Analysis of Optimal Sequential State Discrimination for Linearly Independent Pure Quantum States

Recently, J. A. Bergou et al. proposed sequential state discrimination as a new quantum state discrimination scheme. In the scheme, by the successful sequential discrimination of a qubit state, receivers Bob and Charlie can share the information of the qubit prepared by a sender Alice. A merit of the scheme is that a quantum channel is established between Bob and Charlie, but a classical communication is not allowed. In this report, we present a method for extending the original sequential state discrimination of two qubit states to a scheme of N linearly independent pure quantum states. Specifically, we obtain the conditions for the sequential state discrimination of N = 3 pure quantum states. We can analytically provide conditions when there is a special symmetry among N = 3 linearly independent pure quantum states. Additionally, we show that the scenario proposed in this study can be applied to quantum key distribution. Furthermore, we show that the sequential state discrimination of three qutrit states performs better than the strategy of probabilistic quantum cloning.


Scenario of Sequential State Discrimination.
Let us explain a sequential state discrimination scenario of N linearly independent pure quantum states. A sender Alice prepares a quantum state |ψ i 〉 with a prior probability q i , out of a set of N linearly independent pure quantum states {|ψ 1  Rule 1. Bob discriminates Alice's quantum state using a nonoptimal unambiguous discrimination strategy. However, using an optimal unambiguous discrimination strategy, Charlie mesures Bob's post-measurement state. Rule 2. Classical communication is not allowed between Bob and Charlie. Therefore, the only way for Charlie to obtain information about Alice's quantum state is to measure Bob's post-measurement state.
Therefore, the probability that Bob and Charlie successfully share Alice's quantum state is given by The important problem in sequential state discrimination is in finding the POVM condition of Bob and Charlie, for optimizing the success probability of Eq. (1). If the optimal value of (1) is not zero, Bob and Charlie have a chance of successfully sharing Alice's quantum state. When Alice prepares a qubit state out of two pure qubit states, the success probability is proven to be nonzero 21,[23][24][25] . To date, researchers have considered the optimal success probability only for two qubit states. In this report, proposing the sequential state discrimination for N linearly independent pure quantum states, we provide a method for finding the optimized success probability condition.
Formulation of Optimization Problem. Now . Let us assume that ψ ψ = 〈 | 〉 = G { } i j i j N , 1 is a Gram matrix made by Alice's pure states. When Alice's pure states are linearly independent, G > 0(↔ ∃G −1 ) 28 is satisfied. Therefore, the POVM that can discriminate every Alice's pure state without error can be described by 30  (2) implies that Bob's POVM can correspond to a real vector (α 1 , α 2 , …, α N ) in N−dimensional vector space. Additionally, Bob's POVM set is convex 30 . The convex set is denoted as  ⊂ C N . The conditions under which M 0 is positive-semidefinite are equivalent to 〈ψ|M 0 |ψ〉 ≥ 0 for an N dimensional vector |ψ〉 28 . Because |ψ〉 can be obtained from a linear combination of {|ψ 1 , 1 is non-negative. Every component of real vector α i , which is an element of C, is non-negative. Then, we have the following conjecture about C: Conjecture 1. C, being the set of N dimensional real vectors, is a convex set composed of N + 1 vertices. The edges of convex set C contain N segments satisfying α i = 0.
