Structure of minimum-error quantum state discrimination

The necessary and sufficient conditions that POVMs must satisfy to fulfill optimal discrimination have been obtained from the very beginning when the problem was introduced in [1, 2]. For given states {q_x,ρ_x}^N_x=1 to discriminate among, POVM elements {M_x}^N_x=1 achieve the guessing probability if and only if they satisfy

$$\begin{equation} M_{{x}} (q_{{x}} \rho_{{x}} - q_{{y}} \rho_{{y}}) M_{{y}} = 0 \quad \forall {x}, \;{y}=1,\ldots, N, \end{equation} \tag{ 4 }$$

$$\begin{equation} \sum_{{x} =1 }^{N} q_{{x}} \rho_{{x} } M_{{x} } - q_{{y}} \rho_{{y}} \geqslant 0 \quad \forall\; {y}=1,\ldots, N. \end{equation} \tag{ 5 }$$

That is, for a state discrimination problem, if one finds a set of POVMs that satisfy two of the conditions above, then optimal discrimination is immediately obtained. Note also that for a given problem of state discrimination, optimal measurement is generally not unique.

The optimality conditions can also be derived from a different context, the formalism of convex optimization [21]. The optimization problem in equation (1) with the convex and affine constraints on POVMs can be written as

$$\begin{eqnarray} \max& & \sum_{{x} =1}^{N} q_{{x}}\, \mathrm{tr}[M_{{x}} \rho_{{x}}] \nonumber \\ \mathrm{s.t.} ~\mathrm{} & & \sum_{{x} =1 }^{N} M_{{x}} = I\quad \mathrm{and}\quad M_{{x}} \geqslant 0\quad \forall\; {x} =1,\ldots, N. \end{eqnarray} \tag{ 6 }$$

For convenience, we call equation (6) the primal problem. Note that the primal problem in the above is feasible.

To derive its dual problem, let us first construct the Lagrangian $\mathcal {L}$ as follows:

$$\begin{eqnarray*}&&\fl \mathcal{L}(\{M_{{x}} \}_{{x}=1}^{N}, \{\sigma_{{x}} \}_{{x}=1}^{N},K) = \sum_{{x}=1}^{N} q_{{x}}\, \mathrm{tr}[M_{{x}} \rho_{{x}}] + \sum_{{x}=1}^{N} \mathrm{tr}[\sigma_{{x}} M_{{x}}] + \mathrm{tr}\left[K \left(I - \sum_{{x}=1}^{N} M_{x}\right)\right] \\ &&= \sum_{{x}=1}^{N} \mathrm{tr}[M_{{x}} (q_{{x}} \rho_{{x}} +\sigma_{{x}} -K ) ] + \mathrm{tr}\,{K} ,\nonumber \end{eqnarray*}$$

where {σ_x}^N_x=1 are positive-semidefinite Hermitian operators and K is a Hermitian operator. Note that {σ_x}^N_x=1 and K are called dual parameters. To derive the dual problem, we first have to maximize the Lagrangian over primal parameter {M_x}^N_x=1. If it holds that q_xρ_x + σ_x − K > 0 for some x, then the primal problem becomes not feasible since the maximization may go to +∞. Thus, we have q_xρ_x + σ_x − K ⩽ 0 for all x, where the equality holds if and only if the Lagrangian is maximized. Combined together with the other constraint that σ_x ⩾ 0 for all x, the constraint can be finally rewritten as, K ⩾ q_xρ_x for all x. This also shows that the operator K must be positive. Note that since the Lagrangian is maximized, we now have

$$\begin{eqnarray*} &&\max_{\{ M_{{x}} \}_{{x}=1}^{N}} \sum_{{x} =1}^{N} q_{{x}}\, \mathrm{tr}[M_{{x}} \rho_{{x}}] \leqslant \max_{\{ M_{{x}} \}_{{x}=1}^{N}} \mathcal{L}(\{M_{{x}} \}_{{x}=1}^{N}, \{\sigma_{{x}} \}_{{x}=1}^{N},K) \vert_{q_{{x}} \rho_{{x}} +\sigma_{{x}} -K \leqslant 0}= \mathrm{tr}\, K, \end{eqnarray*}$$

which is called the weak duality.

The dual problem to equation (6) is therefore obtained as follows:

$$\begin{eqnarray} \min & & \mathrm{tr}[K] \nonumber \\ \mathrm{s.t.} ~\mathrm{} & & K \geqslant q_{{x}} \rho_{{x}}\quad \forall~{x}=1,\ldots, N. \end{eqnarray} \tag{ 7 }$$

Putting optimal discrimination problems into this convex optimization framework, the solution, i.e. the guessing probability, is returned efficiently in a polynomial time. In general, solutions from primal and dual problems in the convex optimization are not necessarily the same. It generally holds that either of the solutions is larger than or equal to the other, from the property called weak duality, as is shown in the above. In this problem of state discrimination, it turns out that solutions from both problems coincide each other, which follows from the so-called constraint quantification. This property is called the strong duality. Thus, we have

$$\begin{eqnarray*} &&{P}_{\mathrm{guess}} = \max_{ \{ M_{{x}} \}_{{x}=1}^{N} } \sum_{{x} =1}^{N} q_{{x}}\, \mathrm{tr}[M_{{x}} \rho_{{x}}] = \min_{K\geqslant q_{{x}} \rho_{{x}} } \mathrm{tr}[K]. \end{eqnarray*}$$

That is, the guessing probability is obtained by solving either the primal or the dual problem.

