Geometric bloch vector solution to minimum-error discriminations of mixed qubit states

We investigate minimum-error (ME) discrimination for mixed qubit states using a geometric approach. By analyzing positive operator-valued measure (POVM) solutions and introducing Lagrange operator (cid:2) , we develop a four-step structured instruction to ﬁnd (cid:2) for N mixed qubit states. Our method covers optimal solutions for two, three, and four mixed qubit states, including a novel result for four qubit states. We introduce geometric-based POVM classes and non-decomposable subsets for constructing optimal solutions, enabling us to ﬁnd all possible answers for the general problem of minimum-error discrimination for N mixed qubit states with arbitrary a priori probabilities.


I. INTRODUCTION
Distinguishing between quantum objects is one of the most fundamental tasks in information theory.Because of its particular importance in quantum communication [1,2] and quantum cryptography [3] , discriminating among these objects plays a very important role in quantum information protocols.
The quantum state discrimination problem has been the subject of many researches during recent decades [4].The starting point of these efforts date back to 1978 when Helstrom addressed this issue in his famous book [2].He studied the case of two states and obtained the minimumerror probability in the framework of quantum detection and estimation theory [5].Since then, based on the different approaches on the issue (error-free protocols with a possibility of failure and protocols with minimum-error probability) many developments have been achieved [6][7][8][9][10][11][12][13][14][15][16][17][18][19].Finding which one is the most suitable is highly dependent on the purpose of the process.
The extension to the general case of N possible states {ρ i } N i=1 with associated a priori probabilities {p i } N i=1 is not a straightforward problem.For these problems, however, there are necessary and sufficient conditions on the optimal measurement operators π i known as the Helstrom conditions [2,20,21] where Γ = i p i ρ i π i is a positive Hermitian operator known as Lagrange operator.In fact, these two conditions are not independent and the second condition can be obtained from the first condition [20,21].So, the main condition which is both necessary and sufficient condition for an optimal measurement is Eq.(1).However, we can still use Eq. ( 2) in our procedure.It is also of particular importance to note that all minimum-error measurements that give the optimal probability define a unique Lagrange operator [22].
The Helstrom conditions can be used to check the optimality of a candidate measurement, however, they cannot be used to construct the optimal measurement for a general problem.They can be useful for the problem with some symmetries among states which can be a guide for guessing the form of measurement in such a problem [23][24][25][26][27][28].
Although, for a while, only problems with certain symmetries seemed solvable, in recent years there have been successes in solving problems by employing the conditions (1) and (2).Particularly, for the qubit states, the Bloch representation provides a useful tool to solve the problem of ME discrimination of qubit states.For example, Hunter in 2004 presented a complete solution for pure qubit states with equal a priori probabilities [29].Samsonov in 2009 applied the necessary and sufficient conditions for indicating an algorithmic solution for N pure qubit states [30].Deconinck and Terhal in 2010 used the discrimination duality theorem and the Bloch sphere representation for a geometric and analytic representation of the optimal guessing strategies among qubit states [31].Bae in 2013 represented a geometric formulation for a case with equal priori probabilities in a situation where quantum state geometry is clear [32,33], and in the same year a complete analysis for three mixed states of qubit systems was done by Ha et al. [34].Weir et al. in 2017 used the Helstrom conditions constructively and analytically for solving a problem with any number of qubit states with arbitrary priori probabilities [35] with giving the central role in their approach to the inverse of Γ instead of Γ itself.Later, they used their method to find optimal strategies for trine states [36].
In this paper, we reconsider the problem of ME discrimination of N mixed qubit states with arbitrary a priori probabilities using a different geometric approach.In 2010, Deconinck and Terhal indicated a general algorithm to find the Lagrange operator Γ by using the geometric properties of the Bloch sphere.Our main purpose in this paper is to extent their work and give an analytical method in detail to find all solutions of a general mixed qubit states problem, as well as studying some properties of these solutions.
Our starting point is the Helstrom conditions.Employing these conditions and the Bloch representation for qubit states, one can find all possible ME POVM answers of a given problem with just knowing the Lagrange operator Γ.Generally, we solve the problem first for arbitrary priori probabilities, then, as a special case, for equal a priori probabilities, compare this method with existed results [32].For equal a priori probabilities minimal sphere, circumsphere, plays an important role to find Γ.Furthermore, for a general problem with arbitrary a priori probabilities, we establish a general structured instruction to obtain Γ, with some practical tips involving the polytope of states we introduce two classes of unchanged guessing probability and unchanged measurement operators.This instruction will be used to find Γ for a general ME problem with two, three, and four qubits.A direct solution of the last case of four-qubit states is completely new.Equipped with these tools, we are able to find all possible ME discrimination measurements of the problem.By introducing non-decomposable POVM answers of the problem an alternative way to obtain a general optimal answer is achievable by considering a convex combination of all these non-decomposable POVM sets.
