On the Form of the Optimal Measurement for the Probability of Detection

We consider the problem of maximizing the probability of detection for an infinite number of mixed states. We show that for linearly independent states there exists a unique simple optimal measurement, generalizing thus a result obtained in finite dimension by Y. Eldar (Phys. Rev. A, 68, 052303:1-052303:4 2003).


Introduction
Let ρ 1 , ρ 2 , . . . be (finite or infinite number) quantum states (density matrices) on B(H)the bounded linear operators on Hilbert space H, of arbitrary dimension, which can occur with some a priori probabilities π = (π 1 , π 2 , . . .). We want to find, in an optimal way, the state in which the system really is. To this end we perform a measurement (called also strategy) M, by which is meant a sequence (M 1 , M 2 , . . .) of positive operators from B(H), such that where the series is convergent in the weak operator topology on B(H). A measurement M = (M 1 , M 2 , . . .) for which all M i 's are pairwise orthogonal projections is called simple or sharp (see [1]). Rafał Wieczorek wieczorek@math.uni.lodz.pl 1 Faculty of Mathematics and Computer Science, Łódź University, ul. S. Banacha 22, 90-238 Łódź, Poland If we receive outcome M i , we choose state ρ i . The probability that the true state is ρ i when measurement will give result M j is given by tr(ρ i M j ). Thus tr(ρ i M i ) is the probabilty of guessing correctly state ρ i . If our guess is ρ j while the true one is ρ i , then we pay penalty L(i, j ). Function L is called a loss function. The risk function is defined by the formula The expectation of the risk function is called the Bayes risk, and denoted by r(M, π), i.e.
Consider the concrete loss function of the form Then we have In this case, minimizing Bayes risk is equivalent to maximizing the expression The above expression is the probability of correct guess while performing measurement M, called the probability of detection. We shall denote this probability by P D (M). We want to find a measurement which maximizes the probability of detection.
The existence of an optimal measurement is discussed in [7] in a general setup. In our case, the following result from [7] is sufficient.

Theorem 1 There exists a measurement maximizing the probability of detection.
For two states the solution can be achieved by taking the simple measurement made by the projections on the support of the positive and negative part of the Hermitian operator π 1 ρ 1 − π 2 ρ 2 . Kaniowski [4] did a deeper analysis for two finite dimensional projections with arbitrary a priori probability.
Each state ρ i has the spectral decomposition where λ j i > 0 and m i ∈ {1, 2, . . . , ∞}. In our further considerations we assume that the vectors {ϕ m n } span the Hilbert space H. For arbitrary states it is hard to say anything about the optimal measurement. In the case of finite-dimensional Hilbert space and finite number of a states it is natural to assume linear independence of vectors {ϕ m n }. Then we say that the states are linearly independent. For dim H = ∞ we have a stronger assumption. We say that states are strongly linearly independent if vectors {ϕ j i } are strongly linearly independent, i.e. for each i, j we have ϕ j i ∈ Lin{ϕ m n : n = i, m = j }.
A state ρ is called pure if it has the form |ϕ ϕ| for some unit vector ϕ ∈ H, otherwise a state is called mixed. For dim H < ∞ and pure states Kennedy [5,6] obtained the following result.
It turns out that this result holds also for dim H = ∞.
Theorem (Łuczak [7], 2009) Let pure states ρ 1 , ρ 2 , . . . be strongly linearly independent. Then there exists a unique measurement maximizing the probability of detection and this measurement is simple.
For dim H < ∞ and arbitrary states Eldar [2] obtained the following result.

Theorem
(Eldar [2], 2003) Let the states ρ 1 , ρ 2 , . . . , ρ n be linearly independent. Then there exists a unique measurement maximizing the probability of detection and this measurement is simple.
A natural question is whether the Eldar result can be generalized to infinite dimension. In this paper we show that the answer is positive. Proof We use the method from the proof of Lemma 4 in [7]. Assume that we have e.g., = π 1 tr(ρ 1 (1 − Q)) + π 1 tr

Optimal Measurement
From the above and the optimality of the measurement M we obtain that tr(ρ 1 (1−Q)) = 0. Therefore tr(ρ 1 Q) = 1. This gives This contradicts the relation Before the main theorem we show an interesting result.
is an optimal measurement, then M k is a nonzero uniquely determined projection.
Proof From the Holevo condition we have i =k Therefore for all ξ ∈ H we obtain From the above and assumption (2) N = (N 1 , N 2 , . . .) are two distinct optimal measurements such that M k = N k . From the above M k and N k are projections. Of course 1 2 M + 1 2 N is also an optimal measurement and 1 2 M k + 1 2 N k is a projection. Then a contradiction. Consequently, M k is uniquely determined.
Let the states ρ 1 , ρ 2 , . . . of the form (1) be strongly linearly independent. Our main theorem is Therefore M is the unique simple measurement with the nonzero outcomes.
Suppose now that ρ 1 , ρ 2 , . . . are arbitrary states linearly independent or not. The next theorem shows a relation between the ranges of elements of an optimal measurement and the ranges of the states in question.
Consequently, condition (4) is of the form This implies dimRangeM i ≤ dimRangeρ i .
As a corollary we obtain the main result of [7].

Corollary 1
Let pure states ρ 1 , ρ 2 , . . . be strongly linearly independent. Then there exists a unique measurement maximizing the probability of detection and this measurement is simple. The outcomes of this measurement are rank one operators.
Proof The first part of the corollary is a consequence of Theorem 3. From Theorem 4 the outcomes of the optimal measurement are zero or rank one operators but Lemma 1 implies that the outcomes can't be zero operators.