Apart from solving either the primal or the dual problem, a third approach in the convex optimization, which we are mainly going to consider here, is the so-called complementarity problem that directly analyzes the constraints characterizing the optimal parameters in both primal and dual problems and finds all of the optimal parameters [30]. As both the primal and dual parameters from both convex constraints are taken into account, the approach itself is not considered to be more efficient in numerics than primal or dual problems. Its advantage, however, lies in the fact that the general structure existing in an optimization problem is fully exploited.

Those constraints which characterize optimal solutions are called optimality conditions and can be written as the so-called Karush–Kuhn–Tucker (KKT) conditions. In general, KKT conditions are only necessary since solutions of primal and dual problems can be unequal. As we have mentioned, from the constraint qualification the strong duality holds and consequently the KKT conditions in this case are also sufficient. The list of KKT conditions for the discrimination problem is, then, constraints in equations (6) and (7) and two more conditions in the following.

Lemma 2.2 (Optimality conditions).  For a set of states {q_x,ρ_x}^N_x=1, the optimal ME discrimination is characterized by a symmetry operator denoted by K^⋆ and complementary states {r^⋆_x,σ^⋆_x}^N_x=1, where r^⋆_x ⩾ 0, which satisfy the followings, for x = 1,...,N:

$$\begin{equation} \mathrm{(symmetry}~ \mathrm{operator)}\quad K^{\star} = q_{{x}} \rho_{{x}} + r_{{x}}^{\star} {\sigma}_{{x}}^{\star}, \end{equation} \tag{ 8 }$$

$$\begin{equation} \mathrm{(orthogonality)} \quad r_{{x}}^{\star} \,\mathrm{tr}[M_{{x}}^{\star}{\sigma}_{{x}}^{\star}] = 0, \end{equation} \tag{ 9 }$$

where POVM elements {M^⋆_x}^N_x=1 in the above are an optimal measurement giving the guessing probability.

Conversely, for given states {q_x,ρ_x}^N_x=1, any set of parameters, K, {r_x,σ_x}^N_x=1 and {M_x}^N_x=1 fulfilling KKT conditions, equations (6)–(9), characterize the optimal state discrimination. If parameters satisfy KKT conditions, they are optimal and we write them by K^⋆, {r^⋆_x,σ^⋆_x}^N_x=1 and {M^⋆_x}^N_x=1, throughout.

Note that, with optimal parameters, the guessing probability is given by P_guess = tr[K^⋆] from the dual problem in equation (7).

Let us explain how one can obtain two conditions in equations (8) and (9). In the context of convex optimization, the condition of symmetry operator in equation (8) follows from the Lagrangian stability: $\nabla _{M_{{x}}}\mathcal {L}=0$ for all x. The orthogonality condition in equation (9) is obtained from the complementary slackness [21]. In state discrimination, therefore, those parameters satisfying KKT conditions, equations (6)–(9), are characterized as optimal measurements and complementary states to give guessing probabilities.

We in particular call K^⋆ as the symmetry operator in the ME discrimination of states {q_x,ρ_x}^N_x=1 due to the following reasons. First, the operator is uniquely determined for the ME discrimination of given states, whereas optimal measurement is not unique e.g. [36].¹ Later, the proof is provided in lemma 2.4. This means that for given quantum states, the symmetry operator, rather than optimal measurement, characterizes ME discrimination. Next, due to the uniqueness, the operator preserves the properties concerning the guessing probability in ME discrimination. Note that the guessing probability is of practical importance when exploiting sets of quantum states in a communication task. Then, as we will explicitly show later in examples, the guessing probability does not generally depend on detailed relations (e.g. angles or distances between states) of given states {q_x,ρ_x}^N_x=1 but a single parameter, the operator K^⋆ constructed by given states.

Let us also be precise about the result that the guessing probability does not depend on detailed relations of given quantum states. We point out that the observation from the two-state discrimination [3], where the trace-distance defining the relation between the two states is the parameter dictating the guessing probability, cannot generalize to more than two states. In addition we show, by examples, that it clearly fails to generalize the observation for ME discrimination of more than two states. As is going to be shown in section 4, one of three states can be modified independently, while the guessing probability remains the same. We emphasize that the operator K^⋆ directly dictates and corresponds to the guessing probability, rather than any other parameters in ME discrimination. We also note that this property is along the conclusion in [31] that distinguishability is a global property that cannot be reduced to the distinguishability of each pair of states.

In fact, two distinct sets of quantum states can have the same symmetry operator. Then, since the symmetry operator gives the complete characterization of the ME discrimination such as the guessing probability and complementarity states, the ME discrimination for the two sets is analyzed in terms of the same symmetry operator. This motivates one to construct equivalence classes of sets of quantum states via the symmetry operator in the ME discrimination and shows a general structure in ME discrimination, see section 3.

In the below, we show that optimality conditions in equations (8) and (9) are equivalent to those in equations (4) and (5).

Remark 2.1. KKT conditions in equations (8) and (9) are equivalent to the optimality conditions shown in equations (4) and (5).

Proof. To prove the equivalence, it suffices to show that KKT conditions imply optimality conditions in equations (4) and (5). In the following, we derive equations (4) and (5) from KKT conditions.