This paper is organized as following: In Sec.II, we review the problem of ME discrimination in a general way by reformulating the problem for qubit states using the Bloch vector representation and represent a way to find all optimal answers of a problem using the Lagrange operator Γ by constructing a four-step instruction.Next, in Sec.III we solve a general problem of mixed qubit states with arbitrary priori probabilities for two, three, and four qubits cases.As an example, for three states we resolve the case of trine states to compare it with the known results in [36].Also, another illustration for the case of four qubit states is helpful.In Sec IV we consider some cases of N > 4 including N = 5 and N = 6 states and indicate regions chracterizing by first non-decomposable answer in different colors.We also revisit the special case of N states with equal a priori probabilities.Finally, the last section includes the summary and conclusions.

II. QUANTUM STATE DISCRIMINATION
Suppose that we are given N states ρ i , i = 1, • • • , N , with priori probabilities p i .Our task is to discriminate among these states by performing the optimal measure-ment with minimum probability of error.Such measurement needs to have N outcomes corresponding, respectively, to each of the N states ρ i .A general measurement can be described by a positive operator-valued measure (POVM) [37][38][39], which is defined by a set of operators {π j } satisfying Each of the possible measurement outcomes j is characterized by the corresponding POVM element π j .The probability of observing outcome "j" when the state of system is "i" is Then, the probability of making a correct guess when the given set is {p i , ρ i } N i=1 will be In a minimum-error discrimination approach the aim is to seek the optimal measurement that gives rise to the minimum average probability of error occurred during the process or, equivalently, to the maximum probability of making a correct guess given by P guess = max{P corr }.This is achievable, as we mentioned previously, if and only if the POVM elements satisfy the Helstrom conditions (1) and (2).
To continue, let us reformulate conditions (1) and (2) by summing the second one over j which results in where we have defined ρi = p i ρ i .From these equations ( 6) and ( 7) (and its Hermitian conjugate π i (Γ − ρi ) = 0 ∀i), in the case that Γ − ρi is not a full rank operator, one can easily see that supp{Γ− ρi } ⊥ supp{π i }, or equivalently, supp{Γ− ρi } ⊆ ker{π i }, where supp{•} and ker{•} denotes the support and kernel of an operator, respectively, or in other words, equations ( 6) and (7) imply that both determinants det{Γ − ρi }, det{π i } have to be zero .So far, everything is quite general and no reference has been made to qubit states.In what follows we consider, however, the qubit cases.

A. Discrimination of qubit states
Solving a general problem for minimum-error (ME) discrimination is not an easy task and there is no analytical solution in general.However, there are some approaches for qubit states [29-32, 34, 35].In this work, we consider the problem of the most general case of qubit states using a geometric approach.We use the Bloch vector representation to write ρ i and Γ as Where v i = (v xi , v yi , v zi ) is the Bloch vector corresponding to the state ρ i , σ = (σ x , σ y , σ z ) is a vector constructed by Pauli matrices.Moreover, γ 0 = Tr(Γ) = P guess and γ = (γ x , γ y , γ z ).Non-negativity of ρ i requires 0 ≤ |v i | ≤ 1.The lower and upper bounds are achieved for the extreme cases of maximally mixed and pure states, respectively, so |v i | can be regarded as a kind of purity.Using Eqs. ( 8) and ( 9), one can write where we have defined the sub-normalized Bloch vector ṽi = p i v i associated with the state ρ i and its priori probability p i .But in this paper, for simplicity, we call both vectors v i and ṽi Bloch vectors.It follows from inequality (6) that So, this equation is equal to the main Helstrom condition, Eq. ( 1), and from now on, we will check optimality of our answers by Eq. (11).