In Conjecture 1, convex set C contains a line segment α i = 0, since there exists a case where Bob discriminates unambiguously remaining quantum states except i-th pure state. For N = 3, the geometric argument can be found in refs [31][32][33] . Let us assume that Bob's set of post-measurement states, which corresponds to conclusive result of Bob, is . When Bob discriminates Alice's quantum state using a nonoptimal unambiguous discrimination, information remains in Bob's post measurement state, which can be obtained by Charlie. Additionally, Bob's post-measurement states are linearly independent, which implies that Charlie's POVM can discriminate Bob's post-measurement state without error. The conditions for Charlie's POVM, which can discriminates the Bob's post-measurement state without error, are (C1 . When Bob's post-measurement states are linearly independent, if their Gram matrix is is non-negative, (C1) and (C2) also hold. Then, Charlie's POVM corresponds to the N dimensional real vector α α α . Let the set of the real vectors be In Conjecture 2, the fact that convex set C′ contains a line segment of α′ = 0 i means that there exists a case where Charlie discriminates unambiguously remaining quantum states except i-th state. It can be shown from Figs 1 and 2 that Conjecture 2 is satisfied when N = 2, 3. In the following section, we will explain how Bob's POVM affects C′. In sequential state discrimination, Bob performs a nonoptimal unambiguous discrimination, and (α 1 , …, α N ) belongs to the interior of C, int(C). Meanwhile, Charlie performs an optimal unambiguous discrimination, and a vector α i is located in the boundary of C′, ∂C′. Therefore, the sequential state discrimination problem is equivalent to the following optimization problem. In sequential state discrimination of two pure qubits, the convex sets of Bob and Charlie, C and C′. (a) shows the Bob's convex set C when the overlap between two pure qubits is s = 0.3. In (b), solid(dashed) line displays the boundary ∂C′ of Charlie's convex set when (α 1 , α 2 ) = (0.5, 0.5)((α 1 , α 2 ) = (0.5, 0.7)). ∂C′ relies on (α 1 , α 2 , …, α N ). Now, we prove that Conjecture 1 and Conjecture 2 hold in cases of N = 2 and N = 3. And we investigate the optimization problems of Eq. (4) when N = 2 and N = 3. We expect that our approach can be extended to a case where N > 3.
Examples: sequential state discrimination for two or three pure states. Suppose that Alice prepares a quantum state out of two(or three) pure quantum states 6-9,31-33 .
Two pure states (Qubits). Assume that with a prior probability q i , Alice prepares a qubit state out of two pure qubit states |ψ i 〉 ∈ {|ψ 1 〉,|ψ 2 〉} and sends it to Bob. The overlap between two qubit states is 〈ψ 1 |ψ 2 〉 = sexp (iφ), s ≥ 0. When s ∈ [0, 1), there exists the inverse of the Gram matrix G for Alice's quantum states. This implies the existence of Bob's POVM discriminating Alice's qubit state without error. Bob's POVM can be represented by a two-dimensional real vector (α 1 , α 2 ). (α 1 , α 2 ) ∈ C, satisfying the conditions of Bob's POVM (B1), (B2), (B3), fulfills the following inequality: The inequality is obtained by the non-negativity condition of . Eq. (5) indicates that C is a convex set. One can show that the set C of (α 1 , α 2 ) fulfilling Eq. (5) is convex, in the following way. The strict equality α 2 = f(α 1 )(f(x) ≡ 1 + s 2 (x − 1) −1 ) of Eq. (5) is monotonic decreasing as α 1 varies from 0 to 1−s 2 . In addition, f″(α 1 ) of strict equality is negative-definite. Therefore, C surrounded by strict equality of Eq. (5) and α 1 = 0, α 2 = 0 is convex. Figure 1(a) displays the structure of C when the overlap s between two pure qubits, prepared by Alice is 0.3, which shows that C is convex. Because of the fact that in sequential state discrimination the post-measurement states of Bob should be considered, we need to find the Kraus operator K i that corresponds to M i . When i = 1,2, from singular value decomposition 34 (It is not a general form of the Kraus operator 35 . However, because Charlie performs unambiguous discrimination, it is sufficient to consider a special type of Kraus operator.). Now, let us find the Kraus operator K 0 corresponding to M 0 . K 0 can be thought of as a linear map that sends two of Alice's qubit states to linear independent states ψ ψ | ′ 〉 ′ 〉 and 1 2 , When Bob performs an optimal unambiguous discrimination, s′ becomes one(s′ = 1). This means that Charlie cannot discriminate Bob's post-measurement states. Therefore, Bob should use a nonoptimal unambiguous discrimination. Then, the Gram matrix G′ of Bob's post-measurement states has its inverse. Charlie's POVM, which can discriminate Bob's post-measurement states without error, corresponds to a two-dimensional real vector α α ′ ′ ( , ) 1 2 . The condition for α α ′ ′ ∈ ′ C ( , ) 1 2 is given by This tells us that C′ is a convex set and depends on C. In