First, since the symmetry operator in equation (8) gives the guessing probability, we have

$$\begin{equation} P_{\mathrm{guess}} = \mathrm{tr}[K^{\star}] = \mathrm{tr}\left[\sum_{{x}=1}^{N} q_{{x}} \rho_{{x}} M_{{x}}^{\star}\right]. \end{equation} \tag{ 10 }$$

It follows that $K^{\star } = \sum _{{x}=1}^{N} q_{{x}} \rho _{x} M_{{x}}^{\star }$ with optimal POVMs fulfilling equation (10). From equation (8), noting that r^⋆_xσ^⋆_x ⩾ 0, we have K^⋆ − q_xρ_x ⩾ 0 for all x = 1,...,N. This already proves the condition in equation (5).

Next, from the symmetry operator in equation (8), the equality in equation (10) also implies the following:

$$\begin{eqnarray*} \mathrm{tr}[K^{\star}]\kern1pt &=&\kern-1pt \mathrm{tr}\left[\kern-1pt\sum_{{x}=1}^{N} q_{{x}} \rho_{{x}} M_{{x}}^{\star}\kern-1pt\right]\kern-2pt =\kern-1pt \mathrm{tr}\left[\sum_{{x}=1}^{N} (K^{\star} - r_{{x}}^{\star} \sigma_{x}^{\star} ) M_{{x}}^{\star}\right]\kern-2pt =\kern-1pt \mathrm{tr}[K^{\star}] - \sum_{{x}=1}^{N} \mathrm{tr} [ r_{{x} }^{\star} \sigma_{{x}}^{\star} M_{{x}}^{\star} ] \\ &\Rightarrow &\kern-2pt \sum_{{x}=1}^{N} \mathrm{tr} [ r_{{x} }^{\star} \sigma_{{x}}^{\star} M_{{x}}^{\star} ] = 0.\nonumber \end{eqnarray*}$$

Since tr[r^⋆_xσ^⋆_xM^⋆_x] ⩾ 0 for each x, we conclude that tr[M^⋆_xr^⋆_xσ^⋆_x] = 0 for all x. Moreover, since σ^⋆_x ⩾ 0, M^⋆_x ⩾ 0 and r^⋆_x ⩾ 0, we have r^⋆_xσ^⋆_xM^⋆_x = 0 for all x. We apply this identity to the following, together with the relation q_xρ_x = K^⋆ − r^⋆_xσ^⋆_x in equation (8):

$$\begin{eqnarray*} M_{{x}}^{\star} (q_{{x}} \rho_{{x}} - q_{{y}} \rho_{{y}}) M_{{y}}^{\star} & = & M_{{x}}^{\star} ( (K^{\star} - r_{{x}}^{\star} \sigma_{{x}}^{\star} ) - ( K^{\star} - r_{{y}}^{\star} \sigma_{{y}}^{\star} ) ) M_{{y}}^{\star} \\ & = &M_{{x}}^{\star} ( r_{{y}}^{\star} \sigma_{{y}}^{\star} M_{{y}}^{\star} ) - ( M_{{x}}^{\star} r_{{x}}^{\star} \sigma_{{x}}^{\star}) M_{{y}}^{\star} =0. \end{eqnarray*}$$

Thus, it is shown that the KKT conditions imply the optimality condition in equation (4).   □

In [32], it is shown that the no-signaling principle can generally determine the ME quantum state discrimination. This is shown by proving that if measurement devices are non-signaling, the guessing probability in the state discrimination cannot be larger than that characterized within quantum theory. Here, we briefly sketch the main framework of the derivation and show that the optimality conditions in equations (8) and (9) can be alternatively derived from the fundamental principles.

Two operational tasks in quantum theory are exploited. One is the ensemble steering, which states that there always exists a bipartite quantum state, such that by sharing the state, one party can prepare any ensemble decomposition of the other party's state. This was first asserted by Schrödinger [33] and later formalized as the Gisin–Hughston–Jozsa–Wootters (GHJW) theorem [34, 35]. The other is the no-signaling principle that information cannot be transmitted faster than the speed of light. Then, the result in [32] is that optimal quantum state discrimination is a consequence when two correlations are compatible: (i) from quantum correlations, called ensemble steering, that ensemble decompositions of quantum states can be steered by a party at a distance, and (ii) the no-signaling condition on probability distributions, that probability distributions of input and output random variables at a location cannot be exploited to instantaneous communication. From two fundamental principles, optimality conditions in equations (8) and (9) can be reproduced as follows.

Suppose that two parties share entangled states and one party prepares quantum states (to be discriminated among) to the other far in the distance using ensemble steering. For states {q_x,ρ_x}^N_x=1 for the ME discrimination, the other parties are with the following identical ensemble in different decompositions:

$$\begin{equation} \rho = p_{{x}} \rho_{{x}} + (1-p_{{x}}) \sigma_{{x}}\quad \mathrm{for}\;{x} = 1,\ldots,N \end{equation} \tag{ 11 }$$

for some states {σ_x}^N_x=1 with $q_{{x}} = p_{{x}} / \sum _{{y} = 1}^{N} p_{{y}}$. Notice that an identical ensemble ρ has different decompositions according to indices x = 1...,N. Existence of states {σ_x}^N_x=1 that compose the identical ensemble in the above together with a given set {ρ_x}^N_x=1 follows from the GHJW theorem in [34, 35]. Note that {q_x}^N_x=1 are prior probabilities and thus $\sum _{{x}=1}^{N} q_{{x}} =1$; that {p_x}^N_x=1 are probabilities to steer quantum states {ρ_x}^N_x=1 in the ensembles and thus we do not necessarily have $\sum _{{x}=1}^{N} p_{{x}} = 1$.

