particular quantum mechanical systems, e.g., the binary case [1]. Nevertheless, since the objective is to search a global maximum of a linear function into a convex set, the problem can be faced by means of numerical tools such as SDP. Besides, according to classical results in convex optimization theory [8], in place of the primal problem, it is in general more convenient to consider its corresponding dual problem, since it presents a smaller number of variables and constraints [3, 6]. Dual problem ( DP1 ). Minimize the trace of the optimization operator X over the class P , subject to the constraints X ≥ q$_{i}$ρ$_{i}$ , i = 0 , 1 , . . ., M − 1 . Once found a minimum trace operator X$_{opt}$ , its trace gives the maxi- mum probability of correct detection, P$_{c}$ = Tr ( X$_{opt}$ ) . The equalities ( X$_{opt}$ − q$_{i}$ρ$_{i}$ )Π$_{i}$ = Π$_{i}$( X$_{opt}$ − q$_{i}$ρ$_{i}$ ) = 0 , i = 0 , 1 , . . . , M − 1 , are necessary conditions on the optimal POVM. These conditions become sufficient, once the searched measure operators are constrained to belong to P and to solve the identity on H . Primal problem for symmetric quantum states ( PP2 ). Find the maximum of P$_{c}$ = Tr ( ρ$_{0}$ Π$_{0}$) , with Π$_{0}$ ∈ P and such that ∑ M − 1 i =0 S $^{i}$Π$_{0}$ S − i = I . Proof: This formulation was first given in [6]. It can straightforwardly be proved by using (1) and (2) in PP1 , so that P$_{c}$ = ∑ M − 1 i =0 q$_{i}$ Tr ( ρ$_{i}$ Π$_{i}$) = ∑ M − 1 i =0 1 $_{M}$Tr ( S $^{i}$ρ$_{0}$ S − $^{i}$S$^{i}$Π$_{0}$ S − $^{i}$) = Tr ( ρ$_{0}$ Π$_{0}$) . Moreover, if Π$_{0}$ ≥ 0 then Π$_{i}$ = S $^{i}$Π$_{0}$ S − i ≥ 0 . ■ The optimization problem PP2 is comparatively sim- ple, and, perhaps, this is the reason why no particular attention has been paid in the literature to the study of the dual theorem to obtain an alternative formulation. In the following we investigate this point. Dual problem for symmetric quantum states ( DP2 ). Minimize the trace of the optimization operator X over the class P , subject to the constraints X ≥ 1 $_{M}$ρ$_{0}$ and XS = SX . Once found a minimum trace operator X$_{opt}$ , its trace gives the maximum probability of correct detection, P$_{c}$ = Tr ( X$_{opt}$ ) . Proof: Define ρ ′ i = S $^{i}$( 1 $_{M}$ρ$_{0}$ ) S − $^{i}$. Let V be the fea- sible set according to the general dual problem in Sec- tion II, i.e., the set of PSD operators X such that X ≥ ρ ′ $_{i}$, i = 0 , 1 , . . . , M − 1 , and let V ′ be the set of PSD opera- tors X ′ such that X ′ ≥ ρ ′ 0 and X $^{′}$S = SX $^{′}$. The proof is organized in two steps. In the first step it is shown that V ′ ⊂ V , while in the second step it is proved that for any X ∈ V there exist X ′ ∈ V ′ such that Tr ( X $^{′}$) = Tr ( X ) . Then, the search of X can be confined into V $^{′}$. - Step 1 : If X ′ ∈ V $^{′}$, for the commutativity between X ′ and S we get X ′ = SX $^{′}$S − 1 and recursively X ′ = S $^{i}$X $^{′}$S − $^{i}$. Hence, for any i In quantum detection an important role is played by geometrically uniform symmetry [4–6]. Among the sev- eral generalizations, we consider the basic case of sym- metric mixed quantum states generated from a reference density operator ρ$_{0}$ as where the symmetry operator S is unitary ( SS † = S $^{†}$S = I ) and such that S M = I . The geometry implicitly requires that the mixed states are equiprobable, i.e., q$_{i}$ = 1 /M , i = 0 , 1 , . . ., M − 1 . In [6] it was shown that optimal POVM having the same symmetry can always be found. Hence, without loss of generality, we can assume that where Π$_{0}$ is the reference measure operator. Conse- quently, for a fixed M , the knowledge of ρ$_{0}$ , Π$_{0}$ and S is sufficient to fully describe the state ensemble and the POVM. The density operators (1) have all the same rank, and the same holds for the measure operators (2). As proved in [9], the optimal measure operators can be assumed to have rank no higher than that of the corresponding density operators, namely rank (Π$_{0}$) ≤ rank ( ρ$_{0}$ ) . Moreover, recalling that S = I Being X ≥ ρ ′ i for each i , it follows that since X ′ ≥ ρ ′ 0 for assumption. - Step 2 : For each X ∈ V we consider The specific geometry of the state ensemble can be exploited to get insight into how solving the state dis- crimination problem. In particular, the primal problem PP1 can be rewritten in a simpler form as follows. SX $^{′}$S − 1 = 1 M M − 1 ∑ i =0 S − ( i − $^{1)}$XS i − 1 = X ′ X ′ ≥ 1 M M − 1 ∑ i =0 S − $^{i}$ρ ′ $_{i}$S i = ρ ′ 0 . X ′ = 1 M M − 1 ∑ i =0 S − $^{i}$XSi . X ′ − ρ ′ i = X ′ − S $^{i}$ρ ′ $_{0}$S − i = S $^{i}$X $^{′}$S − i − S $^{i}$ρ ′ $_{0}$S − i = S $^{i}$( X ′ − ρ ′ $_{0}$) S − i ≥ 0 , (3) ρ$_{i}$ = S $^{i}$ρ$_{0}$ S − i , i = 0 , 1 , . . ., M − 1 , (1) Π$_{i}$ = S $^{i}$Π$_{0}$ S − i , i = 0 , 1 , . . ., M − 1 , (2) III. DISCRIMINATION OF SYMMETRIC QUANTUM STATES B. Dual Problem for Symmetric Quantum States A. Primal Problem for Symmetric Quantum States 2