Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ib/paper-3/20f/b/solution
Past exam of the mathematics course of the University of Cambridge 2001 ib Paper 3 20F b Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
In the two-dimensional span of the first two energy eigenstates, the measured observable has matrix . Its square is , and its trace is zero, giving eigenvalues . The orthogonal complement has eigenvalue zero. Thus the full spectrum is .
The normalized eigenstates for and are respectivelyThe lowest measurement outcome prepares up to an irrelevant overall phase. The subsequent unitary time evolution givesBy the Born rule, the probability of measuring the lowest eigenvalue again is the squared projection onto its one-dimensional eigenspace:This is the two-level quantum return probability with . At the probability is one; if the two energies happen to be equal it remains one for every time.
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