Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 67 3 b Solution Created 2026-10-03 Updated 2026-10-07
Replace every CNOT gate by . The resulting quantum circuit contains only Hadamard gates and Controlled-Z gates. Realize each Hadamard gate by a fresh one-bit teleportation link measured at angle zero, so every internal measurement is in the fixed basis. Insert a graph edge between the current wire vertices for each Controlled-Z gate. This gives a suitable graph state because the inputs are , and entangling gates can be moved to resource preparation: they commute with one another and with earlier measurements on vertices no longer used by the gate.
Track a Pauli frame . An angle-zero one-bit teleportation with raw result updates the frame on that wire bybecause equals up to global phase. A Controlled-Z gate updates and , leaving the bits unchanged. These are classical binary updates; no measurement angle depends on them.
All the internal quantum measurements consequently have predetermined bases. Their projectors act on distinct resource vertices and commute, so all internal measurements can be performed simultaneously: the logical measurement depth is at most one. If the output is to be read in the computational basis, those fixed-basis measurements can occur in the same layer; a raw output is relabelled . If quantum outputs are retained, the same calculation gives the desired output in a known Pauli frame. This nonadaptive Hadamard–CNOT measurement pattern concerns measurement depth, not the depth of resource preparation or classical frame computation.