Eq. (7), s′ is a function of (α 1 , α 2 ), being not a function of The fact that Bob Scientific REPORTS | (2018) 8:6515 | DOI:10.1038/s41598-018-24575-w can obtain an information from Alice's quantum state implies that the success probability α i is non-zero. From the relation between s′ and s for non-zero α i , we find a strict inequality s′ > s. As s increases, f(x) decreases and the size of C′ is smaller than that of C. From Fig. 1(b), one can see that C′ is convex and the size of C′ is smaller than that of C. Here, solid line(dashed line) displays ∂C′ when (α 1 , α 2 ) = (0.5, 0.5)((α 1 , α 2 ) = (0.5, 0.7)). As (α 1 , α 2 ) approaches to ∂C, the size of C′ decreases, implying the trade-off phenomena which tells that more information Bob obtains from Alice's pure states, less information Charlie gets. Therefore, we show that Conjecture 1 and 2 hold. Eqs (5) and (7) are the constraints of optimization problem (4). Because Bob(Charlie) uses a nonoptimal(optimal) unambiguous discrimination, Eqs (5) and (7) becomes a strict inequality(strict equality). If Bob does not optimally discriminate Alice's pure qubit, (α 1 , α 2 ) exists in the interior of C. Therefore, (α 1 , α 2 ) satisfies the strict inequality of Eq. (5) 27 . In the similar way, if Charlie optimally discriminate the post-measurement state of Bob, α α ′ ′ ( , ) 1 2 exists on ∂C′. Therefore, α α ′ ′ ( , ) 1 2 satisfies the strict equality of Eq. (7) 27 . Then, Eq. (4) becomes the following optimization problem; , the best strategy for Charlie is to discriminate two of Bob's post-measurement states. This implies that the best strategy for Bob and Charlie is to discriminate every of the two pure quantum state. In this case, the sequential discrimination scenario for Bob and Charlie is summarized as Theorem 1.

Theorem 1.
Suppose that with a prior probability q i , Alice prepares a pure state |ψ i 〉 out of {|ψ 1 〉, |ψ 2 〉} satisfying 〈ψ 1 |ψ 2 〉 = s exp(iφ), s ≥ 0. The sequential state discrimination problem in which Bob and Charlie discriminate two pure states is equivalent to the following optimization problem: The proof of Theorem. 1 is shown in the Methods. Equation (9) is a nonlinear optimization problem and is a difficult to solve analytically. In special cases, the optimization problem can be solved analytically. Because the object function is continuous, when (α 1 , α 2 ) is an optimization point, the gradient of the object function becomes zero. Therefore, the following equations hold at the optimization point: The inverse does not generally hold. However, when two prior probabilities are equal, two equations have an inverse function relationship. Then, the optimization condition becomes α α = = − s 1 1 2 , and the optimized success probability is 2 , which agrees with the result of ref. 21 . The solution of Eq. (9), is located in the interior of C. And the condition of Charlie for providing optimal success probability, α α exists on ∂C′. The fact that optimal condition for sequential state discrimination is α α implies that with the same probability, Bob and Charlie, without error, discriminate i-th pure qubit, with supports the result of ref. 21 .
If two prior probabilities are not equal, it is difficult to analytically prove the argument of ref. 21 . However, it can be shown numerically 36,37 . For example, in the case of q 1 = 0.55, q 2 = 0.45, and s = 0.12, one can numerically find (α 1 , α 2 ) = (0.721987, 0.568364), which is the solution of Eq. (10). Then, by Eq. (26) of Method, one can obtain α α ′ ′ ( , ) 1 2 , which is the same as (α 1 , α 2 ). It should be noted that α α ′ ′ ( , ) 1 2 = (0.721987, 0.568364) is located on ∂C′. In fact, the reason that the success probability of Bob and Charlie is the same is that even though Bob performs a nonoptimal unambigous discrimimination, Charlie should optimally discriminate Bob's post measurement state.
Let us assume that α′ = 0 2 . Then, Charlie discriminates only ψ | ′ 〉 1 | without error. In this case, α′ = − ′ s 1 1 2 is obtained. Likewise, Bob also discriminates only ψ 1 , which is the best strategy (α 2 = 0) for Bob and Charlie. Therefore, the problem of sequential state discrimination is equivalent to the following optimization problem: The optimized condition is . When Bob and Charlie discriminate only |ψ 2 〉(i.e., α 1 = α′ = 0 1 ), the optimized condition becomes . Therefore, the optimized success probability that in sequential state discrimination Bob and Charlie discriminate only one of Alice's two pure qubit states, becomes max{q 1 (1 − s 2 ), q 2 (1 − s 2 )}. The case of q 1 = q 2 contains the result of Pang et al. 23 .
If general prior probabilities ≠ q q 1 2 are considered, one must numerically find (α 1 , α 2 ) satisfying Eq. (10). In this report, a random search method 36 based on Monte Carlo methods is used. By these methods, one can search almost entire (α 1 , α 2 ), fulfilling the constraint of Eq. (9).