To continue, if the inequality strictly holds, i.e. positive definiteness of the determinant of Γ− ρi , inevitably corresponding measurement operators π i have to be zero.But if equality holds, i.e. zero determinant of Γ − ρi , then π i does not have to be zero.So, in the case of nontrivial answer we obtain the following useful formula Regarding γ 0 = P guess , this relation implies that the distance between two vectors γ and ṽi is equal to the difference between two probabilities, i.e. the guessing probability and the priori probability p i associated with state ρ i .The greater p i , the closer are two vectors ṽi and γ.Keeping in mind that Eq. ( 12) does not generally hold for all states.For the sake of simplicity, without loss of generality, we consider that the set of {ρ i } N i=1 and corresponding priori probabilities {p i } N i=1 are arranged in such a way that the first {ρ i } M i=1 are those states that the rank of (Γ − ρi ) is less than 2, so it means that these states are detectable in optimal case and the other states {ρ i } N i=M+1 are states that for them (Γ − ρi ) is a full rank operator, so there is no way to detect them in an optimal way, i.e. π i = 0 for i = M + 1, • • • , N .With this terminology and using the fact that det(π i ) = 0 when Γ − ρi = 0 or full rank, it turns out that the measurement operators π i 's for i = 1, • • • , M must be rank-one and therefore proportional to projectors.As a qubit projector is characterized by its unit vector ni on the Bloch sphere, we have for i = 1, • • • , M , where α i 's are real and range from zero to one, α i = Tr(π i ).The completeness relation (3) requires the following conditions on the parameters α i 's where 0 ≤ α i ≤ 1.On the other hand, inserting Eqs.(10) and (13) in the Helstrom condition (7), and by using Eq. ( 12) we find the unit vector ni as Obviously, knowing vector γ is equivalent to know all the unit vectors ni corresponding to the nonzero measurements π i .
Corollary 1.As a result of Eqs.(14), γ has to be confined in the convex polytope of the points ṽi 's.
Corollary 2. It can be concluded from Eqs. (12) to (14) that by translation of all vectors ṽi 's, with unchanged p i 's, by a fixed vector inside the Bloch sphere the POVM answers will be unchanged, and the only thing that matters is relative locations of ṽi 's.So, if all of ṽi 's move with the same displacement vector a (ṽ i → ṽi +a), we expect the same answer.In other words, a set of {p i , ρ ′ i } has the same answer of measurement operators and guessing probability as the initial set {p i , ρ i }, where This class is characterized by an unchanged guessing probability.
From the above discussion it follows that in order to characterize the optimal measurements, we have to know both the vector γ and the real parameters α i 's.If γ is known, then by using Eq. ( 14) it is an easy task to find all possible sets of parameters {α i }.Each set of {α i } provides a different answer for the optimal POVM.So, finding γ is our priority that exposes all possible answers.As Γ is determined with a scalar γ 0 and a vector γ, we need at most four states to determine Γ satisfying Eq. (12).
On the other hand, if we are given a set of ME discrimination POVM {π i }, we are able to find the Lagrange operator Γ (or equivalently γ) with the help of relation Γ = i ρi π i .Then by using Eq. ( 14) we can obtain all possible optimal answers.

B. Types of states and measurements in an optimal strategy
Similar to [31] we distinguish the states in a general qubit state discrimination problem.We showed that γ has to be confined in the convex polytope of the points ṽi 's.So, based on different representation of γ the states in a discrimination problem can be considered as unguessable, nearly guessable, and guessable states [31].The states for which Γ − ρi is a full rank operator (inequality part of Eq. ( 11)) are called unguessable since their related POVM elements are always null.The nearly guessable states are the states that although for them the operator Γ − ρi is not full rank, they do not appear in any optimal measurement and therefor their related measurement operators are null, and finally guessable states are those states that satisfy Eq. ( 12) and their POVM elements are nonzero for some optimal measurements.
It is important to study under what conditions the optimal measurement of a discrimination problem is unique.First, based on the preceding discussion, we have the following proposition for a discrimination problem with N ≤ 4 Proposition 3. In case of N ≤ 4, the set of parameters {α i } satisfying Eq. ( 14) is unique if the states form a N − 1 simplex inside Bloch sphere.However, it is neither necessarily the case for N > 4, nor for N ≤ 4, where the states do not form a N − 1 simplex, in a sense that each possible set of {α i } leads to a complete solution for the optimal measurement operators.
Proof.The proof simply comes from the theorem 3.5.6 in [40].Based on this theorem, a point in an k-simplex with vertices x 0 , x 1 , • • • , x k can be written in a unique way of the vertices.So, in the problem of N ≤ 4 qubit state discrimination if these states form a simplex the convex combination for γ is unique.
Consequently, in the cases with non-unique solutions we may face with some problems with different POVM measurements that give us the same optimal guessing probability.Thus, one can think about the possibility of constructing different measurements by combining these measurements in a convex way.This fact motivates us to divide optimal measurements in two different families of measurements which we call them decomposable and non-decomposable measurements.
is an optimal measurement for some discrimination problems.According to (13) we are able to define the set of Bloch unit vectors E = {n 1 , n2 , • • • , nm nm nm } such that their convex polytope contains the Origin (Eq ( 14)).If one can find a subset E ′ ⊂ E in such a way that its related convex polytope still contains the Origin, then M is decomposable, otherwise, it is non-decomposable.Based on the three dimensional direction of ni ni ni 's, non-decomposable sets have at most four elements.