If a measurement device is set only to discriminate among states {ρ_x}^N_x=1, the capability of optimal discrimination of {q_x,ρ_x}^N_x=1 cannot work with an arbitrarily high probability. This is because, if the ME discrimination works arbitrarily well, it would violate the no-signaling principle. For instance, suppose that the ensemble decomposition is concluded by the ME discrimination of states {ρ_x}^N_x=1: that is, concluding that ρ_x is found in the discrimination, one guesses that the ensemble corresponds to p_xρ_x + (1 − p_x)σ_x. While no information is announced about the preparation of ensemble decompositions among x = 1,...,N, the no-signaling condition is fulfilled and therefore the ME discrimination of {ρ_x}^N_x=1 must not allow one to gain knowledge about the preparation. Given no information about the preparation, it must be a random guess about ensemble decompositions. Thus, state discrimination can be constrained such that the strategy above gives at its best the random guess.

In [32] it is shown that the ME discrimination must satisfy the following condition to fulfill two constraints in the above:

$$\begin{equation} \sum_{{x}=1}^{N} p_{{x}} P ({x} | {x}, \{ \rho_{{x}}\}_{{x}=1}^{N} ) \leqslant 1, \end{equation} \tag{ 12 }$$

where P(x|x,{ρ_x}^N_x=1) is the probability of giving outcome x when one of the states {ρ_x}^N_x=1 is given i.e. the probability of correctly discriminating among states {ρ_x}^N_x=1. The equality holds if and only if, for an ensemble prepared in equation (11) with some x, a measurement device responds only to quantum states ρ_x but not to states σ_x, that is, P(x|x,{σ_x}^N_x=1) = 0. Note that, with the measurement postulate, this is equivalent to tr[M_xσ_x] = 0 for some POVM M_x, which is indeed the optimality condition in equation (9). Recall that the bound in equation (12) is about the ME discrimination among quantum states {ρ_x}^N_x=1 that are given with prior probabilities $\{q_{{x}} =p_{{x}} / \sum _{{y}=1}^{N} p_{{y}} \}_{{x}=1}^{N}$. All these are summarized as follows.

Lemma 2.3. When quantum states {q_x,ρ_x}^N_x=1 are prepared by ensemble steering, as shown in equation (11), the guessing probability is upper bounded by the no-signaling condition

$$\begin{equation} \fl P_{\mathrm{guess}} \leqslant \frac{1}{ \sum_{{x}=1}^{N} p_{{x}}}~\mathrm{with~ equality~ if ~and ~only~ if} ~P({x} | {x}, \{ \sigma_{{x}}\}_{{x}=1}^{N} ) = 0 ~\mathrm{for}~{x} = 1,\ldots,N, \quad \end{equation} \tag{ 13 }$$

where {p_x}^N_x=1 are from equation (11). With the measurement postulate, the equality holds when POVM M_x does not respond to σ_x but only to ρ_x for each x, that is

$$\begin{equation} \mathrm{tr}[M_{{x}} \sigma_{{x}}] = 0, \end{equation} \tag{ 14 }$$

which reproduces the condition in equation (9). Then, the upper bound is equal to the guessing probability from quantum theory, e.g. equation (7).

Proof. The upper bound can be derived from equation (12). We also recall $q_{{x}} = p_{{x}} / \sum _{{y}=1}^{N}$. Since $\sum _{{x}=1}^{N} q_{{x}} =1$, the upper bound in equation (13) is obtained.

Consequently, two constraints in equations (11) and (14) are optimality conditions for probabilities in the ME discrimination. Optimality conditions in equations (11) and (14) are equivalent to those in lemma 2.2: equation (11) is equivalent to the symmetry operator in equation (8). We also remark that this approach of constraining with the no-signaling principle is valid for generalized probability theories, as long as the steering effect is allowed to do, i.e. that different decompositions of an identical ensemble in equation (11) can be prepared at a distance.

Let us first present a general form of the guessing probability (for quantum states) in the framework of probability-theoretic measures. We first recall from lemma 2.1 that the guessing probability about random variable X given Y is generally expressed as the distance of probability P_X|Y deviated from the uniform distribution. This holds true in general, no matter what physical systems are applied to mediate between the preparation and the measurement. Then, it remains to determine the form of d(X|Y ) once quantum systems are applied.

We recall the KKT conditions of the ME discrimination for {q_x,ρ_x}^N_x=1, as follows:

$$\begin{eqnarray*} &&K = q_{{x}} \rho_{{x}} + r_{{x}} \sigma_{{x}}~\mathrm{for}~{x} = 1,\ldots,N~\mathrm{with~ complementary~ states}~ \{r_{{x}},\sigma_{{x}} \}_{{x}=1}^{N}. \end{eqnarray*}$$

Using the simple identity that $K = \sum _{{x}=1}K/N$, we have

$$\begin{equation} P_{\mathrm{guess}} = \frac{1}{N} \mathrm{tr}\left[ \sum_{{x}} q_{{x}} \rho_{{x}} + r_{{x}} \sigma_{{x}}\right] = \frac{1}{N} + R, \quad \mathrm{where}\; R= \frac{1}{N} \sum_{{x}=1}^{N} r_{{x}}. \end{equation} \tag{ 22 }$$

This compares to equation (3): the distance d(X|Y ) corresponds to the dual parameter R, which now corresponds to the norm of the ensemble average of complementary states.