When Bob can obtain a partial information on Alice's qubit, since the overlap between Bob's post measurement states should be increased, s′ is always greater than s. Therefore, the size of convex set C′ is smaller than that of C. When Bob performs an optimal unambiguous discrimination, because of s′ = 1, the element of C′ is only a zero vector. This implies that Charlie cannot obtain any information from Bob's post-measurement state.
Three pure states(Qutrits). Now let us consider the case of three pure qutrit states. With a prior probability q i , Alice prepares an element |ψ i 〉 out of a set of qutrits {|ψ 1 〉, |ψ 2 〉, |ψ 3 〉} and sends it to Bob. The overlap between those qutrit states is given by 33 . This condition indicates that three dimensional real vector (α 1 , α 2 , α 3 ) ∈ C corresponding to Bob's POVM satisfies the following relationship: Here, α α = − 1 i i and Φ = φ 1 + φ 2 + φ 3 are refered to as a geometric phase 32 . The set of (α 1 , α 2 , α 3 ) satisfying Eq. (12) is convex. Let us obtain Kraus operator K i , corresponding to Bob's POVM. If i = 1, 2, 3, the necessary and sufficient condition for and ψ | ′ 〉 i is a Bob's post-measurement state. For i = 0, let us assume the Kraus operator as follows: Because K 0 is a linear map sending |ψ i 〉 to |ψ i′ 〉, from the condition of . When (α 1 , α 2 , α 3 ) exists in the interior of C, the Gram matrix G′, composed of the post-measurement states of Bob, has an inverse matrix. Therefore, there exists Charlie's POVM, discriminating the post-measurement states of Bob without error. Like Bob's POVM, the conditions of Charlie's POVM are given by i j k The set C′ of three-dimensional real vectors satisfying Eq. (14) is convex and depends on Bob's POVM. Therefore, Conjecture 1 and 2 hold in the sequential state discrimination of the three qutrit states. Because Charlie performs an optimal unambiguous discrimination, α α α ′ ′ ′ ( , 3 exists on the surface of Eq. (14). The sequential state discrimination problem for three qutrit states by Bob and Charlie becomes the following optimization problem: Here, int (C) is a set of (α 1 , α 2 , α 3 ) strictly satisfying Eq. (12). ∂C′ is a set of (α 1 , α 2 , α 3 ) fulfilling =  G det 0 in Eq. (14). The sequential state discrimination of three qutrit states for Bob and Charlie can be categorized by the number of discriminations of qutrit states by Bob and Charlie. (The theorems are proven in the Methods).

Theorem 2.
Alice prepares a element |ψ i 〉, with a prior probability q i , from three pure qutrit states {|ψ 1 〉, |ψ 2 〉, |ψ 3 〉} satisfying 〈ψ i |ψ j 〉 = s k exp (iε ijk φ k ), s k ≥ 0 and sends it to Bob. If in sequential state discrimination, Bob and Charlie discriminate only a single qutrit state (for example, α ) out of three qutrit states, the optimal conditions for the sequential state discrimination are α

Theorem 3.
Suppose that Alice prepares a element |ψ i 〉, with a prior probability q i , from three pure qutrit states {|ψ 1 〉, |ψ 2 〉, |ψ 3 〉} satisfying 〈ψ i |ψ j 〉 = s k exp (iε ijk φ k ), s k ≥ 0 and sends it to Bob. If in sequential state discrimination, Bob and Charlie discriminate only two qutrit states (for example, α α = ′ = 0, 0 ) out of three qutrit states, the solution for the sequential state discrimination is equivalent to that of the following optimization problem: and every index satisfies ≠ ≠ i j k, When (α 1 , α 2 ) is an optimal pint of Eq. (16), the gradient of the object function at the point is zero. Therefore, the optimal condition is given by However, the inverse does not hold. When q i = q j and s 1 = s 2 = s 3 = s, the equations have an inverse function relationship. Then, the optimal condition is obtained by α ). If (α i , α j ) does not fulfill Eq. (16), the best strategy for Bob and Charlie is to discriminate only a single qutrit state out of three pure qutrit states.