An exception for this definition is the existence of a measurement operator π j proportional to unit matrix 1, then, there is an one-element non-decomposable set If M is decomposable then there is at least one subset E ′ ⊂ E which its corresponding POVM, M ′ , is optimal for the same minimum-error discrimination problem {p i , ρ i } N i=1 .In other words, any subset E ′ ⊂ E with at most four elements of ni ni ni 's that contains origin in its convex polytope of its elements, and if we cannot discard any of its elements in a way that origin still be in the convex polytope of the remaining elements, then it is a non-decomposable subset.The proof for this proposition is straightforward according to the definition of decomposable and nondecomposable measurements.
C. Qubit states: Arbitrary priori probabilities Now, we are taking a step further by considering a general problem of a set of qubit states with arbitrary priori probabilities.Before we provide an instruction to find the answers of a general problem {p i , ρ i } N i=1 , consider that one is able to find the Lagrange operator Γ by using Eq.(12).Based on the previous discussion, we know that one of the answers of an ME problem is determined with at most four number of states ρ i 's satisfying Eq. ( 12), so one can test Step (i).-First, if there is a ρ j with the greatest priori probability (p j > p i , ∀i), it must be examined whether Γ is ρj or not.If Γ = ρj , we have from Eqs. ( 8) and ( 9) that γ 0 = p j and γ = ṽj .Then, to satisfy Helstrom condition, Eq. ( 11), it requires the following condition where p ji := p j − p i and dji := |ṽ j − ṽi |.
If the above condition holds for every i, the answer for POVM will be π i = δ ij 1 and P guess = p j .It is called no measurement strategy; simply guessing the state ρ j [41].
Step (ii).-If the above condition is not met, it has to be tested two qubit states together.But first we consider two qubit states ρ l and ρ m with arbitrary priori probabilities p l and p m ( p l ≥ p m ).By applying Eq. ( 12) for states ρ l , ρ m and then subtracting the results we get which defines a hyperbola with its foci at ṽl and ṽm .This hyperbola is characterized with parameters a, b, c 1.The hyperbola of Eq. ( 18), which only the left part of this hyperbola could be an allowed area for γ. (this hyperbola exists if Rlm be positive, Eq. ( 21)) which can be derived from Bloch vectors of states ρ l , ρ m and their corresponding probabilities p l , p m (see Fig. 1) as and It follows from Eq. ( 11) that only the left branch of the hyperbola could provide a candidate answer for γ.
In this case of two-state case the candidate γ is located at the distance of from point ṽl on the connecting line between ṽl and ṽm .As Rlm 's are distances, they are definitely positive.So, possible candidates for γ 0 and γ are and where the primes indicate that they are still candidate answers.Now, the question is: which two states to pick?
In this case we have the following proposition Proposition 5.In a case where a two element optimal POVM exists the answer can be obtained by considering two states which maximize γ ′ 0 lm .Proof.For simplicity, consider that γ ′ 012 is the maximum.Assume that there are states i and j with γ ′ 0ij < γ ′ 012 that satisfy Helstrom conditions.It means that then, writing inequality for l = 1, 2 and summing them gives the second inequality comes from the triangle inequality.So, we end up in γ ′ 0ij > γ ′ 012 which is in contradiction with our first assumption γ ′ 0ij < γ ′ 012 .
To continue, we should find two states which maximizes Eq. ( 22).Then test these two states using the following condition which comes from the Helstrom condition, Eq. ( 11).The cases i = l, m are trivial.If the Eq. ( 22) is maximized for more than a pair of states, we must test Helstrom condition for each of them.Note that ME discrimination might have some answers and what we need is just one of them to find γ.If the condition ( 27) is not met we proceed to the next step.
Step (iii).-Inthis step we consider three-state case.Let us label them by l, m, n. γ ′ is constructed from these three states by drawing three hyperbolas arising from each pair of states l, m, n (lm, ln, mn).The convex polytope in this case is a triangle and it is enough to just calculate the point that two hyperbolas meet (γ ′ ).This point must be inside the triangle.The related probability can be obtained from Eq. (38) .As before, for three states that maximizes this probability we must test Helstrom condition.The cases i = l, m, n are trivial.Likewise the second step, we might have more than one three-state case to maximize Eq. (38).In this case we must test the Helstrom condition for each of them.
Step (iv).-Ifwe still do not have the answer, we have to consider four-state case.This step must reveal the answer since we know that there is always a nondecomposable answer with maximum four number of states.In this case, although there are 4  2 = 6 hyperbolas, it is enough that three hyperbolas, with a common focus point ṽm , meet at one point inside the polytope to obtain γ ′ .The related probability can be obtained from Eq. (50).A four-state case that maximizes this probability must be states we are searching for to find γ.As before, this four-state case that maximizes Eq. (50) might not be unique.Deriving of equation (38) and Eq.(50) for steps (iii) and (iv) respectively will be discussed in next section.