The parameter R in the expression in equation (22) in fact has a geometrical meaning in the state space. Recall from section 2.3 that each complementary state is determined as a vertex of the polytope that is congruent to the polytope constructed by given states, $\mathcal {P} (\{q_{{x}},\rho _{{x}} \}_{{x}=1}^{N} )$. Then, each state r_xσ_x plays the role of the (trace) distance between each vertex q_xρ_x of the polytope and the symmetry operator, i.e.

$$\begin{eqnarray*} &&r_{{x}} = \mathrm{tr} [ r_{{x}} \sigma_{{x}} ]= \mathrm{tr} [ K - q_{{x}} \rho_{{x}} ] = \| K - q_{{x}} \rho_{{x}} \|_{1}, \end{eqnarray*}$$

where we note that K ⩾ q_xρ_x for all x = 1,...,N, from the dual problem in equation (7).

Proposition 3.2. The guessing probability for quantum states {q_x,ρ_x}^N_x=1 constructed from a symmetry operator K is

$$\begin{equation} \fl P_{\mathrm{guess}} = \frac{1}{N} + R (K || \{ q_{{x}} ,\rho_{{x}} \}_{{x}=1}^{N} )\; \mathrm{with}\;R (K || \{ q_{{x}} ,\rho_{{x}} \}_{{x}=1}^{N} ) = \frac{1}{N} \sum_{x=1}^{N} \| K - q_{{x}} \rho_{{x}} \|_{1}, \quad \end{equation} \tag{ 23 }$$

where R(K||{q_x,ρ_x}^N_x=1) shows the averaged trace distance of given states {q_x,ρ_x}^N_x=1 deviated from their symmetry operator K, see also equation (3) in lemma 2.1 for comparison.

The next form of guessing probability (for discrimination among states {q_x,ρ_x}^N_x=1) is obtained by the compatibility between ensemble steering on quantum states and the no-signaling condition on probability distributions of measurement outcomes. Then, as is shown in lemma 2.3, the guessing probability is given by

$$\begin{equation} P_{\mathrm{guess}} = \frac{1}{p_1 + \cdots + p_{N}} \quad \mathrm{with}\; p_{{x}} = q_{{x}} P_{\mathrm{guess}}\quad \mathrm{for}\;{x}=1,\ldots,N, \end{equation} \tag{ 24 }$$

since the bound in equation (13) is tight. From the ensemble steering shown in equation (11), the guessing probability can be interpreted as follows. Each parameter p_x of {p_x}^N_x=1 in the above shows the probability that a party can prepare a quantum state ρ_x to the other party at a distance via the ensemble steering. Therefore, the guessing probability corresponds to the maximal average probability to prepare a set of chosen quantum states to a distant party via the ensemble steering.

Theorem 3.1 While ensemble steering is allowed within quantum theory, the steerability of a quantum state in the ensemble is generally proportional to the guessing probability for those states prepared in the ensemble.

Finally, for the ME discrimination for {q_x,ρ_x}^N_x=1, a huge simplification can be made when prior probabilities are equal i.e. q_x = 1/N for all x. The simplification is shown in the expression of R in equation (23), that it holds r_x = r_y for all x,y = 1,...,N if q_x = 1/N. Note also that since the symmetry operator in equation (8) shows that P_guess = tr[K] = q_x + r_x = q_y + r_y ∀x,y; now since q_x = q_y = 1/N, it holds that r_x = r_y for all x,y. From lemma 3.2, this means that the distances of given states from the symmetry operator are all equal. Thus, for convenience, let us write

$$\begin{equation} r:= r_{{x}}\;~\forall~{x}=1,\ldots,N\quad \mathrm{and~then}\; r = R(K \| \{ 1/N ,\rho_{{x}} \}_{{x}=1}^{N} ) . \end{equation} \tag{ 25 }$$

The problem of ME discrimination becomes even simpler. There is only a single parameter r to find, to solve optimal quantum state discrimination for the uniform prior.

Applying the geometric formulation shown in section 2.3, the geometrical meaning of the parameter r in equation (25) can be found in a simple way. From the KKT condition in equation (8), we have the following relation, see also equation (19):

$$\begin{equation} \frac{1}{N} \rho_{{x}} - \frac{1}{N} \rho_{{y}} = r \sigma_{{y}} - r \sigma_{{x}}\quad \mathrm{and~then}\; r = \frac{ \| \frac{1}{N} \rho_{{x}} - \frac{1}{N} \rho_{{y}} \|_{1} }{ \| \sigma_{{x}} - \sigma_{{y}} \|_{1} }. \end{equation} \tag{ 26 }$$

Then, the parameter r corresponds to the ratio between two polytopes in the state space, the given one $\mathcal {P}(\{ \frac {1}{N},\rho _{{x}} \}_{{x}=1}^{N} )$ constructed from given states and the other one $\mathcal {P}(\{ \sigma _{{x}} \}_{{x}=1}^{N} )$ only of complementary states.

