1967
DOI: 10.1016/s0019-9958(67)90302-6
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Detection theory and quantum mechanics

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Cited by 219 publications
(185 citation statements)
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“…This limiting case corresponds to the standard two-outcome discrimination game [1][2][3], so the minimized cost will be equal to the usual HB. (ii) When k → 0, there is no penalty for non-guesses, so it is always better to decline (p d = 1) to produce min(C) → 0.…”
mentioning
confidence: 99%
“…This limiting case corresponds to the standard two-outcome discrimination game [1][2][3], so the minimized cost will be equal to the usual HB. (ii) When k → 0, there is no penalty for non-guesses, so it is always better to decline (p d = 1) to produce min(C) → 0.…”
mentioning
confidence: 99%
“…However, it is one of the innermost consequences of the laws of quantum mechanics that nonorthogonal states cannot be discriminated with certainty. Optimal detection strategies were first investigated by Helstrom [20,21] and Holevo [22] and a lot of attention has since been devoted to the development of optimal and near-optimal receivers for binary coherent states [23][24][25][26][27][28][29][30][31][32] and for the discrimination of larger signal alphabets [33][34][35][36][37][38][39][40][41]. An overview over different receiver schemes is provided in Appendix B.…”
Section: Binary Coherent-state Cloningmentioning
confidence: 99%
“…An intermediate regime where both erroneous and inconclusive results are allowed has also been considered [29] and the minimal probability of error for a fixed probability of inconclusive results has been derived for pure [51] and mixed states [52]. In the following, we always assume Comparison of the error probability of the homodyne detector (red, dotted), the Kennedy receiver (green, dashed) [24], the optimized displacement receiver (purple, dash-dotted) [29], and the Helstrom bound [20,21].…”
Section: Appendix B: Binary Coherent State Receiversmentioning
confidence: 99%
“…where {P (b), P (e)} are the prior probabilities for Bob's and Eve's bit values, and P (b | e) is the conditional probability for Bob's bit value to be b given that Eve's is e. Alice's bits are equally likely to be 0 or 1, and Eve's conditional error probabilities satisfy [10] …”
Section: +45mentioning
confidence: 99%