It is difficult to analytically solve the optimization problem described by Theorem 4. Unlike the case for two qubits, it is difficult to find an analytic solution for the sequential state discrimination of three qutrit states even in a symmetric case. Therefore, we use a random search method 36 to obtain a numerical solution. Figure 3 displays the optimal success probability when q 1 = q 2 = q 3 and s 1 = s 2 = s 3 = s. Figure 3 shows that it is better for Bob and Charlie to discriminate every three pure qutrit state than it is for them to discriminate one(or two) pure qutrit state(s). This is different from the case involving two pure qubit states. In the case of two pure qubit states prepared with equal prior probability, when the overlap between the two qubit states is greater than − 3 2 2, discriminating only single qubit state is the best method 21,23 . However, for quantum key distribution, receivers should discriminate all quantum states prepared by a sender 21,38 . Therefore, our results for three pure qutrit states demonstrates the advantages of sequential state discrimination in quantum key distribution.
In addition, Hillery and Mimih 26 studied the specific cases of sequential state discrimination. Even though we do not explain in detail, some examples of M. Hillery and J. Mimih 26 can be understood by application of results of this report.
Comparison with an other scenario. Here, we compare the sequential state discrimination of three pure qutrit states with quantum probabilistic cloning strategy 39 . Unlike sequential state discrimination, the method allows a classical communication.
• (Quantum probabilistic cloning) Suppose that Alice prepares a quantum state |ψ i 〉(i ∈ {1, …, N}) with a prior probability q i and sends it to Bob. Bob clones this quantum state probabilistically 39 . When Bob successfully copies the state, the quantum state after Bob's copying becomes |ψ i 〉 ⊗|ψ i 〉. Bob shares |ψ i 〉 ⊗ |ψ i 〉 with Charlie. Bob and Charlie perform an optimal unambiguous discrimination on his own state, respectively. If Bob fails to copy the quantum state of Alice, he tells Charlie that he fails to copy the quantum state of Alice.
, , } N 1 39,44 . Let us assume equal probability for successfully copying a quantum state (γ i = γ ∀i). Then, the probability for successful quantum probabilistic cloning of Bob and Charlie is given by Here, Pr[|ψ i 〉] is a prior probability with which Alice prepares |ψ i 〉. Pr[clone|ψ i 〉] is the probability that Bob successfully copies the quantum state |ψ i 〉 of Alice. Pr[i|clone] is the probability that Bob and Charlie obtain measurement result i. The necessary and sufficient condition for Theorem 5 is that the pure states are linearly independent 39,44 . Therefore, there exists a POVM of Bob and Charlie that can discriminate Alice's quantum states without error.
Suppose that Alice prepares a quantum state out of three linearly independent pure qutrits {|ψ 1 〉, |ψ 2 〉, |ψ 3 〉}. When Bob successfully copies the quantum state of Alice, Bob and Charlie independently discriminate Alice's quantum state. Therefore, the best way to perform a strategy of probabilistic quantum cloning is for Bob and Charlie to perform an optimal unambiguous discrimination. Therefore, the real vector corresponding to an optimal POVM of Bob and Charlie is (α 1 , α 2 , α 3 ). Then, the optimal success probability of Bob and Charlie, when probabilistic quantum cloning is imposed, becomes Here, γ opt is a maximal γ satisfying Theorem 5. The optimal value of γ depends on the overlap between pure states prepared by Alice. The success probability of probabilistic quantum cloning is a convex function of (α 1 , α 2 , α 3 ). Therefore, (α 1 , α 2 , α 3 ) exists on the boundary ∂C of C.
Example. Suppose that the overlap between three pure qutrits {|ψ 1 〉, |ψ 2 〉, |ψ 3 〉} can be given by Suppose that the prior probabilities for the three pure qutrits are q 1 = q 3 = 0.35, and q 2 = 0.3. Let us assume that the geometric phase of the three pure qutrits is φ 1 + φ 2 + φ 3 = 0 + π + π = 2π. Figure 4 displays the optimal success probabilities of probabilistic quantum cloning and sequential state discrimination, where the overlap s is s ∈ [0.4, 0.8]. According to Fig. 4, sequential state discrimination, unlike probabilistic quantum cloning, has a non-zero optimal success probability even when s becomes large. This is different from the case of two pure qubits Figure 4. The optimal success probabilities of probabilistic quantum cloning and sequential state discrimination for three pure qutrit states. Here, the solid line(dash-dot line) shows the optimal success probability of sequential state discrimination based on Eqs (18 and 19), and the dashed line shows that of probabilistic quantum cloning.

Revisiting The Security of Protocol Based on Sequential State Discrimination Scenario.
