to satisfy (9), there must exist cm m=1 with { } (k) 0 cm 1 such that for any m with 1 m M, Thus, to prove (A10), it suﬃces to show cm PkΠmPk = ≤ ≤ ≤ ≤ (k) m m k k (0) c Π . Since P Z holds from (A3) and (A9) (A B P 0 0 A Φm P = cmΠm, denotes that A ≥B is positive semi-deﬁnite), we have≥ | i h | ⊗ − (1) (k) P 1 1 A Φm P = (1 cm)Πm. (A10) Pk Zk cm Πm, k 0, 1 , (A16) | i h | ⊗ − ≥ ≥ ∈{ } (0) (0) where the second inequality follows from (A13). m In contrast, since Φ is a POVM on B , k M { } k H Thus, since P is the orthogonal projection operator, (0) (0) ( ) (k) (k) m=1 Φm = IB holds, where IB is the identity op- cm PkΠmPk = cm Πm holds. Therefore, (A10) holds. k erator on ( ). Thus, from (A10), we have P HB M M (0) cmΠm = P 0 0 Φm P | i h |A ⊗ Appendix B: Proof of Proposition 6 m=1 m=1 ! X X (0) M = P ( 0 0 A IB )P Let Πm m be a measurement with rank-one POVM | i h | ⊗ { } =1 (0) † operators on a 2D (2,N)-space . Assume that Πm = S I S 0 B 0 satisﬁes (13). From Lemmas 4 andH 5, it suﬃces to{ show} † M = S0S0 that there exists cm m satisfying (11). Let { } =1 = Z0, (A11) η0 (1 γ1)ν0 c1 = − − , where the third and ﬁfth lines follow from (A5) and (A8), γ1 respectively. Therefore, cm satisﬁes (11). cm = ν , m 2, 3, ,M . (B1) { } 0 ∈{ ··· } 8

2 We can see that 0 cm 1 for any m with 1 m M. The second equation of (C3) follows from Ps0 = Ps . In ≤ ≤ ≤ ≤ 0 Indeed, since 0 ν0 1, 0 cm 1 holds for any contrast, from P˜ = (P )= U( s s P )U †, (17), and ≤ ≤ ≤ ≤ L | 0i h 0|⊗ m 2. Since η0 ν0 (1 γ1)ν0, which follows from (23), we have γ ≥> 0, c 0 holds.≥ Moreover,≥ − c 1 holds from (13). 1 1 ≥ 1 ≤ From (B1), we obtain D = Ps0 U( s0 s0 P ) | i h |⊗ = Ps0 (sin θ s0 s0 0 0 A IB )( s0 s0 P ) M | i h | ⊗ | i h | ⊗ | i h |⊗ = sin θ s0 s0 (IAB ( 0 0 A IB)P ) cmΠm = c1Π1 + ν0(P Π1) | i h |⊗ | i h | ⊗ − B m=1 = sin θ s0 s0 (( 0 0 A I )P ). (C5) X | i h |⊗ | i h | ⊗ = (c1 ν0)Π1 + ν0P (C3) and (C5) yield − ⊥ ⊥ = (η0 ν0) π π + ν0( π π + π π ) † 2 − | i h | ⊥ ⊥| i h | | i h | U Z˜0U = sin θ s0 s0 (P ( 0 0 A IB )P ) = η0 π π + ν0 π π | i h |⊗ | i h | ⊗ | i h | | i h | 2 = sin θ s0 s0 Z0, (C6) = Z0, (B2) | i h |⊗ where the second line follows from (A7). From (C2) and where the third line follows from Π1 = γ1 π π and (C6), (25) is equivalent to (C1). ⊥ ⊥ | i h | P = π π + π π , and the last line follows from Now, we show that there exists c˜m such that (25) | i h | | i h | (A9). Therefore, cm of (B1) satisﬁes (11). ˜ (1) { } { } and Φ1 = 0 hold. Letc ˜1 = 1 andc ˜m =c ˜2 for any m 3. As shown in the proof of Lemma 5, cm = 1 (i.e., (1)≥ (1) cm = 0) yields Φm = 0 from (A12), which indicates Appendix C: Supplement of (25) and (26) ˜ (1) that Φ1 = 0 holds fromc ˜1 = 1. We have M Assume (26); we will show that there exists c˜m such { } c˜mΠm = Π1 +˜c2(P Π1) ˜ (1) − that (25) and Φ1 = 0 hold. m=1 X In preparation, we show that (25) is equivalent to = (1 c˜ )Π +˜c P − 2 1 2 = (γ + (1 γ )˜c ) π π +˜c π⊥ π⊥ , M 1 − 1 2 | i h | 2 | i h | 2 c˜mΠm = (sin θ)Z0. (C1) (C7) m=1 X where the last line follows from Π1 = γ1 π π and P = π π + π⊥ π⊥ . Thus, from (A9), (C1| )i (i.e., h | (25)) is Premultiplying and postmultiplying both sides of (25) by |equivalenti h | | toi h | U † and U, respectively, yield 2 γ1 + (1 γ1)˜c2 = η0 sin θ, M − 2 † c˜ = ν sin θ, (C8) s s c˜mΠm = U Z˜ U. (C2) 2 0 | 0i h 0|⊗ 0 m=1 2 X so we letc ˜2 = ν0 sin θ. 0 c˜2 1 obviously holds. We can see from (26) that (C8≤) holds;≤ therefore, (25) holds. ˜ ˜ ˜ Let Ps0 = s0 s0 IAB; then, from (24), Z0 = P Ps0 P holds. Thus,| wei h have|⊗ ACKNOWLEDGMENTS † ˜ † ˜ ˜ † U Z0U = U P Ps0 PU = D D, (C3) We thank O. Hirota and K. Kato of Tamagawa Uni- where versity for the useful discussions we had with them. T. S. U. was supported (in part) by JSPS KAKENHI (Grant ˜ D = Ps0 PU. (C4) No. 24360151).

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