Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-57/1/d/solution
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 57 1 d Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-14
Define and . From the branch calculation in part (b), outcome leavesThus the effective input teleported was . A Pauli X gate flips every bit and hence maps a Dicke state with excitations to one with excitations. Since , applying at every receiving party interchanges the two logical basis states and restores . ThereforeIn particular Bob's uncorrected reduced density matrix is , and his correction returns it to the matrix in part (c). The remaining branches can likewise be corrected: implements logical up to the common global phase , while implements logical . Thus the outcome dependence can be removed by local Pauli gates even though the resulting copies remain imperfect.
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