By looking at the Bloch representation of Γ, Eq. ( 9), we see that there are just four unknowns to be determined, a scalar γ 0 and a three dimensional vector γ.To this purpose, one can easily find them by considering four qubit states with their corresponding Bloch vectors (See Eq. (12).So, we do not expect to have more than four states.Furthermore, as a typical ME discrimination problem has, at least, a non-decomposable answer, one can see that these four steps are enough to find γ.Although we introduced a way to find required states which lead to the Lagrange operator, the fourth step seems less probable, because we often expect that to have at least one non-decomposable POVM answer with less number of elements for a given probable.
At this point, it is useful to explicitly express some remarkable results: • According to the above instruction, equation (18) has a key role to find γ, which is unaffected as long as the relative distances between states dlm 's, and also priori probabilities p i 's are consistent.It suggests that by translation or rotation of the polytope, or equivalently all ṽi 's as a whole, with a fixed vector inside the Bloch sphere, P guess is consistent.This class is characterized by an unchanged guessing probability.
• Furthermore, by translation and rescaling of the polytope inside the Bloch sphere, with unaffected p i 's, measurement operators which are defined by Eq. ( 13) will also be unchanged.This class is also characterized by an unchanged measurement operators.
• Based on the mentioned instruction, to have a complete solution of a general qubit state discrimination we need to find all non-decomposable sets, of the problem (It can be simply done, after finding γ, by looking at the geometry of the ni 's; see the explanation after proposition (4)).Then, we are able to obtain all of the possible optimal POVM's using convex combinations of all non-decomposable sets, i.e. k i=1 β i M ′ i .Where k is the number of all non-decomposable sets, 0 ≤ β i ≤ 1, and k i=1 β i = 1 to satisfy completeness relation of a POVM measurement.For this purpose, in the next section, we will solve the problem for these cases analytically.Some of these results like the problem of two states discrimination have been known for a long time.

B. Three States
We now consider a general case of three arbitrary qubit states with priori probabilities p 1 ≥ p 2 ≥ p 3 .To proceed further, note that discrimination of an arbitrary set of three qubit states can be reduced to the discrimination of three qubit states, all embedded in x− z plane, defined by where θ and φ are defined in Fig. 2. For a proof see Appendix A. The corresponding Bloch vectors ṽi 's are given by ṽ1 = (0, ap 1 ), ṽ2 = (bp 2 sin θ, bp 2 cos θ), ṽ3 = (−cp 3 sin φ, cp 3 cos φ), where (x, z) is a simplified representation for (x, 0, z).With the assumption that p 1 is the greatest priori probability, first we have to check whether ρ1 is equal to Γ or not.So, with the help of Eq. ( 17), we need to check the following conditions Where |(x, z) t | = √ x 2 + z 2 .Obviously, if the above conditions are not met, we must calculate P guess from Eq. ( 22) for every two states and then check the Helstrom condition using Eq. ( 27).The explicit form of this conditions to detect one of the pairs of (ρ 1 , ρ 2 ), (ρ 1 , ρ 3 ) or (ρ 2 , ρ 3 ), respectively, are with the corresponding γ ′ ij = (x ′ ij , z ′ ij ) t for these states from Eq. ( 23) where p ij := p i − p j and Rlm is defined in Eq. ( 21).
For any specific problem with known p i , θ and φ, each condition of Eqs.(33) to (35) that is satisfied will be the only solution of the problem.Accordingly, the optimal POVM elements are given by π i = 1 2 (1 + ni • σ), (see Eq. ( 13)), where the corresponding unit vectors of the nonzero measurement elements can be written as Finally, if none of the above conditions were met, the solution for γ must be found while all three states are detectable.It means that three hyperbolas should meet at one point.Note that since three points are embedded in a plane it is enough to find the intersection between two hyperbolas in this plane.So, by using the properties of hyperbolas the guessing probability can be obtained as where ).( 40) To find the vecor γ, we can use the geometric of triangle in Fig. 3.For this purpose, by using the Gram-Schmidt method, we can use two edges of the triangle ṽ2 − ṽ1 and ṽ3 − ṽ1 to construct an orthonormal basis from them for the plain which the triangle lies in.If we show the orthonormal basis by k1 and k2 , then where Hence, γ can be written as So, POVM elements can be obtained using Eqs.( 13) and (15).