Based on B92 protocol 38 , the security of B92 protocol can be obtained when Bob performs an unambiguous discrimination on nonorthogonal quantum states encoded by a sender Alice 21 . However, if an eavesdropper Eve exists between Alice and Bob, then the conclusive result of Bob will contain an error. When a discrepancy occurs from the comparison between Alice's quantum state and Bob's conclusive result, Alice and Bob can notice the existence of eavesdropper Eve.
Note that Alice and Bob are separated in space. Alice should inform Bob of the prior probability, in order for Bob to perform a unambiguous discrimination. In this process, if Alice uses a classical communication, Eve can obtain the prior probability without being noticed. When Eve optimally discriminates two quantum states of Alice, the post-measurement states of Eve are entirely overlapped, which implies that Bob cannot obtain any information from the post-measurement state of Eve. Therefore, Bob's result is always inconclusive. Meanwhile, unless Bob notices Eve, Eve's measurement should be identity. This implies that Eve cannot obtain any information from Alice's quantum state. When Eve obtains a partial information from Alice's quantum state, the overlap between Eve's post-measurement states is larger than the overlap between Alice's quantum states. Therefore, Bob's conclusive result may contain an error with a probability.
In fact, this argument can be applied to the case of sequential state discrimination of three pure qutrit states. Now, we can have the following Conjecture.

Conjecture 3.
In sequential discrimination strategy, when a receiver can obtain a partial information about a sender's quantum state through a POVM, the size of convex set C′ of a second receiver is smaller than that of C of a first receiver.
In fact, SSD strategy provides a protocol for distribution of secure key to multi-parties 21 , based on B92 protocol 38 . When an eavesdropper Eve tries to obtain a partial information between Alice and Bob, the conclusive result of Bob inevitably contains an error. In the similar way, if Eve gains an information between Bob and Charlie, the conclusive result of Charlie should have an error. It implies that whenever Eve obtains an information, Bob and Charlie can notice the existence of eavesdropper Eve. Therefore, Alice, Bob, and Charlie can share a secure key.
In addition, the example of our report shows that the proposed SSD strategy can provide a higher success probability than probabilistic quantum cloning strategy. We should note that in two qubit case, SSD strategy is not better than probabilistic quantum cloning strategy 21 . Therefore, when Alice encodes an information using non-orthogonal pure qutrits, SSD strategy can be more efficient one for QKD than probabilistic quantum cloning. It implies that SSD strategy can be a good candidate for multi-party QKD 27 .
Furthermore, if one uses larger number of linearly independent pure qudits, Alice can share more secure message with multi-parties 26 . Ref. 26 considers only symmetric pure qudits. The result of our report may provide a strategy using general(nonsymmetric) pure qudits, for QKD of multi-parties.

Discussion
In this report, we studied the sequential state discrimination of N pure states. One of advantages of the sequential state discrimination is that a sequential receiver Charlie can determine the pure state of a sender Alice, handling the post-measurement of a previous receiver Bob rather than the pure state of Alice. First, we provided a general formulation of sequential state discrimination for N pure states when Alice prepares a quantum state out of N pure states. Second, we obtained the condition for sequential state discrimination of two qubits with arbitrary prior probabilities. Third, we showed that when Alice prepares three qutrits with the identical prior probabilities, the best way for Bob and Charlie to perform sequential state discrimination of three qutrits is to unambiguously discriminate every three qutrit. This differs from the two qubits case, where the best sequential state discrimination method is not to discriminate every two qubit. In addition, we showed that the sequential state discrimination of three pure qutrit states performs better than probabilistic quantum cloning strategy.
If the three conjectures of our report can hold in the case of N qudits, we have a formulation of sequential state discrimination to N linearly independent qudits. In addition, we may generalize our sequential state discrimination approach to N parties. It is interesting to compare our strategy of N qudits for N parties with other strategies.
Proof of Theorem 4. Suppose that Bob and Charlie discriminate every three pure qutrit state of Alice. The measurement conditions of Charlie for the best sequential state discrimination correspond to the tangential points between the plane α α α α α α ′ + ′ + ′ q q q ( ) ( ) ( ) 1 1 1 2 2 2 3 3 3 and ∂ ′ C . The tangential point can be analytically found when the geometric phase is Φ = 0, π 45 . However, for a general geometric phase, the condition for the plane to be tangent to ∂C′ is given by a 6-th order of algebraic equation 32,33 . For similarity, we assume that Φ = 0. Then, the tangential points become as follows: 32,33 .