Alice verifies each submitted qubit in her recorded secret basis and checks its bit against the revealed number. For a forger who chooses X or Z at random and prepares the corresponding known-bit eigenstate, each bit passes with probability . With independent basis choices,The detection probability depends on Daisy's preparation strategy. The question does not restrict her to legitimate X/Z eigenstates. Once the bit is known, for bit zero the averaged acceptance operator isFor bit one it is . Both have largest eigenvalue . Preparing the corresponding eigenstate, a Bloch-vector bisector of the two candidate encodings, achieves this value; for bit zero it is .
The acceptance operator for the whole known check number is , whose largest eigenvalue is , so entangled preparation cannot improve this no-original-note optimum. Therefore unrestricted preparation instead givesThe random-basis value is the elementary protocol calculation; the second value states the best forging strategy allowed by arbitrary state preparation and no access to the original quantum note.
Yes, access to the original qubits can supply information about the secret encoding basis once the bit is known. For bit zero the possibilities are and , whose overlap is ; for bit one they are and with the same overlap magnitude. A Helstrom measurement for two pure states can identify which candidate was supplied more reliably than a blind basis guess. Measuring and preparing a state based on that evidence can improve an additional counterfeit, although the discrimination is not perfect and the measurement can disturb the original.
The no-cloning theorem rules out a deterministic perfect extra copy of both nonorthogonal candidates; it does not imply that access gives no advantage. The task of producing an additional accepted note should also be distinguished from simply returning the unaltered original, which already passes Alice's test. Requiring both the original and the extra note to survive imposes a further information-disturbance tradeoff.
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