It is of the notice that we first altered the Bloch vectors with a rotation matrix, Appendix A, and a displacement.In case of ρ i and ρ j to be two detectable states, optimal operators are π i = ρ ni and π j = ρ −n i with ni = ṽi −ṽ j dij .So, the answers of the optimal operators {π i } are easily obtained with the vectors of the original problem.But, for the case in which three states are detectable, after obtaining ni 's we must rotate them back, with inverse of rotation matrix, to the original problem and then use the formula π i = ρ ni to get the optimal operators for the original problem.
Here, for more illustration of this case, let us consider discrimination among trine states, i.e. the qubit states associated with equidistant points on the surface of the Bloch sphere.They are defined by Then, based on the previous discussion, it is possible to rotate these states on the Bloch sphere to align them on x − z plane By writing the corresponding density matrices and comparing them with Eq. ( 29), we get a = b = c = 1 and θ = φ = 2π/3.
To continue, we first assume the case with equal priori probabilities.As the three states are pure and located on the surface of the Bloch sphere, finding the answer is straightforward.According to the discussion in Sec.II A (or more straightforward, later, from Appendix.C), the guessing probability is P guess = 2 3 and the corresponding POVM operators for the original problem are π i = 2  3 |ψ i ψ i |, known as the trine measurement [23].Next, we consider the case with arbitrary priori probabilities.We use the same parameterization of Weir et al. [36] for priori probabilities; p 1 = p + δ, p 2 = p − δ, and p 3 = 1 − 2p, where the assumption 0 < p 3 ≤ p 2 ≤ p 1 implies that 1  3 ≤ p ≤ 1 2 and 0 ≤ δ ≤ min{3p − 1, p}.Following our method of Sec.II C, one can see that the "no measurement strategy" cannot be optimal for γ 0 = p 1 and γ = p 1 ẑ.It is obvious by investigating that the condition p 1i ≥ d1i is not satisfied for every i.Equations ( 31) and (32) give the same result as well.So, we must have either two or three POVM measurement elements.For two-state case one can see that Eq. ( 22) is maximum for states ρ 1 and ρ 2 .In this case Eq. ( 33) gives where d12 = 3p 2 + δ 2 .Solving Eq. ( 46) for δ, gives us four roots, which the only valid values for In this region of two-state case the guessing probability is In the region where the POVM measurement has to be a three-element POVM, we can use Eq. ( 18) for three states to find γ = (γ x , γ y , γ z ).By using these equations and some algebraic calculations, the desired γ can be obtained as Using P guess = γ 0 = p i + |ṽ i − γ| or Eq. ( 38) the guessing probability can be calculated The equations ( 47) and (48), and this P guess are in complete agreement with [36] that was obtained with a different approach.

C. Four States
In this section we consider a general problem of four qubit states.According to the previous discussion, we can divide this problem into two cases: (i) When the four states form a two dimensional convex polytope, in this case according to Caratheodory theorem the vector γ can be written as a convex combination of at most three points, therefore the optimal measurement has at most three nonzero elements.In this case, we only need to solve the problem either by using the three steps instruction in the Sec.II C. (ii)When four states form a three dimensional polytope, i.e. a tetrahedron.Based on the type of states a four element optimal POVM may exist.If so, to find the guessing probability and optimal measurement we use the properties of tetrahedrons.Each tetrahedron is composed of four triangle faces, six edges, and four vertices.Based on the location of γ, the number of detectable states in an optimal way can be specified.If γ be on one vertex then an optimal answer is a no measurement strategy and the other three states are unguessable.If it lies on one edge then an optimal measurement will be a POVM with two non-zero elements, i.e. two guessable states.Furthermore, lying γ on one of the faces means that three states can be detected through an optimal measurement and one state will remain unguessable.Finally, if γ be an interior point of tetrahedron then all four states are guessable, i.e. six hyperbolas meet at a single point.To find the guessing probability in this situation, let us show the vector γ as an interior point in the tetrahedron ṽ1 ṽ2 ṽ3 ṽ4 , therefore the guessing probability can be written as (51) we end up with P guess = ζ 2 which we will later see is consistent with result of Corollary 7 for equal a priori probability states, as well as the result obtained in [32].For other cases, based on the values of α and β we can have different optimal answers including two-element and three-element measurements.Fig. 5, for different values of α and β, shows the simplest answer.In this case, since four vectors in (57) constitute a thetrahedron, all answers are unique answers.
Knowing the optimal answers for N ≤ 4, we have everything to solve the general problem of N qubits states.For N > 4 cases since the optimal answer will be fulfilled with a non-decomposable optimal POVM with at most four non-zero elements, we can use the guessing probabilities that were found for N ≤ 4 cases.For this purpose, the guessing probability of a problem of N states ,{p i , ρ i } N i=1 , can be rewritten as where S is a subset of {ρ i } N i=1 with the number of elements N S (N s ≤ 4).P S guess in this relation can be obtained by using equations ( 22), (38) and (51).