Proposition 3.3. The guessing probability for quantum states given with the uniform prior is determined by only a single parameter as

$$\begin{equation} P_{\mathrm{guess}} = \frac{1}{N} + r, \end{equation} \tag{ 27 }$$

where r is the ratio between two polytopes, one from given states and the other from complementary states.

We first suppose that three pure states lying at a half-plane are given with equal prior probabilities 1/3, i.e. {1/3,ρ_x = |ψ_x〉〈ψ_x|}³_x=1. In particular, we suppose that the polytope constructed by the three states forms an isosceles triangle on a half plane of the Bloch sphere. For convenience, we parameterize the three states as follows, for some θ₀:

$$\begin{eqnarray} |\psi_1\rangle & = & \cos \textstyle\frac{1}{2}(\theta_0 + \theta ) |0\rangle + \sin \textstyle\frac{1}{2} (\theta_0 + \theta) |1\rangle, \nonumber\\ |\psi_2\rangle & = & \cos \textstyle\frac{1}{2} \theta_0 |0\rangle + \sin \textstyle\frac{1}{2}\theta_0 |1\rangle, \nonumber\\ |\psi_3\rangle & = & \cos \textstyle\frac{1}{2} (\theta_0 - \theta ) |0\rangle + \sin \textstyle\frac{1}{2} (\theta_0 - \theta) |1\rangle, \end{eqnarray} \tag{ 33 }$$

where let us suppose θ∈[0,π], see figure 2.

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We now apply lemma 4.2 to discriminate among these three states, referring to the geometry shown in figure 2. The maximal polytope which is similar to the given one X₁X₂X₃ is S₁S₂S₃. There are many ways of putting the maximal polytope in the plane, however, it should be put as it is shown in the figure to fulfill the optimality condition equation (19), or equation (8). Let us first compute the guessing probability. We have to find the ratio r between two polytopes. It corresponds to the ratio, X₁X₃/S₁S₃, which is r = (sin θ)/3. Therefore, the guessing probability is

$$\begin{equation} P_{\mathrm{guess}} = \textstyle\frac{1}{3} + \textstyle\frac{1}{3} \sin \theta. \end{equation} \tag{ 34 }$$

In figure 2, the guessing probability for the three states is plotted as angle θ varies. One can easily find that for θ ⩾ π/2, the ratio is r = 1/3 and thus the guessing probability is, 2/3.

Let us then find the optimal measurement. Note that OS₁, OS₂ and OS₃ correspond to complementary states, σ₁, σ₂ and σ₃, respectively. It is shown that σ₂ is not a pure state, i.e. not of rank-one. This means that, to fulfill the optimality condition in equation (9), the optimal POVM element is the null measurement, i.e. M₂ = 0, no-measurement on the state. The other complementary states can be written explicitly as follows:

$$\begin{eqnarray}&&\begin{array}{@{}c@{}} \sigma_{1} = | \varphi_1 \rangle \langle \varphi_1|,\quad |\varphi_1\rangle = \cos \left(\displaystyle\frac{\theta_0}{2} - \frac{\pi}{4}\right) |0\rangle + \sin \left(\displaystyle\frac{\theta_0}{2} - \displaystyle\frac{\pi}{4}\right) |1\rangle, \\ \sigma_{3} = | \varphi_3 \rangle \langle \varphi_3|,\quad |\varphi_3\rangle = \cos \left(\displaystyle\frac{\theta_0}{2} + \frac{\pi}{4}\right) |0\rangle + \sin \left(\displaystyle\frac{\theta_0}{2} + \frac{\pi}{4}\right)|1\rangle \end{array} \end{eqnarray} \tag{ 35 }$$

and the optimal POVM elements are

$$\begin{eqnarray*} &&M_{1} = | \varphi_{3} \rangle \langle \varphi_3|, ~M_{2}=0,~M_{3} = | \varphi_{1} \rangle \langle \varphi_1|,\quad \mathrm{so~that}\; \sum_{{x}=1}^{3} M_{{x}} = I. \end{eqnarray*}$$

One can see that with these POVMs, the guessing probability is obtained

$$\begin{eqnarray*} &&P_{\mathrm{guess}}= \textstyle\frac{1}{3} \mathrm{tr}[ \rho_{1} M_{1}] + \frac{1}{3} \mathrm{tr}[ \rho_{2} M_{2}] + \textstyle\frac{1}{3} \mathrm{tr}[ \rho_{3} M_{3}] = \textstyle\frac{1}{3} + \textstyle\frac{1}{3} \sin\theta \end{eqnarray*}$$

as it is shown in equation (34). Optimal POVM elements show that for discrimination among the three state given with probability 1/3, the optimal strategy corresponds to the measurement setting where the device only responds to the two most distant states out of the three.

The example shown in the previous can be extended to the so-called geometrically uniform states [23, 27], which correspond to the case that θ = 2π/3. It is straightforward to exploit the method shown in the above from lemma 4.2, then the guessing probability is 2/3. In fact, as is shown, the guessing probability is found, P_guess = 2/3 for any three pure (qubit) states given with equal probabilities if the polytope of them in the Bloch sphere contains the origin.

We now consider a set of three pure states given with equal probabilities 1/3 and suppose that the polytope constructed by the three states forms an arbitrary triangle in a half-plane of the Bloch sphere: {1/3,ρ_x = |ψ_x〉〈ψ_x|}³_x=1. As a contrast to the example isosceles triangles shown in the above, we also suppose that the polytope of given states contains the origin of the Bloch sphere, see figure 3.

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To apply the geometric formulation in lemma 4.3, we also refer to figure 3. Note that the three states ρ_x/3 for x = 1,2,3 correspond to OX_x, x = 1,2,3, respectively; thus the polytope of given states is the triangle X₁X₂X₃. Then, the next thing to do is to find a maximal polytope similar to X₁X₂X₃ within the Bloch sphere. The triangle S₁S₂S₃ in figure 3 could be a possibility. The ratio r corresponds to, for instance, X₁X₂/S₁S₂, which is also equal to OX₁/OS₁: therefore r = 1/3, then the guessing probability is

$$\begin{eqnarray*} &&P_{\mathrm{guess}} = \textstyle\frac{1}{3} + r = \textstyle\frac{2}{3}. \end{eqnarray*}$$

Note that this is the guessing probability for many cases of three pure states. This can be generalized to cases where three states have the same purity, as follows.

Remark 4.1. Suppose that three qubit states having an equal purity denoted by f are given with equal prior probabilities 1/3. Here, the purity corresponds to the norm of a Bloch vector. Then, if the polytope of the three states in the Bloch sphere contains the origin, the guessing probability is

$$\begin{eqnarray*} &&P_{\mathrm{guess}} = \textstyle\frac{1}{3} + \textstyle\frac{1}{3}f. \end{eqnarray*}$$

This is independent of the other details of given quantum states, e.g. the angles between them.

The triangle S₁S₂S₃ in figure 3 does not show complementary states yet, since the optimality condition in equation (19), or equivalently equation (8), is not fulfilled. Therefore, one has to put or rotate S₁S₂S₃ to C₁C₂C₃ in figure 3, so that equation (19) holds true: X_xX_y∥C_yC_x for all x,y = 1,2,3, where ∥ means that two lines are parallel. Then, all of the optimality conditions are satisfied.

Complementary states are then those states that correspond to OC_x for x = 1,2,3. It also follows that optimal POVM elements are OP_x for x = 1,2,3 as each of them is orthogonal to its corresponding complementary state, see the optimality condition in equation (9). Thus, optimal POVM elements are {M_x∝OP_x}³_x=1. Since the convex hull P₁P₂P₃ contains the origin, it is also straightforward to have the completeness, $\sum _{{x}=1}^{3} M_{{x}} =I$. Here, it is noteworthy that optimal POVMs can be in vectors which are not parallel to corresponding states: in figure 3, we found that $OX_{x} \nparallel OP_{x}$ for some x.

Finally, let us show how the geometric formulation in lemma 4.3 can be generally applied to a set of arbitrary three qubit states when they are given with equal probabilities, {1/3,ρ_x}³_x=1. We explain this, referring to figure 4. Arbitrary three states are depicted by OX_x for x = 1,2,3 respectively.

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Once they are given, they define a plane of the triangle X₁X₂X₃ in the Bloch sphere and there exists a half-plane parallel to the plane. To find the guessing probability for the states, one has to first find a maximal triangle that is similar to the triangle X₁X₂X₃ in the Bloch sphere. Then, the maximal one must lie on the half-plane parallel to X₁X₂X₃, to fulfill the optimality condition in equation (19), or equivalently equation (8). Then, the maximal one is rotated such that the condition in equation (19) is fulfilled and finally ends up with C₁C₂C₃. Then, complementary states are immediately found as OC_x for x = 1,2,3, from which optimal measurement also follows as those POVM elements orthogonal to complementary states.

Let us first consider pairs of orthogonal states, say {1/4,ρ_x = |ψ_x〉〈ψ_x|}⁴_x=1, where 〈ψ₁|ψ₃〉 = 〈ψ₂|ψ₄〉 = 0. The four states define a plane in the Bloch sphere and then form a rectangle on the plane. To be explicit, they can be written as follows, for some θ₀ and θ:

$$\begin{eqnarray} |\psi_1\rangle & = & \cos \frac{\theta_0}{2} |0\rangle + \sin\frac{\theta_0}{2} |1\rangle, \nonumber \\ |\psi_2\rangle & = & \cos \left(\frac{\theta_0}{2} - \theta\right) |0\rangle + \sin \left(\frac{\theta_0}{2} -\theta\right) |1\rangle, \nonumber \\\\ |\psi_3\rangle & = & \cos \left(\frac{\theta_0}{2} - \frac{\pi}{2}\right) |0\rangle + \sin \left( \frac{\theta_0}{2} - \frac{\pi}{2} \right) |1\rangle, \nonumber \\ |\psi_4\rangle & = & \cos \left(\frac{\theta_0}{2} -\theta- \frac{\pi}{2}\right) |0\rangle + \sin \left( \frac{\theta_0}{2} -\theta- \frac{\pi}{2} \right) |1\rangle,\nonumber \end{eqnarray} \tag{ 36 }$$

which are shown in figure 5. The half-plane in which the four states lie is then defined in the Bloch sphere.