IV. N > 4 CASES
Now that we know how to find answers for the cases N = 2, 3 and 4, we have all we need to solve a general problem of N qubit states.To simplify this task and save our time, we wrote a simple code with the Mathematica.This code is based on our instruction and its primary task is searching for the first optimal answer including γ and P guess [42].
To start, we first consider an example of a five qubit states consisting of four equiprobable symmetric qubit states (57) (p 1 = p 2 = p 3 = p 4 = p), and one more state v 5 = ζ(sin θ cos φ, sin θ sin φ, cos θ) with a different prior probability δ = 1 − 4p.We want to analyze how this state disturbs the optimal answer of four symmetric states which we already know their optimal answer is a four-state case for small δ.For δ ≤ 0.2, one can observe that the only optimal answer can be obtained by a four-element measurement.So, the optimal measurement is still the same as the problem of four equiprobable symmetric qubit states with the same purity, i.e. γ = 0 and not disturbed by the new state.However, in this case the guessing probability decreases by increasing δ (P guess = 2pζ = ζ 1−δ 2 ).As δ exceeds this value (δ > 0.2) other cases will appear as well.Fig. 6 shows the situation for δ = 0.22 and p = 0.195.Based on the location of the state ρ 5 on the Bloch sphere all three cases can happen.As δ increases, the region of four-state case becomes smaller and finally disappears, i.e. γ can be revealed with either two or three-state cases.And for high values of δ, in most cases, a two-state case will give the optimal answer.
To continue, we first reconsider all the five states of the previous example.Then, we change its third and fifth states.The angles of state ρ 3 in spherical coordinate are {3π/4, arccos (−1/ √ 3)}.We replace its azimuthal angle with a variable ω.Then, we can set the polar angle of ρ 5 to π/3.Their prior probabilities are unchanged as before.So, there are three fixed states, with two rotating states on circles.Fig. 7 shows the answers for 0 < φ, ω < 2π.
For another example, we want to consider a more general case of six qubit states with no symmetries or equal The first qubit state is parametrized by two spherical angles φ and θ, so it is free to rotate on a sphere of radius 0.85.Fig. 8 shows the different regions for which the γ can be obtained.Based on this figure, in many cases a three-element measurement will fulfill the guessing probability and there is no need to go to the next step to find γ.Moreover, the region for four-element measurements are small.However, it should be noted that it does not mean there is no optimal four-element measurements for the rest of region where two-state and threestate are answers.For instance, for θ = 1.585948 and φ = 1.288376, there are three possible optimal measurements with unique γ = (0.070525, 0.057738, 0.048073) and P guess = 0.370574.Two of them which are nondecomposable answers can be obtained by considering two three-state sets of {ρ 1 , ρ 2 , ρ 3 } and {ρ 1 , ρ 3 , ρ 4 }.The third one which is decomposable can be obtained by four states {ρ 1 , ρ 2 , ρ 3 , ρ 4 }.We note that the optimal answer in green area can not be obtained either with two-element or three-element measurements.
As the last example, let us consider the special case of N qubit states ρ i with equal a priori probabilities i.e. p i = 1/N for i = 1, • • • , N .We show in appendix C in order to reach P guess , we have to choose a sphere with maximum number of states lying on.It is simply the minimal sphere covering all the v i 's (the same result was obtained in [33] with a different approach).We call it circumsphere, defined by {R, O}, where R = |v i − O| is its radius, and O is its circumcenter (Fig. 9).In view of this and the fact that 0 < R ≤ 1 and P guess = γ 0 , we get Moreover, Eq. ( 15) reduces to A more detailed discussion on the case of equal a priori probabilities can be found in the appendix C.

V. CONCLUSION
In this paper, we have revisited the problem of minimum-error discrimination for mixed qubit states.
For this aim, we employed the necessary and sufficient Helstrom condition in a constructive way to obtain the discrimination parameters for a typical problem of ME qubit state discrimination.Our tools in this way are the representation of qubit states in terms of the Bloch vectors.For the case of arbitrary priori probabilities each two qubit states construct a hyperbola and the desired γ will lie on one of its parts which is next to the more probable state.Using these tools, we introduce an instruction to find the Lagrange operator Γ.Then, with this Lagrange operator, we can find all optimal POVM measurements.
We also discuss some properties of the POVM answers involving the geometric of the polytope of qubit states inside the Bloch sphere, and introduce some classes of answers like the classes of unchanged guessing probability and unchanged measurement operators.
We show that for an optimal strategy, there might be some states that are not detectable i.e. their related POVM elements are zero.So, in the problem of ME discrimination of N qubits {ρ i } N i=1 some states might be undetectable.We indicate them as unguessable, nearly guessable, and guessable states, assuming that there are M number of guessable states in a typical optimal problem (1 ≤ M ≤ N ).
We show that every POVM set M , can be divided into a limit number of non-decomposable POVM subsets E ′ .Finding all of these subsets is an alternative way of constructing a general ME answer of the given problem.They might be also practical in a case that detecting some states ρ j 's are not interested, so we might be able to use those non-decomposable subsets that do not include nj nj nj 's.It is also more reasonable when preparing measurement operators are expensive.
To illustrate the proposed instruction, we need to know the solutions for N ≤ 4. So, we solved the problem for these cases.The case of two states has been known for long time (Helstrom formula).For the case of three qubit states, by applying some rotation and translation we show that the problem can be reduced to the problem of three qubit states in x − z plane and therefore we obtain a full analysis of the problem using our approach.Then, as an specific case, the problem of trine states with arbitrary priori probabilities is considered.We show that the results are in complete agreement with previous findings.Moreover, We also solve the case of four qubit states for the first time, using the geometry of tetrahedron and the intersection of 4  2 = 6 hyperbolas deriving from each two states.
And finally, we use with this instruction to solve some example for the cases with N ≥ 4 including examples of five and six qubit states with non-equal prior probabilities and the general case of N states with equal priori probabilities.In the later case , equal priori probabilities, finding Γ corresponds to finding a sphere with maximum number of states on it.
Appendix A: The Rotation Matrix for three qubit states Discrimination Any qubit state ρ i can be identified with a point inside the Bloch sphere using its Bloch vector v i .The same is true for multiplication of the state by its priori probability i.e. ρi = p i ρ i and ṽi = p i v i .Three states together with their priori probabilities represent three points inside the Bloch sphere.From geometry we know that three non-collinear points determine a plane and each plane can be described by its normal vector which is a vector orthogonal to the plane (i.e.orthogonal to every directional vector of the plane).Having the normal vector in hand, in the next step we can find the rotation matrix which rotates this vector to align it in the y-direction.This rotation matrix then rotates each ṽi in such a way that the y-component of all ṽi become equal.With a translation along y-axis, one can eliminate the y-components of these rotated vectors.Finally, with an additional rotation in x − z plane, we can rotate ṽi in a way that the state with largest priori probability be aligned in the z-direction.Since these rotations and the translation does not affect the relative distances and angles between states, the guessing probability to this new set of states is equal to the original one.The POVM's can be easily related as well, with a rotation, as was explained in the Sec.III B.
To obtain corresponding rotation matrix consider three points P 1 ,P 2 and P 3 in the Bloch sphere optimal measurement operators of the first set { 1 N , ρ i } N i=1 are still optimal answers for the new set { 1 N +K , ρ i } N +K i=1 because defined ni 's from the first problem with N states are still unchanged for the new problem, Eq. (C7).However, this is not the only answer of the new problem.(iii) The guessing probability of the new problem is given in terms of the guessing probability P guess = 1 N (1 + R) of the original one as P N ew guess = N N +K P guess , Eq. (C5).Corollary 8. Consider the case of N equiprobable qubit states with the same purity |v i | located on the circumsphere {R, O}.If any of the following statements be true then the other ones will also be true.In other words, these statements can be interchangeably used in this particular case.
(ii) The radius of the circumsphere is given by R = |v i |.
All implications are trivial and can be inferred by the results given above.A particular case is when there are N equiprobable pure states, |v i | = 1.It follows from the corollary that the guessing probability of N equiprobable pure qubit states with O at Origin is an example which achieves its maximum value P guess = 2/N .This is because the radius of this circumsphere can not be greater than one.
to reach the answer!But, with the following instruction we can find it very quicker.

FIG. 3 .
FIG. 3. Representation of three qubit states problem when a three elements non-decomposable POVM is optimal.

FIG. 5 .
FIG. 5.The answers of four qubit states with the same purity ζ and different prior probabilities depending on two parameters α and β.When α and β intends to zero the answer is obtained by considering all four states.By increasing α and β, γ can be revealed by either two-state or three-state case.

FIG. 6 .
FIG. 6.The answers for five qubit states, all with the same purity ζ.Four equiprobable symmetric states (p = 0.195) with one disturbing state (p5 = δ = 0.22) rotating on a sphere with radius ζ and angles {φ, θ}.In this case, three-state case in most cases reveals γ.
these three points is defined byax + by + cz + d = 0, (A2)where the coefficients a, b and c determine the components of normal vector n vector n is the unit vector orthogonal to every direction vector of the plane, it can be obtained by the following equation n = (P 2 − P 1 ) × (P 3 − P 2 ) |(P 2 − P 1 ) × (P 3 − P 2 )| .(A4)