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Referring to figure 5, we apply the geometric formulation in lemma 4.3 and show the optimal parameters, optimal measurement and complementary states. The polytope of given states is the rectangle X₁X₂X₃X₄. It is straightforward to see that the largest one similar to X₁X₂X₃X₄ is the rectangle having its diagonal in length 2 (in terms of trace-norm). Then, we rotate it so that the relation in equation (19), or the optimality condition in equation (8), is satisfied; then we have the rectangle C₁C₂C₃C₄. The complementary states are found as OC_x for x = 1,2,3,4, from which optimal POVM elements are also obtained {M_x∝OM_x}⁴_x=1. Therefore, we have optimal parameters as follows:

$$\begin{eqnarray}&&\begin{array}{@{}c@{}} M_{1} = \sigma_{3} = | \psi_1 \rangle \langle \psi_1 |,\quad M_{2} = \sigma_{4} = | \psi_2 \rangle \langle \psi_2 |, \\ M_{3} = \sigma_{1} = | \psi_3 \rangle \langle \psi_3 |, \quad M_{4} = \sigma_{2} = | \psi_4 \rangle \langle \psi_4 |.\end{array} \end{eqnarray} \tag{ 37 }$$

From these, the guessing probability can be obtained. Or, applying lemma 4.2, one can see the ratio, for instance,

$$\begin{equation} r = \frac{ X_1 X_2 }{ C_1 C_2} = \frac{OX_1 }{ OC_1} = \frac{1}{4}\quad \mathrm{and~thus} \;P_{\mathrm{guess}} = \frac{1}{4} + r= \frac{1}{2}. \end{equation} \tag{ 38 }$$

We remark that, as it is shown in the above, the guessing probability is equal in the range of angle 0 ⩽ θ ⩽ π/2.

The analysis shown in the above implies that the guessing probability does not depend on detailed relations among given quantum states but on a property assigned by the set of them. This shows the freedom of choosing four states such that the guessing capability about states does not change. More precisely, this can be seen in terms of the symmetry operator, which in this case is

$$\begin{equation} K = \textstyle\frac{1}{4} \rho_{{x}} + r \sigma_{{x}} = \textstyle\frac{1}{2} I \end{equation} \tag{ 39 }$$

consistently to equation (38). That is, all sets of four states of pairs of orthogonal states share the same the symmetry operator in equation (39), meaning that all of them are in the same equivalence class, see definition 3.1.

Remark 4.2. Four states of any two pairs of orthogonal qubit states are in the same equivalence class, $\mathcal {A}_{I/2}$.

Next, let us consider arbitrary four qubit states defined on a half-plane, which are given with the equal probabilities 1/4. For generality, we do not assume any internal symmetry among given states {1/4,ρ_x}⁴_x=1 except the assumption that they are on a half-plane in the Bloch sphere. It is then straightforward to generalize this to arbitrary four states defined on any plane in the sphere.

We now refer to figure 6 to show how the geometric formulation in lemma 4.3 can be applied to those four states. The polytope of four states is shown as the quadrilateral X₁X₂X₃X₄ where each vertex corresponds to ρ_x/4 for x = 1,2,3,4. Then, one has to expand the given quadrilateral so that they are maximal in the Bloch sphere. In this way, the ratio between two quadrilaterals can be found.

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To find complementary states, one has to rotate the obtained maximal quadrilateral so that the optimality condition in equation (19) is fulfilled. The resulting quadrilateral from which complementary states can be found is then, C₁C₂C₃C₄, where OC_x for x = 1,2,3,4 are complementary states. It can happen that, since a given quadrilateral X₁X₂X₃X₄ is arbitrarily shaped, some vertices may not be on the surface of the Bloch sphere. As is depicted in figure 6, suppose that OC₁ cannot lie at the surface of the sphere. This means that complementary state σ₁ is not pure; thus the POVM element for state ρ₁ corresponds to the null measurement, M₁ = 0. This follows from the optimality condition in lemma 4.2 to fulfill equation (9). The optimal discrimination strategy for these states is thus to prepare measurement such that there are three kinds of outcomes from M_x with x = 2,3,4. Note also that, as the convex hull of the POVM elements M₁M₂M₃ contains the origin, one can also construct a complement measurement.

We now consider four qubit states {ρ_x}⁴_x=1 which form a polytope having a volume in the Bloch sphere, see figure 7. Suppose that they are given with probability 1/4 for each and for convenience we assume that the tetrahedron constructed by four states is covered by a sphere. Later, this can be relaxed and generalized to cases where all vertices of the tetrahedron are not touched by a sphere. Here, since the tetrahedron is covered by a sphere, the four states have the same purity, which we denote by f.

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Referring to figure 7, let us show how to find the guessing probability, complementary states and optimal measurement. The guessing probability immediately follows by applying lemma 4.3. The ratio r can be computed by finding the ratio of two diameters of two spheres: one is the sphere covering the given tetrahedron and the other the Bloch sphere covering a maximal tetrahedron similar to the given one. As the purity is given by f, the guessing probability is given by

$$\begin{eqnarray*} &&P_{\mathrm{guess}} = \textstyle\frac{1}{4} + \textstyle\frac{1}{4}f, \end{eqnarray*}$$

which does not depend on detailed relations among given states, such as angles between given states. The above holds true for any four states whose tetrahedron in the Bloch sphere can be covered by a single sphere.

Then, one rotates the maximal tetrahedron within the Bloch sphere such that the resulting one satisfies the optimality condition in equation (19). Since the tetrahedron already fully occupies the Bloch sphere, it is not difficult to see that the resulting tetrahedron has the reflection symmetry to the given tetrahedron, with respect to the origin of the Bloch sphere, see figure 7. Each vertex OC_x for x = 1,2,3,4 corresponds to complementary states; optimal POVM elements are found to be those rank-one operators orthogonal to complementary states